Geotechnical Engineering: Unsaturated and Saturated Soils (Team-IRA) [2 ed.] 1119788692, 9781119788690

Citation preview

Geotechnical Engineering: Unsaturated and Saturated Soils

Geotechnical Engineering: Unsaturated and Saturated Soils Second Edition

Jean-Louis Briaud Texas A&M University TX, USA

Copyright © 2023 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. Edition History John Wiley & Sons, Inc. (1e, 2013) No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www .copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www .wiley.com/go/permission. Trademarks: Wiley and the Wiley logo are trademarks or registered trademarks of John Wiley & Sons, Inc. and/or its affiliates in the United States and other countries and may not be used without written permission. All other trademarks are the property of their respective owners. John Wiley & Sons, Inc. is not associated with any product or vendor mentioned in this book. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Further, readers should be aware that websites listed in this work may have changed or disappeared between when this work was written and when it is read. Neither the publisher nor authors shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our website at www.wiley.com. Library of Congress Cataloging-in-Publication Data Names: Briaud, J.-L., author. | John Wiley & Sons, publisher. Title: Geotechnical engineering : unsaturated and saturated soils / Jean-Louis Briaud. Description: Second edition. | Hoboken, New Jersey : Wiley, 2023. | Includes bibliographical references and index. Identifiers: LCCN 2023007312 (print) | LCCN 2023007313 (ebook) | ISBN 9781119788690 (hardback) | ISBN 9781119788713 (adobe pdf) | ISBN 9781119788706 (epub) Subjects: LCSH: Geotechnical engineering—Textbooks. | Soil mechanics—Textbooks. Classification: LCC TA705 .B75 2023 (print) | LCC TA705 (ebook) | DDC 624.1/51—dc23/eng/20230224 LC record available at https://lccn.loc.gov/2023007312 LC ebook record available at https://lccn.loc.gov/2023007313 Cover Design: Wiley Cover Images: © f8grapher/Shutterstock; Muhammad Fauzul/Getty Images; Argijale/Getty Images; Carol Yepes/Getty Images; imageBROKER/Luca Renner/Getty Images

CONTENTS Acknowledgments CHAPTER 1

CHAPTER 2

xxv Introduction

1

1.1 1.2 1.3 1.4 1.5

1 1 2 3 4 4

Why This Book? Geotechnical Engineering The Past and the Future Geotechnical Engineering Can Be Fun Units Problems and Solutions

Case Histories 2.1

2.2

2.3

2.4

2.5

7

Washington Monument (Shallow Mat Foundation) 2.1.1 The Story 2.1.2 Geology and Soil Stratigraphy 2.1.3 Construction 2.1.4 Geometry and Load 2.1.5 Soil Properties 2.1.6 Bearing Capacity 2.1.7 Settlement Rissa Landslide (Slope Stability) 2.2.1 The Story 2.2.2 The Soil Parameters 2.2.3 Slope Stability Back Analysis Seattle 46 M-High MSE Wall (Retaining Wall) 2.3.1 The Story 2.3.2 The Natural Soil Conditions 2.3.3 The Fill and Wall Construction 2.3.4 The Wall Design The New Orleans Charity Hospital Foundation (Deep Foundation) 2.4.1 The Story 2.4.2 The Soil Conditions 2.4.3 Foundation Design and Construction 2.4.4 Settlement Analysis and Measurements The Eurotunnel Linking France and England (Tunneling) 2.5.1 The Story 2.5.2 Rock Stratigraphy

7 7 7 8 9 9 9 13 15 15 16 16 17 17 17 17 17 19 19 20 20 21 22 22 22 v

vi

CONTENTS

2.6

2.7

2.8

2.9

2.10

2.11

2.12

2.5.3 Tunnel Design 2.5.4 Tunnel Construction The Teton Dam (Earth Dam Erosion) 2.6.1 The Story 2.6.2 The Stratigraphy 2.6.3 The Design of the Earth Dam 2.6.4 Filling of the Dam 2.6.5 An Explanation for the Failure The Woodrow Wilson Bridge (Bridge Scour) 2.7.1 The Story 2.7.2 The Soil Stratigraphy 2.7.3 Scour Depth Calculations 2.7.4 Foundation and Cost San Jacinto Monument (Shallow Mat Foundation) 2.8.1 The Story 2.8.2 Geometry, Weight, Construction, and Loading 2.8.3 Soil Stratigraphy and Parameters 2.8.4 Bearing Capacity Calculations 2.8.5 Settlement Calculations 2.8.6 Subsidence in Houston and Impact on Settlement Data Pointe du Hoc Cliffs (Rock Erosion) 2.9.1 The Story 2.9.2 The Rock Stratigraphy and Properties 2.9.3 The Cliff Erosion Process 2.9.4 Proposed Remediation to Mitigate the Erosion The Tower of PISA (Shallow Foundation) 2.10.1 The Story 2.10.2 Dimensions and Soil Stratigraphy 2.10.3 Bearing Capacity, Settlement, and Inclination 2.10.4 The 2001 Repair The Transcona Silo (Shallow Foundation) 2.11.1 The Story 2.11.2 Dimensions and Weight 2.11.3 Soil Properties 2.11.4 Bearing Capacity, Settlement, and Failure 2.11.5 The Up-Righting of the Silo The Saint John River Bridge Abutment (Slope Stability) 2.12.1 The Story 2.12.2 The Bridge and Right Abutment 2.12.3 The Soil Conditions 2.12.4 The Fill and the Approach Embankment Construction 2.12.5 Water Stress Induced by Embankment Construction 2.12.6 Slope Stability Analysis

23 24 25 25 26 26 28 28 28 28 29 29 31 31 31 31 32 32 32 34 35 35 35 35 36 37 37 38 40 41 42 42 42 43 43 44 46 46 46 46 46 46 48

CONTENTS

2.13

2.14

2.15

2.16

2.17

2.18

2.19

2.20

Foundation of Briaud’s House (Shrink Swell Soils) 2.13.1 The Story 2.13.2 The Soil Conditions 2.13.3 The House Foundation 2.13.4 The Tennis Court Foundation The Eiffel Tower (Deep Foundation) 2.14.1 The Story 2.14.2 The Soil Stratigraphy 2.14.3 The Foundation St. Isaac Cathedral (Mat Foundation) 2.15.1 The Story 2.15.2 Construction, Dimensions, and Load 2.15.3 Soil Data 2.15.4 Bearing Capacity and Settlement Calculations National Geotechnical Experimentation Sites at Texas A&M University 2.16.1 The Story 2.16.2 Tieback Wall at the Sand Site 2.16.3 Spread Footings Tests at the Sand Site 2.16.4 Grouted Anchors Under Tension Load at the Clay Site 2.16.5 Drilled and Grouted Piles in Cyclic Tension at the Clay Site The 827 M-High Burj Khalifa Tower Foundation (Combined Pile Raft Foundation) 2.17.1 The Story 2.17.2 The Soil and Rock Conditions 2.17.3 The Foundation Dimensions, Loading, and Settlement New Orleans Levees and Hurricane Katrina (Overtopping Erosion) 2.18.1 The Story 2.18.2 The Soils and the Levees 2.18.3 Erosion of the Overtopped Levees Three Gorges Dam (Concrete Dam) 2.19.1 The Story 2.19.2 The Dam Dimensions and Construction 2.19.3 Soil and Rock Conditions 2.19.4 Environmental Impact 2.19.5 Simple Calculations The Kansai International Airport (Earth Fill in the Sea) 2.20.1 The Story 2.20.2 Dimensions 2.20.3 Construction 2.20.4 Soil Conditions 2.20.5 Loading and Settlement 2.20.6 Simple Calculations

vii 49 49 49 50 51 51 51 51 52 54 54 54 54 56 57 57 57 58 59 59 60 60 61 62 63 63 63 64 66 66 66 66 67 69 70 70 70 70 70 71 72

viii

CONTENTS

2.21

2.22

CHAPTER 3

CHAPTER 4

CHAPTER 5

The Panama Canal (Excavated Slopes) 2.21.1 The Story 2.21.2 Canal Dimensions and Cross-Section 2.21.3 The Gaillard/Culebra Cut 2.21.4 Stratigraphy and Soil Properties of the Culebra Excavated Slopes 2.21.5 Stability of the Culebra Excavated Slopes The Nice Airport Slope Failure (Slope Stability) 2.22.1 The Story 2.22.2 The Sequence of Events 2.22.3 The Soil Conditions 2.22.4 Failure Explanation: Scenario 1 2.22.5 Failure Explanation: Scenario 2 and Alternative

73 73 73 75 75 76 77 77 78 78 79 80

Engineering Geology

81

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

81 81 82 82 83 83 85 86 87

Definition The Earth Geologic Time Rocks Soils Geologic Features Geologic Maps Groundwater Problems and Solutions

Soil Components and Weight-Volume Parameters

91

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

91 91 93 93 94 94 95 96 97 97 98

Particles, Liquid, and Gas Particle Size, Shape, and Color Composition of Gravel, Sand, and Silt Particles Composition of Clay and Silt Particles Particle Behavior Soil Structure Three-Phase Diagram Weight-Volume Parameters Measurement of the Weight-Volume Parameters Solving a Weight-Volume Problem Problems and Solutions

Soil Classification

109

5.1 5.2 5.3 5.4 5.5

109 110 113 116

5.6 5.7

Sieve Analysis Hydrometer Analysis Atterberg Limits and Other Limits Classification Parameters Engineering Significance of Classification Parameters and Plasticity Chart Unified Soil Classification System Aashto Classification System Problems and Solutions

117 118 120 120

CONTENTS

CHAPTER 6

Rocks 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

CHAPTER 7

127 Rock Groups and Identification Rock Mass vs. Rock Substance Rock Discontinuities Rock Index Properties Rock Engineering Properties Rock Mass Rating Rock Engineering Problems Permafrost Problems and Solutions

127 127 130 130 131 132 133 135 136

Site Investigation, Drilling, and Sampling

143

7.1 7.2 7.3 7.4

143 144 144 144 144 145 146 147 147 148 150 150 152 157 157 158 161 163

7.5

7.6 7.7 7.8 7.9

CHAPTER 8

ix

General Preliminary Site Investigation Number and Depth of Borings and in Situ Tests Drilling Rig and Drilling 7.4.1 Drilling Rigs 7.4.2 Wet Rotary Drilling Method 7.4.3 Auger Drilling Method Sampling 7.5.1 Sample Disturbance 7.5.2 Common Sampling Methods Groundwater Level Field Identification and Boring Logs Soil Names Offshore Site Investigations 7.9.1 Offshore Geophysical Investigations 7.9.2 Offshore Geotechnical Drilling 7.9.3 Offshore Geotechnical Sampling Problems and Solutions

In Situ Tests

171

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11

171 174 176 180 181 184 185 187 187 188 188 188 189 189 190

Standard Penetration Test Cone Penetration Test Pressuremeter Test Dilatometer Test Vane Shear Test Borehole Shear Test Plate Load Test California Bearing Ratio Test Pocket Penetrometer and Torvane Tests Pocket Erodometer Test Compaction Control Tests 8.11.1 Sand Cone Test 8.11.2 Rubber Balloon Test 8.11.3 Nuclear Density/Water Content Test 8.11.4 Field Oven Test

x

CONTENTS

8.12

8.13 8.14

CHAPTER 9

190 191 191 191 193 193 195 196 197 200

Elements of Geophysics

215

9.1 9.2

215 215 215 217 218

9.3

9.4

9.5

CHAPTER 10

8.11.5 Lightweight Deflectometer Test 8.11.6 BCD Test Hydraulic Conductivity Field Tests 8.12.1 Borehole Tests 8.12.2 Cone Penetrometer Dissipation Test 8.12.3 Sealed Double-Ring Infiltrometer Test 8.12.4 Two-Stage Borehole Permeameter Test Borehole Erosion Test Offshore In Situ Tests Problems and Solutions

General Seismic Techniques 9.2.1 Seismic Waves 9.2.2 Seismic Reflection 9.2.3 Seismic Refraction 9.2.4 Cross-Hole Test, Seismic Cone Test, and Seismic Dilatometer Test 9.2.5 Spectral Analysis of Surface Waves Electrical Resistivity Techniques 9.3.1 Background on Electricity 9.3.2 Resistivity Tomography Electromagnetic Methods 9.4.1 Electromagnetic Waves 9.4.2 Ground-Penetrating Radar 9.4.3 Time Domain Reflectometry Remote Sensing Techniques 9.5.1 LIDAR 9.5.2 Satellite Imaging Problems and Solutions

219 220 223 223 224 226 226 226 228 228 228 229 230

Laboratory Tests

235

10.1 10.2

235 236 236 237 237 237 241 243 243 243 243 243 246 246 246

10.3

10.4

General Measurements 10.2.1 Normal Stress or Pressure 10.2.2 Shear Stress 10.2.3 Water Compression Stress 10.2.4 Water Tension Stress 10.2.5 Normal Strain 10.2.6 Shear Strain 10.2.7 Bender Elements Compaction Test: Dry Unit Weight 10.3.1 Saturated Soils 10.3.2 Unsaturated Soils Compaction Test: Soil Modulus 10.4.1 Saturated Soils 10.4.2 Unsaturated Soils

CONTENTS

10.5

10.6

10.7

10.8

10.9

10.10

10.11

10.12

10.13

10.14

10.15

10.16

10.17 10.18 10.19 10.20

CHAPTER 11

Consolidation Test 10.5.1 Saturated Soils 10.5.2 Unsaturated Soils Swell Test 10.6.1 Saturated Soils 10.6.2 Unsaturated Soils Shrink Test 10.7.1 Saturated Soils 10.7.2 Unsaturated Soils Collapse Test 10.8.1 Saturated Soils 10.8.2 Unsaturated Soils Direct Shear Test 10.9.1 Saturated Soils 10.9.2 Unsaturated Soils Simple Shear Test 10.10.1 Saturated Soils 10.10.2 Unsaturated Soils Unconfined Compression Test 10.11.1 Saturated Soils 10.11.2 Unsaturated Soils Triaxial Test 10.12.1 Saturated Soils 10.12.2 Unsaturated Soils Resonant Column Test 10.13.1 Saturated Soils 10.13.2 Unsaturated Soils Lab Vane Test 10.14.1 Saturated Soils 10.14.2 Unsaturated Soils Soil Water Retention Curve (Soil Water Characteristic Curve) Test 10.15.1 Saturated Soils 10.15.2 Unsaturated Soils Constant Head Permeameter Test 10.16.1 Saturated Soils 10.16.2 Unsaturated Soils Falling Head Permeameter Test for Saturated Soils Wetting Front Test for Unsaturated Soils Air Permeability Test for Unsaturated Soils Erosion Test 10.20.1 Saturated Soils 10.20.2 Unsaturated Soils Problems and Solutions

xi

246 246 250 250 250 250 252 252 252 252 252 252 253 253 253 255 255 256 256 256 256 257 257 261 262 262 265 265 265 266 266 266 267 268 268 269 270 272 273 273 273 275 276

Stresses, Effective Stress, Water Stress, Air Stress, and Strains

301

11.1 11.2

301

General Stress Vector, Normal Stress, Shear Stress, and Stress Tensor

301

xii

CONTENTS

11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11.15 11.16 11.17

11.18 11.19 11.20 11.21

CHAPTER 12

Sign Convention for Stresses and Strains Calculating Stresses on Any Plane: Equilibrium Equations for Two-Dimensional analysis Calculating Stresses on Any Plane: Mohr Circle for Two-Dimensional Analysis Mohr Circle in Three Dimensions Stress Invariants Displacements Normal Strain, Shear Strain, and Strain Tensor Cylindrical Coordinates and Spherical Coordinates Stress-Strain Curves Stresses in the Three Soil Phases Effective Stress (Unsaturated Soils) Effective Stress (Saturated Soils) Area Ratio Factors 𝛼 and 𝛽 Water Stress Profiles Water Tension and Suction 11.17.1 Matric Suction 11.17.2 Contractile Skin 11.17.3 Osmotic Suction 11.17.4 Relationship between Total Suction and Relative Humidity 11.17.5 Trees Precision on Water Content and Water Tension Stress Profile at Rest in Unsaturated Soils Soil Water Retention Curve Independent Stress State Variables Problems and Solutions

302 302 303 304 304 304 305 306 307 307 307 308 308 310 310 311 313 313 313 315 316 316 318 318 319

Problem-Solving Methods

333

12.1 12.2 12.3 12.4

333 333 333 334

12.5

General Drawing to Scale as a First Step Primary Laws Continuum Mechanics Methods 12.4.1 Solving a Failure Problem: Limit Equilibrium, Method of Characteristics, Lower and Upper Bound Theorems 12.4.2 Examples of Solving a Failure Problem 12.4.3 Solving a Deformation Problem 12.4.4 Example of Solving a Deformation Problem 12.4.5 Solving a Flow Problem 12.4.6 Example of Solving a Flow Problem Numerical Simulation Methods 12.5.1 Finite Difference Method 12.5.2 Examples of Finite Difference Solutions 12.5.3 Finite Element Method 12.5.4 Example of Finite Element Solution

334 334 336 336 339 339 343 343 344 347 352

CONTENTS

CHAPTER 13

12.5.5 Boundary Element Method 12.5.6 Discrete Element Method 12.6 Probability and Risk Analysis 12.6.1 Background 12.6.2 Procedure for Probability Approach 12.6.3 Risk and Tolerable Risk 12.6.4 Example of Probability Approach 12.7 Regression Analysis 12.8 Artificial Neural Network Method 12.9 Dimensional Analysis ∏ 12.9.1 Buckingham Theorem 12.9.2 Examples of Dimensional Analysis 12.10 Similitude Laws for Experimental Simulations 12.10.1 Similitude Laws 12.10.2 Example of Similitude Laws Application for a Scaled Model 12.10.3 Example of Similitude Laws Application for a Centrifuge Model 12.11 Types of Analyses (Drained–Undrained, Effective Stress–Total Stress, Short-Term–Long-Term) Problems and Solutions

356 356 357 357 360 362 364 365 366 367 367 368 369 369

Soil Constitutive Models

395

13.1

395 395 396 397 397 398 399 400 401 401 402 402 403 403 404 405 407 407

13.2

13.3

13.4

CHAPTER 14

xiii

Elasticity 13.1.1 Elastic Model 13.1.2 Example of Use of the Elastic Model Linear Viscoelasticity 13.2.1 Simple Models: Maxwell and Kelvin-Voigt Models 13.2.2 General Linear Viscoelasticity Plasticity 13.3.1 Some Yield Functions and Yield Criteria 13.3.2 Example of Use of Yield Criteria 13.3.3 Plastic Potential Function and Flow Rule 13.3.4 Hardening or Softening Rule 13.3.5 Example of Application of Plasticity Method Common Soil Models 13.4.1 Duncan-Chang Hyperbolic Model 13.4.2 Modified Cam Clay Model 13.4.3 Barcelona Basic Model 13.4.4 Water Stress Predictions Problems and Solutions

370 370 371 371

Flow of Fluid and Gas Through Soils

419

14.1 14.2

419 419

General Flow of Water in a Saturated Soil 14.2.1 Discharge Velocity, Seepage Velocity, and Conservation of Mass 14.2.2 Heads 14.2.3 Hydraulic Gradient

419 420 420

xiv

CONTENTS

14.2.4 14.2.5 14.2.6 14.2.7 14.2.8

14.3

CHAPTER 15

Darcy’s Law: The Constitutive Law Hydraulic Conductivity Field vs. Lab Values of Hydraulic Conductivity Seepage Force Quick Sand Condition and Critical Hydraulic Gradient 14.2.9 Quick Clay 14.2.10 Sand Liquefaction 14.2.11 Two-Dimensional Flow Problem 14.2.12 Drawing a Flow Net for Homogeneous Soil 14.2.13 Properties of a Flow Net for Homogeneous Soil 14.2.14 Calculations Associated with Flow Nets 14.2.15 Flow Net for Hydraulically Anisotropic Soil 14.2.16 Flow and Flow Net for Layered Soils Flow of Water and Air in Unsaturated Soil 14.3.1 Hydraulic Conductivity for Water and for Air 14.3.2 One-Dimensional Flow 14.3.3 Three-Dimensional Water Flow 14.3.4 Three-Dimensional Air Flow Problems and Solutions

420 421 422 422 423 424 424 424 426 427 428 429 429 431 431 432 434 435 436

Deformation Properties

447

15.1 15.2 15.3 15.4 15.5 15.6 15.7

447 448 448 449 451 451

15.8 15.9 15.10 15.11 15.12 15.13

15.14 15.15 15.16 15.17 15.18 15.19

Modulus of Deformation: General Modulus: Which One? Modulus: Influence of State Factors Modulus: Influence of Loading Factor Modulus: Differences Between Fields of Application Modulus: Modulus of Subgrade Reaction, and Stiffness Common Values of Soil Modulus and the Poisson’s Ratio Correlations with Other Tests Modulus: A Comprehensive Model Initial Tangent Modulus Go or Gmax Reduction of Gmax with Strain: The G/Gmax Curve Preconsolidation Pressure and Overconsolidation Ratio from Consolidation Test Compression Index, Recompression Index, and Secondary Compression Index from the Consolidation Test Time Effect From the Consolidation Test Modulus, Time Effect, and Cyclic Effect from the Pressuremeter Test Resilient Modulus for Pavements Unsaturated Soils: Effect of Drying and Wetting on the Modulus Shrink-Swell Deformation Behavior, Shrink-Swell Modulus Collapse Deformation Behavior Problems and Solutions

452 454 454 457 459 459

461 462 463 465 465 466 470 471

CONTENTS

CHAPTER 16

Shear Strength Properties

485

16.1 16.2

485 485 485 486 486 486 486 487

16.3 16.4

16.5 16.6 16.7

16.8

16.9 16.10 16.11 16.12 16.13 16.14 16.15 16.16

CHAPTER 17

xv

General Basic Experiments 16.2.1 Experiment 1 16.2.2 Experiment 2 16.2.3 Experiment 3 16.2.4 Experiment 4 16.2.5 Experiment 5 16.2.6 Experiment 6 Stress-Strain Curve, Water Stress Response, and Stress Path Shear Strength Envelope 16.4.1 General Case 16.4.2 The Case of Concrete 16.4.3 Overconsolidated Fine-Grained Soils 16.4.4 Coarse-Grained Soils Unsaturated Soils Experimental Determination of Shear Strength (Lab Tests, In Situ Tests) Estimating Effective Stress Shear Strength Parameters 16.7.1 Coarse-Grained Soils 16.7.2 Fine-Grained Soils Undrained Shear Strength of Saturated Fine-Grained Soils 16.8.1 Weak Soil Skeleton: Soft, Normally Consolidated Soils 16.8.2 Strong Soil Skeleton: Overconsolidated Soils 16.8.3 Rate of Loading Effect on the Undrained Strength The Ratio SU ∕𝜎 ′OV and the Shansep Method Undrained Shear Strength for Unsaturated Soils Pore-Pressure Parameters A and B Estimating Undrained Shear Strength Values Residual Strength Parameters and Sensitivity Strength Profiles Types of Analyses Transformation from Effective Stress Solution to Undrained Strength Solution Problems and Solutions

487 488 488 489 490 490 490 491 492 492 494 495 496 496 497 498 498 499 500 502 503 503 504 504

Thermodynamics for Soil Problems

511

17.1 17.2 17.3 17.4 17.5 17.6 17.7

511 511 512 512 513 514 515

General Definitions Constitutive and Fundamental Laws Heat Conduction Theory Axisymmetric Heat Propagation Thermal Properties of Soils Multilayer Systems

xvi

CONTENTS

CHAPTER 18

17.8 Applications 17.9 Thermal Cone Penetrometer Test 17.10 Frozen Soils Problems and Solutions

516 516 517 519

Shallow Foundations

523

18.1 18.2 18.3 18.4

523 523 523

18.5 18.6

18.7 18.8

18.9

18.10

18.11 18.12

18.13

Definitions Case History Definitions and Design Strategy Limit States, Load and Resistance Factors, and Factor of Safety General Behavior Ultimate Bearing Capacity 18.6.1 Direct Strength Equations 18.6.2 Terzaghi’s Ultimate Bearing Capacity Equation 18.6.3 Layered Soils 18.6.4 Special Loading 18.6.5 Ultimate Bearing Capacity of Unsaturated Soils Load Settlement Curve Approach Settlement 18.8.1 General Behavior 18.8.2 Elasticity Approach for Homogeneous Soils 18.8.3 Elasticity Approach for Layered Soils 18.8.4 Chart Approach 18.8.5 General Approach 18.8.6 Zone of Influence 18.8.7 Stress Increase with Depth 18.8.8 Choosing a Stress-Strain Curve and Setting Up the Calculations 18.8.9 Consolidation Settlement: Magnitude 18.8.10 Consolidation Settlement: Time Rate 18.8.11 Creep Settlement 18.8.12 Bearing Pressure Values Shrink-Swell Movement 18.9.1 Water Content or Water Tension vs. Strain Curve 18.9.2 Shrink-Swell Movement Calculation Methods 18.9.3 Step-by-Step Procedure 18.9.4 Case History Foundations on Shrink-Swell Soils 18.10.1 Types of Foundations on Shrink-Swell Soils 18.10.2 Design Method for Stiffened Slabs on Grade Tolerable Movements Large Mat Foundations 18.12.1 General Principles 18.12.2 Example of Settlement Calculations 18.12.3 Two Case Histories Multiple Spread Footings and Interaction 18.13.1 Interaction Effect

526 529 529 529 532 534 536 537 537 540 540 541 542 543 545 545 546 548 548 549 549 551 551 551 551 553 554 555 555 556 560 561 561 562 564 567 567

CONTENTS

18.13.2 Settlement Calculations 18.13.3 Comparison Between Multiple Spread Footings and a Mat Foundation Problems and Solutions CHAPTER 19

xvii 567 568 569

Deep Foundations

591

19.1 19.2 19.3

591 591 593 593 595 598 599 600 601

19.4

19.5

19.6

19.7

19.8

Different Types of Deep Foundations Design Strategy Pile Installation 19.3.1 Installation of Bored Piles 19.3.2 Nondestructive Testing of Bored Piles 19.3.3 Installation of Driven Piles 19.3.4 Pile Driving Formulas 19.3.5 Wave Propagation in a Pile 19.3.6 Wave Equation Analysis 19.3.7 Information from Pile Driving Measurements (PDA, Case, CAPWAP) 19.3.8 Suction Caissons 19.3.9 Load Testing (Static, Statnamic, Osterberg) Vertical Load: Single Pile 19.4.1 Ultimate Vertical Capacity for a Single Pile 19.4.2 Miscellaneous Questions about the Ultimate Capacity of a Single Pile 19.4.3 Settlement of a Single Pile Vertical Load: Pile Group 19.5.1 Ultimate Vertical Capacity of a Pile Group 19.5.2 Settlement of Pile Groups Downdrag 19.6.1 Definition and Behavior 19.6.2 Downdrag on a Single Pile 19.6.3 Sample Downdrag Calculations 19.6.4 LRFD Provisions 19.6.5 Downdrag on a Group of Piles Piles in Shrink-Swell Soils 19.7.1 The Soil Shrinks 19.7.2 The Soil Swells Horizontal Load and Moment: Single Pile 19.8.1 Definitions and Behavior 19.8.2 Ultimate Capacity 19.8.3 Displacement and Maximum Moment: Long Flexible Pile 19.8.4 Displacement and Maximum Moment: Short Rigid Pile 19.8.5 Modulus of Subgrade Reaction 19.8.6 Free-Head and Fixed-Head Conditions 19.8.7 Rate of Loading Effect 19.8.8 Cyclic Loading Effect

605 608 609 612 613 619 623 625 625 627 629 629 629 630 633 634 634 635 635 636 636 636 637 638 639 639 640 641

xviii

CONTENTS

CHAPTER 20

19.8.9 P-y Curve Approach 19.8.10 Horizontal Loading Next to a Trench 19.9 Horizontal Load and Moment: Pile Group 19.9.1 Overturning Moment 19.9.2 Ultimate Capacity 19.9.3 Movement 19.10 Combined Piled Raft Foundation Problems and Solutions

642 643 643 644 645 646 647 649

Slope Stability

679

20.1 20.2 20.3

679 679 680 681 681 681 682 682 683 684 684 685 685 686 687 689 689 691 694 694 696 696 696 696 697 697 697 697 698

20.4 20.5 20.6 20.7 20.8

20.9

20.10

20.11 20.12 20.13 20.14

20.15

General Design Approach Infinite Slopes 20.3.1 Dry Sand 20.3.2 Dry c′ − 𝜑′ Soil 20.3.3 c′ − 𝜑′ Soil with Seepage 20.3.4 c′ − 𝜑′ Soil with Unsaturated Conditions Seepage Force in Stability Analysis Plane Surfaces Block Analysis Slopes with Water in Tensile Cracks Chart Methods 20.8.1 Taylor Chart 20.8.2 Spencer Chart 20.8.3 Janbu Chart 20.8.4 Morgenstern Chart Method of Slices 20.9.1 Ordinary Method of Slices 20.9.2 Bishop Simplified Method 20.9.3 Generalized Equilibrium Method 20.9.4 Critical Failure Circle Water Stress for Slope Stability 20.10.1 Piezometric and Phreatic Surface 20.10.2 Water Stress Ratio Value 20.10.3 Grid of Water Stress Values 20.10.4 Water Stress Due to Loading 20.10.5 Seepage Analysis Types of Analyses Progressive Failure in Strain-Softening Soils Shallow Slide Failures in Compacted Unsaturated Embankments Reinforced Slopes 20.14.1 Reinforcement Type 20.14.2 Factor of Safety Probabilistic Approach 20.15.1 Example 1 20.15.2 Example 2 20.15.3 Example 3

698 699 699 699 699 700 700 701

CONTENTS

CHAPTER 21

20.16 20.17 20.18 20.19 20.20

Three-Dimensional Circular Failure Analysis Finite Element Analysis Stability of Slope Corners Slope Stability Databases Seismic Slope Analysis 20.20.1 Pseudostatic Method 20.20.2 Newmark’s Displacement Method 20.20.3 Post-Earthquake Stability Analysis 20.20.4 Dynamic Finite Element Analysis 20.21 Monitoring 20.22 Repair Methods 20.22.1 Increase the Resisting Moment 20.22.2 Decrease the Driving Moment Problems and Solutions

701 702 703 703 704 704 705 706 706 706 709 709 710 710

Compaction

727

21.1 21.2 21.3 21.4 21.5

727 728 729 730 731 732 733 733 735 737

21.6 21.7

CHAPTER 22

xix

General Compaction Laboratory Tests Compaction Field Tests Compaction and Soil Type Intelligent Roller Compaction 21.5.1 Soil Modulus from Vibratory Rollers 21.5.2 Roller Measurements as Compaction Indices Impact Roller Compaction Dynamic or Drop-Weight Compaction Problems and Solutions

Retaining Walls

743

22.1 22.2

743

22.3

22.4 22.5 22.6 22.7

Different Types (Top-Down, Bottom-Up) Active, at Rest, Passive Earth Pressure, and Associated Displacement Earth Pressure Theories 22.3.1 Coulomb Earth Pressure Theory 22.3.2 Rankine Earth Pressure Theory 22.3.3 Earth Pressure Theory by Mohr Circle 22.3.4 Water in the Case of Compression Stress (Saturated) 22.3.5 Water in the Case of Tension Stress (Unsaturated or Saturated) 22.3.6 Influence of Surface Loading (Line Load, Pressure) 22.3.7 General Case and Earth Pressure Profiles Special Case: Undrained Behavior of Fine-Grained Soils At-Rest Earth Pressure Earth Pressure Due to Compaction Earth Pressures in Shrink-Swell Soils

743 744 744 746 747 748 749 749 749 750 750 752 752

xx

CONTENTS

CHAPTER 23

22.8 Displacements 22.9 Gravity Walls 22.10 Mechanically Stabilized Earth Walls 22.10.1 External Stability 22.10.2 Internal Stability 22.11 Cantilever Top-Down Walls 22.11.1 Depth of Embedment and Pressure Diagram 22.11.2 Displacement of the Wall, Bending Moment, and P-y Curves 22.12 Anchored Walls and Strutted Walls 22.12.1 Pressure Distribution 22.12.2 Pressure vs. Movement 22.12.3 Base Instability 22.12.4 Movement of Wall and Ground Surface 22.12.5 Anchors 22.12.6 Embedment Depth and Downdrag 22.12.7 The P-y Curve Approach and the FEM Approach 22.13 Soil Nail Walls 22.13.1 External Stability 22.13.2 Internal Stability 22.13.3 Wall Movement 22.13.4 Other Issues 22.14 Special Case: Trench Problems and Solutions

760 762 763 764 765 765 767 769 771 771 771 774 776 776 776 778

Earthquake Geoengineering

807

23.1 23.2 23.3 23.4 23.5 23.6 23.7

807 807 809 809 809 813 814

Background Earthquake Magnitude Wave Propagation Dynamic Soil Properties Ground Motion Seismic Hazard Analysis Ground Response Analysis 23.7.1 One-Dimensional Solution for Undamped Linear Soil on Rigid Rock 23.7.2 One-Dimensional Solution for Damped Linear Soil on Rigid Rock 23.7.3 Layered Soils 23.8 Design Parameters 23.8.1 Site Classes A–E for Different Soil Stiffness 23.8.2 Code-Based Spectrum 23.8.3 Hazard Levels 23.9 Liquefaction 23.9.1 Phenomenon 23.9.2 When to Do a Liquefaction Study? 23.9.3 When Can a Soil Liquefy? 23.10 Seismic Slope Stability

753 753 756 757 757 759 760

814 815 815 816 816 817 818 818 818 819 819 823

CONTENTS

CHAPTER 24

xxi

23.11 Seismic Design of Retaining Walls 23.11.1 Seismic Design of Gravity Walls 23.11.2 Water Pressures on Walls during an Earthquake 23.11.3 Seismic Design of MSE Walls 23.11.4 Seismic Design of Cantilever Walls 23.11.5 Seismic Design of Anchored Walls 23.12 Seismic Design of Foundations Problems and Solutions

824 824 825 827 827 827 827 829

Erosion of Soils and Scour Problems

843

24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8

843 844 844 845 849 850 852 853 854

The Erosion Phenomenon Erosion Models Measuring the Erosion Function Soil Erosion Categories Rock Erosion Water Velocity Geometry of the Obstacle Bridge Scour 24.8.1 Maximum Scour Depth (zmax ) Analysis 24.8.2 Maximum Shear Stress at Soil-Water Boundary When Scour Begins 24.8.3 Final Scour Depth (zfinal ) Analysis for Constant Velocity Flow and Uniform Soil 24.8.4 Final Scour Depth (Z final ) Analysis for a Velocity Hydrograph and Layered Soil 24.8.5 The Woodrow Wilson Bridge Case History 24.8.6 The Observation Method for Scour 24.9 River Meandering 24.9.1 Predicting River Meandering 24.9.2 The Brazos River Meander Case History (Park, 2007) 24.9.3 Observation Method for Meanders 24.10 Levee Overtopping 24.10.1 General Methodology 24.10.2 Hurricane Katrina Levee Case History: New Orleans 24.11 Countermeasures for Erosion Protection 24.11.1 Rip Rap 24.11.2 Grass 24.12 Internal Erosion of Earth Dams 24.12.1 The Phenomenon 24.12.2 Most Susceptible Soils 24.12.3 Criterion to Evaluate Internal Erosion Potential 24.12.4 Remedial Measures Problems and Solutions

858 860 861 862 864 864 864 866 868 869 869 870 871 871 872 873 873 874 874 876 877

xxii

CONTENTS

CHAPTER 25

Geoenvironmental Engineering

893

25.1 25.2 25.3 25.4

893 893 894 895 895 897 898 898 898 900 904 904 905 908 909 910 910 912 912 913 914 914 915 916 917

25.5

25.6

25.7

25.8

CHAPTER 26

Introduction Types of Wastes and Contaminants Laws and Regulations Geochemistry Background 25.4.1 Chemistry Background 25.4.2 Geochemistry Background Contamination 25.5.1 Contamination Sources 25.5.2 Contamination Detection and Site Characterization 25.5.3 Contaminant Transport and Fate Remediation 25.6.1 Risk Assessment and Strategy 25.6.2 In Situ Waste Containment 25.6.3 Soil Remediation 25.6.4 Groundwater Remediation Landfills 25.7.1 Waste Properties 25.7.2 Regulations 25.7.3 Liners 25.7.4 Covers 25.7.5 Leachate Collection 25.7.6 Landfill Slopes 25.7.7 Gas Generation and Management Future Considerations Problems and Solutions

Geosynthetics

925

26.1 26.2 26.3

925 925 926 926 929 930 931 932 933 933 934 935 935 936 939 939 940 941

26.4 26.5 26.6

26.7 26.8

General Types of Geosynthetics Properties of Geosynthetics 26.3.1 Properties of Geotextiles 26.3.2 Properties of Geomembranes 26.3.3 Properties of Geogrids 26.3.4 Properties of Geosynthetics Clay Liners 26.3.5 Properties of Geofoams 26.3.6 Properties of Geonets Design for Separation Design of Liners and Covers Design for Reinforcement 26.6.1 Road Reinforcement 26.6.2 Mechanically Stabilized Earth Geosynthetic Walls 26.6.3 Reinforced Slopes 26.6.4 Reinforced Foundations and Embankments Design for Filtration and Drainage Design for Erosion Control

CONTENTS

26.9

CHAPTER 27

Other Design Applications 26.9.1 Lightweight Fills 26.9.2 Compressible Inclusions 26.9.3 Thermal Insulation 26.9.4 Geosynthetics and Landfill Slopes Problems and Solutions

xxiii 942 942 942 943 943 943

Soil Improvement

957

27.1 27.2

957

27.3

27.4

27.5

27.6

27.7

Overview Soil Improvement Without Admixture in Coarse-Grained Soils 27.2.1 Compaction 27.2.2 Dynamic Compaction 27.2.3 Vibrocompaction 27.2.4 Other Methods Soil Improvement Without Admixture in Fine-Grained Soils 27.3.1 Displacement-Replacement 27.3.2 Preloading Using Fill 27.3.3 Prefabricated Vertical Drains and Preloading Using Fill 27.3.4 Preloading Using Vacuum 27.3.5 Electro-Osmosis 27.3.6 Ground Freezing 27.3.7 Hydro-Blasting Compaction Soil Improvement with Replacement 27.4.1 Stone Columns Without Geosynthetic Sock 27.4.2 Stone Columns with Geosynthetic Encasement 27.4.3 Dynamic Replacement Soil Improvement with Grouting and Admixtures 27.5.1 Particulate Grouting 27.5.2 Chemical Grouting 27.5.3 Jet Grouting 27.5.4 Compaction Grouting 27.5.5 Compensation Grouting 27.5.6 Mixing Method 27.5.7 Lime Treatment 27.5.8 Microbial Methods Soil Improvement with Inclusions 27.6.1 Mechanically or Geosynthetically Stabilized Earth 27.6.2 Ground Anchors and Soil Nails 27.6.3 Geosynthetic Mat and Column-Supported Embankment Selection of Soil Improvement Method Problems and Solutions

957 957 957 957 960 961 961 961 963 964 965 965 965 966 966 967 968 969 969 970 970 970 970 971 972 972 973 973 973 973 975 975

xxiv

CONTENTS

CHAPTER 28

Technical Communications

981

28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 28.9 28.10 28.11 28.12 28.13

981 981 982 982 982 984 984 984 985 985 986 986 986

General Emails Letters Geotechnical Reports Theses and Dissertations Visual Aids for Reports Phone Calls Meetings Presentations and Powerpoint Slides Media Interaction Ethical Behavior Professional Societies Rules for a Successful Career

References Index

987 1005

ACKNOWLEDGMENTS FIRST EDITION One of the greatest joys in writing this book was working as a team with all my PhD students. From 2010 to 2013, they contributed tremendously to making this book possible. The leader of the team was Ghassan Akrouch. I thank them all very sincerely for their magnificent help. The beautiful memories of our work together on this huge project will be with me as a source of strength and friendship forever. • • • •

Ghassan Akrouch (Lebanon) Alireza Mirdamadi (Iran) Deeyvid Saez (Panama) Mojdeh Asadollahipajouh (Iran)

• • • • • • • • • • • • •

Congpu Yao (China) Stacey Tucker (USA) Negin Yousefpour (Iran) Oswaldo Bravo (Peru) DoHyun Kim (Korea) Axel Montalvo (Puerto Rico) Gang Bi (China) Mohsen Madhavi (Iran) Seung Jae Oh (Korea) Seok Gyu Kim (Korea) Mohammad Aghahadi (Iran) Yasser Koohi (Iran) Carlos Fuentes (Mexico)

xxv

xxvi

ACKNOWLEDGMENTS

My colleagues also provided advice on many topics:

• • • • •

• Marcelo Sanchez (Texas A&M University) • Don Murff (Exxon) • Jose Roesset (Texas A&M University)

One person stands out as a major helper in this book project by her dedication to the task and her relentless denial of the impossible: my assistant Theresa Taeger, who took care of the hundreds of illustration permission requests in record time. I also want to thank all those who share their knowledge and intellectual property online. Without the Internet as a background resource, this work would have taken much longer. Anna TIMCHENKO

Giovanna Biscontin (Texas A&M University) Chuck Aubeny (Texas A&M University) Zenon Medina Cetina (Texas A&M University) Vincent Drnevich (Purdue University) Chris Mathewson (Texas A&M University)

SECOND EDITION The new chapter on case histories required many figures to be redrawn. Anna Timchenko stands out as the one who single-handedly, patiently, and very reliably prepared dozens of figures for that chapter. Others making significant contributions to the illustrations in the second edition include Jerome Sfeir, Erick Cruz, Tehseena Ali, Anna Shidlovskaya, and Blake Thurman.

Jerome SFEIR

Erick CRUZ

ACKNOWLEDGMENTS

Tehseena ALI

Anna SHIDLOVSKAYA

xxvii

Blake THURMAN

CHAPTER 1

Introduction

1.1

WHY THIS BOOK?

1.2 GEOTECHNICAL ENGINEERING

“Things should be made as simple as possible but not a bit simpler than that.” Albert Einstein (Safir and Safire, 1982) Finding the Einstein threshold of optimum simplicity was a constant goal for the author when writing this book (Figure 1.1). The first driving force for writing it was the coming of age of unsaturated soil mechanics: There was a need to introduce geotechnical engineering as dealing with true three-phase soils while treating saturated soil as a special case, rather than the other way around. The second driving force was to cover as many geotechnical engineering topics as reasonably possible in an introductory book, to show the vast domain covered by geotechnical engineering and its important contributions to society. Dams, bridges, buildings, pavements, landfills, tunnels, and many other infrastructure elements involve geotechnical engineering. The driving force for the second edition was the desire to include case histories to further demonstrate the considerable role played by geotechnical engineers in society and also to update the first edition. The intended audience is anyone who is starting in the field of geotechnical engineering, including university students.

Geotechnical engineering is a young (∼100 years) professional field dealing with soils within a few hundred meters of a planet’s surface for the purpose of civil engineering structures. For geotechnical engineers, soils can be defined as loosely bound to unbound, naturally occurring materials that cover the top few hundred meters of a planet. In contrast, rock is a strongly bound, naturally occurring material found within similar depths or deeper. At the boundary between soils and rocks are intermediate geo-materials. The classification tests and the range of properties described in this book help to distinguish between these three types of naturally occurring materials. Geotechnical engineers must make decisions in the best interest of the public with respect to safety and economy. Their decisions are related to topics such as: • • • • • •

These geotechnical structures or projects are subjected to loads, which include: • • • •

Too complex

Threshold of optimum simplicity

Too simple

Figure 1.1 Einstein threshold of optimum simplicity. (Source: Photo by Ferdinand Schmutzer.)

Foundations Slopes Retaining walls Dams Landfills Tunnels

Loads from a structure Weight of a slope Push on a retaining wall Environmental loads, such as waves, wind, rivers, earthquakes, floods, droughts, and chemical changes, among others

Note that current practice is based on testing an extremely small portion of the soil or rock present in the project area. A typical soil investigation might involve testing 0.0001% of the soil that will provide the foundation support for the structure. Yet, on the basis of this extremely limited data, the

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

1

2

1 INTRODUCTION

Figure 1.2 A rendition of the geotechnical engineering world. (Source: Courtesy of Hayward Baker Inc., Geotechnical Contractor.)

geotechnical engineer must predict the behavior of the entire heterogeneous mass of soil. This is why geotechnical engineering is a very difficult discipline. Yet, as Terzaghi is said to have noted, there is no glory in foundations. Indeed, most of our work is buried (Figure 1.2). For example, everyone knows the Eiffel Tower in Paris, but very few know about its foundation.

1.3

THE PAST AND THE FUTURE

While it is commonly agreed that geotechnical engineering started with the work of Karl Terzaghi at the beginning of the 20th century, history is rich in instances where soils and soils-related engineering played an important role in the evolution of humankind (Kerisel, 1985; Peck, 1985; Skempton, 1985). In prehistoric times (before 3000 BC), soil was used as a building material. In ancient times (3000–300 BC), roads, canals, and bridges were very important to warriors.

In Roman times (300 BC–300 AD), structures started to become larger and foundations could no longer be ignored. The Middle Ages (AD 300–1400) were mainly a period of war, in which structures became even heavier, including castles and cathedrals with very thick walls. Severe settlements and instabilities were experienced. The Renaissance (AD 1400–1650) was a period of enormous development in the arts, and several great artists proved to be great engineers as well. This was the case of Leonardo da Vinci and more particularly Michelangelo. Modern times (AD 1650–1900) saw significant engineering development, with a shift from military engineering to civil engineering. In 1776, Charles Coulomb developed his earth pressure theory, followed in 1855 by Henry Darcy and his seepage law. In 1857, William Rankine proposed his own earth pressure theory, closely followed by Carl Culman and his graphical earth pressure solution. In 1882, Otto Mohr presented his stress theory and the famous Mohr circle, and in 1885 Joseph Boussinesq provided the solution to an important elasticity problem for soils.

1.4 GEOTECHNICAL ENGINEERING CAN BE FUN

From 1900 to 2000 was the true period of development of modern geotechnical engineering, with the publication of Karl Terzaghi’s book Erdbaumechanik (in 1925), which was soon translated into English; new editions were co-authored with Ralph Peck, beginning in 1948. The progress over the past 50 years has been stunning, with advances in the understanding of fundamental soil behavior and associated soil models (e.g., unsaturated soils), numerical simulations made possible by the computer revolution, the development of large machines (e.g., drill rigs for bored piles), and a number of ingenious ideas (e.g., reinforced earth walls, pile driving analyzer, geosynthetics). Geotechnical engineering has transcended the ages because all structures built on or in a planet have to rest on a soil or rock surface; as a result, the geotechnical engineer is here to stay and will continue to be a very important part of humanity’s evolution. The Tower of Pisa is one of the most famous examples of a project that did not go as planned, mostly because of the limited knowledge extant some 900 years ago. Today designing a proper foundation for the Tower of Pisa is a very simple exercise, because of our progress. One cannot help but project another 900 years ahead and wonder what progress will have been made. Will we have: • complete nonintrusive site investigation of the entire soil volume? • automated four-dimensional (4D) computer-generated design by voice recognition and based on a target risk? • tiny and easily installed instruments to monitor geotechnical structures? • unmanned robotic machines working at great depth? (a)

(b)

(d)

3

• significant development of the underground? • extension of projects into the sea? • soil structure interaction extended to thermal and magnetic engineering? • failures down to a minimum? • expert systems to optimize repair of defective geotechnical engineering projects? • geospace engineering of other planets? • geotechnical engineers with advanced engineering judgment taught in universities? • no more lawyers, because of the drastic increase in project reliability?

1.4 GEOTECHNICAL ENGINEERING CAN BE FUN Geotechnical engineering can be fun and entertaining, as the book by Elton (1999) on geo-magic demonstrates. Such phenomena as the magic sand (watch this video: https://www .stevespanglerscience.com/lab/experiments/magic-hydro phobic-sand/), water going uphill, the surprisingly strong sand pile (Figure 1.3e), the swelling clay pie (Figure 1.3d), and the suddenly very stiff glove full of sand will puzzle the uninitiated. Geotechnical engineering is seldom boring; indeed: the complexity of soil deposits and soil behavior can always surprise us with unanticipated results. The best geotechnical engineering work will always include considerations regarding geology, proper site characterization, sound fundamental soil mechanics principles, advanced knowledge of all the tools available, keen observation, and engineering (c)

(e)

Figure 1.3 Soil magic. (Source: Courtesy of David J. Elton.)

4

1 INTRODUCTION

judgment. The fact that geotechnical engineering is so complex makes this field an unending discovery process, which keeps the interest of its adepts over their lifetimes. 1.5

UNITS

In engineering, a number without units is usually worthless and often dangerous. On this planet, the unit system most commonly used in geotechnical engineering is the System International or SI system. In the SI system, the unit of mass is the kilogram (kg), which is defined as the mass of a platinum-iridium international prototype kept at the International Bureau of Weights and Measures in Paris, France. On Earth, the kilogram-mass weighs about the same as 10 small apples. The unit of length is the meter, defined as the length of the path traveled by light in a vacuum during a time interval of 1/299,792,458 of a second. A meter is about the length of a big step for an average human. The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. Watches and clocks often have a hand ticking off the seconds. The unit of temperature is the Kelvin, defined as 1/273.16 of the difference in temperature between the absolute zero and the triple point of water. The degree Celsius (C) is also commonly used; it has the same magnitude as the degree Kelvin but starts at ∼0∘ C (∼273 K) for the freezing point of water and uses ∼100∘ C (∼373 K) for the boiling point of water. There are seven fundamental units in a unit system, but these four (kg, m, s, K) are the most commonly used in geotechnical engineering. The other

fundamental units in the SI system are the mole (substance), the candela (light), and the ampere (electricity). Other geotechnical engineering units are derived from these fundamental units. The unit of force is the Newton, which is the force required to accelerate a mass of 1 kg to 1 m/s2 . 1 N = 1 kg × 1 m∕s2

This force is about the weight of a small apple. Humans typically weigh between 600 and 1000 N. Most often the kilo-Newton (kN) is used rather than the Newton. The kilogram force is the weight of one kilogram mass. On Earth, the equation is: (1.2) 1 kgf = 1 kg × 9.81 m∕s2 The unit of stress is the kN/m2 , also called the kilo-Pascal (kPa); there is about 20 kPa under your feet when you stand on both feet. Note that a kilogram force is the weight of a kilogram mass and depends on what planet you are on and even where you are on Earth. Accepted multiples of units, also called SI prefixes, are: terra giga mega kilo milli micro nano pico (An angstrom is 10–10 meter.)

Problems and Solutions Problem 1.1 How would you decide if you have reached the threshold of optimum simplicity? Solution 1.1 The threshold is not reached if: • • • • •

The solution seems too simple or too complicated. The solution is not used in practice. It costs too much time and money to obtain the solution. The solution leads to erroneous answers. The solution does not contain or address the essential elements of the problem.

The threshold is likely reached if: • • • • •

(1.1)

The solution seems reasonably simple and cannot be simplified further. The solution is used in practice. The cost of obtaining and implementing the solution is consistent with the budget of a large number of projects. The solution leads to reasonable answers. The solution is based on fundamental elements of the problem.

1012 109 106 103 10–3 10–6 10–9 10–12

1.5 UNITS

Problem 1.2 Calculate the pressure under your feet. Solution 1.2 Effective area for one foot ≈ (0.28 – 0.08) × 0.09 = 0.018 m2 (Figure 1.1s). Average weight of a person = 750 N 750 × 10−3 = 20 2 × 0.018

0.2

8m

Pressure under two feet:

0.0

8m

0.0

9m

Figure 1.1s

Feet geometry.

Problem 1.3 Explain the magic behind Figures 1.3b and 1.3c. Solution 1.3 The swelling clay pie (Figure 1.3b) is made of smectite clay, which has a tremendous ability to attract water in the presence of a free water source. This is due to the chemical attraction between the water molecules and the smectite mineral (Al2 Si4 O10 (OH)2 and x interlayers of H2 O). This clay type can swell an amount equal to its initial height or more. This is why the clay pie swelled to twice its height when subjected to a water source (Figure 1.3d) The sand pile in Figure 1.3c fails under the load applied (50 N) because the load exceeds the shear strength of the sand. The sand pile in Figure 1.3e is internally reinforced by sheets of toilet paper that are not visible from the outside. These paper sheets provide enough tension and increased shear strength in the sand for it to resist a much higher load (220 N) than the unreinforced sand pile. Problem 1.4 Are the following equations correct? 1 kgf = 1 kg × 9.81 m∕s2 1 N = 1 kg × 1.0 m∕s2 1 kgf = 9.81 N Solution 1.4

1 kgf = 1 kg × 9.81 m∕s2 ∶ Correct 1 N = 1 kg × 1.0 m∕s2 ∶ Correct 1 kgf = 9.81 N ∶ Correct

Problem 1.5 What is the relationship between a kilopascal (kPa) and a pound per square foot (psf)? ) ( 0.22481 lb Solution 1.5 1000 N × 1N 1 kPa = 1000 N∕m2 = ( ) ) = 20.9 psf ( 3.28 ft 2 2 1m × 1m

5

CHAPTER 2

Case Histories

Good judgment comes from experience, but experience comes from bad judgment. Indeed, we learn a lot from failures. All this is true, but it is better to gain your experience by learning from the bad judgment of others rather than making the mistakes yourself. Recorded and well-described failures and more generally case histories are invaluable in developing good judgment, something that is difficult to teach in the classroom. The following are case histories, which may or may not be failures but can be used in most cases to get the students interested in learning more and to develop engineering judgment. The case histories include (Figure 2.1): 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

Washington Monument (shallow mat foundation) Rissa landslide (slope stability) Seattle airport MSE wall (retaining wall) New Orleans Charity Hospital (deep foundation) Eurotunnel from France to England (tunneling) Teton Dam (dam) Woodrow Wilson Bridge Foundation (bridge scour) San Jacinto Monument (shallow mat foundation) Pointe du Hoc (rock erosion) Tower of Pisa (shallow foundation) Transcona silo (shallow foundation) Saint John’s River Bridge abutment (slope stability) Briaud house (shrink swell soils) Eiffel Tower (deep foundation) St. Isaac Cathedral (mat foundation) National Sites at Texas A&M University (retaining walls, deep and shallow foundations) Burj Khalifa Tower foundation (combined pile raft foundation) New Orleans levees and Hurricane Katrina (overtopping erosion) Three Gorges Dam (concrete dam) Kansai International Airport (consolidation settlement) Panama Canal (excavated slopes) Nice airport slope failure and tsunami (slope stability)

Three resources for case histories and documented failures may be very valuable. The first is the ISSMGE International

Journal of Geoengineering Case Histories, which is a free online journal where you can find many refereed articles related to case histories (http://casehistories.geoengineer .org/). The second is the proceedings of conferences dedicated to forensic engineering and sponsored by the ISSMGE technical committee on Forensic Geotechnical engineering (http://www.issmge.org/committees/technical-committees/ impact-on-society/forensic-). The third is the Proceedings of the Conference on Case Histories started by Shamsher Prakash in 1984 and now organized by the Geo-Institute of ASCE (https://scholarsmine.mst.edu/icchge/). The case histories covered in this chapter are not meant to be an exact rendition of the true case history but rather a general rendition of the case. Some of the data and analyses associated with the case histories have been simplified. Also, the calculations presented are not meant to be a replica of the actual design calculations but an illustration of some of the issues to be addressed in design and associated example calculations; they are simplified calculations. 2.1 WASHINGTON MONUMENT (SHALLOW MAT FOUNDATION) 2.1.1

The Story

George Washington was the first president of the United States of America, and governed from 1789 to 1797. He died in 1799 and it was decided to erect a monument in his memory. It took until 1848 for enough money to be raised and to start building the Monument in Washington, DC. The site was to be in front of the White House and in line with the Capitol. The swampy ground forced the engineers to relocate the structure slightly toward the Capitol where the construction began. The Monument would be a 169.16 m high, classical Egyptian-style, four-sided obelisk (Figure 2.2). Construction started in 1848 and was completed in 1884. 2.1.2

Geology and Soil Stratigraphy

The geology of the site has been impacted by the Potomac River and other stream deposits during significant sea-level

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

7

8

2 CASE HISTORIES

Figure 2.1 Location of the case histories.

variations tied to climate changes. These streams created the Pleistocene terraces on which much of Washington, DC, is built. The site is underlain by bedrock made of decomposed schist of Cambrian age with about 30 m of soil layers on top of it (Figure 2.3). Above the schist are an 11.68 m-thick layer of stiff to very stiff high plasticity blue clay, an 8.30 m-thick layer of very dense sand, gravel, and clay, a 3.67 m-thick layer of very soft sand and clay, and a 6.14 m-thick layer of fill placed toward the end of construction. The water level is found at a depth of 13.55 m below the ground surface. 2.1.3

Figure 2.2 Washington Monument: Washington, DC, USA.

Construction

Construction for Phase I started in 1848 with a 2.34 m deep excavation to the foundation level. The foundation structure is a stair-stepped pyramid made of blocks of blue gneiss (Figure 2.4). The shaft is made of blocks of marble from Baltimore. By the end of 1858, the shaft had risen to a height of 55.5 m when the work stopped because money had run out. For various reasons including the Civil War, construction only resumed in 1879 for Phase II. The US Army Corps of Engineers put Lieutenant Colonel Casey in

2.1 WASHINGTON MONUMENT (SHALLOW MAT FOUNDATION)

0m

10 m

9

20 m Boring 24

Boring 25 Modern-day ground surface

1972 ground surface

Loam, fill, and clay

1848 ground surface

3.76 m

Sand and clay 3.65 m

8.30 m

29.88 m

Sand and gravel and clay

Groundwater level (used in settlement calculations)

11.68 m

Blue clay

Decomposed Wissahickon Schist

Figure 2.3 Washington Monument foundation and soil stratigraphy. (Source: Briaud et al., 2009/ International Journal of Geoengineering Case Histories.)

charge of completing the project. Casey decided to underpin and widen the foundation (Figure 2.4) after a board of engineers determined that in its present state the weaker clay and sand layer immediately under the foundation was almost at the point of bearing capacity failure. The result was a Portland cement concrete foundation shaped like a square ring underneath the existing gneiss foundation. Left inside the new foundation was the intact natural soil. In 1880–1881, a terrace was construct by moving earth around the foundation. This brought the ground level up to the top of the foundation. Then construction of the shaft resumed and was completed to a height of 169.16 m above the top of its pyramidal foundation. The Monument was completed in 1884. 2.1.4

Geometry and Load

The dimensions of the Washington Monument are given in Figure 2.4. The area of the first foundation in contact with the soil was 595 m2 while the underpinning increased that area to 1486 m2 . The weight of the original foundation is 70 MN while the weight of the structure, including the foundation at the end of 1858 when the column was 55.5 m high, is 305 MN. The weight of the new foundation, the old foundation plus underpinning is 153.8 MN. The total weight of the Washington Monument, including the foundation, is 607.7 MN. The pressure on the soil at the end of Phase I was thus 513 kPa and was reduced to 409 kPa at the end of construction because of the increased area created by the underpinning (Figure 2.5).

2.1.5

Soil Properties

Many borings have been drilled around the Monument over the last 150 years with associated soil testing. The soil properties collected include water content, plasticity index, consolidation parameters, undrained shear strength, and SPT blow count. They are presented in Figure 2.6. 2.1.6

Bearing Capacity

Bearing capacity calculations are part of the typical twoprong design of shallow foundations, with settlement calculations being the other one. In the case of the Washington Monument, there are two cases of ultimate bearing pressure to be addressed: the old foundation with the partially completed monument and the underpinned enlarged foundation with the completed monument. For the old foundation, the weight, including the foundation, is 305 MN for an area of 595 m2 . Thus, the applied pressure is 305,000 p= = 513 kPa (2.1) 595 The bearing layer right under the old foundation is listed as sand and clay. If the soil is clay, the ultimate bearing pressure is calculated according to the Skempton equation (see Chapter 18). The depth of embedment D is 7.12 m and the width B is 24.38 m for a D/B ratio of 0.29. For that relative embedment, the N c factor is 7.0. The undrained shear strength data is limited in the bearing layer and Figure 2.6 gives values in the 40 kPa range. Another crude estimate of the undrained shear strength can be derived from correlation with the SPT blow count, which averages about 12 bpf in that bearing layer. The

10

2 CASE HISTORIES

Section View 10.5 m

16.80 m 17.75 m 11.23 m 4.11 m

24.38 m 12.57 m

12.57 m 13.41 m 38.54 m

Plan View 38.54 m 28.99 m

169.16 m

24.38 m 17.75 m

152.4 m

16.80 m 13.41 m 10.50 m

55.58 m

16.80 m

Shaft boundaries Original foundation

11.23 m (est1880

4.11 m

Underpinned foundation

12.57 m

12.57 m 13.41 m

Buttresses

38.54 m

Figure 2.4 Washington Monument dimensions. (Source: Briaud et al., 2009/International Journal of Geoengineering Case Histories.)

Terzaghi and Peck correlation is Su (kPa) = 6.7 N SPT , which gives Su = 80 kPa. A value of 60 kPa may be a reasonable average for that layer. The unit weight is estimated to be 19 kN/m3 . pu = Nc su + 𝛾D = 7 × 60 + 19 × 7.12 = 555.28 kPa (2.2) If the soil is considered to be sand, then the following direct relationship is used (see Chapter 18). pu (kPa) = 60N + 𝛾D = 60 × 12 + 19 × 7.12 = 855.28 kPa (2.3) Both sets of calculations show that the applied pressure is approaching the ultimate bearing pressure with a factor

of safety of 555.28 to 855.28 = 1.08 to 1.67 (2.4) 513 The settlement calculations will confirm that the monument was close to failure at the end of the Phase I construction. For the underpinned and final foundation, the weight is 607.7 MN and the area of the underpinned foundation is 1486 m2 . Thus the applied pressure is Fold =

p=

607,700 = 409 kPa 1486

(2.5)

2.1 WASHINGTON MONUMENT (SHALLOW MAT FOUNDATION)

700 600 Final Phase of Construction

Load (MN)

500 400

Underpinning

300 First Phase of Construction

200 100 0 1845

1850

1855

1860

1865

1870

1875

1880

1885

1890

2007

1890

2007

600

400 300 200 100 0 1845

1850

1855

1860

1865

1870

1875

Final Phase of Construction

First Phase of Construction Underpinning

Pressure (kPa)

500

1880

1885

Completely Distributed Total Pressure

Time (years)

Underpinning-Only Total Pressure

Figure 2.5 Weight and pressure under the Washington Monument. (Source: Briaud et al., 2009/International Journal of Geoengineering Case Histories.)

Plasticity Index

Water Content (%) 20

40

60

80

10

5

5

0

0

–5

Elev. (m)

Elev. (m)

10

0

0

20

40

60

80

–5

–10

–10

–15

–15

–20

–20

Figure 2.6 Soil properties under the Washington Monument. (Source: Briaud et al., 2009/International Journal of Geoengineering Case Histories.)

11

10

10

5

5

0

0 Elev. (m)

Elev. (m)

2 CASE HISTORIES

–5

–5

–10

–10

–15

–15

–20

0

200

400

600

800

–20

1000

0

0.5

10

10

5

5

0

0

–5

1.5

–5

–10

–10

–15

–15

–20

–20 0

0.02

0.04

0.06 Cr

0.08

0.1

0

5

10

15

5

10

10 15 Cv (m2/year)

20

25

5 Elev. (m)

0 –5 –10

0 –5 –10

–15 –20

1 Cc

Elev. (m)

Elev. (m)

σʹp (kPa)

Elev. (m)

12

–15

0

50

100

150

200

–20

0

Undrained Shear Strength (kPa)

Figure 2.6 (Continued)

100

200

300

Blow Count (blows/ft)

400

2.1 WASHINGTON MONUMENT (SHALLOW MAT FOUNDATION)

The sand layer below the underpinned foundation has a blow count averaging about 100 bpf, and the ultimate bearing capacity of that layer is pu (kPa) = 60N + 𝛾D = 60 × 100 + 19 × 11.23 = 6213.37 kPa

(2.6)

However, the sand layer is only about 8 m thick and the bearing capacity failure could take place by punching through the sand layer and into the underlying clay layer. In this case, the ultimate bearing pressure pu is calculated as follows (see Chapter 18): ′ pu Af = pu(clay) Af + (Pinside + Poutside )H × ko 𝜎ov tan 𝜙 (2.7)

where Af is the area of the underpinned foundation, pu(clay) is the ultimate bearing pressure in the underlying clay layer, Pinside and Poutside are the perimeters of the inside and the outside of the underpinned foundation respectively, H is the thickness of the sand layer, ko is the coefficient of horizontal earth pressure at rest, 𝜎 ′ ov is the average vertical effective stress in the sand layer, and 𝜑 is the friction angle of the sand. pu × 1486 = (7 × 100 + 19 × 11.23) × 1486 + (24.38 × 4 + 38.54 × 4) × 8 × 0.5 × (19 × 15.23) tan 40 (2.8) pu = 913.37 + 164.50 = 1077.87 kPa

(2.9)

Since the punching mechanism leads to a smaller ultimate pressure, it is the likely failure mechanism. The factor of safety against the bearing capacity failure of the Washington Monument as it exists today is: 1077.87 = 2.63 (2.10) Funderpinned = 409

2.1.7

200

10

400

Settlement

The settlement of the monument is calculated at the end of Phase I and then after the underpinning. The depth of influence is limited by the Schist bedrock at a depth of 23.74 m below the bottom of the old foundation. The stress increase with depth can be estimated by using charts based on elasticity (see Chapter 18); however, in this case a finite element simulation also in elasticity was used and is presented in Figure 2.7. Table 2.1 summarizes the parameters and settlement calculations for the different layers. The compressible zone below the foundation is divided into three layers: (1) the sand-clay 4 m-thick layer; (2) the sand-gravel-clay 10 m-thick layer; and (3) the blue clay 9 m-thick layer. The effective stress 𝜎 ′ ov is calculated at the center of each layer, then comes the increase in stress Δ𝜎 ′ v , read on Figure 2.7, followed by the preconsolidation pressure 𝜎 ′ p , read on Figure 2.6, then the initial void ratio eo estimated from the consolidation tests, the Recompression Index Cr and Compression Index Cc , read on Figure 2.6 and the calculated compression ΔH of each layer by the equation: ( ′) ( ′ [ )] 𝜎p 𝜎ov + Δ𝜎v′ Ho ΔH = Cr log + Cc log ′ 1 + eo 𝜎ov 𝜎p′ (2.11) This equation assumes that the soil is compressing along the recompression line with the slope Cr until the preconsolidation pressure 𝜎 ′ p is reached and then compressing along the virgin compression line with the slope Cc until 𝜎 ′ ov + 𝛥𝜎 ′ v is reached. The calculated settlement is 1.56 m, which is very large, and confirms that the monument at the end of Phase I was very close to failure, yet the monument did not lean.

Stress Increase (kPa) 0

Stress Increase (kPa) 600

0

200

400

600

10

A 5

5

Depth (m)

B 0

0

–5

–5

–10

–10

–15

–15 1880 1884

1858 flex

–20

13

–20

Figure 2.7 Stress increase with depth below the Monument: left is old foundation. (Source: Briaud et al., 2009/International Journal of Geoengineering Case Histories.)

14

2 CASE HISTORIES

Table 2.1

Settlement calculations for the old foundation

Layer

H o (m)

𝜎 ′ ov (kPa)

Δ𝜎 ′ v (kPa)

𝜎 ′ p (kPa)

eo

Cr

Cc

ΔH (m)

Sand and clay Sand and gravel and clay Blue clay

4 10 9

171 284 369

500 450 340

50 170 450

0.9 0.5 0.5

0.04 0.02 0.03

0.45 0.20 0.45

0.56 0.55 0.45 Σ = 1.56

Table 2.2 Settlement calculations for the underpinned foundation Layer

H o (m)

𝜎 ′ ov (kPa)

Δ𝜎 ′ v (kPa)

𝜎 ′ p (kPa)

eo

Cr

Cc

ΔH (m)

Sand and gravel and clay Blue clay

8.5 10.5

289 348

440 300

764 698

0.5 0.5

0.02 0.03

0.2 0.45

0.045 0.057 Σ = 0.102

The settlement calculations are repeated for the end of construction after underpinning. Table 2.2 summarizes the parameters and settlement calculation results. Note that because the stress created in the soil after underpinning and final construction was less than the stress at the end of the first Phase, the soil was on the reload part of the consolidation curve and Cr is used. In this case, the preconsolidation pressure is equal to the vertical effective

1840 0

1860

1880

1900

stress plus the increase in stress due to the first phase of construction (e.g. for the sand and gravel and clay layer that is 289 kPa + 475 kPa). The final settlement due to underpinning and final construction is calculated as 0.102 m. The measured settlement in 1996 was 0.119 m. The reconstructed settlement versus time curve for the Washington Monument is shown in Figure 2.8 (Briaud et al., 2009).

Settlement Reconstruction 1920 1940

1960

1980

0.2

Settlement (m)

0.4

(1)

Calculated based on consolidation test data

(2)

Inferred from measured creep rate

(3)

Measured

Cv = 10.2 m^2/yr 0.6

(1)

0.8

1

1.2

1.4

(2) (3)

1.6

Figure 2.8 Reconstructed settlement vs. time curve for the Washington Monument. (Source: Briaud et al., 2009/International Journal of Geoengineering Case Histories.)

2000

2.2 RISSA LANDSLIDE (SLOPE STABILITY)

2.2 2.2.1

RISSA LANDSLIDE (SLOPE STABILITY) The Story

The story is best learned by watching the 1982 video found at this site of the Norwegian Geotechnical Institute library: https://www.youtube.com/watch?v=26hooxzCGkY. On April 29, 1978, near the village of Rissa in Norway (Figure 2.9), a major landslide occurred and was captured on film by two amateur photographers. The soil within the sliding mass was a quick clay, a type of clay known to turn into liquid when sheared. A farmer living near the edge of Lake Botnen dug an excavation for the foundation of an extension to his barn and piled the excavated soil near the shore of Lake Botnen (Figure 2.10). This soil mound reached a height of 2 m, which triggered a bearing capacity failure of

Lake Botnen (fresh water)

Community of Rissa

Rissa landslide

Fjord of Trondheim (salt water)

North Sea

City of Trondheim

15

the underlying quick clay. The clay became quick and flowed downhill into the lake. The back of the bearing capacity failure surface created a near-vertical slope too steep for the quick clay strength and another slope failure occurred behind the first one. In turn, the back of the new failure surface created again a near-vertical slope too steep for the quick clay strength and another failure occurred behind the second one. This retrogressive slope failure process continued over a retrogression length of 1350 m and over a period of about one hour (Figure 2.11) until the back of the last slope reached a soil with a high enough strength to stand at the newly created slope angle. The two photographers captured this process on film, but the film is interrupted at one point as one of the photographers had to run to safety because the size of the slides was endangering his position. The final size of the slide covered 330,000 m2 and about 5,500,000 m3 (Figure 2.11). On the video, houses are seen at one point floating down the quick clay stream into Lake Botnen at about 30 km/hr (Figure 2.12). All in all, 12 structures were destroyed, and one person died. Geoengineers from the Norwegian Geotechnical Institute got involved and decided to remove the remaining quick clay deposits, which were considered unstable. To do so, they used dynamite and purposely triggered additional slope failures. The restoration of the area took place the following winter so that the equipment could operate on the frozen ground. The farmers continued to use the land after safety was restored.

Time = zero

Time ~ 20 minutes

Time ~ 40 minutes

Time ~ 60 minutes

Time ~ 65 minutes

Time ~ 70 minutes

5 km

Figure 2.9 Location of Rissa landslide. (Source: after Google Maps.)

Figure 2.10 Initial excavation and soil mound on the shore of Lake Botnen. (Source: Courtesy of NGI.)

Figure 2.11 Progression of the regressive slope failure (final extent about 330,000 m2 area and 5,500,000 m3 volume). (Source: Courtesy of NGI.)

16

2 CASE HISTORIES

Description 0

Water content, W (%) 10

20 30 30

Undrained vane shear strength, su (kN/m2) 10 20 30

DRY CRUST CLAY, silty

WP

WL

W QUICK CLAY, silty

Undisturbed

Figure 2.12 Houses floating down the quick clay stream. (Source: Courtesy of NGI.)

Depth, m

10

CLAY, silty

20

2.2.2

The Soil Parameters

The clay was likely deposited under a sea water environment at a time where the sea level in the Fjord of Trondheim was much higher than today. The salt in the sea water minimized the repulsion forces between particles and helped to develop bonds at the contacts. The sea retreated and the clay was exposed. Meanwhile rainwater, ice melt, and runoff created the freshwater Lake Botnen. The freshwater runoff from the nearby mountains seeped through the clay into the lake. This leaching process decreased the salt content between particles and weakened the clay strength, which had a high void ratio structure, like a house of cards. This process led the clay to be metastable with a water content higher than the liquid limit and the possibility of turning into a liquid through total collapse if disturbed. It became a quick clay. Some of the properties of the quick clay are shown in Figure 2.13 (Gregersen, 1981; Liu et al., 2021). As can be seen, a clay crust exists near the surface underlain by the quick clay. The water content averages about 32% and is higher than the liquid limit, which averages about 24%. The plasticity index is quite low with a range of 4–7 because the clay is very silty, and the unit weight averages 18.6 kN/m3 . The undrained shear strength measured by vane testing decreases from about 40 kPa at the surface down to 20 kPa at a depth of 4 m and as low as 8 kPa at a depth of 8 m. The sensitivity ratio of undisturbed undrained shear strength to

Remolded

30

Figure 2.13 Rissa clay properties. (Source: After Gregersen, 1981.)

disturbed undrained shear strength, is about 100 and the clay when tested in a triaxial test exhibits significant post-peak softening. The slope geometry and soil stratigraphy are shown in Figure 2.14. 2.2.3

Slope Stability Back Analysis

The slope was long and very flat; it is estimated to have a slope angle 𝛽 = 3 degrees along the sliding direction (Gregersen, 1981). Note that Figure 2.14 is not in the direction of the slope failure. It seems therefore appropriate in this case to use the solution for an infinite slope. Recall from the chapter on slope stability (Chapter 20) that for an infinite slope in a c′ – 𝜑′ soil with seepage, the factor of safety F is F=

(𝛾 − 𝛾w ) tan 𝜑′ c′ + sat 𝛾sat h sin 𝛽 cos 𝛽 𝛾sat tan 𝛽

Elevation, m

Road

30

30

20 10 0 –10

20 τp=0.1p′0

τH=0.2p′0

Quick CLAY

Insensitive CLAY

τa=0.3p′0

10 0 –10

Figure 2.14 Cross-section of the Rissa slope perpendicular to the mountain direction. (Source: After Gregersen, 1981.)

(2.12)

2.3 SEATTLE 46 M-HIGH MSE WALL (RETAINING WALL)

17

where c’ is the effective stress cohesion intercept, 𝜑’ is the effective stress friction angle, 𝛾 sat is the saturated soil unit weight, h is the depth of the failure surface, 𝛽 is the slope angle, and 𝛾 w is the unit weight of water. The Rissa landslide is a case of an undrained slope failure and in this case Eq. (2.12) becomes su F= (2.13) 𝛾sat h sin 𝛽 cos 𝛽 where su is the soil’s undrained shear strength. Using a lower su value from the profile on Figure 2.13 of 10 kPa, a unit weight of 18.6 kN/m3 , a depth to the failure surface h of 10 m (Figure 2.14) and a slope angle of 3 degrees leads to a factor of safety equal to 10 = 1.03 (2.14) 18.6 × 10 sin 3 cos 3 This shows that the slope was near failure but did not fail until a trigger mechanism initiated it at the bottom of the slope, thereby starting the progressive retrogression failure. The bearing capacity of the clay under the spoil pile can be estimated as follows. Recall from the shallow foundations chapter (Chapter 18) that the bearing capacity pu of a long footing on the surface of a clay with an undrained shear strength su is (2.15) pu = 5.2 su

Figure 2.15 MSE walls required for the new runway of the Seattle-Tacoma international airport. (Source: Courtesy of ASCE and RECO.)

F=

The pressure imposed by the 2 m-high spoil pile made of excavated material can be estimated as p = 𝛾z = 18.6 × 2 = 37.2 kPa

(2.16)

Since failure occurred, then p = pu and the back calculated value of su would be 5.2 su = 37.2

or

su = 7.15 kPa

(2.17)

This value is close to the minimum value found on the undrained shear strength profile (Figure 2.12) and the dumping of the excavated soil may have increased the dynamic load.

2.3 SEATTLE 46 M-HIGH MSE WALL (RETAINING WALL) 2.3.1

The Story

This is a case history about a very high mechanically stabilized earth (MSE) wall constructed in 2005 for a new runway at the Seattle-Tacoma international airport also known as SEATAC (Sankey et al., 2008). The problem was that the terrain around the existing airport was quite uneven, and to create a 2600 m-long runway required building several retaining walls (Figure 2.15). Because of the economy of using MSE walls for tall walls and because MSE walls are more resilient in earthquake-prone zones (such as Seattle), MSE walls were chosen for this purpose. This case history is about the West MSE wall design and construction.

2.3.2

The Natural Soil Conditions

The natural soil at the site of the wall prior to construction was made of soft peat from the ground surface down to 4 m mixed with interbedded layers of loose to medium dense silty sand and sandy peat. Below that was glacial dense to very dense, slightly gravelly, silty to very silty sand (Stuedlein et al., 2010). The groundwater level was shallow and found at the surface of the sand layer; a creek ran perpendicular to the center of the wall. Because the high wall would create significant pressure on the natural soil, soil improvement was considered. In the end, soil improvement was found to be insufficiently reliable and the layer of soft peat was excavated. It was replaced with compacted fill made of clean sandy gravel and gravelly sand with less than 3% fine particles to ensure reasonable drainage. 2.3.3

The Fill and Wall Construction

The wall construction took one year and was completed in 2005 (Figure 2.16). The wall cross-section is shown in Figure 2.17. After replacing the underlying 4 m-thick layer of peat by compacted fill, the first tier of wall was constructed in 0.3 m-thick lifts. The reinforced fill for the wall backfill had to adhere to very stringent requirements including gradation with no fines and electrochemical content to minimize corrosion and environmental impact. Direct shear tests gave a mean friction value of 41 degrees with a coefficient of variation of 0.06, indicating very good uniformity. The design value of the friction angle was 37 degrees. The fill would be compacted to a minimum of 92% of Modified Proctor dry density within +/− 2% of optimum moisture content. The in-place unit weight averaged 22 kN/m3 . 2.3.4

The Wall Design

The design of an MSE wall includes internal stability and external stability criteria. Both criteria are satisfied through calculations of length, density, and strength of the reinforcement strips. The overall guide for the length of the reinforcing strips is to use 0.7H, where H is the height of the wall. In this

18

2 CASE HISTORIES

γ=22 kN/m3 φ=37° αmax=0.47

35 m 0.3H

H 2

W

ΔPAE

Pi

46 m

PA H 3

0.6H

Figure 2.18 Design diagram for SEATAC MSE wall.

design earth pressure at the bottom of the wall. The pressure 𝜎 h is given by 𝜎h = kr 𝜎v = 1.2Ka 𝛾h (2.19)

Elevation in meters

Figure 2.16 MSE wall construction. (Source: Courtesy of Dr. Armin Stuedlein.)

130 125 120 115 110 105 100 95 90 85 80 75 70 65 60 55

-LEGEND-

3rd Runway centerline ~120 m Structural fill Finished grade

Survey point Strain gages Sondex ring

Limits of reinforcement

Inclinometer w/ piezometer

Common embankment fill

Stiff, sandy PEAT

M. dense, silty to v. silty SAND and stiff sandy PEAT and SILT Subgarde improvement zone

Very dense, slightly gravelly, silty to very silty SAND (Glacially overridden)

–40 –30 –20 –10

0

10

20

30

40

50

60

Distance from toe in meters

Figure 2.17 MSE wall cross-section. (Source: Stuedlein et al., 2010, courtesy of ASCE and RECO.)

case, this would be 0.7 × 46 = 32.2 m. A design diagram for the SEATAC wall is shown in Figure 2.18. For internal stability, the reinforcement must satisfy the pull-out resistance and tensile capacity criteria. For the pull-out resistance, the reinforcement must be long enough to absorb the static and dynamic forces. The static force is due to the active earth pressure and the dynamic force is due to the horizontal inertia force of the active wedge. The following are simplified calculations for the bottom strip. The horizontal static load T max(static) to be resisted by one strip at the bottom of the wall is Tmax (static) = sv sh 𝜎h

(2.18)

where sv is the vertical spacing of the strips (0.24 m) and sh is the horizontal spacing of the strips (0.14 m) and 𝜎 h is the

where kr is a coefficient of earth pressure, 𝜎 v is the vertical effective stress which is the same as the vertical total stress since there is no water in the fill by design, K a is the coefficient of active earth pressure, 𝛾 is the total unit weight, and h is the height of the wall. Given the parameters mentioned earlier for the fill, it comes out as 𝜎h = 1.2 ×

1 − sin 37 × 22 × 46 = 302 kPa 1 + sin 37

(2.20)

And the load T max(static) is Tmax(static) = 0.24 × 0.14 × 302 = 10.1 kN

(2.21)

The horizontal dynamic force Pi(wedge) is due to the horizontal inertia force generated by the earthquake and acting on the failing wedge shown in Figure 2.18. This wedge is subjected to a horizontal acceleration ac during the earthquake. By design ac is 0.46g for the MSE wall site. Thus Pi(wedge) is given by Pi(wedge) =

ac = 0.46(23 × 0.3 × 46 + 0.5 × 23 W g wedge × 0.3 × 46) × 22 = 4818 kN∕m (2.22)

If there are three strips per meter of height and three strips per meter of width, the number of strips over the height of the wall for a meter of width of wall is: N = 3 × 46 × 3 = 414 strips∕m of width of wall

(2.23)

Therefore, the load per strip is: Tmax(wedge) =

4818 = 11.6 kN 414

(2.24)

The strip length is then determined by the limit state equation 𝛾S Tmax(static) + 𝛾E Tmax(wedge) = 𝜑Tpullout

(2.25)

where 𝛾 S is the static load factor, taken as 1.35, 𝛾 E is the extreme event load factor, taken as 1.0, 𝜑 is the resistance

19

2.4 THE NEW ORLEANS CHARITY HOSPITAL FOUNDATION (DEEP FOUNDATION)

factor, taken as 0.9, and T pullout is the ultimate pull-out load one strip can resist. 1.35 × 10.1 + 1 × 11.6 = 0.9 × 2fmax bLe

(2.26)

where f max is the maximum friction stress that can be resisted by the strip, b is the width of one strip (5 cm) and Le is the embedment length required. The value of f max is given by: fmax = f ∗ 𝜎v

(2.27)

where f* is the friction coefficient equal to tan 𝜑 in this case and 𝜎 v is the vertical effective stress at the depth considered. fmax = tan 37 × 22 × 46 = 763 kN∕m2

(2.28)

Then Eq. (2.26) becomes 1.35 × 10.1 + 1 × 11.6 = 0.9 × 2 × 763 × 0.05Le

(2.29)

and Le = 0.367 m

(2.30)

The total length of the strip at the top of the wall would have to be Le + 0.3H where H is the height of the wall but not at the bottom. For external stability, the MSE wall had to satisfy the sliding and overturning criteria. For the sliding criterion, the static active earth pressure load per unit width of wall is given by 1 1 1 − sin 37 K 𝛾H 2 = × × 22 × 462 = 5819 kN∕m 2 a 2 1 + sin 37 (2.31) The increase in active earth pressure load per unit width of wall due to the earthquake for the design wall mass acceleration (ac /g = 0.46) is given by a ΔPAE = 0.375 c 𝛾H 2 = 0.375 × 0.46 × 22 × 462 g = 8030 kN∕m (2.32) PA =

Or L = 35 m

(2.36)

For the overturning criterion, moments are taken around the bottom front of the wall, which gives ) ( L H H = 𝜑HL𝛾 𝛾S PA + 𝛾E ΔPAE × 0.6H + 0.5Pi × 3 2 2 (2.37) 46 1.35 × 5819 × +1 3 ) ( 46 × 8030 × 0.6 × 46 + 0.5 × 16293 × 2 L2 = 22 0.9 × 46 × 2 = 1162.6 m2 (2.38) Or L = 34 m (2.39) As can be seen in this case, the external stability controls the design of the strip length.

2.4 THE NEW ORLEANS CHARITY HOSPITAL FOUNDATION (DEEP FOUNDATION) 2.4.1

The Story

Apparently, in 1735, a French sailor and shipbuilder called Jean-Louis died and left in his will a grant to build the first Charity Hospital in New Orleans. Fast forward 200 years and five hospitals later, a sixth Charity Hospital was built in 1939 on Tulane Avenue (Figure 2.19). It was a very large hospital for the time with over 3000 beds. Difficulties were encountered during the construction of this 20-story structure because of an inadequate foundation, which is the subject of

Then the inertia force of the wall mass is given by a Pi = c 𝛾HL = 0.46 × 22 × 46 × 35 = 16293 kN∕m (2.33) g where L is the width of the MSE wall reinforced mass and thus the strip length assumed to be 35 m as an initial estimate (see Figure 2.17). The design load for sliding and overturning stability is chosen as PA +ΔPAE +0.5 Pi . The reason for using only 0.5Pi and not Pi is that ΔPAE and Pi do not occur at the same time. The ultimate limit state for sliding is 𝛾S PA + 𝛾E (ΔPAE + 0.5Pi ) = 𝜑 W tan Φ

(2.34)

where 𝛾 S is the static load factor taken as 1.35, 𝛾 E is the extreme event load factor taken as 1.0, 𝜑 is the resistance factor taken as 0.9, W is the weight of 1 m wide reinforced wall mass, and 𝛷 is the backfill friction angle. 1.35 × 5819 + 1(8030 + 0.5 × 16293) = 0.9 × 46 × L × 22 × tan 37

(2.35)

Figure 2.19 Charity Hospital in New Orleans. (Source: Infrogmation/Wikimedia Commons.)

20

2 CASE HISTORIES

the following sections. In 2005, Hurricane Katrina delivered a final blow to the 66-year-old hospital, which will not be renovated. However, the structure will be used for developing residential units, retail shops, and restaurants. The source of information for this case history comes mostly from notes and information shared with me by Gordon Boutwell of STE in Baton Rouge and Bill Gwyn of Eustis Engineers in New Orleans. Additional documents include the Works Progress Administration (WPA) study of foundations in New Orleans (1935–1942) and letters from Terzaghi and Hardy Cross regarding the settlement (originals at the Terzaghi Library in Oslo, Norway, and copies at Eustis Engineers in New Orleans, USA). 2.4.2

The Soil Conditions

The geology of the site is associated with the deposition of soft sediments in the New Orleans area over millions of years brought about by the Mississippi River. As such, the soil in New Orleans is typically made of soft normally consolidated clay with some intermittent layers of sand and a high groundwater level. The soil layering at the hospital site (Figure 2.20) can be represented by a simplified stratigraphy of soft clay from the ground surface to a depth of 12.5 m, a 2.5 m-thick layer of medium dense sand, a 60 m-thick layer of soft to medium clay with sand seams, and a dense sand layer below that. The soil parameters for each of those layers are shown in Table 2.3.

0m SOFT CLAY

–10 m

SAND –20 m –30 m

A

B SOFT CLAY

–40 m C –50 m –60 m

D

Table 2.3 Soil properties below the New Orleans Charity Hospital Layer Soft clay

Depth (m)

Soil properties

0–12.5

Average undrained shear strength Su = 20 kPa Unit weight 𝛾 t = 18 kN/m3 Initial void ratio eo = 0.8

Medium dense sand

12.5–15

Soft clay

15–75

Compression index Cc = 0.095 SPT blow count = 30 bpf Unit weight 𝛾 t = 20 kN/m3 Average undrained shear strength Su = 30 kPa Unit weight 𝛾 t = 19 kN/m3 Initial void ratio eo = 0.7 Compression index Cc = 0.065

Dense sand

2.4.3

75+

Foundation Design and Construction

The foundation would be made of 7 m-long 0.3 m average diameter timber piles, which were readily available in the area. The piles would be driven through the soft clay layer and reach the sand layer. The capacity of one pile can be calculated by the API RP2A method (driven piles in clay) for the side friction and by the Briaud-Tucker method (driven piles in sand) for the point capacity. 1. Find the depth z at which 𝜎 ′ ov = su or 𝛾z – uw = su or 18 z – 9.8 z = 20 or z = 2.44 m. 2. Because the pile cap is located at a depth of 5.5 m, it is considered that 𝜎 ′ ov > su and that the second API-RP2A criterion for side friction in clay is satisfied. 3. Find the friction shear stress f u on the side of the pile. f u = 0.5 (su /𝜎 ′ ov )-0.25 su . The average value of 𝜎 ′ ov over the 7 m length of pile is 𝜎 ′ ov = 𝛾z – uw where z is the depth of the pile cap (5.5 m) plus half the length of the pile (3.5 m) or z = 9 m. Then 𝜎 ′ ov = 18 × 9 – 9.8 × 9 = 73.8 kPa and f u = 0.5 (20/73.8)−0.25 20 = 13.9 kPa. 4. Find the ultimate point bearing pressure pu of the pile. pu (kPa) = 1000 N0.5 = 1000 × 300.5 . So pu = 5477 kPa. 5. Calculate the ultimate capacity of one pile Ru = f u As + pu Ap = 13.9 × π × 0.3 × 7 + 5477 × π × (0.3/2)2 = 91.7 + 387 = 478.7 kN.

–70 m DENSE SAND

Figure 2.20 Charity Hospital sketch and simplified soil stratigraphy.

A few vertical load tests were conducted to measure the ultimate capacity of the piles and a final allowable load of 150 kN per pile was selected. Since the building weighed about 1500 MN, a total of about 10,000 piles would be driven (Figure 2.21). A major question arose and had to be answered.

2.4 THE NEW ORLEANS CHARITY HOSPITAL FOUNDATION (DEEP FOUNDATION)

2.4.4

Figure 2.21 Plan view and timber piles locations for Charity Hospital. (Source: Courtesy of W.W. Gwyn and Eustis Engineering.)

Would the 10,000 piles carry 10,000 times the load carried by one pile? It is reported that to address this issue a vertical load test on a group of four piles was conducted and gave an ultimate capacity equal to four times the capacity of one pile. The pile driving was undertaken with timber piles driven on a spacing of 0.9 m center to center. The pile cap was constructed, and the first few levels of the building were erected. This is when the settlement started to develop and soon became a major concern. Terzaghi was called to help and got involved. He used this case history to verify his consolidation settlement theory. By today’s practice the group effect would be checked by the block failure analysis of the pile group as follows. To simplify the problem, it is assumed that the 10,000 piles form a group that is 75 m by 75 m in plan-view and 7 m deep, which is the length of the piles. The ultimate capacity Ru-group of this pile block is

Settlement Analysis and Measurements

Another important aspect of this case history is the settlement analysis of this large group of piles. In the early days of geotechnical engineering, settlement analysis of deep foundations was not often carried out. This changed in the second half of the twentieth century, in part under the influence of the development of offshore platform foundations. There are at least two major cases where settlement of deep foundations must be included in the design: one is the settlement of large pile groups, as in this case, and the other is the settlement of deep foundations subjected to downdrag. The settlement analysis for Charity Hospital is addressed below and follows the general procedure. The pile group is replaced by an equivalent footing with the same plan view dimensions, in this case 75 m × 75 m and located on top of the sand layer where the pile group transfers the load. The depth of influence would be two times the width of the 75 m by 75 m equivalent footing. That depth would be 150 m but there is a dense sand layer starting at 75 m, thus the depth of influence goes from a depth of 12.5 m (top of the sand layer) to a depth of 75 m (depth of the next sand layer). The settlement of both sand layers is considered to be very small compared to the compression of the clay layer between the two sand layers; thus, the settlement of the sand layers is neglected. The clay layer sandwiched between the two sand layers is 60 m thick and is broken down into a set of four layers 10 m thick, 15 m thick, 15 m thick, and 20 m thick (see Figure 2.20). The calculations proceed by calculating the initial vertical effective stress σ′ ov in the middle of each layer before construction, then the increase in vertical stress Δσ′ v in the middle of each layer due to the building (Table 2.4). The average pressure under the building is the weight of the building (1500 MN) divided by the area (75m × 75m) or p = 266.7 kPa. The compression of each layer is then computed by using the consolidation settlement magnitude equation for this normally consolidated clay.

Ru−group = Ru−group side + Ru−group base = 2(Lg + Bg )Lp su + Lg Bg Nc su

(2.40)

where Ru-group side is the ultimate friction resistance in shear along the perimeter of the pile group, Ru-group base is the ultimate point resistance at the base of the pile group, Lg and Bg are the length and width of the pile group, Lp is the length of the piles, su is the undrained strength of the clay, and N c is Skempton’s bearing capacity factor for a relative embedment depth of 12.5 m (foundation depth) over 75 m (width of the pile group). Ru−group = 2 (75 + 75) 7 × 20 + 75 × 75 × 6.4 × 20 = 42000 + 720,000 = 762,000 kN

(2.41)

Since the building weighs 1500 MN, one can readily see that the block failure capacity is insufficient and contributed to the early problems faced during construction.

21

s = Ho

𝜎 ′ + Δ𝜎 ′ Cc log ov ′ v 1 + eo 𝜎ov

(2.42)

where s is the compression of the layer, H o the thickness of the layer, Cc the compression index, eo the initial void ratio, 𝜎 ′ ov the initial vertical effective stress in the middle of the layer before construction, and Δ𝜎 ′ v the increase in vertical stress in the middle of the layer due to the building. A value of 0.544 m settlement is obtained (Table 2.4). The soil settlement became a concern after construction of several stories. It was decided to measure the settlement starting in May 1938 until June 1940 (Figure 2.22) when the settlement reached 0.356 m. Since the settlement prior to starting monitoring must have been significant and since Figure 2.22 indicates that the settlement would continue beyond June 1940, the calculation of a total settlement equal to about 0.544 m may be reasonable.

22

2 CASE HISTORIES

Table 2.4 Settlement calculations for Charity Hospital, New Orleans Layer

H o (m)

𝜎 ′ ov (kPa)

Δ𝜎 ′ v (kPa)

eo

Cc

ΔH (m)

1 2 3 4

10 15 15 20

170 237.5 305 395

0.96 × 266.7 = 256 0.88 × 266.7 = 234.7 0.71 × 266.7 = 189.4 0.52 × 266.7 = 138.7

0.7 0.7 0.7 0.7

0.065 0.065 0.065 0.065

0.153 0.171 0.120 0.100

Total settlement

Σ = 0.544

METERS 0.40 0.375

N

0.35 0.325 A G

0.30

A

A

B

B

0.275

G

0.25

C

0.225

B

D

0.20 0.175

G

0.15

C D N

0.125 0.10 0.075 0.05 0.025 June June 10, 1940 July

May

April

March

February

January, 1940

December

October

November

September

July

August

May

June

April

March

February

January, 1939

November

December

October

August

September

July

June

April April 21, 1938 May

0

DATE OF RECORDINGS

Figure 2.22 Settlement measurement for Charity Hospital. (Source: After W.W. Gwyn and Eustis Engineering.)

2.5 THE EUROTUNNEL LINKING FRANCE AND ENGLAND (TUNNELING) 2.5.1

The Story

The link between France and England relied for a long time on ferries going back and forth from one coast to the other. The idea of constructing a permanent and more convenient link seems to have started around 1800, including a proposal from Napoleon but at that time, it was felt that it would compromise the British national security. An attempt in the late 1800s took place on the British side but failed. In the early 1980s, Margaret Thatcher and François Mitterrand decided that it would happen but with private funds. Five different solutions, including bridges and tunnels, were proposed but the tunneling approach won the job for economic reasons. The project started in 1988 and opened in 1994 (Figure 2.23).

2.5.2

Rock Stratigraphy

The stratigraphy of the crossing is shown in Figure 2.24. Below the water surface, which is about 50 m deep at the deepest point, is a 20–40 m-thick layer of chalk underlain by a 30 m-thick layer of chalk-marl. Chalk is limestone rock with a high percentage of carbonates including fossils. Marl also has a high content of lime but has a much higher content of clay than chalk and is darker in color than the chalk. Both are soft sedimentary rocks and are well suited for tunneling. Because the chalk-marl was more impermeable than the chalk, it was decided to tunnel through the chalk-marl to minimize water infiltration issues. The depth of the tunnel would be about 30 m below the sea bottom near the coasts, reaching 70 m below the sea bottom or 120 m below the water level toward the center of the crossing. The associated water pressure could therefore be quite high.

2.5 THE EUROTUNNEL LINKING FRANCE AND ENGLAND (TUNNELING)

23

Dover

Folkestone Terminal Folkestone CALAIS M20

A2 6

Coquelles Terminal

The Channel

ay ilw Ra

on

nd

Lo

ris

Pa

ay

To

ilw

Ra

A1

6

To

Figure 2.23 Eurotunnel plan view. (Source: Courtesy of CAD 2000 MLS International, www.cady2k.com.)

Castle Hill Holywell 350 300 UK sector French sector

Shakespeare Cliff

250 Fosse Dangeard tributary Sea-level UK cross-over Middle Chalk Point M French cross-over White Chalk

Upper Chalk Middle Chalk White Chalk Lower Chalk Grey Chalk Chalk Marl Glauconitic Marl Zone 6A Gault Lower Greensand

200 CTHD

Chalk Marl

150

Grey Chalk Zone 6A 15

20

Lower-Greensand 25 30

Gault Clay 35

100

40

Glauconitic Marl 45

50

55

50 Chainage km 0

CTHD - Channel Tunnel height daturn. Mean sea-level equivalent to 200 CTHD

Figure 2.24 Geologic cross-section of the Eurotunnel. (Source: Courtesy of The Geologic Society of London, https://www.geolsoc.org.uk/GeositesChannelTunnel.)

2.5.3

Tunnel Design

It was decided that no car would travel through the tunnel to avoid air pollution issues and that electric-powered trains would be the mode of transportation. Cars, trucks, passengers would be transported in train cars from one side to the other. The tunnel cross-section is shown in Figure 2.25. The two main tunnels where the trains would travel would be 7.6 m

in diameter, 30 m apart, and about 50 km in length. A 4.8 m diameter service tunnel located between the 7.6 m tunnels and parallel to them would be drilled first to learn about the tunneling process (Figure 2.25). A set of 3.3 m diameter tunnels perpendicular to the train tunnels would connect those tunnels every 375 m (Figure 2.25). They would provide access for service and maintenance during routine operations and

24

2 CASE HISTORIES

English Channel White & gray chalk

Piston relief duct Cross-passage

Chalk marl

Gault clay North Running Tunnel Service Tunnel

Figure 2.25 Tunnel overview and cross-section. (Source: Courtesy of Eurotunnel and ICE, https://www .youtube.com/watch?v=2zu_u0xDSks.)

serve as passenger escape tunnels in case of emergency. This was the case when fires broke out in 1996, 2006, and 2008; no one was injured as the escape process worked well but the fires did damage the tunnels, which had to be repaired before reopening. Piston relief ducts would connect the two train tunnels every 250 m (Figure 2.25). Two large undersea cross-over caverns would allow for flexibility of train movement.

Disc Cutter

Cutterhead

Forward Shield

Telescopic Shield

Thrust Cylinder

Gripper Cylinder

2.5.4

Tunnel Construction

There are different ways of boring tunnels. For deep, large diameter tunnels, the most common and efficient way is to use tunnel-boring machines (TBM). These enormous machines can be 200 m long, weigh 150 MN or more, and cost tens of millions of dollars. The way they work is shown in Figure 2.26. The cutterhead is at the front and cuts the

Tail Shield

Gripper Shoe

Probe Drill

Operator’s Cabin

Installed Segments

TBM Conveyor

Ring Beam Erector

Figure 2.26 Principles of TBMs. (Source: Courtesy of Robbins Company.)

2.6 THE TETON DAM (EARTH DAM EROSION)

soil/rock by rotation. It is pushed forward by reaction against the tail shield which has gripper shoes expanded into the sides of the tunnel. The telescopic shield between the cutterhead and the tail shield allows the cutterhead to advance while the tail shield is stationary. Once the cutterhead has advanced, the gripper shoes are retracted, the tail shield is advanced and re-grips the tunnel sides to permit a new sequence of progression of the cutterhead. As the tunnel progresses, the ring beam erector at the back of the tail shield grabs tunnel liner segments and installs them one after the other. Once a set of liner rings is placed, the gap between the back of the liner and the tunnel wall is grouted with cement. The spoils generated by the drilling are carried from the cutterhead to the back of the TBM by a conveyor belt and all the way to the beginning of the tunnel where they are disposed of. Eleven TBMs were used to drill the tunnels, 6 on the English side and 5 on the French side. On the French side, earth pressure balance TBMs were used because water infiltration was expected when drilling through the fractured chalk in the descent toward the chalk marl. These TBM have a pressure chamber behind the cutting head to be able to counterbalance the water pressure, which could be as high as 1000 kPa in this case. On the English side, major water infiltration was not anticipated and simpler, open-faced TBM were used. Water infiltration was encountered and was mitigated by grouting ahead of the tunneling process. The tunneling rate varied quite a bit but established some remarkable world records including 75 m per day on the English side. In all, about 10 million m3 of spoil were generated. The huge number of liner segments came from pre-casting factories. Advanced surveying techniques, including gyrotheodolites, were used to control the tunnels’ direction. The tunnels did meet near the middle and within very satisfactory precision of the order of a few centimeters (Figure 2.27). On the money side, the project was financed by 10 construction companies and 5 banks. It was awarded for about $7 billion and ended up costing about $14 billion. Today, some 20 million passengers use the Eurotunnel every year, which also helps transport 20 million tonnes of freight every year. While the decision of building the Eurotunnel was fraught

25

with controversy and opposition, the result is a magnificent success for the economies of both countries and Europe.

2.6 THE TETON DAM (EARTH DAM EROSION) 2.6.1

The Story

The site is located on the Teton River near Idaho Falls, Idaho. In the fall of 1972, construction of this 100 m-high earth dam started. The construction was completed in the fall of 1975 when filling of the reservoir was initiated. On June 3, 1976, as the reservoir level approached full capacity, a leak developed 400 m downstream of the dam toe at the right abutment. On June 4, 1976, the leak had moved backward toward the dam and was now 60 m from the dam toe at the right abutment. On June 5, 1976, at 7 a.m., muddy water was flowing at about 1 m3 /s from the downstream face near the embankment toe. Around 9:45 a.m. that same day, a wet spot appeared in the middle of the downstream face of the dam (Figure 2.28). At 10:30 a.m. a loud burst was heard, and the wet spot became larger and regressed toward the crest of the dam. At 11:20 a.m. two bulldozers were sent to backfill the hole, the bulldozers ended up falling into the hole, but the operators escaped in time. At 11:32 a.m., a second hole appeared on the downstream face (Figure 2.28). At 11:57 a.m., the Teton Dam was breached, and the reservoir emptied its contents. Eleven people died, 13,000 head of cattle died, many lawsuits took place, the dam was never rebuilt but what is left of the dam is still in place (Figure 2.29). An independent review panel was formed and delivered a report (Chadwick et al., 1976), which forms the basis of this summary. 9:45 am 5 June 1975

11:20 am 5 June 1975

11:25 am 5 June 1975

11:32 am 5 June 1975

11:50 am 5 June 1975

11:57 am 5 June 1975

Figure 2.28 Teton Dam failure sequence. (Source: Photos by Mrs. Eunice Olson. Courtesy of Arthur G. Sylvester.) a - Teton Dam site today (looking upstream)

Figure 2.27 Meeting of the two tunnels and two tunneling teams. (Source: Courtesy of Eurotunnel, (https://www.youtube.com/ watch?v=SGikebYtAZc.)

b - Teton Dam cross-section today (from left abutment)

Figure 2.29 Teton Dam site today: (a) overall site; (b) cross-section.

26

2 CASE HISTORIES

After reviewing the drilling logs and after a pilot grouting program, it was decided that the alluvium layer would be removed; this led to a 30 m-deep key trench at the bottom of the dam. It was also decided that on both riverbanks, it would be more economical to remove the upper 20 m of rock than to conduct an extensive grouting program necessary to seal the highly jointed rock. A 20 m-deep key trench was excavated above El 5100 ft (Figure 2.31) through the highly jointed rock in both abutments, with a three-row grout curtain below that.

Aeolian silt sediments about 10 m thick on plateau

2.6.3

Alluvium deposited in the channel About 30 m thick

There are two types of dams: arch dams and earth dams. Arch dams are typically used when the valley is narrow, and the mountain sides are strong. Arch dams are typically made of concrete and resist the push from the water they retain by transferring the load to the mountain sides through the arch shape. Earth dams are typically used when the valley is wide; they resist the water push by gravity and friction resistance against sliding. The Teton Dam is an earth dam with a height of 100 m and a crest length of 1000 m. The low permeability core of the dam was made of the silt from the plateau (Zone 1 in Figures 2.32a and 2.32b). The silt had the following average properties: liquid limit 26, plasticity index 3, 80% fines ( 50 b/0.3m

Dense Sand End of Pile Tips

Pier

su = 10–20 kPa w = 45% PI = 41

Gray and brown dense to very dense sand with silt, gravel, and clay lenses, cobbles and boulders

The Soil Stratigraphy

The Potomac River at the location of the Woodrow Wilson Bridge (WWB) is made of two flow channels (Figure 2.34). The main channel is about 14 m deep and is located close the right bank. This is where the two large bascule piers were located (V1 and M1 on Figure 2.34). At those pier locations, the top layer is a 20 m-thick soft organic clay with an undrained shear strength between 10 and 20 kPa. Below that is a thin layer of sand underlain by a hard clay with an undrained shear strength varying between 100 and 250 kPa. The sand layer is very thick (∼25 m) in the secondary channel but practically inexistent in the main channel. The problem was to estimate the depth of the scour hole that would form around the bascule pier location should a 500-year storm come through the river.

End of Pile Tips

2.7.3

Hard Clay

su = 100–250 kPa Hard gray and low and high plasticity clay w = 24% with some lenses of sand and gravel PI = 72

(c)

Figure 2.34 Soil stratigraphy at the site of the new Woodrow Wilson Bridge: (a) diagram to scale; (b) diagram with vertical scale significantly exaggerated); (c) key to the soil layers.

Scour Depth Calculations

Borings were drilled from barges near the location of the bascule piers. The collected samples were tested in the laboratory (Kwak et al., 2002) to obtain the erosion function of the two main layers: the soft organic clay and the hard clay. Figure 2.35 shows the erosion functions obtained from the soft clay and the underlying very stiff to hard clay. Note

Scour Rate (mm/hr)

Scour Rate (mm/hr)

100 50

0

5

τc = 5.09

(N/m2)

10

20

15

25

60 40

τc = 0.16 N/m2

20 0

10

20

30

40

Shear Stress (N/m2)

(a)

150

100

100 50 vc = 1.35 m/s 0

80

0

30

Shear Stress (N/m2)

(a)

(b)

(b) – Hard clay (su = 130 kPa) 100

Scour Rate (mm/hr)

Scour Rate (mm/hr)

(a) – Soft clay (su = 22 kPa) 150

0

0

1

29

2

3

Velocity (m/sec)

80 60 40

0

4 (b)

vc = 0.2 m/s

20 0

1

2 3 4 Velocity (m/sec)

Figure 2.35 Measured erosion functions.

5

30

2 CASE HISTORIES

Direction of Flow

Pedestal

3.7m

31.7m

Approx. River Bottom

5.7m

1.37m diam. open-ended steel pipe piles filled with concrete 4.8m

27.7m

31.3m

Figure 2.36 Dimensions of the bascule pier and foundation.

M1: Earthquake

EL. 4 m (500 yr) EL. 3 m (100 yr) EL. 0.6 m (Ave.)

12000 kN

M2: Vessel Collision M3~M10: Wind Load

618 kN

River

5.9 m

EL. -10.5 m (River Bottom) Alluvial Soil

13.0 m

Glacial Sand 500 yr scour depth

45.1 m

Cretaceous Clay

Open-Ended Steel Pipe Pile (D = 1.8 m, L = 64 m)

EL. -68.6 m (Estimated Pile Tip)

Figure 2.37 Pile dimensions and loads. (Source: After Stan Davis, Maryland DOT.)

2.8 SAN JACINTO MONUMENT (SHALLOW MAT FOUNDATION)

(a) – Static load test

(b) – Statnamic load test

make their predictions, the results came out mostly around a 20 m scour depth except for one prediction, which was about 10 m. The predictions of 20 m depth were all based on the assumption that the soil was fine sand including some flume tests simulations. The prediction, which ended up with a 10 m depth, was the only one based on testing on the clay from the site. A 13 m scour depth was finally retained. 2.7.4

Figure 2.38 Pile load tests. (Source: Courtesy of Stan Davis, Maryland DOT.)

that the critical velocity is inversely proportional to the undrained shear strength in this case. This shows that the critical velocity, which represents the erosion “strength” of the soil is not necessarily proportional to the mechanical shear strength of the soil. A group of experts was formed with each one of them asked to predict, for the 500-year flood, the depth of the scour hole that would form around the large 31.3 m-wide bascule piers (Figure 2.36). While the experts used different approaches to

31

Foundation and Cost

The design load on each pile was 12,000 kN vertically and 618 kN horizontally (Figure 2.37). The horizontal load was due to a superposition of vessel collision load, wind load, and earthquake load. Load tests were carried out both by static load testing and by Statnamic testing (Figure 2.38). Open-ended steel pipe piles filled with concrete were finally selected with a length of 64 m and a diameter of 1.8 m (Figure 2.39). The following costs are in 2009 prices. While the entire project including the approaches was more than $2 billion, the cost of the bridge between abutments was $625 million with about $125 million for the foundation. 2.8 SAN JACINTO MONUMENT (SHALLOW MAT FOUNDATION) 2.8.1

The Story

In 1830, what are now the states of Texas, New Mexico, Arizona, Utah, Nevada, California, and parts of Colorado belonged to Mexico. In 1836, the Texans decided that they wanted their independence from Mexico, but the Mexican government refused to give it to them. Santa Anna was president of Mexico and the general of the Mexican Army. On March 6, 1836, Santa Anna and 1800 of his soldiers won a memorable battle at the Alamo in San Antonio. General Sam Houston retreated toward what is now called the city of Houston while recruiting as many Texans as possible on the way. By the time he got to the San Jacinto River, he had about 800 soldiers ready to fight the Mexican Army. Using the element of surprise on April 21, 1836, Sam Houston and his makeshift Texan Army defeated Santa Anna and his Mexican Army. The battle of San Jacinto is classified as a very important victory as it freed the States of Texas, New Mexico, Arizona, Utah, Nevada, California, and parts of Colorado. In 1936, the San Jacinto Monument was constructed near Houston to celebrate the 100-year anniversary of that most important Texas victory. This case history is about the mat foundation of the San Jacinto Monument and its recorded settlement over the last 75 years (Briaud et al., 2007c, 2015). 2.8.2

Figure 2.39 Pile driving.

Geometry, Weight, Construction, and Loading

The dimensions of the San Jacinto Monument are presented in Figure 2.40. The foundation is a 37.8 m by 37.8 m mat with an area of 1397.3 m2 (cutoff corners) and founded at a depth of 4.6 m below the natural ground surface. The total weight of the Monument is 312.7 MN, therefore the average pressure

Pressure, in kPa

32

2 CASE HISTORIES

191.5 Load on Soil

95.8 0

Nov. Jan. Mar. May Jul. Sept. Nov. Jan. Mar. May Jul. 1936 1937 1938

aquifers took place between 1935 and 1975. The Houston ship channel is very close to the Monument and, while the current (2020) groundwater level is deeper, the long-term stable groundwater level is probably around 5 m below the ground surface. 2.8.4

13.7 m

110.9 m Observation floor 144.5 m

FIRST FLOOR PLAN 74.2 m

The ultimate bearing pressure under the large mat can be calculated using various methods: undrained shear strength, CPT and PMT methods. The undrained shear strength method is based on the equation proposed by Skempton (1951):

167.4 m

153.7 m

37.8 m

14.3 m

Bearing Capacity Calculations

pu = Nc su + 𝛾D

where pu is the ultimate bearing pressure, N c is the bearing capacity factor, su is the undrained shear strength, γ is the total unit weight of the soil, and D is the embedment depth. The relative embedment depth for the mat is 4.6/ 37.8 = 0.12 and the corresponding value of N c is 6.3.

Fill

pu = 6.3 × 100 + 20 × 4.6 = 722 kPa

Elevation =0m

9.5 m

2.7 m

1.8 m

(2.43)

(2.44)

The CPT method is based on the following equation:

4.6 m 9.1 m

1.8 m 14.3 m 37.8 m 74.2 m 110.9 m

pu = kc qc + γD

(2.45)

where kc is the CPT bearing capacity factor and qc is the CPT point resistance. The bearing capacity factor kc is taken as 0.38 for a relative embedment of 0.12. The CPT qc value for a shallow depth is taken as 2500 kPa from Figure 2.41.

Figure 2.40 San Jacinto Monument dimensions. (Source: Briaud et al., 2007c, 2015, courtesy of ASCE.).

pu = 0.38 × 2500 + 20 × 4.6 = 1042 kPa

(2.46)

The PMT method is based on the following equation: on the soil below the mat is 223.8 kPa. The mat foundation itself weighs 133 MN representing 42.5% of the total weight. The weight of the 4.6 m depth of excavated soil is 115.7 MN. The construction started in the Fall of 1936 and by Fall 1937, the Monument was completed. The loading history is shown at the top of Figure 2.40.

pu = kp pL + 𝛾D

where kp is the PMT bearing capacity factor and pL is the PMT limit pressure. The bearing capacity factor kp is taken as 0.9 for clay. The PMT pL value for a shallow depth is taken as 1000 kPa from Figure 2.41. pu = 0.9 × 1000 + 20 × 4.6 = 992 kPa

2.8.3

Soil Stratigraphy and Parameters

The soil stratigraphy at the site is relatively simple. It consists primarily of a deep layer of stiff to very stiff clay with intermittent layers of sand. The unit weight is about 20 kN/m3 , the water content averages 25% and the plasticity index 40%. The depth of influence of the 37.8 m square foundation is theoretically 75.6 m but the CPT sounding stops at 16 m depth and the pressuremeter sounding at 37 m depth (Figure 2.41). Consolidation tests were performed in 1953 and then again in 2007 and the results are summarized in Figure 2.42. Unfortunately, the undrained shear strength was not measured but estimates can be obtained from correlations to the CPT point resistance and the PMT limit pressure (Figure 2.41). It appears that an undrained shear strength value of 100 kPa is a cautious estimate at shallow depth. The groundwater level has fluctuated widely (90 m) over the last 100 years as pumping for drinking water from deep

(2.47)

(2.48)

Since the actual pressure under the mat foundation is 223.8 kPa, these calculations indicate that the mat foundation has sufficient safety against bearing capacity failure. 2.8.5

Settlement Calculations

The long-term settlement of this large mat can be calculated by at least two methods: the modulus method and the consolidation method. Both methods rely on a measure of the stress strain relationship for the soil. The modulus method is based on the elastic equation: pB (2.49) E where s is the long-term settlement, I s , I e , and I h are factors quantifying the effect of shape, of embedment depth and of the presence of a hard layer within the depth of influence of the foundation respectively, 𝜈 is the Poisson’s ratio, p is s = Is Ie Ih (1 − 𝜈 2 )

2.8 SAN JACINTO MONUMENT (SHALLOW MAT FOUNDATION)

0

γt (kPa) 20

40

0

ω (%) 20

33

PI (%) 0 20 40 60 80

40

0

0.0 Very Stiff Clay

15.0 18.0

10 Depth (m)

Silty Sand

Stiff to Very Stiff Clay

20 30 40 50 60

63.0 70.0

Very Stiff Silty Sand

70

Cone penetrometer

Pressuremeter

Tip Resistance (kPa) 5000 0 10000

Limit Pressure, PL (kPa)

0

0

4000

Pressuremeter Reload Modulus, Er (kPa) 0 100000 200000

0

4

5

6

10

8

15

Depth (m)

Depth (m)

2

2000

Pressuremeter First Load Modulus, E0 (kPa) 0 5000 100000

10 12

20 25

14

30

16

35

18

40

Figure 2.41 Soil stratigraphy and soil properties at the San Jacinto Monument. Cr

Cc 0

0.25

0.5

0.75

0

0.2

σ′p (kPa) 0.4

0

1000

2000

1E-09

cv (m2/s) 1E-07

1E-05

0 10

Depth (m)

20 30 40 50 60 70 80

Boring in 1953 Fugro Boring in 2007

Boring in 1953 Fugro Boring in 2007

Boring in 1953

Boring in 1953

Fugro Boring in 2007

Fugro Boring in 2007

Figure 2.42 Soil properties from consolidation tests. (Source: Briaud et al., 2007c, 2015, courtesy of ASCE.)

the foundation pressure, B is the foundation width and E is the long-term soil modulus. Here I s was taken as 1.12 for the center of a flexible square foundation; the flexible case is selected following the comment in Chapter 18 based on the work of Focht et al. (1978). The influence factor I e is taken as 0.988 because the expression for this factor is 1 – 0.1D/B and I h is taken as 1 as no hard layer is found

within the depth of influence of the foundation. The drained value of the Poisson’s ratio is estimated to be 0.35 and the pressure is taken as the net pressure for this settlement calculation (p – γD = 223.8 – 20 × 4.6). This assumes that the heave of the soil at the bottom of the excavation due to the weight of soil removed will be compensated by the recompression of the soil under the nearly equal weight of

34

2 CASE HISTORIES

the foundation. Thus, it is assumed that the settlement is only due to the increase in pressure beyond the pressure of the excavated soil. The modulus is taken as a reasonable average of the PMT first load modulus profile (Figure 2.41); a value of 20,000 kPa is selected for this simple and global calculation. (223.8 − 20 × 4.6) × 37.8 s = 1.12 × 0.988 × 1(1 − 0.352 ) 20000 = 0.242 m (2.50) The consolidation settlement method is based on the equation: ( ′) ( ′ ( )) 𝜎p 𝜎ov + Δ𝜎v Ho ΔH = Cr log + Cc log 1 + eo 𝜎ov ′ 𝜎p ′ (2.51) where ΔH is the settlement, H o is the thickness of the compressing layer, eo is the initial void ratio, Cr is the recompression index, 𝜎 ′ p is the preconsolidation pressure, 𝜎 ′ ov is the effective stress in the center of the compressing layer before loading, Cc is the compression index, and Δ𝜎 v is the increase in vertical stress in the middle of the compressing layer. This equation applies when the effective stress after loading 𝜎 ′ ov + Δ𝜎 v exceeds the preconsolidation pressure 𝜎 ′ p . If 𝜎 ′ ov + Δ𝜎 v is less than 𝜎 ′ p , then the following equation applies. ( ′ ) 𝜎ov + Δ𝜎v Ho ΔH = C log (2.52) 1 + eo r 𝜎ov ′ Because Δ𝜎 v changes with depth, it is necessary to perform the settlement analysis by first dividing the soil into layers within the depth of influence. For this, the soil within the depth of influence (2B) below the San Jacinto Monument foundation is divided into four layers each 18.9 m thick (B/2). Because the vertical stress after loading does not exceed the preconsolidation pressure, the calculations proceed by applying Eq. (2.52) to each layer, as shown in Table 2.5. The calculated settlement is the sum of the compression of the layers and gives 282.3 mm. Dawson, who was in charge of the foundation design, installed a number of targets on top of the mat foundation Table 2.5

0.00 Minimum Settlement

SETTLEMENT, m

0.05

Maximum Settlement

0.10

Average Settlement

0.15 0.20 0.25 0.30 0.35 0.40 1936

1946

1956

1966

1976

1986

1996

2006

YEAR

Figure 2.43 Settlement of the San Jacinto Monument (record collected by Dawson, then McClelland, then Fugro).

right after its construction to be able to record the settlement of the Monument with respect to a benchmark. The record has been kept for the last 83 years first by Dawson, then by McClelland Engineers and then by Fugro. The settlement versus time is presented in Figure 2.43 with the most recent value being 328 mm. 2.8.6 Subsidence in Houston and Impact on Settlement Data The city of Houston got its water from 1938 to 1975 by pumping from deep aquifers under the city imprint. This induced a drastic lowering of the water table documented by a well next to the Monument and presented on Figure 2.44. This lowering of the groundwater level created a general subsidence of the Houston area by as much as 3 m in some places. Because subsidence is such a large-scale phenomenon, one would think at first glance that it would equally impact the Monument and the benchmark about 100 m away. Yet a more detailed analysis has to recognize that, at the same depth, the mean stress and deviator stress levels are higher under the Monument than under the benchmark. The higher mean confining stress would lead to a higher modulus, thus decreasing the

Calculations of consolidation settlement for San Jacinto Monument

H o (m)

Depth to center of layer (m)

Effective stress 𝜎 ′ ov * (kPa)

Net foundation pressure (kPa)

Stress increase factor

Stress increase Δ𝜎 v (kPa)

Recompression index Cr

Preconsolidation pressure 𝜎 ′ p (kPa)

Initial void ratio eo **

Compression of each layer ΔH (mm)

18.9 18.9 18.9 18.9

14.05 32.95 51.85 70.95

190.5 379.5 568.5 759.5

131.8 131.8 131.8 131.8

0.88 0.50 0.25 0.15

116.0 65.9 33.0 19.8

0.08 0.08 0.08 0.08

1000 1000 1000 1000

0.67 0.67 0.67 0.67

187.0 63.0 22.2 10.1 Σ = 282.3

*

Assumes that the groundwater table is at 5 m depth. Calculated using 𝛾 t = 20 kN/m3 , w = 25% and Gs = 2.65.

**

2.9 POINTE DU HOC CLIFFS (ROCK EROSION)

35

TIME (years) 1930 1940 1950 1960 1970 1980 1990 2000 2010 2020 0

DEPTH OF WATER BELOW GROUND SURFACE (m)

10

?

20

?

85M DRAWDOWN

30 40 50 60

? ? ?

70 80 90

Measured data Extrapolated data

100

Figure 2.44 Groundwater level at the San Jacinto Monument over time. (Source: Adapted from USGS National Water Information System.)

measured settlement while the higher deviator stress would lead to a lower modulus, thus increasing the measured settlement. Briaud et al. (2015) conducted numerical simulations to investigate the subsidence effect and to find out which of the two opposite influences would dominate. These simulations indicated that the net effect is to increase the measured settlement by 33 mm. This shows that the increase in deviator stress has more impact than the increase in confinement. Since the measured settlement was 328 mm including subsidence, the true settlement of the Monument without the differential subsidence effect between the Monument and the benchmark would have been 295 mm. 2.9 2.9.1

POINTE DU HOC CLIFFS (ROCK EROSION) The Story

On June 6, 1944, the Allied forces landed 160,000 soldiers in Normandy on the northwest coast of France to push back the German Army, which had invaded France. The perilous landing operation took place at the same time on five beaches and one cliff site called La Pointe du Hoc (Figure 2.45). The cliffs at Pointe du Hoc are about 25 m high and 200 Rangers, under the command of General Earl Rudder, were to climb the cliffs. Only 90 of the 200 Rangers assaulting the cliffs under gunfire made it to the top. The author will be forever grateful for those young men who sacrificed their lives to come and deliver his parents! During the 60 years from 1944 to 2004, the cliffs eroded significantly, and some of the war monuments were in danger of falling into the sea. In 2007, US Congressman Chet Edwards asked Texas A&M University to research the erosion process and suggest a repair scheme to arrest the erosion. This case history looks at the erosion process for the cliffs and suggests how to mitigate the process to save the monuments.

Figure 2.45 Pointe du Hoc and Observation Post. (Source: s4svisuals/Adobe Stock.)

2.9.2

The Rock Stratigraphy and Properties

The rock stratigraphy is shown in Figure 2.46 (Briaud, 2018). It consists of an 8 m-thick layer of silty clay over interbedded layers of limestone and sandstone down to a depth of 24 m and underlain by a layer a black marl. During coring of the rock, no drop of tool was experienced that might have indicated the existence of internal caverns in the rock mass. However, the water circulation was lost at times because of the high permeability of the fractured rock. Indeed, during high rain events, the water drained from the bottom of the craters and was seen gushing out of the cliff surface. The groundwater level matched the sea level quite well because of the highly pervious rock. The recovery ratio (RR) varied from 50% to 100% and averaged 85% while the rock quality designation (RQD) varied from 25% to 100% and averaged 63%. The average tensile strength measured in split “Brazilian” tests was 3.36 MPa for the limestone, 4.55 MPa for the sandstone and 4.52 MPa for the gray marl. The unconfined compression strength of the intact rock cores varied significantly from 316 kPa to 68800 kPa while averaging 20900 kPa. This unconfined compression strength was found to correlate well with the total unit weight of the cores, which varied from 20 to 26 kN/m3 . 2.9.3

The Cliff Erosion Process

Between 1944 and 2004, the cliff at Pointe du Hoc retreated by about 10 m with an average erosion rate of 0.019 mm/hr. The Observation Post (OP) (Figure 2.46) on top of which the Rudder Memorial is erected was closed to visitors in 2005 because it was considered to be in danger of falling into the sea. As a first step in studying the erosion process, the rock cores were subjected to erosion testing in the laboratory where water was circulated on top of the cores at chosen velocities. The measured erosion rates were comparable to the cliff retreat rate. For example, at a flow velocity of 3.6 m/s, the measured erosion rate on the core averaged 0.025 mm/hr. However, this slow erosion process of the rock material was not consistent with the erosion mechanism observed at the

36

2 CASE HISTORIES

0m

10 m the Observation Post (O.P.)

10 m

Estimated Section Profile in 1944

8m 2006

Silty Clay 4m

Fractured Limestone

3m

Very Hard Sandstone

22 m High Tide plus Storm Wave 6m

9m

High Tide

Fractured Limestone and Sandstone

7m Low Tide Very Stiff Gray Marly Limestone

Figure 2.46 Rock stratigraphy and erosion process.

Figure 2.47 Cavern and overhang at the bottom of the cliffs.

Figure 2.48 Massive cliff failure at Pointe du Hoc.

site and described later, but it did explain that any rock block falling off the cliff would slowly erode to much smaller sizes such as gravel and even sand in the long run. The erosion process observed at the site was a sequence starting with the continuous attack by the waves at the bottom of the cliffs generating caverns (Figures 2.46 and 2.47). When the caverns became deep enough, the rock mass held up above the caverns was increasingly loaded in cantilever and if the tensile strength of the rock mass was exceeded, the rock mass would fail in massive failures (Figure 2.48). No cavern depth larger than 4 m could be found at the site, thus it was concluded that cavern depths larger than 4 m would likely stress the rock mass beyond its mass tensile capacity. Numerical simulations were performed to find out what tensile stress would be generated for a cavern depth of 4 m. The answer was about 40 kPa (Figure 2.49). This showed that the tensile strength of this fractured rock mass was 100 times less

than the intact rock core tensile strength, which was about 4000 kPa. This ratio between the strength of the intact rock and the fractured rock mass is not unusual. 2.9.4

Proposed Remediation to Mitigate the Erosion

The goal of the remediation scheme was to slow down the erosion to the point where the war monuments could be saved and, in particular, the Observation Post for a long period of time. To minimize the probability of failure while minimizing the funding necessary to actuate the countermeasure, a redundancy principle was adopted. This meant to look for two sets of measures that would complement each other in case one of them failed. The first and most obvious measure was to backfill the caverns; however, the request was that the backfill should be esthetically blending in the landscape (Figure 2.50). The second measure was to place the OP on micropiles so that the OP would still be safe if the cliff eroded in a sudden

2.10 THE TOWER OF PISA (SHALLOW FOUNDATION)

37

fashion (Figure 2.51). The countermeasure was constructed in 2010 and the OP was reopened to the public in 2011 with many dignitaries from both countries being present.

Horizontal Stress Distribution Compression Tension 30 Elevation from Sea Level (m)

25

2.10 THE TOWER OF PISA (SHALLOW FOUNDATION)

20

2.10.1

15 10 1m Erosion 2m Erosion 3m Erosion 4m Erosion 5m Erosion

5 0 –60

–40

–20

0

20

40

60

Horizontal Stress (kPa)

Figure 2.49 Stress distribution in overhanging rock mass from numerical simulation.

BEFORE

AFTER

Figure 2.50 Caverns before and after repair.

0m

The Story

The Tower of Pisa is the most successful failure in geotechnical engineering history. Indeed, its leaning has brought tremendous fame and tourism to the city of Pisa while bringing geotechnical engineering to the mind of the public (Figure 2.52). In the early 1100s, the people of Pisa decided to build a great cathedral complex which would consist of a cathedral, a baptistry, a bell tower, and a cemetery on the Piazza dei Miracoli. Construction of the bell tower started in 1173 but was halted in 1178 due to ongoing wars after four stories were constructed. Construction resumed in 1272 and three additional stories were constructed. The construction stopped again in 1284 as the Tower was already leaning significantly. The final stage consisted of adding the bell chamber, which was constructed in 1360 and increased the tilt of the Tower. Various attempts have been made over time to reverse the movement of the Tower, but all made matters worse except for the last one. Indeed, in 1990, the decision was made to close the Tower to visitors and to work on a repair scheme. The Tower reopened to the public in 2001 after the successful repair was completed. The geotechnical team that accomplished this very stressful task was made up of Michele Jamiolkowski, John Burland, and Carlo Viggiani (Jamiolkowski, 2001).

10 m Motion Sensors

Walkway and Rail

10 m

8m

Capping Beam Microphiles on perimeter of O.P.

Silty clay

Estimated Section Profile in 1944 4m

Fractured limestone

3m

Very hard sandstone

22 m High Tide plus Storm Wave Shotcrete Backfill

9m

Fractured limestone and sandstone

Monitoring/maintenance of erosion each year

6m High Tide 7m Low Tide

Very stiff gray marly limestone

Figure 2.51 Repair scheme.

38

2 CASE HISTORIES

(a)

(b)

Figure 2.52 Tower of Pisa photos. (a) Piazza dei Miracoli with Tower of Pisa on right. (Source: Guglielmo Giambartolomei/Wikimedia Commons/CC BY-SA 4.0); (b) Classic tourist picture.

2.10.2

Dimensions and Soil Stratigraphy

The Tower of Pisa is 56 m high and weighs 142 MN. The depth of embedment of the foundation is about 5 m; the foundation has a doughnut shape with an outside diameter of 19.6 m and an inside diameter of 4.5 m (Figure 2.53). The foundation area is therefore 285.7 m2 and the pressure on the soil 497 kPa. The tilt is about 5.5 degrees, the height of the center of gravity is 22.6 m, and the overturning moment is 327 MN.m.

C'

8th Cornice

7th Cornice

6th Cornice

55°

≈ 60 m

5th Cornice 4th Cornice

This overturning moment leads to a much higher soil pressure on the lower side. The alluvial soils below the Tower (Figure 2.54) were deposited by the Arno River during the Holocene-Pleistocene Age. Overall, the soil can be classified as a high plasticity silty clay with the following average index properties: 𝛾 = 18.5 kN/m3 , w = 37.1%, wL = 53.4%, PI = 28.4%. The initial void ratio eo calculated from those parameters and using a unit weight of solids equal to 27 kN/m3 is 1. The groundwater level is approximately at the bottom of the foundation. For the mechanical properties (Figure 2.55), the cone penetrometer sounding shows the strength profile with several distinct layers. From the foundation level down to 8 m is a layer of silt with a varying CPT point resistance. Below this silt layer is a layer of clay down to 21 m where the average CPT point resistance is about 1150 kPa. The CPT point resistance averages 1620 kPa in the layers below that. The shear wave velocity averages 160 m/s from the surface to a depth of 19 m followed by an average of 220 m/s down to 37 m. The compression index averages 0.75 down to 18 m and then 0.4 down to 35 m. The undrained shear strength measured in undrained triaxial tests on high-quality samples (Figure 2.55) gave an average of 50 kPa down to 18 m

3rd Cornice 2nd Cornice

S

N

Catino (1838)

CF 7.54 m

4.5 m

7.54 m

19.60 m Figure 2.53 Diagram of the Tower of Pisa. (Source: Courtesy of Michele Jamiolkowski.)

Elevation a. m.s.l. in meters

1st Cornice

10 0 –10

G.W.L. Sandy and clayey silts Upper sand Upper clay

–20

Intermediate clay

–30

Lower clay

–40

Sea level

Intermediate sand

Lower sand

Figure 2.54 Soil stratigraphy below the Tower of Pisa. (Source: Courtesy of Michele Jamiolkowski.)

2.10 THE TOWER OF PISA (SHALLOW FOUNDATION)

qc (MPa) +3.0 m 0

GWL

B1

B2 B3

Elevation below m.s.l. (m)

A

0

2

4

6

8

σ′vo, σ′p (kPa)

OCR 10 12 14

1 2 3 4

0

8

9 10

0

–4

–4

–4

–8

–8

–8

–12

–12

–12

–16

–16

–16

–20

–20

–20

–24

–24

–24

–28

–28

–28

–32

–32 –36

–40

200

400

–36

–40

–40

–44

–44

600

σ′p

–32 Tests 1971 Tests 1986

B4 –36

0

σ′vo

qc > 25 MPa

C –44

Data collected by G. Calabresi (University of Rome) Lancellotta and Pepe, 1990 Compression index Cc

Undrained shear strength Su (kPa) 0 0

50

100

150

200

0.0 0

0.5

A1

A1 –5

–5 A2

A2

B1

–10

B1

–10

B2 –15

B2 –15

B3

B3

B5 B6

–25 B7 B8

–30

B4 Elevation (m a.s.l.)

Elevation (m a.s.l.)

B4 –20

–20

B5 B6

–25 B7 B8

–30

B9 –35

B10

B9 –35

B10 C1

C1 –40

–45

–50

1.5

1.0

–40 Osterberg piston sampler Mechanical piston sampler Laval sampler

–45

Osterberg piston sampler Mechanical piston sampler Laval sampler (Cc1)

–50

Laval sampler (Cc2)

Figure 2.55 Soil properties below the Tower of Pisa. (Source: Lo Presti et al., 2003.)

800

39

40

2 CASE HISTORIES

Table 2.6

Calculations of consolidation settlement for the Tower of Pisa

Compression Depth to Effective Foundation Relative Stress Stress ComRecom- Initial of each pressure depth increase increase pression pression void layer with H o center of stress (m) layer (m) 𝜎 ′ ov (kPa) (kPa) d/B factor Δ𝜎 v (kPa) index Cc index Cr ratio eo Cc ΔH (mm)

Compression using Cr in first layer ΔH (mm)

9.8 9.9 9.8 19.7 9.8 29.5 9.8 39.3

2284 1217 294 151

427 1217 294 151

𝚺 = 3976

2089

134.1 217.4 300.7 384.0

497 497 497 497

0.25 0.75 1.25 1.75

0.88 0.5 0.25 0.15

437.4 248.5 124.2 74.5

followed by an average of 80 kPa down to 37 m. Further soil properties information can be found in Lo Presti et al. (2003). 2.10.3

Bearing Capacity, Settlement, and Inclination

The pressure under the foundation is 497 kPa. The ultimate bearing capacity of the soil below the Tower can be estimated as pu = Nc su + 𝛾D (2.53) where pu is the ultimate bearing pressure, N c is the bearing capacity factor, su is the undrained shear strength, 𝛾 is the total unit weight of the soil, and D is the embedment depth. The relative embedment depth for the foundation is 5/19.6 = 0.255 and the corresponding value of N c is 6.7. The undrained shear strength in the shallower layer is 50 kPa, thus pu is: pu = 6.7 × 50 + 18.5 × 5 = 427.5 kPa

(2.54)

Another way to estimate pu is to use the CPT method: pu = kc qc + 𝛾D

(2.55)

where kc is the CPT bearing capacity factor and qc is the CPT point resistance. The bearing capacity factor kc is taken as 0.39 for a relative embedment of 0.25. The CPT qc value at shallow depth is taken as 1150 kPa from Figure 2.54. pu = 0.39 × 1150 + 18.5 × 5 = 541 kPa

(2.56)

These two estimates show that the pressure under the Tower of Pisa is close to the ultimate bearing capacity of the soil. The settlement of the Tower can be estimated using the theory of consolidation as follows ) ( ′ 𝜎ov + Δ𝜎v Ho ΔH = (2.57) Cc log 1 + eo 𝜎ov ′ where ΔH is the settlement, H o is the thickness of the compressing layer, eo is the initial void ratio, 𝜎 ′ ov is the effective stress in the center of the compressing layer before loading, Cc is the compression index, and Δ𝜎 v is the increase in vertical stress in the middle of the compressing layer. Because Δ𝜎 v changes with depth, it is necessary to perform the settlement analysis by first dividing the soil into layers within the depth of influence. For this, the soil within the depth of influence (2B = 39.2 m) below the Tower of Pisa is divided into four layers each 9.8 m thick (B/2). The calcu-

0.75 0.75 0.4 0.4

0.11

1.0 1.0 1.0 1.0

lated settlement (Table 2.6) is the sum of the compression of the layers and gives 3976 mm. This value is larger than the observed settlement very likely because the first 9.8 m thick layer below the foundation is overconsolidated and one should use the recompression index, as shown in Table 2.6, instead of the compression index. Another way to estimate the settlement is to use the modulus approach together with the cone penetrometer point resistance value. This simple and global calculation is based on the equation: pB s = Is Ie Ih (1 − 𝜈 2 ) (2.58) E where s is the long-term settlement, I s , I e , and I h are factors quantifying the effect of shape, of embedment depth and of the presence of a hard layer within the depth of influence of the foundation respectively, 𝜈 is the Poisson’s ratio, p is the foundation pressure, B is the foundation width, and E is the long-term soil modulus. Here I s is taken as 1.12 for the center of a flexible circular foundation; the flexible case is selected here following the comment in Chapter 18 based on the work of Focht et al. (1978). The influence factor I e is taken as 0.974 because the expression of this factor is 1 – 0.1D/B and I h is taken as 1 as no hard layer is found within the depth of influence of the foundation. The drained value of the Poisson’s ratio is estimated to be 0.35 and the pressure is taken as the net pressure for this settlement calculation (p – 𝛾D = 497 – 18.5 × 5). This assumes that the heave of the soil at the bottom of the excavation due to the weight of soil removed during excavation for the foundation is compensated by the recompression of the soil under the nearly equal weight of the foundation. Thus, it is assumed that the settlement is only due to the increase in pressure beyond the pressure of the excavated soil. The modulus is taken as 4qc as recommended by Bahmani and Briaud (2020) and the overall average qc value is taken as 1200 kPa. (497 − 18.5 × 5) × 19.6 s = 1.12 × 0.974 × 1(1 − 0.352 ) 4 × 1200 = 1581 mm (2.59) The settlement of the Tower was not measured over time but the dishing of the 10 m deep interface between the

2.10 THE TOWER OF PISA (SHALLOW FOUNDATION)

41

Figure 2.56 Inclination vs time and tower weight vs time. (Source: Courtesy of Professor Jamiolkowski.)

shallow silt layer and the clay layer below was measured to be 2.54 m (Jamiolkowski, 2001). If the estimated compression of the silt layer between the bottom of the foundation and the 10 m depth is added, the final average settlement comes out to be about 3 m. Note that the difference in settlement between the high side and the low side of the foundation is about 1m. A heave of about 0.4 m was measured on the south side of the foundation indicating a low factor of safety against bearing capacity failure. The Tower of Pisa is leaning because of a combination of phenomena: 1. The high pressure, which approaches the ultimate bearing capacity of the soil. 2. The large settlement, which is about 3 m. 3. The leaning instability, which is due to the high slenderness ratio of the tower. This slenderness ratio is the ratio of the height of a tower divided by the diameter of its foundation; for the Pisa Tower this is 56/19.6 = 2.86. This is a very important contributing factor. For example, the Baptistry on the other side of the Cathedral (Figure 2.52a) has a height of 54 m and a diameter of 34.1 m. Its slenderness ratio is much smaller than the Tower at 54/34.1 = 1.58 and it only leans 0.6 degrees compared to the 5.5 degrees of the Tower. The Cathedral between the Tower and the Baptistry is a beautiful squat building with a slenderness ratio much below one; it also settled but is not leaning. If the Tower of Pisa was built today, it would require a much larger mat foundation to bring the pressure on the soil down to a safe pressure. For a factor of safety of 3, a foundation diameter equal to 33 m would be required. However, even with this larger foundation the settlement could be

excessive by modern standards and a pile foundation might be considered. 2.10.4

The 2001 Repair

Because the combination of distress in the Tower masonry due to high compression stresses and the ever-increasing leaning angle of the Tower versus time (Figure 2.56), the Tower was closed to the public in 1990. At that time the inclination was 5.5 degrees. Shortly thereafter an international commission chaired by Professor Jamiolkowski was organized by the Italian government with the goal of reopening the Tower under safer conditions. Two repair schemes were used. The first one was a temporary scheme that consisted of loading the high side of the foundation with lead weights (Figure 2.57). The moment applied by the lead weights was 45 MN.m in the

Figure 2.57 Lead weights placed on the high side as a temporary measure. (Source: Courtesy of Michele Jamiolkowski.)

42

2 CASE HISTORIES

Massive underexcavation Extraction hole - Section B-B'

+3.00 ~20°

0.00

Ingots Catino

+2.49 +1.86 –0.59–0.87

Tower plinth +1.86

0.00

Casing, OD=219mm, L=14.00m

–3.85

Casing, OD=168mm, Ground extraction, L=6.15m

Tower cent re

Guide tube OD=273mm, L=2.00m Beam

CL

Figure 2.58 Excavation of soil under the high side. (Source: Courtesy of Michele Jamiolkowski.)

opposite direction of the 327 MN.m moment the Tower is subjected to. This temporary measure did reverse the leaning trend of the Tower by 60 seconds of arc (0.017 degree). The second repair scheme consisted of excavating soil from under the high side of the Tower (Burland et al., 2009) (Figure 2.58). Indeed, while the first scheme had worked, the lead weights were not esthetically appealing and an inconspicuous and safe solution was sought. A preset series of 40 casings were lined up to guide the soil removal process (Figure 2.58). The soil was drilled out and a total of 70 metric tons or 38 m3 of soil was removed. The high side of the Tower moved down 180 mm while the low side did not go down. The inclination was reversed and with an inclination of 5 degrees instead of 5.5 degrees, the Tower is back to where it was in 1838 (Figure 2.56).

and in 1913, the 55 m-high work-house and the 30 m-high bin-house were completed. The bin-house suffered a foundation failure on October 18, 1913, as it was reaching full grain storage capacity. However, the bin-house failed while remaining nearly structurally intact (Figure 2.59) and was jacked back up into position to resume functionality (Figure 2.60). This case history addresses the bearing capacity issue as well as the up-righting repair. 2.11.2

Dimensions and Weight

The Transcona silo is located outside of Winnipeg in Canada (Figure 2.59). Construction started in the winter of 1912

The Transcona silo is made of a drier-house, a work-house, and a bin-house. The 55 m-high work-house is on a mat foundation which is 21 m by 29 m while the 30 m high bin-house is on a 0.6 m thick reinforced concrete mat which is 23.5 m by 59.5 m. The surface area for the mat foundation of the bin-house is therefore 1398 m2 ; it is embedded 3.7 m below the ground surface. The weight of the empty bin-house is estimated at 196 MN and the 875,000 bushels of wheat at the time of the failure added 255 MN for a total weight of 451 MN. The wheat has a unit weight of about 8.3 kN/m3 . The mean pressure under the bin-house is therefore 322.6 kPa.

Figure 2.59 The Transcona silo after failure. (Source: Courtesy of the Canadian Geotechnical Society, https://www.cgs.ca/pdf/110928 %20Introduction%20and%20papers.pdf.)

Figure 2.60 The Transcona silo after up-righting. (Source: Courtesy of the Canadian Geotechnical Society, https://www.cgs.ca/pdf/ 110928%20Introduction%20and%20papers.pdf.)

2.11 THE TRANSCONA SILO (SHALLOW FOUNDATION) 2.11.1

The Story

43

2.11 THE TRANSCONA SILO (SHALLOW FOUNDATION)

Untrained shear strength (kPa) El. 235.12 0 meters

0

25

50

75

100

125

150

Particle size distribution (%)

Moisture content (%) 175

0

0

20

40

60

80

100

0

0

20

60

40

80

100

Yellow silt 3

3

Grayish tan silty clay, joints and cracks, small calcareous silt pockets, few pebbles.

6

6

3

Plastic limit

9

Liquid Limit

9

9

12

12

15

15

% silt

6

% sand

Moisture

% clay

Depth (m)

Tan and gray clay, slickensided, faintly laminated, small calcareous silt pockets.

12

% gravel

Gray silty clay, tan calcareous silt pockets, few pebbles. su from unconfined compression test (kPa) su from remolded unconfined compression test (kPa)

Tan silty gravel (limestone chips) few clay pockets. 15 Refusal

Figure 2.61 Soil stratigraphy, index and strength properties. (Source: Adapted from Baracos, 1957; Peck and Bryant, 1953.)

Soil Properties

While no borings were taken prior to the construction of the silo, several borings were drilled in 1951 (Peck and Bryant, 1953) and again in 1952 (Baracos, 1957). The soils in the area are post-glacial lacustrine deposits. Below a surface fill, there is a layer of brown silty clay down to a depth of 8.5 m followed by a 5.5 m-thick layer of gray silty clay. Figure 2.61 shows the stratigraphy with an unconfined compression strength profile and some index properties. The water content averages 50% and the specific gravity of solids was 2.7, leading to a total unit weight of 17.2 kN/m3 assuming saturation. A consolidation test was also performed from a sample at a depth of 9 m (Figure 2.62); additional consolidation tests were performed later (Baracos, 1957). Below the silty clay layers is a hard layer of glacial till with bedrock below it. The groundwater level is at the bottom of the foundation. 2.11.4

Bearing Capacity, Settlement, and Failure

The strength profile indicates that there is a stronger clay layer from the 3.7 m foundation depth to a depth of 8.5 m (Figure 2.63); in that layer the unconfined compression strength averages 110 kPa or an undrained shear strength su of 55 kPa. Below this layer is a weaker clay layer down to a depth of 14 m where the unconfined compression strength averages 65 kPa (su = 32.5 kPa). The average pressure under the mat of the bin-house is 451,000 = 322.6 kPa (2.60) p= 1398

The ultimate bearing pressure of the soil below the bin-house can be estimated as pu = Nc su + 𝛾D

(2.61)

where pu is the ultimate bearing pressure, N c is the bearing capacity factor, su is the undrained shear strength, 𝛾 is the total unit weight of the soil, and D is the embedment depth. The relative embedment depth for the foundation is 3.7/23.5 = 0.157 and the corresponding value of N c is 6.5. The undrained shear strength in the shallower and stronger layer is 55 kPa and in 2 1.8 1.6 Void ratio, e

2.11.3

1.4

eb ea

Boring No.1 Sample:10-4 Depth (m): 10 w (%) = 64 LW = 95

1.2 1 0.8 0.6 10

σ′ov σ′ov+Δσv 100 Pressure, kPa

1000

Figure 2.62 Consolidation test result. (Source: After Peck and Bryant, 1953.)

44

2 CASE HISTORIES

29°

After failure Oct. 18,1913

Original position

Clay fill

3.7 m

Brown clay su = 55 kPa

4.8 m

su = 32 kPa

5.5 m

Gray clay

Glacial silt till Bedrock

Figure 2.63 Simplified Transcona silo bin-house and soil layers.

the weaker layer below, it is 32.5 kPa. A weighted average of the strength is 43 kPa, thus pu is: pu = 6.5 × 43 + 17.2 × 3.7 = 343.1 kPa

(2.62)

Another way to estimate the ultimate bearing pressure is to consider that the failure mechanism consisted of the bin-house foundation punching through the stronger upper layer and failing the lower clay layer in bearing capacity. The equation would be (see Chapter 18) pu BL = 2(B + L)Hsu(upper) + BL(Nc su(lower) + 𝛾D)

(2.63)

where B and L are the width and length of the foundation, H is the thickness of the upper stronger clay layer with an undrained shear strength su(upper) , N c is the bearing capacity factor for an embedment corresponding to the depth of the lower weaker clay layer (D/B = 8.5/23.5 = 0.36 for which N c = 7)) with an undrained shear strength su(lower), 𝛾 is the total unit weight of the soil and D is the depth to the lower weaker clay layer (D = 8.5 m). pu 23.5 × 59.5 = 2(23.5 + 59.5)4.8 × 55 + (7 × 32.5 + 17.2 × 8.5)23.5 × 59.5 pu = 405 kPa

(2.64) (2.65)

This value is larger than the previously calculated value and the straight down punching mechanism is not consistent with the observed failure mechanism. Nevertheless, it is clear that the foundation pressure was very close to the ultimate bearing capacity of the soil and explains the failure. The settlement of the silo, if failure had not occurred, can be estimated by using the consolidation curve as follows: e − ea (2.66) ΔH = Ho (𝜀b − 𝜀a ) = Ho b 1 + eb where ΔH is the settlement, H o is the thickness of the compressing layer, 𝜀b is the vertical strain corresponding to

the soil state before loading, 𝜀a is the vertical strain corresponding to the soil state after loading, eb is the void ratio corresponding to the soil state before loading, ea is the void ratio corresponding to the soil state after loading. The void ratios ea and eb are obtained directly from the consolidation curve at stresses corresponding to the initial effective vertical stress 𝜎 ′ ov and the effective vertical stress long after loading (𝜎 ′ ov + Δ𝜎 v ) in the middle of the compressible soil layer, which is 10.3 m thick. In this case, the effective stress 𝜎 ′ ov is 100.7 kPa (8.85 × 17.2 − 5.15 × 10). The increase in stress Δ𝜎 v is 258 kPa calculated as the product of the stress bulb influence factor 0.8 for a relative depth of 8.85/23.5 = 0.362 by the foundation pressure of 322.6 kPa. The void ratio eb corresponding to 𝜎 ′ ov =100.7 kPa is read as 1.60 from Figure 2.62 and the void ratio ea corresponding to 𝜎 ′ ov + Δ𝜎 v = 358.7 kPa is 1.27. ) ( 1.60 − 1.27 = 1.31 m (2.67) ΔH = 10.3 × 1 + 1.6 Allaire (1916) indicates that on October 18, 1913, within an hour after the bin-house being loaded with 875,000 bushels of wheat, a uniform settlement of 0.3 m was noted followed by a tilt that grew to 29 degrees over 24 hours, after which the bin-house came to rest. 2.11.5

The Up-Righting of the Silo

The bin-house had suffered little structural damage during the soil failure and moved as a rigid block. Evaluation of the structure and economic estimates showed that it would be advantageous to up-right and save the bin-house, which weighed 196 MN empty. The process consisted of several steps (Figure 2.64, Allaire, 1916). 1. Remove the grain from the bin-house. 2. Shore the work-house and underpin it to rock under each column to prevent failure of the work-house. 3. Drill through the mat of the bin-house 14 bored concrete piles with casing to rock under the low side of the bin-house. 4. Remove soil from under the high side of the bin-house mat and start jacking upward the low side against the piles. 5. Drill through the mat under each bin, rows of 14 bored concrete piles with casing to rock. 6. Continue jacking gradually until the structure was level. 7. Concrete the space between the top of the bored piles and the mat of the bin-house In the end, the bin-house was rotated 29 degrees and moved upward 3.7 m. It stands today 4.3 m below its original position.

2.11 THE TRANSCONA SILO (SHALLOW FOUNDATION)

Figure 2.64 Up-righting of the Transcona silo. (Source: Allaire, 1916, courtesy of ASCE.)

45

46

2 CASE HISTORIES

2.12 THE SAINT JOHN RIVER BRIDGE ABUTMENT (SLOPE STABILITY) 2.12.1

The Story

The Saint John River flows through the city of Fredericton in the Province of New Brunswick in Canada (Figure 2.65). In 1973, a new bridge across the river was to be built. One of the issues was to design the right abutment of the approach embankment, which included a slope stability problem as the river valley is about 20 m below the city level at that location. The design included several scenarios for the longitudinal cross-section of the embankment with various encroachments into the river main channel. Indeed, embankments are typically less expensive per meter of length than bridges and it is economically advantageous to have long embankments and short bridges. However, short bridges constrict the river flow and can lead to scour problems at the bridge supports. A compromise had to be found. At the time, the author was a graduate student at the University of New Brunswick where his advisor, Professor Landva, had been asked to lead the stability study. The author ran over 150 methods of slices analyses by hand at a rate of $10/hr. and felt very rich! 2.12.2

The Bridge and Right Abutment

The St. John River flows through Fredericton in New Brunswick, Canada, on its way to the Atlantic Ocean in the Bay of Fundy. The Westmoreland Street bridge crosses the River in Fredericton and was constructed starting in the late 1970s; it opened in 1981. The bridge is 0.75 km long between abutments and carries four lanes of traffic and a pedestrian/bicycle lane across the river. The right abutment included a fill to elevation +14.64 m, the bridge abutment and wall placed on piles, a four-lane riverfront drive along the river and a bicycle and pedestrian walkway also along the river (Figure 2.65). One of the proposed cross-section to be

analyzed is shown in Figure 2.66. The height of the slope is 17.38 m and the desired factor of safety is 1.5. 2.12.3

The Soil Conditions

The geology of the area was strongly influenced by the presence of glaciers some 10,000 years ago. The movement of these glaciers and subsequent melting created the deeper and thick layer of overconsolidated silty clay found at the site (Figure 2.66). The shallow layer of sand above the silty clay was deposited by the river. The mean water level of the river is at elevation +1.5 m so, for the purpose of the slope stability analysis, the natural soil layers are below water level and considered saturated. The total unit weight of the sand layer and of the silty clay layer is 19.2 kN/m3 . The effective stress shear strength properties were measured in consolidated drained triaxial tests with water stress measurements (Figure 2.67). For the sand, the effective stress cohesion was zero (c′ sand = 0 kPa) and the average effective stress friction angle was 30o (𝜑′ sand = 30o ). For the overconsolidated silty clay, also called glacial till, the average value of c′ clay was 14.4 kPa and that of 𝜑′ clay was 25o (Figure 2.68). 2.12.4 The Fill and the Approach Embankment Construction The sandy fill for the approach embankment was compacted in layers and had a total unit weight of 20.4 kN/m3 . The effective stress cohesion was zero and the effective stress friction angle was 32o (𝜑′ fill = 32o ). The embankment was built in three stages (Figure 2.66). Stage 1 consisted of building Embankment 1 to elevation +7.6 m uniformly. Stage 2 consisted of building Embankment 2 which would support the riverfront drive roadway; this embankment would rise to elevation +11 m. Stage 3 consisted of building Embankment 3, which would carry the Westmoreland Street approach; this embankment would rise to 14.6 m. One of the major concerns was the buildup of excess water compression stresses (excess pore pressure), which would delay the buildup of effective stresses and lead to a lower factor of safety right after or even during construction. 2.12.5 Water Stress Induced by Embankment Construction

Figure 2.65 Right abutment of the Westmoreland Street bridge in Fredericton, NB, Canada. (Source: Courtesy of the City of Fredericton, Canada, https://frederictonenvtimeline.ca/smartgrowth/# magnific/0/.)

The low permeability of the silty clay would no doubt lead to a buildup of excess compression stresses in the water during construction and it was necessary to calculate the magnitude of those water stresses. The first step was to calculate the increase in vertical stress Δ𝜎 v at various locations in the silty clay due to the embankment. An influence chart to obtain the increase in vertical stress under an embankment was used (Figure 2.69). Today it would be easier to use a finite element program and simulate the embankments load to obtain the increase in vertical stress in the silty clay layer. The second step was to calculate the increase in pore pressure (Δuw ) generated by the increase in vertical stress (Δ𝜎 v ).

2.12 THE SAINT JOHN RIVER BRIDGE ABUTMENT (SLOPE STABILITY)

+18

Embankment 3

+12 1st stage

Elevation (m)

+6 Sand

0

+10.9 m

+14.6m 3rd 1 sta ge 2

2

γ = 20.4 kN/m φ = 32° c = 0 kPa

Embankment fill

Embankment 2

+7.6 m

3

2nd stage 1 1st stage1

Embankment 1 2

+1.5m

γ = 19.2 kN/m φ = 30° c = 0 kPa

3

–6

2 –2.7m

River bed

γ = 19.2 kN/m3 φ = 25° c = 14.4 kPa

Silty clay

–18

1

Previous ground level

–12

47

–24 Bedrock

–30

–42 –36 –30 –24 –18 –12 –6

0

+12 +18 +24 +30 +36 +42 +48

+6

Chainage (m)

Figure 2.66 Right abutment cross-section. 300

The equation is

Stress (kPa)

250

Δuw = B × Δ𝜎v σ′1–σ′3

200

where B is a pore pressure parameter. This parameter was back-calculated from the triaxial test results and a conservative (high) value of 0.5 was finally selected. The decision was made to conduct the slope stability analysis using B = 0.5, ensure a satisfactory factor of safety, install piezometers founded in the silty clay layer as the embankment was being built, monitor the increase in pore pressure during construction, and be prepared to pause the embankment construction if the excess pore pressure Δuw was exceeding 0.5 × Δ𝜎 v where the excess pore pressure was being measured, and resuming construction once Δuw had receded to a safe level.

150 100

σ′3 = 138 kPa

50 u 0 0

5

10

20

15

(2.68)

Strain (%)

Figure 2.67 Example of consolidated drained triaxial test with water stress measurement. 300

Shear stress (kPa)

250

200 φ′=25° 150

100

50 c′ =14.4 kPa

0 0

50

100

150

200

250

300

350

400

450

Normal effective stress (kPa)

Figure 2.68 Results of triaxial tests on the overconsolidated silty clay.

500

550

600

48

2 CASE HISTORIES

2.12.6

Slope Stability Analysis

In 1976, the slope stability analysis was conducted by hand using effective stresses analysis and effective stress shear strength parameters. This approach was used to simulate the undrained behavior of the clay right after construction together with a B value equal to 0.5. The analysis was carried out for the three stages of construction. In 2020, the analysis was conducted using modern slope stability software. The increase in vertical stress due to the embankment was estimated at the bottom of each slice as the pressure generated by the embankment placed on the ground surface. The corresponding excess water stress right after construction was calculated at the bottom of each slice as 0.5 times the increase in vertical stress. That excess water stress distribution was used in the slope stability analysis. The critical circle and associated factor of safety for stage 3 are shown in Figure 2.70. As can be seen, the factor of safety is acceptable. The bridge abutment was constructed, and no slope stability issue was encountered.

Figure 2.69 Embankment stress influence chart. (Source: Courtesy of ISSMGE.)

20

779 kN

0

2340 kN

2379 kN

86 kN 775 kN –20

ELEVATION (m)

40

60

1.880

Pore Pressure = 265 kPa

0

20

40

60 80 CHAINAGE (m)

100

Figure 2.70 Slope stability analysis results using Rocscience software.

120

140

2.13 FOUNDATION OF BRIAUD’S HOUSE (SHRINK SWELL SOILS)

2.13 FOUNDATION OF BRIAUD’S HOUSE (SHRINK SWELL SOILS) 2.13.1

The Story

The house was to be built in 1989 (Figure 2.71). A boring indicated that the soil had a high potential for shrinking and swelling (see Figure 2.72) and houses in the area had shown signs of distress. Clearly the foundation would have to be designed with this issue in mind. Since I was supposedly an expert on foundations, I would be the one designing the foundation and any failure would have a negative impact on my reputation! The pressure was on and I learned a lot in the process of designing the foundation for this light two-story

49

building with an imprint area of 250 m2 . I am happy to say that 30 years later the house is still standing and that only the kitchen door sticks from time to time. 2.13.2

The Soil Conditions

The soil boring results are reproduced in Figure 2.72. It shows a silty clay from 0 to 2 m, a clayey sand from 2 to 2.7 m, and a clay with silt seams from 2.7 to 7.6 m depth. The water content at the time of the boring was about 13% in the top layer up to 30% in the bottom layer. The highest plasticity index was 61% found near the top of the bottom clay layer. The undrained shear strength estimated as 0.3 times the pocket penetrometer value averaged 128 kPa. Considering that the pressure under

Figure 2.71 The house, pool, and tennis court.

Figure 2.72 Soil boring for house.

50

2 CASE HISTORIES

a house is typically less than the pressure under your feet, there was no bearing capacity or settlement problem for the structure. The issue was the swelling and shrinking of the soil from one season to the next near the edges of the house. This would induce bending of the structure and potential cracking of rather rigid materials such as bricks, cement, and stucco. 2.13.3

Figure 2.73 Beam cross-section for the stiffened slab on grade.

The House Foundation

There are four major types of foundation solutions for light buildings on shrink swell soils (see Chapter 18). The best foundation solutions are those that let the soil move as it wishes yet prevent the structure from bending excessively because of the foundation stiffness. The foundation that was chosen because of its optimum balance between economy and stiffness was a stiffened slab on grade (Figures 2.73 and 2.74). The beams would be 0.3 m wide and 0.91 m deep for the exterior beams and 0.76 m deep for the interior beams. The spacing between beams in both directions averaged 3 m (Figure 2.74); this type of foundation is often called a waffle slab on grade because it is in the shape of a waffle. In addition, a thick compacted fill was placed before constructing the

Figure 2.74 Foundation plan including beam spacing and distribution.

51

2.14 THE EIFFEL TOWER (DEEP FOUNDATION)

2.13.4

The Tennis Court Foundation

The foundation of the swimming pool was made of a 0.3 m-thick 2.4 m-deep reinforced concrete tub and worked well. The foundation of the tennis court required a different way of thinking. Indeed, in this case there is no structure that can break on top of the foundation. So minor longwavelength bending of the foundation is not detrimental, but cracking must be avoided to provide a smooth playing surface. A thin post-tension slab was selected in this case with post-tensioning cables every 1 m in both directions loaded to 150 kN after concrete curing. The edges of the tennis court have moved up and down with the seasons as shown on Figure 2.75 (Abdelmalak, 2018). One end of the court had a tree next to it and the court movement has been more severe next to that tree, as shown in Figure 2.76. However, even in that case, the distortion expressed as the inverse of the differential movement over the length over which it occurs is 6000 mm / 12 mm = 500, which is acceptable. The cracking of this 30-year-old court has been minimal and limited to hairline cracks. Heave of the lawn has been significant as it has been watered regularly for 30 years; this heave reaches 150 mm in several areas.

Relative Movements (mm)

2 1 0 –1 –2 –3

0

2

4

6

8

10

12

14

16

18

Y-Coord. (Vertical Distance from the West Edge), m 22-Jun-05

8-Oct-05

10-Jan-06

17-Apr-06

5-Oct-06

9-Feb-07

14-May-05

29-Jul-05

27-Aug-05

10-Nov-05

15-Dec-05

13-Feb-06

15-Mar-06

17-May-06

12-Jul-06

18-Aug-06

2-Nov-06

14-Dec-06

Figure 2.75 Edge movement of the tennis court in the short direction. (Source: Abdelmalak, 2018.)

3

0 Relative Movements (mm)

foundation so the house would be elevated well above grade to provide positive water drainage away from the foundation. Drainage of the property away from the house was also assured. Large trees were kept reasonably far away from the house although they could have been further. The house was built in 1989 and to this day (2021) no cracks have appeared in the walls inside or outside or in the floor tiles. From time to time, the door from the kitchen to the outside sticks but because the frame of that door is made of wood, a few strikes of a chisel and a coat of paint solve the problem.

–3

–6

–9

–12

–15 20

22

24

26

28

30

32

34

36

38

X-Coord. (Horizontal Distance from the North Edge), m 14-May-05

22-Jun-05

29-Jul-05

27-Aug-05

8-Oct-05

10-Nov-05

15-Dec-05

10-Jan-06

13-Feb-06

15-Mar-06

17-Apr-06

17-May-06

12-Jul-06

18-Aug-06

5-Oct-06

2-Nov-06

14-Dec-06

9-Feb-07

Figure 2.76 Edge movement of the tennis court near a large tree. (Source: Abdelmalak, 2018.)

2.14 2.14.1

THE EIFFEL TOWER (DEEP FOUNDATION) The Story

The Eiffel Tower was built by about 300 workers from December 1887 to May 1889 for the 10th World Exhibition in Paris, France. The Tower would be the centerpiece of the Exhibition grounds which would stretch from the Palais de Chaillot across the River Seine on one side to the Invalides on the other side. Gustav Eiffel presented his project in 1886 but was clever to present it not only as a showcase for the Exhibition but also as a tower for strategic observation, for physics experiments and for transmission. This, he thought, would prevent it from being destroyed 20 years later as was originally planned. Today the Eiffel Tower houses close to 100 transmission antennas and remains at 324 m, the tallest building in Paris (in 2021) (Figure 2.77). The shape of the tower is an exponential function and its elegance has stood the test of time although some consider it to be but a big drill rig. While many know what the Eiffel Tower looks like, very few know the story about its foundation; this is the topic of this section (Lemoine, 2006). 2.14.2

The Soil Stratigraphy

The bank of the River Seine is about 200 m from the Eiffel Tower (Figure 2.78). The soil stratigraphy is related to the river deposits, which are deeper for piers 1 and 4 closer to the river than for piers 2 and 3, which are 100 m further away.

52

2 CASE HISTORIES

Figure 2.77 The Eiffel Tower. (Source: Luc Lombarda/Adobe Stock.)

2.14.3

PLAN VIEW PIER 2 10 m

6m

10 m

10 m 6m

PIER 3

PIER 1 141 m A

m

in

≈2

e

m

Se

10 0

ve r

A

0 10

Ri

14 m 6m

ALL PIERS HAVE THE SAME PLANAR DIMENSIONS BUT DIFFERENT DEPTHS AS SHOWN IN CROSS-SECTION AA.

00 50

m

≈1

The Foundation

6m

PIER 4

m Fl

ow

Figure 2.78 Plan view of the Eiffel Tower site.

Today the average water depth of the river is about 9 m as it crosses Paris. Starting around the year 1300, retaining walls were built to contain the River Seine, and to place fill behind the walls to allow for more construction. As such, the top layer at the site is a fill below which are interbedded alluvial layers of sand, clay, and gravel (Figure 2.79). The water level at the site is 7 m deep.

The Eiffel Tower weighs 99 MN and rests on four piers numbered 1 through 4 (see Figure 2.78). The angle of each leg with the horizontal at the ground surface is 54 degrees. The foundation under each pier is made of four elements with three of them being 6 m × 10 m in plan view and the back element being 6 m × 14 m (see Figure 2.78). The depth of these foundation elements is 13 m for piers 1 and 4, which are close to the river and 8 m for piers 2 and 3, which are away from the river. Indeed for piers 2 and 3, the stronger soil was found at shallower depth. For piers 1 and 4, founded at a depth of 13 m, the water table was encountered at a depth of 7 m. Therefore, piers 1 and 4 were excavated and constructed under air pressure to prevent the water from invading the excavation (Figure 2.80). The foundation elements were made of concrete and massive limestone blocks with embedded anchors to connect to the legs of the tower (Figure 2.81). The pressure under the foundation due to the tower itself is 99,000 = 93.75 kPa (2.69) 4(3 × 6 × 10 + 6 × 14) However, this pressure does not include the weight of the foundation, which is estimated as follows. The weight of the two large foundation elements under piers 1 and 4 is p=

53

2.14 THE EIFFEL TOWER (DEEP FOUNDATION)

CROSS-SECTION A-A

54°

4m 1m

8m

13 m

54°

1m 4m 2m

7m 10 m 10 m

Pier 1 7m 5m 2m 2m

Ground surface 13 m Sandy Clay

Limestone

Pier 3

Pier 4 Fill

13 m

8m

Sand and Gravel 6m

Plastic Clay

7m 4m

6m 6m

Figure 2.79 Cross-section of the Eiffel Tower site along section AA in Figure 2.78.

Figure 2.80 Metal containers used to excavate under air pressure. (Source: Lemoine, 2006, Courtesy of the Musée d’Orsay.)

Figure 2.81 Foundation elements with connection anchors. (Source: Lemoine, 2006, Courtesy of the Musee d’Orsay.)

calculated as the volume times an estimated unit weight of 25 kN/m3 .

The weight of the two large foundation elements under piers 2 and 3 is calculated as the volume times an estimated unit weight of 25 kN/m3 .

Q1&4large = 2 × 25(14 × 7 + 3.6 × 6 + 0.5 × 2.6 × 3.6

Q2&3large = 2 × 25(14 × 2 + 3.6 × 6 + 0.5 × 2.6 × 3.6

+ 0.5 × 5.4 × 8.6) × 6 = 44250 kN

(2.70)

+ 0.5 × 5.4 × 8.6) × 6 = 23250 kN

(2.72)

The weight of the six smaller foundation elements under piers 1 and 4 is calculated as the volume times an estimated unit weight of 25 kN/m3 .

The weight of the six smaller foundation elements under piers 2 and 3 is calculated as the volume times an estimated unit weight of 25 kN/m3 .

Q1&4small = 6 × 25(10 × 7 + 3.6 × 6 + 0.5 × 2.6 × 3.6 + 0.5

Q2&3large = 6 × 25(10 × 2 + 3.6 × 6 + 0.5 × 2.6 × 3.6 + 0.5

× 5.4 × 8.6) × 6 = 107550 kN

(2.71)

× 5.4 × 8.6) × 6 = 62550 kN

(2.73)

54

2 CASE HISTORIES

So the total weight of the Eiffel Tower foundation is Qfoundation = 44250 + 107550 + 23250 + 62550 = 237600 kN

(2.74)

The foundation is 2.4 times heavier than the tower itself and the soil under the foundation must carry both the tower weight and the foundation weight or 99 + 237.6 = 336.6 MN. The total pressure under the foundation is therefore: ptotal =

336600 = 318.75 kPa 4(3 × 6 × 10 + 6 × 14)

(2.75)

This is the pressure to use to check against bearing capacity failure. The net pressure is the one used for simplified settlement calculations. The net pressure is the total pressure ptotal minus the soil pressure removed by excavation pexcav . Since an average of 10.5 m of soil was excavated for the foundation and assuming a soil unit weight of 20 kN/m3 , pexcav is 20 × 10.5 = 210 kPa and the net pressure pnet is 318.75 – 210 = 108.75 kPa. No excessive settlement or tilt has been reported for the Eiffel Tower. 2.15 2.15.1

ST. ISAAC CATHEDRAL (MAT FOUNDATION) The Story

St. Isaac Cathedral is located in St. Petersburg, Russia. Peter the Great decided in 1703 to create a new capital city called St. Petersburg on the delta of the Neva River. He wanted many outstanding monuments to decorate the new city and today these historical monuments are a wonderful sight. The first St. Isaac Cathedral, built in 1707, was relatively modest, was made of wood, was too close to the Neva River, and was destroyed by floods. Two more cathedrals were built on the site but later destroyed until the current St. Isaac cathedral was built from 1818 to 1858 (Figure 2.82). This majestic and very heavy structure sat on the soft deposits

of the Neva River delta. Today the differential settlement of this combined pile raft foundation has led to damage of the structure, including cracks in the columns. This case history deals with the foundation of this famous cathedral and its behavior over the last 162 years (Shidlovskaya et al., 2017). 2.15.2

Rinaldi and later Brenna worked on the third St. Isaac Cathedral but it ended up being not as glamorous looking as Tsar Alexander I wanted it. Alexander I decided to destroy the building and asked August de Montferand to take charge of the new project. Montferrand decided to reuse the foundation under the third cathedral, which consisted of 10,756 timber piles 0.26 m in diameter, 10.4 m long under the columns and 6.4 m long elsewhere. Montferrand started by opening a 5 m-deep excavation and driving 13,000 additional timber piles 10.5 m long under the columns and 8.4 m elsewhere. Starting at the bottom of the excavation, Montferrand built a 7.5 m-thick mat made of massive granite blocks under the most heavily loaded elements, with limestone masonry for the remainder of the foundation. So, the foundation consists of an early version of what is now called a combined pile raft foundation (Figure 2.83) with a 7.5 m-thick mat on top of 23,756 timber piles with an approximate 1 m center-to-center spacing. The piles act as friction piles as the end bearing is small compared to the friction capacity. The plan view and front view of the cathedral are shown in Figure 2.83. Using a unit weight of the mat material of 25 kN/m3 , the weight of the mat is estimated to be 997 MN (Sotnikov, 1986). The weight of the piles is estimated to be 127 MN. The total height of St. Isaac Cathedral is 100 m. The overall length of the Cathedral with porticoes is 100 m, and its width is 92 m. The total weight of the Cathedral structure above the mat is 2158 MN for a total weight on the soil below the piles of 3282 MN. This makes St. Isaac Cathedral the heaviest building in St. Petersburg. The outside dimensions of the mat are 100 m long by 92 m wide. The mat is in the shape of a cross (Figure 2.83) and has a total area of 7600 m2 . The average pressure on the soil below the foundation is therefore 432 kPa. 2.15.3

Figure 2.82 St. Isaac Cathedral.

Construction, Dimensions, and Load

Soil Data

The top layers are post-glacial and glacial deposits of quaternary age underlain by a much older clay moraine. From the surface down to 10 m are coarse grain fills and organic swamp soils (Figure 2.84), followed by very fine and fine clay and sand of marine and lacustrine origin down to 20 m, followed by a moraine layer consisting of silt and lean clay with some gravel down to 40 m. Below the moraine is the Vendian clay, which is a pre-quaternary soil. The groundwater is 2 m deep. The available soil data includes a CPT sounding (Figure 2.84), water content profiles, plasticity index profiles, permeability profiles, and undrained shear strength and modulus profiles from unconsolidated undrained triaxial tests (UU tests). The

2.15 ST. ISAAC CATHEDRAL (MAT FOUNDATION)

55

16 m

60 m

100 m

16 m GWL 7.5 m 25 m

50 m

25 m

Figure 2.83 Dimensions of St. Isaac Cathedral. TIP RESISTANCE qc (MPa) 1

2

3

4

5

6

7

8

9 10 11

2

qc = RANGE 2000–6000 kPa MEAN = 4000 kPa

4 6

qc = RANGE 3000–21000 kPa MEAN = 12000 kPa

8 10 12

qc

14

fs

qc = RANGE 1000–3500 kPa MEAN = 2000 kPa

18 20 22 24 26 28 30

qc = RANGE 2000–14000 kPa MEAN = 3000 kPa

32 34 36 38 40 0 10 20 30 40 50 60 70 80 90 100 110 SLEEVE FRICTION fs (kPa)

Figure 2.84 Stratigraphy and CPT data. (Source: After Trest GRII, www.grii.ru.)

GLACIAL LEAN CLAY (MORAINE)

DEPTH (m)

16

SAND WITH ORGANIC MATTER

0

LACUSTRINE DEPOSITS

0

56

2 CASE HISTORIES

UNDRAINED SHEAR STRENGTH (kPa) 0

20

40

60

80

100

120

140 0

0

MODULUS OF DEFORMATION (kPa) 500 1000 1500 2000

2500

0

5

10

SILTY SAND

PILE FOUNDATION 10 VARVED CLAY 15 30 LEAN CLAY (MORAINE)

DEPTH (m)

DEPTH (m)

20

PILE FOUNDATION

20

VARVED CLAY

25

40

30 50

LEAN CLAY (MORAINE) HARD CALY

35

60

40

Figure 2.85 Undrained shear strength and modulus profile (UU triaxial tests).

water content varies and averages about 30%, the plasticity index averages 13%, the hydraulic conductivity 3x10-5 cm/s, and the undrained shear strength and modulus increase with depth, as shown in Figure 2.85. Note that at a depth of 45 m, a much harder layer of clay is encountered; thus the settlement calculation will focus on the compression of the top 45 m below the cathedral. 2.15.4 Bearing Capacity and Settlement Calculations The foundation is a combined pile raft with relatively shallow piles compared to the width of the structure (see Figure 2.83). Thus, the foundation behavior can be considered as that of an equivalent mat foundation in first approximation. The raft or mat depth is 5 m and the pile length below that varies from 6.4 to 10.4 m. Thus, the equivalent mat would be located at a depth of about 13.4 m (5+8.4). This would bring the bearing capacity issue into the lacustrine deposits of Figure 2.85. The ultimate bearing capacity pu of a square shallow foundation can be estimated by using the cone penetrometer point resistance qc . The equation is (see Chapter 18): pu = kc qc + 𝛾D

(2.76)

where kc is the bearing capacity factor, 𝛾 the total unit weight of the soil within the embedment depth, and D the embed-

ment depth. A lower bound of qc is selected from Figure 2.84 as 1500 kPa. If the lacustrine deposits are considered to be clay, the bearing capacity factor kc is 0.4. If the lacustrine deposits are considered to be sand, the bearing capacity factor kc is 0.2. Using an average for kc equal to 0.3 gives a bearing capacity pu of 0.3 × 1500 + 20 × 13.4 = 718 kPa. The pressure under the equivalent mat of the cathedral was 432 kPa, which gives a factor of safety against bearing capacity failure of 718/432 = 1.66. The bearing capacity can also be estimated using the undrained shear strength su . An undrained analysis can be carried out even 150 years after construction as the bearing capacity failure, if it were to occur today, would occur in a rapid undrained fashion. The equation is: pu = Nc su + 𝛾D

(2.77)

where N c is the bearing capacity factor depending on the foundation shape and the relative embedment depth. The relative embedment depth for the mat is 13.4/92 = 0.146 and the corresponding value of N c = 6.6 (see Chapter 18). The su value at a depth of 13.4 m is about 20 kPa (Figure 2.85), which gives a bearing capacity pu of 6.6 × 20 + 20 × 13.4 = 400 kPa. The su value of 20 kPa seems to be quite low. Indeed, if the qc value is 1500 kPa, the back-calculated N k value (su = (qc – 𝜎 ov )/N k ) would be 61.6, which is unusually high. Sample disturbance may be the reason for this low su value. So, it appears that the

2.16 NATIONAL GEOTECHNICAL EXPERIMENTATION SITES AT TEXAS A&M UNIVERSITY

57

ignores the reload settlement after excavation. Thus, ) ( 332 × 92 13.4 × 0.28 × (1 − 0.352 ) × E = 1 × 1 − 0.1 × 92 1 = 7392 kPa (2.79)

Figure 2.86 Settlement measurement.

soil under the Cathedral has enough bearing capacity to carry the load. The settlement is the next concern. Settlement measurements were made as follows (Figure 2.86). At the time of construction, the 7.5 m thick mat was 2.5 m above the ground surface. Today the mat was measured to be 1.8 m above the ground surface. This indicates that the Cathedral has sunk 0.7 m into the soil. Furthermore, the ground surface at the base of the Cathedral is lower than the street level by 0.35 m. These simple measurements allow one to conclude that the total settlement of the edge of the Cathedral is of the order of 1 m. It is possible to use this settlement estimate to backcalculate the soil modulus, which would give 1 m of settlement. Settlement calculations start by obtaining the depth of influence zi , which is defined as the depth at which the increase in total vertical stress due to the weight of the structure is equal to 10% of the pressure under the foundation. This is typically taken as 2 times the width of the foundation for a square foundation, which would be 2 × 92 m or 184 m for the Cathedral. However, with the assumption that the hard Vendian clay below the moraine is much stronger than the surface layers, zi is taken as corresponding to a depth of 45 m below the ground surface. The elastic equation is (see Chapter 18): pB (2.78) s = Is Ie Ih (1 − 𝜈 2 ) E where s = settlement, I s = shape factor (0.56 for the edge of a square foundation), I e = embedment factor (1 – 0.1 × 13.4/92), I h = depth to hard layer factor (for H/B = (45 – 5)/92 = 0.435, I h = 0.28), 𝜈 = the Poisson’s ratio taken as 0.35 for drained behavior, p = net pressure, B = width of the foundation, E = average soil modulus. Considering that the bottom of the mat is founded at a depth of 5 m, the overburden pressure at that depth is 5 × 20 = 100 kPa. Therefore, the net pressure for the Cathedral is 432 – 100 = 332 kPa. This pressure is used in the settlement equation; using this pressure

Briaud (2013) using a pressuremeter database suggested that the ratio of the long-term modulus E to the cone penetrometer point resistance qc is about 2.5 for clays. Using this ratio and the value of 7392 kPa for E gives an equivalent qc of 2957 kPa. This is well within the range of value of the qc profile (Figure 2.84). However, the drained long-term modulus of 7392 kPa is much higher than the modulus values obtained in the unconsolidated undrained triaxial tests (see Figure 2.85). Sample disturbance may be the reason. 2.16 NATIONAL GEOTECHNICAL EXPERIMENTATION SITES AT TEXAS A&M UNIVERSITY 2.16.1

The Story

I came to Texas A&M University in 1978 and I had read a quote from Terzaghi that said, “The action is in the field.” I really liked it, so I proceeded with developing full-scale experiments on infrastructure elements at a site which was selected as a National Geotechnical Experimentation Site in 1992. There are two sites, one where the top 12.5 m are made of medium dense silty sand (Figure 2.87) and one where the top 12.5 m are made of stiff to very stiff medium plasticity clay (Figure 2.88). Over the last 40 years, many projects have been conducted including retaining walls, deep foundations, shallow foundations, anchors, nails, embankments, culverts, truck impact tests on barriers, new in situ tests, and more. These expensive tests are usually performed in parallel with extensive numerical simulations that, once calibrated, can extend significantly the value of the full-scale investment. For each project, practical guidelines were developed and published (Briaud, 2021a, 2021b). A few major projects have been selected and are described as mini case histories in the following sections. 2.16.2

Tieback Wall at the Sand Site

A top-down tieback retaining wall was constructed and instrumented at the sand site; it was 50 m long and 10 m high. A row of 22 piles were installed then the soil was excavated in front of the piles while placing wood lagging and setting inclined anchors (Figure 2.89). The measured data is shown in Figure 2.90, including the lateral movement, the bending moment with anchor loads, and the axial load on the piles in the wall. One lesson learned was that the vertical capacity of the wall impacts the horizontal behavior of the wall. Indeed, the retained soil mass creates downdrag

58

2 CASE HISTORIES

Figure 2.87 Soil properties at the clay site. (Source: After Briaud, 1997b.)

Figure 2.88 Soil properties at the sand site. (Source: After Briaud, 1997b.)

on the wall facing and, if the vertical capacity is not sufficient to resist this load, the wall facing settles and the wall rotates around the anchor points, thus leading to horizontal deflection. 2.16.3

Spread Footings Tests at the Sand Site

Five square spread footings were constructed, and load tested to 150 mm of penetration at the sand site; the sizes were 1m, 1.5m, 2.5m, 3m, and 3m and the thickness was 1.2m. The

load test set up is shown in Figure 2.91. The vertical load and the settlement of the footings were measured and gave the curves on Figure 2.92. In addition, the vertical movement of the soil at various depth was obtained by using telltales under the center of the footings and the horizontal movement by using inclinometers on the side of the footings. The guidelines derived from this series of tests included the load settlement curve method, and some of the direct bearing capacity equations (both in Chapter 18).

2.16 NATIONAL GEOTECHNICAL EXPERIMENTATION SITES AT TEXAS A&M UNIVERSITY

59

POST CONSTRUCTION GROUND LEVEL WALE 1.8 m

2.7 m

3m

4.8 m

2.7 m FINAL EXCAVATION LEVEL SOLDIER BEAM 4 @2.4=9.6 m

12 @2.4=31.2 m

4 @2.4=9.6 m GROUND SURFACE

1.8 m

LOAD CELL

5.05 m TENDON BONDED LENGTH

WALE TUBE 3.0 m BRACKET

7.3 m TENDON BONDED LENGTH 2.7 m

FINAL EXCAVATION LEVEL PRESSURE INJECTED TIEBACK

SOLDIER BEAM

INCLINOMETER CASINGS

206 kN

185 kN

0

0

–1

–1

206 kN

–2

–2

–3

–3

–4 –5

DEPTH (m)

0 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –45

DEPTH (m)

DEPTH (m)

Figure 2.89 Wall dimensions.

185 kN

–6 –7

–35

–25

–15

–5

5

–5

185 kN

–6 –7 –9

–9

DEFLECTION (mm)

–4

–8

–8 –10 –40

206 kN

–20

0

20

40

60

–10 –400 –300 –200 –100

BENDING MOMENT (kN·m)

0

100

AXIAL LOAD (kN)

Figure 2.90 Wall response: (a) deflection; (b) bending moment; (c) axial load.

2.16.4 Grouted Anchors Under Tension Load at the Clay Site

2.16.5 Drilled and Grouted Piles in Cyclic Tension at the Clay Site

Ten drilled and grouted anchors were constructed vertically at the clay site (Figure 2.93). Their diameter was 0.3 m and the length 13.8 m. A steel tendon was placed in the center of the borehole and grout was poured in the annulus between the steel tendon and the borehole wall. The load was applied to the tendon at the ground surface and the grout, once cured, transferred the load from the tendon to the soil. In 1991, all anchors were load tested to failure in tension. The ratio of the ultimate friction to the undrained shear strength was back-calculated to be 0.52 on average. The same anchors were retested 22 years later (2013) and the same coefficient had increased to 0.81.

Two drilled and grouted piles were constructed; they were 0.2 m in diameter and 12.2 m in length. the borehole was drilled, a casing was lowered in the center of the hole and grout filled the annulus between the casing and the borehole wall. They were loaded in tension and subjected to cyclic loading (Figure 2.94). Briaud (2013) proposed the following equation for the influence of N cycles on the ultimate load carried by a pile: QUN = N −a (2.80) QU1 where QUN is the ultimate capacity after N cycles and a is the cyclic exponent. The value of a depends on a number of

60

2 CASE HISTORIES

Settlement LVDT Beam Jack

ANCHORS 1 2 3 4 5 6

ANCHORS 7 8 9 10

8.5 m Load Cell

0.5B

UNBONDED LENGTH

B SAND

4.6m Telltales

2B

6.5m

V. STIFF CLAY su=110 kPa

9.2m

GWL

0.75m

Inclinometer Casings

V. STIFF CLAY su=140 kPa

9.2m TENDON

5.25m

10.7 m

SAND & GRAVEL

BOND LENGTH 4.6m

B 15 m

Dywidag bars only. No Concrete

CLAY SHALE

Figure 2.93 Drilled and grouted anchors at the clay site.

LOAD, (kN)

CLAY SHALE

–160

0

200

400

600

800

1000

1200

1400

Drilled Shaft (Concrete + Bars)

7.6 m

Steel Plate

2.7 m

2.7 m

Figure 2.91 Load test set-up for the spread footing tests.

1.8

Pressure (MPa)

1.2

1m

2.5m

1.5

TOP MOVEMENT, (mm)

–180 –200 –220 –240 –260 –280

3m 1.5m

0.9

–300

0.6

3m

Figure 2.94 Cyclic load test in tension on a drilled and grouted pile at the clay site.

0.3 0.0 0.00

0.02

0.04

0.06

0.08

0.10

Settlement / Width, S/B

Figure 2.92 Normalized load settlement curves for the five footings.

factors associated with the cyclic loading; a was back-calculated from the load test as follows: Q 1150 log U1 log QUN 1040 = 0.044 a= = (2.81) log N log 10 This value is below the trend line proposed by Briaud (2013); see Chapter 18.

2.17 THE 827 M-HIGH BURJ KHALIFA TOWER FOUNDATION (COMBINED PILE RAFT FOUNDATION) 2.17.1

The Story

Tall buildings are iconic structures in our world (Figure 2.95) and the race to be the tallest has led to some marvelous and gravity-defying structures. The current (2022) tallest structure is the Burj Khalifa tower in Dubai, UAE, at 827 m. The Kingdom tower in Jeddah, Saudi Arabia, is under construction and should reach slightly over 1000 m in the future. The construction of the Burj Khalifa tower started in 2004 and ended in 2009 (Figure 2.96). While everyone is amazed at the height of the tower, the foundation is a very important part of

2.17 THE 827 M-HIGH BURJ KHALIFA TOWER FOUNDATION (COMBINED PILE RAFT FOUNDATION)

61

Figure 2.95 Tallest buildings in the world in 2020. (Source: Courtesy of the Council on Tall Buildings and Urban Habitat, https://www.ctbuh.org.) (a)

(b)

Figure 2.96 Construction of the Burj Khalifa. (Source: (a) Soonadracula/Wikimedia Commons; (b) Aheilner/Wikimedia Commons.)

its successful completion and is the subject of this case history (Poulos, 2009; Poulos and Bunce, 2008). The Soil and Rock Conditions

The city of Dubai is built in a sand desert area where the parent material is calcareous rock. A total of 33 borings were drilled at the site and indicated that the top 15 m below the ground surface was made of medium dense silty sand with some sandstone bands. Below that depth is the parent rock formation, which is a very weak to weak calcareous sandstone and siltstone with gypsum layers. Zones of perched groundwater were found starting at 2.5 m depth. Sixty pressuremeter tests were carried out (Figure 2.97) along with shear wave velocity tests, resonant column tests, and triaxial stress path tests. The pressuremeter first load modulus averaged about 350 MPa below the foundation level.

1000

2000

3000

4000

5000

6000

7000

8000

9000 10000

0 –10 –20 –30 DEPTH (m)

2.17.2

YOUNG'S MODULUS (MPa) 0

–40 –50 –60 –70 –80 –90

Press Init AVE Press First AVE Press Second AVE Seismic (x 0.2) TRXL 0.01% TRLX 0.1% RC Initial RC 0.0001% RC 0.001% RC 0.01% adopted profile for drained modulus E’

Figure 2.97 Soil and rock modulus data. (Source: After Poulos and Bunce, 2008.)

62

2 CASE HISTORIES

tests. The behavior of the combined pile raft was analyzed by numerical simulations (Figure 2.99) and indicated that the maximum working loads were 37 MN and located at the end of the three wings. The working loads below the center of the foundation were 12.5 MN. The settlement was predicted to be 77 mm and was measured as 43 mm at the end construction (Figure 2.100). Interestingly the elastic equation can be used to calculate the settlement of an equivalent circular mat foundation without piles:

2.17.3 The Foundation Dimensions, Loading, and Settlement The weight of the structure is estimated to be 4413 MN. The base of the structure is in the form of a three-wing star (Figure 2.98) with an area of 3400 m2 , which leads to an average equivalent pressure on the rock of 1298 kPa. The foundation of the structure is at the base of a 15 m-deep excavation where a 3.7 m- thick mat was poured. Below the mat are 194 bored piles 0.9 to 1.5 m in diameter and 35 to 55 m in length. The ultimate capacity of the larger piles was calculated to be 60 MN and verified by 7 full-scale pile load

s = I(1 − 𝜈 2 )

pB E

Figure 2.98 Mat and pile foundation. (Source: Left, Courtesy of Missouri University of Science and Technology; right, courtesy of N. Tagge, https://commons.wikimedia.org/wiki/File:Burj_Khalifa_foundation_ structure.gif.) Legend 50

Tower raft foundation outline

40

30

20

10

0

–10

–20 Pile Location

–30 –50

–40

–30

–20

–10

0

10

20

30

40

37000 36000 35000 34000 33000 32000 31000 30000 29000 28000 27000 26000 25000 24000 23000 22000 21000 20000 19000 18000 17000 16000 15000 14000 13000 12000 11000 10000

50

Figure 2.99 Pile locations and working loads in kN. (Source: Poulos and Bunce, 2008, courtesy of Missouri University of Science and Technology.)

(2.82)

2.18 NEW ORLEANS LEVEES AND HURRICANE KATRINA (OVERTOPPING EROSION)

63

Distance along wing cross-section (m) 0 0

10

20

30

40

50

60

70

80

–10 27-Jun-06 16-Jul-06 16-Aug-06 18-Sep-06 16-Oct-06 14-Nov-06 19-Dec-06 16-Jan-07 19-Feb-07 18-Mar-07 Design

–20

Settlement (mm)

–30 –40

End of construction Oct-2009

–50 –60 –70 –80 –90

Figure 2.100 Measured and predicted settlement. (Source: Courtesy of H. Poulos and G. Bunce.)

The influence factor is for the center of a circular flexible foundation (I = 1.0), the long-term Poisson’s ratio is taken as 0.35, the average pressure is 1298 kPa, the equivalent diameter is 65.8 m = (4 × 3400/π)0.5 and the pressuremeter modulus is 350,000 kPa. 1298 × 65.8 s = 0.79(1 − 0.352 ) = 169 mm (2.83) 350,000 This demonstrates the value of the piles, which decreased the settlement by a significant factor. 2.18 NEW ORLEANS LEVEES AND HURRICANE KATRINA (OVERTOPPING EROSION) 2.18.1

The Story

On August 29, 2005, Hurricane Katrina (Figure 2.101) made landfall near New Orleans with 200 km/h winds and devastated the city. Hurricanes are huge rotating masses of rain and wind. The high winds on the outside edge of the rotating mass push the offshore waters in front of them and generate a storm surge. The storm surge is a very long wave that can, for several hours, overtop levees and other infrastructures upon arrival on the shoreline. Hurricane Katrina was about 400 km in diameter and traveled at about 40 km/h; therefore, the time spent by the hurricane in one place is 10 hours. The associated storm surge reached 8.5 m in height along the coast, overtopping levees which are typically 5 m high. Residents who did not evacuate faced serious flooding as some places in New Orleans are 6 m below the top of the levees; about 1,500 people died and the economic losses rose to about $120 billion.

Figure 2.101 Hurricane Katrina makes landfall on 29 August 2005. (Source: Courtesy of NASA Goddard Space Flight Center Scientific Visualization Studio, as quoted on https://64parishes.org/entry/ hurricane-katrina.)

This was one of the worst natural disasters in US history. This case history discusses levee overtopping and soil resistance to such erosion (Briaud, 2008). 2.18.2

The Soils and the Levees

The levees in New Orleans were built over the last 300 years. It started as farmers pushed soil against the banks of the Mississippi River to keep their farmland and cattle from

64

2 CASE HISTORIES

flooding. In 1802, the US Army Corps of Engineers (USACE) was created but it was not until 1927 that the U.S. Congress gave the USACE the responsibility of designing and constructing levees throughout the Mississippi River Valley. Today there are about 300 km of levees around New Orleans with about 130 km of flood walls on top of them. The soil type inside the levees is extremely variable and only known for the recent levees. In New Orleans, the levee cross-sections vary but a reasonable average may be a 5 m-high levee with 5 to 1 slopes on either side. The natural soil in New Orleans is made of recent normally consolidated clays and sands deposited by the Mississippi River and nearby lakes. Figure 2.102 shows a cross-section stratigraphy of New Orleans from the Mississippi River to Lake Pontchartrain. As can be seen, most of New Orleans is below the water level. 2.18.3

A team from Texas A&M University collected Shelby tube samples from the levee surfaces at many locations around New Orleans (Figure 2.104). During the visit, there was overwhelming evidence of levee overtopping as some ships were found perched on top of the levees and floodwalls after the storm receded. The samples were brought back to the laboratory and were tested for erodibility in the Erosion Function Apparatus as well as for index properties. The soil type varied widely from loose uniform sand to high plasticity stiff clay. The EFA results are shown in Figure 2.105. The first observation is that the erodibility varies widely. The erosion functions from the samples associated with eroded levees and non-eroded levees were identified. They banded into two very distinct groups (Figure 2.106) and this made it possible to prepare a levee overtopping design chart (Figure 2.107). Post Katrina, the USACE was given funding to build a storm surge barrier near the confluence of the Gulf Intracoastal Waterway (GIWW) and the Mississippi River Gulf Outlet (MRGO) to improve the protection of New Orleans (Figure 2.108).

Erosion of the Overtopped Levees

The storm surge overtopped some of the levees around New Orleans. Some of the overtopped levees held and some of them were eroded to various degrees (Figure 2.103).

Human-Made Levee

20 10 0 –10 –20

Gentilly Ridge

Natural Levee Deposits French Quarter

Mississippi River

Elevation (Feet)

30

St. Claude Ave. I-10

Gentilly Blvd. I-610

Gentilly Mirabeau Ave.

University of New Orleans Flood Wall

Leon C. Simon Drive

Artificial Fill Lake Ponchartrain

Sea Level Marsh/Swamp (Organic-Rich Clays)

Lake Deposits Pine Island Barrier/Beach Sands

Filled Distributary Channels Point Bar/Splay/Deltaic Deposits

Figure 2.102 New Orleans cross-section between the Mississippi River and Lake Pontchartrain. (Source: Courtesy of Steve Nelson, Tulane University: www.tulane.edu/~sanelson/Katrina/katrina_images.htm.)

Figure 2.103 Overtopped levees. (Source: Courtesy of ASCE.)

2.18 NEW ORLEANS LEVEES AND HURRICANE KATRINA (OVERTOPPING EROSION)

65

Figure 2.104 Sample locations. (Source: Adapted from USGS Map.)

100000

100000 10000

Very High Erodibility I

High Erodibility II

Medium Erodibility III

1000

10000

Medium Erodibility III Low Erodibility IV

100 Very Low Erodibility V

10 1 0.1 0.1

High Erodibility II

1000

Low Erodibility IV

EROSION 100 RATE (mm/hr)

Very High Erodibility I

Non-Erosive VI 1.0

10.0

100.0

VELOCITY (m/s)

Figure 2.105 Erosion test results. (Source: Courtesy of ASCE, Briaud, 2008.)

EROSION RATE 10 (mm/hr)

Very Low Erodibility V

1 0.1 0.1

Non-Erosive VI

1.0

10.0

100.0

VELOCITY (m/s) Legend: Levee breaches No levee damage

Figure 2.106 Erosion test results for levees with breach and levees with no damage. (Source: Courtesy of ASCE, Briaud, 2008.)

66

2 CASE HISTORIES 100000 10000

Very High Erodibility I

1000

High Erodibility II

Failure 2 days

dam but not the highest dam. The dam provides a remarkable amount of electricity but also flooded cities and created landslides. The dam, its foundation and its environmental impact are discussed in this case history (Liu et al., 2003a, 2003b).

Medium Erodibility III Low Erodibility IV

100 EROSION RATE 10 (mm/hr)

Very Low Erodibility V

Failure 2 hrs

Resist 2 days

1

Resist 2 hrs

0.1 0.1

1

10

Non-Erosive VI

100

VELOCITY (m/s)

Figure 2.107 Erosion chart. (Source: After Bernhardt et al., 2011.)

2.19.2

The Three Gorges Dam is a concrete gravity dam. It is 2335 m long and 175 m high and required 28 Mm3 of concrete to build. Construction started in 1993, in 1997, the river was diverted, in 2006, the dam body was completed, in 2008, the reservoir reached capacity, in 2012, all 32 turbines were operational and the last component of the project, the ship lift, was completed in 2015. The project cost was $32 billion, the dam generates 22,500 Megawatts and consists of five parts all made of concrete (Figures 2.110 and 2.111): the right power plant, the spillway dam in the center of the river, the left power plant, the ship lift, and the permanent ship lock. The reservoir is about 660 km long and averages 1.1 km in width. Over 1 million residents were relocated as a result of the dam construction, 60,000 workers were employed by the project with 25,000 working on the dam itself. 2.19.3

Figure 2.108 New storm surge barrier across Lake Borgne. (Source: US Army Corps of Engineers.)

2.19 2.19.1

THREE GORGES DAM (CONCRETE DAM) The Story

The Three Gorges Dam in China is located on the Yangtze River, the longest river in Asia, near the city of Taipingxi about 1300 km upstream of Shanghai, where the river flows into the East China Sea. It is called the Three Gorges Dam because over the 400 km upstream of the dam, the river goes through three distinct gorges (Figure 2.109). This hydroelectric dam also aims to enhance the navigation on the Yangtze River and to control flooding. The dam construction started in 1993 and was completed in 2006. It is the world’s largest

The Dam Dimensions and Construction

Soil and Rock Conditions

The Three Gorges Dam is built on plagioclase granite, an igneous rock in various stages of weathering. The site investigation was extensive and included many different types of geotechnical, geophysical, hydrogeological, and rock mechanics tests to capture the complexity of the rock mass and allow for a comprehensive dam design. The rock mass can be divided into four groups of weathering zones with depth; the completely weathered zone (Zone IV) is the shallowest, the highly weathered zone (Zone III) is next, followed by the moderately weathered zone (Zone II) on top of the slightly weathered zone (Zone I). The bottom of the dam foundation is located near the top of the slightly weathered Zone I. That elevation varies from about 40 m above sea level close to the riverbanks to about 18 m in the flood plain and to about 11 m in the main channel. The dam cross-section through the spillway at the center of the dam is shown in Figure 2.112, the height is 175 m and the width at the foundation level on top of the rock is 133 m.

Qutang Gorge

Wu Gorge

Xiling Gorge

(a)

(b)

(c)

Figure 2.109 The three gorges upstream of the Three Gorges Dam. (Source: (a) Tan Wei Liang Byorn/ Wikimedia Commons; (b) Photnart/Wikimedia Commons; (c) Fredlyfish4/Wikimedia Commons.)

67

2.19 THREE GORGES DAM (CONCRETE DAM)

N ane nt

ship l

ock

Sh ip

lift

Tem

po

rar y

shi

plo

ck

Ri gh tp ow er pl an t

Sp illw ay

da m

Le ft po we rp la nt

Per m

e Yangtz

River

Scale

1 km

Figure 2.110 Plan view of the Three Gorges Dam. (Source: Freer, 2001, courtesy of ICE.)

185.0 m 180.4 175

145 120 m

Figure 2.111 The Three Gorges Dam. (Source: Freer, 2001, courtesy of ICE, http://www.simonfoucher.com/McGill/MIME221 %20Ethics/notes/3gordesdam1.pdf.)

94 m 90 m 83.1 m

The faults and the joints in the rock mass were carefully mapped and studied. Recall that a fault has experienced movement in its past while a joint has not. Both were found to have mostly steep dip angles which was beneficial to the sliding stability of the dam. The slightly weathered Zone I had the following mechanical properties: unit weight 27 kN/m3 , compressive strength 85 to 100 MPa, friction angle 52 to 57o , cohesion 1.4 to 1.8 MPa, the Poisson’s ratio 0.22, the modulus of deformation 20 to 40 GPa. However, the mechanical parameters of the joints provided reduced friction resistance with a friction angle of 31 to 45o depending on the roughness and a cohesion of 0.13 to 0.6 MPa. 2.19.4

62 m 56-57 m

11 m

Environmental Impact

A huge dam such as the Three Gorges Dam has many impacts on the surrounding environment. On the positive side, they include generation of a lot of electricity, flood protection,

Figure 2.112 Cross-section through the spillway. (Source: Freer, 2001, courtesy of ICE.)

68

2 CASE HISTORIES

Figure 2.113 Example of landslide near Three Gorges Dam. (Source: Courtesy of the Asia-Pacific Journal / Japan Focus, Bosshard, 2009, https://apjjf.org/-Peter-Bosshard/3262/article .html.)

120

112

170

100

160

80

150

60

51 36

140 1114 2

36 17 16 9 6

40

new landslide water level 1 1

4 3 3

3 3 3 1 2 1 1

1 2 1

2

1 1

4

2 1

2

2

3

1 1

2

20

2 1 2

120

Date (yy/mm/dd)

Figure 2.114 Water level and number of landslides per month at Three Gorges Dam. (Source: Courtesy of the Journal of Rock Mechanics and Geotechnical Engineering, Yin et al., 2016, https://www.researchgate .net/publication/307998774_Reservoir-induced_landslides_and_risk_control_in_Three_Gorges_Project_ on_Yangtze_River_China.)

2014/07/01

2014/05/01

2014/03/01

2014/01/01

2013/11/01

2013/09/01

2013/07/01

2013/05/01

2013/03/01

2013/01/01

2012/11/01

2012/09/01

2012/07/01

2012/05/01

2012/03/01

2012/01/01

2011/11/01

2011/09/01

2011/07/01

2011/05/01

2011/03/01

2011/01/01

2010/11/01

2010/09/01

2010/07/01

2010/05/01

2010/03/01

2010/01/01

2009/11/01

2009/09/01

2009/07/01

2009/05/01

2009/03/01

2009/01/01

2008/11/01

0 2014/09/01

130

2008/09/01

Water level ASL (m)

180

Number of landslide events per month

irrigation of farmlands, improved navigability of the river, creating jobs. On the other side, it includes flooding of many cities and archeologic sites, floating wastes during filling of the reservoir, disturbance of wildlife, sedimentation at the bottom of the reservoir, and creating many landslides (Figure 2.113) (Bosshard, 2009). Slope failures have increased since the dam was constructed, and the reservoir filled (Yin et al., 2016). The filling started in 2003 when the water rose from elevation 75 m to elevation 135 m. In 2006, the water level reached 156 m and, in 2008, the water level reached its design level of 175 m. Associated with the management of the reservoir level is a required 30 m yearly water level fluctuation. Indeed, the water level is decreased by 30 m in the summer to control winter flooding. This fluctuation induces cyclic stresses in the surrounding rock and soil which lead to landslides and potential increase in earthquake frequency. The landslides are due to a loss of water tension and associated decrease in shear strength upon first filling and later cyclic strength degradation during the yearly fluctuations (Figures 2.113 and 2.114). As can be seen the number of landslides increased dramatically during the first filling peaking at 112 per month and decreased significantly after a year of activity. This cyclic variation of stresses in the rock and soil mass may also exercise the two faults on which the dam sits and lead to earthquakes (Figure 2.115). Sedimentation of the reservoir is also a problem to be considered. Indeed, the dam slows the river flow dramatically and 340,000 MN of sediments arrive in the reservoir annually, thus decreasing the reservoir capacity and its ability to regulate flooding. This decrease in capacity can reach 1% per year for some dam reservoirs. This accumulation also starves the river downstream of the dam which impacts its geomorphology.

2.19 THREE GORGES DAM (CONCRETE DAM)

69

Figure 2.115 Number of earthquakes at the Three Gorges Dam location. (Source: Courtesy of Elsevier, Tang et al., 2019, https:// www.sciencedirect.com/science/article/pii/S0013795219285789.)

2.19.5

Simple Calculations

A project like the Three Gorges Dam requires an enormous number of very complex engineering considerations and calculations to greatly minimize the probability of failure. The following are a couple of very simple calculations. A factor of safety must be established against sliding of the dam. This requires the selection of a sliding plane below the dam together with the shear strength properties on the sliding plane. Let’s select the base of the dam as the sliding plane and the rock shear strength as the sliding plane properties. The maximum resistance to sliding is R = cAc + (W − uAu ) tan 𝜑

(2.84)

where R is the maximum resistance to sliding per m of dam width, c is the soil or rock cohesion (1400 kPa from Section 2.19.3), Ac is the area per meter of dam width over which the cohesion is applied (taken as the area at the bottom of the dam: 133 × 1 m2 from Section 2.19.3), W is the weight of the dam per meter of dam width (simplified cross-section area from Figure 2.112 × 1 m = 17268 m2 times the unit weight of concrete at 25 kN/m3 ), u is the water pressure under the dam (estimated at 770 kPa under the front of the dam from Figure 2.116), Au is the area per meter of dam width over which the water pressure is applied (estimated at 6 m2 from Figure 2.116), and 𝜑 is the rock friction angle (52o from Section 2.19.3). Then if the failure sliding plane is in the rock, the resistance is Rrock : Rrock = 1400 × 133 + (17268 × 25 − 770 × 6) tan 52 = 186200 + 427080 × 1.28 = 723862 kN

(2.85)

If the dam base is assumed to slide on a joint, then the cohesion and friction angle are reduced to c = 130 kPa and 𝜑 = 31o from Section 2.19.3. Thus, if the failure sliding plane is along a joint at the base of the dam, the resistance is Rjoint : Rjoint = 130 × 133 + (17268 × 25 − 770 × 6) tan 31 = 17290 + 427080 × 0.60 = 273965 kN

(2.86)

Note that the area of the sliding plane is likely longer than the horizontal plane assumed for those calculations because

Figure 2.116 Water level and water pressure below the Three Gorges Dam. (Source: Courtesy of the Journal of Rock Mechanics and Geotechnical Engineering, Fan et al., 2011, https://www .researchgate.net/publication/228823225_Key_issues_in_rock_ mechanics_of_the_Three_Gorges_Project_in_China.)

the joint plane is unlikely to be perfectly horizontal. The driving force, on the other hand, is the horizontal push F water generated by the water behind the dam minus the one in front of the dam expressed as (see Figure 2.112): Fwater =

) 1 ( 2 𝛾w H1 − H2 2 2

(2.87)

where 𝛾 w is the unit weight of water (9.81 kN/m3 ) and H 1 and H 2 are the water height on the upstream side and downstream side of the dam respectively (169 m and 62 m from Figure 2.112). Then Fwater =

1 × 9.81 × (1692 − 622 ) = 121237 kN 2

(2.88)

The factor of safety (FoS) against sliding can be calculated as R FoS = (2.89) Fwater which gives FoS = 5.97 for the sliding plane through the rock and FoS = 2.26 for the sliding plane through the joint. Note that the shear strength parameters selected for the sliding plane were the lower bound of the range and thus so are the factors of safety. These factors of safety are acceptable but only address one aspect of the design.

70

2 CASE HISTORIES

The settlement of the dam can also be calculated using very simple means. The elastic equation for settlement is: pB (2.90) E where s is the settlement, I s , I e , and I h are the factors for the influence of the foundation shape, the relative depth of embedment and the depth to the hard layer respectively, 𝜈 is the Poisson’s ratio, p is the pressure applied, B is the width of the foundation and E is the rock mass modulus. The foundation shape is a large rectangle with the length L of the dam (2335m) and the width B of the dam (133m) for an aspect ratio L/B = 17.56. For this ratio, I s is equal to 2.5, I e is equal to 1 because the dam is assumed to sit on top of the rock surface, and I h is taken as 1 as it is conservatively assumed that the rock is at least as deep as the depth of influence of the dam. The Poisson’s ratio is taken as 0.22 and the modulus as 30 GPa (average from Section 2.19.3). The average pressure p under the dam is taken as the weight of the dam (17268 × 25 = 431700 kN/m) divided by the dam foundation area (133 m2 /m). Then s = Is Ie Ih (1 − 𝜈 2 )

431700 133 2

s = 2.5 × 1 × 1(1 − 0.22 )

× 133 = 0.034 m (2.91)

30000000 According to those calculations, the settlement of the dam would be 34 mm.

2.20 THE KANSAI INTERNATIONAL AIRPORT (EARTH FILL IN THE SEA) 2.20.1

The Story

The Kansai airport was built into the sea (Figure 2.117) because Japan has a limited amount of land available. Kansai is a region of Japan that includes the cities of Osaka, Kyoto, and Kobe. The water depth of the Osaka Bay at the location of the airport islands is about 20 m and the soils are soft marine sediments. The construction of two very large human-made islands 4 km into the Osaka Bay brought about the issues

Figure 2.117 Kansai region, Osaka Bay and Kansai airport in Osaka Bay. (Source: Ankou1192/Wikimedia Commons.)

of settlement and bearing capacity of high fills on very soft marine sediments. 2.20.2

Dimensions

The Kansai airport was built in two phases. Phase I consisted of building Island I, 4 km away from the shore; Island I would house Terminal I and a 3500 m-long runway. The construction of Phase I started in 1987 and was completed in 1994. Island I is 4.5 km long, 1.3 km wide, and required 180 Mm3 of fill. The original plan was to place a 30 m-thick fill but it ended up being 37 m thick to accommodate for the settlement and the 4 m required free board. Phase II consisted of building Island II parallel to Island I further out into the bay. Island II would house Terminal II and a 4000 m-long runway. The construction of Phase II started in 1999 and was completed in 2007. Island II is 4.9 km long, 1.3 km wide at the north-east end, 0.65 km wide at the south-west end, and required 250 Mm3 of fill. The final thickness of the fill is 43 m. 2.20.3

Construction

The first construction step was to place vertical sand drains at the bottom of the bay to minimize the amount of time required for the soil consolidation process to take place. Each sand drain was a 0.4 m diameter column of sand which averaged 20 m in length. Over 2 million sand drains were installed to speed up the drainage process and therefore the settlement of the bay soil. Then the sea wall was constructed along the perimeter of the islands (Figure 2.118). Once the sea wall was constructed, the area enclosed inside the sea wall was backfilled with soil that came from several sources including leveling a few mountains. The total cost including both phases reached over $20 billion, 1 million workers contributed to the project, and 80 ships were mobilized. 2.20.4

Soil Conditions

The depth to bedrock at the location of the Kansai airport is estimated to be about 1.2 km (Kagawa et al., 2004). The top 400 m of the soil deposits are quaternary deposits with, at the surface, a 20 m-thick 10,000-year-old Holocene soft clay layer. The islands have a width of the order of 1 km and a length to width ratio of about 4. Such a rectangular area would lead to a zone of influence (depth where the stress increase is down to 10% of the surface applied stress) of 3.5 times the width (see Chapter 18). Such simple calculations indicate that the zone being compressed is extremely deep. At the same time the stiffness of the sediments increases with depth because the vertical effective stress increases naturally with depth and because of the geologic history. Therefore, the depth within which most of the settlement takes place is much smaller than 3.5 km, is limited to the depth of bedrock of about 1.2 km and likely concentrated within the top 200 m or so. A simplified soil stratigraphy is presented in Figure 2.119. The top clay layer also called Ma13 is the softest layer and contributed the most to the settlement of the islands. The

2.20 THE KANSAI INTERNATIONAL AIRPORT (EARTH FILL IN THE SEA) Cross-section Diagram of Slow Grade Sea Wall of Stonework

Upper mound Lower mound Sand bed (behind the upper mound 2) Sand bed (behind the lower mound 1) Sand bed (after consolidation)

Upper concrete Upper blocks Wave-dissipating blocks Cladding stones

Sea level

71

Mound Artificial island

Surface of the seabed

Alluvial clay layer

Sand drain-installed ground Paving sand

Pleistocene layer

Figure 2.118 Cross-section of the island and construction sequence. (Source: Courtesy of Kansai Airports, http://www.kiac.co.jp/en/tech/safety/natural/natural_b/index.html.)

Sea level

2nd-Phase Zone

1st-Phase Zone

Depth: –20 – 18 m (Average: 19.5 m)

Depth: –19 – 16.5 m (Average: 18 m)

Unconfined compressive strength, qu(kPa) –20

Seabed Holocene clay layer (Hardness roughly equivalent to that of toy clay)

Upper Pleistocene (Hard enough that the clay is slightly dented if pressed)

29 ~ 20 m Average: 24 m

210 ~ 140 m Average: 180 m

0

100

200

300

before construction (1982) before lower rubble (2001) before upper rubble (2001) before reclamation (2003)

21 ~ 15 m Average: 18 m

160 ~ 100 m Average: 140 m

–25 About 200 m

Lower Pleistocene (Hard enough that the clay is barely dented if pressed) About 800 m (900 m or deeper) Tertiary (Hard as concrete)

Granite (Very Hard rock)

Figure 2.119 Soil stratigraphy. (Source: Courtesy of Springer Netherlands, Puzrin et al., 2010.)

undrained shear strength of that normally consolidated layer increases linearly with depth and averaged 25 kPa initially but increased significantly as the island fill was constructed (Figure 2.120). Other average properties for Ma13 include (Funk, 2013): water content w = 86%, unit weight 𝛾 = 15 kN/m3 , initial void ratio eo = 2.3, compression index Cc = 2, normally to lightly overconsolidated, cv = 2.8 m2 /yr. It is underlain by a layer of sand, which would allow the Ma13 layer to drain. The average effective unit weight of the fill was 19.6 kN/m3 above water and 11.8 kN/m3 below water. The unit weight of the sea water is taken as 10 kN/m3 . 2.20.5

Loading and Settlement

The measured settlement at various locations of Islands I and II is shown in Figure 2.121 (Furudoi, 2010; Mesri and Funk,

Elevation (mCDL)

500 ~ 300 m

–30

–35

–40

Figure 2.120 Unconfined compression strength. (Source: After Furudoi, 2010.)

2015). In both cases the loading consisted of back-filling slowly for the first half of the loading and more rapidly afterwards while monitoring the pore pressures. This is the reason for the double shape of the settlement vs. time curves. The settlement of Island I was an average of 12 m in 2010 with 15 m being reached in some places. A bit more settlement is

72

2 CASE HISTORIES

Time

Settlement (m)

1986 0

1990

1994

1998

2002

2006

2010

2014

2018

2022 0

2

2

4

4

6

6

8

8

10

10

12

12

14

14

16

16

18

1st Island Mean Settlement

18

2nd Island Mean Settlement 20 1986

1990

1994

1998

2002

2006

2010

2014

2018

20 2022

Figure 2.121 Settlement of Kansai airport Iislands I and II. (Source: After Furudoi, 2010, data source http:// www.kansai-airports.co.jp/efforts/our-tech/kix/sink/sink3/sink3_c.html.)

likely judging from the shape of the curve. The settlement rate fell from 0.5 m/yr during 1994 to 0.07 m/yr in 2008. The settlement of Island II was an average of 14 m in 2009 and extrapolation of the settlement vs. time curve indicates that a final settlement of 20 m is not unlikely. The loading of the seabed by the islands’ fill created a significant pressure at the bottom of the bay. This pressure is calculated as follows for Island I and Island II. In the case of Island I, the final dimensions were 37 m of fill in 18 m of water with an estimated final settlement of 15 m. Therefore, if the fill had been placed instantly (before settlement), the effective stress increase on the seabed would have been Δ𝜎 ′ instant = (37-18) × 19.6 + 18 × 11.8 = 584.8 kPa (2.92) However, if a final settlement of 15 m is considered, 15 more meters of fill end up below the sea level and the effective stress increase on the bottom of the bay is Δ𝜎 ′ long term = (37-18-15) × 19.6 + (18 + 15) × 11.8 = 467.8 kPa

(2.93)

In the case of Island II, the final dimensions were 43 m of fill in 20 m of water with an estimated final settlement of 20 m. Therefore, if the fill had been placed instantly (before settlement), the effective stress increase on the seabed would have been Δ𝜎 ′ instant = (43 − 20) × 19.6 + 20 × 11.8 = 686.8 kPa (2.94)

However, if a final settlement of 20 m is considered, 20 more meters of fill end up below the sea level and the effective stress increase on the bottom of the bay is Δ𝜎 ′ long term = (43 − 20 − 20) × 19.6 + (20 + 20) × 11.8 = 530.8 kPa 2.20.6

(2.95)

Simple Calculations

Simple global calculations can be undertaken as follows. First, without sand drains, the time required for 90% primary consolidation can be estimated by considering the surface soft layer Ma13. That layer is 20 m thick with the fill above providing drainage and a sand layer below also providing drainage. Considering this two-way drainage, a drainage length of 10 m is used. The average coefficient of consolidation cv for that clay is given as 2.8 m2 /yr, therefore the time for 90% primary consolidation is: t90 = T90 H 2 ∕cv = 0.85 × 102 ∕2.8 = 30 years

(2.96)

The sand drains were 0.4 m in diameter and installed on a 2.5 m spacing in a square pattern. The average horizontal coefficient of consolidation ch was given as 4.4 m2 /yr (Funk, 2013). The Barron-Hansbo equation (Hansbo, 1981; see Chapter 27) can be used to estimate the time required for 90% primary consolidation with the sand drains. ( ( ) ) ( ) d2 dw 1 t = w Ln − 0.75 + Fs Ln (2.97) 8ch de 1 − Uh

2.21 THE PANAMA CANAL (EXCAVATED SLOPES)

where dw is the sand drain influence diameter taken as 1.15 times the spacing for a square arrangement: dw = 1.15 × 2.5 = 2.875 m, de is the sand drain diameter: de = 0.4 m, F s is a coefficient varying from 0 to 2 to take into consideration the effect of a potential smearing zone, it is taken equal to 1 for an average influence, and U h is the percent consolidation taken as 0.9 for 90% consolidation. Then, ( ( ) ) ( ) 2.8752 2.875 1 Ln − 0.75 + 1 Ln t90 = 8 × 4.4 0.4 1 − 0.9 = 1.20 yr (2.98) Therefore, and according to those calculations, the time required for 90% primary consolidation was decreased by a factor of 25 (30/1.2) by installing the sand drains. The top clay layer called Ma13 is 20 m thick and its compression under the load from the fill contributed significantly to the settlement of the islands. The compression of Ma13 can be estimated as follows for Island I and Island II. The consolidation settlement magnitude equation is ( ′ ) 𝜎ov + Δ𝜎v′ Cc s = Ho log (2.99) ′ 1 + eo 𝜎ov For Island I, the thickness H o of the layer is 20 m, the clay layer average properties as mentioned earlier are Cc = 2 and eo = 2.3. Since the unit weight of the soft clay layer is 15 kN/m3 , the effective stress in the center of the layer (10 m below the seabed) prior to loading is 10 × (15 – 10) = 50 kPa, and the increase in stress at the surface is taken as the average of 584.8 kPa and 467.8 kPa, as calculated in Section 2.20.5 (0.5 (584.8+467.8) = 526.3 kPa). Since the depth is very shallow compared to the width of the island, the increase in stress at the surface is considered to also exist at the center of the compressing layer. The settlement corresponding to 90% primary consolidation of the 20 m thick soft clay surface layer is: ) ( 2 50 + 526.3 = 12.87 m (2.100) × log s = 20 × 1 + 2.3 50 Since the total settlement at the seabed is likely to reach 15 m, this leaves about 2.1 m of settlement contributed by the lower layers and by the secondary consolidation of the Ma13 layer. For Island II, the compression of Ma13 can also be estimated using the same soil properties but the loading is now the average of 686.8 kPa and 530.8 kPa, as calculated in Section 2.20.5 (0.5 (686.8 + 530.8) = 608.8 kPa). Since the depth is very shallow compared to the width of the island, the increase in stress at the surface is considered to also exist at the center of the compressing layer. The settlement corresponding to 90% primary consolidation of the 20 m thick soft clay surface layer for Island II is: ) ( 2 50 + 608.8 = 13.57 m (2.101) × log s = 20 × 1 + 2.3 50 Since the total settlement at the seabed is likely to reach 20 m, this leaves about 6.4 m of settlement contributed by the

73

lower layers and by the secondary consolidation of the Ma13 layer. Placing a high fill on such a soft soil (su = 25 kPa) brought about the issue of bearing capacity associated with edge failure. The bearing capacity of the soil can be calculated as pu = Nc su + 𝛾D = 5.2 × 25 = 130 kPa

(2.102)

This is much smaller than the pressures applied on the seabed which exceeds 500 kPa. This was mitigated by using shallow slopes at the base of the seawall and giving some time for the soil to increase in strength as the seawall and fill were placed on top of the sand drains (Figure 2.118). The undrained strength increased from 25 kPa to 100 kPa, which would give a bearing capacity somewhat higher than 500 kPa.

2.21 2.21.1

THE PANAMA CANAL (EXCAVATED SLOPES) The Story

There are two major human-made isthmus crossings on our planet: the Suez Canal and the Panama Canal. Both save a tremendous amount of time and money to the shipping industry by providing much shorter routes. For example, the Panama Canal saves about 15,000 km for a ship that would have to go around South America. In 1880, France undertook to build the canal and Ferdinand de Lesseps, the builder of the Suez Canal, was put in charge. His idea was to build a sea-level canal, but he encountered many problems including very deep unstable excavations and diseases. France abandoned the project in 1888. The US took over the project in 1903 and planned for a canal that would have locks to elevate the ships above ocean level, thus minimizing the depth of excavation. Another very important part of the solution was to eradicate diseases including malaria and yellow fever brought about by mosquitoes. The Panama Canal opened in 1914. In 2007, the Panama Canal Expansion project added a new, wider, and deeper lane for larger ships with one lock on the Pacific side and one lock on the Atlantic side. The expansion project was opened in 2016. The main geotechnical problem discussed here is the stability of the excavated slopes forming the banks of the canal. 2.21.2

Canal Dimensions and Cross-Section

A plan view and a longitudinal cross section of the Panama Canal are shown in Figure 2.122. From the shores of the Pacific Ocean to the shores of the Atlantic Ocean, the canal is 65 km long. The width of the canal varies significantly, and the minimum water depth is about 12.2 m. The 1914 locks are 305 m long, 33.5 m wide, with a water depth of 12.8 m. When going from the Pacific Ocean to the Atlantic Ocean, the ship first encounters the Miraflores Locks, which lifts the ship 16.5 m above sea level in two steps associated with two lock chambers. The ship then sails inland through the small Miraflores Lake to the Pedro Miguel Lock, which lifts the

74

2 CASE HISTORIES

Figure 2.122 Panama Canal plan view. (Source: Thomas Römer/Wikimedia Commons/CC BY-SA 2.0.)

ship another 9.5 m to the level of Gatun Lake. The first part of the canal through Gatun Lake had to be excavated through clay shale mountains to provide for a free passage for the ships. This is the Gaillard Cut, later renamed the Culebra Cut, which created most of the geotechnical slope stability problems. Once past the Gaillard Cut, the ship sails over to the other side of the Gatun Lake. This lake was created by building two dams on the Chagres River, which crosses Panama: the Madden Dam upstream and the Gatun Dam downstream. On the Atlantic side of the Gatun Lake, the ship goes through the Gatun Locks, which lower the ship 26 m in three steps down to the level of the Atlantic Ocean. The 2016 locks are 427 m long, 55 m wide, with a water depth of 18.3 m. Larger ships, called NeoPanamax (Figure 2.123), coming from the Pacific Ocean, first arrive at the Cocoli Locks, which lift the ships 26 m in three steps to the Gatun Lake level. After sailing through the Gatun Lake, the ships come to the Agua Clara Locks, which lower the ships 26 m in three steps to the Atlantic Ocean level.

Figure 2.123 A NeoPanamax ship passing through the Agua Clara lock. (Source: Mario Roberto Duran Ortiz/Wikipedia Commons/CC BY-SA 4.0.)

2.21 THE PANAMA CANAL (EXCAVATED SLOPES)

75

Figure 2.124 Excavation during the French work and later during the American work. (Source: Courtesy of Fred Henstridge, http://henstridgephotography.com/FHP%20Volume%203%20Issue%209/.)

The Gaillard/Culebra Cut

The Gaillard Cut, later renamed the Culebra Cut, was necessary to create a passage for the canal through the mountain range called the Continental Divide, which rises about 102 m above sea level at the Gaillard Cut zone. For the sea-level solution, the depth of excavation would have been about 115 m deep (102 m + 13 m water depth). The locks raised the ship 26 m above sea level, so the locks decreased the excavation depth to 89 m. The French effort between 1880 and 1888 excavated about 38 m of the mountain or 14.5 million m3 (Figure 2.124). The American effort between 1903 and 1914 excavated the additional 51 m or 73.4 million m3 , 20% of which was the removal of slope failures that occurred during construction. One of the main slope failures site within the Gaillard Cut stretch is the Culebra failures which occurred on the east side and on the west side of the canal.

2.21.4 Stratigraphy and Soil Properties of the Culebra Excavated Slopes Before excavation, the stratigraphy at the Culebra site can be simplified as 100 m of volcanic rock (basalt) underlain by 60 m of Cucaracha shale clay (Figure 2.125). The basalt is relatively strong, but the Cucaracha clay behaves like an overconsolidated clay with a high peak shear strength and post-peak softening to a much lower shear strength (Figure 2.126). The largest part of the failure surface developed in the Cucaracha clay layer (Figure 2.125). The index properties of that clay shale vary significantly (possibly +/-50%) with the following average values: water content 18%, liquid limit 86%, plasticity index 55%, dry density 17 kN/m3 . The shear strength

Figure 2.125 Cross-sections in 1913 and 2020 at the Culebra Cut. (Source: Tony Waltham 2020/John Wiley & Sons.) 1200

1000 SHEAR STRESS (kPa)

2.21.3

800 600

400

σ′ = 957.6 kPa

200 σ′ = 143.6 kPa 0 0

2

4

6

8

10

12

14

16

HORIZONTAL DISPLACEMENT (mm)

Figure 2.126 Examples of drained direct shear test results on samples of Cucaracha clay shale. (Source: After Banks et al., 1975.)

76

2 CASE HISTORIES 1200

SHEAR STRENGTH, s (kPa)

1000

800

UPPER LIMITS c′p = 478.80 kPa, φ′p = 28° for σ′n > 287.28 kPa

E

UR AIL (5)

EF

L ITT

c′p = 0, φ′p = 65° for σ′n < 287.28 kPa

BR

(7) RE

(5) 600

TIC AS PL

(9)

(10) (7)

(7)

ILU (10)

FA

(5)

400

LOWER LIMITS c′p = 47.88 kPa, φ′p = 28°

(9) (3) 200

(10)

(9) (3)

WMS-1 WMS-PIE and WMS-PIF WCSE-1 EXIBITED LARGE DROP IN SHEAR STRESS AFTER FAILURE SPECIMEN CONTAINED HARD PIECES OF MATERIAL IN SHEAR PLANE

(3) 200

400

600

800

1000

1200

1400

1600

EFFECTIVE NORMAL STRESS, σ′n (kPa)

Figure 2.127 Shear strength envelope of the Cucaracha clay shale. (Source: After Banks et al., 1975.)

parameters were obtained from drained direct shear tests performed by the US Army Corps of Engineers in the mid-1970s (Banks et al., 1975). The results varied significantly (Figure 2.126) with peak c’ values from 50 to 500 kPa and peak friction angles from 20o to 28o . The effective stress level has a significant impact on those values and the 1975 Corps of Engineers report gives two envelopes of shear strength, as shown in Figure 2.127. The residual shear strength was obtained by repeated drained direct shear tests with residual c’ value equal to 0 and effective stress residual friction angles varying from 3.6o to 10.8o and averaging 7.3o . The overburden pressure of samples tested in consolidation tests varied from 200 to 1100 kPa and the preconsolidation pressure was always higher that the overburden pressure varying from 1100 to 9800 kPa. One factor that contributes to the variability of the strength parameters is the erratic presence of slickensides in the clay shale. These small slick surfaces, likely due to a series of volume expansion and volume contraction over time or defoliation in these high plasticity clays, can significantly reduce the shear strength of the soil locally. Considering the complexity of the stability of the Culebra excavation slopes, the best way to evaluate the soil strength parameters is to back-calculate strength values from observed slope failures. Even at that, such failures probably involve softened values and average mobilized values over the length of the failure surface. Nevertheless, such back-calculated values have been accumulated by the Panama Canal Authority and serve as the basis for slope stability analysis for the canal. 2.21.5

Stability of the Culebra Excavated Slopes

The excavation angle of the sides of the Panama Canal at Culebra were originally set at 45o to 50o . This is in part because at the start of the project in 1880, much before the beginning of modern geotechnical engineering, the shale was

Figure 2.128 Plan view of slope failures at Culebra. (Source: Tony Waltham 2020/John Wiley & Sons.)

considered a strong soft rock, and thus would be able to stand at such a steep angle. In 1913, at the end of construction after repeated slope failures (Figures 2.128 and 2.129) and associated debris removal, the angle was down to about 20o . Today, after more failures, the angle is about 7o . This is a strong indication that the excavation gradually brought the clay shale to its residual friction angle through a series of decompressions and progressive failures. During this progressive failure process (see Chapter 20), the soil mass experiences shearing displacement and some zones of the failure plane reach strains beyond the peak strain corresponding to the peak shear strength and the soil starts losing its strength on its way to the residual value. To compensate for this strength

2.22 THE NICE AIRPORT SLOPE FAILURE (SLOPE STABILITY)

1911

1912

1914

1915

1923

1986

77

Figure 2.129 Some Panama slope failures over the last 100 years. (Source: Courtesy of J. David Rogers, 2011.)

loss, the soil reaches to zones of the failure plane where strains are in the pre-peak domain, which then go beyond that threshold due to the higher demand on their resistance. Gradually, the failure plane reaches its capacity. An important trigger of the landslides is the rainfall and particularly long and sustained periods of rain rather than short intense rainfall (De Puy and Alfaro, 2010). The approach adopted by the Panama Canal Authority engineers is to accept that slope movements will occur and that a cautious monitoring program can make a major difference. Such a Landslide Monitoring Program (LMP) was put in place over 40 years ago and consists of frequent monitoring of slope movements, rapid response when significant movements are detected, and excavation of debris as soon as they occur. The LMP has been very effective; indeed, while the Canal was closed due to slope failures 11 times between 1914 and 1986, it has not been closed once since the LMP has been in place (Duncan, 2010).

2.22 THE NICE AIRPORT SLOPE FAILURE (SLOPE STABILITY) 2.22.1

The Story

On October 16, 1979, at 13h 57m 30s, a major tsunami and slope failure occurred on the coast near the city of Nice, located in the South of France. The city was completing work on a new airport and a new harbor which were contiguous (Figure 2.130). As part of the project, a dike was being constructed into the Mediterranean Sea to protect the harbor and increase its capacity. When the dike slope failed and the tsunami occurred, many people died and 30 km of the coast of the Bay of Angels around Nice flooded up to 150 m inland. Several groups of experts were formed and gave their opinion on the geotechnical aspects of the accident. As you will see, the answer is still in doubt as several explanations have been formulated and none can be ascertained because of the limited precision of the data collected, including eyewitness accounts. One of the key elements is to know whether a

78

2 CASE HISTORIES Good Definitive Testimony

r Var Rive

PaillonRiver

Billardello Picard Schafter Pinotti

Bay of Angels rt

o irp

1355

A

Poggio Ardisson Venon Renaud 1400-1401

New runway rt

Po

Failed embankment

Figure 2.130 Nice airport and the new port.

submarine landslide occurred first and created the tsunami, which in turn induced the dike slope failure or whether the dike slope failure occurred first and created a larger submarine landslide, which in turn induced the tsunami. It is a bit like the problem of the chicken and the egg: which came first?

Normal Water Level

Projected Time Sequence 1355 Corroborative Witnesses Duclas Mac Dowell Burgeaud Guiliano Pitavy

2.22.2

The Sequence of Events

Because the dike slope failure occurred next to the airport, the staff in the airport control tower were able to record the time corresponding to the initiation of the dike slope failure with great precision. The recorded times were 13h 57m 30s when the first crane on the dike construction site disappeared and 14h 01m 30s when the last crane disappeared. The precision of these time observations is estimated at 15 seconds. This leads to a time of 4 to 5 minutes for the disappearance of the dike into the sea. The tsunami that occurred started by creating a depression of the sea level of about 2.5 m along the coast of the Bay of Angels (Figure 2.131). This initial depression was observed by several eyewitnesses and can possibly be placed at 13h55m, but the precision of the time at which the depression started is estimated to be of the order of minutes and not seconds like the control tower observations. This depression was followed by a series of waves with a period of about 8 minutes and a trough-to-crest amplitude of 3.4 m; these waves penetrated the land over some 150 m. The tsunami decayed over a 12-hour period and created loss of life and significant material damage. A submarine landslide, also called an underwater avalanche, took place and created a large mass of soil flowing very far into the Mediterranean Sea at an average of 7 km/hr. The evidence includes the rupture of two cables located 90 and 120 km away from the coast in the evening of October 16, 1979. 2.22.3

Simoncini Bonnenfont Lonergan

1359 7 min

1402

1406

1410

8 min

1414

1418

8 min

1422

1426

8 min

1400 1405 1405 1406 1407 1415 1423 1423

Figure 2.131 Reconstructed sea-level fluctuation based on witnesses’ observations. (Source: Courtesy of R.B. Seed, from Seed et al., 1988.)

of the Var Canyon show evidence of large submarine landslides, therefore the possibility of another large submarine landslide is not unreasonable. The Var Canyon is one of many canyons found along the coast and the walls of those canyons dug in shale can be quite steep (> 60o ). The soil stratigraphy

The Soil Conditions

The Nice airport is built in 15 m of water on land reclaimed from the Mediterranean Sea just east of the mouth of the Var River. Over the years the Var River has deposited fine sediments, creating a delta on the continental shelf followed by the Var Canyon further offshore (Figure 2.132). The walls

Figure 2.132 Canyons off the shore of the Nice airport. (Source: Courtesy of P. Habib, 1994, and Revue Française de Géotechnique.)

79

2.22 THE NICE AIRPORT SLOPE FAILURE (SLOPE STABILITY)

Figure 2.133 Soil stratigraphy at Nice airport. (Source: Courtesy of R.B. Seed, from Seed et al., 1988.).

Figure 2.134 Failure surface. (Source: Courtesy of R.B. Seed, from Seed et al., 1988.)

2.22.4

The second event in this scenario is the tsunami created by the submarine landslide; this tsunami starts by decreasing the water pressure in the soil along the coast, including the sand layers under the dike of the airport. The third event is

600

500 loading Dr = 38% Slow - drained Dr = 31% conditions

400

300

200

Failure Explanation: Scenario 1

Several failure scenarios have been suggested (Habib, 1994); the two main ones are discussed next. They relate to the sequence between the three components of the event: the submarine landslide, the dike slope failure, and the tsunami. Failure scenario 1 suggests that the submarine landslide occurred first and created a tsunami that increased the water level leading to the dike slope failure. The large submarine landslide off the coast of Nice is consistent with the submarine landslide scars observed in the canyons along the coast.

Nice Silty Sand

For all tests: σ3=196 kPa Kc=2

DEVIATOR STRESS, σ1- σ3 (kPa)

offshore of the airport is presented in Figure 2.133. The airport fill is 20 m thick. Below the fill is a 25 m-thick layer of clayey silt followed by a 35 m-thick layer of silty and clayey sand underlain by sandy gravel. The bottom sandy gravel layer is connected to a higher freshwater table upstream and exhibits an artesian pressure of 5 m above sea level. The lower silty and clayey sand layer has thin sand seams made of loose sand with relative densities as low as 15–20%. The sea bottom surface after the failure was at the interface between the clayey silt layer and the silty and clayey sand layer (Figure 2.134). The unit weight and strength properties of the soil layers are shown in Figure 2.134 and indicate that the effective stress cohesion is zero and the effective stress friction angle is larger than 30o . Since the average slope angle of the failure scarp is less than 10o with a maximum slope angle of 26.5o over a short distance within that reach, static loading alone cannot explain the failure. One argument remains, which is to know if the properties of the soil making up the failing slope mass were the same as those of the material left behind on which the soil properties were based.

Dr = 38% Relatively rapid loading Dr = 31% - undrained conditions

100

0 0

2

4

6

8

10

12

14

16

18

20

AXIAL STRAIN, (%)

Figure 2.135 Stress-strain curves for drained and undrained triaxial tests on Nice silty sand. (Source: Courtesy of R.B. Seed, from Seed et al., 1988.)

80

2 CASE HISTORIES

the dike slope failure, which needs to be explained based on the increase in weight of the embankment created by the decrease in water level due to the tsunami. Seed and his colleagues (Seed et al., 1988) showed how this was possible. They ran triaxial tests with an initial state of stress equal to the best estimate of the one at the bottom of the failure surface of the dike slope failure. The confining pressure was 200 kN/m2 and the initial deviator stress was also 200 kN/m2 (Figure 2.135). These calculations considered the depth of the failure surface and the presence of the 5 m high artesian pressure. With this initial stress state, Seed et al. (1988) showed that, under undrained conditions, very little additional deviator stress was necessary to create failure by static liquefaction. The increase in deviator stress required was shown to be consistent with the 2.5 m lowering of the sea level at the beginning of the tsunami.

2.22.5 Failure Explanation: Scenario 2 and Alternative Failure scenario 2 suggests that the dike failure occurred first and created the tsunami. This scenario considers that the dike failure started at larger depth and proceeded by regression until it failed the dike embankment. The factors of safety of the dike are in the range of 1.3 and higher, thus making this scenario relatively unlikely. A variation of this scenario suggests that a submarine landslide occurs first, creates the tsunami, and that the submarine landslide while flowing downslope to larger depth into the Mediterranean Sea also regresses to the dike location and takes it out. The suggested scenarios all agree that the construction activities at the Nice airport and harbor, including dynamic compaction, are unlikely to have caused the failure.

CHAPTER 3

Engineering Geology

T

his chapter is intended to give readers a general overview of engineering geology. More detailed information should be sought in textbooks and other publications (Waltham, 1994; Bell, 2007).

3.1

DEFINITION

Geology is to geotechnical engineering what history is to humankind. It is the history of the Earth’s crust. Engineering geology is the application of the science of geology to geotechnical engineering in particular and engineering in general. The same way we learn from history to avoid repeating mistakes in the future, we learn from engineering geology to improve geotechnical engineering for better design of future structures. Engineering geology gives the geotechnical engineer a large-scale, qualitative picture of the site conditions. This picture is essential to the geotechnical engineer and must always be obtained as a first step in any geotechnical engineering project.

3.2

THE EARTH

The age of the universe and of the Earth is a matter of debate. The most popular scientific views are that the universe started with a “big bang” some 15 billion years ago and that the Earth (Figure 3.1) began to be formed some 4.5 billion years ago (Dalrymple, 1994), when a cloud of interstellar matter was disturbed, possibly by the explosion of a nearby star. Gravitational forces in this flat, spinning cloud caused its constituent material to coalesce at different distances from the Sun, depending on their mass density, and eventually to form planets. The Earth ended up with mostly iron at its center and silicates floating toward the surface as they are lighter. The Earth has a radius of approximately 6400 km (Jefferis, 2008). The first layer, known as the crust (Figure 3.2), is about 100 km thick and is made of plates of hard silica rocks. The next layer, called the mantle, is some 2800 km thick and made

Figure 3.1 The Earth. (Source: NASA.)

of hot plastic iron silicates. The core is the third and last layer; it has a radius of 3500 km and is largely made of molten iron. Early on, the planet was very hot and all earth materials were melted as they are on the Sun today. The cooling process started right away and has been progressing ever since. The temperature gradient close to the surface (100 km) is much steeper than the gradient at larger depth because the surface is in contact with the cooler atmosphere. The average temperature gradient within those 100 km is about 15∘ C per km. This gradient varies significantly on the surface of the Earth and the locations, and much higher gradients are ideal for the development of geothermal energy. Below the first 100 km the temperature gradient is much smaller and averages 0.635∘ C per km. The gravity field is governed by the acceleration due to gravity (9.81 m/s2 ) on average). This gravity field generates an increase in stress versus depth, which leads to an enormous pressure at the center of the Earth of about 340 GPa. The Earth’s magnetic field is created by magma movement in the core and varies between 30 and 60 microteslas; it is strongest near the poles, which act as the two ends of the Earth dipole. The Earth is a dynamic medium that changes and evolves through major events such as plate tectonics and earthquakes. The rock plates (about 100 km thick) that “float” on the semi-liquid and liquid layers below accumulate strains at various locations where they run into each other. The

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

81

82

3 ENGINEERING GEOLOGY

Temperature (°C)

Free space

3000

0

Mass density (kg/m3)

Pressure (GPa) 200

6000 0

400 0

4000

8000

Atmosphere

12000

0

15°C/km Down to 100 km

Mantle Core

4000

5500°C

0.63°C/km Down to 6400 km

Depth (km)

2000

Crust

6000

Figure 3.2 Earth temperature, pressure, and density.

Europe Asia Africa

South America

Figure 3.3 South America and Africa fit. (Source: Courtesy of John Harvey.)

movement of the rock plates is due to the movement of the semi-liquid below them, and the drag force generated at the interface. When the stress buildup is released abruptly, the result is an earthquake. Earthquakes and other movements allow the plates to move slowly (centimeters per year) yet significantly over millions of years. For example, on today’s world map, South America still looks like it could fit together with Africa—because in the distant past they were in fact joined (Figure 3.3).

The Earth crust is 95% silica—and when silica cools, it hardens. This cooling creates the first kind of rocks: igneous rocks. Igneous rocks (e.g., granite, basalt, gneiss) are created by the crystallization of magma. Sedimentary rocks (e.g., sandstone, limestone, clay shales) are made of erosional debris on the Earth surface which was typically granular and recemented; they are created by wind erosion and water erosion, and are recemented by long-term high pressure or by chemical agents such as calcium. Metamorphic rocks (e.g., schist, slate) are rocks that have been altered by heat and/or pressure. The strength of rocks varies greatly, from 10 times stronger than concrete (granite) to 10 times weaker

GEOLOGIC TIME

1000

Figure 3.4 Geologic time (eras).

oz en C

es M 100

Millions of years

c

65.5

oi

oi

c

251

oz

ic

542

zo

am

br ia

n 10,000

4500

Pr ec

Ea o N

o N

rth

15000

un

iv

er se

Geologic time is a scale dividing the age of the Earth (4600 million years) into five eras (Figure 3.4): Precambrian (4600 million years ago [MYA] to 570 MYA), Paleozoic (570 MYA to 245 MYA), Mesozoic (245 MYA to 65 MYA), Tertiary (65 MYA to 2 MYA), and Quaternary (2 MYA to the

100,000

ROCKS

eo

3.3

3.4

Pa l

North America

present) (Harland et al. 1989). Each era is subdivided into periods and then into epochs (Figure 3.5). The Quaternary era, for example, is divided into the Pleistocene period and the Holocene or Recent period. Typically, the older the earth material, the stronger it is. The last Ice Age occurred about 10,000 years ago at the beginning of the Holocene period. Glaciers, some of them 100 meters thick, covered the Earth from the North Pole down to about the 40th parallel (St. Louis in the USA) and preloaded the soil. Because of this very heavy preloading, called overconsolidation or OC, those soil types (e.g., till) are very stiff and strong and do not settle much under load, but may erode quickly (as in the Schoharie Creek bridge failure disaster in 1987). When the glaciers melted, the soil surface rebounded; in some places this movement is still ongoing at a rate of about 10 mm per year.

10

3.6 GEOLOGIC FEATURES

Cenozoic ERA Periods

Epoch

Quaternary (Present–2.6 My)

Neogene (2.6–23.0 My)

Mesozoic ERA

Paleozoic ERA

Periods

Periods

Holocene (present–0.01 My) Pleistocene (0.01–2.6 My) Pliocene (2.6–5.3 My)

Permian (251–299 My)

Cretaceous (65.5–145.5 My)

Silurian (416–444 My) Jurassic (145.5–201.6 My)

Tertiary (2.6–65.5 My) Paleogene (23–65.5 My)

Carboniferous (299–359 My) Devonian (359–416 My)

Miocene (5.3–23.0 My) Oligocene (23–33.9 My)

83

Ordovician (444–488 My)

Eocene (33.9–55.8 My)

Paleocene (55.8–65.5 My)

Triassic (201.6–251 My)

Cambrian (488–542 My)

Figure 3.5 Geologic time (periods and epochs).

than concrete (sandstone). Older rocks are typically stronger than younger rocks. Figure 3.6 shows some of the main rock types.

3.5

SOILS

Soils are created by the exposure of rocks to the weather. This weathering can be physical (wetting/drying, thermal expansion, frost shatter) or chemical (solution, oxidation, hydrolysis). The elementary components of rocks and soils are minerals such as quartz and montmorillonite. Some minerals are easier to break down (montmorillonite) than others (quartz). As a result, the coarse-grained soils (sand, gravel) tend to be made of stable minerals such as quartz, whereas the fine-grained soils (silt and clay) tend to be made of less stable minerals such as montmorillonite. Organic soils may contain a significant amount of organic matter (wood, leaves, plants) mixed with the minerals, or may be made entirely of organic matter, such as the peat often found at the edges of swamps. Figure 3.7 shows some of those soils categories. Note that what the geotechnical engineer calls soil may be called rock by the engineering geologist; this can create confusion during discussion and interpretation.

3.6 GEOLOGIC FEATURES The ability to recognize geologic features helps one to assess how the material at the site may be distributed. These features (Waltham, 1994; Bell, 2007) include geologic structures (faults, synclines, anticlines), floodplains and river deposits (alluviums, meander migration), glacial deposits (glacial tills and boulders left behind by a glacier), arid landforms (dunes, collapsible soils, shrink-swell soils), and coastal processes (shoreline erosion, sea-level changes). The following list identifies some of the most common and important geological features that can affect geotechnical engineering projects. Faults (Figure 3.8) are fractures in a rock mass that has experienced movement. They can lead to differences in elevation at the ground surface, differential erosion, contrasting visual appearance, and weaker bearing capacity of the fault material compared to the parent rock. Outcrops show up at the ground surface when the rock layers are inclined. The area on the ground surface associated with an outcrop depends on the thickness of the layer and its dip or angle with the horizontal. Escarpments are asymmetric hills formed when an outcrop is eroded unevenly or when the edge of rock layers is not flat. A cliff is an extreme case of an escarpment.

84

3 ENGINEERING GEOLOGY

Igneous rocks

Basalt

Gabbro

Granite

Obsidian

Pumlce

Sedimentary rocks

Breccia Conglomerate Limestone Sandstone

Shale

Metamorphic rocks

Figure 3.8 Example of rock fault. (Source: USGS U.S. Geological Survey.) Gneiss

Marble Metaquartzite Schist

Slate

Minerals

Biotite Copper (black mica)

Halite

Magnetite

Diamond

Quartz

Feldspar

Silver

Gold

Talc

Figure 3.6 Main categories of rocks. (Source: Courtesy of EDUCAT Publishers)

Folds (Figure 3.9) are created when rock layers are curved or bent by earth crust movement. Synclines are concave features (valleys), whereas anticlines are convex features (hills). Folds are best seen on escarpments. Inliers and outliers are the result of erosion. Older rocks are typically below younger rocks. When an anticline erodes, the old rock appears at the surface between two zones of younger rocks (inlier). When a syncline erodes, it can lead to the reverse situation (outlier) Karst is the underground landscape created when limestone is eroded or dissolved by groundwater. This process leads to holes in the limestone, called sinkholes, which can range from 1 meter to more than 100 meters in size and may become apparent while drilling during the site investigation (Figure 3.10).

Figure 3.7 Main soil categories (crushed rock, gravel, sand, silt, clay).

3.7 GEOLOGIC MAPS

85

Figure 3.11 Example of meander migration.

Figure 3.9 Example of anticline–syncline combination. (Source: Photo by R.W. Schlische.)

Figure 3.10 Example of sinkhole. (Source: Courtesy of International Association of Certified Home Inspectors, Inc.)

Subsidence refers to settlement of the ground surface over large areas (in the order of square kilometers). Subsidence can be caused by pumping water out of the ground for irrigation or drinking purposes (Houston, Mexico City), pumping oil, digging large tunnels and mines, the presence of sinkholes, melting of the permafrost, and wetting of certain soils that collapse in the presence of water (called collapsible soils). Meander migration occurs because rivers are dynamic features that change their course by lateral erosion, particularly around bends or meanders. The soil forming the bank on the outside of the meander is eroded and is sent to the inside of the meander by the helical current of the river as it takes the meander turn. The inside of the meander then forms a sand bar (Figure 3.11). Flood plain deposits occur when rivers experience flooding and the water spills over from the main channel into the floodplain. The main channel is a high-energy deposition environment, and typically coarse-grained soils heavy enough not to be transported away are found there. In contrast, floodplains are a low-energy deposition environment where fine-grained soils are found. Floodplains and main channels can end up being buried or abandoned as the river migrates laterally and vertically. Abandoned floodplains are called river terraces. Alluvium and alluvial fans are soil deposits transported to the bottom of a steep slope by the erosion of a river flowing down that steep slope (Figure 3.12).

Figure 3.12 Example of an alluvial fan. (Source: Mike Norton/ Wikimedia/CC BY-SA 3.0.)

Colluvial fans are deposits that form by gravity at the bottom of steep slopes when the slope fails. Dunes are wind-blown sediments that accumulate over time to form a hill. Permafrost is a zone of soil that remains frozen year-round.

3.7 GEOLOGIC MAPS Geologic maps are very useful to the geotechnical engineer when evaluating the large-scale soil and rock environment to be dealt with in a project. These maps typically have a scale from 1:10,000 to 1:100,000 and show the base rock or geologic unit and major geologic features such as faults. Each rock area of a certain age is given a different color (Figure 3.13); soil is usually not shown on those maps. These maps can provide useful information regarding groundwater and hydrogeology, landslide hazards, sinkhole susceptibility, earthquakes, collapsible soils, flood hazards, and karst topography. Remember that what the geotechnical engineer calls soil may be called rock by the engineering geologist; to avoid confusion during discussion and interpretation, it is best to clarify the terminology.

86

3 ENGINEERING GEOLOGY

Precipitation Infiltration Evaporation

Artesian pressure

GWT Clay Sand

Figure 3.13 Example of geologic map. (Source: National Park Service, NPS.)

3.8

GROUNDWATER

Another important contribution of engineering geology to geotechnical engineering is a better understanding of how the groundwater is organized at a large scale. This field involves aquifer conditions, permeability of the rocks, and weather patterns (Winter et al., 1999). If you drill a hole in the ground, at some point you are likely to come to a depth where there is water. This water is called groundwater and it comes from infiltration from rain, rivers, springs, and the ocean. It may be

Figure 3.14 Groundwater.

stationary or flow slowly underground. If you go very deep (about 3 km or more), you will get to a point where there is no more water and the rocks are dry. The groundwater table (Figure 3.14) is the surface of the water within the soil or rock where the water stress is equal to the atmospheric pressure (zero gage pressure). Under natural conditions and in the common case, the groundwater table is close to being flat. The phreatic surface, also called the piezometric surface, is the level to which the water would rise in a tube connected to the point considered in the soil mass. Most of the time, the groundwater table (GWT) and the phreatic surface are the

Figure 3.15 Example of flow due to artesian pressure. (Source: USGS U.S. Geological Survey.)

3.8 GROUNDWATER

same. In some cases, though, they are different: artesian pressure refers to the case where the pressure in the water at some depth below the groundwater table is higher than the pressure created by a column of water equal in height to the distance between the point considered and the groundwater table. This can occur when a less permeable clay layer lies on top of a more permeable sand layer connected to a higher water source (Figure 3.14). Indeed, if you were to drill a hole through the soil down to a zone with artesian pressure, the water would rise above the level of the ground surface and could gush out into a spring (Figure 3.15). Perched water is a zone of water in the soil where the water appears at a certain depth in a boring and then disappears at

87

a deeper depth; it acts as a pocket of water in the ground. Aquifers are typically deeper reservoirs of water that are supplied by surrounding water through a relatively porous rock. Aquifers are often pumped for human consumption. Their depletion can create kilometers-wide zones of settlement called subsidence, and in some instances the settlement can reach several meters in depth. In geotechnical engineering, it is very important to know where the groundwater table is located, as it often affects many aspects of the project. Furthermore, it is important to identify irregularities in groundwater, such as artesian pressure or perched water.

Problems and Solutions Problem 3.1 Calculate the pressure at the center of the Earth. Solution 3.1 To calculate the pressure at the center of the Earth, we will use Newton’s law of universal gravitation. The force between two masses, m1 and m2 , separated by a distance r, is: m •m F = G• 1 2 2 r where G is the gravitational constant = 6.67 ∗ 10−11 N m2 kg−2 The density of soil layers varies with depth; the average density value for each layer is given in the following table: Layer

Thickness (km)

Average density (kg/m3 )

Crust Mantle Core

100 2800 3500

2700 5000 12000

Consider a small element of Earth dr thick and rd𝜃 wide at a depth such that the distance from the center of the Earth is r (Figure 3.1s). This small element has a mass dm1 . The force acting on that element consists of three gravitational force components: the force due to mass Ma, which pulls the element away from the center; the force due to mass Mb, which pulls the element toward the center, and the force due to mass Mc, which also pulls the element toward the center. Newton showed that the forces due to mass Ma and Mb are equal and opposite so that the only force acting on the element is the force due to mass Mc. Ma

Earth

dr

Mc

r

Mb

Figure 3.1s Parameters definition.

88

3 ENGINEERING GEOLOGY

Therefore, the pressure p is p =

F A

where A is the area of the element, so:

dm1 G ⋅ m2 G ⋅ m2 𝜌 ⋅ dV G ⋅ m2 𝜌 ⋅ dr ⋅ A G ⋅ m2 = = = 𝜌 ⋅ dr ⋅ ⋅ ⋅ ⋅ 2 2 2 A A A r r r r2 m2 4 3 p= 𝜌 ⋅ G ⋅ 2 ⋅ dr, where m2 = πr 𝜌 ∫ 3 r 4 2 P= π⋅G⋅ 𝜌 ⋅ r ⋅ dr ∫ 3

dp =

Because the density of the Earth’s layers is not constant (see Figure 3.2), the pressure at the center of the Earth is: ) ( 3500×1000 6300×1000 6400×1000 4 −11 2 2 2 P = 𝜋 × 6.67 × 10 12000 rdr + 5000 rdr + 2700 rdr ∫3500×1000 ∫6300×1000 ∫0 3 ( ) | | | 6 6 6 N −4 2 |3.5 × 10 2 |6.3 × 10 2 |6.4 × 10 P = 2.79 × 10 72r | = 3.44 × 1011 2 = 344 GPa + 12.5r | + 3.645r | |0 |3.5 × 106 |6.3 × 106 m | | | Note that in geotechnical engineering we calculate the pressure, also called vertical total stress, at a given depth z as: ∑ p= 𝛾i 𝛥Zi where 𝛾i is the unit weight of the 𝛥Zi thick ith layer within the depth z. This is an approximation, as the unit weight 𝛾 = 𝜌g is not constant and depends on the depth z (since g is a function of z). This approximation is very acceptable for the usual depth involved in a geotechnical project (a few hundred meters at most); indeed, this approximation only makes a difference of a small fraction of a percent. Problem 3.2 Calculate the temperature at the center of the Earth. Solution 3.2 The temperature gradient is 15∘ C per km in the crust and 0.63∘ C per km in the mantle and the core. Therefore, the temperature at the center of the Earth is: T = 15 × 100 + 0.63 × 6300 = 5469∘ C center

Problem 3.3 What is the depth of interest for most geotechnical engineering projects? Solution 3.3 The depth of interest for most geotechnical engineers is a few hundred meters. Problem 3.4 List the Tertiary and Quaternary epochs. Solution 3.4 Holocene Pleistocene Pliocene Miocene Oligocene Eocene Paleocene

0 to 10,000 years ago 10,000 to 1.8 million years ago 1.8 to 5.3 million years ago 5.3 to 23.8 million years ago 23.8 to 33.7 million years ago 33.7 to 54.8 million years ago 54.8 to 65 million years ago

3.8 GROUNDWATER

89

Problem 3.5 What happened about 10,000 years ago on the Earth? What are some of the consequences for soil and rock behavior today? Solution 3.5 An Ice Age occurred about 10,000 years ago, at the beginning of the Holocene period. At that time, glaciers about 100 meters thick covered the Earth from the North Pole down to about the 40th parallel and loaded the soil. This very heavy loading increased the density, stiffness, and strength of the soils below the glaciers. When the glaciers melted, they left behind these very dense, overconsolidated soils, called glacial tills. These soils do not settle much as long as the pressure does not exceed the pressure exerted by the Ice Age glacier. (The glaciers also carried within them very large and heavy rocks, and deposited these boulders along their paths when they melted.) When the glaciers melted, the soil surface rebounded, and in some places this movement still goes on today at a rate of about 10 mm per year. An example of this is the landmass in England. Problem 3.6 What are the three main categories of rocks, and what is the origin of each category? Solution 3.6 The three main categories of rocks are: • Igneous rocks, which come from the solidification and crystallization of magma. Common igneous rocks are granite, basalt, and gneiss. • Sedimentary rocks, which are composed of rocks previously eroded through wind and hydraulic erosion and recemented by long-term high pressure or chemical agents (e.g., calcium). Common sedimentary rocks are sandstone, limestone, and clay shales. • Metamorphic rocks, which have been altered by heat and/or pressure. Common types of metamorphic rocks are schist and slate. Problem 3.7 What are the four main categories of soil sizes? How were each of these soils generated? Solution 3.7 Soil class

Soil type

Size (by USCS)

Coarse-grained soil

Gravel Sand Silt Clay

75 mm to 4.75 mm 4.75 mm to 0.075 mm 0.075 mm to 2 μm < 2 μm

Fine-grained soil

Soils are generated by the exposure of rocks to the weather and other altering mechanisms. The weathering can be physical (wetting/drying, thermal expansion, frost shatter) or chemical (solution, oxidation, and hydrolysis). Erosion and deposition are other mechanisms responsible for soil formation. Problem 3.8 What engineering geology features should you look for when you visit a site for a geotechnical engineering project? Solution 3.8 • • • • •

Geologic structures (faults, synclines, anticlines) Floodplains and river deposits (alluviums, meander migration) Glacial deposits (glacial tills and boulders left behind after glacier melting) Arid landforms (dunes, collapsible soils, shrink-swell soils) Coastal processes (shoreline erosion, sea-level changes)

90

3 ENGINEERING GEOLOGY

Problem 3.9 Why are geologic maps useful to the geotechnical engineer? Solution 3.9 Geologic maps help geotechnical engineers to identify and evaluate the soil and rock in an area and to find specific geologic features such as faults. Problem 3.10 Define the following terms: groundwater level, perched water, phreatic surface, aquifer. Solution 3.10 • Groundwater level: the level at which water is found in an open borehole. • Perched water: a zone of water in the soil where the water appears at a certain depth in a boring and then disappears at a deeper depth; it acts as a pocket of water in the ground. • Phreatic surface: the level where the water would rise in a tube connected to the point considered in the soil mass. Most of the time, the groundwater table and the phreatic surface are the same. Some exceptions include artesian pressure and water flow. • Aquifer: a deep reservoir of water created by the infiltration of the surrounding water through a porous soil or rock. Drinking water may come from an aquifer.

CHAPTER 4

Soil Components and Weight-Volume Parameters

4.1

PARTICLES, LIQUID, AND GAS

4.2 PARTICLE SIZE, SHAPE, AND COLOR

Soils are made of particles, gas (most often air), and fluid (most often water). Particles are also called grains. The space between the particles makes up the voids, sometimes also called pores. If the voids are completely filled with air, the soil is called dry. If the voids are completely filled with water, the soil is called saturated. If the soil is filled partly with air and partly with water, the soil is called unsaturated. Figure 4.1 shows a soil sample and its graphical representation (the three-phase diagram discussed later in this chapter). Note that in some cases, there is a subtle distinction between saying that a soil is dry and saying that a soil has no water. If a small sample of wet soft clay is left in the sun or in a low-humidity laboratory, it will become “dry” after a while and at the same time much stronger than when it was wet. This “dry” clay still has a tiny bit of water firmly bound between the particles. This water is in tension and sucks the particles together through a phenomenon called suction (explained in Chapter 11 on effective stress). This suction is responsible for the increase in strength of the clay. If the dried clay is ground into individual particles and placed in an oven at 100∘ C, then it will have no water and no tensile strength. Thus, it becomes important to make a distinction between dried and no water; for example, a dried clay is a hard block of soil whereas a clay with no water may simply be a dry powder.

VT

Vv

Vs (a)

Va

Air

Wa

Vw

Water

Ww

Soil

Ws

WT

(b)

Figure 4.1 Three-phase diagram representation.

Depending on their size, soil particles are called gravel-size particles, sand-size particles, silt-size particles, or clay-size particles. Gravel particles, sand particles, and the coarser silt particles are typically made of quartz and are more rounded in shape. They can be seen with the naked eye or a simple microscope. Clay particles and the finer silt particles are too small to be seen with the naked eye; they are visible only with the use of electron microscopy or X-ray diffractometry. Figure 4.2 shows photos of soil particles. Ranges of particle sizes are defined as: Gravel-size particles: 20 mm to 4.75 mm Sand-size particles: 4.75 mm to 0.075 mm Silt-size particles: 0.075 mm to 0.002 mm Clay-size particles: less than 0.002 mm These ranges indicate a huge difference in size between a sand-size particle and a clay-size particle. For example, if the clay particle were a postage stamp, the sand particle would be a very large airplane. Soil particle sizes are so dramatically different that showing them on a natural scale is not very helpful (Figure 4.3); instead, a logarithmic scale is used that allows the very small particles to appear on the scale as well as the very large ones. Figure 4.4 shows such a scale and summarizes the main differences between soil particles. There is also a big difference in shape between the gravel-size and sand-size particles, on the one hand, and the silt-size and clay-size particles, on the other. Gravel particles, sand particles, and the larger silt particles tend to be rounded, whereas clays and the smaller silt-size particles tend to be rod-like or plate-like. This is because minerals such as quartz, which form the larger particles, are much more stable and resistant to weathering than the minerals, such as kaolinite (baby powder), that form the smaller particles. The surface of sand particles and gravel particles can present various degrees

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

91

92

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

Figure 4.2 Examples of cobbles, gravel-, sand-, silt-, and clay-size particles. Particle size (mm) 0 0.075

2

4 4.75

6

8

Sand size

10

12

Gravel size

Silt and clay size

Figure 4.3 Particle sizes on a natural scale. Particle size Particle size in mm on a log scale Log of particle size on an arithmatic scale Seeing the particle

Silt size

Clay size 10–4 10–3 –4

10–2

10–1

–2

–1

–3

log (0.002) 5 22.7 Electron microscope

Sand size 100 0

log (0.075) 5 21.125

Microscope

Gravel size 101 102 1

2

log (4.75) = 0.677

Hand lens

Naked eye

Shape

Plate-like—rod-like

Sphere-like

Minerals

Smectite-montmorillonitebentonite-illite-kaolinite

Mostly quartz: some felspar and mica

Forces involved

Electostatic-electromagneticVan der Waals intermolecular-gravity

Gravity

Structure Other factors

Flocculated-dispersed Cation exchange capacitysuction

Loose-dense Surface roughness-suction

Figure 4.4 Particle sizes on logarithmic scale and some characteristics of each size.

of roughness. At one end of the spectrum are the angular particles (freshly broken from the parent rock, for example) and at the other end are the smooth, rounded particles (eroded by water over a long period of time, for example). Clays and silts are typically much smoother to the touch than sands and gravels. Soil particles are gray, tan, brown, or reddish. The

brown or reddish color may come from the presence of iron. The wetter the soil is, the darker the color will be; this may help in determining the location of the groundwater level when retrieving samples from a boring. A darker color may also indicate the presence of organic matter, although a foul smell is another and possibly better indicator.

4.4 COMPOSITION OF CLAY AND SILT PARTICLES

4.3 COMPOSITION OF GRAVEL, SAND, AND SILT PARTICLES Soil particles are made of mineral or organic matter. Mineral matter is inert matter such as silica, whereas organic matter is of biological origin (basically, anything that lives or has lived). Organic particles include leaves, plants, grasses, fibers, tree trunks, shells, and fossils. Most soil particles are made of minerals, which have a crystalline structure. The most common mineral is silica; indeed, silica makes up 95% of the Earth’s crust. Minerals are to particles what bricks are to houses: they are the building blocks of the particle. The most stable minerals are framework minerals, which are resistant to erosion and weathering, and form the larger particles (gravel and sand). The least stable minerals are the sheet minerals which make up the clay particles. The most common constituent mineral in gravel, sand, and the coarser silt particles is quartz (SiO2 ), but feldspar (KAlSi3 O8 ) and mica (SiO2 , Al2 O3 ) are also encountered. The behavior of gravel particles, sand particles, and the coarser silt particles is determined by the weight of the particle and associated friction. Other phenomena, such as electromagnetic and intermolecular forces, do exist, but in these coarser particles their effects are negligible compared to the particle weight. This is not the case for extremely small particles, such as clay particles or the finer silt particles.

4.4 COMPOSITION OF CLAY AND SILT PARTICLES Note that silt particles are listed in the title of this section and the last section. The reason is that silt particles straddle the properties of coarse-grained particles and clay particles. Three major minerals make up clay particles. In decreasing order of size, they are kaolinite, illite, and smectite (Mitchell and Soga, 2005). Montmorillonite and bentonite are subgroups of the smectite minerals. These minerals are composed of elementary sheets, which are the silica sheet (SiO2 ), the gibbsite sheet (Al2 (OH)6 ), and the brucite sheet (Mg3 (OH)6 ). The mineral kaolinite (Al2 Si2 O5 (OH)4 ) is made of a stack of a silica sheet and a gibbsite sheet. Kaolinite makes up the larger clay particles with length on the order of 1000 nanometers (Figure 4.5), a thickness of about 100 nanometers, and a specific surface (particle surface per unit mass) of 10 m2 /g. Kaolinite is commonly used in baby powder. Smectite (Al2 Si4 O10 (OH)2 and x interlayers of H2 O) is made of a gibbsite sheet sandwiched between two silica sheets. Smectite makes up the smaller clay particles with length on the order of 100 nanometers (Figure 4.5), a thickness on the order of 1 nanometer, and a specific surface (particle surface per unit mass) of 800 m2 / g. This remarkably high specific surface allows the smectite particle to absorb a significant amount of water between the elementary sheets. This leads to extreme swelling and shrinking potential for these clays

Montmorillonite particle 1 nm = 10–9 m

93

20 nm 3 nm

100 nm

20 nm

Kaolinite particle 40 nm 100 nm

1000 nm

Figure 4.5 Approximate dimensions of montmorillonite and kaolinite particles.

(Figure 4.6). Montmorillonite and bentonite are subgroups of the smectite mineral group. Bentonite is sold commercially for drilling mud applications because it can form a nearly impervious cake on the wall of the borehole and keep groundwater from penetrating the borehole (see Chapter 7 on site investigation). The mineral illite has properties intermediate between those of kaolinite and smectite. Cations are positive ions that are attracted to the surface of clay particles. Silicium (Si4+ ) is a very common cation in soils. Because Si4+ has a high valence, a negative charge will be generated if it is replaced by another cation such as Al3+ or Mg2+ or Na+ . This cation exchange is called isomorphous substitution because the exchange cation has the same shape (isomorphous means “same shape” in Greek), allowing it to fit in the crystalline lattice, but a lower valence. This substitution will occur if an exchange cation is available when a Si4+ cation is not. The cation exchange capacity or CEC is a measure of how many cations a clay particle can catch; it is measured in milliequivalents per unit mass (meq/100g). The milliequivalent is a unit of amount of substance and is related to the mole, the SI unit used to quantify the amount of substance. Kaolinite has a smaller CEC (∼5 meq/100g) than montmorillonite (∼ 80 meq/100 g). As a result of isomorphous substitution, the surface of clay particles is negatively charged except at the ends of the particles, where positive charges may appear due to broken bonds. In this case, clay particles can be thought of as little magnets that attract or repel each other. The negative and sometimes positive electrical charges on the surface of clay particles influence the way the structure of the clay mass develops (flocculated or dispersed). The water next to the clay particle surface is made of molecules that can be thought of as electrical dipoles (H+ and OH– ). The H+ end of the dipole is attracted to the negative charges on the clay particle surface and the water molecule adheres to the surface. Cations such as Na+ may also be present in the water and will be attracted to the surface in an effort to neutralize the negative charge. The sodium adsorption ratio or SAR gives an indication of how much

94

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

Before water supply

During water intake

Figure 4.6 Absorption of water in bentonite. (Source: Courtesy of Komine and Ogata (2004), American Society of Civil Engineers.) uw

Double layer water Adsorbed water layer

– +

– +

– –

+ –

– –

+ –

+ –

+ –

– +

– +

– – Clay particle – – + –

+ –

– +

– +

– +

– –

– +

– –

+ –

fci

fci C.G

+ –

W

fei fei

Electrical dipoles(H+(OH)−)or cations

fei

C.G fci

Figure 4.7 The electrical double layer of clay particles.

fei

C.G

fei

C.G fci C.G

fci

C.G

sodium is available around the particles. It is defined as: [Na+ ]

(4.1) )0.5 [Ca2+ ] + [Mg2+ ] 2 where the value within brackets [] is the concentration of cations in meq/liter. This layer of bound water is called the electrical double layer (Figure 4.7) and its thickness is on the order of 1–50 nm, with the higher values found in very active clay particles such as montmorillonite and bentonite. The layer of water most closely bound to the particle surface within the electrical double layer is called the adsorbed water layer (Figure 4.7). The attraction between clay particles is attributed to the Van der Waals forces that overcome the repulsion between two negatively charged particles. Van der Waals forces are intermolecular forces that give water its tensile strength, for example. The other important source of cohesion in a clay is the attraction between water and silica, which sucks the particles together. This phenomenon, called suction, is discussed in Chapter 11 on effective stress. SAR = (

4.5

PARTICLE BEHAVIOR

Gravels and sands are called coarse-grained soils, while silts and clays are called fine-grained soils. The weight of soil particles varies tremendously; for example, a gravel-size particle is about 10 billion times heavier than a clay-size particle. Coarse-grained soil particles tend to behave according to their weight. In contrast, the behavior of fine-grained, clay-size particles is significantly influenced by the electrostatic and electromagnetic forces that exist at the particle surface. These forces create attraction and repulsion much like small magnets would do. They give clays their consistency, which

Soil Water

fci = Forces at contacts between particles fei = Electrical forces between particles C.G. = Center of Gravity W = Weight of particle uw = Water pressure around particle

Figure 4.8 Forces acting on a soil particle.

you might wish to think of as stickiness. The behavior of silt-size particles is intermediate between that of gravel and sand, on the one hand, and that of clay on the other. In addition to the weight of the particle and the electrostatic/electromagnetic forces affecting the particles, water can strongly influence the behavior of an assembly of particles (Figure 4.8). First, the water can create buoyancy if the particle is below the groundwater level. This buoyancy reduces the effective weight of the particle (like when you go into a swimming pool) and therefore reduces the friction that it can generate when rubbing against other particles. Second, even above the groundwater level, water is still present in the voids because of two fundamental phenomena: the attraction between water and the clay minerals (e.g., water is attracted to silica, which leads to capillary suction) and the attraction between water and salt (osmosis). Both phenomena allow the water to stay in the voids, go into tension, and suck the particles together. This “glue” between particles influences the behavior of the particles, contributes to the soil consistency, and is responsible for the strength of a dry clay. This topic is developed in Chapter 11 on effective stress. 4.6

SOIL STRUCTURE

The structure of a soil refers to the arrangement of the soil grains. Loose or dense structures are found in coarse-grained

4.7 THREE-PHASE DIAGRAM

(a) – Flocculated structure

Loose 90°

95

(b) – Dispersed structure

Figure 4.11 Flocculated and dispersed clay structures.

Figure 4.9 Sphere organized in a loose structure and associated rosette of contacts.

soils, whereas flocculated and dispersed structures exist in fine-grained soils. A loose soil structure is similar to the arrangement of the spheres shown in Figure 4.9. In this case, the contacts between particles are mostly at 90 and 180 degrees on the rosette of contacts. Shearing the mass would lead to a loss of volume of the mass, as the particles will tend to move toward a more stable arrangement. This soil would be called contractive. Such loose structures are found, for example, when the soil settles under water in a very low-energy environment and without vibration. This can be the case with hydraulic fills. A dense soil structure is similar to the arrangement of the spheres shown in Figure 4.10. In this case, the contacts between particles are mostly at 45 and 135 degrees on the rosette of contacts. Shearing the mass would lead to an increase in volume of the mass, as each particle will tend to ride on top of the next one. This soil would be called dilatant. Such dense structures are found in compacted soils for dams or pavements that are densified during placement by a combination of pressure and vibration. In a dispersed structure, the particle arrangement is like a deck of cards (Figure 4.11). Such structures tend to be very stable and exhibit high stiffness. However, they have little strength against shearing that takes place in the direction of the “cards.” The stacks of particles can, however, be organized in different ways within a single soil, and that will influence the overall behavior. In a flocculated structure, the particle arrangement looks like a card castle (Figure 4.11). Such structures tend to be unstable and can easily collapse. When a flocculated clay derives the strength of its particle contact from salt bonding, a quick clay may be formed. These quick clays (such as found in Norway and Canada) may liquefy if the salt is leached from the contacts by exposure to freshwater and/or if an event triggers the breaking of the bonds. The Rissa

Dense 60°

60°

Figure 4.10 Sphere organized in a dense structure and associated rosette of contacts.

Figure 4.12 Examples of clay and sand structures. (Source: From Terzaghi et al. (1996). Reproduced with permission of John Wiley & Sons.)

event in Norway was a remarkable quick-clay landslide in which the clay literally turned into liquid—to the extent that houses floated down the hill (see Chapter 2). Most natural clays exhibit a mixture of dispersed and flocculated structures. Examples of clay and sand structures are shown in Figure 4.12 (Terzaghi et al., 1996). Composite structures are associated with mixtures of coarse particles and fine particles. In a matrix structure, the fine particles are predominant and the coarse particles do not touch each other. In a void bound structure, the coarse particles touch each other and are bound together by the fine particles, which effectively act as a glue.

4.7 THREE-PHASE DIAGRAM The three-phase diagram is a graphical representation of the soil components. Figure 4.1(a) shows a soil sample in its natural state, with particles, gas (most often air), and liquid (most often water) all mixed together. All the air can be regrouped into one volume V a , and all the water can be regrouped into one volume V w . The sum of V a and V w is the volume of voids V v . Once all the air is in V a and all the water is in V w , then what is left are the particles regrouped into one volume V s . This particle volume has no voids, because they have been sucked out into V a and V w ; therefore, the volume V s is a solid piece of rock with no voids. The unit weight of this solid piece of rock made with the particulate material has a unit weight called the unit weight of solids, 𝛾 s , a ratio of the weight of solids W s over the volume of solids V s . The unit weight of solids varies depending on the mineral or organic matter of the particles, but for mineral matter it is in the range of 25.5 to

96

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

27 kN/m3 and for organic matter it is 9 to 13 kN/m3 . The specific gravity of solids, GS , is the ratio between the unit weight of solids 𝛾 s and the unit weight of water 𝛾 w .

4.8

WEIGHT-VOLUME PARAMETERS

Some of the most important parameters describing the volume and weight of soils are the unit weight, the water content, the void ratio, the porosity, the degree of saturation, the specific gravity of solids, and the density index (Table 4.1). The natural unit weight (total weight of soil W t over total volume of soil V t ) is the unit weight of the soil as it is found in its natural environment. The natural unit weight is also called total unit weight or simply unit weight. Numbers between 17 and 22 kN/m3 are common. The dry unit weight is the unit weight of the dry soil (weight of solids W s over total volume of soil V t ). Numbers between 14 and 18 kN/m3 are common. The saturated unit weight is the ratio of the weight of the soil when the voids are full of water or liquid over the total volume of the soil. The weight of the saturated soil is the weight of solids plus the weight of water necessary to fill the voids Table 4.1

(Ws + Vv 𝛾w ). Numbers between 18 and 22 kN/m3 are common. The submerged unit weight is the difference between the saturated unit weight and the unit weight of water. The effective unit weight is equal to the total unit weight for a point in the soil mass above the groundwater level and equal to the submerged unit weight for a point below the groundwater level. The unit weight of solids 𝛾 s is the unit weight of the particle itself. It is the ratio of the weight of solids W s over the volume of solids V s . The unit weight of solids varies depending on the composition of the particles (mineral or organic matter), but for mineral matter it is in the range of 25.5 to 27 kN/m3 and for organic matter it is in the range of 9 to 13 kN/m3 . The specific gravity of solids Gs is the ratio between the unit weight of solids 𝛾 s and the unit weight of water 𝛾 w . The water content w, also called gravimetric water content, is the ratio of the weight of water W w over the weight of solids W s or weight of dry soil. Although the water content is a ratio and should be used as such in most formulas, it is most often quoted as a percentage. Numbers around 10 to 40% are common, but the water content can be 0 for a dry soil and can reach 400%, as in the Mexico City silt, or even 2000% for

Weight-volume parameters and typical values

Parameter

Symbol Definition

Typical range

Observation

Total unit weight Dry unit weight Maximum dry unit weight Minimum dry unit weight Unit weight of solids

𝛾t 𝛾d 𝛾 d max 𝛾 d min 𝛾s

17–22 kN/m3 14–18 kN/m3 15–19 kN/m3 13–17 kN/m3 25.5–27 kN/m3 for mineral

Unit weight of soil in natural state Unit weight of dry soil Densest state Loosest state Unit weight of particles

Specific gravity of solids

Gs

Saturated unit weight Submerged unit weight Effective unit weight

𝛾 sat 𝛾 sub 𝛾 eff

Unit weight of water Water content (gravimetric) Volumetric water content Degree of saturation Porosity Void ratio Maximum void ratio Minimum void ratio Density Index

𝛾w w 𝜃w S n e e max e min I d or Dr

Wt ∕Vt Ws ∕Vt Ws max ∕Vt Ws min ∕Vt Ws ∕Vs

𝛾s ∕𝛾w (Ws + Vv 𝛾w )∕Vt 𝛾sat − 𝛾w 𝛾 t if above GWL 𝛾sat − 𝛾w if below GWL Ww ∕Vw Ww ∕Ws Vw ∕Vt Vw ∕Vv Vv ∕Vt Vv ∕Vs Vv max ∕Vs Vv min ∕Vs (e max − e)∕(e max − e min )

9–13 kN/m3 for organic 2.6–2.7 for mineral 0.9–1.3 for organic 18–22 kN/m3 8–12 kN/m3 See 𝛾 t and 𝛾 sub 9.81 kN∕m3 10–40% 5–30% 50–100% 25–50% 0.4–1 0.6–1.2 0.3–0.9 20–90%

Dimensionless Voids are full of water Buoyancy force accounted for

0–∞ theoretical range 0–1 theoretical range 0–100% theoretical range 0–100% theoretical range 0–∞ theoretical range Loosest state Densest state 0–100% theoretical range

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

some peaty soils (soil near swamps, made up mostly of grass and plants). Indeed, there is no theoretical upper limit to the water content. The gravimetric water content is the water content measure most widely used in geotechnical engineering. Sometimes for unsaturated soils, the volumetric water content 𝜃 w is used; 𝜃 w is defined as the ratio of the volume of water V w over the total volume V t . Numbers between 5 and 30% are common; 𝜃 w is zero for a dry soil and approaches 100% for extremely wet soils such as peat. The degree of saturation S is the volume of water V w over the volume of voids V v . Although the degree of saturation is a ratio and should be used as such in most formulas, it is most often quoted as a percentage. Numbers from 0 to 100% are found, although most soils below the groundwater level and some distance above it are saturated or nearly saturated. In many cases soils near the surface are unsaturated. The porosity n is the ratio of the volume of voids V v over the total volume V t . Although the porosity is a ratio and should be used as such in most formulas, it is most often quoted as a percentage. Numbers in the range of 25 to 50% are common, and the porosity is always between 0 and 100%. The void ratio e is the ratio of the volume of voids V v over the volume of solids V s . It is most often quoted as a number. Numbers from 0.4 to 1 are common. Although the theoretical limits of the void ratio are 0 and infinity, the practical limits for a given soil are the minimum void ratio e min and the maximum void ratio e max . The minimum void ratio corresponds to the densest state of a given soil, and the maximum void ratio corresponds to the loosest state for a given soil. Both e min and e max are particularly useful in the case of coarse-grained soils and lead to the definition of the density index I d (also designated as Dr ), which is quoted as a percentage and expresses the density of a coarse-grained soil as a percentage between the two extreme states of density (Id = (e max − e)∕(e max − e min )). Also associated with the densest and loosest states are the maximum and minimum dry densities 𝛾 d max and 𝛾 d min . Note that 𝛾 d max corresponds to e min and that 𝛾 d min corresponds to e max . The density index can be expressed in terms of 𝛾 d , 𝛾 d max , and 𝛾 d min as ( ) 𝛾d max 𝛾d − 𝛾d min Id = (4.2) 𝛾d 𝛾d max − 𝛾d min 4.9 MEASUREMENT OF THE WEIGHT-VOLUME PARAMETERS To obtain the natural or total unit weight of a soil, the sample is trimmed into a simple geometrical shape, the dimensions are measured to obtain the volume, the weight is measured, and the weight over volume is calculated (ASTM 2005b [ASTM D2937]). This test is possible only if the sample can keep a geometric shape long enough for the measurements to be made. If this is not possible, as in the case of a dry sand or gravel, for example, then the unit weight is typically obtained by correlation with other measurements such as the blow count during a standard penetration test (SPT) (ASTM 2005a

97

[ASTM D1586]). The water content is obtained by taking a small piece of the sample and measuring its wet weight (W t ), drying it in an oven at 100 ∘ C for 24 hours, and obtaining its dry weight W s , and then calculating the water content w = (Wt − Ws )∕Ws (ASTM 2005c [ASTMD4959]). These two measurements, natural unit weight and water content, are the two most common measurements on a soil sample. Unless the sample is dry or saturated, a third input parameter is necessary to obtain all the weight-volume parameters for a soil. This parameter is often the specific gravity of solids Gs . If it is known that the soil particles are mineral and not organic, then a reasonable assumption can be made for Gs , such as Gs = 2.65. If the composition is not known, or if a more precise value for Gs is needed, then Gs is determined by the specific gravity test (ASTM 2005d [ASTM D854]). This test consists of drying the soil in an oven, pulverizing it by grinding, placing the ground-up material in a container, and filling the container with water up to a chosen level. The container with water plus soil is weighed. Then the container is emptied, cleaned, and filled up to the same chosen level with water only and weighed. The weight of the container with water plus soil minus the weight of the container with water only gives the weight of the buoyant soil. The buoyancy force is the difference between the weight of the buoyant soil and the dry soil. The ratio of the dry weight over the buoyancy force is the specific gravity of the solids. If the unit weight of the soil, its water content, and the specific gravity of solids are known, all other weight-volume parameters can be obtained by calculations (see Section 4.10), including the dry unit weight, the saturated unit weight, the submerged unit weight, the effective unit weight, the degree of saturation, the porosity, and the void ratio. Finding the density index of a coarse-grained soil requires two special tests in addition to the determination of the natural dry unit weight 𝛾 d : one test to obtain the maximum dry unit weight 𝛾 d max (ASTM D4253) and one test to obtain the minimum dry unit weight 𝛾 d min (ASTM D4254). The maximum dry unit weight is obtained by pouring the dry sand or dry gravel into a container of known volume, placing a standard weight on top of the sample surface, and vibrating the soil and the container for a standard time. During the vibrations, the soil volume decreases and reaches equilibrium at the maximum dry unit weight. Measurements of weight and volume at that time allow one to calculate the maximum dry unit weight. The minimum dry unit weight is obtained by very gently pouring a dry sand or gravel sample into a container of known volume, measuring the weight, and calculating the dry unit weight. Once 𝛾 d , 𝛾 d max , and 𝛾 d min are known, the density index I d (or Dr ) is calculated according to Eq. 4.2.

4.10

SOLVING A WEIGHT-VOLUME PROBLEM

Geotechnical engineers often encounter problems where some information related to the weight or volume of a

98

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

soil is known but different weight-volume properties are required. The best way to solve such problems is to follow these steps: 1. Draw a three-phase diagram and indicate the known quantities. If the soil is dry or saturated, then only a two-phase diagram is necessary. 2. If no quantity is given (for example, you are given a unit weight but not a weight or a volume), assume a volume of solid of 1 m3 . 3. Using the information in the specific problem case, complete the weight and volume values for the different phases. If some information is missing, make reasonable assumptions (e.g., the unit weight of solids). Also realize that the unit weight of water is known (9.81 kN/m3 ). 4. Complete the calculations to derive the weight-volume parameters required. The assumptions made in step 2 have no impact on the answers as long as the answers are in the form of ratios (unit

weight, void ratio, porosity, degree of saturation); if a different volume of solids were assumed, the final answer would be the same. Although the step-by-step procedure described here is foolproof, it might be faster in some cases to use the relationships existing between weight-volume parameters. Table 4.2 shows some of these. Table 4.2 Useful relationships between weight-volume parameters n = e∕(1 + e) e = n∕(1 − n) e = (𝛾s − 𝛾d )∕𝛾d Se = Gs w Ws = Wt ∕(1 + w) 𝛾t = 𝛾d (1 + w) 𝛾t = 𝛾w (Gs (1 − n) + Sn) 𝛾t = 𝛾w (Gs + Se)∕(1 − n) 𝛾d = 𝛾w Gs (1 − n)

Problems and Solutions Problem 4.1 A sample of clay is brought back from the field, extruded from the Shelby tube, and trimmed to the following dimensions: height = 150 mm, diameter = 75 mm. It weighs 13.2N. The water content has been determined to be 25% and the soil does not exhibit any signs of the presence of organic matter (e.g., the soil is not very dark and does not smell foul). Find the following parameters for the clay: a. b. c. d. e. f.

Natural unit weight Degree of saturation Porosity Void ratio Dry unit weight Saturated unit weight

Solution 4.1 πD2 π × 0.0752 ×h= × 0.15 = 6.62 × 10−4 m3 4 4 Wt 13.2 × 10−3 The weight of the solid is: Ws = = = 10.6 × 10−3 kN 1+w 1.25 The weight of the water is: Ww = Wt − Ws = 0.0132 − 0.0106 = 2.64 × 10−3 kN

The volume of the sample is: Vt =

Assuming that the unit weight of the solids is 𝛾s = 27 kN∕m3 , the volume of solid is: Ws 0.01056 = = 3.91 × 10−4 m3 𝛾s 27 W 0.00264 The volume of water is: Vw = w = = 2.69 × 10−4 m3 𝛾w 9.81 The volume of air is: Va = Vt − Vw − Vs = 6.62 × 10−4 − 3.91 × 10−4 − 2.69 × 10−4 = 2.48 × 10−6 m3 . Vs =

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

Based on these results, the three-phase diagram of this sample is shown in Figure 4.1s.

Air

VA = 2.48 × 10–6m3

Water

VW = 2.69 × 10–4m3

WA = 0 KN

WT = 0.0132 KN

WW = 2.64 × 10–3 KN WS = 0.0106 KN

Soil

VV = 2.72 × 10–4m3 VT = 6.62 × 10–4m3 VS = 3.91 × 10–4m3

Figure 4.1s Three-phase diagram.

a. Natural unit weight 𝛾t =

WT kN 13.2 N = = 19.92 × 103 3 = 19.92 3 V m m π × 0.0752 × 0.15 4

b. Degree of saturation S=

Vw 2.69 × 10−4 = = 0.991 × 100% = 99.1% Vv 2.72 × 10−4

n=

Vv 2.72 × 10−4 = = 0.409 × 100% = 40.9% VT 6.627 × 10−4

c. Porosity

d. Void ratio e= e. Dry unit weight 𝛾d =

Vv 2.72 × 10−4 = = 0.694 Vs 3.91 × 10−4

𝛾t 19.92 kN = = 15.94 3 1 + w 1 + 0.25 m

f. Saturated unit weight 𝛾sat =

Ws + (Vv × 𝛾w ) 0.0106 + (2.72 × 10−4 × 9.81) kN = 19.95 3 = −4 V m 6.627 × 10

Problem 4.2 a. The sample from problem 4.1 shrinks by 10% when it dries. What is the difference between the dry unit weight and the unit weight of the dry soil? b. The sample from problem 4.1 is placed under water and has swollen by 15% when it reaches its swell limit. What is the difference between the saturated unit weight and the unit weight of the soil at the swell limit? Solution 4.2 a. Shrinking case The volume of the sample is 6.627 × 10−4 m3 . The volume of the sample after the 10% reduction due to shrinkage is VT(Shrink) = 6.627 × 10−4 m3 × 0.90 = 5.96 × −4 10 m3 . W kN 10.6 N 𝛾dried soil = S = = 17.78 × 103 3 = 17.78 3 V m m 5.96 × 10−4 m3 Based on these results, the unit weight of the dry soil is higher than the dry unit weight and the difference is 17.78 − 15.94 = 1.84 kN . m3

99

100

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

b. Swelling case The volume of the sample after the 15% volume increase due to swelling is 6.627 × 10−4 m3 × 1.15 = 7.62 × 10−4 m3 . It is assumed that during the swelling process the soil becomes completely saturated. Therefore, the volume of air in the original sample is replaced by a volume of water. The increase in weight of the sample is equal to the weight of water corresponding to an increase in the volume of water equal to (7.62 − 6.627) × 10−4 m3 plus the volume of water necessary to fill the air voids in the original sample. Additional weight of water: (7.62 − 6.627) × 10−4 × 9.81 + 2.48 × 10−6 × 9.81 = 9.98 × 10−4 kN = 0.998 N WT kN 13.2 + 0.998 N = 18.63 × 103 3 = 18.6 3 3 = V m m 7.62 × 10−4 m3 Based on these results, the unit weight of the swollen soil at the swell limit is lower than the unit weight of the saturated soil in the previous problem and the difference is: 𝛾swollen soil =

19.95 − 18.63 = 1.32

kN . m3

Problem 4.3 A farmer wants to buy a 10 kg bag of fertilizer (organic soil). He has the choice between two merchants. Merchant A sells the 10 kg bag for $10 and the bag indicates that the fertilizer is completely dry. Merchant B sells the 10 kg bag for $8 and the bag indicates that the fertilizer has a water content equal to 20%. If the farmer wishes to buy the least expensive solid constituents, which merchant should he buy from? Show your calculations. Solution 4.3 Case 1 Merchant A (fertilizer in completely dry condition). The three-phase diagram for the fertilizer from merchant A is shown in Figure 4.2s. Completely dry MA = 0

Air Merchant A:

MT = 10 kg Solid

Ms = 10 kg

Figure 4.2s Three-phase diagram for the fertilizer from merchant A.

The unit price for the solid constituents of merchant A is

10$ 10 kg

= 1$∕kg

Case 2 Merchant B (fertilizer with water content = 20%). The three-phase diagram for the fertilizer from merchant B is shown in Figure 4.3s. Air Merchant B: Water Solid

MA = 0 Mw = 0.2 x

MT = 10 kg

Ms = x

Water content = 20%

Figure 4.3s Three-phase diagram for the fertilizer from merchant B.

Assume that the mass of the solids is x; then the mass of the water is 0.2x. The total mass of the fertilizer bag is 1.2x, which is equal to 10 kg. So the mass of the solids can be obtained from the following equation: 1.2x = 10. So x = 8.33 kg The unit price for the solid constituents in merchant B’s bag is: So, the farmer should buy the fertilizer from merchant B.

8$ 8.33kg

= 0.96$∕kg

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

101

Problem 4.4 An airport runway is being extended into a bay and requires a 10 m-high embankment above the bottom of the bay. Calculations indicate that, once constructed, the long-term settlement of the soil beneath the embankment will be about 1 m. The sand used to build the embankment is taken from a pit where the sand has a relative density of 40% (Figure 4.4s). The maximum void ratio is 0.7; the minimum void ratio is 0.4. Once compacted in the embankment, the sand will have a relative density of 90%. What height of sand must be obtained from the borrow pit so that, a long time after completion, the embankment will be 10 m above the initial position of the bottom of the bay before construction started? Embankment Dr = 90%

10 m

11 m

Settlement = 1 m embankment

Pit Dr = 40% Pit

Figure 4.4s Illustration of embankment and pit.

Solution 4.4 The void ratio of the soil in the pit is obtained with the equation: Dr =

e max − e40 0.7 − e40 = = 0.4 e max − e min 0.7 − 0.4

Therefore, e40 = 0.58. For the soil after compaction: Dr =

e max − e90 0.7 − e90 = = 0.9 e max − e min 0.7 − 0.4

Vv = Vse40

Air Water

Vs

Solid

VT = A*H90 = 1 + e90

VT = A*H40 = 1 + e40

Therefore, e90 = 0.43 The three-phase diagram for the soil in both conditions is shown in Figure 4.5s. ΔH = H40–H90 Vv = Vse90

Soil from pit

Air Water

Vs

Solid

Soil in embankment after compaction

Figure 4.5s Three-phase diagram for the soil in the two conditions.

Based on the three-phase diagram in Figure 4.5s, we can write: V (1 + e40 ) H40 = s H90 Vs (1 + e90 ) Knowing that the long-term settlement of the soil in the bay beneath the embankment will be 1 m, the total height of soil necessary is 11 m. We have to calculate the height of soil H 40 that should be taken from the pit with a 40% relative density such that, when compacted to 90% relative density, the height H 90 will equal 11 m. H40 1 + 0.58 = or H40 = 12.15 m 11 1 + 0.43 The height of the soil that must be taken from the pit is 12.15 m.

102

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

Problem 4.5 A shrink test is performed on a sample of clay. At time zero, the sample is 25 mm high, 75 mm in diameter, weighs 2.2 N, and is saturated. The sample is left on a laboratory table; this laboratory is at 20 ∘ C and 50% relative humidity. The sample dries and shrinks. It is weighed and the dimensions are measured with digital calipers as a function of time. At the end of the test, the sample is placed in the oven to obtain its dry weight, which comes out to be 1.8 N. The results of the test are shown in the following table. Time (hr) Height (mm) Diameter (mm) Weight (N)

0

1

2

3

5

8

12

24

25 75 2.200

24.932 74.497 2.160

24.662 73.987 2.115

24.490 73.470 2.079

24.315 72.946 2.034

24.138 72.414 1.989

23.958 71.874 1.944

23.958 71.874 1.872

Plot the curve of water content versus decrease in volume. Comment on the shape of that curve. Solution 4.5 The water content and decrease in volume (Figure 4.6s) at each reading are given by the following equations: w% = 𝛥V =

(WTotal − Wdry ) Wdry 𝜋D20 4

H0 −

× 100

𝜋D2 H 4

Water content (%)

25 20 15

Shrinkage limit

10 5 0 –14.0 –12.0 –10.0 –8.0 –6.0 –4.0 –2.0

0.0

2.0

4.0

Volume change (cm3)

Figure 4.6s Water content versus decrease in volume.

We can make two observations: 1. The curve is almost linear for most of the test. This indicates that within that range of water content, the relative change in volume of the soil is linearly proportional to the change in water content. 2. At a water content of 8%, further drying does not lead to further reduction in volume. The soil has reached its shrinkage limit, which is 8% in this case. Note that this shrinkage limit is the shrinkage limit of the undisturbed soil, not the shrinkage limit of the Atterberg limit that would be obtained from a remolded sample. Problem 4.6 A 2.2 N sample of clay is 25 mm high and 75 mm in diameter and has a water content of 22.2% (same sample as in problem 4.5). It is placed in a stainless steel ring that has the same dimensions as the sample, so the sample cannot expand laterally. The sample is inundated and allowed to swell vertically for several weeks until it reaches equilibrium. The height and time measurements are shown in the following table. At the end of this free swell test, the sample is taken out of the steel ring and weighed; it weighs 2.40 N. Plot the relative increase in volume of the sample versus time and calculate the swell limit for this clay. The swell limit is the water content at which the soil can no longer absorb any additional water. Time (hr) Height (mm)

0 25

100 28

200 29.5

300 30.25

400 30.75

500 31

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

103

Solution 4.6 The volume, relative volume, and relative increase in volume are calculated in the following table. Time (hr)

0

Height (mm) Volume (cm3 ) Relative volume V∕V0 Relative increase in volume 𝛥V∕V0 t∕(𝛥V∕V0 )

100

25 110.45 1 0

200

28 123.70 1.12 0.12 833

300

29.5 130.33 1.18 0.18 1111

30.25 133.64 1.21 0.21 1428

400 30.75 135.85 1.23 0.23 1739

500 31 136.95 1.24 0.24 2083

As can be seen in Figure 4.7s, the sample did not reach the swell limit because at the end of the test its volume is still increasing slightly. One way to solve this issue is to use the hyperbolic extension method. To do so, we assume that the curve in Figure 4.7s is a hyperbola with an equation: t 𝛥V = V a + bt

Relative volume V/V0

1.3 1.25 1.2 1.15 1.1 1.05 1

0

100

200

300

400

500

600

Time (hr)

Figure 4.7s Relative volume variation with time.

To determine the constants a and b, we write:

t (𝛥V∕V0 )

versus t and fit a straight line through the data, as shown in Figure 4.8s. 2500 y = 3.128x + 500.4 R2 = 0.9987

2000 t/(V/V0)

Then we plot

t = a + bt (𝛥V∕V)

1500 b

1000

1

500

a

0 0

100

200

300

400

500

Time (hr)

Figure 4.8s Graph showing the parameters a and b.

600

104

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

According to this graph, a = 500.4 hr and b = 3.13. The extended swell test curve is shown in Figure 4.9s.

Relative volume (∆V/V0)

0.35 0.3 0.25 0.2 1/b

0.15 0.1 0.05 0 0

1000

2000

3000

4000

5000

Time (hr)

Figure 4.9s Extended swell test curve.

When t goes to infinity, 𝛥V∕V goes to 1∕b.

) ) ( 1 𝛥V t = lim = t→∞ t→∞ a + bt V b So, the limit value of 𝛥V∕V = 1∕b = 0.319. The asymptotic volume of the sample at the swell limit is: Vswell limit = 1.319 or Vinitial 7.52 Vswell limit = 1.319 × 2.5 × × 𝜋 = 145.67 × 10−6 m3 4 The volume of solids in the sample is: (

lim

Wdry = 1.8 N Ww = 2.2 − 1.8 = 0.4 N 0.4 × 106 = 40.77 × 10−6 m3 Vw = 9810 Total volume

⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 752 1 ×𝜋× − 40.77 = 69.67 × 10−6 m3 4 1000 At the swell limit, the water content of the sample is: Vs = 25 ×

Vw = Vt − Vs = 145.67 − 69.67 = 76 × 10−6 m3 Ww = 76 × 10−6 × 9810 = 0.745 N W 0.745 w% = w × 100 = × 100 = 41.3% Ws 1.8 Problem 4.7 A silty sand is compacted in a mold. The volume of the mold is 9.46 × 10−4 m3 . The weight of compacted soil in the mold is 18.9 N and the water content is 8%. Assume that Gs is 2.65 and calculate the dry unit weight and the degree of saturation. Solution 4.7 The volume of the sample is: Vt = 9.46 × 10−4 m3 The weight of the sample is: Wt = 18.9 N

The weight of the solids is: Ws =

Wt 18.9 = = 17.5 N 1 + 𝜔 1.08

105

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

The weight of the water is: Ww = Wt − Ws = 18.9 − 17.5 = 1.40 N Assuming that the density of solids is Gs = 2.65, the volume of solids is: Vs =

Ws 17.5 = = 6.74 × 10−4 m3 Gs × 𝛾w 2.65 × 9810

The volume of the water is: Vw =

Ww 1.40 = = 1.43 × 10−4 m3 𝛾w 9810

The volume of air is: Va = Vt − Vw − Vs = (9.46 − 6.74 − 1.42) × 10−4 = 1.29 × 10−4 m3 The degree of saturation can be calculated from the following formula: S=

Vw 1.43 × 10−4 = × 100 = 53% Vv (1.29 + 1.43) × 10−4

or w ⋅ Gs = S ⋅ e 1.29 + 1.43 6.74 S = 53%

0.08 × 2.65 = S ×

The dry unit weight of the sample is: 𝛾d =

Ws 17.5 kN = = 18.5 3 −4 Vt m 9.46 × 10

Figure 4.10s shows the above calculations in a three-phase diagram.

Vv = 2.69 × 10–4 m3

VA = (9.46 – 1.43 – 6.74) x 10–4 = 1.29 × 10–4 m3

Air

Vw = 1.4/9810 = 1.43 × 10–4 m3

Water

V = 9.46 × 10–4 m3

Vs = 17.51/9810/2.65 = 6.74 × 10–4 m3

WA = 0 Ww = 0.08 × 17.51 = 1.40 W = 18.91 N Ws = 18.91/1.08 = 17.51 N

Soil

Figure 4.10s Three-phase diagram.

Problem 4.8 A consolidation test is performed on a sample of soft clay that is 25 mm high and 50 mm in diameter. The test consists of placing a disk of soil in a steel ring and applying load on the sample in a series of steps. The steps last 24 hours and measurements of vertical compression are obtained at the end of each step. The following table shows the time, load, and compression results of the test. Calculate the pressure and vertical strain for the sample at the end of each load step and plot the curve that links the pressure to the vertical strain (stress-strain curve). Why does this curve indicate an apparently surprising result, in that the more load applied to the sample, the stiffer the sample becomes? Can the sample fail? Time (days) Load (N) Stress (kN∕m2 ) Height (mm) Displacement (mm) Strain

0

1

2

3

4

5

6

0 0 25 0 0

294 149.73 24.62 0.38 0.0152

589 299.98 24.25 0.75 0.03

1177 599.45 23.62 1.38 0.0552

2453 1249.31 22.87 2.13 0.0852

4906 2498.61 21.47 3.53 0.1412

9812 4997.21 19.7 5.3 0.212

106

4 SOIL COMPONENTS AND WEIGHT-VOLUME PARAMETERS

Solution 4.8 The stress-strain curve (Figure 4.11s) indicates that the soil become stiffer as the stress increases. Indeed, the ratio of stress increment to strain increment becomes increasingly larger. The reason is that as the stress increases, the influence of the steel ring becomes more important in providing confinement to the sample. With increased confinement, the sample becomes stiffer. The sample cannot fail unless the steel ring fails. 6000

Stress (kPa)

5000 4000 3000 2000 1000 0

0

0.05

0.1

0.15

0.2

0.25

Strain

Figure 4.11s Stress-strain curve.

Problem 4.9 A silt has a unit weight of 20 kN∕m3 and a water content of 26%. What is the specific gravity of the particles? Solution 4.9 The information given in this problem is not sufficient to solve for specific gravity; thus, the exact answer is that it is not possible to solve this problem. However, if we make one assumption, then it becomes possible. For example, assume that the sample is saturated. w = 0.26, 𝛾 = 20 kN∕m3 , 𝛾w = 10 kN∕m3 , S = 1 G 𝛾 (1 + w) 𝛾= s w 1+e S ⋅ e = Gs ⋅ w and S = 1; therefore e = Gs ⋅ w G 𝛾 (1 + w) G × 10(1 + 0.26) 𝛾= s w → 20 = s → Gs = 2.7 1 + Gs ⋅ w 1 + Gs ⋅ 0.26 Problem 4.10 A 5 m-high embankment is made of a sand that has a void ratio of 0.55 and the following boundary void ratios e max = 0.6, e min = 0.4. The embankment is subjected to an earthquake that creates a settlement of 0.32 m due to vibration without a change in lateral dimensions. Calculate the void ratio and the relative density of the sand after the earthquake. Solution 4.10 Let’s assume a reference volume of solids equal to 1 m3 : e=

Vv V = v = Vv . Vs 1

The total height of the embankment, H, is proportional to 1 + e, so that H (before earthquake) 5 1 + 0.55 = = H (after earthquake) 5 − 0.32 1 + e (after earthquake) (5 − 0.32) × (1 + 0.55) − 1 = 0.45 Therefore e (after earthquake) = 5 The relative density is Dr =

e max1 − e 0.6 − 0.45 × 100(%) = × 100 = 75% e max − emin 0.6 − 0.4

4.10 SOLVING A WEIGHT-VOLUME PROBLEM

107

Problem 4.11 Find the relationship between the dry unit weight, the void ratio, and the specific gravity of solids for a soil (Figure 4.12s). Va Vv VT

Vw

Vs

Air

Wa

Water

Ww

Soil

Ws

WT

Figure 4.12s Three-phase diagram.

Solution 4.11 Definition: S = Vw ∕Vv , e = Vv ∕Vs , Gs = Ws ∕(Vs •𝛾w ) w = Ww ∕Ws , 𝛾t = WT ∕VT , and 𝛾d = Ws ∕VT Using the definition of S and e, the volume of water and air can be rewritten as: Va = Vv − Vw = e•Vs − S•e•Vs = (1 − S)•e•Vs Vw = S•e•Vs Using the relationships and definitions, the dry unit weight is: W Ws Gs •𝛾w •Vs 𝛾d = s = = VT Vs + Vw + Va Vs + Vw + Va Gs .yw .Vs G •𝛾 .V G •𝛾 = = s w s = s w Vs + S•e•Vs + (1 − S)•e•Vs Vs •(1 + e) 1+e Problem 4.12 Demonstrate that S•e = Gs •w. Solution 4.12 Definition (Figure 4.13s).: S = Vw ∕Vv , e = Vv ∕Vs , Gs = Ws ∕(Vs •𝛾w ), and w = Ww ∕Ws Ws Ww Ww V •𝛾 V • Gs •w = = = w w = w Vs •𝛾w Ws Vs •𝛾w Vs •𝛾w Vs Vw Vv Vw • S •e = = Vv Vs Vs • • S e = Gs w Va Vv VT

Vs

Vw

Air

Wa

Water

Ww

Soil

Ws

WT

Figure 4.13s Three-phase diagram.

CHAPTER 5

Soil Classification

T

o classify a soil, tests are performed according to the American Society for Testing and Materials (ASTM) standards, and the results of these tests are used in a classification system recommended by the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The tests are the sieve analysis, the hydrometer analysis, and the Atterberg limits. The classification system is called the Unified Soil Classification System (USCS).

5.1

The purpose is to wash out the fine particles that may adhere to the larger particles or form clumps. Sieves are stacked in order of increasing opening, with the largest-opening sieve at the top. The dry soil is placed on the top sieve, which is then covered so that no soil is ejected during shaking. The stack is shaken in a vibrator for a given period of time. At the end of shaking, each sieve is weighed with the soil retained on it.

SIEVE ANALYSIS

Sieve analysis is used for the classification of gravels and sands, which are coarse-grained soils. It consists of taking a given weight of dry soil (W t ), breaking the clumps of soil down to individual particles (using a mortar and rubber-tipped pestle), washing the soil through the smallest sieve (sieve #200), drying what remains on the sieve #200 to get the weight W t (washed), and then sieving that remainder by shaking it through a stack of sieves of decreasing openings (Figure 5.1 and see Table 5.1), the last one being a retaining pan. Recording all the weights involved during this process (see Table 5.2) leads to the percent of soil finer than a given particle size by weight versus the particle size; this is the particle size distribution curve (Figure 5.2). A typical set of sieve numbers and sieve openings is given in Table 5.1. The sieve number corresponds to the number of openings per 25 mm. For example, the no. 200 sieve—the smallest sieve commonly used—has 200 openings per 25 mm; however, each opening is not equal to 25 mm divided by 200 because of the thickness of the wire between openings. In fact, the opening of the no. 200 sieve is 0.075 mm. This opening corresponds to the boundary between sand-size and silt-size particles; this is why sieve analysis is limited to the classification of gravels and sands. The sieve analysis proceeds as follows. First, each sieve is weighed empty. Then the dry soil sample is weighed, soil clumps are broken down, and the soil sample is placed on a sieve #200. The sample is washed under a gentle stream of water and the soil left on the sieve #200 is dried in the oven.

Figure 5.1 Stack of sieves and shaker.

Table 5.1 Sieve numbers and sieve openings Sieve number

Sieve opening

#1 #4 #10 #20 #40 #80 #200

25.4 mm 4.75 mm 2 mm 0.85 mm 0.425 mm 0.18 mm 0.075 mm

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

109

110

5 SOIL CLASSIFICATION

Table 5.2

Initial weight of dry soil Weight of dry soil retained on #200 after washing through #200 Weight of dry soil washed through #200 Dry weight retained on #4 Dry weight retained on #10 Dry weight retained on #40 Dry weight retained on #200 Dry weight retained on bottom pan Percent finer than #4 (4.75 mm) Percent finer than #10 (2 mm) Percent finer than #40 (0.425 mm) Percent finer than #200 (0.075 mm)

Clay 0.002 mm

Percent finer (%)

100

WT

Sieve analysis calculations Wt W t (washed)

Wf

W 40

Wp ((W10 + W40 + W20 + Wp + Wfines )∕Wt ) × 100 ((W40 + W200 + Wp + Wfines )∕Wt ) × 100 ((W200 + Wp + Wfines ) ∕Wt ) × 100 ((Wp + Wfines )∕Wt ) × 100

Silt

Sand 0.075 mm

10–2

Sieve no. 40 (0.425 mm)

W200

Sieve no. 200 (0.075 mm) WP

Pan

Figure 5.3 Dry weight retained on a stack of sieves.

Gravel 4.75 mm

Clay fraction 10–3

W40

W 200

Sand fraction

0 10–4

Sieve no. 10 (2.00 mm)

W 10

Gravel fraction

Silt fraction

W10

W4

60

20

Sieve no. 4 (4.75 mm)

Wfines = Wt − Wt (washed)

80

40

W4

10–1 100 Particle size (mm)

101

102

Figure 5.2 Particle size distribution curve.

Because the total weight of the dry sample is known, the proportion of the soil sample on each sieve is calculated as the weight of that sieve plus soil minus the weight of the empty sieve divided by the total weight of the sample. With this data, the particle size distribution curve can be obtained. This curve is a plot of the percent finer by weight (sum of the weight of soil passing a certain sieve divided by the total weight of the sample, expressed as a percentage) on the vertical axis and

the sieve opening taken as the particle size on the horizontal axis (log scale). Figure 5.3 shows the sieves and the dry weight retained on each sieve. The sieve analysis calculations are shown in Table 5.2. Figure 5.4 gives examples of particle size distribution curves. Note that the particle size determined by sieving through a given sieve is the second largest dimension of the particle that can pass through the sieve opening.

5.2

HYDROMETER ANALYSIS

A hydrometer is an instrument made of glass (Figure 5.5) with a graduated stem on top of a bulb ballasted with lead beads so that it can float upright. The hydrometer reading R is obtained at the liquid surface (level of flotation of the hydrometer) on the graduated scale placed on the stem of the hydrometer. If the liquid being tested is very dense, the hydrometer does not sink very deep into the liquid, and vice versa. Therefore, the higher readings are at the bottom of the stem. Hydrometer analysis is used to obtain the particle size distribution curve of fine-grained soils: silts and clays (ASTM D7928 and D422-63). The test consists of taking a given weight of dry soil (e.g., 40 grams), breaking it down into individual particles if clumps exist, mixing it with a dispersing agent (liquid), placing the wet mixture in a graduated cylinder, filling the container with water up to a known volume (e.g., 1000 ml), shaking the cylinder to reach a uniform mixture, letting the soil particles settle, and recording the fall velocity at which the particles settle. The dispersing agent is used to ensure that the fine particles remain individually separated and do not form clusters. The fall velocity is obtained by measuring the unit weight of the soil-water mixture at a given depth z and at a given time t with a hydrometer. This unit weight decreases with time as the particles settle to the bottom of the container (Figure 5.6). Stokes’s law relates the diameter of a sphere to its fall velocity in a liquid: ) ( 𝛾s − 𝛾f (5.1) D2 v= 18𝜇

111

5.2 HYDROMETER ANALYSIS

Well-graded 10

–3

Clay

100

–2

–1

0

10 10 10 Particle size (mm) Silt

Sand

10

1

10

Percent finer (%)

60 40

Fine

0

10–4

10–3 Clay

100

10–2 10–1 100 Particle size (mm) Silt

Sand

101

102

40

10

Cc>1.0 –4

10

–3

–2

–1

0

10 10 10 Particle size (mm)

10

1

10

2

–2

–1

0

10 10 10 Particle size (mm) Silt

Sand

10

1

10

40 20

Coarse 10–3 Clay

10–2 10–1 100 Particle size (mm) Silt

Sand

101

60 40

Cu>>1.0 –4

10

10

–3

–2

–1

0

10 10 10 Particle size (mm)

10

1

10

2

–2

–1

0

10 10 10 Particle size (mm) Silt

Sand

101

102

Gravel

80 60 40 20

Cc #200 Gravels (> #4) > ½ (> #200)

Sands (> #4) < ½ (> #200) (% < #200) < 5

Cu > 6 and 1 < Cc 3

5 < (% < #200) < 12 Dual symbol required (e.g. SW-SM) work out (% < #200) < 5 and (% < #200) > 12

(% < #200) > 12

(% < #200) < 5

Above Below “A” “A” in line line or hatched and below zone above hatched hatched zone zone

SP

SM

SM-SC

Cu > 4 and 1 < Cc 3

GW

Sands

5 #4) (% #200) Dual symbol required (e.g. GW-GM) work out (% < #200) < 5 and (% < #200) > 12

Above Below “A” “A” in line line or hatched and below zone above hatched hatched zone zone

GP

GM

GM-GC

Fine-grained soils more than ½ < #200 Liquid limit less than 50 Below “A” line and below hatched zone ML-OL

Liquid limit greater than 50

In hatched zone

Above “A” line and above hatched zone

Below “A” line

Above “A” line

ML-CL

CL

MH-OH

CH

Figure 5.17 Flowchart to classify a soil by the USCS.

AASHTO soil classification system based on AASHTO M 145 (or ASTM D 3282) GENERAL CLASSIFICATION GROUP CLASSIFICATION

GRANULAR MATERIALS (35 percent or less of total sample passing No. 200 sieve (0.075 mm) A-2

A-1 A-1-a

A-1-b

50 max. 30 max. 15 max.

50 max. 25 max.

A-3

A-2-4

A-2-5

A-2-6

A-2-7

35 max.

35 max.

35 max.

35 max.

SILT-CLAY MATERIALS (More than 35 percent of total sample passing No. 200 sieve (0.075 mm) A-7 A-4 A-5 A-6 A-7-5, A-7-6

Sieve analysis, percent passing: No. 10 (2 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Characteristics of fraction passing No 40 (0.425 mm)

51 min. 10 max.

36 min.

36 min.

36 min.

36 min.

40 max. 41 min. 41 min. 40 max. Liquid limit 40 max. 41 min. 40 max. 41 min. 10 max. 11 min. 11 min.* 10 max. 10 max. 10 max. 11 min. 11 min. Plasticity index 6 max. NP Usual significant Fine Stone fragments, Silty or clayey gravel and sand Silty soils Clayey soils constituent gravel and sand sand materials 0 0 0 4 max. 8 max. 12 max. 16 max. 20 max. Group Index** Classification procedure: With required test data available, proceed from left to right on chart; correct group will be found by process of elimination. The first group from left into which the test data will fit is the correct classification. *Plasticity Index of A-7-5 subgroup is equal to or less than LL minus 30. Plasticity Index of A-7-6 subgroup is greater than LL minus 30 (see Fig 4-5). **See group index formula (Eq. 4-3). Group index should be shown in parentheses after group symbol as: A-2-6(3), A-4(5), A-7-5(17), etc.

Figure 5.18 AASHTO soil classification system. (Source: Courtesy of FHWA, FHWA-NHI-06-088, 2006.)

GC

119

120

5 SOIL CLASSIFICATION

70

AASHTO CLASSIFICATION SYSTEM

The American Association of State Highway and Transportation Officials (AASHTO) has developed another soil classification system for highway construction purposes starting in 1929 (Hogentogler and Terzaghi, 1929). This classification is quite different from the ISSMGE USCS starting with the AASHTO classification designation being A-1 to A-7 instead of the dual letter symbols of USCS. A-1, A-2 and A-3 are coarse-grained soils and A-4, A-5, A-6, A-7 are fine-grained soils. The distinction between coarse-grained soils and fine-grained soils is set at 35% finer than 0.075 mm (Sieve #200) instead of 50% for USCS. Then further distinction between A-1, A-2, and A-3 as well as subsets of these three main coarse-grained soil categories is based on the percent passing 2 mm (Sieve #10), 0.425 mm (Sieve #40) and 0.075 mm (Sieve #200) as well as the Atterberg limits for A-2 (Figure 5.18). The classification of fine-grained soils is based on a modified plasticity chart, as shown in Figure 5.19. The notes under Figure 5.18 indicate how to proceed with the classification and also give further details on the distinction between categories of clayey soils.

60 Plasticity index

5.7

50 40 A-7-6

30

A-2-6 A-6

20

A-2-7 A-7-5

10 0

A-2-4 A-4

0

10

20

A-2-5 A-5

30

40

50

60

70

80

90

100

Liquid limit

Figure 5.19 AASHTO plasticity chart. (Source: Courtesy of FHWA, FHWA-NHI-06-088, 2006.)

Problems and Solutions Problem 5.1 Calculate the thickness of the wire in the no. 200 sieve. Solution 5.1 The sieve number corresponds to the number of openings per 25 mm. For the sieve #200, the width of any opening is 0.075 mm; therefore, the total width of the openings in 25 mm of the #200 mesh is 200 × 0.075 = 15 mm. The total thickness of the wires in 25 mm of the #200 mesh is (25 − 15) = 10 mm, so the thickness of the wires in a sieve #200 is 10∕200 = 0.05 mm (about the diameter of a human hair). Problem 5.2 A dry sample of soil weighs 5N. It is shaken on a set of sieves: No. 4 (4.75 mm), No. 40 (0.425 mm), No. 200 (0.075 mm), and a pan. The weight retained on No. 4 is 2 N, on No. 40 is 1.5 N, and on No. 200 is 1 N. Calculate: • • • • •

The percent of coarse grain size particles by weight The percent of gravel-size particles by weight The percent of sand-size particles by weight The percent of fine grain-size particles by weight The coefficient of uniformity and the coefficient of curvature

Based on these results, what would you call the soil? Solution 5.2 2 + 1.5 + 1 × 100 = 90% The percent of coarse grain size particles is = 5 2 The percent of gravel-size particles is = × 100 = 40% 5 1.5 + 1 × 100 = 50% The percent of sand-size particles is = 5 0.5 × 100 = 10% The percent of fine grain-size particles is = 5

5.7 AASHTO CLASSIFICATION SYSTEM

Retained soil on sieve Accumulated Accumulated weight (N) weight %

Weight (N) No. 4 (4.75 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Pan

2 1.5 1 0.5

2 3.5 4.5 5

121

Passing through sieve Weight Accumulated (N) weight %

40 70 90 100

3 1.5 0.5 0

60 30 10 0

From these results, D60 = 4.75 mm, D30 = 0.425 mm, and D10 = 0.075 mm. Cu =

D60 4.75 = = 63 D10 0.075

Cc =

D2 30 0.4252 = = 0.5 D10 × D60 0.075 × 4.75

Based on these results, the soil has 90% coarse fraction, therefore, the soil is a coarse-grained soil; furthermore, 50% of the soil is retained between sieves #40 and #200, so the soil is sand. Problem 5.3 Why is the particle size of the particle size curve plotted on a log scale? Plot the particle size curve of problem 5.2 as percent finer vs. particle size on a log scale and then as percent finer vs. log of particle size. Determine by calculations the position of a particle size equal to 0.075 mm and 4.75 mm on the particle size (log scale) axis and on the log of particle size axis. Solution 5.3

100

100

90

90

80

80 Percent finer (%)

Percent finer (%)

The range of particle sizes in soils is very large, so we use the logarithmic scale because this scale stretches out the particle size distribution in the very small range. This allows us to distinguish the small sizes as well as the large sizes. Figure 5.1s shows the particle size curve as percent finer vs. particle size on a log scale. Figure 5.2s shows the particle size curve as percent finer vs. log of particle size. For the 0.075 mm particle, log 0.075 = −1.125; this point can easily be found on the linear scale of Figure 5.2s. The position of this point is the same on the scale of Figure 5.1s. The same approach applies to the 4.75 mm particle: log 4.75 = 0.677.

70 60 50 40 30

70 50 40 30

20

20

10

10

0 0.01

0.1

1

Particle size (mm) in log scale

Figure 5.1s Percent finer vs. particle size on a log scale.

10

log (4.75) = 0.677

60

0 –2.0

log (0.075) = –1.125

–1.5

–1.0

–0.5

0.0

0.5

1.0

Log (particle size (mm))

Figure 5.2s Percent finer vs. log of particle size on a normal scale.

122

5 SOIL CLASSIFICATION

Problem 5.4 Calculate how fast a particle of soil will settle in water if its equivalent diameter is 0.075 mm and then if its equivalent diameter is 0.002 mm. Solution 5.4 Assume that: • • • • •

Water temperature = 20∘ C Specific gravity of particles is 2.65 Viscosity of water is 𝜇 = 10−3 N • s∕m2 Unit weight of water is 𝛾w = 9.79 kN∕m3 Unit weight of soil particles 𝛾s = 2.65 × 9.79 kN∕m3 = 25.95 kN∕m3

The fall velocity of a soil particle in water can be calculated using Stokes’s law: ( ) 𝛾s − 𝛾f v= D2 18𝜇 where 𝛾s = 25.95 kN∕m3 and 𝛾f = 𝛾w = 9.79 kN∕m3 . For particles with D = 0.075 mm ( v=

25.95 − 9.79 18 × 10−6

)

( ×

0.075 1000

)2 = 0.0051(m∕ sec) = 5.1(mm∕ sec)

For particles with D = 0.002 mm: ) ( ( ) 25.95 − 9.79 0.002 2 = 3.59 × 10−6 (m∕ sec) = 0.00359(mm∕ sec) × v= −6 1000 18 × 10 Problem 5.5 A cylindrical hydrometer has a radius of 20 mm and weighs 2N. It is lowered into water mixed with fine soil particles (Figure 5.3s). If the hydrometer sinks and comes to floating equilibrium when it is 100 mm in the liquid, calculate the ratio of soil solids by volume that exists in the liquid. Assume that Gs = 2.65 if needed. R = 20 mm

0.62 m3 100 mm

Water

1.0 m3

15.92 kN 0.38 m3

Soil

Figure 5.3s Hydrometer and three-phase diagram.

Solution 5.5 Fbuoyancy = W V × 𝛾mixture = W 𝜋 × 0.04 × 0.1 × 𝛾mixture = 2 4 𝛾mixture = 15.92 × 103 N∕m3 = 15.92 kN∕m3 2

5.7 AASHTO CLASSIFICATION SYSTEM

Assuming 1 m3 of the mixture and Gs = 2.65: WW + WS = 15.92 kN VW + VS = 1 m3 𝛾w VW + 𝛾S VS = 15.92 kN 9.85 × (1 − VS ) + (2.65 × 9.81) × VS = 15.92 kN ∴ VS = 0.38 m3 ,

VW = 0.62 m3

The volumetric percent of solids in the mixture is 38%. Problem 5.6 Explain the hydrometer analysis in your own words. Develop the equations necessary. Solution 5.6 See Section 5.2 in this chapter. Problem 5.7 A soil has a natural water content of 22% and the following limits. • • • •

Shrinkage limit = 13% Plastic limit = 25% Swell limit = 36% Liquid limit = 55%

Calculate the following: • Plasticity index • Liquidity index • Shrink-swell index Solution 5.7 • Plasticity index: PI = LL − PL = 55 − 25 = 30 • Liquidity index: LI = (w − PL )∕PI = (22 − 25)∕30 = −0.1 • Shrink-swell index: I ss = swell limit-shrinkage limit = 36 – 13 = 23 Problem 5.8 Classify the following soils:

#1 #4 #10 #40 #200 w L w P 0.03 from hydrometer

S1 (% finer)

S2 (% finer)

S3 (% finer)

S4 (% finer)

S5 (% finer)

100 52 38 18 8 17 11 -

100 52 38 18 2 NP NP -

100 63 56 42 4 NP NP -

100 98 90 47 20 32 26 -

100 100 97 82 70 48 34 9

123

124

5 SOIL CLASSIFICATION

Solution 5.8 The soils are classified based on the following criteria: • • • • •

Coarse grain-size particles: retained on the no. 200 sieve (0.075 mm) Gravel-size particles: retained on the no. 4 sieve (4.75 mm) Sand-size particles: passing no. 4 sieve, retained on the no. 200 Fine grain-size particles: passing no. 200 Plastic and liquid limit: Coefficient of uniformity Cu = Coefficient of curvature Cc =

D60 D10 D2 30 D10 × D60

The particle size distribution curves are drawn on Figures 5.4s to 5.8s and the classification of the five soils is presented in the table below. S1 (% finer) Sieve opening (mm) 25 4.75 2 0.425 0.075 0.03 from hydrometer

100 52 38 18 8 —

Other properties WL Wp Ip Coarse fraction (%) Fine fraction (%) Gravel fraction (%) Sand fraction (%) D10 (mm) D30 (mm) D60 (mm) Cu Cc Classification

S

S3 (% finer) Percent finer 100 63 56 42 4 —

100 52 38 18 2 —

17 11 6 92 8 48 44 0.11 1.05 7 63.6 1.4 GW-(GC-GM)

M

100

S2 (% finer)

NP NP NP 98 2 48 50 0.19 1.05 7 36.8 0.8 SP

NP NP NP 96 4 37 59 0.098 0.23 3.2 32.7 0.2 SP

G

Percent passing (%)

Percent passing (%)

60

D60

40

D30

20

D10

0 0.01

0.1 1 Particle size (mm)

10

Figure 5.4s Percent finer vs. log of particle size of sample S1.

100 98 90 47 20 —

100 100 97 82 70 9

S

48 34 14 30 70 0 30

ML

G

Sample 2

Sample 1 80

S5 (% finer)

32 26 6 80 20 2 78 0.036 0.16 0.7 19.4 1.0 SM

M

100

S4 (% finer)

80 60

D60

40

D30

20

D10

0 0.01

0.1 1 Particle size (mm)

10

Figure 5.5s Percent finer vs. log of particle size of sample S2.

125

5.7 AASHTO CLASSIFICATION SYSTEM

M

100

S

G

M

100

80 60

D60

40

D30

20

D10

0 0.01

S

G

Sample 4 Percent passing (%)

Percent passing (%)

Sample 3

0.1 1 Particle size (mm)

D60

40

D30

20

D10 0.1 1 Particle size (mm)

10

Figure 5.7s Percent finer vs. log of particle size of sample S4.

M

100

60

0 0.01

10

Figure 5.6s Percent finer vs. log of particle size of sample S3.

80

S

G

Percent passing (%)

Sample 5 80 60 40 20 0 0.01

0.1 1 Particle size (mm)

10

Figure 5.8s Percent finer vs. log of particle size of sample S5.

Problem 5.9 In a hydrometer test, 40 g of dry soil are placed in 1000 ml of water and dispersing agent mixture with a unit weight of 10 kN/m3 . The reading on the 152H hydrometer after a time equal to 1 hour is 8 g/l and the sinking depth z of the hydrometer is 15 cm. The specific gravity of the soil particles is 2.7 and the temperature of the fluid is 20∘ C. Knowing that in this case the value of the constant K is 0.01344 and the hydrometer reading correction factor a is 1, calculate the particle diameter and the corresponding percent finer. Solution 5.9 The diameter of the soil particle is given by

√

D=K

z = 0.01344 t

√

15 = 0.0067 mm 60

And the corresponding percent finer is given by P=

aR152 1×8 × 100 = × 100 = 20% Ws 40

126

5 SOIL CLASSIFICATION

Problem 5.10 Classify soils S2 and S5 of Problem 5.8 according to the AASHTO classification system. Solution 5.10 Soil S2 The percent passing sieve #200 is 2%, thus A-1, A-2, A-3 are possible. The percent passing sieve #40 is 18%, thus only A-1 and A-2 are possible. The percent passing sieve #10 is 38%, thus A-1 and A-2 are still possible. The Atterberg limits were not possible, thus only A-1 can be retained. The first note under Figure 5.18 indicates that, when several classifications are possible, the first one from the left is the correct one. Therefore, S2 is classified as A-1-a. Soil S5 The percent passing sieve #200 is 70%, thus A4, A5, A6, A7 are possible. The liquid limit is 40%, thus A5 and A7 are possible. The plasticity index is 14%, thus only A-7 is possible. The second note under Figure 5.18 states that when PI < (wL – 30), the soil is A-7-5. This is the case for S5 since the PI is 14 and (wL – 30) = 18. Therefore, S5 is classified as A-7-5.

CHAPTER 6

Rocks

I

n many instances geotechnical engineers work on rock problems. For example, locating the depth of bedrock is often an important part of any soil investigation. Rock slopes, rock tunneling, rock excavations, rock fill in dams, and foundations on rock are other examples of projects requiring the expertise of the geotechnical engineer. This chapter is intended to give the reader an overview of rocks, rock properties, and rock engineering. Further information and more detailed coverage of the topic should be sought in textbooks and other publications such as Goodman (1989).

6.1

ROCK GROUPS AND IDENTIFICATION

A rock is a mixture of minerals (Sorrell and Sandström, 2001). You may wish to think of minerals as being the building blocks of the various rocks. The primary mineral groups forming rocks are silicates (e.g., feldspar and mica), oxides (e.g., quartz), carbonates (e.g., dolomite and calcite), and sulfates (e.g., gypsum). Some of the rare minerals are topaz, jade, and emerald (silicate); ruby and sapphire (oxides); and turquoise (phosphate). Diamond is pure carbon, so it is a basic element rather than a mineral. From the point of view of their origin, rocks are classified as igneous, sedimentary, or metamorphic. Igneous rocks (Figure 6.1) are formed by the cooling process of magma (i.e., granite and basalt). Granite is formed when viscous lava cools slowly. It is light in color and contains large elements such as quartz and feldspar. Basalt is formed by the rapid cooling of fluid lava. It is dark-colored and contains fine-grained elements undetectable by the naked eye. Sedimentary rocks (Figure 6.2) are formed by the weathering of a parent rock, when the weathered materials are transported and redeposited into a different setting and lithified back into rock by some form of cementation, pressure, or heat process. They are divided into clastic rocks (rocks made from particles of other rocks) and nonclastic rocks (rocks formed by chemical precipitates, often calcite). Sandstone, siltstone, mudstone, marl, and shale are clastic rocks, whereas limestone, dolomite, gypsum, lignite, and coal are nonclastic rocks.

Metamorphic rocks (Figure 6.3) are formed when the constituents of sedimentary and igneous rocks are changed by tremendous heat and pressure, with the possible influence of water and gases. The two main types of metamorphism processes involve temperature and pressure or temperature alone. Pressure alone is uncommon. In order of decreasing strength, marble, gneiss, slate, and schist are all metamorphic rocks. For identification purposes, the charts in Figures 6.4 and 6.5a–6.5c are very useful. Figure 6.4 helps in identifying the minerals that form a rock. It proceeds through a series of testing steps, including use of a hand lens to observe the rock-forming mineral; use of a knife and one’s fingernail to test the strength; and observation of the cleavage, the color, and the luster. Figures 6.5a–6.5c help in identifying the rock itself. They distinguish between rocks with a crystalline texture, rocks that have no grains visible and are uniformly smooth, and rocks with a clastic texture.

6.2 ROCK MASS VS. ROCK SUBSTANCE Rock mechanics makes a major distinction between rock substance and rock mass. Rock substance refers to a piece of intact rock with no fissures; rock mass refers to the entire mass of rock, including fissures and joints. There is usually a big difference between the tensile strength of an intact piece of rock (rock substance), and a weathered mass of rock (rock mass). In most cases the rock mass is much weaker than the rock substance. Therefore, a description of the joint pattern is very important, and should include joint spacing (less than 50 mm for very fractured rock to more than 3 m for solid rock), joint width, joint roughness, joint direction (using a rose diagram), and joint strength. Although it is easiest to measure the properties of the rock substance through laboratory testing, it is often more important to determine the behavior of the rock mass. This is the case for rock slopes, foundations on or in rock, and seepage through rock. An exception is the behavior of rock fill and rip rap, where the properties of the rock substance are critical.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

127

128

6 ROCKS

Granite

Felspar

Basalt

Figure 6.1 Igneous rocks: (a) granite, (b) feldspar, (c) basalt. (Source: Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.) Sandstone

Slate

Limestone

Siltstone

Figure 6.2 Sedimentary rocks: (a) sandstone, (b) slate, (c) limestone, (d) siltstone. (Source: Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.) Gneiss

Marble

Figure 6.3 Metamorphic rocks: gneiss, marble. (Source: Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.)

6.2 ROCK MASS VS. ROCK SUBSTANCE

129

Examine mineral in rock specimen using hand lens

Hardness:

Can be scratched by fingernail

One perfect cleavage

Cleavage:

Color and luster:

Glassy or sugary

Black

Name:

Gypsum

Graphite

CaSO4 • 2H2O

Cannot be scratched by a knife

Can be scratched by a knife but not by a fingernail

One perfect cleavage

Light colored

Green

Chlorite Muscovite

Carbon

Three good cleavages at 75° and 105°

Two good cleavages at 90°

No cleavage

Two good cleavages at 60°, 120°

Dark colored

Glassy or white

Glassy, gray, or white

White, gray, or pink

Dark glassy, or pearly

Dark glassy, or pearly

Biotite

Calcite, dolomite

Quartz

Feldspar

Pyroxene

Amphibole

CaCO3, Ca, Mg {CO3}2

SiO2 Network silicate

Sheet silicate

Chain silicates

Plagioclase

Orthoclase

NaAlSi3O8 CaAlSi2O8

KaAlSi3O8

Network silicates

Figure 6.4 Identifying rock minerals. (Source: Goodman (1989). Reproduced with permission of John Wiley & Sons, Inc.) Crystalline texture

Isotropic structure

Anisotropic structure

Softer than knife blade

Calcite

Calcite and dolomite

Halite

Gypsum

Anhydrite

Limestone

Dolomitic limestone

Rock salt

Gypsum

Anhydrite

Very dense, Green with calcite or sheared dolomite surfaces Marble*

Harder than knife blade Fine, uniform crystal size distribution Light colored Dark colored

Aplite Diabase

Mixed sizes; coarse with fine or very fine crystal sizes

Coarse, uniform crystal size distribution

Parallel needle shaped grains

Rhyolite Latite Andesite Basalt

Pegmatite Granite Grandodiorite Gabbro Peridotite

Amphibole schist and amphibolite

*May be anisotropic in hand specimen.

Mica absent

Green without sheared surfaces

Serpentinite*

Greenstone*

Altered peridotite

Hydrothermally altered diabase

Bands of light and dark layers

Parallel platey minerals

Gneiss

Schist

Mica is Continuous Chlorite disseminated mica Mica schist

Green schist

Figure 6.5a Identification of rocks with crystalline texture. (Source: Goodman (1989). Reproduced with permission of John Wiley & Sons, Inc.)

130

6 ROCKS

No grains visible, uniformly smooth

Isotropic structure

Anisotropic structure

Harder than knife blade

Softer than knife blade

No associated volcanic features

Associated volcanic features

Spheroidal weathering

Soluble

Hornfels or granulite

Feliste (light colored)

Claystone Siltstone Mudstone

Finegrained limestone

Weak Vitreous luster, Razor sharp edges after conchoidal fissile cleavage fracture structure Shale

Siliceous shale and cbert

Silvery sheen, no visible mica

Slate

Phyllite

Mica absent

Finely divided mica

Trap rock (dark colored)

Figure 6.5b Identification of rocks with no grains visible. (Source: Goodman (1989). Reproduced with permission of John Wiley & Sons, Inc.) Clastic texture Isotropic or anisotropic

Mainly volcanic pebbles and cobbles

Mainly nonvolcanic pebbles and cobbles

Agglomerate

Conglomerate

Angular blocks

Sand grains

Breccia Sandstone

Quartzite Greywacke Uniform quartz grains

Dirty sand with rock grains

Mainly volcanic sand (lapilli) and ash Tuff

Arkose Quartz feldspar (some mica and other minerals)

Figure 6.5c Identification of rocks with elastic texture. (Source: Goodman (1989). Reproduced with permission of John Wiley & Sons, Inc.)

6.3

ROCK DISCONTINUITIES

Rocks usually exhibit a network of discontinuities that significantly affect the mass behavior. Many words exist to refer to these discontinuities: fissures, cracks, fractures, joints, and faults (Priest, 1993). Fissures are the smallest and faults are the largest (Figures 6.6 and 6.7). Nonetheless, the two main types are joints and faults. Joints are created over geologic

time by bending of the rock mass, by vertical expansion, by horizontal stress relief (e.g., cliffs), by temperature differences, and sometimes by chemical action. Joints tend to exhibit a pattern. Faults are due to the movement of rock plates on a large scale and tend to be singular elements. These discontinuities introduce nonlinearities in behavior, stress dependency and anisotropy in properties, and weaknesses with regard to deformation and strength. Cementation in clastic rocks also significantly influences a rock’s properties; often the properties of the binder control the behavior of the rock mass, much as cement controls the behavior of concrete. If the network of joints is random (rare), it weakens the rock evenly, but if the joints are directional (common), the weakness is accentuated in the direction of the joints in shear and reduces the shear strength to the strength of the joint surfaces. The tensile strength of the rock mass perpendicular to the joint direction is reduced to a small fraction of the intact rock strength. Compression perpendicular to joints increases deformation compared to the intact rock but has little influence on strength. Another type of discontinuity is cavities and voids in the rock mass. These cavities most commonly form in limestone, dolomite, gypsum, and salt. Sinkholes in limestone occur in karst regions and can reach impressive dimensions. 6.4

ROCK INDEX PROPERTIES

Rock index properties include the dry unit weight of the rock substance and the porosity of the rock substance. The dry unit weight of the rock substance varies from a possible 21 kN/m3 for a shale or a limestone to a possible 27 kN/m3 for a marble or a granite. The most common values are between 25 and 26 kN/m3 . The porosity of rock substance is at most

6.5 ROCK ENGINEERING PROPERTIES

(a)

(b)

(c)

(d)

131

Figure 6.6 Fissures and joints. (Source: a: Lupin, CC BY-SA 3.0. c: Courtesy of Charles DeMets, University of Wisconsin-Madison. d: Alex Brollo, CC BY-SA 4.0.)

a few percent; exceptions include shale, sandstone, and schist, for which the porosity can reach that of soils at several tens of percent. The degree of weathering significantly affects the rock mass unit weight and porosity, with the lowest unit weights and highest porosities for the highest degree of weathering.

6.5 ROCK ENGINEERING PROPERTIES

Figure 6.7 A fault. (Source: Courtesy of The United State Geological Survey USGS, USA.)

Engineering properties of the rock substance include durability, hardness, permeability, modulus, and strength (Waltham, 1994). Although it is generally more important to know the properties of the rock mass, the first step is to find out the properties of the rock substance. An exception to this “rule” is when rip rap or rock fill has to be used for protection, as in scour or stability in rock-fill dams. The durability of a rock is measured by a test called the slaking durability test. Ten pieces of rock are weighed and placed in a rotating drum lined with a 2 mm opening mesh. The drum is slowly rotated through a water bath for 10 minutes and the rock pieces remaining after the test are weighed again. The ratio in percent of the weight after and before the

132

6 ROCKS

test is the slaking durability index I sd . Rocks typically have I sd values in excess of 90%. Values below 70 are undesirable for rip-rap applications. Hardness is a measure of how hard a surface is. For rocks, it may refer to the hardness of the parent mineral or the rock surface. Talc is one of the softest minerals, whereas diamond is the hardest known mineral. On Mohs scale of hardness, talc has a rating of 1, gypsum 2, quartz 7, and diamond 10. The hardness of a rock surface can be measured by using a Schmidt hammer. The Schmidt hammer generates an impact on the rock surface and the mass that impacts the surface rebounds to a measured height. The rebound height divided by the maximum height is called the rebound value R. The rebound value has been correlated to the unconfined compression strength and the modulus of rocks. The hydraulic conductivity, k, of a rock can be measured in the laboratory on an intact sample or in the field on the rock mass. The results are usually extremely different, with the field values being 10 to 100,000 times (or even more) larger than the laboratory values depending on the extent of the network of discontinuities in the rock mass. The densest intact rocks will have k values in the 10–10 to 10–15 m/s range, but volcanic intact rocks can have hydraulic conductivities in the range of 10–3 m/s. In the field, the hydraulic conductivity is drastically increased compared to the intact rock, as water could be gushing out of the joints of the rock mass. The k value can exhibit significant anisotropy depending on the direction of the joints. The modulus of deformation, E, of the rock substance is measured on samples in the laboratory, most commonly using the unconfined compression test. In the field, the plate test, the half-cylinder test, or the pressuremeter test can be used. Values of E for intact rock or rock substance are in the range of 2000 MPa to 100,000 MPa (concrete is around 20,000 MPa). The softer rocks include chalk and shale; the stiffer ones include granite and marble. The Poisson’s ratio of rocks is relatively small, with values ranging from 0.15 to 0.3. The strength of the intact rock, as measured by unconfined compression tests, can vary from more than 200 MPa for very hard rock to less than 10 MPa for very soft rock. Concrete has an unconfined compression strength of 20 MPa. Therefore, concrete is a soft to medium rock. The ratio between the rock modulus of deformation E and the unconfined compression strength qu is in the range of 150–600, with an average of 350. The lower values are found for the softer rocks (sandstone, shale), while the higher values are found for the harder rocks (marble, granite). The tensile strength of a rock can be measured indirectly by using a special splitting test called the Brazilian test. The values range from less than 1 MPa for a shale up to about 15 MPa for granite. The shear strength of intact rocks leads to cohesion intercepts in the range of 5 to 40 MPa and friction angles in the range of 30 to 50 degrees.

6.6

ROCK MASS RATING

Rock masses are rated by using indices that help in evaluating the relationship between the rock substance properties and the rock mass properties. Samples of rock are obtained by coring the rock, a process that consists of rotating an open steel tube or barrel with a coring bit (diamond) on the end of the steel tube wall. The tube is rotated into the rock at high speed while water is simultaneously injected for lubrication and cooling. Cores are retrieved and placed in core boxes. The recovery ratio (RR) is the ratio expressed in percent of the length of the core recovered divided by the length cored. The rock quality designation (RQD) is the ratio of the length obtained by adding all the pieces of core longer than 100 mm over the length cored. The velocity index I v is also a useful index to evaluate the difference between the rock substance properties and the rock mass properties. It is defined as the ratio of the square of the compression-wave velocity of the rock mass in the field to the square of the compression-wave velocity of the intact rock in the laboratory. Rock mass quality is excellent when the RQD is higher than 90% and the velocity index is higher than 0.8. Rock mass quality is very poor when the RQD is less than 25% and the velocity index is less than 0.2. The Unified Rock Classification System or URCS (Williamson, 1984) was developed to parallel the Unified Soil Classification System (USCS). It provides a systematic and reproducible method of describing rock weathering, strength, discontinuities, and density in a manner directly usable by engineers. The URCS is described in ASTM D5878. In 1989, Bieniawski proposed the rock mass rating (RMR) by combining several indicators of rock mass features. They include the strength of the rock substance (qu ), the rock quality designation (RQD), the joint spacing, the joint condition, the joint orientation, and the groundwater conditions. Table 6.1 shows the RMR categories. The RMR value is obtained by adding the ratings defined in each category. Rock mass classes I through V correspond to RMR values between 80–100, 60–80, 40–60, 20–40, and 0–20, respectively. A class I rock mass would be labeled a very good rock, whereas a class V rock mass would be considered very poor rock. Such classes can be correlated to estimated values of rock mass strength and safe bearing pressures, for example. Another and similar rock mass rating system exists and is called the Norwegian Q system. This system, created in 1974, is credited to Barton, Lien, and Lunde of the Norwegian Geotechnical Institute (1974). It is based primarily on the analysis of tunneling case histories and uses six parameters to assess the rock mass quality. The parameters are the RQD, the joint set number J n , the roughness of the joints J r , the degree of alteration and filling of the joints J a , the water inflow J w , and the stress reduction factor SRF. Using these six parameters, the Q factor is derived with the following equation: RQD × Jr × Jw Q= Jn × Ja × SRF

133

6.7 ROCK ENGINEERING PROBLEMS

Table 6.1

Rock Mass Rating (RMR) geomechanics system

Parameter Intact rock USCS, MPa rating RQD % rating Mean fracture spacing rating Fracture conditions rating Groundwater state rating Fracture orientation rating

Assessment of values and rating >250 15 >90 20 >2 m 20 Rough tight 30 Dry 15 Very favorable 0

100–250 12 75–90 17 0.6–2 m 15 Open 5 mm 0 Flowing 0 Very unfavorable –25

6.7

ROCK ENGINEERING PROBLEMS

Some common rock engineering problems include allowable pressures for foundations, ultimate side shear and ultimate point pressure for bored piles, slope stability, tunneling, excavations, blasting, rippability, and scour. The allowable pressure for shallow foundations in rock is sometimes used prescriptively. These values vary significantly and depend on the quality of the rock mass, which can be described by the RMR classes of the geomechanics system. Estimates of these allowable values are from 6000 to 10,000 kPa for a very good rock (Class I), 4000 to 6000 kPa for a good rock (Class II), 1000 to 4000 kPa for a fair rock (Class III), 200 to 1000 kPa for a poor rock (Class IV), and less than 200 kPa for a very poor rock (Class V). The settlement associated with these allowable pressures is usually calculated using elasticity theory; the main issue is obtaining the right modulus of deformation for the rock mass. The columns of the World Trade Center towers were on shallow foundations on rock (Figures 6.9–6.11). The towers weighed approximately 4500 MN each, were 417 m high, and had a footprint of 62 m × 62 m. The mica schist bedrock was found at a depth of about 21 m and exhibited inclined joints. Excavation took place so the shallow foundations could rest directly on the rock. The rock substance modulus was determined through laboratory tests and averaged 80,000 MPa. The rock mass was tested by a full-scale footing test which gave a

Structure Intact/ massive

Blocky

Very blocky Blocky/ disturbed/ seamy

Very poor

Poor

Decreasing surface quality 90

N/A 80 70 60 50 40 30 20

Disintegrated

Laminated/ sheared

Fair

Good

Very good

Gelogical strength index for jointed rocks (Hoek and Marinos, 2000)

Decreasing interlocking of rock pieces

In 1994, Hoek introduced the geologic strength index (GSI) to rate jointed rock masses. The GSI takes into consideration the interlocking of rock pieces or “blockiness” of the rock mass on the one hand and the condition of the rock surfaces or joints on the other (Figure 6.8). Then the GSI is used to extrapolate from the intact rock strength and modulus to the strength and modulus of the rock mass.

Surface conditions

Source: Waltham (1994)/Taylor & Francis. Rock mass rating (RMR) is the sum of the six ratings. Note that orientation ratings are negative.

10 N/A

Figure 6.8 Geologic strength index (GSI) for jointed rock masses. (Source: Adapted from Marinos and Hoek (2000).)

rock mass modulus equal to 1400 MPa or 1/57 of the rock substance value, due to the presence of joints. The design pressure for the footings was approximately 3000 kPa and the maximum pressure applied during the full-scale footing test was well over 3000 kPa.

134

6 ROCKS A

2.90 m

A

63.14 m

B

B

2.59 m 63.14 m

2.59 m

2.90 m Plan 2.29 m Fl. slab

1.68 m

2.90 m

Fl. slab

Sound rock Section A-A

Section B-B

Figure 6.9 Foundation plan for the World Trade Center. (Source: Courtesy of the Port Authority of New York and New Jersey.)

Figure 6.10 Photo of the foundation for the World Trade Center. (Source: The Port Authority of New York and New Jersey.)

The shallow foundations are shown in Figures 6.9 and 6.10; the total area was 1426 m2 . The calculated settlement for the foundation elements using elasticity theory and the measured modulus was very small, varying from 6 to 12 mm. Most of this settlement is likely to have happened during construction. The cliffs at the Pointe du Hoc site in Normandy, France, are made of interbedded layers of limestone and sandstone (see Chapter 2). These cliffs are eroded at their base by wave action from the sea; caverns develop at the base as a result of this wave action (Figure 6.12). When the caverns become deep enough, the overhanging rock mass fails. These failures allow back-calculation of the tensile strength of the rock mass. The tensile strength of the rock substance tested in the

Figure 6.11 Photo of the World Trade Center twin towers. (Source: The Port Authority of New York and New Jersey.)

laboratory by the Brazilian test (Figure 6.13, ASTM D3967) gave an average tensile strength of 3400 kPa in the limestone and 4500 kPa in the sandstone. The average tensile strength of the rock mass back-calculated from the overhang failures (Figure 6.14) indicated 40 kPa tensile strength, or about 1/100 of the rock substance value. Recommendations for the ultimate side shear values for a bored pile socketed in rock range from 300 kPa for a weak, fractured, decomposed rock (say, RQD = 20%) to 3000 kPa for a massive competent rock. Common formulas equate the ultimate side shear with the square root of the unconfined compression strength diminished by additional factors that take the rock mass quality into consideration. The ultimate bearing pressure at the bottom of a bored pile or a driven pile to rock can range from 4000 kPa for a poor rock mass

6.8 PERMAFROST

Figure 6.12 Jointed rock mass and caverns at Pointe du Hoc.

135

Figure 6.14 Massive collapse of rock cliff at Pointe du Hoc.

direction of sliding over the driving moment or driving force in the direction of sliding. Other rock engineering problems include tunneling, excavations, blasting, rippability, and scour.

6.8 PERMAFROST In areas of the Earth where the mean annual temperature of the air does not get above 0∘ Celsius, the soil may be permanently frozen down to a certain depth. These areas include the North Pole, the South Pole, and any mountain above about 5000 m high (Figure 6.15). The permafrost can

Figure 6.13 Brazilian tension test on Pointe du Hoc limestone.

quality to 400,000 kPa for a massive competent rock with a high strength. Of course, the ultimate bearing capacity may be limited by the strength of the pile itself. The ultimate bearing pressure is usually given as proportional to the unconfined compression strength of the intact rock diminished by a coefficient that takes the rock mass quality into consideration. In rock slope stability, the main influencing factors are the direction of the joints compared to the direction of the potential failure surface, the shear strength of the joints, and the water pressures in the joints. Failure analyses usually use planar surfaces and wedges following the joints’ contours. The failing mass is analyzed using fundamental laws and constitutive laws to give a factor of safety. The factor of safety is defined as the ratio of the resisting moment or resisting force in the

Figure 6.15 Zones of permafrost in the Northern hemisphere. (Source: Courtesy of NSIDC.)

136

6 ROCKS

be shallow (a few meters) or deep (several hundred meters). Because permafrost is rich in ice, its properties are very much tied to the properties of the ice. This implies that, much like ice, the strength and modulus of permafrost increase when the temperature decreases and are rate dependent. Note that the influence of temperature is much more important than the influence of rate effect. Permafrost also exhibits creep under sustained loading. Because ice is the binder that strengthens permafrost, like cement for concrete, the higher the degree of saturation of permafrost, the stronger the permafrost is. For construction on permafrost, it is best to isolate the building or structure from the permafrost so as to minimize the temperature changes induced by the structure in the permafrost. Indeed, when permafrost melts, it loses tremendous strength. This is why buildings and pipelines are elevated above permafrost ground through the use of piles (Figure 6.16).

Figure 6.16 Elevated structure on permafrost. (Source: Courtesy of Anadarko Petroleum Corporation.)

Problems and Solutions Problem 6.1 Answer the following questions: a. b. c. d.

What are the three main categories of rocks? Are granite, feldspar, and basalt igneous, metamorphic, or sedimentary rocks? Are sandstone, shale, limestone, and lignite igneous, metamorphic, or sedimentary rocks? Are gneiss and marble igneous, metamorphic, or sedimentary rocks?

Solution 6.1 a. b. c. d.

From the point of view of their origin, rocks are classified as igneous, sedimentary, or metamorphic. Granite, feldspar, and basalt are igneous rocks. Sandstone, shale, limestone, and lignite are sedimentary rocks. Gneiss and marble are metamorphic rocks.

Problem 6.2 Is diamond a rock or a mineral? Is there any rock harder than diamond? Is there anything harder than diamond? Solution 6.2 Diamond is pure carbon, so it is a basic element and is classified as a mineral. Diamond is the hardest mineral on the Mohs scale of mineral hardness, based on its resistance to scratching. There is no rock or other natural material harder than diamond. A few human-made materials—all made of carbon—have been claimed to be harder than diamond. Problem 6.3 What is the difference between rock mass and rock substance, and how does this difference affect the engineering properties? Solution 6.3 Rock substance refers to a piece of intact rock with no fissures; rock mass refers to a large volume of rock, including the fissures and joints. In most cases, the rock mass is much weaker than the rock substance. The presence of fissures and joints weakens the rock mass and affects all its engineering properties.

6.8 PERMAFROST

137

Problem 6.4 What is the typical range of values for the dry unit weight and porosity of a rock substance? Solution 6.4 The dry unit weight of rock substance varies from about 21 kN/m3 (e.g., a shale or a limestone) to about 27 kN/m3 (e.g., a marble or a granite). The most common values are between 25 and 26 kN/m3 . The porosity of rock substance is very low except for shale, sandstone, and schist, for which the porosity can reach that of soils (several tens of percent). Problem 6.5 How is the durability of a rock substance measured? Describe the test. Solution 6.5 The durability of a rock is measured by a test called the slaking durability test. Ten pieces of rock are weighed and placed in a rotating drum lined with a 2 mm opening mesh. The drum is slowly rotated through a water bath for 10 minutes and the dry weight of the rock pieces remaining after the test is measured again. The ratio in percent of the weight after and before the test is the slaking durability index I sd . Rocks typically have I sd values in excess of 90%. Values below 70% are undesirable for rip-rap applications. Problem 6.6 What are the typical range of modulus and Poisson’s ratio values for rock substance? How do this range and these values compare with concrete? Solution 6.6 Values of the modulus of deformation for rock substance range from 2000 MPa to 100,000 MPa and the Poisson’s ratio values of rock substance range from 0.15 to 0.3. By comparison, concrete has a modulus of 20,000 MPa, equivalent to that of a soft to medium rock. Problem 6.7 What is the typical range of unconfined compression strength for rock substance and its ratio to the rock substance modulus? What is the typical range of tensile strength for rock substance? How does this compare with concrete? Solution 6.7 The typical range of unconfined compression strength for rock substance is from 10 MPa for very soft rock to 200 MPa for very hard rock. By comparison, concrete has an unconfined compression strength of 20 MPa, equivalent to the strength of a soft rock. The ratio between the rock modulus of deformation E and the unconfined compression strength qu is in the range of 150 to 600, with an average of 350. The typical range of tensile strength for rock substance is 1 MPa to 15 MPa. The average tensile strength for concrete is about 2.5 MPa, so concrete is equivalent to a soft rock. Problem 6.8 Explain what the RR, the RQD, and the I v are and use the words excellent, good, fair, poor, or very poor to qualify ranges of values of these various indices. Solution 6.8 The recovery ratio (RR) is the ratio of the length of the core recovered divided by the length cored, expressed in percent. The rock quality designation (RQD) is the ratio of the length obtained by adding all the pieces with length longer than 100 mm over the length cored, expressed in percent. The velocity index I v is the ratio of the square of the compression-wave velocity of the rock mass in the field over the square of the compression-wave velocity of the intact rock in the laboratory.

138

6 ROCKS

Rock quality

RR

RQD (%)

Iv

Excellent Good Fair Poor Very poor

97–100 90–97 67–90 35–67 22500

Table 8.4

B A E0 (kPa) ER (kPa) P*L (kPa) qc (kPa) fs (kPa) N (bl/30cm)

E0 (kPa) 1 8 0.125 0.87 0.0174 0.0026

Correlations for sand

Column A = number in table × row B ER P* L qc (kPa) (kPa) (kPa) 0.125 1 0.0156 0.16 0.0032 0.00046

8 64 1 9 0.182 0.021

1.15 6.25 0.11 1 0.02 0.0021

fs (kPa) 57.5 312.5 5.5 50 1 0.104

N (bl/30 cm) 383 2174 47.9 479 9.58 1

180

8 IN SITU TESTS

Table 8.5 Column A = number in table × row B ER B E0 (kPa) (kPa) A E0 (kPa) ER (kPa) P* L (kPa) qc (kPa) fs (kPa) Su (kPa) N (bl/30cm)

1 3.6 0.071 0.40 0.079 0.010 0.0015

0.278 1 0.02 0.077 0.0038 0.0033 0.0005

Correlations for clay

P* L (kPa)

qc (kPa)

fs (kPa)

su (kPa)

14 50 1 5 0.25 0.133 0.02

2.5 13 0.2 1 0.05 0.037 0.0056

56 260 4 20 1 0.625 0.091

100 300 7.5 27 1.6 1 0.14

The applications of the PMT include the design of deep foundations under horizontal loads, the design of shallow foundations, the design of deep foundations under vertical loads, and the development of a modulus profile and the determination of other soil properties. The PMT is not very useful for slope stability and retaining structures. The advantages of the PMT are that it can be performed in most soils and rocks; that it stresses a larger soil mass than other tests; that it gives a complete stress-strain curve of the soil in situ, including cyclic loading and long-term loading; that it is relatively inexpensive; and that the quality of the test can be judged by the shape of the curve obtained. One drawback of the PMT is that the quality of the borehole influences the PMT parameters, in particular the first load modulus Eo .

8.4

DILATOMETER TEST

The dilatometer test (DMT) was developed in Italy in the mid-1970s and can be credited to Silvano Marchetti. The

N (bl/30cm) 667 2000 50 180 10.7 6.7 1

DMT (Marchetti, 1975; Briaud and Miran, 1992b; ASTM D6635) consists of pushing a flat blade located at the end of a series of rods (Figure 8.15) into a soil to a desired depth. The blade is 230 mm long, 95 mm wide, and 15 mm thick. Once the testing depth is reached, the operator uses gas pressure to expand horizontally into the soil a circular membrane located on one side of the blade. The membrane is 60 mm in diameter and expands 1.1 mm into the soil. Two pressures are recorded: po and p1 . The pressure po is the pressure on the blade before expansion, and p1 is the pressure required to produce the 1.1 mm expansion into the soil. A number of soil parameters are obtained from the DMT by using the formulas and correlations shown in Table 8.6. The applications of the DMT include the design of foundations, the determination of soil properties, and soil classification (Figure 8.16). The advantages of the DMT include that it is fast, economical, easy to perform, and reproducible, giving a wealth of soil properties through correlations. A drawback is that it cannot be used in soils that are difficult

Figure 8.15 Dilatometer test and equipment. (Source: Courtesy of Dr. Sylvano Marchetti, www.marchettidmt.it.)

8.5 VANE SHEAR TEST

Table 8.6

181

Soil parameters from dilatometer test

Symbol Description

Basic reduction formulae

p0

Corrected first reading

p0 = 1.05(A − Zm + ΔA) − 0.05 (B − Zm − ΔB)

p1 ID KD ED

Corrected second reading Material index Horizontal stress index Dilatometer modulus

p1 = B − Zm − ΔB ID = (p1 − p0 )∕(p0 − u0 ) ′ KD = (p0 − u0 )∕𝜎V0 ED = 34.7(p1 − p0 )

K0

Coefficient of Earth pressure in situ Overconsolidation ratio Undrained shear strength Friction angle Coefficient of consolidation Coefficient of permeability Unit weight and description Vertical drained constrained modulus

K0,DMT = (KD ∕1.5)0.47 − 0.6

u0 = pre-insertion pore pressure ′ 𝜎V0 = pre-insertion overburden stress ED is not a Young’s modulus E. ED should be used only after combining it with KD (stress history). First obtain MDMT = RM ED , then (e.g.) E′′ 0.8 MDmt for ID < 1.2

OCRDMT = (KD ∕1.5)0.47 − 0.6 ′ Cu,DMT = 0.22 𝜎V0 (0.5 KD )1.56 𝜙safe, DMT = 28 + 14.6 log Kd − 2.1 log2 Kd Ch, DMTA ≈ 7cm2 ∕Tflex

for ID < 1.2 for ID < 1.2 for ID > 1.8 Tflex from A–log t DMTA–decay curve

OCR cu 𝜙 ch kh 𝛾 M

U0

Equilibrium pore pressure

Z m = Gage reading when vented to atmosphere. However, if ΔA and ΔB are measured with the same gage used for current readings A & B, set Zm = 0 (Z m is compensated)

kh = Ch 𝛾w ∕Mh (Mh ≈ K0 MDMT ) (see chart) MDMT = RM ED If (ID ≤ 0.6)RM = 0.14 + 2.36 log Kd If (ID ≥ 3)RM = 0.5 + 2 log Kd If (0.6 < ID < 3)RM = RM,0 + (2.5 − RM,0 ) log Kd where RM,0 = 0.14 + 0.15(ID − 0.6) If Kd > 10 RM , = 0.32 + 2.18 log Kd If RM < 0.85, set RM = 0.85 U0 = p2 ≈ C − Zm + ΔA

In freely draining soils

(Source: Courtesy of Dr. Sylvano Marchetti, www.marchetti-dmt.it.)

to penetrate by pushing. Sample profiles are presented in Figure 8.17. 8.5

VANE SHEAR TEST

The vane shear test (VST) can be traced back to 1919 when it was first used in Sweden, but it is unclear if it can be credited to one person (Richards, 1988). The VST (Figure 8.18) is used to determine the undrained shear strength of fine-grained soils (clays and silts). It can be performed either in the field with

a field vane (ASTM D2573; Figure 8.19) or on the sample with a mini vane or a hand vane (ASTM D4648, Figure 8.20). The vane is made of two perpendicular blades, each having a 2-to-1 height-to-width ratio. The width of the field vanes varies from 38 to 92 mm; the larger vanes are used in softer soils. The width of the lab vanes varies from 10 to 20 mm. The VST consists of pushing a vane at the end of a rod into the soil until the desired depth is reached. For the lab vane, the top of the vane should be at least one vane height below the sample surface. Once the testing depth is reached, the vane is rotated

182

8 IN SITU TESTS

Soil description and estimated γ / γw 2000

500

Clay

5 2.1

Silt

1.8

1.8

1.8

1.7

1.7

* 1.7

1.6

D

50

2

2.1

1.9

1.9

100

Silty

5 1.9

Silty

5 2.0

200

Sandy

A B C D

n 1.737 2.013 2.289 2.584

m 0.585 0.621 0.657 0.684

Clayey

1000

Dilatometer modulus ED (bar)

Sand

Equation of the lines: ED = 10(n+m log ID)

C B

20

* 1.6

A

12 Mud 10 and/or Peat 5 0.1

0.8 1.2

0.33 0.6

Mud

3.3 1.8

(′) If PI>50, reduce γ by 0.1

1.5

1 2 0.5 Material index ID

0.2

5

Figure 8.16 Soil classification using the DMT. (Source: Courtesy of Dr. Sylvano Marchetti, www.marchetti-dmt.it.)

at a slow rate (less than 1 degree per minute) while measuring the torque developed and the rotation angle (Figure 8.21). The peak value of the torque is recorded as T max . Then the blade is rotated at least 10 times rapidly and a new maximum torque value, T res , is measured. The VST is used in saturated fine-grained soils to obtain the undrained shear strength su . The reason is that these soils z (m) .1 0

Material index

.5 1 Clay

SILT

z (m) 5 10 0 0

Constrained modulus

10

SAND

z (m) 20 0 0

have a low permeability and do not allow appreciable drainage during a test that typically lasts less than 10 minutes. The field vane is particularly useful in gassy sediments because it avoids the sample decompression associated with sample extrusion before testing. Therefore, in these saturated fine-grained soils, it is reasonable to assume that the shearing process is undrained and that the undrained shear strength su is the parameter being measured. For a rectangular vane, the following equation gives su from T max : ) ( H D (8.11) Tmax = 𝜋su D2 + 2 6 where D is the diameter of the vane and H is the height of the vane. Proof of this equation is shown in the solution to problem 8.4. The residual undrained shear strength sur is obtained from the same formula using T res : ) ( H D (8.12) Tres = 𝜋sur D2 + 2 6 The VST can be used in coarse-grained soils, but no useful result can be obtained. These soils drain fast enough that one would not be measuring the undrained shear strength, but instead the drained or partially drained shear strength. Back-calculating the shear strength parameters from this test would require knowledge of the normal effective stress on the plane of failure in addition to T max . This is not measured during the VST. The advantages of the VST include that it is fast, simple, economical, and useful for obtaining the undrained shear strength of fine-grained soils. A drawback is that it is limited to fine-grained soils where other methods are commonly used to obtain su . One exception is in offshore applications, where obtaining samples is very expensive and the decompression of methane-rich sediment samples can alter the true undrained strength of the soil in situ; in this case the VST is extremely useful. Undrained shear strength

40

z (m) 80 0 0

Horizontal stress index

2

4

6

z (m) 8 0 0

4

4

4

4

4

8

8

8

8

8

12

12

12

12

12

16

16

16

16

16

20

20

20

20

20

24

24

24

24

24

28

28

28

28

28

32

32

32

32

32

36

.6

1.8 Id

36

0

10 M (MPa)

20

36

0

40 Cu (kPa)

80

36

0

2

4 Kd

6

8

36

0

Shear wave velocity

200

400

200 Vs (m/s)

400

Figure 8.17 Example of dilatometer test results. (Source: Courtesy of Dr. Sylvano Marchetti, www .marchetti-dmt.it.)

8.5 VANE SHEAR TEST

Vane rods

Push in vane at bottom of borehole

Torquemeter

Lower vane to bottom of prebored hole Four-bladed vane shear Device: D = 62.5 mm H = 130 mm e = 2 mm

H = blade height

B = borehole diameter

df ≈ 4 B

blade width = D blade thickness = e

1. Insertion of vane

Vane shear test (VST)

2. Within 1 minute, rotate 3. Perform an vane at 6 deg./minute; additional 8 to measure peak torque, Tmax 10 revolutions

4. Measure residual torque Tres for remolded case

per ASTM D 2573:

Undrained shear strength: Suv = 6 T/(7πD3) For H/D = 2 In-situ sensitivity: St = Suv (peak)/Suv (remolded)

Figure 8.18 The vane shear test. (Source: From Mayne et al., 2002. Courtesy of Professor Paul Mayne, Georgia Institute of Technology.)

Figure 8.19 Field vane shear test. (Source: Courtesy of Dr. Dimitrios P. Zekkos.)

183

184

8 IN SITU TESTS

Figure 8.20 Laboratory vane shear test. (Source: a: Courtesy of ELE International, b: Courtesy of Impact Test Equipment Ltd.)

Failure surface

Torque (kN.m)

Vane

0

once at the testing depth, are pushed horizontally against the wall of the borehole under a chosen total stress 𝜎h . After a proper time for dissipation of the pore pressures generated by the application of 𝜎h , the device is pulled upward to shear the soil along the side of the borehole. The upward force applied increases and is measured as a function of time; the peak force

Tmax 10 rapid rotations TRes

θmax

Rotation angle (°)

Figure 8.21 Vane shear test results.

8.6

BOREHOLE SHEAR TEST

The borehole shear test (BST) was developed in the USA in the 1960s and is credited to Richard Handy (Handy, 1975, 1986). The BST (Figures 8.22 and 8.23) consists of drilling a borehole, removing the drilling tool, and inserting the borehole shear test probe down to the testing depth. The device is made of two diametrically opposed grooved plates, which,

Figure 8.22 Borehole shear test device. (Source: Courtesy of In-Situ Soil Testing, L.C.)

8.7 PLATE LOAD TEST Rod clamp Ring gear Bearing

Torque arm Crank

Shear gage

Dynamometer

Fill port

Pull rod

Not to scale Gas lines

Pull strap

Shear plate

c (kPa)

Shear force divided by area of plates τf (kPa)

Figure 8.23 Schematic of the borehole shear test device. (Source: Courtesy of Professor Richard L. Handy, Handy Geotechnical Instruments, Inc.)

τf3

Failure envelope given by stage BST

τf2 τf1 φ8 σh1

σh2

the friction angle of sands by direct measurements in the field. One drawback is that it is difficult to know exactly what pore pressures are generated. A pore pressure sensor on the plates helps in that respect. The phicometer developed by Philiponat (Philiponat, 1986; Philiponat and Zerhouni, 1993) is a similar tool.

8.7 PLATE LOAD TEST

Base plate

Shear head

185

σh3

Pressure applied on wall of borehole σN (kPa)

Figure 8.24 Results of a borehole shear test.

generated divided by the plates area gives the shear strength of the soil 𝜏f . If the shearing part of the test is performed slowly enough to ensure that no excess pore pressures arise, and if the soil has no effective stress cohesion intercept (c = 0), the ratio 𝜏f ∕𝜎h is equal to tan 𝜙′ and 𝜙′ can be measured with the BST. If the shearing part of the test is performed slowly enough to ensure that no excess pore pressures arise, and if the soil has an effective stress cohesion intercept (c′ > 0), a stage test can be performed where a second test at a higher value of 𝜎h follows the first one. The two tests give enough information to back-calculate c′ and 𝜙′ for the soil (Figure 8.24). If, however, the test is performed rapidly, and does not allow any drainage to take place in the soil, an undrained shear strength su of the soil is obtained. The advantages of the BST are that it is simple, economical, and one of the best tools—if not the only tool—to obtain

The plate load test or PLT (Figure 8.25; ASTM D1196 and D1195) is one of the simplest and oldest in situ tests. It consists of placing a circular plate with a diameter D on a prepared soil surface and loading the plate in steps while recording the settlement of the plate. The plate diameter is usually on the order of 0.3 m. Sometimes one or more unload-reload loops are performed during the test. All load steps and associated pressure p are held for the same period of time, during which readings of the plate settlement s are made as a function of time t. Loading the plate to soil failure is often desirable but not always possible. The load is measured with a load cell and the settlement is measured by using dial gages or electronic displacement devices (e.g., a linear variable differential transformer [LVDT]) attached to a settlement beam. It is critical that the supports of the settlement beam be far enough from the plate influence zone. Five plate diameters on each side seem appropriate. The result of the test is a load Q versus displacement s curve (Figure 8.26), which can also be presented in normalized form as the ratio of the average pressure p under the plate over a measure of soil strength SS versus settlement of the plate s over the plate diameter D. The soil strength SS can be the ultimate bearing pressure under the plate pu , the pressuremeter limit pressure pL , the cone penetrometer point resistance qt , the undrained shear strength su , the SPT blowcount N, or another measure of soil strength. The ultimate bearing pressure pu is often defined as the pressure reached when settlement of the plate is equal to 10% of the plate diameter. The advantage of plotting the results in this fashion (p/SS versus s/D) is that the results of the test become a property of the soil within the zone of influence of the plate and do not depend on the plate size or embedment (Briaud, 2007). The soil modulus E0 as measured during a plate test is obtained from the initial loading portion (O to A on Figure 8.26) or from the slope of the reloading part of the unload-reload loop Er (B to C on Figure 8.26). The equations to be used for E0 and Er , if the plate can be assumed to be rigid, are: (1 − v2 )𝜋pD (8.13) 4s (1 − v2 )𝜋 ΔpD (8.14) Er = 4Δs where E0 is the initial modulus from a plate load test, v is the Poisson’s ratio (to be taken as 0.5 if the plate test is fast enough that no drainage can take place during the test and 0.35 if the E0 =

186

8 IN SITU TESTS

Figure 8.25 Plate load tests. (Source: a: Photo by David Wilkins. Courtesy of Raeburn Drilling and Geotechnical (Northern) Limited; www.raeburndrillingnorthern.com. b: Courtesy of GEMTECH Limited, Fredericton, New Brunswick.) 1.6

120

1.4 Pressure (MPa)

Load (kN)

100 80 60 C

40

A

20

1.2 1.0 0.8 0.6 0.4 0.2

O

0

B

0

5

10

15

20

25

0.0 0.00

30

0.02

0.06

0.08

0.10

160

0.08 91.9 kN 84.1 kN 72.3 kN

62.8 kN 53.4 kN

91.9 kN 84.1 kN 72.3 kN

140

62.8 kN 53.4 kN

120

0.06 Displacement (mm)

Log displacement (S/S1)

0.04

Displacement over width, s/B

Displacement (mm)

0.04

100 80 60 40

0.02

20 0

0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Log time (t/t1)

0

5

10

15

20

25

30

35

Time (min)

Figure 8.26 Results of load test for 0.3-m-diameter plate on medium dense silty sand.

test is drained), p is the average pressure under the plate corresponding to the settlement s, D is the diameter of the plate in contact with the soil surface, Er is the reload modulus from a plate load test, and Δp is the pressure increment during the reload loop corresponding to the settlement increment ΔΔs. In addition to obtaining the soil modulus, sometimes the modulus of subgrade reaction is calculated from the plate test, as follows: K=

p in kN∕m3 s

(8.15)

Note that K is not a soil parameter, since it depends on the size of the plate according to Eq. 8.13: K=

4E0 (1 − v2 )𝜋D

(8.16)

Therefore, the modulus of subgrade reaction K measured with a plate of a given diameter D cannot be used for plates or footings that have diameters significantly different from D. It is also useful to plot the settlement of the plate s versus the time t for each load step on a log-log plot (Figure 8.26).

187

8.9 POCKET PENETROMETER AND TORVANE TESTS

The plot of log s versus log t is remarkably linear in most cases within the working load range. The slope of that line is called the viscous exponent n and allows one to predict by extrapolation the displacement at much longer times than the time taken to run the plate test, based on Eq. (8.17): ( )n s1 t = 1 (8.17) s2 t2 where s1 is the settlement after a time t1 and s2 is the settlement after a time t2 and n is the slope of the log s versus log t curve for the load step corresponding to s1 . Alternatively, the soil modulus E0 or Er can be written as: ( )−n E1 t = 1 (8.18) E2 t2 The advantage of the plate load test is that it is very simple and economical to perform. The drawback is that it only tests a zone of soil near the ground surface (one to two plate diameters deep), although larger depths can be reached by performing the test at the bottom of open pits.

of the plate on the reference soil (crushed California limestone) has been measured to be 6900 kPa. So, the reference pressure is 6900 kPa and the CBR number is a percentage given by: 100 × p(kPa) (8.19) CBR = 6900 This test is used primarily for pavement design, where the depth of influence of the plate is similar to the depth of influence of a truck tire. If the CBR value is less than 3%, the soil is too soft for road support without modification, values between 3% and 5% are average, and values from 5% to 15% are good. Crushed rock values are around 100%. Several correlations have been developed to link the CBR to soil properties, such as: (8.20) Mr (kPa) = 10,000 × CBR su (kPa) = 11 × CBR

(8.21)

where M r is the resilient modulus and su is the undrained shear strength. 8.9 POCKET PENETROMETER AND TORVANE TESTS

8.8

CALIFORNIA BEARING RATIO TEST

The California bearing ratio test (CBR) is a form of plate test (Figure 8.27). It can be performed in the field or in the lab. In the field (ASTM D4429), it consists of placing a 254 mm diameter plate weighing 44.5 N on the ground surface and loading it until the settlement s is 2.5 mm. The load Q corresponding to a settlement s of 2.5 mm is divided by the plate area to get the pressure p. The California bearing ratio is the ratio between p and the pressure necessary to reach a settlement s of 2.5 mm on a reference soil (crushed California limestone). The pressure necessary to create 2.5 mm of settlement

Figure 8.27 CBR test in the field. (Source: Courtesy of A F Howland Company.)

A number of simple tests can be performed on the sample in the field as soon as it is retrieved from the borehole. They are typically performed on the end of samples taken with a Shelby tube. These tests include the pocket penetrometer, the torvane, and the hand vane tests. The pocket penetrometer test (PPT) (Figure 8.28a) consists of pushing by hand the end of a spring-loaded cylinder 6.35 mm in diameter until the ultimate bearing pressure is reached. The compression of the spring increases as the force increases and a floating ring on the body of the pocket penetrometer (PP) indicates how much force is exerted. The ultimate pressure is reached when the cylinder penetrates without further increase in the PP reading. The PP number ranges from 0 to 4.5 and gives an estimate of the unconfined compression strength of the soil in tsf. The PP number has been correlated with the undrained shear strength of clays (su (kPa) ∼ 30 PP), but the scatter in this correlation is very large—not to mention the fact that the mass of soil tested is extremely small. The advantage of the PPT is that it is a very simple test that gives a quick indication of the soil strength. The drawback is that it tests only a very small zone of soil and thus must not be used in design. The torvane test (TVT) (Figure 8.28b) consists of pushing a set of vanes about 6.5 mm into the face of the sample and then rotating the spring-loaded cap until the spring releases because the shear strength of the soil has been reached. A maximum value indicator stays at the maximum reading reached during the rotation and indicates the shear strength of the soil. The hand vane shear test (VST) (Section 8.5, Figure 8.20) is also a simple and quick test that can be performed on the end of a Shelby tube sample. These three simple tests are mostly used on silts and clays. Of the three, the hand vane described in Section 8.5 is the most reliable.

188

8 IN SITU TESTS

(a)

(b)

Figure 8.28 Pocket penetrometer and torvane: (a) Pocket penetrometer (Encyclopedia.com, www .encyclopedia.com/video/PBoOUDVWhSo-hand-penetrometer-test.aspx, last accessed under September 07, 2022). (b) Torvane (Encyclopedia.com, www.encyclopedia.com/video/PBoOUDVWhSo-hand-penetro meter-test.aspx, last accessed under September 07, 2022).

8.10

POCKET ERODOMETER TEST

The pocket erodometer test (PET) (Figure 8.29, Briaud, Bernhardt, and Leclair, 2011) is to erosion resistance what the pocket penetrometer test is to shear resistance. The pocket erodometer (PE) is a regulated mini-jet-impulse-generating device. The water jet comes out of the nozzle at 8 m/s and is aimed horizontally at the vertical face of the sample. Verification that the velocity is 8 m/s when leaving the nozzle is achieved by aiming the jet from a height H (Figure 8.29), measuring the distance x where the water reaches the floor, and using the following equation: x v0x = √

(8.22)

2H g

where v0x is the velocity at the nozzle and g is the acceleration due to gravity. The depth of the hole in the surface of the sample created by 20 impulses of water is recorded. The depth of the hole is entered in the erosion chart (shown in Figure 8.30) to determine the erodibility category of the soil. This erosion category allows the engineer to make preliminary decisions in erosion-related work. The advantage of the

PET is its simplicity; its drawback is that it tests a very small portion of the soil.

8.11

COMPACTION CONTROL TESTS

Soil compaction is one of many techniques of soil improvement and is discussed in Chapter 21. In short, the soil to be used at the site is tested in the laboratory where compaction tests are performed. The results of these tests are used to establish the target values (dry unit weight, modulus, water content) to be achieved during the compaction process in the field. In the field it becomes necessary to verify that the target value has been reached. These in situ tests include tests to measure the dry unit weight (e.g., sand cone method, rubber balloon method, nuclear density probe), water content (e.g., nuclear density probe, field oven test), and soil modulus (e.g., BCD, falling weight deflectometer). 8.11.1

Sand Cone Test

The sand cone test (SCT; Figure 8.31) consists of digging a hole in the ground, obtaining the weight and the volume of the soil excavated, drying the soil and obtaining the dry weight,

75 mm

100 mm

H

X

d/2

d/2

Figure 8.29 Pocket erodometer test.

8.11 COMPACTION CONTROL TESTS

189

100000 Very high erodibility I > 75 mm

Erosion rate (mm/hr)

10000 1000 100

High erodibility II 75-15 mm

Medium erodibility III 15-1 mm

MH CL

SP ML

10

CH

SM 1

Air valve

Low erodibility IV < 1 mm

Rock

Very low erodibility V

Flat surface

No noticeable erosion

0.1 0.1

1

10

Fill

100

Velocity (m/s)

Figure 8.30 Erosion chart for various erosion depths from the PET.

Water Rubber balloon

Figure 8.32 Field unit for testing weight and water content by rubber balloon. (Source: b: Courtesy of Humboldt Mfg. Co.)

volume of the soil excavated is measured in a different way. The rubber balloon device is a cylinder filled with water up to a level indicated on a graduated scale. At the bottom of the cylinder is a rubber balloon that can be expanded into the hole below by pumping water into it. When the balloon fills the hole, the reading on the graduated scale on the cylinder gives the volume of the hole. The data reduction is the same as for the sand cone test. 8.11.3 Figure 8.31 Field unit weight and water content by sand cone test. (Source: b: Courtesy of Durham Geo Slope Indicator.)

and calculating the water content and the dry unit weight. More specifically, a standard steel plate with a 172 mm diameter hole through it is placed on the ground surface. A hole is dug into the ground through the hole in the steel plate to a depth of about 150 mm. The excavated soil is weighed, then dried, then weighed again. This gives the water content of the soil that was in the hole. As soon as the hole is excavated, an inverted funnel in the form of a cone is placed on top of the opening in the base plate and a bottle full of sand of known unit weight is connected to the top of the funnel. (The weight of the bottle full of sand is measured beforehand.) The valve between the bottle and the funnel is then opened and the sand of known unit weight flows out of the bottle until the hole in the ground and the funnel above it are full. The valve is closed, the bottle is disconnected, and the bottle is weighed again. The difference in weight of the bottle before and after filling the hole, divided by the known unit weight of the sand, gives the volume of the hole plus the funnel. Because the volume of the funnel is known, the volume of the hole can be deduced, and the dry unit weight is obtained from the dry weight and the volume of the soil in the hole. 8.11.2

Rubber Balloon Test

The rubber balloon test (RBT; Figure 8.32) follows exactly the same procedure as the sand cone method except that the

Nuclear Density/Water Content Test

The nuclear density/water content test is a device to measure indirectly the density and water content of a soil at the soil surface. It consists of sending radiation from a source into the soil and counting the amount of radiation coming back to a detector. In the case of the nuclear density test, a source generating medium-energy gamma rays is used. These gamma rays send photons into the soil (photons are particles of light; see Section 9.4.1). These photons go straight to the detector, or bump into the soil particles (Compton scattering) and deflect to arrive or not arrive at the detector. The gamma rays arriving at the detector are counted, and the gamma count is inversely proportional to density. In the case of the water content test, a source generating high-energy neutrons is used. The principle is that when a high-energy neutron hits a much heavier atomic nucleus, it is not slowed down significantly. However, if it hits an atomic nucleus that is about the same weight as the neutron, then the neutron is slowed down significantly. The hydrogen atom has a nucleus that is very comparable in weight to the neutron, and therefore is very good at slowing neutrons down. Because water has a lot of hydrogen, counting the number of slow neutrons coming back to a detector will indicate how much water is in the soil. The test can be done in direct transmission or in backscatter mode. In the direct transmission mode, the source rod penetrates into the soil anywhere from 75 mm to 220 mm (Figure 8.33); the detector is on the bottom side of the nuclear gage. This mode is preferred for density measurements. In the backscatter mode, the nuclear gage sits on the soil surface and the source and detectors are on the bottom side of the gage (Figure 8.33). This is the mode used for water content

190

8 IN SITU TESTS

Nuclear test

Gage

Gage

Detectors

Detectors Source

Source Direct transmission

Figure 8.34 Field oven test (FOT) for water content. Backscatter

Figure 8.33 Nuclear density probe test for unit weight and water content.

determination. The nuclear gage is calibrated by the manufacturer initially and after any repair. The calibration consists of placing the gage on a sufficiently large block of material of known density and known water content. 8.11.4

Field Oven Test

The field oven (Figure 8.34) is a device that is used to determine the water content of a soil in the field. A small piece of soil is carved from the soil surface; the sample is placed between the two plates of the field oven, which looks like a waffle maker. A load cell located below the heating pad gives the weight of the sample. Then the two plates are closed and the oven dries the soil sample. After a few minutes, the soil is dry and the heating plates are opened. The load cell records the dry weight of the sample and the water content is displayed.

8.11.5

Lightweight Deflectometer Test

The lightweight deflectometer (LWD) test (Figure 8.35) (ASTM E2583) consists of dropping a weight guided along a rod from a chosen height onto a plate resting on the ground surface. The typical values for the LWD are a weight of 100 N, a drop height of 0.5 m, and a plate diameter of 0.2 m. During the impact, a load cell located above the plate measures the force versus time signal and a geophone attached to the plate measures the deflection of the plate. The soil modulus E is back-calculated using the theory of elasticity with the peak force F and the peak deflection Δ. 4F (8.23) 𝜋DΔ where E is the soil modulus measured by the LWD, f is a plate rigidity factor (1 for flexible plates and 0.79 for rigid plates), v is the Poisson’s ratio (range from 0.3–0.45, depending on soil type), F is the maximum force on the force versus time E = f (1 − v2 )

Guide rod Drop weight Shock absorber Load cell Geophone Loading plate (a)

(b)

Figure 8.35 Falling weight deflectometer for soil modulus: (a) Principle. (b) Equipment. (Source: b: Courtesy of Minnesota Department of Transportation.)

8 7 6 5 4 3 2 1 0

0.6 Load Deflection

0.45 0.3 0.15

0

5

10

15

20

25

30

35

40

45

50

55

Deflection, ∆ (mm)

Load, F (kN)

8.12 HYDRAULIC CONDUCTIVITY FIELD TESTS

0 60

Time, t (ms)

Figure 8.36 Falling weight deflectometer data.

plot, D is the plate diameter, and Δ is the maximum displacement on the displacement versus time plot. For example, referring to the flexible plate LWD test in Figure 8.36, the modulus would be calculated as: E = 1(1 − 0.352 ) 8.11.6

4 × 7.5 = 76.3 MPa (8.24) 𝜋 × 0.2 × 0.55 × 10−3

BCD Test

A modulus E can also be obtained with a device called the BCD (Figure 8.37). It consists of a 150 mm diameter, 2 mm-thick flexible steel plate at the bottom of a rod with handles—a kind of scientific cane. Strain gages are mounted on the back of the plate to record the bending that takes place during the loading test. When the operator leans on the handle, the load on the plate increases and the plate bends. If the soil is soft (low modulus), the plate bends a lot. If the soil is hard (high modulus), the plate does not bend much. The amount of bending is recorded by the strain gages and is correlated to the modulus of the soil below. The test is called the BCD test or BCDT (Briaud, Li, and Rhee, 2006) and is performed as follows. First, the BCD plate is placed on top of the ground surface (Figure 8.37). Then the operator leans on the handles of the BCD and the vertical load increases. When the load goes through 223 N, a load sensor

600 mm

Acquisition processing & readout

triggers the reading of the strain gages. The device averages the strain gage values, uses the internal calibration equation linking the strains to the modulus, and displays the modulus E. This evaluates the level of compaction achieved at that location. The modulus obtained with the BCD corresponds to a reload modulus, to a mean stress level averaging about 50 kPa within the zone of influence, to a strain level averaging 10–3 within the zone of influence, and to a time of loading averaging about 2 s. The BCD test can also be performed in the laboratory on top of the compaction mold to obtain the modulus versus water content curve in parallel with the dry density versus water content curve (see Chapter 21, Section 21.2).

8.12

HYDRAULIC CONDUCTIVITY FIELD TESTS

The purpose of these hydraulic conductivity in situ tests is to measure the hydraulic conductivity k (m/s) of the soil. The soil can be either below the groundwater level (saturated), or above the groundwater level (saturated by capillary action or unsaturated). For saturated soils below the GWL, several tests exist, including the borehole tests (falling head test, rising head test, constant head tests), the pumping test, and the cone penetrometer dissipation test. For soils above the GWL, the tests include the sealed double-ring infiltrometer (SDRI) test and the two-stage borehole permeameter. 8.12.1

Borehole Tests

Borehole tests consist of drilling a borehole, changing the water level in the borehole, and recording the movement of the water level in the borehole as a function of time. Sometimes the borehole is cased to help in keeping the borehole stable. The data collected are used to back-calculate the hydraulic conductivity k. The equations to calculate k are based on developing the governing differential equation for the problem and then solving it while satisfying the boundary conditions. This is where the problem becomes quite complicated and requires charts or software. The following examples are cases in which the geometry is simple. When the soil layer is deep and uniform, when the casing goes down to the bottom of the borehole, and when the water is bailed out so that the water level starts far below the groundwater level outside of the casing (Figure 8.38), the hydraulic conductivity k is obtained from the equation: h

Load-cell

2 mm

150 mm (a)

191

k=

Strain-gage instrumented plate

(b)

Figure 8.37 BCD test for soil modulus: (a) Principle. (b) Equipment.

2𝜋r Ln h1 2

11(t2 − t1 )

(8.25)

where r is the radius of the casing, h1 and h2 are the distances from the groundwater level in the soil deposit outside of the casing to the level of the water in the casing, and t1 and t2 are the times at which h1 and h2 are measured. This equation applies when the depth D as shown in Figure 8.38 is between 0.15 m and 1.5 m. In the case where the pervious soil layer to be tested is underlain by an impervious layer, where the uncased boring

192

8 IN SITU TESTS

Phreatic surface Q Initial Phreatic during pumping surface R

2r GWL D

h1

h2

Casing

Drawdown r1 Sands

r2

Well

Figure 8.38 Inflow well test in deep uniform soil. (Source: Adapted from Hunt (1984).)

H h1

h2

Impervious

Figure 8.40 Pumping test in sand layer using three borings. (Source: Adapted from Hunt (1984).)

Phreatic surface Q Initial Phreatic surface during pumping r R

r

Drawdown

R

H h

Sand

Aquiclude (Clayey Soil)

Well

H

h

Impermeable

Well

Aquifer (Sands)

Figure 8.39 Pumping test in sand layer using one boring. (Source: Adapted from Hunt (1984).)

(or screened boring) penetrates through the entire pervious layer all the way to the impervious layer, and where the water level is maintained constant by pumping at a flow rate Q (as shown in Figure 8.39), the hydraulic conductivity k is obtained from the equation: k=

Q Ln Rr 𝜋(H 2 − h2 )

(8.26)

where Q is the flow rate pumped out of the well to maintain the water level constant in the well, r is the radius of the borehole, R is the radius of the zone of influence where the water table is depressed, H is the vertical distance between the bottom of the boring (impervious layer) and the groundwater level at or further than R, and h is the vertical distance between the bottom of the boring and the water level in the borehole. Note that for this equation to apply, a steady-state flow must be reached; this may take a time related to the hydraulic conductivity itself. Finding the value of R requires some borings down to the groundwater level away from the test boring. To improve the precision of this test, observation borings can be drilled at radii r1 and r2 from the test boring and the vertical distances h1 and h2 between the bottom of the boring (impervious layer) and the water level in the observation borings recorded (Figure 8.40). Then Eq. (8.26) becomes: r

Q Ln r2 k = ( 2 1 2) 𝜋 h2 − h1

(8.27)

In the case where the pervious layer to be tested is sandwiched between two impervious layers, where the uncased

D

Figure 8.41 Pumping test in confined aquifer. (Source: Adapted from Hunt (1984).)

boring (or screened boring) penetrates through the first two layers and stops at the top of the second impervious layer, and where the water level is maintained constant by pumping at a flow rate Q (as shown in Figure 8.41), the hydraulic conductivity k is obtained from the equation: Q Ln Rr

(8.28) 2𝜋D(H − h) where Q is the flow rate pumped out of the well to maintain the water level constant in the well, r is the radius of the borehole, R is the radius of the zone of influence where the water table is depressed, H is the vertical distance between the bottom of the boring (top of the second impervious layer) and the groundwater level at or further than R, and h is the vertical distance between the bottom of the boring (top of the second impervious layer) and the water level in the borehole. Note that for this equation to apply, a steady-state flow must be reached; this may take a time related to the hydraulic conductivity itself. Finding the value of R requires some borings down to the groundwater level away from the test boring. To improve the precision of this test, observation borings can be drilled at radii r1 and r2 from the test boring and the vertical distances h1 and h2 between the top of the second impervious layer and the water level in the borehole recorded (Figure 8.42). Then Eq. (8.28) becomes: k=

r

k=

Q Ln r2 1

2𝜋D(h2 − h1 )

(8.29)

8.12 HYDRAULIC CONDUCTIVITY FIELD TESTS

r

(spherical stress tensor); the other is due to the water stress response Δud to the shearing of the soil element (deviatoric stress tensor). When the soil element is subjected to an all-around mean pressure, Δus is always positive, but when the soil element is subjected to a shear stress, Δud can be positive or negative depending on the change in volume of the element during shearing. If the soil element decreases in volume during shearing, it is called contractive, Δud is positive, and both Δus and Δud decrease as a function of time (Figure 8.43a). If, however, the soil element increases in volume during shearing, it is called dilatant, and Δud is negative. The combination of Δut decreasing with time and Δud increasing with time (becoming less negative) leads to a bump on the decay curve (Figure 8.43b). The initial pore pressure when recording starts is ui . Note that two ui values exist depending on the location of the pore-pressure measuring device. In the case of a monotonic decay and for the pore-pressure measurement right behind the cone point (shoulder), Parez and Fauriel (1988) proposed a correlation between t50 and the hydraulic conductivity k (Figure 8.44), which is well represented by the equation: )1.25 ( 1 (8.30) k(cm∕s) = 251 t50 (s)

R r1

Aquiclude (Clayey Soil)

r2

H

h h1

Aquifer (Sands)

h2

D

Well

Figure 8.42 Pumping test in confined aquifer using three borings.

Solutions for more complicated geometries are found in Mansur and Kaufman (1962) and in Cedergren (1967). The advantages of these tests are that they give a large-scale value of k in the field which includes the mass features of the soil deposit. Some of the drawbacks are the lack of control over problems such as filter cake development around the wall of the borehole, and quick conditions development in high-gradient situations. 8.12.2

Cone Penetrometer Dissipation Test

The cone penetrometer dissipation test (CPDT) is performed during a CPT sounding and makes use of the cone point equipped with a pore pressure measuring sensor: a piezocone. The piezocone is pushed to a depth below the groundwater level where the measurement of k has to be made, the penetration stops, the initial excess pore pressure is read, and then the decay of excess pore pressure versus time is recorded. Two situations can arise: heavily overconsolidated soil or normally to lightly overconsolidated soil. In the case of normally consolidated to lightly overconsolidated soil, the decay of excess pore pressure will be monotonic (Figure 8.43a). In the case of heavily overconsolidated soils, the response shows first an increase in excess pore pressure followed by a decrease (Figure 8.43b). The reason for this dual behavior is that the total excess pore pressure Δut has two components: one is due to the water stress response Δus to the mean all-around compression of the soil element

where k is the hydraulic conductivity in cm/s and t50 is the time in seconds to reach a decrease in water stress equal to 50% of the total decrease in water stress. A typical example is shown in Figure 8.43a for a lightly overconsolidated clay. The time to 50% dissipation is found halfway between the initial value ui (t = 1s in Figure 8.43a) and the equilibrium value corresponding to the hydrostatic pressure u0 . In the case of a decay curve exhibiting a rise followed by a decay (highly overconsolidated soil), obtaining the hydraulic conductivity k from the dissipation curve is more complicated (Burns and Mayne, 1998). 8.12.3

600 Normally consolidated to lightly overconsoilidated

400 ∆ us

∆ut

200

u ∆ ud

0 1

u0 10

100

1000

–200 Time t(s)

(a)

10,000

100,000

Pore pressure u(kPa)

Pore pressure u(kPa)

Sealed Double-Ring Infiltrometer Test

The sealed double-ring infiltrometer test (SDRIT) was developed in the late 1970s in the USA and is credited to Steve

600 ui

193

400

ui ∆ut

200

–400

u

u0

0

–200

Heavily overconsolidated

∆us

1

∆ud

10

100

1000

Time t(s)

(b)

Figure 8.43 Decay of excess pore pressure in piezocone dissipation test.

10,000

100,000

194

8 IN SITU TESTS

Hydraulic conductivity, k (cm/s)

1.E-01

Trautwein and David Daniel (1994). The SDRIT aims at measuring the hydraulic conductivity at shallow depth in soils above the groundwater level. A typical situation is testing to obtain the hydraulic conductivity k of a 1 m-thick clay liner above a free-draining layer of sand and gravel. The test setup starts by placing a square outer ring about 4 m in size on the soil surface and embedding and grouting the walls of the ring about 0.45 m below the surface (Figure 8.45). Then an inner ring is placed in the center of the outer ring and the walls are embedded and grouted about 0.15 m into the ground. The outer ring is open to the atmosphere while the inner ring is sealed. A tube goes from the inner ring to a deformable plastic bag, where it can be easily connected and disconnected. The bag is filled with water and weighed, and the entire system is saturated with water. The SDRIT is often used to test soils that are not saturated, in which case tensiometers are placed at different depths to measure the tension in the water within the layer being tested (see Chapter 11 on water stress for an explanation of how tension occurs in the soil water and Chapter 10 on laboratory tests for an explanation of how tensiometers work). As the water seeps through the unsaturated soil layer

Sand and gravel

1.E-02

Sand

1.E-03

u2 Silty sand to sandy silt

1.E-04 1.E-05

Silt 1.E-06 Clay

Parez & Fauriel (1988)

1.E-07 1.E-08 0.1

1

10

100 t50 (sec)

1000

10000

Figure 8.44 Relationship between t50 and hydraulic conductivity for piezocone dissipation test. (Source: From Mayne, Christopher, Berg, and DeJong, 2002. Courtesy of Professor Paul Mayne, Georgia Institute of Technology.)

Dial gages for heave measurement

Reference for swelling measurement

4.0 m 1.5 m

Outer ring

Tensiometer H

Inner ring

Tubing Plastic bag

0.15 m

0.45 m

Dw Wetting front

D Bentonite grout

Low k layer (e.g. compacted clay) Drainage layer (Pressure head = 0)

Figure 8.45 Sealed double-ring infiltrometer. (Source: Courtesy of Professor Xiaodong Wang, University of Wisconsin.)

8.12 HYDRAULIC CONDUCTIVITY FIELD TESTS

below the SDRI, the water fills the voids in the soil, thereby saturating the soil; a wetting front advances and the plastic bag loses water. The volume of water Q leaving the plastic bag and entering the soil is measured by weighing the bag as a function of time. Reducing the data of an SDRIT requires knowledge of water flow through saturated and unsaturated soils (see Chapter 14). Obtaining the hydraulic conductivity k from the SDRIT requires some assumptions: (1) steady-state seepage; (2) vertical, one-dimensional flow; and (3) saturated conditions. If the soil is unsaturated to start with, it will take time for the water to permeate through the soil layer thickness and saturate the soil. This time can be several weeks. To obtain the hydraulic conductivity k from the SDRIT data, the following equations are used: v=ki

(8.31)

This is called Darcy’s law and is explained in Chapter 14 on flow through soils; v is the discharge velocity; and i is the hydraulic gradient, defined as the loss of total head Δht of the flowing water per distance traveled Δz. Δht Δz Conservation of mass leads to: i=

Vf = vd At

(8.32)

(8.33)

where V f is the volume of water that has infiltrated the soil in a time t, A is the plan view area of the inner ring, and vd is the discharge velocity. This leads to an expression for k: Vf

k=

At Δht Δz

(8.34)

If the test is run long enough that the whole layer becomes saturated, then Δht is the vertical distance from the bottom of the layer to the level of the water in the outer ring and Δz is the thickness of the layer. The tensiometer readings help in deciding when this stage has been reached. If this assumption is made but the wetting front has not penetrated the whole layer, then i will be underestimated and the k value obtained will be lower than the true k value. If the test does not reach this stage and the waterfront has penetrated to a depth Dw below the top of the soil surface, the value of Δz is Dw and the value of Δht is: Δht = H + Dw + hp

(8.35)

where hp is the tension in the water on the wetting front expressed in height of water. This value can be obtained from the tensiometer readings. Here two assumptions can be made: (1) hp is given by the tensiometers, or (2) hp = 0. In practice, the second assumption seems to give more acceptable results, especially as the test is often run to prove that the hydraulic conductivity of the soil layer is lower than 10–9 m/s (clay liner for waste disposals). Indeed, with assumption 2(hp = 0), Δht is underestimated and k is overestimated.

195

When the layer being tested swells, it is necessary to take the swelling into account. In this case, some of the water leaving the plastic bag is stored in the swelling process while some of the water is seeping through the soil. Ignoring the swelling component would give an overestimated value of V f and therefore an overestimated value of k. The volume of water V s used to increase the volume of the soil through swelling is measured as follows: A reference beam is set up above the SDRI (Figure 8.45) and the vertical movement of the inner ring is recorded with respect to that beam (using dial gages, for example). The volume V s corresponding to the vertical movement of the inner ring is subtracted from the volume of water V t leaving the plastic bag to obtain the true volume V f . flowing through the soil. 8.12.4

Two-Stage Borehole Permeameter Test

The two-stage borehole permeameter test (TSBPT) was developed in the USA in the 1980s and is credited to Gordon Boutwell (Boutwell and Derick, 1986). The TSBPT aims at measuring the vertical and horizontal hydraulic conductivity at shallow depth in soils above the groundwater level. A typical situation is testing to obtain the hydraulic conductivity k of a 1 m-thick clay liner above a free-draining layer of sand and gravel. The test takes place in two stages. Stage 1 consists of drilling a hole (for example, 0.5 m deep and 0.1 m in diameter), inserting a permeameter (e.g., open PVC 75 mm diameter pipe with graduated cylinder above, Figure 8.46a) in the open hole, sealing the permeameter to the walls of the borehole by grouting, and keeping the bottom of the boring open and intact. Once the borehole is sealed, the test consists of filling the permeameter with water and letting the water seep into the soil through the bottom of the casing. The drop in water level in the graduated tube is recorded as a function of time. The hydraulic conductivity k1 from stage 1 is calculated from the following equation (Hvorslev, 1949): k1 =

h 𝜋d2 Ln 1 11D(t2 − t1 ) h2

(8.36)

where d is the diameter of the graduated tube above the permeameter, D is the diameter of the permeameter, and h1 and h2 are the heights of water above the bottom of the casing recorded at times t1 and t2 respectively. The k1 values are plotted as a function of time until steady state is reached. Note that this equation assumes that the material below the casing is uniform to a large depth. It is prudent to use it only if the depth to the next layer is at least five borehole diameters below the bottom of the boring. Stage 2 consists of deepening the borehole (for example, 0.2 m deeper and 75 mm in diameter), and repeating the permeability test (falling head test). The hydraulic conductivity k2 from stage 2 (Figure 8.46b) is calculated from the following equations (Hvorslev, 1949): k2 =

A h1 Ln B h2

(8.37)

196

8 IN SITU TESTS

Casing

d D

H

≥ 5D Grout

Casing

d D

H

≥ 5D Grout

L ≥ 5D Compacted clay

≥ 5D Compacted clay

(a)

(b)

Figure 8.46 Two-stage borehole permeameter: (a) Stage 1; (b) Stage 2. (Source: Third picture: Courtesy of Craig Benson, University of Wisconsin.)

Ratio of hydraulic conductivity kh/kv

9

In Eq. (8.41), all quantities are known except m, which can therefore be obtained. Alternatively, m can be found by using Figure 8.47, which presents k2 /k1 versus kh /kv for L/D ratios of 1, 1.5, and 2. Once m is known, kh and kv can be found as follows (Daniel, 1989):

L/D = 1.5

7 L/D = 1 5

L/D = 2

3

1

1

1.2 1.4 1.6 1.8 Ratio of hydraulic conductivity k2/k1

2

Figure 8.47 Relationship between k1 /k2 and m for two-stage borehole permeameter. (Source: Adapted from Daniel (1989).)

with

(

)

√

( )2 L L + 1+ D D ) ( L B = 8L(t2 − t1 ) 1 − 0.562e−1.57 D A = d2 Ln

(8.38) (8.39)

Note that A is in m2 while B is in m.s. The k2 values are plotted as a function of time until steady state is reached. Then the anisotropy can be taken into account by using the ratio k2 /k1 and relating it to the ratio kh /kv . This is done by first defining m as: √ kh (8.40) m= kv where kh and kv are the hydraulic conductivity in the horizontal and vertical directions, respectively. Then k2 /k1 is related to m through: ) ( √ ( ) Ln k2 =m k1

L D

+

( Ln

√ mL D

+

L D

1+

1+

(

mL D

2

)2

)

(8.41)

kh = mk1

(8.42)

kv = k1 ∕m

(8.43)

The analysis of both stage 1 and stage 2 presented here makes a number of limiting assumptions that may or may not be verified in the field (Daniel, 1989).

8.13

BOREHOLE EROSION TEST

The Borehole Erosion Test (BET) was developed in 2013 (Briaud et al., 2016; Briaud et al., 2017a, Figure 8.48). It is an in situ erosion test that consists of drilling a hole approximately 100 mm in diameter by the wet rotary method to a depth covering the zone of interest for the erosion problem at hand. Once the hole is drilled, the rods and drill bit are removed, and a borehole caliper is lowered to the bottom of the hole (Figure 8.49). The diameter of the borehole is logged by pulling the borehole caliper up the hole to obtain the zero-reading borehole diameter profile. The caliper is pulled out of the borehole and the rods without the drill bit are re-inserted until the bottom of the rods is 150 mm above the bottom of the borehole. Water is circulated at a chosen velocity down the rods, around the bottom of the rods, and up the annulus between the rods and the wall of the borehole. This erodes the walls of the soil borehole if the water velocity is larger than the critical velocity. After a set time (say, 10 minutes), the flow of water is stopped, and the rods are removed from the hole. The borehole caliper is reinserted to the bottom of the borehole and the diameter of the borehole is logged again to obtain the eroded borehole diameter profile. Several caliper profiles are recommended to improve the precision. The increase in radius that occurred

8.14 OFFSHORE IN SITU TESTS

(a) Concept

(b) Test in progress

Water flow

Diameter, D

Water flow velocity, V

during the flow divided by the flow duration is the erosion rate associated with the flow velocity. It gives one point on the erosion function curve (see Chapter 24). The caliper is removed from the borehole and the rods without the bit are reinserted to the bottom of the borehole. A higher velocity is chosen, and the flow is maintained during the set time (10 minutes) to repeat the process. Velocity after velocity, the erodibility profile of the soil layers is obtained (Figure 8.50). One advantage of the BET is that it gives a complete profile of erosion rates and allows for the evaluation of all soil layers within the depth of the borehole in one single test.

8.14

Figure 8.48 The Borehole Erosion test (BET).

Figure 8.49 A mechanical caliper.

197

OFFSHORE IN SITU TESTS

The in situ tests most commonly used offshore are the cone penetrometer test and the vane shear test. Other in situ tests used offshore include the pressuremeter test, the dilatometer test, and a number of geophysical tests (see Chapter 9). The offshore CPT is used for stratigraphy, classification, undrained shear strength in fine-grained soils, and friction angle and relative density in coarse-grained soils. It is performed from the seabed or down a borehole. The seabed systems (Figure 8.51) are lowered to the seabed and provide the vertical reaction against which to push the CPT. A total push of 100 kN can be expected from these units. The rods are prestrung on the seabed unit. The downhole systems (Figure 8.52) consist of lowering the CPT system through the drill string that drilled the borehole, latching the CPT system to the bottom of the drill string, and pushing the CPT into the soil below by using the mud pressure in the drill string. The drill string is typically steadied by clamping the drill string to an external mass resting on the seabed. The offshore vane shear test is used to measure the undrained shear strength of fine-grained soils. Like the CPT, the VST can be performed from a downhole tool (Figure 8.52) or from a seabed platform (Figure 8.53). Although samples can be taken, obtaining the undrained shear strength from such samples in the laboratory suffers from the decompression of the sample when it is brought back to the surface. In gassy soils, this decompression can be very significant and reduce the undrained shear strength by up to 40% (Figure 8.54; Denk et al., 1981). The VST measures the undrained shear strength in situ and therefore does not allow decompression. As a result, the value obtained is much more reliable.

8 IN SITU TESTS

(a)

0

2

3

(b)

Borehole Diameter (in.) 4 5 6

7

1

Before flushing After flushing Reading 1 Reading 2 Reading 3

2

2

4

Borehole Diameter (in.) 6 8 10

12

3

4

5

Before flushing After flushing Reading 1

6

Depth (ft)

Depth (ft)

198

8

7

Reading 2

9

10 11 12 13 14

Figure 8.50 Example of BET results.

Lift lines Rod support guide

Cone rod Friction sleeve Weight Pore pressure sensor

Wheel drive Frame

CPT cone

Figure 8.51 Seabed units to deploy the CPT offshore. (Source: a and b: Courtesy of Swan Consultants Ltd., Copyright EFS Danson 2005.)

8.14 OFFSHORE IN SITU TESTS

199

Hoisting sensor

Drilling fluid pressure Overshot knob

Signal cable

Metering cylinder Extension piece

Borehole

Drill pipe

Locking strips

Landing ring

Remote memory unit

Electronics

Drilling fluid

Battery case Skirt Vane assembly

Cone rod Drill bit

Figure 8.53 Seabed system for vane shear test. (Source: Courtesy of Swan Consultants Ltd., Copyright EFS Danson 2005.) Soil formation

Cone tip

Figure 8.52 “Dolphin” downhole system to deploy the CPT offshore. (Source: Courtesy of FUGRO Inc.)

120 Non-decompressed sample (480 kN/m2 downhole pressure)

Shear strength, τ (kPa)

105 90 75 60

Decompressed sample (0 kN/m2 ) 45 30 15 0

0

3

6

9

12

15

18

21

24

Rotation (degrees)

Figure 8.54 Influence of sample disturbance on vane shear results. (Source: Adapted from Denk et al. (1981).)

200

8 IN SITU TESTS

PROBLEMS AND SOLUTIONS Problem 8.1 Assume that the blow count profile shown in Figure 8.4 is an uncorrected blow count profile obtained for a silty sand. Assume further that the energy recorded during these SPT tests was 332 J, that the groundwater level was at the surface, and that the soil has a significant amount of silt. Create the corrected profile for energy level N 60 , the corrected profile for stress level N 1 , and the corrected profile for silt content N ′ . Then create the combined corrected profile for energy, stress level, and silt ′ . content, N1(60) Solution 8.1 The corrections of the SPT values are shown in Table 8.1s and are based on the following formulas: (

) Emeasured (J) 285 (J) ( )0.5 100 Correction for stress level ∶ N1 = Nmeasured × 𝜎 ′ v0 (kPa) ( ) Nmeasured − 15 ′ Correction for silt content ∶ N = 15 + 2 ( )0.5 ⎛ ⎞ 100 − 15 ⎟ ⎜ N60 × 𝜎 ′ 0v ′ Combined corrections ∶ N1(60) = 15 + ⎜ ⎟ 2 ⎜ ⎟ ⎝ ⎠

Correction for energy level ∶ N60 = Nmeasured ×

Table 8.1s Depth

Measured N measured

Energy level Emeasured N 60

Corrected SPT values

𝛾 sat

Stress level ′ 𝜎ov

N1

Silt N′

Combination N ′ 1 (60)

m

bpf

J

bpf

kN/ m3

kPa

bpf

bpf

bpf

1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18 19.5

15 20 17 12 18 21 24 28 31 30 32 29 31

332 332 332 332 332 332 332 332 332 332 332 332 332

17 23 20 14 21 24 28 33 36 35 37 34 36

19 19 19 19 19 19 19 19 19 19 19 19 19

14 28 41 55 69 83 96 110 124 138 152 165 179

40 38 26 16 22 23 24 27 28 26 26 23 23

15 18 16 14 17 18 20 22 23 23 24 22 23

31 30 23 17 20 21 22 23 24 22 23 21 21

8.14 OFFSHORE IN SITU TESTS

201

The corrections of the SPT values are plotted on the graph shown in Figure 8.1s. SPT Blow Count (blow/0.3 m)

0

0

10

20

30

40

50

Depth (m)

5

10

Nmeasured N60

15

N1 N′ N′1(60)

20

Figure 8.1s Corrected SPT values.

Pressure on cavity wall (kPa)

Problem 8.2 A pressuremeter test gives the test curve shown in Figure 8.2s. Calculate the first load modulus E0 , the reload modulus of the first loop Er 1 , the yield pressure py , the horizontal pressure poh corresponding to the reestablishment of the horizontal in situ stress, and the limit pressure pL . What do you think each parameter can be used for? 1500 1250 1000 750 500 250

0 10 20 30 40 50 Relative increase in probe radius ΔR (%) R0

Figure 8.2s Pressuremeter test results.

Solution 8.2 According to the test results shown in Figure 8.2s, the following parameters are obtained: = 12423 kPa • First load modulus E0 = (1 + 0.35) (0.181500 − 0.017)

202

8 IN SITU TESTS

• • • •

The reload modulus of the first loop Er1 = (1 + 0.35) (0.121500 = 28928 kPa − 0.05) The yield pressure py = 700 kPa The horizontal pressure p0h = 120 kPa The limit pressure pL = 1200 kPa

The applications of the PMT include the design of deep foundations under horizontal loads, the design of shallow foundations, the design of deep foundations under vertical loads, and the determination of a modulus profile and other soil properties. The PMT is not very useful for slope stability and retaining structures. The first load and reload modulus can be used in settlement/displacement analysis. The yield pressure can be used as an upper limit for the allowable foundation pressures. The limit pressure can be used to calculate the ultimate capacity of the foundation. Problem 8.3 Use the CPT profiles of Figure 8.8 to identify the main soil layers. Then classify the soil in each layer according to the CPT classification systems of Figure 8.10 and Figure 8.11. Solution 8.3 A total of 10 layers are identifiable from the CPT profiles of Figure 8.8 and are shown in Figure 8.3s and Table 8.2s. Furthermore, the porewater pressure profile can be extended back to zero pressure and may indicate that the water level would be at a depth of 2.5 m below the ground surface. The classifications of the soil layers based on Figures 8.10 and 8.11 are presented in Table 8.2s, Figure 8.4s, and Figure 8.5s.

Sleeve Friction fs (kPa)

Tip Resistance qT (MPa) 0

10

20

30

40

0

200

400

Porewater Pressure u2 (kPa) 600 –500

0

500

1000 0

Friction Ratio FR (%) 2 4 6 8

10

0

1 2

5

3 4 5 6

Depth (m)

10

7 8 15

9

20

10

25

Figure 8.3s CPT profiles and soil layers. (Source: Courtesy of Professor Paul Mayne, Georgia Institute of Technology.)

8.14 OFFSHORE IN SITU TESTS

Table 8.2s

1 2 3 4 5 6 7 8 9 10

0.0–1.0 1.0–5.0 5.0–7.0 7.0–8.8 8.8–9.5 9.5–11.3 11.3–12.7 12.7–14 14–24.2 24.2–25

(MPa)

(Bar)

fs (kPa)

FR (%)

Figure 8.10

Figure 8.11

4.0 0.8 10.0 26.0 13.0 26.0 37.0 28.0 0.9 10.0

40 8 100 260 130 260 370 280 9 100

100 10 60 210 100 200 250 200 40 160

2.50 1.25 0.60 0.81 0.77 0.77 0.68 0.71 4.44 1.60

Sandy silts & silt Sandy silts and silts Sands Sands Sands Sands Sands Sands Clay Silty sands

Sandy silt Clayey silt Sand Sand Sand Sand Gravelly sand Gravelly sand Clay Silty sand

Robertson & Campanella (1983) 1000

Cone Bearing, qt (bar)

Layer

qt

Depth (m)

Classification of soil layers

8 4, 6 and 7 Sands 5 3

100

10 Silty sands 1 Sandy silts & silts

10

9

Clayey silts

2 Clays

0 0

1

2 3 4 Friction Ratio, FR = fs/qt (%)

5

6

Figure 8.4s Robertson and Campanella 1983 soil classification based on CPT results. (Source: Courtesy of Professor Paul Mayne, Georgia Institute of Technology.)

203

204

8 IN SITU TESTS

Robertson et al. (1986)

1 = Sensitive clay

Zone 12 - sand to clayey sand

2 = Organic soil

1000

Cone bearing, qt (bar)

Zone 10

Zone 11 very stiff fine-grained soil

9 8

100

3 = Clay 4 = Silty clay

7

5 = Clayey silt

6 5

6 = Sandy silt

4 10

7 = Silty sand Zone 3 - clay

Zone 1 sensitive clay

8 = Sand to silty sand 9 = Sand

Zone 2 - organic

1 0

1

2

3

4

5

6

7

8

10 = Gravelly sand

Friction ratio, FR = fs qt (%)

Figure 8.5s Robertson et al. 1986 soil classification based on CPT results. (Source: Courtesy of Professor Paul Mayne, Georgia Institute of Technology.)

Problem 8.4 For a rectangular vane, develop the equation that links the maximum torque T max to the undrained shear strength su of a fine-grained soil. Solution 8.4 The failure surface around the vane is a cylinder with a diameter D and a height H (Figure 8.6s). The torque generated from the sides of the cylinder is: T1 = 𝜋DHsu

D 2

T

H

D

Figure 8.6s Vane subjected to torque.

The torque generated by the top and bottom of the cylinder (ignoring the area occupied by the rod) is: D ( 3 )D 2 r 2 D3 2 ⋅ 𝜋 ⋅ r ⋅ su ⋅ r ⋅ dr = 2 ⋅ 𝜋 ⋅ su = 𝜋 ⋅ su T2 = ∫0 3 12

8.14 OFFSHORE IN SITU TESTS

205

D3 D T = T1 + 2T2 = 𝜋DHsu + 2𝜋 ⋅ su ⋅ 12 )2 ( D 2 H + T = 𝜋su ⋅ D 2 6 For the common case of vanes with H = 2D, the equation becomes: T1 =

7 𝜋s D3 6 u

Problem 8.5 Why is the vane test not used in coarse-grained soils? Develop a way, including placing instrumentation on the vane, that would allow the vane test to give the effective stress friction angle of a sand with no effective stress cohesion intercept. Solution 8.5 The vane test gives one measurement: the torque at failure. It can easily be used to obtain the undrained shear strength of a fine-grained soil because in this case the strength is represented by one parameter, su . The vane test cannot be used easily to obtain the drained or effective stress parameters (c and 𝜑) because we need three equations to solve for the three parameters involved: 𝜎 ′ , c, and 𝜑. The shear strength equation is: 𝜏f = c + 𝜎 ′ tan 𝜙 If c = 0, the shear strength equation becomes: 𝜏f = 𝜎 ′ tan 𝜙 To get 𝜑 from the vane test in this case, it is necessary to make two separate measurements. This can be accomplished by placing a pressure sensor on one of the blades, as shown in Figure 8.7s. A free-body diagram of a quadrant of the failing soil mass gives the following equations: dFy

σ

τ

τ

σ

dFx

r

Pressure sensor

y

dθ

T θ

x pavg.r

Figure 8.7s Applied stresses on vane.

{

dFy = 𝜎 ⋅ r ⋅ d𝜃 ⋅ sin 𝜃 + 𝜏 ⋅ r ⋅ d𝜃 ⋅ cos 𝜃 dFx = 𝜎 ⋅ r ⋅ d𝜃 ⋅ cos 𝜃 + 𝜏 ⋅ r ⋅ d𝜃 ⋅ sin 𝜃

Based on these equilibrium equations: p ⋅ r = Fy = p=𝜏 +𝜎

𝜋 2

∫0

𝜋

|2 | (𝜎 ⋅ r ⋅ sin 𝜃 + 𝜏 ⋅ r ⋅ cos 𝜃)d𝜃 = −𝜎 ⋅ r ⋅ cos 𝜃 + 𝜏 ⋅ r ⋅ sin 𝜃 | = (𝜏 + 𝜎)r | |0

At failure: 𝜏f = 𝜎 ′ tan 𝜙 𝜎 ′ = p − 𝜏f → 𝜏f = (p − 𝜏f ) tan 𝜙 → 𝜏f =

tan 𝜙 p 1 + tan 𝜙

206

8 IN SITU TESTS

From problem 8.4, we have: D D3 D3 + 𝜋𝜏top + 𝜋𝜏bottom 2 12 12 tan 𝜑 D3 D D3 ′ T = 𝜋DH p + 𝜋𝛾 z tan 𝜑 + 𝜋𝛾 ′ (z + H) tan 𝜑 1 + tan 𝜑 2 12 12 3 tan 𝜑 D D2 T=𝜋 H p + (2z + H) 𝛾 ′ 𝜋 tan 𝜑 2 1 + tan 𝜑 12 ( ) 3 3 1 D D + 𝜋pD2 H − T tan 𝜑 − T = 0 (2z + H) 𝜋𝛾 ′ tan2 𝜑 + (2z + H) 𝜋𝛾 ′ 12 12 12 √ −B + B2 + 4AT tan 𝜙 = 2A 3 ′D A = (2z + H) 𝜋𝛾 12 3 1 ′D B = (2z + H) 𝜋𝛾 + 𝜋pD2 H − T 12 2

T = 𝜋DH𝜏side

T: D: H: 𝜑: p: Y: z:

torque applied to the vane diameter of the vane height of the vane internal friction angle of sand pressure on the blade of the vane (which is measured by a sensor) unit weight of soil depth of top of the vane

Problem 8.6 A borehole shear test is performed in a saturated clay below the water level. The test is performed fast enough to ensure no drainage. When the horizontal pressure is applied, the plates penetrate 4 mm into the soil of the borehole wall (Figure 8.8s). How long should the plates be for the end effect created by the resistance of the wedge at the leading edge of the plates to represent less than 10% of the shear force measured?

T = 2(p + Su.l) Failure wedge p N

W

τf

Su

L

B 0 < f < Su i y

P X

Figure 8.8s Borehole shear test.

8.14 OFFSHORE IN SITU TESTS

207

Solution 8.6 ⎧∑ 𝜏f •B ⎪ Fx = 0 → N sin i = fB + cos i cos i ⎪ ⎨ 𝜏f •B ⎪∑ sin i ⎪ Fy = 0 → p = N sin i + W + cos i ⎩ f ⋅ B 𝜏f B N= + sin i sin i 1 W = 𝛾B2 tan i ) (2 𝜏f B f ⋅ B 𝜏f B 1 + cos i + 𝛾B2 tan i sin i p= sin i sin i 2 cos i 𝜏f B f ⋅B 1 p= + + 𝛾B2 tan i + 𝜏f B tan i tan i tan i 2 Because B, the penetration of the blades into the soil, is typically very small (say, less than 10 mm), and because the weight of wedge W is a function of B2 , it is reasonable to neglect the influence of the weight of the wedge in calculating p: 𝜏f B f ⋅B + + 𝜏f B tan i tan i tan i By assuming i = 45∘ + 𝜙∕2 and using Mohr-Coulomb theory (Figure 8.9s), we have: p∼

p∼

(

f .B

tan 45 + p∼

(

𝜑 2

f .B

tan 45 +

𝜑 2

)+

( su B cos 𝜑 𝜑) ) + su B cos 𝜑 tan 45 + ( 2 tan 45 + 𝜑2

) + 2su B

If Φ = 30∘ for upper and lower limits of f , we will have: ⎧ f = 0 → p = 2su B ( √ ) ⎪ ⎨ 3 su .B ⎪ f = su → p = 2 + 3 ⎩ ( √ ) 3 su .B 2su B < p < 2 + 3 p is the force needed to fail the wedge of soil above the borehole shear device (Figure 8.9s). If this force must be less than 10% of the force measured by the borehole shear device, then: Tmeasured = 2(p + su .l) →

2p 0.9p < 10% → p < 0.1(p + su .l) → l > T 0.1su Shear stress on the failure surface

τ

Su τf = Sucosφ ϕ σ

Figure 8.9s Stress envelope.

208

8 IN SITU TESTS

This assumes that the borehole shear device is associated with a plane strain failure, which is a simplifying assumption. In this case, the requirements on the length of the BSD to ensure that the end effect is less than 10% of the measured value are: f = 0 → l > 18B f = su → l > 23.2B In the worst condition, which is (f = su ), the length of plates must be longer than 23.2B. If B = 4 mm, for example, then l > 92.8 mm. Problem 8.7 A plate test gives the load settlement curve shown in Figure 8.26. The plate is 0.3 m in diameter and the test is performed at the ground surface. Calculate the soil modulus from the early part of the plate test curve. Would you use this modulus to calculate the settlement of a 3 m by 3 m square footing? Explain. Solution 8.7 The pressure versus displacement/width curve is shown in Figure 8.10s. 1.6 Pressure (MPa)

1.4 1.2 1.0 0.8 0.6 0.4 A 0.2 0.0 0.00

0.02

0.04

0.06

0.08

0.10

Displacement over width, s/B

Figure 8.10s Pressure versus displacement/width curve.

The soil modulus is calculated based on point A in Figure 8.10s using the following equation: E=

𝜋(1 − v2 )pB 𝜋(1 − v2 )p 𝜋(1 − 0.352 ) × 0.36 = = = 62 MPa 4s 4 × 0.004 4 × Bs

The soil modulus obtained in this fashion from the plate test is 62 MPa. I would not use this soil modulus to calculate the settlement of a 3 m by 3 m footing without checking the soil stratigraphy first. The plate-bearing test can only give the response of the soil down to a depth of about twice the plate diameter, which is 0.6 m in this case. It cannot reflect the soil properties beneath the 3 m by 3 m square footing unless they are the same. Problem 8.8 Use the elastic settlement equation for a plate test to explain why the modulus of subgrade reaction K is not a soil property while the soil modulus E is. Which one would you rather use, and why? Solution 8.8 The elastic settlement equation for a plate load test is: s=

I(1 − v2 )pB E

Here, I is the shape factor, E is the soil modulus, p is the average pressure under the footing, B is the plate diameter, and v is the Poisson’s ratio. The modulus of subgrade reaction K is calculated as the ratio between the pressure and the settlement: p p E K = = I(1−v2 )pB = s I(1 − v2 )B E

8.14 OFFSHORE IN SITU TESTS

209

Therefore, the modulus of subgrade reaction K is a function of the soil modulus E and the foundation size B. The larger the foundation is, the smaller the modulus of subgrade reaction is. I would prefer to use the soil modulus E because it is a true soil property, whereas K is not. Indeed, as shown here, K depends on E and B. Any K value determined from a given size foundation test cannot be used directly for a different size without paying attention to the scale effect. Problem 8.9 Calculate the settlement of a footing on sand after 50 years under a pressure of 100 kPa if the settlement after 1 hour under a pressure of 100 kPa during a load test is 10 mm. The soil has a viscous exponent n = 0.04. Solution 8.9 Based on Eq. (8.17), the settlement s2 of a footing after t2 = 50 years under a pressure of 100 kPa based on the settlement s1 of the same footing after t1 = 1 hour is: ( )n s1 t = 1 s2 t2 With s1 = 10 mm, t1 = 1 hr, t2 = 50 years = 50 × 365 × 24 = 438,000 hr, and n = 0.04: s 10 s2 = ( 1)n = ( )0.04 = 16.8 mm t 1 438000

1

t2

So, the calculated settlement of the footing after 50 years under a pressure of 100 kPa is 16.8 mm. Problem 8.10 Pocket erodometer tests (PETs) are performed on the end of Shelby tube samples retrieved from a levee. The average depth of the PET holes is 6 mm and the standard deviation is 2 mm. Estimate the rate of erosion if the mean velocity overflowing the levee will be 5 m/s. If the levee is subjected to overtopping for 2 hours (hurricane), how much erosion is likely to take place? Solution 8.10 Using Figure 8.30 and a PET hole depth of 6 mm, the soil erosion category is category III or medium erodibility. For this category, the PET hole varies between 1 mm and 15 mm, corresponding to erosion rates of 3 mm/hr and 2000 mm/hr respectively. For 6 mm, the erosion rate is estimated to be near the middle of the range on the logarithmic scale and an erosion rate of 80 mm/hr is selected (Figure 8.11s). With 2 hours of overtopping at this rate, 160 mm of erosion is estimated. 100,000

Very high erodibility I > 75 mm

10,000 1000 Erosion rate (mm/hr)

Medium erodibility III

High erodibility II 75-15 mm

15-1 mm SP

100

SM

10

Low erodibility IV 2.5 pF

5 4 3 2

Sensitivity starts below 2.5 pF

1 0 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 Filter paper water content, w

239

a high relative humidity, little evaporation takes place, the cooling process is limited, and the difference in temperature is smaller. The difference in temperature given by the two thermometers is related to the relative humidity, which in turn is related to the water tension or total suction. In the pores of a soil, there has to be a balance between the water tension in the air phase and in the water phase. See Chapter 11 for more details on these relationships. Because psychrometers work on the basis of precise temperature measurements, any exterior fluctuation in temperature will lead to poor precision. Therefore, psychrometers are not well suited for in situ measurements, because of the daily temperature cycle. It also takes a fair amount of time for equilibrium to be reached between the psychrometer and the air in the soil pores. Chilled Mirror Psychrometers

Figure 10.5 Filter paper method for water tension measurement (Bulut et al., 2001). (a) Filter paper (matric suction only). (b) Filter paper (total suction). (c) High-precision scale with hood. (d) A calibration curve. (Source: a-c: Courtesy of Dr. Bulut)

Thermocouple Psychrometers Thermocouple psychrometers (psykhros means “cold” in Greek) can be used to give the total suction of a soil by measuring the relative humidity in the air phase of the soil pores or the region near the soil (Figure 10.6). They measure the total suction because the evaporation process creates pure water, while the water in the soil pores is not pure water. Hence, the osmotic suction is realized. Psychrometers give the relative humidity by measuring the difference in temperature between a nonevaporating surface and an evaporating surface. Imagine two thermometers, one with a dry bulb and the other with a wet bulb. The dry-bulb thermometer measures the ambient temperature, but the wet-bulb thermometer measures a temperature lower than ambient because the evaporation of the water on the bulb cools the bulb. The thermometers can be replaced by transistors in transistor psychrometers. If the air phase has a low relative humidity, the evaporation is faster, the cooling process is high, and the difference in temperature is larger. If the air phase has

Teflon filling

Chilled mirror psychrometers can be used to give the total suction of a soil (Figure 10.7). Much like the thermocouple psychrometers, they measure the relative humidity and then relate the relative humidity to the suction. The relative humidity in a chilled mirror psychrometer is obtained as follows: The soil is inserted into a small chamber that is sealed off from the outside air and has a mirror present. Facing the mirror is a camera able to detect when dew forms on the mirror. The air in the chamber comes to relative humidity equilibrium with the air in the soil sample. Then the mirror is chilled down to the

Ceramic bulb Thermocouple

Chamber

6 mm

Temperature sensor

Figure 10.6 (a) Cross-section of a thermocouple psychometer. (b) Thermocouple psychometer. (Source: b: Courtesy of Wescor-Elitechgroup.)

Figure 10.7 Chilled mirror psychrometer. (Source: Courtesy of Decagon Devices, Inc.)

240

10 LABORATORY TESTS Mercury manometer

Bourdon vacuum gage

Pressure transducer

Manual observation

To chart recorder for continuous observation

Manual observation Ground surface

Porous cup

Figure 10.8 Tensiometers. (a) Tensiometer with pressure-vacuum gage. (b) Types of tensiometers. (c) Tensiometer with pressure transducer. (Source: a: Courtesy of Envco Global. c: Courtesy of STEP Systems GmbH, www.stepsystems.de.)

point where dew forms on the mirror and the temperature of the mirror at that point is recorded. The temperature of the soil is also recorded and the difference in temperature between the mirror at the dew point and the soil is related to the relative humidity in the soil. The suction is then obtained through its relationship with the relative humidity (see Chapter 11).

Air-tight chamber

Air pressure supply Soil specimen High air-

entry disk

Distilled water reservoir

Tensiometers Tensiometers can be used to measure the water tension or matric suction in a soil (Figure 10.8). A tensiometer consists of a high air entry porous ceramic tip (also called a ceramic cup) that is saturated with water and placed in good contact with the soil. In the tensiometer, the space behind the ceramic tip is filled with de-aired water and connected with a negative pressure measuring device. The stress slowly equalizes between the water tension in the tensiometer and the water tension in the soil pores. That tension is then measured either through a water-mercury manometer, a Bourdon vacuum tube, or an electrical pressure transducer. The water tension that can be measured in a tensiometer is limited to approximately negative 90 kPa (2.95 pF) due to the possibility of water cavitation in the tensiometer above such a value. Pressure Plate Apparatus (PPA) The pressure plate apparatus (PPA) is a closed pressure chamber that can be used to increase the air pressure in the soil pores to the point where the air chases the water out of the pores (Figure 10.9). The sample is placed in the chamber on a high air entry ceramic disk. This disk, which is saturated with water, has the property of letting water go through but not air, up to a certain rated pressure, known as the air entry value of the disk. The air pressure is increased and the stress in the water is increased accordingly (decrease in tension). When the water tension becomes equal to zero, the water comes out and at that point, the air pressure is equal to the water tension. This technique is called the axis translation technique because

Figure 10.9 Pressure plate apparatus: (a) 500kPa pressure plate. (b) 1500 kPa pressure plate.

it simply translates the origin of reference by applying an air pressure equal to the water tension (Figure 10.10). The PPA can be used to determine the natural water tension or to generate a soil-water retention curve. If the soil sample is placed at its natural water content in the PPA, the air pressure that starts the water flow is the natural water tension. If the soil specimen starts as a saturated sample and

Tension Uw

0

Compression

Ua

Uw

Natural state Axis translation Ua

Pressure plate apparatus

Figure 10.10 Axis translation for water tension determination.

10.2 MEASUREMENTS

241

Cover

Porcelain plate

Soil specimen

Salt solution

Figure 10.11 Salt solution equilibrium containers for water tension determination.

the air pressure is increased in steps, each pressure step will drive water out of the sample until equilibrium is reached, and this will give the water tension corresponding to the water content of the sample. This water content can be measured separately by stopping the test or inferred from the water loss read on the burette connected to the PPA. The air pressure is increased in steps and each step gives the water tension and the corresponding water content. The soil-water retention curve (SWRC) is thus obtained. The range of application of the PPA is from 0 to about 1500 kPa (4.17 pF). Salt Solution Equilibrium (SSE) The salt solution equilibrium (SSE) is a water tension measurement technique which relies on the fact that salt solutions have significant osmotic suction. As explained in Chapter 11, osmotic suction comes from the fact that water molecules are attracted to salt molecules: more salt, more attraction. A closed chamber with a salt solution at its lower part (Figure 10.11) will generate a certain relative humidity in the air above it. The higher the salt concentration, the lower the relative humidity above the solution in the chamber will be. If a soil sample is suspended in the air above the salt solution, it will dry and the water tension in the soil sample will come to equilibrium with the ambient relative humidity. At equilibrium, the water tension is given by the relative humidity in the air of the chamber. This relative humidity depends on the salt concentration in the solution and can be calculated from it (see Chapter 11). This relationship depends on the type of salt, the molality, and the temperature. Table 10.3 gives the osmotic suction for different salts and different molalities. Molality, in this case, is the number of moles of salt per kilogram of water. Note that in most cases,

molarity is different from molality because molarity is the number of moles per liter of solvent. The range of application for the SSE technique is very wide, from 0 to close to 100,000 kPa (6pF). It is also a very inexpensive and very reliable technique. Hence, it is used as a reference to calibrate many other techniques. The drawback is that it is quite time-consuming: The time necessary for equilibrium to be reached between the water tension in the soil sample and the relative humidity in the surrounding air can be a couple of weeks. 10.2.5

Normal Strain

A normal strain 𝜀 is defined in one direction as the change in length Δz divided by the initial length z between two points. A normal strain is measured either by measuring a displacement and a length (Δz/z) or by using a strain gage (𝜀). Measurements of length are done with a ruler or a set of calipers (Figure 10.12). Displacements are measured with mechanical devices such as dial gages (Figure 10.12) or electrical devices such as LVDTs, DCDTs, and potentiometers. A linear variable differential transformer (LVDT) (Figure 10.13) has three solenoid coils arranged like three side-by-side donuts. A small metallic rod is attached to the point where the displacement is to be measured and the solenoids are attached to an immobile reference point. The small rod passes through the center of the three solenoids without touching them. An alternating current through the center solenoid creates a voltage in the side solenoids. The movement of the metallic rod creates a change in voltage that is linearly proportional to the movement of the rod. The change in voltage is transformed into a displacement

242

10 LABORATORY TESTS

Table 10.3

Molality (mol/kg)

NaCl

0.001 0.002 0.005 0.010 0.020 0.050 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.200 1.400 1.500 1.600 1.800 2.000 2.500

5 10 24 48 95 234 463 916 1370 1824 2283 2746 3214 3685 4159 4641 5616 6615 N/A 7631 8683 9757 12556

Osmotic suction in kPa of some salt solutions at 25∘ C Osmotic suction in kPa at 25∘ C KCl NH4 Cl Na2 SO4 5 10 24 48 95 233 460 905 1348 1789 2231 2674 3116 3562 4007 4452 5354 6261 N/A 7179 8104 9043 11440

5 10 24 48 95 233 460 905 1348 1789 2231 2671 3113 3558 4002 4447 5343 6247 N/A 7155 8076 9003 11366

7 14 34 67 129 306 585 1115 1620 2108 2582 3045 3498 3944 4384 4820 N/A N/A 6998 N/A N/A 9306 11901

CaCl2

Na2 S2 O3

MgCl2

7 14 34 67 132 320 633 1274 1946 2652 3396 4181 5008 5880 6799 7767 N/A N/A 13391 N/A N/A 20457 29115

7 14 34 67 130 310 597 1148 1682 2206 2722 3234 3744 4254 4767 5285 N/A N/A 7994 N/A N/A 11021 14489

7 14 35 68 133 324 643 1303 2000 2739 3523 4357 5244 6186 7187 8249 N/A N/A 14554 N/A N/A 22682 32776

(Source: Adapted from Bulut et al., 2001.) Primary coil

Core

Figure 10.12 Mechanical devices to measure displacement: (a) Calipers. (b) Dial gage.

measurement through calibration. A direct current differential transformer (DCDT) is an LVDT in which the current passing through the solenoids is a direct current instead of an alternating current. A potentiometer or pot is a resistor with three terminals. Two are fixed and one moves between the two fixed terminals. By sliding the moving terminal, the resistance offered by the potentiometer varies and so does the voltage. The rod connected to the point where the movement is to be measured is tied to the sliding terminal. The change in voltage induced by the movement of the rod is related to the movement through calibration.

Secondary coils

Figure 10.13 Linear variable differential transformers: (a) Principle. (b) Device.

Strain gages are of two main types: foil strain gages and vibrating wire strain gages. A foil gage is a thin sheet of metal (copper-nickel alloy is common) with a pattern (Figure 10.14) glued to the material that is deforming. Actually, a layer of flexible insulating material is first glued to the deforming material and then the foil gage is glued onto the

243

10.3 COMPACTION TEST: DRY UNIT WEIGHT Direction of elements trip and soil particle movement

+ voltage

– voltage

Direction of shear wave propagation Bearing plate

Figure 10.14 Foil and vibrating wire strain gages. (a) Foil strain gage. (b) Model 4000 Vibrating Wire strain gage. (Source: b: Courtesy of Geokon, Inc.)

insulator, so that the current passing through the gage only travels through the gage. When the material deforms, the foil length changes and so does its resistance. The voltage changes accordingly and the strain is related to the change in voltage through calibration. Vibrating wire strain gages consist of two small anchor blocks solidly connected to the material that is deforming. Between these two anchors is a high-tensile-strength wire brought taut to a chosen initial load. Around the wire is a cylinder that protects the wire and contains a permanent magnet and a plucking coil. When the wire is plucked, it vibrates at its natural frequency. If the material deforms, the end blocks move and the natural frequency of the vibrating wire changes. The change in natural frequency of the wire is related to the normal strain by theory and calibration. 10.2.6

Shear Strain

A shear strain 𝛾 is defined for two perpendicular directions (x and y as shown in Figure 10.15). When the shear strain is small enough, the shear strain is equal to the change in angle 𝛾 expressed in radians between the two perpendicular directions due to the shearing process. Obtaining shear strain is most easily done by measuring the normal strain in two perpendicular directions (Figure 10.15). It can be shown (Chapter 11) that the shear strain in this case is given by: 𝛾xy = 𝜀1 − 𝜀2

Figure 10.16 Bender elements: (a) Principle. (b) Device.

plates, when a voltage is driven through the plates, one shortens and the other lengthens; this forces the plates to shake in bending. If the small plates are buried in the soil, the repeated lateral motion of the plates generates a wave that propagates in shear through the soil. This is the wave generation function of a bender element. At the other end of the sample, a similar bender element is also buried in the soil and acts as a receiver. This receiver senses the arrival of the shear wave because that wave forces the two plates to move sideways. This bending movement shortens one and lengthens the other; this alternating tension and compression create an electrical signal that can be measured. When the top bender element generates a shear wave, the wave travels through the soil and reaches the bottom bender element, which detects its arrival. Knowing the length of travel (sample length) and the time necessary for the wave to propagate from the generating bender element to the receiving bender element, one can calculate the shear-wave velocity vs . Theory on shear-wave propagation in an elastic body tells us that the shear modulus G of the soil from measurement of shear-wave velocity vS is given by: G = 𝜌(vS )2

where 𝜌 is the mass density of the soil sample. Note that the shear modulus measured in this fashion is associated with very small shear strains.

(10.1) 10.3

10.2.7

Bender Elements

10.3.1

A bender element (Figure 10.16) is a small electromechanical device used to generate or sense bending waves. It is made of two thin piezoceramic plates glued together. Between the two plates and on the outside of the two plates are conducting surfaces. Because of the different polarizations of the two

Y

(10.2)

X

2 1

Figure 10.15 Getting shear strain from two normal strain gages.

COMPACTION TEST: DRY UNIT WEIGHT Saturated Soils

Most of the time, the soil in a compaction test is unsaturated. 10.3.2

Unsaturated Soils

The compaction test dates back to the work of Ralph Proctor, an American civil engineer, in the early 1930s. Today, the test is actually two tests: the Standard Proctor Compaction Test (SPCT; ASTM D698) and the Modified Proctor Compaction Test, (MPCT; ASTM D1557). Proctor developed the SPCT, but in the late 1950s, as compaction machines became much bigger than in the 1930s, the MPCT was developed to better correspond to the higher energy generated by the larger roller compactors. In both cases, the result of the test is the dry unit weight 𝛾 d vs. water content w curve (Figure 10.17).

244

10 LABORATORY TESTS

Dry unit weight γd (kN/m3)

20

25 γdmax

Table 10.4 Compaction requirements for standard proctor compaction test

γdmax

20 C 15

19

B

A

18

0

D

10 5

Wopt 4

8 12 16 Water content w (%)

20

0

Wopt 0

4

8 12 16 Water content w (%)

20

Figure 10.17 Compaction curve.

The first step in the SPCT is to take a soil sample, dry it, break the clumps of soil down to individual particles (e.g., with a mortar and rubber-tip pestle), and measure its weight Wd . Then, calculate the weight of water Ww that must be added to the dry soil sample to reach a chosen water content w: (10.3) Ww = wWs = wWd

152 mm diameter 116 mm high mold

3 soil layers

3 soil layers

25 blows per soil layer

56 blows per soil layer

Hammer weight 24.5 N

Hammer weight 24.5 N

Hammer drop height 305 mm Volume 9.43 × 10–4 m3 Total energy 600 kN.m/m3

Hammer drop height 305 mm Volume 21.2 × 10–4 m3 Total energy 600 kN.m/m3

where 𝛾 d is the dry unit weight, W t is the total weight of the soil sample in the mold, V t is the total volume of the sample, and w is the water content of the sample. The combination of 𝛾 d and w gives one point on the compaction curve. By repeating the SPCT for different water contents, the compaction curve is described point by point (see Figure 10.17). Note that this curve has a well-defined bell shape because the vertical scale is concentrated around the range of values within which the dry unit weight varies. If the same curve is plotted at the full scale of the unit weight, the curve still has a bell shape but shows that the dry unit weight is not very sensitive to the water content. The reason for this bell curve is that at point A on Figure 10.17 the soil is relatively dry and it is difficult for a given compaction energy to bring the particles closer together. At point B the water content is such that water tension exists between the particles and hinders the effectiveness of the compaction process. At point C, the water

Add the water to the soil and mix thoroughly. Weigh the empty compaction mold to be used for the test. Using the prepared soil mixture, place a first layer in the compaction mold (Figure 10.18) and compact that layer of loose soil by dropping a standard compaction hammer a standard number of times. The blows should be distributed evenly across the soil layer to reach uniform compaction. Repeat this process for all layers and aim for the last layer to coincide with the top of the mold. Two mold sizes are used; Table 10.4 gives the detailed requirements. At the end, weigh the mold plus soil and calculate the total soil weight W t . The dry unit weight is obtained by: W 1 𝛾d = t (10.4) Vt (1 + w)

(a)

102 mm diameter 116 mm high mold

(b)

(c)

Figure 10.18 Compaction equipment and test: (a) Compaction mold. (b) Compaction test. (c) Compaction hammer. (Source: (a) and (c): Courtesy of Forney LP, Hermitage, PA.)

10.3 COMPACTION TEST: DRY UNIT WEIGHT

SGs 𝛾w 𝛾d = S + Gs w

=

S+

Vw Ww Ws Ww Ws Vs

s

v

SGs 𝛾w = S + Gs w

s

(10.6)

Equation (10.6) shows that the relationship between the dry unit weight and the water content for a given degree of saturation S is a hyperbola. This hyperbola is called the saturation line and corresponds to S (Figure 10.19). The saturation line for S = 1 is a bounding envelope for all compaction curves for that soil, called the zero air void line. It is also possible to draw the lines of equal water tension on the same graph as the compaction curve, as shown in Figure 10.20. In 1958, a second compaction test, the Modified Proctor Compaction Test (MPCT), was developed as an ASTM standard. A higher compaction standard was necessary to better correspond to the larger and heavier compaction equipment, such as large vibratory compactors and heavier steam rollers. 21 S = 0.6 S = 0.8

S=1

Lines of equal degrees of saturation

Dry unit weight, γd (kN/m3)

20 γ dmax γd =

SGS S+GSW

γw

Modified Proctor, 3 2700 kN*m/m 18 γ dmax

17

16

0

4

16

wopt

8 12 Water content, w (%)

wopt 16

Line of equal water tension

15 Increasing water tension 14 10

14

18

22

26

30

Water content w (%)

The MPCT is very similar to the SPCT except for the different requirements listed in Table 10.5. The data reduction is the same and the result is also the 𝛾 d vs. w curve. The difference is that, due to the higher compaction effort (2700 kN.m/m3 compared to 600 kN.m/m3 ), the curve for the MPCT is located higher than the curve for the SPCT (see Figure 10.19). The peak of the curve has the coordinates γdmax and wopt , called the maximum dry density and the optimum water content respectively. The specifications for field applications usually require that the water content be within ± x% of the optimum water content and that the dry density be at least y% of the maximum dry density. Then these requirements are checked by field testing at the compaction site (see Chapter 8 on in situ testing). Note that the dry unit weight is used on the vertical axis of the compaction curve and not the total unit weight. The reason is best explained through the example of Figure 10.21. Both soil A and soil B have a total unit weight of 20 kN/m3 , yet soil A has a dry unit weight of 17.5 kN/m3 whereas soil B has a dry unit weight of 19 kN/m3 . Soil B has more solid constituents per unit volume and is therefore more compact. The selection of soil B over soil A can be made on the basis of the dry unit weight (19 vs. 17.5) but not on the basis of the total unit weight (20 vs. 20). Table 10.5 Compaction requirements for Modified Proctor Compaction Test

19

Standard Proctor, 3 600 kN*m/m

17

Figure 10.20 Compaction test and water tension lines.

Ws 𝛾s Vs G𝛾 SGs Yw = = s wV = V V v Vt Vv + Vs 1+ V S + Vw Vv SGs 𝛾w

Saturation line S = 1.0

(10.5)

This relationship can be demonstrated as follows: 𝛾d =

18 Dry unit weight, γd (kN/m3)

tension loses its effect, and the primary role of the water is to lubricate the contacts between particles, thereby allowing the given compaction effort to reach a low void ratio and a high dry density. At point D the soil is nearing saturation and the added water simply increases the volume of the voids, which negates the benefit of the compaction. The compaction curve is bounded on the right side by the saturation line for a degree of saturation equal to 1. Indeed, the relationship between the dry unit weight 𝛾 d and the water content w is a function of the degree of saturation:

245

20

Figure 10.19 Compaction curve for Standard and Modified Proctor Compaction Tests.

102 mm diameter 116 mm high mold

152 mm diameter 116 mm high mold

5 soil layers 25 blows per soil layer Hammer weight 44.5 N Hammer drop height 457 mm Volume 9.43 × 10–4 m3 Total energy 2700 kN.m/m3

5 soil layers 56 blows per soil layer Hammer weight 44.5 N Hammer drop height 457 mm Volume 21.2 × 10–4 m3 Total energy 2700 kN.m/m3

246

0.1

1

Soil B Weight (kN)

Air

Water

0.25

3

Volume (m )

0

2.5

20

0.19

Air

0.1

Water

Solids

0.65

0

1

20 0.71

Solids

20

17.5

γt = 20 kN/m3 γd = 17.5 kN/m3

COMPACTION TEST: SOIL MODULUS

10.4.1

Saturated Soils

Most of the time, the soil in a compaction test is unsaturated. 10.4.2

Unsaturated Soils

The compaction test described in Section 10.3 yields the dry unit weight 𝛾 d vs. water content w curve. The soil modulus also plays a very important role in the field of compaction. Indeed, one of the major goals of compaction is to minimize deformation, so a sufficiently high modulus should be reached for compaction to be adequate. A modulus E vs. water content w curve can be generated in parallel with the 𝛾 d vs. w curve by using a device called the BCD (Figure 10.22). It consists of a 150 mm diameter thin and flexible steel plate at the bottom of a rod with handles—a kind of scientific cane. Strain gages are mounted on the back of the plate to record the bending that takes place during the loading test. When the operator leans on the handle, the load on the plate increases and the plate bends. If the soil is soft (low modulus), the plate bends a lot. If the soil is hard (high modulus), the plate does not bend (a)

(b)

600 mm

Acquisition, processing & readout

Load-cell Strain-gage instrumented plate

2 mm

20

50

16

40

150 mm

Figure 10.22 BCD apparatus to get soil modulus during a Proctor compaction test: (a) BCD principle. (b) BCD on Proctor mold.

12

30 8

20

4

BCD modulus (MPa) 3 Dry unit weight (kN/m )

10 0

γt = 20 kN/m3 γd = 19 kN/m3

Figure 10.21 Three-phase diagram showing the usefulness of dry unit weight.

10.4

60 Weight (kN)

Modulus, E (MPa)

Soil A 3

Volume (m )

0

2

4 6 8 10 Water content, w (%)

12

0 14

Dry unit weight, γd (kN/m3)

10 LABORATORY TESTS

Figure 10.23 Compaction curves for dry density and modulus.

much. The amount of bending is recorded by the strain gages and is correlated to the modulus of the soil below. This test, called the BCD test (Briaud et al., 2006), consists of the following steps. First, the BCD plate is placed on top of the sample in the 150 mm diameter compaction mold (Figure 10.22a). The operator then leans on the handles of the BCD and the vertical load increases. When the load goes through 223 N, a load sensor triggers the reading of the strain gages. The device averages the strain gage values, uses the internal calibration equation linking the strains to the modulus, and displays the modulus E. By performing the BCD on top of the 150 mm diameter compaction mold filled with soil (Figure 10.22b), the modulus of the soil sample can be measured. This gives one point on the modulus vs. water content curve. By repeating this test for different water contents when the SPCT or MPCT is performed, a complete E vs. w curve can be obtained (Figure 10.23). The modulus obtained with the BCD corresponds to a reload modulus, to a mean stress level averaging about 50 kPa within the zone of influence, to a strain level averaging 10–3 within the zone of influence, and to a time of loading averaging about 2 s. 10.5 10.5.1

CONSOLIDATION TEST Saturated Soils

The consolidation test dates back to the early 1900s, and it may be appropriate to attribute its early development to Terzaghi, around 1925, with Cassagrande and Taylor making significant contributions as well. The consolidation test (ASTM D2435) is used mostly for determining the compressibility of saturated fine-grained soils. It consists of placing a disk of soil approximately 25 mm high and 75 mm in diameter in a steel ring of the same diameter and applying a vertical load on the sample while recording the decrease in thickness of the sample (Figure 10.24). Filter stones are placed at the top and bottom of the sample to allow the water squeezed out of the sample to drain at both ends. There are several loading procedures: incremental loading, constant rate of strain, and constant gradient. The incremental loading procedure is the most popular and consists of placing a load on the sample for 24 hours

10.5 CONSOLIDATION TEST

(a)

1

Dial gage or LVDT

2

3

4

σ

σ

σ

247

Loading

Porous stone Soil specimen

σ

0

σ

σ

σ

uw

0

uw

½ uw

0

σ′

0

σ – uw

σ – ½ uw

σ

Stress, σ

Porous stone

2

1

(b)

σ

3

4

(c)

σ′

uW

24 hrs Time, t

Settlement, s

end of load step

Figure 10.24 Consolidation test and equipment: (a) Principle. (b) Sample in ring. (c) Complete setup. (Source: (b) Courtesy of Lev Buchko, P.E.//Timely Engineering Soil Tests, LLC. (c) Courtesy of Humboldt Mfg. Co.)

σ = Total stress Draining water

σ′

uw : Water stress

Water Soil skeleton feels the effective stress σ′ Real sample

Rheological model

Figure 10.25 Consolidation model.

while recording the decrease in sample thickness. The load creates a constant total normal stress 𝜎 on the surface of the sample. When 𝜎 is applied, the water stress uw goes up because the water has difficulty escaping from the small soil pores quickly enough (Figure 10.25). It takes some time for the water stress uw to decrease and come back to its original value. This decrease in uw is associated with a corresponding increase in effective stress (𝜎 ′ = 𝜎 − uw in this case because the soil is saturated) and settlement of the soil; this is the process of consolidation (Figure 10.26). The 24-hour loading step is considered to be sufficient in general for the water stress uw to decrease back to zero. Therefore, it is assumed that at the end of each 24-hour loading step, the water stress is back to zero and the total normal stress 𝜎 is equal to the effective normal stress 𝜎 ′ . The loads and associated pressures are applied in a sequence

24 hrs

Immediate compression

Time, t

Figure 10.26 Consolidation process.

where the load is doubled each time. A typical sequence is 12, 25, 50, 100, 200 kPa for 𝜎. The last point at the end of the 24-hour loading step curve (displacement vs. time, Figure 10.26) gives one point (vertical effective stress 𝜎 ′ and vertical strain 𝜀) on the consolidation test stress-strain curve (stress vs. strain, Figure 10.27a). The upward curvature of this stress-strain curve and the lack of maximum stress or failure stress or strength are due to the steel ring that confines the soil sample. The more load that is applied to the sample, the more the steel ring contributes to the resistance. Note that this curve is often presented as the void ratio e versus the decimal logarithm of the effective stress log 𝜎 ′ (Figure 10.27b). The compression index Cc is defined as the slope of the linear portion of the e-log 𝜎 ′ curve past the initial rounded part of that curve (Figure 10.27). As such, Cc is: Δe (10.7) Cc = Δ log 𝜎 ′ During each 24-hour loading step, the decrease in sample height ΔH is recorded as a function of time t to be able to develop the ΔH vs. t curve. The vertical strain 𝜀 is obtained by dividing the change in height ΔH by the original height Ho of the sample. Figure 10.28 shows the 𝜀 vs. t curve for three loading steps. The coefficient of consolidation cv can be obtained from the 𝜀 vs. t curve of each load step through the formula: H2 (10.8) t where T is the time factor, H the drainage length, and t the time elapsed. The drainage length H is equal to the height cv = T

248

10 LABORATORY TESTS

6000

0

Time (min) 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 Stress, σ′

0 0.01

5000

Strain, ε

4000 3000

0.02 0.03

Strain, ε

0.04 0.05

2000

0.06 0.05

1000 0 0.02

0.04

0.08 0.06 Strain, ε

0.1

0.12

0.06

0.14

Strain, ε

0

(a) 0.6

Stress, σ′

Effective vertical stress, σ′ (kPa)

7000

0.07

Strain, ε

0.08 0.55

0.09 0.084 1

Stress, σ′

0.086

Cc

0.45

0.088 Strain, ε

Void ratio e

0.5

0.4

0.09

Strain, ε

0.092

0.35

0.094 10

100 1000 Effective vertical stress, σ′ (kPa)

10000

(b)

Figure 10.27 Consolidation test results: (a) Stress-strain curve. (b) e-log 𝜎 ′ curve.

H o of the sample if there is drainage only on one side of the sample (top only or bottom only) and equal to half the height of the sample, H o/2 , if there is drainage on the top and bottom of the sample. The time factor T comes from the solution of the governing differential equation for the one-dimensional consolidation theory (see Chapters 12 and 15 for more on consolidation theory). This time factor is linked to the average percent consolidation U, defined as: ue s (t) =1− (10.9) U= smax ue max where s(t) is the settlement at time t, smax is the settlement at a time equal to infinity, ue is the average excess water stress or pore pressure at time t, and ue max is the maximum excess water stress. The theoretical curve linking the average percent consolidation U to the time factor T is shown in Figure 10.29. This curve describes the normalized displacement vs. time curve for the sample according to the one-dimensional consolidation theory. It represents a normalized version of the settlement vs. time curve under a given load. A value of cv can be obtained for each load step by choosing a value of the percent consolidation U (50% or 90%, for example) and finding the corresponding time on the 𝜀 vs. t curve. Two methods are available to do this: the log time method developed by Cassagrande (1938) and the

0.096

Figure 10.28 Vertical strain vs. time for three consolidation test loading steps. 0 Percent consolidation U (%)

1

20 40 60 80

100

0

0.2

0.4

0.6

0.8 1 Time factor, T

1.2

1.4

1.6

1.8

0 Percent consolidation U (%)

0.3

20 40 60 80 100 0.001

0.01

0.1

1

Time factor, T

Figure 10.29 Percent consolidation vs. time factor.

square root of time method developed by Taylor (1948). The log time (Cassagrande) method requires that 𝜀0 and 𝜀100 be found on the 𝜀 vs. t curve (Figure 10.30). Note that 𝜀0 is not necessarily zero, as 𝜀0 refers to zero percent consolidation, not zero deformation. This is a subtle distinction, as the first part of the deformation may be elastic in nature and does not

249

10.5 CONSOLIDATION TEST 0

0 ε0

0.01

x x t1

t2 = 4t1

ε50

t50

0.02

0.03

Strain, ε

Strain, ε

0.02

A

0.01

0.04 ε100 0.05

0.03

B

End of primary consolidation

C 0.04

0.06

0.1

1

10 Log time (min)

100

1000

0.05

x

Figure 10.30 Log time method to obtain the coefficient of consolidation. (Source: Adapted from Cassagrande 1938.)

0.15x

√ t 90 0.06

𝜀0 = 𝜀1 − (𝜀2 − 𝜀1 ) = 2𝜀1 − 𝜀2

(10.10)

The basis for this technique is that, according to the theory, the beginning of the 𝜀 vs. t curve is a parabola, so that the beginning of the parabola satisfies Eq. (10.10). Once 𝜀0 is known, 𝜀100 is found at the intersection of the two straight lines drawn on the 𝜀 vs. log t curve as shown in Figure 10.30. Then 𝜀50 is read on the curve halfway between 𝜀0 and 𝜀100 . The time t50 is read as the time corresponding to 𝜀50 on the curve. Once t50 is obtained, Eq. (10.8) is used to calculate cv . All other quantities are known, including T 50 = 0.197, and the drainage length as described previously. The square root of time (Taylor) method consists of plotting the 𝜀 vs. t curve √ as 𝜀 vs. t curve (Figure 10.31). Then a straight line is fitted to the early part of the curve (AB on Figure 10.31). A straight line with a slope equal to 1/1.15 times the slope of the first line is then drawn through point A (AC√ on Figure 10.31). The t curve corresponds intersection of line AC with the 𝜀 vs. √ to t90 . Once t90 is known, Eq. (10.8) is used to calculate cv . All other quantities are known, including T 90 = 0.848, and the drainage length as described previously. The preconsolidation pressure 𝜎p′ is another important soil parameter that can be obtained from the consolidation test. It is the effective vertical stress before which the deformation of the soil is small and after which the deformation of the soil increases more rapidly. It can be thought of as a vertical yield stress, although failure does not necessarily happen at 𝜎p′ . This effective stress corresponds to the highest long-term effective stress that the soil has been subjected to. The following procedure is recommended to obtain 𝜎p′ from the consolidation test (Figure 10.32). Choose the point of highest curvature on the

0

10 20 30 Square root of time, (min1/2)

40

Figure 10.31 Square root of time method to obtain the coefficient of consolidation. (Source: After Taylor, 1948.)

𝜀 vs. log 𝜎 ′ curve (Point A on Figure 10.32); then draw a horizontal line through that point and a line tangent to the curve at that point. Then draw the bisectrice of the angle formed by these two lines. Draw the straight line that best fits the portion of the 𝜀 vs. log 𝜎 ′ curve past the 𝜎p′ value. The intersection between this best-fit straight line and the bisectrice is a point that defines the preconsolidation pressure 𝜎p′ (Figure 10.32). The constant rate of strain procedure consists of the same procedure as the incremental loading procedure but with the following differences. The water is allowed to drain from the top of the sample but not from the bottom of the sample, where the water stress is measured. The sample is then deformed at a constant rate of displacement with time. This rate is chosen in such a way that the increase in water stress Δuw at the bottom 0.6

σ′p

0.55 A

0.5 Void ratio e

correspond to water being expelled from the pores (consolidation). Cassagrande proposed the following way to find 𝜀0 (Figure 10.30): Plot the 𝜀 vs. t curve as 𝜀 vs. log t; choose a point near the beginning of the 𝜀 vs. log t curve with coordinates 𝜀1 and log t1 ; find the point with coordinates 𝜀2 and log t2 = log 4 t1 ; calculate the difference 𝜀1 – 𝜀2 ; and find 𝜀0 as:

α α

0.45

0.4

0.35

0.3

σ′p 1

10

100 1000 Effective vertical stress, σ′ (kPa)

10000

Figure 10.32 Method to determine the preconsolidation pressure from the consolidation test.

250

10 LABORATORY TESTS

of the sample is kept at 5–10% of the vertical stress 𝜎 applied on the sample. The constant gradient procedure consists of the same procedure as the constant rate of strain procedure but with the following differences. When the load is applied, a water stress (pore pressure) Δuw develops throughout the sample. Soon the excess water stress at the top of the sample decreases to zero because drainage is allowed but the bottom water stress remains close to Δuw because the sample is not allowed to drain at the bottom. This creates a gradient between the top and bottom of the sample. This gradient is maintained constant as the load on the sample is slowly increased. However, at the end of each loading step, the water stress is allowed to dissipate to obtain an equilibrium compression of the soil. Advantages of the consolidation test include its relative simplicity and its yield of the response of a soil sample to one-dimensional confined compression. A drawback is that the confinement provided by the steel ring around the sample prevents lateral deformations and may not represent the true deformation of the soil in the field. 10.5.2

Unsaturated Soils

If the soil is unsaturated, the test procedures are unchanged. However, the water is in tension initially, when the sample is placed in the consolidometer. The increase in vertical stress on the sample as the test proceeds may create enough of an increase in water stress that it goes from tension to compression. If the soil is saturated, it is implicitly assumed that at the end of each 24-hour loading step in the loading step procedure, the water stress is zero; that way the effective stress on the sample can be calculated for each step. In the case of unsaturated soils, it becomes more difficult to calculate the effective stress on the sample. The following expression can be used if the air stress is zero (see Chapter 11): 𝜎 ′ = 𝜎 − 𝛼 uw

(10.11)

where 𝜎 is the effective stress, 𝜎 the total stress, 𝛼 the water area ratio coefficient, and uw the water tension stress. The coefficient 𝛼 can be estimated as the degree of saturation S, but the error can be ±40% of the correct value. A better estimate consists of using the air entry value, as shown in Chapter 11. Either way, obtaining 𝜎 ′ requires that the water tension uw be measured during the test. Most of the time, a soil in the saturated state with the water in compression is more compressible than the same soil in the unsaturated state with the water in tension. One exception is collapsible soils; with such materials, an unsaturated soil can experience significant and sudden compression when inundated (see Section 10.8). ′

10.6 10.6.1

to measure how much swelling takes place when a soil has access to water. Consider a sample of dry, clean gravel in a container: When you add water to it, the water will fill the voids, but when the voids are full, no more water will be absorbed by the gravel. Clean gravel does not swell during wetting. Now consider a dry piece of montmorillonite clay with a high dry density and place it at the bottom of a glass of water. The first thing that you will see through the wall is tiny explosions at the surface of the clay sample. The reason is that the water is drawn into the voids, but these voids are full of air that cannot escape because the water is coming in. This pressurizes the voids. The pressure increases until it overcomes the tensile strength of the dry clay, and a series of mini explosions is created. After a while the air finds a way to escape and the water enters the voids. The amount of swelling then depends on what the soil particles are made of. Montmorillonite minerals have a tremendous ability to attract water, so the swelling can be very significant for such clays and the sample may more than double in height. Swelling soils have very fine, highly plastic clay particles and are relatively dense. If they are located in regions where the water content of the soil varies significantly from one season to the next, they can create a lot of damage to structures, particularly light ones like houses, as they swell or shrink unevenly and distort those structures. If the water in the voids is in compression (below the groundwater level), then no swelling will take place. If the water in the voids is in tension (above the groundwater level), then more water will be attracted into the voids. Thus, the swell test is more useful for soils above the groundwater level. These soils may be saturated or unsaturated. The procedure for the swell test is the same for both saturated soils and unsaturated soils and is described in Section 10.6.2. 10.6.2

Unsaturated Soils

The swell test (Figure 10.33; ASTM 4546) consists of placing a soil sample in a snug-fitting cylindrical container (consolidometer ring), inundating the soil by placing it in a bath

SWELL TEST Saturated Soils

When soils absorb water, they may swell; some soils swell more than others. This is why it is important in many cases

Figure 10.33 Swell test equipment.

10.6 SWELL TEST

9

B Swell pressure

7

Vertical strain, ε (%)

Vertical strain, ε (%)

8

6 5 4 3 2

E H O

Prevent swell

D

F I

A

1

2

3 4 5 Time, t (days)

6

7

8

C J

G K

0

Increase in volume

Swell

1 0

251

Collapse

Vertical total stress, log σ

Figure 10.35 Shrink-swell test results: Vertical strain vs. vertical total stress.

Figure 10.34 Swell test results: Vertical strain vs. time.

of water, and measuring the vertical swell movement (vertical strain) as a function of time (Figure 10.34). The vertical strain is the change in height of the sample divided by the initial height of the sample. Water access to the sample is provided by porous disks placed at the top and bottom of the sample. The swelling can take days or even weeks. If the top of the sample is not subjected to any vertical load, the test is called a free swell test (path AB on Figure 10.35, path CD on Figure 10.36). If a vertical load is applied, the test is simply called a swell test (path DE on Figure 10.35, path CE on Figure 10.36). Note that after swelling, a regular consolidation test can be performed on the sample (paths BC and EF on Figure 10.35).

The free swell test gives the swell limit, which is the water content of the sample at the end of the free swell test (point B on Figure 10.35). The swell limit represents an upper limit of the water content that the soil can reach in the undisturbed state. When a vertical load is applied, it is usually applied before water is added on top of the sample and swelling starts. The magnitude of the load influences the swelling. It is often advantageous to apply a vertical stress on the sample equal to the stress that the soil will experience in the field (under the planned structure, for example). Sometimes the pressure is high enough that no swelling can take place, and settlement takes place instead. W

Swell limit W (free swell test) Wsw

Gravel, clean sand W Dirty sand, siltless Iss = 0–10 % Iss = 0

D E

WSH shrink limit (free shrink test)

Shrinkswell index C γw/γd 1 B Unsaturated

Iss = (ΔW)max

d

ate tur

Sa

Decrease in volume

W

(ΔV/V)max

0

ΔV/V ΔV/V Clay, low plasticity W Clay, high plasticity Iss = 10–30 % Iss = 30–70 %

ΔV/V

ΔV/V (b)

(a) W (%)

ΔV/V

W (%) Bentonite

76

Kaolinite

56%

38 24%

20

14 –0.15 0 0.17

ΔV/V

–0.3

0

0.5

(c)

Figure 10.36 Shrink-swell test results: Water content vs. relative change in volume. (a) Idealized behavior. (b) Typical ranges. (c) Low-plasticity clay example.

252

10 LABORATORY TESTS

Another way to run the swell test is to add the water first so that swelling can start and to increase the vertical stress on the sample gradually to prevent any swelling (path OI on Figure 10.35). During this test, the volume of the sample is maintained constant and equal to its initial volume. When the vertical stress reaches an equilibrium value, that stress is called the swelling pressure. Swelling pressures can reach 1000 kPa or higher for high-plasticity clays. 10.7

SHRINK TEST

10.7.1

Saturated Soils

The shrink test (Figure 10.37) consists of trimming a sample of soil into a cylindrical shape, measuring its dimensions, and recording its weight. The initial volume V o and the initial weight W o are recorded. Then the sample is left to dry while the dimensions and the weight are measured as a function of time. This gives the volume V(t) and weight W(t). When the sample is air-dried, it is placed in the oven to obtain the oven dry weight W s . The average water content w of the sample at any time during the test is (W(t) – W s )/W s . The results of the test consist of a plot of the relative change in volume (V(t) – V 0 )/V 0 = ΔV/V 0 as a function of time t (Figure 10.38), and the water content w as a function of the relative change in volume ΔV/V 0 (Figure 10.36). The undisturbed sample shrinkage limit wSH is the water content corresponding to

the point where the sample first stops decreasing in volume (point B on Figure 10.36 (a)). As in the case of the swell test, the shrink test can be performed without any vertical load applied (free shrink test) or with vertical load applied (shrink test). The free shrink test is much more common. 10.7.2

The shrink test applies equally to saturated soils and unsaturated soils. In both cases the water is in tension throughout the test. The soil may start as a saturated soil, but, as it dries, it goes through the air entry value uwae , at which point it becomes unsaturated. The shape of the relative volume change vs. time curve for the free shrink test (see Figure 10.38) is similar to the shape of the relative volume change vs. time curve for the free swell test (see Figure 10.34). During the free shrink test, the weight of the sample is measured as a function of time, so it is possible to plot the water content as a function of relative volume change (BCD on Figure 10.36a). This curve indicates where the undisturbed shrink limit wSH occurs. Note that the undisturbed shrink limit is more obvious for low-plasticity soils than for high-plasticity soils. The undisturbed shrink limit is different from the Atterberg shrink limit, which is obtained on a remolded sample. 10.8 10.8.1

Relative change in volume, ΔV/V (%)

Figure 10.37 Free shrink test for shrinkage limit of undisturbed sample. 0 2 4 6 8

10 0

10

20

30 40 Time, t (hrs)

50

60

Figure 10.38 Free shrink test result: relative change in volume vs. time.

Unsaturated Soils

COLLAPSE TEST Saturated Soils

Consider a natural sample of dry silt with a low dry density and a reasonable strength. Place it in a steel ring and place some weight on top of the sample. In the dry state, the sample has no problem carrying the load without much deformation. Now add water on top of the sample: You will likely see a significant amount of compression take place due to collapse of the soil skeleton. What happens is that the small amount of water tension that exists at the contacts between the silt particles is lost when the water enters the voids and the loose structure of the silt collapses. It is important to check if a soil is collapsible; you can imagine the distress associated with any structure built on such soils if a significant amount of water permeates below the foundation. Collapsible soils consist of loose, dry, low-density materials (say, less than 16 kN/m3 ) that decrease in volume (collapse and compact) with the addition of water. These soils are often found in arid regions, specifically in areas of windblown silty sediments (loess), young alluvial fans, and debris flow sediments. Soil collapse occurs within soils above the groundwater level. The process of saturation weakens or eliminates the clay bonds holding the soil grains together through water tension. 10.8.2

Unsaturated Soils

The collapse test (ASTM D5333) is the same for saturated and unsaturated soils. It is performed with the sample confined

10.9 DIRECT SHEAR TEST

0 2 Vertical strain, ε (%)

Collapse due to inundation 4 6 8 10 12

1

10

100

1000

10000

Vertical total stress, σ(kPa)

Figure 10.39 Collapse test: vertical strain vs. vertical stress.

in a consolidometer ring. Typically, it consists of loading the soil sample to a vertical stress equal to the vertical total stress that the soil will experience at a chosen depth, recording the vertical strain vs. time curve (consolidation test), and then (once the compression is complete) inundating the sample while continuing to record the vertical strain vs. time curve. Once the collapse is completed, the consolidation test can be resumed by increasing the vertical stress. A sample vertical strain vs. vertical stress curve is shown in Figure 10.39.

10.9 10.9.1

DIRECT SHEAR TEST Saturated Soils

The direct shear test (ASTM D3080) is a simple test used to obtain the shear strength of a soil. A disk of soil is placed in a steel cylinder split horizontally at mid-height (Figure 10.40). The steel split cylinder is made of two rings stacked on top of each other. One filter stone is placed on top and one at the bottom of the sample so that the water can drain from the sample during the test. A vertical load is applied to the top of the sample and maintained constant during the test. This vertical load creates a total normal stress 𝜎. Then the soil sample is sheared horizontally by pushing on the bottom ring while holding the top ring. This forces a shear plane to develop around the mid-height of the sample. During the shearing process, the shear force is measured with a load cell, the horizontal displacement with an LVDT or dial gage, and the vertical displacement with an LVDT or dial gage. The result of a direct shear test is a shear stress vs. horizontal displacement curve and, if the vertical movement is also measured, a vertical movement vs. horizontal movement curve (Figure 10.41). During the first part of the direct shear test, the soil sample is allowed to consolidate under the vertical stress applied, if such a stress is applied. The consolidation is monitored by recording the vertical movement of the sample as a function

253

of time. When the settlement stops or becomes very small, it is assumed that the water stress has returned to zero and the shearing part of the test can start. During the second part of the test, the sample is sheared and shearing takes place along a thin horizontal band at mid-height of the sample near the junction between the two steel rings. The shear stress versus horizontal movement curve is obtained point by point. The shear strength is the maximum shear stress on the shear stress versus horizontal movement curve. This shear strength is the undrained shear strength if the shearing part of the test is run quickly enough that water does not have time to drain; it is the drained shear strength if the test is run slowly enough that the water stress remains zero. It is best also to measure the pore pressure or water stress, but that is not common with this simple test. The shear strength measured in an undrained direct shear test is the undrained shear strength su . This undrained shear strength corresponds to the effective stress 𝜎 ′ generated at the end of the consolidation phase. This undrained shear strength also corresponds to the stress path followed in a direct shear test. The shear strength measured in a drained direct shear test provides one point on the shear strength envelope. This envelope links the shear strength to the effective stress 𝜎 ′ normal to the plane of failure. As described in Chapter 16 on shear strength, the envelope is represented by the following equation: (10.12) s′ = c′ + 𝜎 ′ tan 𝜑′ This equation has two soil parameters: the effective stress cohesion c′ and the effective stress friction angle 𝜑′ . Because the drained direct shear test gives only one point on the envelope, it is necessary to run at least two direct shear tests to obtain c′ and 𝜑′ for a given soil (Figure 10.42). When soils are subjected to shearing, they can increase in volume (dilate), decrease in volume (contract), or not change volume (critical state). If a soil dilates during shear, the shear strength increases compared to a soil that does not change in volume. The increase in shear strength is reflected by the dilation angle 𝜓 (see Chapter 16 for more details). The dilation angle 𝜓 can be estimated from a direct shear test as the slope of the curve linking the vertical movement z to the horizontal movement x. Because this curve is rarely a straight line, the equation is written in an incremental fashion. tan 𝜓 = Δz∕Δx

(10.13)

Advantages of the direct shear test include that it is easy to perform and gives a shear strength of the soil. A drawback of the direct shear test is that it cannot give the shear strain of the soil as it is sheared, because the thickness of the shearing zone is not known. 10.9.2

Unsaturated Soils

If the soil is unsaturated, or if the soil is saturated but the water in the voids is in tension (e.g., above the groundwater level), then the direct shear test requires measurement of the

254

10 LABORATORY TESTS

0 Water grooves

Loading yoke

Water inlet

Pressure pad

Porous stone Sample Horizontal force

Loading cell

Porous stone

Large container

(a)

Water grooves

(c)

(b)

Figure 10.40 Direct shear test and equipment. (a) Principle. (b) Sample. (c) Complete setup. (Source: (b) Courtesy of Lev Buchko, P.E.//Timely Engineering Soil Tests, LLC.) τ

τ

Peak

150 Critical state

𝛿v

𝛿h (mm) 𝛿v

τ

σ′

Residual

Dilates

𝛿h Contracts

𝛿h (mm)

Shear stress τ (kPa)

σ′

𝜎 ′ = 𝜎 − 𝛼 uw

(10.14)

Right 50 0 c′ 0

Figure 10.41 Direct shear test results stress-displacement curve.

50 100 150 200 Normal effective stress σʹ (kPa)

250

150 Shear stress τ (kPa)

water tension stress (suction) to obtain the effective stress shear strength parameters c′ and 𝜙′ . Indeed, although the test procedure is the same for a soil with water in compression and for a soil with water in tension, the assumption that the water stress is zero when the test is performed slowly is not valid when the water is in tension. The reason is that if the water is in compression at the beginning of the direct shear test, the water compression stress is very small compared to the general stress level; in contrast, if the water is in tension, the water tension stress can be very large when the degree of saturation is low. The water tension stress uw can be measured by any one of the methods described in Section 10.2.4, but it is most often done with a tensiometer during the shear test. Once the water tension stress is known, the effective stress (assuming the air stress ua is zero) is calculated as:

φ′ 100

uw3 100 uw2

≠φ′

Wrong

uw1

50 ≠c′

0 0

200 50 100 150 Normal total stress σ (kPa)

250

Figure 10.42 Example of direct shear test strength results for saturated soils.

where 𝜎 ′ is the effective stress, 𝜎 the total stress, 𝛼 the water tension coefficient, and uw the water tension stress. The coefficient 𝛼 can be estimated as the degree of saturation S, but

10.10 SIMPLE SHEAR TEST

255

Shear stress τ (kPa)

300 Vertical stress

200 σtan φ′

Brass rings

Right

100 Base pedestal

Capp = (–αuw–βua) tanφ′ c′

0 0

c′ 200 100 300 400 Normal effective stress σʹ (kPa)

300 Shear stress τ (kPa)

Top cap

φ′

(a)

Figure 10.44 Simple shear test equipment: (a) Principle. (b) Complete setup. (Source: (b) Courtesy of GDS Instruments.)

–α2uw2 ≠φ′ –α1uw1

Wrong

100 ≠c′ 0 100 200 300 400 Normal total stress σ (kPa)

500

Figure 10.43 Example of direct shear test strength results for unsaturated soil.

the error can be as large as ±40% of the correct value. A better estimate can be obtained by using the correlation to the air entry value uwae as shown in Chapter 11. It is assumed here that the air stress remains zero during the test. The results are then plotted as shear strength vs. effective normal stress, as shown in Figure 10.43. If the results of direct shear tests on soils where the water is in tension are plotted as shear strength vs. total stress, the cohesion intercept will be much larger, as it includes the effect of the water tension on the soil strength (Figure 10.43). The apparent cohesion capp is equal to: (10.15) capp = −𝛼 uw However, capp is not a constant for a given soil, because uw depends on the water content of the sample. The apparent cohesion is called apparent rather than true cohesion because it is due to the effective stress created by the water tension and because it disappears if the soil is inundated (water tension goes to zero). In contrast, the parameter c′ is a characteristic of the soil that is constant and independent of the water content.

10.10 10.10.1

(b)

500

–α3uw3

200

0

Shear stress

SIMPLE SHEAR TEST Saturated Soils

The simple shear test (ASTM D6528) can be traced back to the mid-1960s with a publication by Bjerrum and Landva (1966). A disk of soil is placed in a flexible membrane with a porous stone on the top and on the bottom of the disk (Figure 10.44). A vertical load is applied to the top of the

sample and maintained constant during the test. This vertical load creates a total normal stress 𝜎. Then the soil is sheared by holding one of the two platens and pushing the other one horizontally. The major difference between the direct shear test and the simple shear test is that in the direct shear test, the shearing takes place along a predetermined thin band of soil near the middle of the sample. In the simple shear test, the shearing takes places over the entire height of the sample. Therefore, the shearing strain 𝛾 can be measured in the simple shear test as: 𝛾 = Δx∕ho

(10.16)

where Δx is the difference in horizontal movement between the top and the bottom of the sample and ho is the initial height of the sample. The shear stress 𝜏 is measured as the shear force divided by the cross-sectional area of the sample. Thus, the simple shear test gives the shear stress-shear strain curve for the sample and therefore a shear modulus G. During the first part of the simple shear test, the soil sample is allowed to consolidate (through drainage) under the vertical stress applied if such a stress is applied. The consolidation is monitored by recording the vertical movement of the sample as a function of time. When the settlement stops or becomes very small, it is assumed that the water stress has returned to zero and the shearing part of the test can start. During the second part of the test, the sample is sheared. The shear stress vs. shear strain curve is obtained point by point (Figure 10.45). The shear strength 𝜏 f is the maximum shear stress on the shear stress vs. shear strain curve. This shear strength is the undrained shear strength if the shearing part of the test is run without allowing water to drain out of the sample; it is the drained shear strength if the test is run slowly enough that the water stress remains zero or if the water stress (pore pressure) is measured. The shear strength 𝜏 f is obtained in the same fashion as for the direct shear test, including the shear strength parameters c′ and 𝜑′ . The shear modulus G is the slope of the τ vs. 𝛾 curve. Because the curve is typically nonlinear, G varies with 𝛾 and a G vs. 𝛾 curve can be generated. Therefore, an advantage of the simple shear test is that it can give the shear modulus G as a function of shear strain, in addition to the shear strength of the soil sample.

256

10 LABORATORY TESTS

τ

σ′ τ τ

Peak

Critical state γ

𝛿v γ

εv

σ′

Dilates

γ

Contracts

Figure 10.45 Simple shear test results.

When soils are subjected to shearing, they can increase in volume (dilate), decrease in volume (contract), or not change volume (critical state). If a soil dilates during shear, the shear strength increases compared to a soil that does not change in volume. If a soil contracts during shear, the shear strength decreases compared to a soil with no change in volume. The increase or decrease in shear strength is reflected by the dilation angle 𝜓 (see Chapter 16 on shear strength for more details). The dilatancy angle 𝜓 can be estimated from a simple shear test as the slope of the curve linking the change in vertical movement Δz to the change in horizontal movement Δx (Eq. (10.13)). 10.10.2

su = qu ∕2

Unsaturated Soils

If the soil is unsaturated, or if it is saturated but the water is in tension, the testing procedure is unchanged except for measurement of the water stress. The tensile stress in the water will typically require the use of a different measuring device, such as a tensiometer. The data reduction requires calculation of the effective stress, as discussed for the direct shear test.

10.11 10.11.1

times the diameter. The ratio h/d is about 2, to ensure that the oblique shear plane that typically develops during failure can propagate through the entire sample without intersecting the top or bottom platen. The sample remains unconfined during the test; therefore, the minor principal stress 𝜎 3 is zero. A vertical load is applied to the sample by pushing up on the bottom platen at a constant rate of displacement while holding the top platen in a fixed position. The vertical total stress 𝜎 is calculated by dividing the vertical load by the cross-sectional area of the sample. Because it is assumed that there is no shear between the top of the sample and the bottom of the top platen, that stress is the major principal stress 𝜎 1 . The sample compresses and the vertical displacement Δh is measured with an LVDT or a dial gage. Knowing the initial height h of the sample, the vertical strain 𝜀 can be obtained as 𝜖 = Δh∕h. The result of an unconfined compression test is a complete total stress 𝜎 vs. strain 𝜀 curve for the soil sample under zero lateral confinement (Figure 10.47). The maximum stress on the curve is the unconfined compression strength qu . Because the test is rather rapid, the shearing process is considered to be undrained for fine-grained soils. The undrained shear strength su is equal to qu /2, as shown in Chapter 16.

UNCONFINED COMPRESSION TEST

(10.17)

An unconfined compression modulus of deformation E can also be obtained from this test as: E = 𝜎1 ∕𝜖

(10.18)

Because the curve is often nonlinear, several moduli can be obtained depending on the chosen strain level. Advantages of the unconfined compression test are its simplicity and the fact that it gives both an undrained shear strength and a modulus of deformation for fine-grained soils.

Saturated Soils

The unconfined compression test (ASTM D2166) (Figure 10.46) is one of the simplest tests to perform if the soil can stand up under its own weight. In this test, the sample is a cylinder with a diameter d and a height h equal to about 2

Load transducer

Displacement transducer

10.11.2

Unsaturated Soils

If the soil is unsaturated, the test procedure is unchanged. Because the water stress is not measured in this test, there is also no difference in measurement and data reduction. One interesting observation is that the water tension can be estimated from the unconfined compression strength qu . Indeed, the shear strength equation for unsaturated soils when the air stress ua is assumed to be zero is: s = c′ + (𝜎 − 𝛼 uw ) tan 𝜙′

Soil sample

In the unconfined compression test, the horizontal total stress is zero, therefore: 𝜎h = 0 = 𝜎 ′ h + 𝛼uw and 𝜎 ′ h = −𝛼uw

(a)

(10.19)

(b)

Figure 10.46 Unconfined compression test equipment: (a) Principle. (b) Complete setup. (Source: (a) Adapted from Ian Smith. (b) Courtesy of ELE International.)

(10.20)

Meanwhile, the vertical total stress at failure is equal to qu : 𝜎v = qu = 𝜎 ′ v + 𝛼uw and 𝜎 ′ v = qu − 𝛼uw

(10.21)

The shear strength s is given by the point of tangency between the effective stress Mohr circle and the shear strength

10.12 TRIAXIAL TEST

160

257

qu

Axial stress, σ (kPa)

140 120 100 80 E

60 40 20

εf

0 2

Shear stress, τ (kPa)

0

4 6 Axial strain, ε(%)

8

10

100 80

su

60 40

su = qu /2

qu

20 40

80

120

160

Total stress, σ (kPa)

Figure 10.47 Unconfined compression test results.

envelope (Figure 10.48). Triangle ACD on Figure 10.48 is such that: (( ) ( )) 0.5 q − 𝛼u − −𝛼u u w w CD = sin 𝜑′ = (( ) ( )) ′ AD 0.5 qu − 𝛼uw + −𝛼uw + tanc 𝜑′ (10.22) which leads to uw =

0.5qu (sin 𝜑′ − 1) + c′ cos 𝜑′ 𝛼 sin 𝜑′

(10.23)

Equation (10.23) gives the water tension at failure in the unconfined compression test. If it is further assumed that c′ =

Uw = – Shear stress τ

qu for c = 0, φ′ = 30° 2s φ′ C

B c′ A

c′ O –αuw tanφ′

D qu–αuw Normal effective stress, σ′

Figure 10.48 Water tension and unconfined compression strength relationship. In the figure s is the degree of saturation.

0, 𝜙′ = 30∘ , and 𝛼 = S, then Eq. (10.23) becomes: q uw = − u 2S 10.12 10.12.1

(10.24)

TRIAXIAL TEST Saturated Soils

The triaxial test (ASTM D5311) (Figure 10.49) is similar to the unconfined compression test except that a chosen confining pressure is applied to the sample before compression takes place. The sample has a height equal to about two times the diameter to ensure that the oblique shear plane that typically develops during failure in compression can propagate through the entire sample without intersecting the top or bottom platen. Typical diameters range from 30 to 75 mm. First, porous disks (also called filter stones) are placed at the top and bottom of the sample. Then the sample is fit in an impervious rubber membrane and set on the pedestal of the triaxial cell. The top platen is placed, and the top of the triaxial cell is brought down to cover the sample. The shaft of the piston is lowered in contact with the top platen on one side and connected to the load cell or proving ring on the other. The cell is filled with liquid (water or oil) and the confining pressure is applied. Sometimes the cell is not filled with liquid and only air pressure is used. The triaxial cell is placed in a frame and the load is applied by moving the bottom of the frame upward and at a constant rate of displacement against the stationary

258

10 LABORATORY TESTS

(a)

(b)

Load

Piston Cap Fluid inside chamber

Sealing ring

Perspex cylinder Soil specimen Membrane

Porous disc Pore-pressure measurement and drainage

Cell pressure measurement

Figure 10.49 Triaxial test equipment: (a) Principle. (b) Equipment. (Source: (b) Courtesy of Geotechnical Testing Equipment Ltd., UK.)

top of the frame. Pore-pressure measurements are an option and are typically made by placing a saturated porous stone at the base of the sample and measuring the pressure in the water through a pressure transducer tied to the base platen. Measuring the change in volume of the sample is also an option. The movement of the sample is typically obtained by measuring the movement of the shaft applying the load with respect to the triaxial cell. For more advanced testing, the movement measurements are taken between two rings directly tied to the sample. Measuring the change in diameter of the sample is rarely done but measuring the change in volume of the sample is sometimes done. More recently, efforts have been made to develop techniques based on photogrammetry to measure

the full displacement field by placing targets on the sample and taking regular photos with a camera (Zhang et al., 2015, Figure 10.50). For dynamic measurements of displacement, an array of cameras is mounted on rings around the sample (Xia et al., 2022). There are many different types of triaxial tests because of the possible combinations related to drainage and type and sequence of stress applications. However, nearly all triaxial tests start with a consolidation phase followed by a shearing phase. The consolidation phase is designed to bring the sample to a desired state of stress that is often intended to match the stress conditions that the sample would face in the field under the project conditions. During the consolidation

(b) (a)

(c)

Figure 10.50 (a) System setup for the photogrammetry-based method; (b) Design of a coded target for automatic recognition; (c) Reconstructed 3D model for the triaxial testing system; (d) Strain localization in the soil specimen. (Source: Courtesy of Professor Xiong Zhang, Missouri University of Science and Technology.)

(d)

10.12 TRIAXIAL TEST

phase, the cell pressure is increased to a chosen value of the confining pressure. This pressure confines the sample hydrostatically and represents the minor principal stress 𝜎 3 . During this phase of consolidation, drainage may or may not be allowed. If drainage is not allowed, the word “unconsolidated” is used in describing the triaxial test and the letter U is used in the acronym. If drainage is allowed and the water stress (pore pressure) generated by the application of 𝜎 3 is allowed to dissipate back to zero, the word “consolidated” is used to describe the test and the letter C is used in the acronym. During the shearing phase of the test, the vertical load Q on the sample is increased gradually and the stress in the vertical direction increases. This stress is the major principal stress 𝜎 1 : 𝜎1 = 𝜎3 + Q∕A

(10.25)

where 𝜎 3 is the confining pressure, Q is the vertical load and A is the cross-section of the sample. If drainage is not allowed during the shearing phase, the word “undrained” and the letter U are used. If drainage is allowed and the excess water stress (pore pressure) is kept equal to zero (very slow loading), then the word “drained” and the letter D are used. So, in the end, the following triaxial tests are possible: 1. UU test: unconsolidated undrained test. 2. CU test: consolidated undrained test. 3. CD test: consolidated drained test. A UD test is not possible, because allowing drainage during the shearing phase would also allow some consolidation under 𝜎 3 . UU tests are commonly performed to obtain the undrained shear strength, particularly in offshore studies where recompressing the sample to the high bottom pressures is important; UU tests are also simpler and faster than the other two. CD tests are quite time-consuming, as loading must be slow enough not to generate water stresses (pore pressures), but they are simple to run. CU tests with water stress (pore pressure) measurements are faster to run, but require more sophisticated equipment because water stress (pore pressure) must be measured. Both CD tests and CU tests with water stress measurements are used to obtain the effective stress shear strength parameters c′ and 𝜑′ . The result of a triaxial test is a stress-strain curve that typically links the deviator stress (𝜎1 − 𝜎3 ) to the vertical strain (𝜀 = Δh/h) where h is the initial height of the sample and Δh is the change in height of the sample. Figure 10.51 shows some results for two categories of soils: overconsolidated or dense soils, on the one hand, and normally consolidated or loose soils, on the other. The first category exhibits a clear peak stress (maximum strength), followed by strain softening to reach a residual strength. The second category exhibits strain hardening, with the strength being reached at larger strain. The peak stress value on this curve is the failure deviator stress (𝜎 1f – 𝜎 3 ). This failure stress, along with information on the water stress, is used to obtain the effective stress shear

259

strength parameters c′ and 𝜑′ . This process requires use of the Mohr circle (see Figure 10.52 and Chapter 16). A Mohr circle is a circle in the shear stress vs. normal stress set of axes that describes the state of stress at a point when the principal stresses reduce from three stresses to two stresses. This is the case in the triaxial test where 𝜎1′ and 𝜎3′ are different and 𝜎3′ is equal to 𝜎2′ . The points corresponding to the principal stresses 𝜎1′ and 𝜎3′ plot on the horizontal axis because they exist on planes with zero shear stress. The circle representing the state of stress in the triaxial sample at failure is drawn (Figure 10.52). Because the failure envelope is described by two parameters c′ and 𝜙′ (Eq. (10.12)), a minimum of two triaxial tests at two different confining pressures (𝜎 3 ) must be performed to obtain the effective stress cohesion intercept c′ and the effective stress friction angle 𝜙′ . Figure 10.53 shows the difference between the Mohr circles in the effective stress set of axes and in the total stress set of axes. A modulus of deformation E can also be obtained from the stress-strain curve as follows: E = (𝜎1 − 2v𝜎3 )∕𝜀

(10.26)

where E is the total stress modulus of deformation of the soil, 𝜎 1 and 𝜎 3 are the major and minor principal total stresses, respectively, v is the Poisson’s ratio, and 𝜀 is the vertical strain. Note that because the stress-strain curve is rarely linear, many different moduli can be obtained depending on the strain level among other factors. The modulus defined in terms of effective stress is typically more useful and more fundamentally rooted: ) ( (10.27) E′ = 𝜎1′ − 2v𝜎3′ ∕𝜀 where E′ is the effective stress modulus of deformation of the soil, and 𝜎1′ and 𝜎3′ are the major and minor principal effective stresses, respectively. The stress path describes the evolution of certain stresses during the test. Specifically, it tracks the path described by the points with p, q stress coordinates where p and q are defined as follows: 𝜎 + 𝜎h 𝜎 + 𝜎3 or p = v (10.28) p= 1 2 2 𝜎 − 𝜎h 𝜎 − 𝜎3 or q = v (10.29) q= 1 2 2 where 𝜎 v and 𝜎 v are the vertical and horizontal total stresses in a triaxial test, for example. The most useful stress paths are plotted in terms of effective stresses (p′ and q′ ): 𝜎′ + 𝜎′h 𝜎′ + 𝜎′3 p′ = 1 or p′ = v (10.30) 2 2 𝜎′ − 𝜎′h 𝜎′ − 𝜎′3 q′ = 1 = q or q′ = v = q (10.31) 2 2 where 𝜎v′ and 𝜎h′ are the vertical and horizontal total stresses in a triaxial test, for example. Examples of effective stress paths are shown in Figure 10.54 for different types of tests. In any lab test, it is most desirable to match the effective stress path followed by the soil in the field during the project construction and the project life.

10 LABORATORY TESTS

200

200 Normally consolidated or loose

Deviator stress: σ1–σ3 (kPa)

Overconsolidated or dense

150

150

100

0

100

Maximum or peak strength

50

0

2

4

Residual strength 6

50

10

8

75

0

0

2

4

Water stress (pore pressure), uw (kPa)

50

25

25

0

2

4

6

10

8

–25

0

0

2

4

Deviator stress: σ1–σ3 (kPa)

Normally consolidated or loose

150

150

100

100

Maximum or peak strength

50

0

2

4

Residual strength 6

50

10

8

4

0

0

2

4

6

10

8

4 Overconsolidated or dense

Volumetric strain, εν (%)

10

8

200 Overconsolidated or dense

Normally consolidated or loose

2 0

6

–25

200

0

10

8

Normally consolidated or loose

50

0

6

75 Overconsolidated or dense

2

0

2

4

6

10

8

0

–2

–2

–4

–4

0

2

4

6

8

10

Figure 10.51 Triaxial test results (example stress-strain curves): (a) Overconsolidated or dense. (b) Normally consolidated or loose.

250

250 σ′3f = 100 kPa

200 150

200 Shear stress σ (kPa)

Deviatoric stress σ1–σ3 (kPa)

260

σ′3f = 60 kPa

100 σ′3f = 30 kPa

50 0

2

4

6

Axial strain ε (%)

8

100 50

c = 10 0

Φ = 30°

150

10

c = 10 kPa

0

0

100

200

300

Effective normal stress σʹ (kPa)

Figure 10.52 Triaxial test results: Examples of Mohr circles and strength envelope for saturated soils.

400

10.12 TRIAXIAL TEST

2. CU test: consolidated undrained test. For unsaturated soils, both air and water are allowed to drain during the consolidation phase. During the shearing phase, both are prevented from draining, so both pressures must be measured. Typically, the air stress and the water stress increase (decrease in water tension) during the shearing phase because the soil volume decreases (except for dilatant soils). 3. CD test: consolidated drained test. Both the air and the water are permitted to drain. The water tension can therefore be held constant throughout the test. The strain rate must be sufficiently slow to allow for flow of water from the soil through the high air entry disk. 4. CWC test: constant water content test. For unsaturated soils, it is also possible to conduct a test where the air can drain but not the water. Air drains much faster than water, so a judiciously chosen strain rate can achieve this condition.

Shear stress τ (kPa)

150 φ′

100 Right 50

c′

0 0

50

250 150 100 200 Normal effective stress σʹ (kPa)

300

Shear stress τ (kPa)

150

≠φ′

100

Wrong 50 ≠c′

0 0

uw1

50

uw2

uw1

100 150 200 Normal total stress σʹ (kPa)

uw2

250

300

Figure 10.53 Triaxial test results: Mohr circles and strength envelope for saturated soils.

10.12.2

The data reduction changes as well. The effective stress must be calculated according to the following formula (instead of 𝜎 ′ = 𝜎 − uw ): 𝜎 ′ = 𝜎 − 𝛼 uw − 𝛽 ua

Unsaturated Soils

If the soil is unsaturated, or if it is saturated and the water in the voids is in tension, the test procedure does not change, but the water and air stress measurements change. The water stress can be measured with a tensiometer and the air stress with a pressure transducer. The meaning of the tests that were described for saturated soils changes as well: 1. UU test: unconsolidated undrained test. For unsaturated soils, UU means that both the air and water are prevented from draining from the beginning to the end of the test. The air stress increases as the air compresses and the water stress increases (decrease in the absolute value of the water tension).

Half deviator stress q = q′ = (σ1 – σ3)/2 (kPa)

261

(10.32)

where 𝜎 ′ is the normal effective stress, 𝜎 the normal total stress, 𝛼 the water area ratio parameter, uw the water stress, 𝛽 the air area ratio parameter, and ua the air stress. This difference will affect the location of the Mohr circle on the shear stress 𝜏 vs. effective normal stress 𝜎 ′ graph. If instead the results are plotted in the shear stress 𝜏 vs. total normal stress 𝜎 graph, then the effective stress shear strength parameters c′ and 𝜙′ cannot be obtained. The cohesion intercept c in the shear stress 𝜏 vs. total normal stress 𝜎 graph is much larger than c′ , as it includes the effect of the water tension on the soil strength (Figure 10.55). The apparent cohesion capp is equal to: capp = −𝛼 uw − 𝛽 ua

Effective stress path (lightly overconsolidated) Effective stress path (normally consolidated) Effective stress path (heavily overconsolidated)

Total stress path

Mean effective stress p′ = (σ′1 + σ′3)/2 (kPa)

Figure 10.54 Triaxial test results: stress paths.

(10.33)

262

10 LABORATORY TESTS

10.13

Shear stress τ (kPa)

300

10.13.1

φʹ 200 Right σtanφʹ 100 capp = (–αuw –βua) tanφʹ

0 cʹ 0

cʹ 100 200 300 400 Normal effective stress σ′ (kPa)

600

500

Shear stress τ (kPa)

300 ≠φʹ 200 Wrong 100 –α1uw1

≠cʹ 0

–α2uw2

100

–α1uw1

–α2uw2

200 300 400 Normal total stress σ (kPa)

500

600

Figure 10.55 Triaxial test results: Mohr circles and strength envelope for unsaturated soils.

This cohesion is called apparent cohesion rather than true cohesion because it is due to water tension and because it disappears if the soil is inundated (water and air stresses go to zero).

RVDT

RVDT guide pins

Torsional accelerometer

Mounting post

The resonant column test (ASTM D4015) is used to determine the dynamic small strain properties of a soil. Such results are useful in earthquake engineering and machine vibration, for example. A cylinder of soil with a height-to-diameter ratio of about 2 is placed in a cell where a confining pressure can be applied. The base of the sample is fixed to the bottom platen, which does not move. The top of the sample is mounted with a top platen having a mass m and able to generate cyclic torsion (Figure 10.56). The test consists of applying a sinusoidal torque T(𝜔) to the top of the sample. This torque is generated through an electromagnetic drive system that controls the angular frequency 𝜔 of the sinusoidal torque application. The response of the sample is monitored by measuring (through LVDTs, for example) the rotation of the top of the sample. The water stress (pore pressure) is sometimes also measured during this test. In a first step, a confining pressure is applied to the sample. Then the top of the sample is subjected to a chosen torque. The torque applied gives the shear stress 𝜏 imposed on the sample and the rotation 𝜃 is used to obtain the shear strain 𝛾 of the sample. The response is presented in term of loops linking 𝜏 to 𝛾. The frequency of the sinusoidal torque is increased gradually while recording the strain in the sample. Resonance occurs when the frequency of the soil vibrations matches the frequency of the torque application (Figure 10.57). This

LVDT casing LVDT core

Fluid bath

Drive coil

Saturated Soils

Suspending spring

Support stand RVDT guide bracket

RESONANT COLUMN TEST

Permanent magnet Top drive plate

Top cap inner Soil specimen

Containment cell Rubber membrane 0-ring

Porous disc Base pedestal

Drainage lines

Base plate (a)

(b)

Figure 10.56 Resonant column test: (a) Principle. (b) Equipment. (Source: (b Courtesy of Geotechnical Research Lab, Dept. of Civil Engineering, University of British Columbia.)

263

Amplitude

Amplitude

10.13 RESONANT COLUMN TEST

1 0.707

1

0.707

ln A1 δ=1 n An+1 f1 fn f2

D=

Frequency

Figure 10.57 Rotation amplitude vs. frequency of induced vibration.

frequency is 𝜔n . At that point the sample rotation reaches its maximum value. The data are used as follows to obtain the soil shear modulus G when the sample is fixed at the bottom and free at the top where the torque is applied. The mass polar moment of inertia of the sample J s is: Js = Ms d2s ∕8

(10.34)

where M s is the sample mass and ds is the sample diameter. The mass polar moment of inertia of the mass on top of the sample Jm is: (10.35) Jm = Mm d2m ∕8 where Mm is the mass of the mass on top of the sample and dm is the diameter of that mass. By using fundamental and constitutive equations, it can be shown that: ( ( ) ) J Js 𝜔L 𝜔n L 2𝜋fn L 2𝜋fn L = n tan tan or s = Jm vs vs Jm vs vs (10.36) where J s and J m are the polar moments of inertia of the sample and of the mass on top of the sample respectively, 𝜔n is the resonant angular frequency, vs is the shear-wave velocity in the sample, L is the length of the sample, and f n is the natural frequency of the soil. In Eq. (10.36), J s , J m , and L are known, f n is measured in the test, and vs can be back-calculated. Then the shear modulus is obtained from: (10.37) G = 𝜌 v2s For the case where there is no mass at the top, Jm = 0, then 2𝜋fn L∕vs = 𝜋∕2, and then G = 𝜌 v2s = 16 𝜌f2n L2

fn f2 Frequency (f –f ) D= 2 1 2fn

(a)

(b)

Figure 10.58 Method to obtain damping ratio from resonant column test: (a) Logarithmic decrement. (b) Half-power bandwidth.

amount of damping that results in the sample returning to its original position without oscillation. The damping ratio can be obtained from the decay curve (Figure 10.58) as follows. The amplitude of the first cycle is x1 and the amplitude of the nth cycle is xn , which is smaller than x1 . It can be shown that: Lnx1 − Lnxn 2𝜋D (10.39) =√ n−1 1 − D2 In Eq. (10.39), all quantities are known except for D, the damping ratio. The damping obtained by this method includes the damping of the device, which must be accounted for separately. This method also requires stopping the test, and the strain level decreases during the vibration decay. Another way to obtain the damping ratio is to use the half-power bandwidth method. This method makes use of the amplitude vs. frequency plot (Figure 10.58) obtained during steady-state torsional vibration of the sample: D = (f2 − f1 )∕2fn

(10.40)

where f 2 , f 1 , and fn are defined in Figure 10.58. This method is best applied when the system is linear. These curves can also be obtained from direct measurements of the shear stress and the shear strain. The maximum shear stress 𝜏 generated during the cycles is calculated as an average of the shear stress generated on the sample cross section. This shear stress is zero at the center of the sample (𝜏 center = 0) and maximum at the edge (𝜏 edge ) (Figure 10.59). τmax

(10.38)

Eq. (10.38) gives the shear modulus G for a given shear strain amplitude 𝛾. There are several ways to obtain the damping ratio, and each way has its own advantages and limitations. One way is to stop the excitation and let the sample vibration die out while recording the sample rotation as a function of time. This is called the logarithmic decrement method. The damping ratio D is defined as the ratio of the damping coefficient to the critical damping coefficient. The critical damping is the minimum

f1

δ (2π)2 + δ2

θ

r ρ

L

r γ

dρ (a)

(b)

Figure 10.59 Shear stress and shear strain in a resonant column torsion test: (a) Shear stress. (b) Shear strain.

10 LABORATORY TESTS

Shear stress, τ

Shear stress, τ 1 Gmax

γʹ2

1

γʹ1

Shear strain, γʹ Ac

D=

1 G

τ G5 γ

Ae

Shear strain, γʹ

Shear stress, γʹ Ac 4πAe

G 1 (a)

(b)

7

10

6

Shear Modulus, G (MPa)

(a) 12

5

8

4 6

Shear modulus Pore pressure increase

3

4

2

2

Pore pressure increase, Δu (kPa)

Figure 10.60 Shear stress-strain loops in resonant column test and damping ratio calculation: (a) Evolution of stress-strain loop. (b) Calculation of shear modulus and damping ratio.

1

0 0.0001

0.001

0 0.01

Shear strain, γ (%) (b)

10 9 8

Damping ratio, D (%)

264

7

Steady state Amplitude decay

6 5 4 3 2 1 0 0.0001

0.001

0.01

Strain amplitude (%)

Figure 10.61 Shear modulus vs. shear strain and damping coefficient vs. shear strain: (a) Shearing modulus. (b) Damping ratio.

265

10.14 LAB VANE TEST

The mean shear stress 𝜏 is related to the maximum torque T as follows: (10.41) 𝜏 = 2T∕𝜋 r3e

Motor for torque application

where re is the equivalent radius, which can be anywhere from 0.6r to 0.8r where r is the radius of the sample. The maximum shear strain during the cycle exists at the edge of the sample (𝛾 edge ), while the shear strain is zero along the axis of the cylindrical sample (𝛾 center = 0) (Figure 10.59). The mean shear strain in the sample is usually taken as.

where again re is the equivalent radius, often taken as 0.8r where r is the radius of the sample. A typical 𝜏 vs. 𝛾 curve is shown in Figure 10.60. The shear modulus G is calculated as the slope of the line joining the two extremities of the loop. Alternatively, this curve can be generated by calculating the shear strain first, obtaining the shear modulus by the resonant frequency method, and then calculating the shear stress as G𝛾. The damping ratio D is defined from the curve as the ratio of the energy necessary to perform one cycle of torsion to the elastic energy expanded to load the sample to the peak of the cycle (Figure 10.60): D = Ac ∕4𝜋Ae

(10.43)

where D is the damping ratio, Ac is the area inside the cycle, and Ae is the area inside the triangle shown in Figure 10.60. The previous discussion identifies how G, D, and 𝛾 can be obtained for a given amount of torque applied at the top of the sample. This torque can then be increased to create a larger shear strain in the sample. The test is repeated and a new set of values of G, D, and 𝛾 are obtained. Point by point, the G vs. 𝛾 curve and the D vs. 𝛾 curve are described (Figure 10.61). The G vs. 𝛾 curve and the D vs. 𝛾 curve are the two results of a resonant column test. The strain that can be tested with this test typically ranges from 10–6 to 10–3 . 10.13.2

Unsaturated Soils

If the soil is unsaturated, or if the soil is saturated but the water is in tension, neither the test procedure nor the data reduction changes. Indeed, the water stress is rarely measured during the resonant column test.

10.14 10.14.1

Soil sample

(10.42)

LAB VANE TEST Saturated Soils

The lab vane test or VST (Figure 10.62) is used to determine the undrained shear strength of fine-grained soils (clays and silts). It can be performed either in the field with a field vane (ASTM D2573) or on the sample with a mini vane or a hand vane (ASTM D4648; Figure 10.62). The lab vane is made of two perpendicular blades, each having a 2-to-1 height-to-width ratio. The width of the blades varies from 12 to 25 mm; the larger vanes are used in softer soils. The vane

Vane

(a)

(b)

Figure 10.62 Lab vane test equipment: (a) Principle. (b) Equipment. (Source: (a) Adapted from BS 1377-7: 1990.)

Vane Tmax Failure Surface

Torque (kN.m)

𝛾 = re 𝜃∕L

10 rapid rotations

TRes

0

θmax

Rotation angle (°)

Figure 10.63 Lab vane test results.

is pushed perpendicularly into the end of a sample until the top of the blades is one blade height below the surface of the sample. Then the vane is rotated at a slow rate (less than 1 degree per minute) while the torque applied is measured as well as the rotation angle (Figure 10.63). The peak value of the torque is recorded as T max . The blade is then rotated at least 10 times rapidly and a new maximum torque value, T res , is measured. The VST is used in saturated fine-grained soils to obtain the undrained shear strength su . The reason is that these soils have a low permeability and do not allow appreciable drainage during the test, which typically lasts less than 10 minutes. Therefore, for these saturated fine-grained soils, it is reasonable to assume that the undrained shear strength su is the parameter being measured. For a rectangular vane, the following equation gives su from T max : ) ( H D (10.44) Tmax = 𝜋su D2 + 2 6

266

10 LABORATORY TESTS

where D is the diameter of the vane and H is the height of the vane. Proof of this equation is shown in the solution to problem 8.4. The residual undrained shear strength sur is obtained from the same formula using T res : ) ( H D (10.45) Tres = 𝜋sur D2 + 2 6 The VST can be performed in coarse-grained soils, but no useful result can be obtained. These soils drain fast enough that one would not measure the undrained shear strength, but instead the drained or partially drained shear strength. Back-calculating the shear strength parameters from this test would require knowledge of the normal effective stress on the plane of failure in addition to T max . This is not measured during the VST. The VST has the advantages of being fast, simple, economical, and useful for obtaining the undrained shear strength of fine-grained soils. Its drawbacks include that it is limited to fine-grained soils. 10.14.2

Unsaturated Soils

If the soil is unsaturated, or if the soil is saturated but the water is in tension, neither the test procedure nor the data reduction changes. Water stress is not measured during the vane test.

10.15 SOIL WATER RETENTION CURVE (SOIL WATER CHARACTERISTIC CURVE) TEST 10.15.1

Saturated Soils

The soil water retention curve (SWRC), also known as the soil water characteristic curve, is a property of the soil much like the shear strength parameters (Figure 10.64). It is a plot of the water content of the soil as a function of the water tension stress (suction) in the soil pores. It depends on many factors, including the particle size distribution, pore size distribution, soil structure, and soil texture. During the drying process from a saturated state, the water tension in the soil will increase until it becomes large enough to force air into the soil pores. This water tension value is called the air entry value uwae . Beyond the air entry value,

30

w = Cwlog uw

Cw

20

1 10

Saturated

Unsaturated

(10.46)

where w is in percent, uw is in kPa (positive), and Cw is the slope of the SWRC. The gravimetric water content is the most commonly used water content definition in geotechnical engineering, but for the SWRC the volumetric water content often is used. These are defined in the following equations: Gravimetric water content ∶ w = Ww ∕Ws

(10.47)

Volumetric water content: 𝜃w = Vw ∕V

(10.48)

where Ww and Vw are the weight and volume of water respectively, Ws is the weight of solids, and V is the total volume. Example SWRCs are presented in Figure 10.65. It stands to reason that different soils will have different SWRCs: A sand will not retain water the same way a clay would. Imagine that you insert a straw into a sand; it would not take much sucking to get the water out of the sand. Now imagine that your straw is inserted into a clay; in this instance it would take a lot of sucking to get a little bit of water out. The suction or water tension that you would have to exert through the straw would be much higher for the clay than for the sand. This phenomenon is what the SWRC characterizes. Soils under the groundwater level (GWL) are generally saturated and the water is in compression. Soils above the GWL can be saturated or unsaturated, but in both cases the water is in tension (suction). The SWRC is a property of a soil where the water is in tension. As such, the SWRC for a saturated soil refers to the case where the soil is saturated above the GWL by capillary action and other electrochemically based phenomena such as the affinity between water and clay minerals. If a saturated soil sample is placed on a table top and is strong enough to stand by itself, it is likely held together by water tension unless it has some cementation (effective stress

Volumetric water content θ* (%)

Air entry stress uwt

40 Water content w (%)

dw = Cw d(log10 uw )

60

50

0

the decrease in water content is well approximated by a linear relationship between the water content and the log of the water tension, which can be written as:

Free draining water uwt

50 40

uwt

30

Water unavailable to plants

Water available to plants

uwt = air entry water tension *θ =

uwt

20

Silt

Volume of water Total volume

Clay

Sand

10 0

1

10 100 1000 Water tension stress uw (kPa)

Figure 10.64 Soil water retention curve.

10,000

1

10

100 1000 10,000 100,000 1,000,000 Water tension stress uw (kPa)

Figure 10.65 Examples of soil water retention curve.

10.15 SOIL WATER RETENTION CURVE (SOIL WATER CHARACTERISTIC CURVE) TEST

10.15.2

Unsaturated Soils

There are essentially two methods for obtaining the SWRC (ASTM D6836). The first consists of taking a saturated soil sample and measuring the water tension and the water content of the sample as a function of time as it dries up. The water content measurement was described in Section 4.9; the water tension measurement was described in Section 10.2.4 and summarized in Table 10.2. In this case, the two most common methods to measure water tension for the SWRC are the filter paper method and the chilled mirror psychrometer. For lower values of water tension, the hanging column method can also be used (ASTM D6836). As a guide, and for tests performed in an air-conditioned laboratory environment where the relative humidity is around 50%, a 25 mm high, 75 mm diameter sample is likely to become air-dry in about 24 hours. In these circumstances, a water content and water tension measurement every 1 to 2 hours is suitable to get a good description of the SWRC. The second method of obtaining the SWRC is to use a saturated soil sample and force the sample to come to equilibrium at a selected series of water tension (suction), while measuring the water content for each one of those water tension values. The pressure plate apparatus can be used in this case (Figure 10.9); it makes use of the axis translation technique (Figure 10.10) and increases the air pressure to push the water out of the soil pores. The air pressure is equal to the water tension in the sample when the water starts moving out of the pores. The water content of the soil sample is measured when the water stops flowing. Such measurements are made at increasingly higher air pressures so as to describe the complete SWRC. Yet another way is to use the salt solution equilibrium technique, in which “identical” samples are placed in different salt solution chambers (Figure 10.11) and left in the chamber until the water tension in the sample comes into equilibrium with the relative humidity created by the salt solution at the bottom of the chamber; reaching this equilibrium may take 1 or 2 weeks. The salt concentration in each chamber is different and is chosen to create a series of values of the relative humidity and therefore water tension, which gives a good description of the SWRC. After equilibrium is reached, the soil water content is measured in each chamber and the SWRC can be plotted. The SWRC describes the fact that the water tension increases when the water content decreases but recognizes that this relationship is not the same when the soil is drying

40.0

Drying

35.0 Water content, w (%)

cohesion). As the soil dries, it initially shrinks while remaining saturated. The water tension increases and at a given water tension stress (suction), air enters the pores. The water tension at this point is called the air entry value (uwae ). From this point on during the drying process, the soil is unsaturated. The procedure to determine the SWRC is the same below (saturated) and above (unsaturated) the air entry value. This procedure is detailed in Section 10.15.2 related to unsaturated soils.

267

Wetting

30.0 25.0 20.0 15.0 10.0 5.0 0.0 1

10

100 1000 10,000 Water tension, uw (kPa)

100,000 1,000,000

Figure 10.66 Drying and wetting hysteresis loop in the SWRC.

as when it is wetting; this is called the hysteresis in the SWRC. Figure 10.66 shows the difference between the drying curve and the wetting curve. It is likely that this difference decreases as the number of drying and wetting cycles between the same values increases. The hysteresis effect may be attributed to several causes: the geometric nonuniformity of the individual pores, the pore fluid contact angle, entrapped air, and swelling, shrinking, or aging. The geometric nonuniformity of the pores can be explained as follows (Figure 10.67). When the soil is drying, the water level in the conduits formed by the voids between particles can drop down through a larger void cross-section, as shown in Figure 10.67. However, if the soil is wetting, there is a limit to how large a cross-section the water can move up, as the capillary force is limited. As a result, the loss of water is larger during drying than the gain of water during wetting and thus the wetting curve is below the drying curve (Figure 10.66). Several stages are identifiable in the drying or wetting process, as shown in Figure 10.68. During drying, at first the soil is saturated (S = 1) until the air entry value of the water tension uwae is reached; then a linear semilog relationship exists between the water content and the water tension; and then the soil reaches a residual stage (S = Sr ) where the water no longer forms continuous conduits in the pores, but rather exists only at the contacts between particles.

Drying

Wetting

Figure 10.67 Geometrical explanation of drying and wetting hysteresis in the SWRC.

268

10 LABORATORY TESTS

100

Desaturation Zone

1.2

Residual Stage

wPI = % Passing #200*PI

1.0

Residual air content

80 60

Wetting path Drying path Residual conditions

40

0.8 0.6 wPI = 0.1 3

0.4 D60 = 1 mm

0.2

Air entry value uwac

20

Degree of saturation

Degree of saturation, S (%)

Saturation Zone

0.1

10

10 5 15

D60 = 0.1 mm

0.0

0 1

wPI = 50 40 20 30

100 1000 10000 100000 1000000 Soil suction, uw (kPa)

The effective degree of saturation Se is defined for a given degree of saturation S as: S − Sr (10.49) Se = 1 − Sr During the wetting process, a similar progression takes place in reverse and after the saturation phase, where again there is a linear semilog relationship between water content and water tension. The soil reaches a residual air content when the air is occluded and cannot be chased out of the voids through normal means. Various empirical models have been proposed to describe the SWRC. Among the most common are: ) ( ( 1if ) uw ≤ uwae Brooks and Corey (1964) Se = uw if uw ≤ uwae uwae (10.50) )m ( 1 van Genuchten (1980) Se = ( )n 1 + 𝛼uw

Fredlund and Xing (1994) 𝜃 ( )m ( ) 1 = C uw 𝜃s ( ( ))n Ln e+ uw ∕a

10

100 1000 10000 Matric suction (kPa)

100000

1000000

Figure 10.69 SWRC as a function of percent passing #200 and plasticity index. (Source: Courtesy of NCHRP.)

Figure 10.68 Various stages in the SWRC.

with m = 1 − 1∕n

1

(10.51)

(10.52)

where Se is the effective degree of saturation; uw is the water tension (kPa); uwae is the air entry value of the water tension (kPa); 𝜆 is a fitting parameter mostly influenced by the pore size distribution of the soil; 𝛼, n, and m are fitting parameters; 𝜃 is the volumetric water content (volume of water over total volume); 𝜃 s is the volumetric water content at saturation; C(uw ) is a correction factor that forces the model through a prescribed water tension value of 106 kPa at zero water content; a is a fitting parameter; and e is the logarithmic constant (Ln e = 1). More details on these models can be found in Lu and Likos (2004). ARA-ERES (2000) proposed a set of SWRCs (Figure 10.69) predicted on the basis of D60

in mm, the particle size for which 60% by weight is finer, and an index called the wPI. The wPI is defined as the product of the percent passing sieve number 200 as a decimal (ratio not percentage) and the plasticity index as a percent.

10.16 10.16.1

CONSTANT HEAD PERMEAMETER TEST Saturated Soils

The constant head permeameter (CHP) (ASTM D2434; Figure 10.70) is used to obtain the coefficient of hydraulic conductivity k of saturated coarse-grained soils. The soil sample is placed in a cylinder about 75 mm in diameter and 150 mm high, with one filter stone at the top and another at the bottom. The top of the sample is connected by tubing to a container in which the water level is kept constant through an overflow regulator. The bottom of the sample is connected to another container in which the water level is also kept constant. The bottom container is kept lower than the top container and the flow Q (m3 /s) out of the bottom container is measured. The measurement simply consists of weighing the amount of water collected in the overflow container during a corresponding time. Typical results are shown in Figure 10.71. Often manometer tubes are connected to the side of the sample container at two points to give the water stress (pressure) at those two locations. (See Chapter 14 on flow through soils for an explanation of the following equations and parameters.) Darcy’s law gives: v = ki

(10.53)

where v is the discharge velocity through the sample, k is the hydraulic conductivity, and i is the hydraulic gradient. The hydraulic gradient in this case is given by: i = h∕l

(10.54)

where h is the loss of total head through the flow distance l (Figure 10.70). Also, conservation of mass gives: Q=v A

(10.55)

269

10.16 CONSTANT HEAD PERMEAMETER TEST

Filter stone

Level maintained constant 1

k=

Water uw =

hp2

4V  

tπD2

uw

l ht2 – ht3

×

hp2 + hp3

2

× γo> 0

2 ht2 ht3

ht1

Saturated soil

hp3

l

3

Filter stone Volume of water V collected in time t

V

Datum

(a)

(b)

Figure 10.70 Constant head permeameter equipment: (a) Principle. (b) Equipment. (Source: (b) Courtesy of ELE International.)

Velocity, v (m/s)

Volume, V (mm3)

500,000 400,000 300,000 q = 10–5m3/s

200,000 1

100,000 0

0

5

10

15 20 25 Time, t (s)

30

35

40

0.008 0.006 k = 2.6 × 10–4m/s

0.004 1

0.002 0

0

0.5

1 1.5 2 2.5 Hydraulic gradient, i = Δh/L

3

Figure 10.71 Constant head permeameter test results.

where Q is the flow out of the sample, v is the discharge velocity, and A is the cross-sectional area of the sample. Note that A is the total cross-sectional area of the sample, not just the area of the pores through which the water is flowing. As a result, v is not the actual speed of the water molecules flowing through the pores (seepage velocity) but rather an equivalent speed called the discharge velocity. Combining Eqs (10.53) through (10.55) gives the value of k: k = Q 1∕h A

(10.56)

3

where Q is the discharge (m /s), l is the flow length between two points in the sample, h is the loss of total head between the same two points, and A is sample cross-sectional area. The discharge Q is the volume V collected in a time t divided by t. The cross-sectional area A is 𝜋D2 /4 where D is the sample diameter. The loss of total head h is ht2 – ht3 , as shown in Figure 10.70. Therefore, the hydraulic conductivity k is: k=

l 4V t𝜋D2 ht2 − ht3

(10.57)

The advantage of the constant head permeameter test is that it is a very simple test to run; the drawback is that it is limited to measuring the hydraulic conductivity k of coarse-grained soils at a small scale. The k values typically measured with this test range from 10–1 to 10–6 m/s.

10.16.2

Unsaturated Soils

If the soil is unsaturated, things are quite different. The first thing to realize is that the hydraulic conductivity of an unsaturated soil is less than that of a saturated soil: Water goes through an unsaturated soil more slowly than through a saturated soil. The reason is that air is in the way of the flow, and the water is attracted to the walls of the tiny conduits formed by the particles. Of course, one must remember that in the equation giving the water velocity v (m/s) from the flow discharge Q (m3 /s) (Eq. (10.55)), A is the total cross-sectional area of the sample, not the actual water flow area. Because the flow area is significantly reduced in the case of unsaturated flow, the actual water velocity is quite a bit higher than the velocity given by Eq. (10.55). The steady-state permeameter test for unsaturated soils consists of the same equipment except for two differences: (1) The measurements of water compression are changed to measurements of water tension, and (2) a tube is connected to the center of the sample to control the air pressure in the sample. The measurement of the water tension is made at two locations, using tensiometers or other appropriate devices. Figure 10.72 shows the diagram for an unsaturated steady-state permeability test. The water level is maintained on the upstream side (point 1) and the water starts flowing.

270

10 LABORATORY TESTS

Level maintained constant

Unsaturated soil

1

k=

Water uw =

2 hp2

D

l

ht1

4V 2

tπD

×

hp2 + hp3

2

1.E-02

l ht2 – ht3

+

× γw < 0

Air 4 pressure ua –

3 hp3

5 ht2

High air entry disk

ht3

hp4

V

Volume of water V collected in time t

1.E-03

uw Hydraulic conductivity k (m/s)

High air entry disk

ksat

1.E-04 1.E-05 1.E-06

ksat

1.E-07

Fine grain soil

1.E-08 1.E-09 Coarse grain soil

1.E-10 1.E-11 1.E-12

Datum

1

10

100

1000

10,000 100,000

Water tension u (kPa)

Figure 10.72 Constant head permeameter test for unsaturated soils.

Figure 10.73 Hydraulic conductivity as a function of water tension for unsaturated soils.

It arrives at the high air entry disk. This disk lets the water go through but not the air; that is a property of that disk. Then the water goes through the soil voids. One would think that it would flood the voids as it is attracted by the water tension (suction) in the water phase. But the air is in the way, and it has no way to escape because there is another high air entry disk at the other end of the sample. So the water is forced to flow through the continuous water phase around the air phase. The water tension is greater at point 3 than at point 2 (Figure 10.72) because the water loses energy as it drives through the soil. A friction force arises between the water molecules and the soil particles as the water drags through the voids. This force is called the seepage force. As a result of this force, there is an associated loss of pressure between points 2 and 3. Because the pressure at point 2 is negative (water tension), the pressure at point 3 is even more negative (hp3 < hp2 < 0). The hydraulic conductivity depends on the water tension (Figure 10.73). As the water tension increases, the amount of water in the soil decreases, and it becomes harder and harder for the water to percolate through the soil: There is less room for the water to flow and a higher attraction between the water and the soil particles. The effect of the water tension on the hydraulic conductivity can be documented in this test by changing the air pressure through port 4. Applying an air pressure ua different from zero changes the water tension uw . This allows one to run the permeability test at different water tensions and establish the relationship between hydraulic conductivity and water tension. From the calculations point of view, the hydraulic conductivity k is obtained as:

points where the total heads ht2 and ht3 are measured. Note that Eq. (10.58) is the same as Eq. (10.57). The difference is that ht2 and ht3 are different, because the soil is unsaturated. The average water tension uw associated with the hydraulic conductivity k of Eq. (10.58) is:

k=

4V l ( ) t𝜋D2 ht2 − ht3

(10.58)

where V is the volume of water collected in a time t, D is the sample diameter, and l is the distance between the two

hp2 + hp3

85 %

T (mN/m) Ua air

Continuous air uw ≠ 0 ua = 0 σ' = σ – αuw S < 85%

Figure 11.14 Effective stress equation for various common situations.

(11.61)

The analogy is tempting, but it must be said that the ratio of areas Aw /Av is not likely equal to the ratio of volumes V w /V v , because the plane that cuts through the contacts in Figure 11.13 does not represent the general situation in the soil volume. As a result, there is quite a bit of scatter in the correlation between 𝛼 and S (see Figure 11.16). Uw water

Saturated uw ≠ 0 ua = 0 σ' = σ – uw S = 100%

D Ua Uw water

Uw

Figure 11.15 Pressure difference across an air bubble boundary.

1 Silt, drained test (Donald, 1961)

Water area ratio, α

0.8

Silt, constant water content test, (Donald, 1961)

0.6

Madrid clay sand (Escuario and Juca, 1989) Madrid silty clay (Escuario and Juca, 1989)

0.4

Madrid gray clay (Escuario and Juca, 1989)

0.2

0

0

20 40 60 80 Degree of saturation, S (%)

309

100

Figure 11.16 Water area ratio 𝛼 vs. degree of saturation S. (Source: Lu and Likos, 2004/John Wiley & Sons.)

310

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Water area ratio, α

1.0

Soil state

0.8

Unsaturated 0.6 0.4

Saturated 0.2

GWL 0.0

0

2

4 6 8 10 Water tension ratio, uw /uwae

12

14

Figure 11.17 Water area ratio 𝛼 vs. water tension ratio. (Source: Adapted from Khalili and Khabbaz, 1998.)

Khalili and Khabbaz (1998) proposed a better relationship to predict 𝛼 (Figure 11.17): ) ( (ua − uw ) −0.55 (11.62) 𝛼= (ua − uw )ae which can be simplified without much loss of accuracy when ua is zero as: √ uwae (11.63) 𝛼= uw where ua is the air stress, uw is the water stress, and (ua − uw )ae refers to the difference between ua and uw at the air entry value. At the beginning of the drying process of a saturated sample of soil, the water tension uw increases (becomes more negative) as the water is evaporating out of the soil and into the surrounding air, but the soil remains saturated. As the drying continues, uw continues to become more negative and gets to a point where air first enters the pores. This value of the water tension is called the air entry value uwae . As the drying continues, the water tension continues to become more negative. The area ratio for air is 𝛽, and because 𝛼 + 𝛽 is equal to 1 (Ac ∼ 0), once 𝛼 is known, so is 𝛽. 11.16

WATER STRESS PROFILES

The water normal stress can be positive (pore pressure, compression) or negative (suction, tension). In the field, the groundwater level (GWL) is found at some depth below the ground surface (Figure 11.18). In some cases, that depth is very large (deserts); in others, it is very shallow (regions close to oceans, lakes, or rivers). At the GWL, the water stress is zero. Below the GWL, the water is in compression (pore pressure) and, in the most common case, the water stress profile shows a linear increase with depth (hydrostatic pressure) and can be calculated as 𝛾 w z where 𝛾 w is the unit weight of water and z is the depth below the GWL. Sometimes the water stress profile below the GWL is complicated by the presence of perched aquifers (water

Saturated

Figure 11.18 Groundwater level (GWL) and zones above the GWL. (Source: Adapted from a photo by Art Koenig, reproduced with permission.)

bodies sandwiched between dry soil) or artesian conditions (water body connected to a pressure higher than the local hydrostatic pressure). Figure 11.19 shows examples of such conditions. Above the GWL, the water is in tension (suction). In the zone above the GWL and deep enough to be unaffected by the weather at the ground surface, the water stress is linear and given by (–𝛾 w z) where z is the absolute value of the vertical distance above the water table. In the zone above the GWL and close enough to the ground surface that the weather can influence the water stress profile by evapotranspiration and rainfall (generally a few meters), the water stress profile becomes curved to reach an equilibrium between the weather and the soil (Figure 11.20). This part of the water stress profile is very difficult to calculate and varies daily with the weather.

11.17

WATER TENSION AND SUCTION

Water tension is the tension in the water expressed in kN/m2 . Suction is the potential that the water has to achieve a certain water tension; it is also expressed in kN/m2 . This suction potential is not always realized. If the suction potential is fully realized, the suction is equal to the water tension. If the suction is not fully realized, the suction is higher than the water tension. It is a bit like standing on top of a building but not jumping: you have potential energy, but you are not transforming it into velocity because you are not jumping. Later we will discuss cases in which the suction is not transformed into water tension. Although the suction is important, the water stress is the one that enters into most calculations. Note that suction is often defined as the difference between the air stress and the water stress (ua – uw ). Because the air stress is often zero in the field (continuous air voids), suction is defined here as uw . Note further that with ua – uw , suction is positive, whereas with uw , suction is negative. Suction and water tension will always be negative in the rest of this book,

11.17 WATER TENSION AND SUCTION

0

uw

0

uw

GWL

(a) General case

uw

0

GWL

311

GWL

(b) Perched water

(c) Artesian pressure

Figure 11.19 Examples of water stress profiles below the groundwater level. uw

0

uw

uw

0

0

Unsaturated Saturated

Unsaturated Saturated

GWL

(a) Very wet

(b) Common case

(c) Very dry

Figure 11.20 Examples of water stress profiles above the groundwater level.

as compression has been chosen as the positive sign convention for stresses. The water tension and suction come from two different sources: attraction of water to the minerals in the soil particles and attraction of distilled water to salty water. The first one is called matric suction; the second one osmotic suction. 11.17.1

Matric Suction

Matric suction is due to the attraction between water molecules and the minerals in soil particles. If the mineral is silica, the phenomenon is called capillary action. The attraction between water and silica generates a force of 73 mN/m. Other minerals, such as smectite (Al2 Si4 O10 (OH)2 ), can generate much higher attraction forces and therefore much higher water tension. Let’s discuss capillary attraction first. The force of 73 mN/m is given per unit of length because it exists along the contact line of the meniscus interface between the water, the air, and the silica. Recall that one Newton is about the weight of a small apple, so 73 mN is a very small force, yet this force is responsible for some major phenomena when dealing with very small scales. For example, when a very-small-diameter glass (silica) tube open at both ends is placed in water, that force lifts the water in the tube like one would pull up a sock. If the glass tube is small enough, the water can rise more than 10 m in the tube. Note that if the tube were made of a different mineral, the water would not rise to the same level. Also, if the tube were made of glass, but instead of water you had mercury in the container, the mercury would actually go down in the

small tube rather than up, because there is a basic repulsion between mercury and silica. The water rising in the silica tube does so up to a height where the volume of water lifted in the tube has a weight equal to the vertical component of the attraction force at the top of the column times the contact length of the meniscus. Equating the weight of the column of water to the vertical component of the attraction force leads to the height of the water column or capillary rise (Figure 11.21): hc =

𝜋d2 𝛾 = 𝜋dT cos 𝛼 4 w

(11.64)

4T cos 𝛼 𝛾w d

(11.65)

Therefore, hc =

It is clear that the capillary rise depends on the diameter of the tube; the capillary rise will be high in small-diameter tubes and small in larger-diameter tubes. If the tube has a diameter equal to the size of clay particles—say, 0.001 mm—Eq. 11.65 gives a height of capillary rise equal to 29.2 m (height of a 10-story building). The continuous voids in a soil play the role of the tiny glass tube because, like glass, many soil particles are made of silica. Continuous clay voids are similar to tiny tubes and the water can saturate the clay high above the groundwater level (15 m or more). In sands, the height to which the water can rise is more limited. Let’s study the water stress profile in the capillary tube (Figure 11.21). Below the water level, in the big container, the water is in compression and the water stress is positive. Above that level, in the tiny glass tube, the water is in

312

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Glass T α Contractile α T skin

hcγw

0 kPa –1,000 kPa

d hc

Water

Water

h

Atmospheric pressure 0

–

+

uw

hγw

Figure 11.21 Capillary tube experiment.

tension because the water is pulled up into the tube by the force πdTcos𝛼. The water tension increases (becomes more negative) linearly with the height in the tube, as shown in Figure 11.21. At the top of the column, the water tension is maximum and equal to -hc 𝛾 w . Yet in the air immediately above the water level in the small tube, the pressure is atmospheric or zero gage pressure. It is not possible for such a discontinuity to exist between two fluids unless there is a membrane separating the water from the air: this membrane is the contractile skin. It is similar to a car tire: the pressure in the tire is much higher than outside the tire, and this is made possible by the membrane represented by the tire. We will discuss the contractile skin a bit later. Consider now two soil particles in the form of spheres (Figure 11.22). The soil is allowed to dry and the water between the particles evaporates. When the water is almost gone, the water is only found around the contact between the two particles (Figure 11.22). The water is in tension and the air is at atmospheric pressure. The contractile skin allows a large stress difference to exist between the two fluids. Now let’s calculate the force at the contact. We draw a free-body diagram of the upper particle and show the forces imparted

by the water and the contractile skin on the particle. The water is under a tension stress uw , so the water pulls on the particles above and below the contact area A with a force uw A. The contractile skin is also in tension and pulls on the particle at an angle 𝛼. The calculations are shown in Figure 11.22. The force is a compression force equal to 10–6 N. Remember that 1 N is about the weight of a small apple, so the force is extremely small—yet the stress is very large (1000 kPa). These stresses develop when the soil dries and are the reasons why dry fine-grained soils are a lot harder than saturated soils. The preceding discussion focused on the case of water attraction to silica and the water tension that can be generated due to this phenomenon. Some clay minerals, such as smectite (Al2 Si4 O10 (OH)2 ), can generate much higher attraction forces and therefore water tension, which can reach 100,000 kPa or even 1,000,000 kPa (Figure 11.23). These water tension values correspond to soils that are very dry yet have a little bit of water between particles. It is not clear in these cases whether the water is still in liquid form, or in viscous form, or possibly approaching solid form.

Soil grain D

Soil grain Contractile skin, T (kN/m)

Water tension, uw (kPa)

T

T uw

α

Soil grain Soil grain If uw = –1000 kPa, T = 73 mN/m, D = 0.001 mm, α = 45° F = uwπD2/4 + (Tcosα) πD = 10–9 kN σ = F/A = 103 kN/m2

Figure 11.22 Water tension at the contact between two spherical particles.

11.17 WATER TENSION AND SUCTION

Water tension 200 kPa

contractile skin. (So, it is possible to walk on water, at least for the water strider.) This water membrane is able to generate 73 mN of force for every meter of linear contact with silica. This represents a very small force, but the membrane is extremely thin. Its thickness is estimated at 20–30 nanometers; therefore, the stress is very large.

Water tension 100,000 kPa Smectite Al2Si4O20 (OH)2

Quartz SiO2 Water Quartz SiO2

Contractile skin

11.17.3 Water

Figure 11.23 Water tension between soil particles.

Figure 11.24 Water strider resting on contractile skin. (Source: KKPCW/Wikimedia Commons.)

11.17.2

Contractile Skin

The membrane called the contractile skin exists at the interface between the water and the air. The existence of this membrane is rooted in the Van der Waals forces, which are elementary attractive forces between molecules. In the water, these forces act in all three directions and give water its tensile strength. This tensile strength can be measured by placing water in a cylinder and pulling on the piston until the water breaks in tension, at about 20 MPa. This is remarkably large, approaching the strength of concrete in compression. At the interface between the water and the air, the molecules of water attract those that are below the interface but are unable to attract water molecules above the surface, as there are none available. Instead, the water molecules enhance their attraction in the horizontal direction, thereby creating a membrane. Figure 11.24 shows a water strider resting on that

Osmotic Suction

There is a second reason why water can go into tension in a soil: osmotic suction. Osmotic suction is due to the basic attraction that exists between water and salt. The phenomenon can be explained as follows. Imagine a container with two sides (Figure 11.25). On one side is distilled water, and on the other side is water with salt in it. Imagine that there is an imaginary screen separating the two sides that allows water molecules to travel across it but not salt molecules. This imaginary screen therefore prevents the two water bodies from mixing. In this experiment, the distilled water will be attracted to the salt water and therefore a difference in elevation will be generated, as shown in Figure 11.25. This difference is a suction potential called the osmotic suction. To help you remember that the distilled water goes toward the salt water, just remember that when you eat salty food, you get thirsty! Osmotic suction depends on the salt concentration in the water on the right side and on the type of salt in that water. Osmotic suction exists in a soil if the soil contains dissolved salts. This suction exists as a potential and is realized into a water tension if there is a change in salt concentration between two locations. This can happen when a sprinkler system is installed in the backyard of a home. In the majority of real situations, the osmotic suction is much smaller than the matric suction. The sum of the matric suction plus the osmotic suction is the total suction. Figure 11.26 shows values of total suction or water tension for a range of conditions. If the salt concentration is high, as would be the case in a prepared solution, the osmotic suction can be very high. Table 11.1 shows the values of osmotic suction associated with various concentrations and various salt types. Note that osmotic suction exists in saturated soils as well as in unsaturated soils, as it is related only to the chemistry of the pore fluid. 11.17.4 Relationship between Total Suction and Relative Humidity If you place water at the bottom of a container with air above it and then you close the container, the humidity of the air in After time t initial time

h = osmotic suction

Initial state after time t

Pure water

313

Salt water

Figure 11.25 Osmotic suction experiment.

314

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Water state

Tension

Water stress cm kPa

Examples

pF

Degree of saturation

Water content

0

0

Oven dry

7

–107 –106

Air dry

6

–106 –105

Shrinkage limit

4

–104 –103 Near 100 % 8 to 15 %

Swell limit

2

–102 –101

0 Large river Compression Deepest offshore platforms Bottom of deepest ocean

0

0

103

102

105

104

109

108

Swell

Shrink

No Yes

25 to 50 %

Yes

100 % No

Figure 11.26 Range of water tension and water compression for various conditions.

Table 11.1

Osmotic suction in kPa of some salt solutions at 25 ∘ C Osmotic suction in kPa at 25 ∘ C

Molality (mol/kg)

NaCl

KCl

NH4 Cl

Na2 SO4

CaCl2

Na2 S2 O3

MgCl2

0.001 0.002 0.005 0.010 0.020 0.050 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.200 1.400 1.500 1.600 1.800 2.000 2.500

5 10 24 48 95 234 463 916 1370 1824 2283 2746 3214 3685 4159 4641 5616 6615 N/A 7631 8683 9757 12556

5 10 24 48 95 233 460 905 1348 1789 2231 2674 3116 3562 4007 4452 5354 6261 N/A 7179 8104 9043 11440

5 10 24 48 95 233 460 905 1348 1789 2231 2671 3113 3558 4002 4447 5343 6247 N/A 7155 8076 9003 11366

7 14 34 67 129 306 585 1115 1620 2108 2582 3045 3498 3944 4384 4820 N/A N/A 6998 N/A N/A 9306 11901

7 14 34 67 132 320 633 1274 1946 2652 3396 4181 5008 5880 6799 7767 N/A N/A 13391 N/A N/A 20457 29115

7 14 34 67 130 310 597 1148 1682 2206 2722 3234 3744 4254 4767 5285 N/A N/A 7994 N/A N/A 11021 14489

7 14 35 68 133 324 643 1303 2000 2739 3523 4357 5244 6186 7187 8249 N/A N/A 14554 N/A N/A 22682 32776

1 All suction values are in kPa. (Source: After Bulut et al., 2001/American Society of Civil Engineers.)

11.17 WATER TENSION AND SUCTION

the container will increase or decrease until it comes to an equilibrium. This equilibrium depends on the pressure and temperature in the container. At atmospheric pressure and at a temperature of 25 ∘ C, dry air consists of nitrogen (∼78% by volume), oxygen (∼21% by volume), and a few other gasses. If such a dry air is in the container, there is plenty of room for water molecules to become part of the air, thereby increasing the relative humidity of the air. Part of the liquid water at the bottom of the container will become vaporized, and join the air phase by fitting vaporized water molecules between the molecules of nitrogen and oxygen. This process will continue until an equilibrium is reached. Each gas component in the air has a partial pressure, and the partial pressures add up to the total pressure, according to the ideal gas law: Pair = Pnitrogen + Poxygen + Pwater + · · · •

(11.66)

At a certain relative humidity, the air has a corresponding partial water vapor pressure Pwater . At 100% relative humidity, the partial water vapor pressure pwater equals the saturated water vapor pressure Pwater,sat . This pressure is 3.17 kPa for conditions of atmospheric pressure (101.3 kPa) and a temperature of 25 ∘ C. The general equation for the saturated partial vapor pressure of water in air Pwater,sat at atmospheric pressure for different temperatures is (Tetens, 1930): Pwater,sat (kPa) = 0.611

( ) o C) 17.27 T(oT( C)+237.2

(11.67)

where T is the temperature in degree Celsius. The relative humidity of the air is defined as the ratio: RH =

Pwater Pwater,sat

(11.68)

The relationship between the relative humidity RH of the air in the void of an unsaturated soil and the suction potential 𝜓 is given by Kelvin’s equation (Fredlund and Rahardjo, 1993; Lu and Likos, 2004): (11.69)

where RH is taken as a fraction. This equation is shown in Figure 11.27. It indicates, among other interesting observations, that a humidity room at 95% relative humidity has a water tension potential of almost 7000 kPa and therefore is a drying room. 11.17.5

(a)

1

(b)

0.8 0.6 0.4 0.2 0 100

1000

10000 100000 Total suction, uw (kPa)

1000000

Trees

Water is drawn up to the top of trees through suction. Osmotic suction in the tree is due to the difference in mineral concentration of the water in the tree and of the water in the soil. Capillary suction is due to the very small size of the tiny tubes (xylem conduits) that exist through the stem or tree trunk; water is attracted to the walls of the xylem conduits much like water is attracted to the glass (silica) wall of a capillary tube. In trees, suction or water tension can reach 2000 kPa. Evaporation takes place from the leaf surfaces and a continuous flow of water is generated in this fashion. This flow can reach 1 m3 per day. The tree absorbs carbon dioxide (CO2 ) from the air and pumps water (H2 O) from the ground. It then uses the energy from the sun (photosynthesis) to combine the carbon dioxide with the water to make sugar (C6 H12 O6 ) and release oxygen (O2 ). Sugar is the essential basis for all plant growth. Trees and plants in general are extremely important to humankind because they absorb what we exhale (CO2 ) and produce what we inhale (O2 ).

Relative humidity, RH

𝜌w RT LnRH M

where 𝜓 is the suction potential in Pa, 𝜌w is the mass density of the water (1000 kg/m3 ), M is the molecular weight of water (0.01802 kg/mol), T is the absolute temperature in Kelvin, R is the universal gas constant (8.314 N m/mol K), and RH is the relative humidity expressed as a ratio rather than a percent. This suction potential in the void of the unsaturated soil can develop into a water tension, in which case: 𝜌 RT uw = w LnRH (11.70) M At 20 ∘ C and given the same constants used earlier, the relationship is: (11.71) uw (kPa) = 135,000LnRH

Relative humidity, RH

Ψ=

315

1

0.95

0.9

0.85

0.8

0

5000

10000 15000 20000 25000 30000 35000 Total suction, uw (kPa)

Figure 11.27 Water tension vs. relative humidity: (a) Relative humidity 0–100%. (b) Relative humidity 80–100%.

316

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS n ∑ 𝛾ti hi 𝜎ov =

11.18 PRECISION ON WATER CONTENT AND WATER TENSION

(11.73)

i=1

Water tension is more complicated to measure than water content. Water content also typically varies much less than water tension. A typical range of water content variation is 5 to 50%, whereas the typical range for water tension is −10 to −1,000,000 kPa. In an experiment conducted by Garner (2002, unpublished), three samples were sent to eight laboratories in Texas requesting that the water content and the suction be measured. Most laboratories used the filter paper method for the suction determination. The results were collected, and an error band was created for each sample. The results are shown in Figure 11.28. They confirm that the arithmetic value of the suction varies a lot more than the water content. They also indicate that the error band for identically prepared samples is much larger for the determination of suction than for water content. If the log of the suction is used instead of the arithmetic value, then the error band of log(suction) approaches the error band of water content (Figure 11.28).

where 𝛾 ti is the total unit weight of layer i and hi is the thickness of layer i. Note that if there is water above the ground surface, the water must be included as a layer to calculate the total vertical stress. This is the case with a river, a lake, or an ocean. At the bottom of the deep oceans, the total vertical stress is very large and compresses any object tremendously. For example, a Styrofoam coffee cup going to 3000 m of water depth comes back the size of a thimble. Below the groundwater level, the water stress at rest uwo is calculated under normal circumstances as: uwo = 𝛾w zw

where 𝛾 w is the unit weight of water and zw is the depth below the GWL. Note that this water stress acts equally in all directions (hydrostatic), as it is assumed that water has no shear strength. This stress is a compressive stress. If an artesian condition exists, then information about the water stress in the artesian layer must be known or inferred from the global aquifer analysis. If a perched GWL condition exists, then information must be gathered where the groundwater layer ends. Above the GWL, in the zone saturated by capillary action, the water stress uw is calculated as:

11.19 STRESS PROFILE AT REST IN UNSATURATED SOILS The total vertical stress at rest 𝜎 ov at any depth z in a uniform soil is equal to the total unit weight of the soil 𝛾 t times the depth z: (11.72) 𝜎ov = 𝛾t z

uwo = −𝛾w zw

Water content, % Sample 1 Sample 2 Sample 3 5

10

15

20

25

30 %

4

5 log kPa

Water tension log (uw, kPa) Sample 1 Sample 2 Sample 3 0

1

0

15000

2

3

Water tension, kPa Sample 1 Sample 2 Sample 3

(11.75)

where zw is positive and represents the vertical distance above the GWL. The water stress in this case is a tensile stress. Close to the surface, the water tension no longer exhibits a linear profile

If the soil above the depth z is made of n layers, the total vertical stress at rest 𝜎 ov at depth z is:

0

(11.74)

30000

45000

60000 kPa

Figure 11.28 Error bands for suction and water content determination by eight different laboratories in Texas for three identically prepared samples. (Source: Garner, D. 2002. “Comparison of Suction and Water Content Determination.” Personal communication.)

11.19 STRESS PROFILE AT REST IN UNSATURATED SOILS

(Figure11.29). In that zone there is a power struggle between the soil particle minerals, which tend to attract the water, and the low relative humidity in the soil pores caused by the sun, which tends to draw the water away from the particles. The water is pulled hard in both directions, so high tensile stresses develop. Quantifying the variation of uw with depth within that region requires advanced computations and depends on many factors, including rainfall, wind speed, solar radiation, temperature, soil hydraulic conductivity, extent of the cracks in the soil, and so on. Such computations are beyond the scope of this book. Once the total vertical stress at rest 𝜎 ov is known, and once the water stress at rest uwo is known, the vertical effective stress at rest 𝜎 ′ ov is calculated as: ′ = 𝜎ov − 𝛼uwo 𝜎ov

Eq. 11.63 (𝛼 will be 1 under the GWL and in the zone saturated by capillary action). 8. Calculate the effective vertical stress at rest, 𝜎 ′ ov , using Eq. 11.76. 9. Plot the values of 𝜎 ′ ov on a graph at the depths corresponding to the discontinuities and join these points by straight lines. Figure 11.29 is an illustration of this step-by-step procedure under the following conditions: 1. The soil is uniform. 2. The total unit weight of the soil is equal to 20 kN/m3 , and the unit weight of water is taken as 10 kN/m3 . 3. The GWL is at a depth of 5 m. 4. The points of discontinuity are the bottom of the profile (z = 7 m), the GWL (z = 5 m), the top of the capillary zone (z = 3 m), and the ground surface (z = 0 m). 5. The total vertical stress at rest at the bottom of the profile is equal to 𝜎 ov = 20 × 7 = 140 kN/m2 . At the top of the profile, it is 𝜎 ov = 0. Because there is no discontinuity in total unit weight between these two discontinuities, the profile is a straight line between the two values. 6. The water stress at rest at the bottom of the profile is equal to uwo = 2 × 10 = 20 kN/m2 . At the GWL, the water stress uwo is zero. At the top of the capillary zone, the water stress is uwo = –2 × 10 = –20 kN/m2 . The profile of water stress in the unsaturated zone above the top of the capillary zone is estimated as shown in Figure 11.29. 7. The water area ratio is equal to 1 in the zone where the soil is saturated. In the unsaturated zone, a linear decrease of 𝛼 from 1 at the top of the capillary zone to 0 at the ground surface is assumed in this case. 8. The effective stress is calculated according to Eq. 11.76. For the point at the bottom of the pro′ file, 𝜎ov = 140 − 20 = 120 kN∕m2 . For the point at ′ the GWL, it is 𝜎ov = 100 − 0 = 100 kN∕m2 . For the

(11.76)

where 𝛼 is the water area ratio estimated as the degree of saturation or obtained from Eq. 11.63. One of the important initial steps in solving a geotechnical problem is to prepare the profile of vertical stresses at rest for the site. This is done in the following steps: 1. Identify the layers and their thicknesses for the deposit considered. 2. Determine the total unit weight of each layer. 3. Determine the location of the GWL and any irregularity associated with the water regime (artesian pressure, perched water table). 4. Identify the points of discontinuity versus depth. These points include boundaries between two layers and depth to the GWL. 5. Calculate the total vertical stress at rest, 𝜎 ov , at each discontinuity using Eq. 11.73. 6. Calculate the water stress at rest, uw , at each discontinuity using Eq. 11.74 below the GWL and Eq. 11.75 above the GWL. 7. Calculate the water area ratio 𝛼 at each discontinuity, estimated as the degree of saturation or by using σov (kPa) 0

0 1

Depth (m)

2

7

–100

1

–50

0

σ'ov (kPa) 50

0

40

s (kPa)

80 120

–100

20 40 60 80

72

120

Unsaturated 1

Uniform soil 100

γw = 10 kN/m3

αuwo (kPa)

α

–400

Top of –20 60 Capillarity zone

5 6

uwo (kPa) –400 –200 0 –200 0

γt = 20 kN/m3

3 4

80 160

G.W.L. 0

–20

Saturated by capillarity 1

80

48

100

0

60

Saturated 140

20

317

1

20

Figure 11.29 Stress profiles in a soil deposit.

120

72

318

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

point at the top of the saturated capillary zone, it is ′ = 60 − 1 × (−20) = 80 kN∕m2 . In the unsaturated 𝜎ov zone above the capillary zone, the profile is obtained by using Eq. 11.76. 9. The values of effective stress are plotted in Figure 11.29. Note that the effective stress decreases linearly as the depth decreases when the soil is saturated but increases as the depth continues to decrease in the unsaturated zone. We can then calculate the shear strength of the soil on horizontal planes by multiplying the vertical effective stress by the tangent of the friction angle, assuming that the soil has no ′ tan 𝜑. Therefore, the shear effective stress cohesion: s = 𝜎ov strength profile has the same shape as the effective stress profile. The increase in effective stress, and therefore strength close to the surface due to higher water tension, often leads to a crust that can be a few meters thick.

11.20

SOIL WATER RETENTION CURVE

The soil water retention curve (SWRC), also known as the soil water characteristic curve, is a property of the soil much like the shear strength parameters (Figure 11.30). It is a plot of the water content of the soil as a function of the water tension stress (suction) in the soil pores. Figure 11.30 is a SWRC on a semilog plot; the water content is on a natural scale while the water tension is on a log scale. From point A to point B on Figure 11.30, the soil remains nearly saturated while the water tension increases. At the air entry value (point B), the water content decreases while the water tension increases. Up to point C on Figure 11.30, the water content is usually well represented by a straight line and the slope of that line is the coefficient Cw : u Δw = Cw log w (11.77) uwae 50

Water content w (%)

where Δw is the change in water content, Cw is the slope of the SWRC, uw is the water tension, and uwae is the air entry value of the water tension. From C to D, the water content continues to decrease while the water tension continues to increase, but at a much higher rate. If a saturated soil sample is placed on a table top and is strong enough to stand by itself, it is likely held together by water tension unless it has some cementation (effective stress cohesion). As the soil dries, it initially shrinks while remaining saturated. The water tension increases, and at a given water tension stress (suction), air enters the pores. This water tension is called the air entry value (uwae ). From this point on during the drying process, the soil is unsaturated. By definition, the water content at the air entry value is the undisturbed shrinkage limit because, during the shrinkage process, it is the last water content where the soil is saturated. The gravimetric water content is the water content definition most commonly used in geotechnical engineering, but for the SWRC, the volumetric water content is often used. They are defined as follows:

Air entry stress uwae

40 A

B

30

(11.78)

Volumetric water content ∶ 𝜃w = Vw ∕V

(11.79)

When the term water content is used in this book, it means gravimetric water content. Example SWRCs are presented in Figure 11.31. Different soils have different SWRCs; for instance, a sand will not retain water the same way a clay would. Imagine that you insert a straw into a sand. It would not take much sucking to get the water out of the sand. Now imagine that your straw is inserted into a clay. In this case it would take a lot of sucking to get a little bit of water out. The suction or water tension that you would exert through the straw would be much higher for the clay than for the sand. This phenomenon is what the SWRC characterizes. Soils under the groundwater level are generally saturated and the water is in compression. Soils above the GWL can be saturated or unsaturated, but in both cases the water is in tension (suction). The SWRC is a property of a soil where the water is in tension. Thus, the SWRC for a saturated soil refers to the case where the soil is saturated above the GWL by capillary action and other electrochemically based phenomena such as the affinity between water and clay minerals (point A to B on Figure 11.30). Beyond point B, the soil is unsaturated.

w = Cwlog uw Cw

20

11.21 1

Saturated 1

Unsaturated

INDEPENDENT STRESS STATE VARIABLES

Effective stress 𝜎 ′ , as defined in Eq. 11.54, is:

10 0

Gravimetric water content ∶ w = Ww ∕Ws

𝜎 ′ = 𝜎 − 𝛼 uw − 𝛽 ua C

10 100 1000 Water tension stress uw (kPa)

Figure 11.30 Soil water retention curve.

D 10000

(11.80)

Effective stress is defined on the basis of three stresses (𝜎, uw , ua ) and two soil properties (𝛼, 𝛽). Therefore, it depends not only on the state of stress in a soil, but also on the soil properties. Hence, it cannot be considered an independent stress variable, even if it is a very useful stress in

11.21 INDEPENDENT STRESS STATE VARIABLES

Volumetric water content θ* (%)

60

Free draining water

50

Water available to plants

319

Water unavailable to plants

uwt

40

uwt = air entry water tension

uwt

30

Volume of water Total volume

*θ =

uwt

20

Silt Sand

10

Clay

0 0

10

100

1000

10000

100000

1000000

Water tension stress uw (kPa)

Figure 11.31 Examples of soil water retention curve.

solving many soil problems. Equation 11.80 can be rewritten as follows:

two stress tensors, which include all the stress information necessary to solve an unsaturated soil problem.

𝜎 ′ = 𝜎 − 𝛼 uw − 𝛽 ua − ua + ua

𝜏xy 𝜏xz ⎤ ⎡𝜎xx − ua 𝜎yy − ua 𝜏yz ⎥ = ⎢ 𝜏yx 1 ⎥ ⎢ 𝜏zy 𝜎zz − ua ⎦ ⎣ 𝜏zx

(11.82)

0 0 ⎤ ⎡ua − uw 0 ⎥ ua − uw =⎢ 0 2 ⎥ ⎢ 0 ua − uw ⎦ ⎣ 0

(11.83)

∑

= (𝜎 − ua ) − 𝛼 uw + (1 − 𝛽)ua = (𝜎 − ua ) + 𝛼(ua − uw )

(11.81)

∑

In this form, it becomes clear that two independent stress state variables are necessary to describe the effective stress: the net normal total stress in excess of air stress (𝜎 – ua ) and the net water tension with respect to the air stress (ua – uw ). In terms of total stresses, the stress tensor at a point is defined in Eq. 11.3. This stress tensor does not include information on the water stress or the air stress. Keeping in mind that shear stresses are unaffected by water or air stress, the stress state in a soil can be fully described by the following

In the case of a saturated soil, only one tensor is necessary: ∑ 1

𝜏xy 𝜏xz ⎤ ⎡𝜎xx − uw ⎢ 𝜏 𝜎 − u 𝜏yz ⎥ = yx yy w ⎥ ⎢ 𝜏zy 𝜎zz − uw ⎦ ⎣ 𝜏zx

(11.84)

Problems and Solutions Problem 11.1 A wedge has applied stress vectors on two faces as shown in Figure 11.1s(a) and Figure 11.1s(b). Calculate the stress on the third face in both cases. Hint: You can compose forces, but you cannot compose stresses unless they act on the same area. σ3 = ?

σ1 = 10 kPa

A3 30° A2

A1 = 1 m2

A1 = 1 m2

τ3 = ?

A2

τ2 = 10 kPa

σ1 = 10 kPa

30° σ2 = 10 kPa (a)

A3

τ3 = ? σ3 = ? (b)

Figure 11.1s Stress vectors on wedge faces.

320

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Solution 11.1 Part a: F1 = 𝜎1 •A1 = 10 (kN) A 1 A2 = 1 = = 1.732 (m2 ) tan 𝜃 tan 30∘ F2 = 𝜎2 •A2 = 10 ∗ 1.732 = 17.32 (kN) A 1 = 2 (m2 ) A3 = 1 = sin 𝜃 sin 30∘ 3 ∑ Fxi = 0 → Fx3(shear) • cos 30 + Fx3(normal) • sin 30 − 10 = 0 → (I) i=1 3 ∑ Fyi = 0 → Fy3(shear) • sin 30 − Fy3(normal) • cos 30 + 17.32 = 0 → (II) i=1

{

(I)&(II) →

FS3 = 0 (kN) → FN3 = 20 (kN)

{

𝜏3 = 0 (kPa) 𝜎3 = 10 (kPa)

Part b: F1 = 𝜎1 •A1 = 10 (kN) A 1 A2 = 1 = = 2 m2 sin 𝜃 sin 30∘ A 1 A3 = 1 = = 1.732 (m2 ) tan 𝜃 tan 30∘ 3 ∑ Fxi = 0 → −F1 + t2 A2 + 𝜏3 A3 = 0 → 𝜏3 = −5.77 kPa i=1 3 ∑ Fy = 0 → 𝜎3 = 0 i=1

Problem 11.2 In a triaxial test, the confining stress (minor principal stress) 𝜎 3 is 50 kPa, and the vertical stress (major principal stress) 𝜎 1 is 150 kPa. a. Form the total stress tensor shown in Eq. 11.3. Decompose the tensor into the deviatoric and spherical tensor forms shown in Eq. 11.4. b. The soil is saturated, and under the given stresses, the water stress is 20 kPa. Form the stress tensor in terms of effective stress. c. The soil is unsaturated, and under the given stresses, the air stress is 30 kPa and the water tension is –1000 kPa. Form the two tensors describing the state of stress in the sample in terms of independent stress state variables. Solution 11.2 a. ∑

0 ⎤ ⎡𝜎xx 𝜏xy 𝜏xz ⎤ ⎡50 0 = ⎢ 𝜏yx 𝜎yy 𝜏yz ⎥ = ⎢ 0 50 0 ⎥ (kPa) ⎥ ⎢ ⎢ ⎥ ⎣ 𝜏zx 𝜏zy 𝜎zz ⎦ ⎣ 0 0 150⎦

𝜎M = ∑

(𝜎xx + 𝜎yy + 𝜎zz ) 3

⎡𝜎xx 𝜏xy = ⎢ 𝜏yx 𝜎yy ⎢ ⎣ 𝜏zx 𝜏zy

(50 + 50 + 150) = 83.33 (kPa) 3 𝜏xz ⎤ ⎡𝜎M 0 0 ⎤ ⎡𝜎xx − 𝜎M 𝜏xy 𝜏xz ⎤ 𝜏yz ⎥ = ⎢ 0 𝜎M 0 ⎥ + ⎢ 𝜏yx 𝜎yy − 𝜎M 𝜏yz ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 𝜎zz ⎦ ⎣ 0 0 𝜎M ⎦ ⎣ 𝜏zx 𝜏zy 𝜎zz − 𝜎M ⎦ =

11.21 INDEPENDENT STRESS STATE VARIABLES

∑

0 0 ⎤ ⎡50 − 83.33 0 0 ⎡83.33 ⎤ ⎥ (kPa) 83.33 0 ⎥+⎢ 0 50 − 83.33 0 =⎢ 0 ⎢ ⎥ ⎢ ⎥ 0 83.33⎦ ⎣ 0 0 150 − 83.33⎦ ⎣ 0

∑

0 0 ⎤ ⎡−33.33 0 0 ⎤ ⎡83.33 83.33 0 ⎥+⎢ 0 −33.33 0 ⎥ (kPa) =⎢ 0 ⎢ ⎥ ⎢ ⎥ 0 83.33⎦ ⎣ 0 0 66.67⎦ ⎣ 0

321

b. 𝜎′ = 𝜎 − u ∑ c.

′ ′ 𝜏xy 𝜏xz′ ⎤ ⎡50 − 20 0 0 0 ⎤ ⎡𝜎xx ⎤ ⎡30 0 ′ ′ ⎢ ⎥ = ⎢ 0 30 0 ⎥ (kPa) 50 − 20 0 = 𝜏yx 𝜎yy 𝜏yz′ ⎥ = ⎢ 0 ⎢ ′ ⎥ ⎢ ⎥ ⎢ ⎥ 0 150 − 20⎦ ⎣ 0 0 130⎦ ⎣ 𝜏zx 𝜏zy′ 𝜎zz′ ⎦ ⎣ 0

𝜎 ′ = (𝜎 − ua ) + 𝛼(ua − uw ) 𝜏xy 𝜏xz ⎤ ⎡50 − 30 0 0 ⎡𝜎xx − ua ⎤ ∑ ⎥ (kPa) 𝜎yy − ua 𝜏yz ⎥ = ⎢ 0 50 − 30 0 = ⎢ 𝜏yx 1 ⎥ ⎢ ⎢ ⎥ 0 150 − 30⎦ 𝜏zy 𝜎zz − ua ⎦ ⎣ 0 ⎣ 𝜏zx 0 ⎤ ⎡20 0 = ⎢ 0 20 0 ⎥ (kPa) 1 ⎢ ⎥ ⎣ 0 0 120⎦

∑

0 0 ⎤ ⎡30 − (−1000) 0 0 ⎡ua − uw ⎤ ⎥ (kPa) 0 ⎥=⎢ ua − uw 0 30 − (−1000) 0 =⎢ 0 2 ⎥ ⎢ ⎢ ⎥ 0 ua − uw ⎦ ⎣ 0 0 30 − (−1000)⎦ ⎣ 0

∑

0 0 ⎤ ⎡1030 1030 0 ⎥ (kPa) =⎢ 0 2 ⎢ ⎥ 0 1030⎦ ⎣ 0

∑

Problem 11.3 A simple shear test is performed in a plane strain condition (Figure 11.2s). The vertical normal stress on the plane of failure is 80 kPa, the horizontal normal stress is 40 kPa, and the shear stress is 30 kPa on the horizontal plane. The Poisson’s ratio for the soil is 0.35. Form the total stress tensor (Eq. 11.3). Decompose this tensor into the deviatoric and spherical tensor forms shown in Eq. 11.4. z

σzz = 80 kPa τzx = 30 kPa τzx = 30 kPa σxx = 40 kPa

x

y

Figure 11.2s Stresses during the simple shear test.

Solution 11.3 Eq. 11.4: ∑

𝜏xy 𝜏xz ⎞ ⎛𝜎m 0 0 ⎞ ⎛𝜎xx − 𝜎m ⎛𝜎xx 𝜏xy 𝜏xz ⎞ 𝜎yy − 𝜎m 𝜏yz ⎟ kPa = ⎜ 𝜏yx 𝜎yy 𝜏yz ⎟ = S + D = ⎜ 0 𝜎m 0 ⎟ + ⎜ 𝜏yz ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 𝜏zy 𝜎zz − 𝜎m ⎠ ⎝ 0 0 𝜎m ⎠ ⎝ 𝜏zx ⎝ 𝜏zx 𝜏zy 𝜎zz ⎠

where 𝜎m = 13 (𝜎xx + 𝜎yy + 𝜎zz ).

322

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

From the problem statement: 𝜎xx = 40 kPa, 𝜎zz = 80 kPa, 𝜏zx = 30 kPa, and 𝜏xy = 𝜏yz = 0 due to the plane strain condition. The value of 𝜎 yy is found using the plain strain condition: 𝜀yy = 0 =

1 (𝜎 − v(𝜎xx + 𝜎zz )) E yy

so that 𝜎yy = v(𝜎xx + 𝜎zz ) = 0.35(40 + 80) = 42 kPa Therefore, 𝜎m =

1 (40 + 42 + 80) = 54 kPa. 3

The deviatoric and spherical tensor forms are: ∑

⎛𝜎xx 𝜏xy 𝜏xz ⎞ ⎛54 0 0 ⎞ ⎛−14 0 30⎞ = ⎜ 𝜏yz 𝜎yy 𝜏yz ⎟ = S + D = ⎜ 0 54 0 ⎟ + ⎜ 0 −12 0 ⎟ kPa ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 0 26⎠ ⎝ 𝜏zx 𝜏zy 𝜎zz ⎠ ⎝ 0 0 54⎠ ⎝ 30

Problem 11.4 A sample of cohesionless silt is tested in a direct shear test. At failure, the vertical normal stress is 100 kPa and the shear stress on the horizontal plane where failure occurs is equal to 40 kPa. The water stress is 20 kPa. a. Calculate the effective principal stresses by using the equilibrium equations approach in two dimensions. b. Calculate the effective principal stresses by using the Mohr circle approach in two dimensions. Solution 11.4 a. The effective principal stresses are related to the shear and normal stress on the failure plane through the equilibrium equations (Eqs. 11.10 and 11.11): 𝜎′1 + 𝜎′3 𝜎′1 − 𝜎′3 + cos 2𝛼 2 2 ′ ′ 𝜎 −𝜎 3 sin 2𝛼 𝜏=− 1 2 Because the sample is at failure and the silt is cohesionless, the shear strength equation can be written as: 𝜎′ =

𝜏 = 𝜎 ′ tan 𝜑′ This also means that: sin 𝜑′ =

𝜎1′ − 𝜎3′ 𝜎1′ + 𝜎3′

The last four equations, together with the given values of 𝜎 ′ = 80 kPa and 𝜏 = 40 kPa, give the values of the four unknowns: 𝜑′ , 𝜎1′ , 𝜎3′ and 𝛼. The solution is 𝜑′ = 26.6∘ , 𝜎1′ = 144.4 kPa, 𝜎3′ = 55.4 kPa, and 𝛼 = 59o . b. The effective principal stresses can be found using the Mohr circle as shown in Figure 11.3s. 90 Shear stress τ (kPa)

80

27°

70 60 50 40 30 20

2α =117°

10

63°

55.4 kPa

20

40

60

80

144.4 kPa

100

120

140

Effective stress σʹ (kPa)

Figure 11.3s Mohr circle for direct shear test.

160

11.21 INDEPENDENT STRESS STATE VARIABLES

323

Problem 11.5 For problem 11.4, use the Pole method to locate the principal stresses planes. Solution 11.5 First, we draw the failure stress point on the shear stress vs. effective normal stress set of axes (𝜏 = 40 kPa, 𝜎 = 80 kPa). This point is on the failure envelope, and because the soil has no cohesion intercept, the failure envelope can be drawn through the origin and the failure point. The Mohr circle is found tangent to the failure envelope at the failure stress point. According to the Pole method, the line parallel to the plane on which the stresses act (horizontal plane) intersects the Mohr circle at two points: the stress point and the Pole. This allows us to find the Pole (Figure 11.4s). Knowing the Pole, we draw the lines that join the Pole to the two principal stress points 𝜎1′ and 𝜎3′ . These lines define the directions of the planes on which the principal stresses are acting (Figure 11.5s). 90 Shear stress τ (kPa)

80 27°

70 60 50

Pole

40 30

59°

20 10

55 kPa

20

40

144 kPa

80 60 100 120 Effective stress σ′ (kPa)

140

Figure 11.4s Pole method. σ′ = 80 kPa

τ′ = 80 kPa 31°

59°

Fu

Fu

σ3′ = 55 kPa

σ1′ = 144 kPa

Figure 11.5s Principal planes in direct shear test.

Problem 11.6 For the sample in Figure 11.6s, a. Find the stresses on the plane shown. b. On what plane does the maximum shear stress exist?

σ1 = 40 kPa

τ σ3 = 20 kPa

30°

σ σ3 = 20 kPa

Figure 11.6s Stress state.

160

324

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Solution 11.6 a. We can solve this problem with the equilibrium equations or with the Mohr circle. Recall Eqs. 11.10 and 11.11 from the text: 𝜎 + 𝜎3 𝜎1 − 𝜎3 𝜎= 1 + cos 2𝛼 2 2 𝜎1 − 𝜎3 𝜏=− sin 2𝛼 2 By using these equations, we obtain: 40 + 20 40 − 20 + cos(2 × 30∘ ) = 35 kPa 𝜎= 2 2 40 − 20 𝜏=− sin(2 × 30∘ ) = −8.67 kPa 2 We then confirm the solutions by use of the Mohr circle (Figure 11.7s).

Shear stress, τ (kPa)

20

10 Pole 0

20

10

40

30° 30

τ = –8.67 –10

Normal stress, σ (kPa)

σ = 35

Figure 11.7s Mohr circle.

b. To find the plane where the maximum shear stress acts, we use the Pole method. We first find the Pole by drawing a line parallel to the plane where 𝜎 1 acts (horizontal). That line intersects the Mohr circle at two points: the 𝜎 1 stress point and the Pole. Then we join the Pole to the largest shear stress point of the Mohr circle. That line is a 45-degree line and gives the plane on which the highest shear stress acts (Figure 11.8s).

Shear stress, τ (kPa)

20

10 Pole 0 10

45° 20

30

40

Normal stress, σ (kPa)

–10

Figure 11.8s Pole method.

Problem 11.7 What happens to the Pole method when the diagram of a stress element in space is rotated by an angle θ? Does the Mohr circle change? Do the stresses on any plane change? Does the Pole location change?

11.21 INDEPENDENT STRESS STATE VARIABLES

325

Solution 11.7 When the diagram of the stress element in space is rotated by an angle 𝜃, the Mohr circle does not change because the principal stresses do not change; accordingly, the stresses on any plane do not change either. However, the location of the Pole on the Mohr circle rotates with the diagram to maintain the rule of parallelism. Problem 11.8 In a simple shear test, the horizontal displacement at the top of the sample is 1 mm and the vertical displacement is a reduction in height of 0.5 mm. The original height of the sample is 25 mm. a. Calculate the shear strain and the vertical normal strain. b. Is the sample dilating or contracting? Solution 11.8 a. The shear strain and the vertical normal strain are (Figure 11.9s): 1 𝛾shear = tan−1 = 0.04 or 4% shear strain 25 0.5 = 0.02 or 2% compression normal strain 𝜀normal = 25 b. The sample is contracting. 1 mm 0.5 mm 25 mm

γ

Figure 11.9s Normal and shear strain.

Problem 11.9 In a triaxial test, the sample has an initial height of 150 mm and an initial diameter of 75 mm. During the loading in the vertical direction, the vertical displacement is 3 mm and the increase in diameter is 2 mm. a. Calculate the normal strains 𝜀zz and 𝜀rr . b. Form the strain tensor. c. Calculate the shear strain on a 45-degree plane. Solution 11.9 a. The normal strains 𝜀zz and 𝜀rr are:

3 2 = 0.02, 𝜀rr = − = −0.027 150 75 ] [ [ ] 𝜀rr 21 𝛾rz −0.027 0 = 1 0 0.02 𝛾zr 𝜀zz

𝜀zz = b. The strain tensor is:

2

c. The shear strain on a 45-degree plane is: 𝛾45 = (𝜀zz − 𝜀rr ) sin 2𝛼 = (0.02 − (−0.027)) ∗ sin(90) = 0.047 Problem 11.10 Consider the sphere-shaped soil particles shown in Figure 11.10s. The degree of saturation S is 1, the porosity n is 0.4, and the ratio between the sum of the contact areas and the total area (Ac /At ) is 0.01. Calculate the following quantities and show the relationship between the total stress and the effective stress if the water stress is +40 kPa.

326

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

a. The average effective normal stress. b. The average normal stress at the contacts. c. The average normal total stress.

1 mm

1 mm

1 mm

0.1 N

0.15 N

0.05 N

water

3 mm

water

1 mm

u = 40 kPa Plan view

Side view

Figure 11.10s Sphere-shaped soil particles.

Solution 11.10 At = 3 × 1 × 10−6 = 3 × 10−6 m2 The average effective normal stress is: ′ 𝜎aver =

(0.1 + 0.15 + 0.05) × 10−3 3 × 10−6

= 100 kPa

The average normal stress at the contacts is: 𝜎c−aver =

(0.1 + 0.15 + 0.05) × 10−3 3 × 10−6 × 0.01

= 10000 kPa

The average normal total stress is: 𝜎aver =

(0.1 + 0.15 + 0.05) × 10−3 + 40 × (1 − 0.01) × 3 × 10−6

3 × 10−6 By definition, the relation between the total stress and the effective stress is:

= 139.6 kPa

𝜎 = 𝜎 ′ + u = 100 + 40 = 140 ≈ 139.6 kPa Problem 11.11 The surface tension of water is T = 73 mN/m, the diameter of a glass tube plunged into water is 0.002 mm, and the contact angle between the wall of the clean glass and the water is 𝛼 = 10 degrees. Find the height to which the water will rise in the small tube. Solution 11.11 d = 0.002 mm T = 73mN∕m 𝛼 = 10 deg hc =

4 × 73 × 10−6 × cos(10) 4T cos 𝛼 → hc = = 14.37 m d𝛾w 0.002 × 10−3 × 10

Problem 11.12 Consider the sphere-shaped soil particles shown in Figure 11.11s. The porosity n is 0.4, the ratio between the sum of the contact areas and the total area (Ac /At ) is 0.01, and the ratio between the sum of the areas of water and the total area (Aw /At ) is 0.1.

11.21 INDEPENDENT STRESS STATE VARIABLES

327

Calculate the following quantities and show the relationship between the total stress and the effective stress if the water stress is –6000 kPa. a. The average normal effective stress. b. The average normal stress at the contacts. c. The average normal total stress. 1 mm

1 mm

1 mm 3 mm

air

1N

air

1.5 N

1 mm 0.5 N

water

u = –6000 kPa Plan view

Side view

Figure 11.11s Sphere-shaped soil particles.

Solution 11.12 At = 3 × 1 × 10−6 = 3 × 10−6 m2 a. The average normal effective stress is: ′ 𝜎aver =

(1 + 1.5 + 0.5) × 10−3 3 × 10−6

= 1000 kPa

b. The average normal stress at the contacts is: 𝜎c−aver =

(1 + 1.5 + 0.5) × 10−3 3 × 10−6 × 0.01

= 100000 kPa

c. The average normal total stress is: 𝜎aver =

(1 + 1.5 + 0.5) × 10−3 − 6000 × 0.1 × 3 × 10−6 3 × 10−6

= 400 kPa

The relation between the total stress and the effective stress is: 𝜎 ′ = 𝜎 − 𝛼u = 400 − 0.1 × (−6000) = 1000 kPa Problem 11.13 A soil has a degree of saturation of 92%. The air is occluded, and the bubbles are 1 mm in diameter. Knowing that the water tension can reach 73 mN/m, what is the maximum difference in pressure that can exist between the water stress and the air stress? Solution 11.13 It seems reasonable that the air is occluded, as the degree of saturation of the soil is 92%, which is larger than 85%. Based on the equilibrium of the free-body diagram of half the bubble, and knowing that the water tension is 73 mN/m, we have (Eq. 11.59): 4 × 73 × 10−6 kN∕m 4T = = 0.29 kPa ua − uw = D 1 × 10−3 m Problem 11.14 A soil has a degree of saturation of 35%, an air entry value of −150 kPa, and a water tension stress of −1500 kPa at a depth of 2 m. Estimate the vertical effective stress at rest at a depth of 2 m below the ground surface, assuming that the unit weight of the soil is 19 kN/m3 .

328

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Solution 11.14 The soil has a degree of saturation of 35%, which means the soil is unsaturated and the relationship between the effective stress and total stress is: 𝜎 ′ = 𝜎 − 𝛼 uw Here, 𝛼 can be obtained based on Khalili and Khabbaz (1998): √ √ uwae −150 𝛼= = = 0.316 uw −1500 Therefore, at the given depth of 2 m below the ground surface, the vertical effective stress at rest is: 𝜎 ′ = 𝜎 − 𝛼 uw = 𝛾 Z − 𝛼 uw = 19 × 2 − 0.316 × (−1500) = 512 kPa Problem 11.15

( ) ′ Draw the three profiles 𝜎ov , uo , 𝜎ov for the layered system shown in Figure 11.12s. 2m

Capillary zone

Clay (saturated) γd = 18 kN/m3 Sand γd = 20 kN/m3

4m

Figure 11.12s Soil profile.

Solution 11.15 σ0v

σʹ0v

u0 –2 × 9.81 = –19.6 kN/m2

19.6 kN/m2

(–)

2m

2 × 18 = 36

kN/m2

36 kN/m2

Capillary zone

4m (+) 36 + 4 × 20 = 116 kN/m2

4 × 9.81 = 39.2 kN/m2

76.8 kN/m2

Figure 11.13s Stress profiles.

Problem 11.16 ( ′ ) Draw the effective stress profiles 𝜎ov for the layered system shown in Figure 11.14s. σov (kPa) 0 1

Depth (m)

2

7

1 –100 Unsaturated

kN/m3

Top of –20 60 Capillarity zone

1

Uniform soil

Saturated by capillarity 100

5 6

α

–400

γt = 20

3 4

uwo (kPa)

0 80 160 –400 –200 0 –200 0

G.W.L. 0

1

γw 5 10 kN/m3

Saturated 140

20

Figure 11.14s Soil and stress profile.

1

11.21 INDEPENDENT STRESS STATE VARIABLES

Solution 11.16 ′ The effective vertical stress 𝜎ov profile is shown in Figure 11.15s. σov (kPa) 0

0 1

Depth (m)

2

γt = 20

7

0

σʹov (kPa)

50 0

40 80 120

–100

120

Unsaturated

kN/m3

Top of –20 60 Capillarity zone

–20

1

Uniform soil

80

Saturated by capillarity 0

G.W.L.

100

5 6

αuwo (kPa) –100 –50

1

–400

3 4

α

uwo (kPa)

80 160 –400 –200 0 –200 0

1

γw = 10

0

100

Saturated

kN/m3

140

20

120

20

1

Figure 11.15s Stress profiles.

Problem 11.17 ( ) ′ Draw the three profiles 𝜎ov , uw , 𝜎ov below the center of the river for the layered system shown in Figure 11.16s. Clay Sand

7m

River Clay

4m 9m

Sand

Figure 11.16s River profile.

Solution 11.17 ′ The stress profile for 𝜎ov , uw , 𝜎ov is shown in Figure 11.17s:

0

0m

4m

Water γ = 10 kN/m3

σ0v (kPa)

40

uw (kPa)

0

40

0

σʹ0v (kPa)

0

Clay γ = 20 kN/m3 220

13 m

130

200

Sand γ = 18 kN/m3

Figure 11.17s Stress profile in the river.

20

90

329

330

11 STRESSES, EFFECTIVE STRESS, WATER STRESS, AIR STRESS, AND STRAINS

Problem 11.18 An insect has 4 legs and is able to walk on water. The depression created under each foot is a sphere, as shown in Figure 11.18s. What is the maximum possible weight of the insect?

FT

θ

35° W/4

FT cosθ

FT cosθ

35° W/4

FT

θ

55°

3 mm Contractive skin

Figure 11.18s Free-body diagram of insect.

Solution 11.18 The contact radius of the insect foot with the contractile skin is: r = 3 mm × sin 35∘ = 1.72 mm For a water temperature of 20 ∘ C, the surface tension (𝜎 T ) is 73 mN/m. ∑ Fv = 0 FT × 2𝜋r cos 𝜃 − W∕4 = 0 W = 8𝜋rFT cos 𝜃 W = 8𝜋(0.00172m)(73 mN∕m) cos 55∘ = 1.79 milliNewtons Problem 11.19 A soil has a water content of 42% and an air entry value of –8 kPa. If the slope of the soil water retention curve is 0.2 per log cycle of water tension in kPa, calculate the water tension for a water content of 10%. Solution 11.19 Given values: Cw = −0.2 uwae = −8 kPa w1 = 42% w2 = 10% 𝛥w = 0.42 − 0.10 = −0.2 log ( ) −8 0.32 log = = −1.6 uw −0.2 −8 10−1.6 = uw −8 uw = −1.6 10 uw = −318.5 kPa

(

−8 uw

)

Problem 11.20 A tree’s root system occupies a volume equal to 1000 m3 . How much water is available to that tree if it is rooted in the three soils described by the retention curves of Figure 11.31?

11.21 INDEPENDENT STRESS STATE VARIABLES

Solution 11.20 From the portion in the graph referring to “Water available to plants” in Figure 11.31: Change in volumetric water content (Θ): 1. Clay: ΔΘ = 0.50 - 0.34 = 0.16. 2. Silt: ΔΘ = 0.40 - 0.02 = 0.38. 3. Sand: ΔΘ = 0.15 - 0 = 0.15. ΔΘ × Volume of soil in root system = Water available to the tree. 1. Clay: 0.16 × 1000 m3 = 160 m3 . 2. Silt: 0.38 × 1000 m3 = 380 m3 . 3. Sand: 0.15 × 1000 m3 = 150 m3 .

331

CHAPTER 12

Problem-Solving Methods

12.1

GENERAL

There are three main types of problems in geotechnical engineering: failure load problems, deformation problems, and flow problems. Each problem can be solved by performing experimental modeling, by doing theoretical modeling, or by using experience. The best solutions are those that have a theoretical framework, are calibrated against and correlated with experimental measurements, and are verified by experience at full scale. Experience is obtained by years of practice. As the saying goes, good judgment comes from experience, but experience comes from bad judgment. An attempt can be made at teaching experience by letting engineers, who have been practicing successfully for a long time, discuss case histories—including failures—in a classroom environment. Theoretical modeling includes continuum mechanics closed-form solutions, numerical simulations, dimensional analysis, probabilistic analysis, and risk analysis. Experimental modeling includes the use of scaled models, centrifuge models, and/or full-scale models. In all cases, fundamental laws and constitutive laws help in solving the problem.

are often drawn too steep. By making a drawing to scale, you give yourself a better chance of recognizing some of the problems associated with the physical dimensions of the project. Always make a drawing to scale as a first step in solving an engineering problem.

12.2

12.3

DRAWING TO SCALE AS A FIRST STEP

One very important first step in solving a geotechnical engineering problem (or any engineering problem in general) is to always start by making a drawing to scale of the problem. If this step is not taken, the engineer may not get a proper sense of the issues at hand. For example, if one is designing a pile foundation under a building with the pile tips bearing into a sand layer, making a drawing to scale helps the engineer evaluate whether the sand layer is thick enough to prevent serious compression of the layers below. Failing to make that drawing properly, and instead drawing only a sample single pile bearing into the sand layer, may give the false impression of a thick sand layer (Figure 12.1). Also, if you draw a driven pile as a thick, short vertical member instead of the actual slender member, the issue of buckling will not come to your attention. Embankments typically have side slopes of 2 to 1 or 3 to 1, yet when they are sketched on a piece of paper, these slopes

Building Soft clay Sand

Soft clay

Sand

Soft clay

Figure 12.1 Make a drawing to scale.

PRIMARY LAWS

Two main types of laws are used to solve problems: fundamental laws and constitutive laws. Fundamental laws are valid no matter what material is being considered. They apply equally to soil, concrete, steel, or marshmallow. Fundamental laws include, for example, force and moment equilibrium, conservation of energy, and conservation of mass. The constitutive laws describe the behavior of the material. They are different for each material, whether it is soil, concrete, steel, or peanut butter. Constitutive laws include, for example, elasticity, plasticity, and viscoelasticity. Shear strength laws such as the Mohr-Coulomb criterion belong to the class of constitutive laws. Most theoretical problems are solved by making use of a combination of fundamental laws and constitutive laws. Other laws exist, such as the similitude laws used in dimensional analysis and the probabilistic laws used in risk analysis.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

333

334 12.4

12 PROBLEM-SOLVING METHODS

CONTINUUM MECHANICS METHODS

The basic and general steps in developing a theoretical solution to a soil problem are to describe the problem precisely, identify the variables, write the applicable equations, and solve for the unknowns. If there are more equations than unknowns, then one must choose which equations are most important to satisfy. If there are more unknowns than equations, it is time to make reasonable assumptions to generate new equations. The reasonableness of the assumptions should then be verified by experimentation at model scale or (even better) at full scale. In soil mechanics, there are three main types of problems: failure problems, deformation problems, and flow problems. 12.4.1 Solving a Failure Problem: Limit Equilibrium, Method of Characteristics, Lower and Upper Bound Theorems A typical solution to the problem of finding a failure load (e.g., ultimate bearing capacity of a footing) or a failure moment (e.g., slope stability) is to use the limit equilibrium analysis. In such a failure analysis, the step-by-step process advances as follows: 1. Assume a reasonable failure mechanism. If such a failure mechanism is not obvious, an experiment can be performed to observe the failure mechanism. 2. Draw a free-body diagram of the failing body (soil mass) and identify the external forces and external moments applied to the failing body. 3. Write the applicable fundamental equations. These are equations that are valid for all problems and independent of the type of material involved. They include equilibrium equations (three forces, three moments), conservation of mass, and conservation of energy, among others. 4. Write the applicable constitutive equations. These are the equations describing the behavior of the material under load. Constitutive laws include, for example, elasticity, plasticity, and viscoelasticity. The shear strength equation for soils, which states that the shear strength is a function of the effective stress on the failure plane, is another example of a constitutive law. 5. Count the number of equations and the number of unknowns. If there are as many equations as there are unknowns, proceed to the next step. If there are more equations than unknowns (rare), choose which equations are most important to satisfy. If there are more unknowns than equations, formulate assumptions that lead to additional equations. These assumptions should be based on engineering judgment, or experience, or experimental observations. The reasonableness of these assumptions should be verified by comparing the solution to observed full-scale behavior. 6. Combine all equations and solve for the unknown (failure load or failure moment).

There can be as many solutions as there are assumed failure mechanisms, so obviously it becomes important to choose the failure mechanism that most closely duplicates the real one. This is where observation of full-scale failures becomes very useful. Another solution is to use the method of characteristics. Characteristics are lines in the physical soil mass where the partial differential equation collapses into an ordinary differential equation. The equilibrium equations at the element level typically lead to partial differential equations. The method of characteristics simplifies these equations to the point where the problem is easier to solve. The method of characteristics can help to calculate failure loads for simple geometries. Yet another solution is to use the bound theorems and apply them to soil masses. There are two such theorems: the lower bound theorem and the upper bound theorem. The lower bound theorem states that if any stress distribution throughout the soil mass can be found, which is everywhere in equilibrium internally, does not violate the yield condition, and balances the external loads, the soil mass will safely carry the external loads. The upper bound theorem states that if an estimate of the failure load of a soil mass is made by equating internal rate of energy to the rate at which external forces do work in any postulated but kinematically admissible mechanism of deformation of the soil mass, the estimate will be either high or correct. In short, the lower bound theorem involves guessing a stress field that leads to a lower bound of the failure load; the upper bound theorem involves guessing a velocity or displacement field that leads to an upper bound estimate of the failure load. 12.4.2

Examples of Solving a Failure Problem

The first example problem is to find the ultimate pressure pu that a strip footing of width B (Figure 12.2) can exert on the surface of a saturated clay that has a shear strength s equal to the undrained shear strength su because the loading is rapid. The steps described in Section 12.4.1 for the limit equilibrium method are followed. 1. A cylindrical failure surface, as shown in Figure 12.3a, seems reasonable. This failure mechanism has been observed in many old silo failures. 2. The failing soil mass is the half cylinder shown in Figure 12.3b together with its free-body diagram. All Q Strip Footing

Li B

ery sv

ge

lar

Saturated Clay, Su

Figure 12.2 Strip footing dimensions.

335

12.4 CONTINUUM MECHANICS METHODS

Qu pu

o B

B

o B

B W su

(a) Failure mechanism

su

ΠB

L= (b) Free body diagram of failing mass

Figure 12.3 Failure load for a strip footing.

external forces and stresses are shown on the diagram, including the weight of the mass. Note that the weight is always an external force. 3. The most useful fundamental equation in this case is moment equilibrium around point O on Figure 12.3b. B M@O = 0 = pu B − s 𝜋BB (12.1) 2 4. The constitutive equation in this case is the shear strength equation, which states that the shear strength s is equal to the undrained shear strength of the clay. s = su

W

Dry sand γ, φ

P

H

P

θ

φ θ

(a) Problem

T

N

(b) Free body of failing mass

Figure 12.4 Example of a wall moving away from the backfill.

(12.2)

5. There are two unknowns (pu and s) and two equations, so the problem can be solved. 6. Now we combine the equations and obtain: pu = 2𝜋su

Wall movement

(12.3)

Other failure mechanisms are plausible and would lead to slightly different estimates of pu . The second problem is the one of a vertical wall with a height H supporting a clean, dry sand backfill with a friction angle 𝜑 (Figure 12.4). It is assumed that there is no friction between the wall and the backfill. The wall exerts a horizontal load P against the sand. As the wall moves very slightly away from the sand, the load P decreases and there is a point where the sand behind the wall starts to fail. At that point, the load is Pa and the question is to find the load Pa corresponding to impending failure of the sand. Note that the problem is a plane strain problem; therefore, all the loads will be line loads expressed in kN∕m. The steps described in Section 12.4.1 for the limit equilibrium method are followed: 1. The soil is assumed to fail as a wedge making an angle 𝜃 with the horizontal, as shown in Figure 12.4. This failure mechanism has been observed in model scale and centrifuge experiments. 2. The failing soil mass is the wedge; its free-body diagram is shown in Figure 12.4. All external forces are shown on the diagram, including the action of the wall P, the weight of the soil mass W, the normal force N, and the shear force T on the failure plane. Note that the shear force T is acting uphill because the wedge is

falling down along that plane and the soil outside of the wedge is resisting that tendency. 3. The most useful fundamental equations in this case are vertical and horizontal equilibrium of forces: ∑ (12.4) Fv = 0 = W − N cos 𝜃 − T sin 𝜃 ∑ Fh = 0 = P − N sin 𝜃 + T cos 𝜃 (12.5) 4. The constitutive equations in this case are the shear strength equation of the sand and the expression of the weight of the wedge. The shear strength equation states that the ultimate shear force T comes from the friction generated by the normal force N on the failure plane. The weight of the wedge is equal to the area of the wedge times the unit weight of the sand 𝛾: T = N tan 𝜑

(12.6)

𝛾H (12.7) 2 tan 𝜃 5. There are four unknowns (W, N, T, P) and four equations, so the problem can be solved. 6. Now we combine the equations and obtain: ( ) 𝛾H 2 sin 𝜃 cos 𝜃 − tan 𝜑 cos2 𝜃 P= (12.8) 2 sin 𝜃 cos 𝜃 + tan 𝜑 sin2 𝜃 2

W=

There is one more issue to resolve. The load P depends on 𝜃, yet there is a unique value of 𝜃 associated with the failure load Pa . This is the load at which the wedge fails behind the wall, and this load corresponds to the 𝜃 value that maximizes P (Figure 12.5). In other words, the wedge that needs the maximum support will fail first. The maximum value of P

336

12 PROBLEM-SOLVING METHODS

P (θ)

P (θ)

Wall movement

Wall movement

Pa 0 φ π ϕ + 4 2

Pp

θ

π 2

Pa π ϕ + 4 2

π ϕ – 4 2

Figure 12.5 Load P as a function of the wedge angle 𝜃; wall moves away from backfill.

is obtained by setting dP = 0. This derivative is: d𝜃 ( ) dP 𝛾H 2 sin 𝜃 cos 𝜃 − sin(𝜃 − 𝜑) cos(𝜃 − 𝜑) = =0 d𝜃 2 sin2 𝜃cos2 (𝜃 − 𝜑) (12.9) There are two solutions to Eq. 12.9: one is 𝜑 = 0, which is not realistic, and the other one is: 𝜋 𝜑 𝜃= + (12.10) 4 2 The load Pa can then be obtained from Eq. 12.8. ( ) 𝛾H 2 1 − sin 𝜑 Pa = (12.11) 2 1 + sin 𝜑 This problem is repeated but now with the wall being pushed into the sand (Figure 12.6) instead of pulled away from the sand, as in the previous case. The question is to find the load Pp that corresponds to the failure of the soil mass. Only one thing changes in the derivation: the direction of the shear force T on the failure plane. Because the wedge will now move up along the failure plane, the soil outside the wedge will exert a shear force acting toward the bottom of the wedge. Therefore, in the equations, T is replaced by −T and the problem leads to the situation shown in Figure 12.7. The failure load Pp is the load corresponding to the value of 𝜃 that minimizes P; that is, the wedge offering the minimum resistance is the wedge that will fail first. 𝜋 𝜑 (12.12) 𝜃= − 4 ( 2 ) 𝛾H 2 1 + sin 𝜑 (12.13) Pp = 2 1 − sin 𝜑

P

θ

Figure 12.7 Load P as a function of the wedge angle 𝜃; wall moves into backfill.

12.4.3

Solving a Deformation Problem

A typical solution to a deformation problem proceeds through the following steps: 1. Zoom in at the infinitesimal element level. This element has dimensions expressed in differential lengths. 2. The knowns and unknowns (e.g., loads, displacements, stresses, and strains) are identified on the element, including their variation from one side of the element to the other. This variation involves derivatives expressing the rate of change of the variable in one direction over a small distance. 3. The fundamental equations are written using the knowns and unknowns identified in step 2. These are equations that are true for all materials. They include equilibrium equations (three forces and three moments), conservation of mass, and conservation of energy, among others. 4. The constitutive equations are written using the knowns and unknowns identified in step 2. These equations describe the behavior of the material involved in the deformation. They include elasticity equations, plasticity equations, and viscosity equations, among others. 5. All equations are regrouped into the governing differential equations (GDEs). 6. The boundary conditions are expressed mathematically. If the problem is a dynamic problem, the boundary conditions involve both space and time. 7. The GDEs are solved in closed-form solutions if they are simple enough and through numerical solutions, such as the finite difference method if they are too complicated. The boundary conditions are used to define the constants involved in the solution. 12.4.4

γ, φ

θ π – ϕ 2

ϕ π – 4 2

W

Wall movement Dry sand

Pp

H

T

P θ

N

φ

Figure 12.6 Example of a wall moving toward the backfill.

Example of Solving a Deformation Problem

The example problem is to find the horizontal displacement y(z) of a pile as a function of z if the pile is loaded in overturning by a horizontal load H o and an overturning moment M o applied at the ground surface (Figure 12.8). For this simple example, the influence of the axial load will be ignored.

12.4 CONTINUUM MECHANICS METHODS

z (m)

dy dz

P y= k

y' =

y (mm)

y' (rd)

0

2

0

2

M = EI

dy dz2

M (kN-m) –3

10

0

40

V=

dM dz

V (kN) 0

P=

dV dz

P (kN/m)

50

0

337

40

H0 = 89 kN

5 Sand K = 25,000 kN/m2 0.61 m pipe pile Wall 9.5 mm thick

10

15

35 Zeros Local maximums Figure 12.8 Horizontally loaded pile example.

Y Z

dz

M V

O

Pdz

3. The fundamental equations that are most useful in this case are the horizontal equilibrium and the moment equilibrium. Let’s write the horizontal equilibrium first (Figure 12.9): ∑ (12.14) FH = 0 = Pdz + V − (V + dV) or P=

V + dV M + dM

Figure 12.9 Element of horizontally loaded pile.

The solution proceeds by following the steps described in Section 12.4.3: 1. Zoom in at the element level. In this case, we will select an element of the pile that is dz long (Figure 12.9). 2. The forces and moments acting on the element are shown on the element (Figure 12.9). These actions are the shear V (kN) and moment M (kN.m) at both ends of the element, and the soil resistance P(kN∕m) as a line load. Some of these quantities change by a little bit from one end of the element to the other. This little bit is expressed mathematically as 𝜕V dz for the shear force 𝜕z V, for example, expressing that the change is equal to the rate of change of V times the distance dz. Because V is dependent only on z, 𝜕V dz can be simply written 𝜕z as dV.

dV dz

(12.15)

So, the horizontal equilibrium states that the line load P on a pile is equal to the first derivative of the shear V. Now let’s write moment equilibrium around point O (Figure 12.9). Again, because M is only a function of z, 𝜕M dz can be simply written as dM. 𝜕z ∑

dz dz − (V + dV) 2 2 (12.16) Neglecting the higher-order term, we are left with: M@ o = 0 = M + dM − M − V

V=

dM dz

(12.17)

So, the moment equilibrium states that the shear in a pile is equal to the first derivative of the bending moment. 4. The constitutive equations describe the behavior of the pile and of the soil. The pile behavior is described by relating the bending moment M applied to the pile to the curvature generated in the pile. This curvature is expressed by the second derivative of y. The proportionality constant between the moment and the curvature is

338

12 PROBLEM-SOLVING METHODS

the bending stiffness Ep I, where Ep is the modulus of the pile material and I is the moment of inertia of the pile cross section around the axis of the moment. Again, 2 because y is only a function of z, 𝜕𝜕zy2 can be more simply written as

12.21. Applying boundary condition from step 6a gives c = d = 0. Applying boundary conditions 6b and 6c requires that the expressions of V and M be derived using Eqs. 12.17 and 12.18: ( ) z z −z (12.23) y(z) = e lo a sin + b cos lo lo ( ) dy z z 1 −z = e lo −(a + b) sin + (a − b) cos dz lo lo lo (12.24) ( ) 2 z d y 2 − z z = − 2 e lo −b sin + a cos (12.25) 2 l l dz lo o o ( ) d3 y 2 − lzo z z = 3e (a − b) sin + (a + b) cos lo lo dz3 lo (12.26) ( ) 4 z d y 4 − z z = − 4 e lo a sin + b cos (12.27) 4 l l dz lo o o

d2 y . dz2

M = Ep I

d2 y dz2

(12.18)

2

Note that the unit of ddz2y is 1∕m because d2 y is a little piece of y and dz2 is the square of a little piece of n z. This extends to the nth derivative; the unit of ddzny is 1∕m(n−1) because dn y is still a little piece of y while dzn is the nth power of a little piece of z. For the constitutive equation describing the soil behavior, a simple linear relationship is used between the line load P(kN∕m) characterizing the soil resistance and the deflection y (m) of the soil-pile interface. The proportionality constant is a spring constant K (kN∕m2 ) that characterizes the stiffness of the soil: P = −Ky

(12.19)

The minus sign is there because P and y are in opposite directions (Figure 12.9). 5. The governing differential equations can now be assembled by regrouping the fundamental and constitutive equations: P = −Ky =

d4 y dV d2 M = Ep I 4 = 2 dz dz dz

or

d4 y

4 = − 4y lo dz 4

(

Ep I d4 y

The lo parameter is required because of the need to match the factor Ep I∕K in the differential equation

(12.28)

which, compared to Eq. 12.21, leads to:

(12.20)

=0 (12.21) K dz4 6. The boundary conditions are stated for both ends of the pile. To simplify the solution of the differential equation, it is assumed that the pile is infinitely long and that the deflection is zero at the infinite end. At the top of the pile, the horizontal force and the overturning moment are known. The boundary conditions are: a. z = infinity, y = 0 b. z = 0, M = Mo , c. z = 0, V = Ho 7. Now we need to solve the differential equation. The solution y(z) has to be a function that becomes the same function when differentiated four times. This means a combination of exponential and trigonometric functions. The solution is therefore of the general form: ( ) z z − lz o a sin + b cos y (z) = e lo lo ( ) z z z lo +e c sin + d cos (12.22) lo lo y+

It can be seen from Eqs. 12.23 and 12.27 that:

lo =

4Ep I

) 14

K

(12.29)

Now boundary condition from step 6b can be written as: ( ) Mo d2 y 0 0 2 − l0o = e + a cos −b sin (12.30) = − lo lo lo2 dz2 @z=0 Ep I and a=−

Mo lo2 2Ep I

(12.31)

Then: d3 y dz

3

=

Ho Ep I

=

2 − l0o e lo3

@z=0

( ) 0 0 (a − b) sin + (a + b) cos lo lo (12.32)

and a+b=

Ho lo3 2Ep I

(12.33)

so b=

Ho lo3 Mo lo2 + 2Ep I 2Ep I

(12.34)

339

12.4 CONTINUUM MECHANICS METHODS

Now we can build the equations for the deflection y, the slope y′ , the bending moment M, the shear force V, and the line load P: ( ( ) ) Ho lo3 Mo lo2 Mo lo2 z z − lz o − sin + + cos y(z) = e 2Ep I lo 2Ep I 2Ep I lo (12.35) But Kl4 (12.36) Ep I = o 4 Therefore, finally: ( ) z 2Ho − z 2M z z z y(z) = e lo cos + 2 o e− lo cos − sin lo K lo lo lo lo K (12.37) ( ) z z 2H − 4M z z z y′ (z) = 2 o e lo cos + sin − 3 o e− lo cos lo lo lo lo K lo K (12.38) ( ) z z z −z −z (12.39) M(z) = Ho lo e lo sin + Mo e lo cos + sin lo lo lo ( ) 2Mo − z z z z −z e lo sin V(z) = Ho e lo cos − sin − lo lo lo lo (12.40) where y(z) is the pile displacement at a depth z, y′ (z) is the pile slope at z, M(z) is the bending moment at z, and V(z) is the shear force at z; H o and M o are the shear and moment at the ground surface, K is the soil spring constant, and lo is the transfer length given by Eq. 12.29. The profiles of y(z), y′ (z), M(z), and V(z) corresponding to Eqs. 12.37 to 12.40 are shown in Figure 12.8 as a function of depth for a real pile. 12.4.5

Solving a Flow Problem

A typical solution to a flow problem proceeds through the following steps: 1. Zoom in at the infinitesimal element level. This element has dimensions expressed in differential lengths. 2. The knowns and unknowns (flow velocities, volumes, total head, and water stress, for example) are identified on the element, including their variation from one side of the element to the other. This variation involves derivatives expressing the rate of change of the variable in one direction over a small distance. 3. The fundamental equations are written using the knowns and unknowns identified in step 2. These equations are true for all materials. The most useful in this case is the conservation of mass equation. 4. The constitutive equations are written using the knowns and unknowns identified in step 2. These equations describe the behavior of the material involved in the flow. The main equation in this case is Darcy’s law in the three dimensions. 5. All equations are regrouped into the governing differential equations.

6. The boundary conditions are expressed mathematically. These boundary conditions are usually in the form of total head or flow conditions. If the problem is a transient flow problem, the initial conditions are also stipulated. 7. The governing differential equations are solved in closed-form solutions if they are simple enough, and through numerical solutions such as the finite difference method if they are too complicated. The boundary conditions are used to define the constants involved in the solution. 12.4.6

Example of Solving a Flow Problem

One example of a flow problem is the flow of water out of a saturated soil layer when it is loaded by a long embankment (Figure 12.10a). Before loading, the layer is under an at-rest ′ and an initial state of stress with a vertical effective stress 𝜎ov ′ water stress uwo . Both 𝜎ov and uwo vary with the depth z. When the vertical stress is increased by Δ𝜎 due to the embankment loading, the water stress increases by an amount called the excess pore pressure uwe . The excess pore pressure uwe is high at first and decreases as a function of time while the water drains out. The settlement takes place as a result of this water drainage (Figure 12.10b). The problem is to predict the variation of the excess pore pressure uwe as a function of time t and the settlement ΔH of the embankment as a function of time t. The following simplifying assumptions are made: a. The soil is saturated with water. b. The water is incompressible. c. The soil skeleton is linear elastic (linear stress-strain relation). d. The soil particles are incompressible. e. Darcy’s law governs the flow of water through the soil. f. The water drains at the top and at the bottom of the layer. g. The flow is in the vertical direction only.

Δσ = γh

Random soil element

0

uw

h

Fill

Flow of water

Clay Sand

(a) uwe,max = Δσ

0 uwe (z) at t = ∞

uwe

uw0

H0 (b)

Ho

uw uwe (z) at t = t uwe (z) at t = 0

uwe,max = ∆σ uw0 = γwHo z

Figure 12.10 Embankment example.

340

12 PROBLEM-SOLVING METHODS vz dy

Volume of water dV squeezed out in a time dt 0

and by differentiation

dx

dz

dht = dhe + dhp

z

0

0

x y

0

(12.44)

Note that for the element, the elevation head he is constant and therefore dhe = 0. Note also that, by definition: u + uwe (12.45) hp = wo 𝛾w Because uwo is constant:

vz + dvz

Figure 12.11 Element of soil under the embankment.

dhp =

duwe 𝛾w

(12.46)

Combining the previous observation, we get: h. The increase in stress Δ𝜎 in the layer due to the embankment is constant within the layer. i. The excess water stress uwe increases by Δ𝜎 when the embankment is placed. j. No lateral soil movement takes place. With these assumptions, the solution proceeds as follows: 1. Zoom in at the element level. In this case, we will select an element of soil with an elementary volume V equal to dx dy dz (Figure 12.11). 2. Considering the element of Figure 12.11, the water velocity in the z direction is vz when it enters the element and vz + dvz when it exits the element. It is assumed that the water does not flow in the y direction because of the plain strain condition induced by the infinitely long embankment. It is also assumed that there is no flow in the x direction because the total head gradient is much higher in the z direction than in the x direction. Since the water velocity is proportional to the total head gradient (Darcy’s law), most of the water goes in the vertical direction. Also shown on the element is the change of volume dV of the element during a time dt. This change of volume corresponds to the water loss and also to the compression of the element, given that the soil is saturated. 3. The fundamental equation in this case is the conservation of mass equation expressing that, during a time dt, the volume of water entering the element plus the water squeezed out of the element due to the stress applied is equal to the volume of water exiting the element. Use is made of the flow equation (Qdt = vAdt): vz dx dy dt + dV = (vz + dvz )dx dy dt (12.41) dvz dV = (12.42) Vdt dz Another fundamental equation is conservation of energy, which leads to the relationship between the total head ht , the elevation head he , and the pressure head hp . Note that the velocity head is neglected because water flows very slowly through soils: ht = he + hp

(12.43)

dht = dhp =

duwe 𝛾w

(12.47)

The effective stress in the element is: 𝜎 ′ = 𝜎 − (uwo + uwe )

(12.48)

By differentiation and noting that both 𝜎 and uwo are constant during the loading and subsequent drainage: d𝜎 ′ = −duwe

(12.49)

4. The first constitutive equation describes how fast the water flows through the soil (Darcy’s law): vz = ki = −k

dht dz

(12.50)

and by taking the first derivative of vz with respect to z: dvz d2 h (12.51) = −k 2t dz dz The second constitutive equation describes how much the soil compresses under stress (stress-strain relationship): dV d𝜎 ′ = M (12.52) V The strain in this case is the volumetric strain (𝜀v = dV∕V) and M is the constrained modulus because the soil is not allowed to expand laterally. 5. By regrouping Eqs. 12.42, 12.47, 12.49, 12.51, and 12.52, the governing differential equation is obtained: 1 d𝜎 ′ 1 duwe 1 dV dvz = = =− V dt dz M dt M dt 2 d2 ht k d uwe = −k 2 = − 𝛾w dz2 dz 2 duwe k M d uwe = dt 𝛾w dz2

(12.53) (12.54)

The coefficient of consolidation cv is expressed in m2 ∕s and is defined as: cv =

k M 𝛾w

(12.55)

12.4 CONTINUUM MECHANICS METHODS

and the governing differential equation for this problem is: duwe d2 uwe (12.56) = cv dt dz2

U =1− T=

6. Now we need to organize the space and time boundary conditions. The space boundary conditions state that the excess water stress uwe at the ground surface is zero because the water can drain freely at that location. Also, the excess water stress uwe is zero at the bottom of the layer because the water can drain freely at that depth: uwe @ z=0 = 0 at any time t

(12.57)

uwe @ z=H0 = 0 at any time t

(12.58)

(12.59)

uwe @ t=∞ = 0 at any depth z

(12.60)

uwe (max)

cv t

(12.63)

Hd 2

dU d2 U (12.64) = dT dZ2 The solution to this partial differential equation, together with the space and time boundary conditions, is a Fourier series expansion of the form (Terzaghi, 1943): ∑ 2 ) ( sin (MZ) exp −M 2 T M m=0 𝜋 with M = (2m + 1) (12.65) 2 The graphical representation of U as a function of Z and T is shown in Figure 12.12. m=∞

7. This is the step where we solve the governing differential equation (12.56) and apply the boundary conditions. To simplify the mathematical process, it is convenient to use the following transformation into dimensionless variables: z Z= (12.61) Hd

U =1−

uwe/uwe,max 0

0

0.2

0.4

0.6

0.8

0.2

1

T=0

0.4 0.6

T = 0.7

Z/Hd

0.8

T = 0.9

1

T = 0.3

T = 0.1

T = 0.5

1.2 1.4 1.6 1.8 2

1

0.8

(12.62)

where z is the depth below ground surface, H d is the longest drainage path, U is the degree of consolidation at depth z and time t, uwe is the excess water stress at depth z and time t, uwe(max) is the maximum excess water stress at time t = 0 at any depth taken as equal to Δ𝜎, T is the time factor, and t is the time. Note that the maximum drainage length is equal to the layer thickness H o if the water can only drain on one side (top or bottom of the layer), but is equal to 0.5 H o if the water can drain at both ends. With these transformed variables, the GDE (Eq. 12.56) becomes:

The time boundary conditions state that the excess water stress uwe is equal to the increase in total stress Δ𝜎 at time t = 0 and then equal to 0 at time t = infinity: uwe @ t=0 = Δ𝜎 at any depth z

uwe

341

0.6

0.4

0.2

Uz

Figure 12.12 Degree of consolidation and excess water stress as a function of depth and time.

0

342

12 PROBLEM-SOLVING METHODS

uwe uwe,max

0

z Hd 1

At

A0

2 1

The average degree of consolidation represents the ratio of the area under the excess water stress profile at time t over the same area at time t = 0 (Figure 12.13). The graphical representation of U av as a function of T is shown in Figures 12.14 and 12.15. Equation 12.52 indicates that the volumetric strain dV∕V in the layer is linearly proportional to the increase in effective stress d𝜎 ′ . Because the soil is assumed not to move laterally, the volumetric strain is also the vertical strain dH∕H. Furthermore, because the total stress is constant, the increase in effective stress is equal to the decrease in excess water stress (Eq. 12.49). Therefore, the average degree of consolidation U av can be rewritten as:

1

U=1–

uwe uwe,max

0

Average degree of consolidation Uav in percent

Figure 12.13 Definition of the average degree of consolidation.

Uav = 1 − 0 20

M ΔH Δ𝜎 ′ (av) ΔH = = ΔHH = ′ max Δ𝜎 max (av) M ΔHmax H

C3

40

uwe(max ) (av) − uwe (av) uwe (average) = uwe(max ) (average) uwe(max ) (av)

C1

(12.67)

60 80

100

This means that U av represents the settlement of the structure divided by the maximum settlement. In contrast, because T is a function of the time t, the complete settlement vs. time curve (ΔH∕ΔHmax vs.t) can be constructed by using the U av vs. T curves. An example is shown in Figure 12.16. The following equations have been proposed to approximate Eq. 12.66: For Uav < 0.6:

C2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

Time factor T

Curve C1 Curve C2 Curve C3 Two way drainage

Curve C1 Curve C2 Curve C3 One way drainage

𝜋 T = Uav 2 4

Figure 12.14 Average degree of consolidation U av vs. time factor T on natural scale for different stress increase profiles.

Uav = 1 −

uwe

⎛ − ⎜1 − 10 ⎜ ⎝

(

cv t +0.085 Hd 2 0.933

)

⎞ ⎟ ⎟ ⎠

C3 C1

40 60

C2

80

100 0.01

(12.68)

(12.69)

0 20

cv t 𝜋

) ( T = −0.933 log 1 − Uav − 0.085 or ΔH = ΔHmax

∑ 2 ( ) exp −M 2 T =1− 2 dz M max m=0 𝜋 with M = (2m + 1) (12.66) 2 m=∞

Average degree of consolidation Uav in percent

H

√

For Uav > 0.6:

It is also useful to define the average degree of consolidation U av : ∫0 uwe dz

or

2 ΔH = ΔHmax Hd

0.1

1

10

Time factor T

Figure 12.15 Average degree of consolidation U av vs. time factor T on semilog scale.

343

12.5 NUMERICAL SIMULATION METHODS

y

Time (day)

Settlement (m)

0

0

20

40

60

80

100

Slope y′i

120

yi + 1

0.05

yi – 1

0.1

yi

yi + 2

Finite difference estimate of y′i yi = (y′i + 1 – y′i – 1)/2h

yi – 2

0.15 0.2

h h h h zi – 2 zi – 1 zi zi + 1 zi + 2 z

0.25 0.3

Figure 12.16 Example of settlement vs. time curve obtained from the U av vs. T curve.

12.5

NUMERICAL SIMULATION METHODS

Numerical solutions typically require the use of a computer because of the complexity and amount of the mathematical computations involved. They tend to work as follows. The soil space or the foundation is discretized into many small elements (linear, surface, or volume). The points forming the geometry of these elements are the nodes. The unknowns (e.g., stresses, strains, displacements, forces, moments, flow velocity, head) have to be calculated at all the nodes. The governing differential equations are transformed into algebraic equations that must be written as many times as there are nodes in the discretized soil space. This usually yields a large number of equations organized in matrix form. From this matrix equation, the unknowns must be extracted and solved for; this often requires an inversion process of the main matrix and can only be done by computers. The output of these numerical solutions is in the form of large tables that give the calculated values of the unknowns at each node within the soil mass. Numerical methods (Jing and Hudson, 2002; Bobet, 2010) include the finite difference method (FDM), the finite element method (FEM), the boundary element method (BEM), and the discrete element method (DEM). 12.5.1

Finite Difference Method

The finite difference method is very powerful in solving differential equations. The main idea is to replace the differential equation by incremental algebraic equations. This is done by using algebraic expressions of the derivatives of the functions involved in the governing differential equation. In Figure 12.17, the function y(z) has values yi−2 , … , yi+2 corresponding to values of z equal to zi−2 , … , zi+2 respectively. The values of z are separated by a constant distance h. The first derivative of y evaluated at z = zi can be expressed as the slope of the tangent at zi : yi+1 − yi−1 dy = y′i = dz @zi 2h

(12.70)

Figure 12.17 Derivative expressed by the central finite difference formulation.

This expression is called the central difference expression of the derivative, as it balances the influence of both sides of the function with respect to point i (Figure 12.17). The forward difference would be: yi+1 − yi dy = y′i = , (12.71) dz @zi h and the backward difference would be: y − yi−1 dy = y′i = i dz @zi h

(12.72)

The second derivative can be expressed using the same approach. Indeed, the second derivative is the first derivative of the first derivative. This gives the following expression, using a forward and a backward formulation for y′ to end up with a centered formulation of y′′ . y′ i+1 − y′ i dy′ d2 y ′′ = = y = i dz @zi h dz2 @zi yi+1 −yi yi −yi−1 − h yi+1 − 2yi + yi−1 (12.73) = h = h h2 Using the same process, the third derivative can be expressed as: d3 y

yi+2 − 2yi+1 + 2yi−1 − yi−2 dy′′ = y′′′ i ⊢= 3 dz 2h3 @zi dz @zi (12.74) and the fourth derivative: =

d4 y dz4 @zi

=

dy′′′ dz @zi

=⊢ y′′′′ ⊢= i

yi+2 − 4yi+1 + 6yi − 4yi−1 + yi−2 h4 (12.75)

A typical finite difference solution proceeds through the following steps: 1. The structure or soil mass involved is broken down into small elements of chosen finite dimensions. Each element has a number and each node at the boundaries of these elements has a number.

344

12 PROBLEM-SOLVING METHODS

M0

2. The knowns and unknowns (loads, displacements, stresses, strains, velocities, and heads, for example) are identified for each node and given a subscript corresponding to the node number. 3. The governing differential equation is written in algebraic finite difference form as many times as there are nodes in the structure or soil mass. 4. The space and time boundary conditions are also expressed in terms of the algebraic expressions of the variables. 5. All equations are regrouped into a matrix equation. 6. The matrix equation is solved to extract the unknown quantities. This usually requires that the matrix be inverted. Considering the size of these matrices, a computer is required for this step. 7. The solution is presented in the form of a table that gives the sought quantities at all the nodes. 12.5.2

Examples of Finite Difference Solutions

Ep I d4 y K dz4

=0

(12.76)

The solution to this problem is the function y(z) describing the horizontal deflection of the pile as a function of the depth z. The process consists of the following steps: 1. The pile is discretized into elements as shown in Figure 12.18. The displacement at node i is yi . There are a total of n + 1 unknown values of the horizontal displacement y (y0 to yn ). 2. The GDE is written for each node using the expressions of the derivatives presented in Section 12.5.1: yi +

–2 –1 0 1 2 | |

Mode numbers

|

i–2 i–1 i i+1 i+2 | |

n–1 n n+1 n+2

Figure 12.18 Pile discretized into numbered elements and nodes.

The example is to solve the governing differential equation by using the FDM for the problem of Section 12.4.4: a pile subjected to a horizontal force H o and an overturning moment M o applied at the ground surface. The GDE is (Eq. 12.21): y+

H0

Ep I d4 y

K dz4 @zi ) ( Ep I yi+2 − 4yi+1 + 6yi − 4yi−1 + yi−2 = yi + =0 × K h4 (12.77) ( ) 4 Kh yi − 4yi−1 + yi−2 = 0 yi+2 − 4yi+1 + 6 + Ep I (12.78)

Because there are n + 1 nodes along the pile (0 to n), Eq. 12.78 theoretically could be written n + 1 times. That is not the case here, because Eq. 12.78 involves 5 nodal values of the displacement y, so in fact Eq. 12.78 can only be written n − 3 times. Because there are n + 1 values of the horizontal displacement y, we are missing four equations. Can the boundary conditions help us?

3. The boundary conditions are that the horizontal load is H o at the ground surface and zero at the bottom of the pile and that the moment is M o at the ground surface and zero at the bottom of the pile. To express these four boundary conditions, additional and fictitious nodes are created. These are nodes –1 and –2 at the top of the pile and nodes n + 1 and n + 2 at the bottom of the pile (Figure 12.18). The fact that the moment is M o at the ground surface and zero at the bottom of the pile is written as: d2 y M@z=0 = Ep I 2 dz @z=0 ( ) y1 − 2y0 + y−1 = Ep I (12.79) = Mo h2 d2 y M@z=L = Ep I 2 dz @z=L ( ) yn+1 − 2yn + yn−1 = Ep I =0 (12.80) h2 The fact that the shear force is H o at the ground surface and zero at the bottom of the pile is written as: V@z=0 = Ep I

V@z=L

d3 y

dz3 @z=0 ( ) y2 − 2y1 + 2y−1 − y−2 = Ep I (12.81) = Ho h3 d3 y = Ep I 3 dz @z=L ( ) yn+2 − 2yn+1 + 2yn−1 − yn−2 = Ep I = 0 (12.82) h3

The boundary conditions lead to four new equations, but we have also created four new unknowns (y−2 , y−1 , yn+1 ,

345

12.5 NUMERICAL SIMULATION METHODS

and yn+2 ). Thus, the new count is n + 5 unknowns and n + 1 equations. The extra four equations are created because the additional nodes allow the GDE to be written four more times. Now we have n + 5 unknowns and n + 5 equations. These n + 5 equations are written in matrix form as: [K][Y] = [C]

Node 0

K1 = 0 K0 = 0

Node 2

[Y] = [K]−1 [C]

Node 4

(12.84)

This solution is illustrated by solving for the deflection and pressure distribution for a retaining wall, as shown in Figure 12.19. The units for this problem are not stated, because as long as the units are consistent, the solution is independent of the units. The bending stiffness of the wall is 10,000 and the element height is 1. The soil reaction curves at each node must be prepared (Figure 12.20). The reaction curves represent the relationship between the line load P on the wall and the horizontal displacement y of the wall. A number of simplifying assumptions will be made to facilitate the solution. At node 0, the reaction curve shows that the line load Po is equal to zero for all y values: Po = 0

(12.85)

At node 1, the reaction curve is taken as a constant equal to 60. In fact, the reaction curve at node 1 should reflect the mobilization of the active pressure if the wall moves away from the soil and of the passive pressure if the wall moves into the soil. However, because the active pressure is the pressure that will be mobilized considering the problem, and because the active pressure requires very little movement to be mobilized, it is reasonable to assume that the movement will be large enough that the line load on the wall will correspond to the active pressure for a large range of lateral displacements: P1 = 60 (12.86)

4h

h

60

y

y

(12.83)

where [K] is an n + 5 by n + 5 matrix of the coefficients of the yi values in the algebraic equations corresponding to the GDE and the boundary conditions, [Y] is a n + 5 long column matrix of the y values (y−2 to yn+2 ), and [C] is a n + 5 column matrix of the constants in the n + 5 GDE equations. Because the y values are the unknowns to be solved for, the [K] matrix must be inverted and the solution is:

–2 –1 0 1 2 3 4 5 6

Node 1 P

P

2h

Figure 12.19 Wall discretized into numbered elements and nodes.

P

K2 = 0

Node 3

P

120 y

y K3 = 1000

P

y K4 = 1500

Figure 12.20 Reaction curves for the wall at each node.

The same reasoning applies to the reaction curve at node 2, where the pressure is twice as high and the line load is equal to 120: (12.87) P2 = 120 At node 3, the reaction curve is as shown in Figure 12.20. It indicates that the line load is linearly proportional to the lateral displacement of the wall. Again, this reaction curve should reflect the influence of the active and passive pressures on both sides of the wall. The simplifying assumption in this case is that the passive resistance dominates the behavior of the wall below the excavation level. Knowing that it takes much larger displacements to mobilize the passive resistance than the active pressure, it is likely that below the excavation depth the wall will be in the range of displacement where a linear assumption is reasonable. Therefore, the reaction curve at node 3 is characterized by: P3 = −K3 y = −1000y

(12.88)

The reason for the minus sign is that when the deflection increases to the right (Δy > 0), the line load decreases (ΔP < 0). The same reasoning applies for the reaction curve at node 4, but with a higher stiffness K 4 , as node 4 is deeper in the soil and therefore likely stiffer: P4 = −K4 y = −1500y

(12.89)

As you can see, this problem has been greatly simplified compared to the real problem. The reason is that without such simplifications, the mathematics would become quite complicated.

346

12 PROBLEM-SOLVING METHODS

⎡y−2 ⎤ ⎡ 0 ⎤ ⎢y ⎥ ⎢ 0 ⎥ ⎥ ⎢ −1 ⎥ ⎢ ⎢ y0 ⎥ ⎢ 0 ⎥ ⎢ y1 ⎥ ⎢0.006⎥ × ⎢ y2 ⎥ = ⎢0.012⎥ ⎥ ⎢ ⎥ ⎢ ⎢ y3 ⎥ ⎢ 0 ⎥ ⎢ y4 ⎥ ⎢ 0 ⎥ ⎢ y5 ⎥ ⎢ 0 ⎥ ⎥ ⎢y ⎥ ⎢ ⎣ 6⎦ ⎣ 0 ⎦

Now the problem is clearly defined and we can proceed with the step-by-step procedure: 1. The wall has been discretized as shown in Figure 12.19. 2. The line loads and the horizontal displacements are numbered from 0 at the top of the wall to 4 at the bottom of the wall. 3. The GDE is the same as the one for the horizontally loaded pile (Eq. 12.21): P − Ep I

d4 y dz

4

=0

(12.90)

Expressed in finite difference formulations, it becomes: ( Pi − Ep I

yi+2 − 4yi+1 + 6yi − 4yi−1 + yi−2 h4

) =0

(12.91) 4. The boundary conditions are that the moment and the shear force are zero at both ends of the wall. This requires adding two fictitious nodes at both ends of the wall, as shown in Figure 12.19. The equations for the shear and moment are: ( ) yn+1 − 2yn + yn−1 =0 (12.92) M = Ep I h2 ( ) yn+2 − 2yn+1 + 2yn−1 − yn−2 V = Ep I =0 h3 (12.93)

The first two equations in the matrix equation are the two equations for the moment and shear boundary conditions at the top of the wall; then there are five GDEs written at five nodes; and the last two equations are the two equations for the moment and shear boundary conditions at the bottom of the wall. Now it is time to invert the matrix to obtain the [Y] matrix as the solution to the problem. 6. The computer does that for us, and the deflections y at each node are calculated. The line loads on the wall are obtained by using the relationship between the load and the deflection given by the reaction curves of Eqs. 12.85–12.89. The results of this finite difference solution are presented in Table 12.1. The deflection profile y(z) and the line load profile P(z) are shown in Figure 12.21. The profile P(z) shows that the wall is in horizontal equilibrium because the area under the left side Table 12.1 Results of the finite difference solution for the simulated wall

5. Now all the equations can be written and assembled in a matrix: ⎡−1 2 0 −2 1 0 0 0 ⎢ 0 1 −2 1 0 0 0 0 ⎢ 1 −4 6 −4 1 0 0 0 ⎢ 0 0 ⎢ 0 1 −4 6 −4 1 ⎢ 0 0 1 −4 6 −4 1 0 ⎢ ⎢ 0 0 0 1 −4 6.1 −4 1 ⎢ 0 0 0 0 1 −4 6.15 −4 ⎢ 0 0 0 0 0 1 −2 1 ⎢ 0 0 0 0 −1 2 0 −2 ⎣

.

1

0⎤ 0⎥ ⎥ 0⎥ 0⎥ 0⎥ ⎥ 0⎥ 1⎥ 0⎥ 1⎥⎦

1 2 3 4 5

Deflected shape

.

Horizontal deflection y of wall

Line load P on wall

2.72 1.95 1.18 0.42 −0.32

0 60 120 −420 480

Pressure distribution

0 1

.

2 3 . 4

Node number (depth)

.

0

(12.94)

2 420

–

+ 120

3 4 +

Figure 12.21 Wall deflection and line load.

480

12.5 NUMERICAL SIMULATION METHODS

Initial mesh

Deformed mesh

(a)

(b)

347

Figure 12.22 Example of finite element mesh: (a) Initial mesh. (b) Deformed mesh.

of the profile is equal to the areas under the right side of the profile. This is the way it should be, as horizontal equilibrium was one of the fundamental equations we used in setting up the solution. 12.5.3

Finite Element Method

The finite element method (FEM) is another powerful numerical method to solve geotechnical problems (Clough, 1960; Desai and Abel, ; Zienkiewicz et al., 2005). The output, like most numerical methods, will be in the form of tables giving the unknown quantities at discrete locations in the soil mass. The general steps in developing a solution to a finite element simulation are as follows: 1. Discretize the soil mass into finite elements connected by nodes. 2. Choose the functions describing the variation of the unknowns across each element and between its nodes. 3. Write the strain-displacement equations. 4. Write the stress-strain equations for the soil. 5. Derive the equations governing the behavior of the soil element. 6. Assemble the element equations into the global matrix equation. 7. Introduce the boundary conditions into the global matrix equation. 8. Solve the global matrix equation for the unknowns.

a number of small elements (Figure 12.22). The sides of the elements intersect at the nodes. Each element and each node are numbered in sequence. The size of the elements is influenced by a number of factors, including how fast the stress changes from one point to another of the soil mass (stress gradient). Various shapes of elements exist: lines, triangles, quadrilaterals, parallelepipeds, or brick elements. One of the big advantages of the FEM is that irregular boundaries do not present a big problem. 2. Choose the functions describing the variation of the unknowns across each element and between its nodes. These are called interpolation functions or shape functions. The solution of the FEM will give the answers at the nodes (Figure 12.23), but we need to be able to calculate the unknowns everywhere in the mass to establish the general equations. The interpolation functions relate, for example, the displacement anywhere in the element to the displacements at the nodes. These interpolation functions are typically in the form of polynomials. It is more convenient, however, to write them in the following form:

uy2 (2)

uy1

ux1

(1)

Element uy3

Each step is discussed in more detail here. (3)

1. Discretize the soil mass into finite elements connected by nodes. In this step the soil mass is subdivided into

ux2

ux3

uy4

ux4

(4)

Figure 12.23 Element in plane strain.

348

12 PROBLEM-SOLVING METHODS

y

∑

s

#nodes

ux (x, y) = H1 ux1 + H2 ux2 + H3 ux3 + H4 ux4 =

Hi uxi

i=1

1 (x1, y1)

2 (x2, y2)

(12.95) ∑

Hi uyi

i=1

(12.96) where ux (x, y) is the displacement in the x direction of any point within the element with coordinates x and y, uxi is the displacement in the x direction of node i, uy (x, y) is the displacement in the y direction of any point within the element with coordinates x and y, uyi is the displacement in the y direction of node i, and the H i s are the interpolation functions. Equations 12.95 and 12.96 would be for an element with four nodes and plain strain condition in the z direction. They describe the shape of the displacement surface across the element. In matrix form:

[ ] [ ux H1 0 H2 0 H3 0 H4 = uy 0 H1 0 H2 0 H3 0

⎡ux1 ⎤ ⎢u ⎥ ⎢ y1 ⎥ ] ⎢ux2 ⎥ 0 ⎢uy2 ⎥ H4 ⎢ux3 ⎥ ⎢ ⎥ ⎢uy3 ⎥ ⎢ux4 ⎥ ⎢u ⎥ ⎣ y4 ⎦

or [u] = [H][ui ]

(12.97)

Note that the [u] matrix is the matrix of displacements as variables, whereas the [ui ] matrix is the matrix of displacements at the nodes. The [H] matrix is called the shape function matrix. Note also that these matrices are written for the element and not for the entire soil mass. Regarding the coordinates x and y, it is more convenient to use natural coordinates (Figure 12.24). As can be seen, regardless of the element’s original shape, the transformation leads to a set of coordinates varying from -1 to +1 along each face. Also, the element has become a square. The interpolation functions for a four-node element in natural coordinates are: 1 (1 + r)(1 + s) 4 1 H2 = (1 − r)(1 + s) 4 1 H3 = (1 − r)(1 − s) 4 1 H4 = (1 + r)(1 − s) 4 H1 =

(12.98) (12.99) (12.100) (12.101)

where r and s are the natural coordinates (Figure 12.24).

1 (1, 1)

x

#nodes

uy (x, y) = H1 uy1 + H2 uy2 + H3 uy3 + H4 uy4 =

2 (–1, 1)

r

3 (x3, y3)

3 (–1, –1)

4 (x4, y4) (a)

4 (1, –1) (b)

Figure 12.24 Finite element representation in real and natural coordinates: (a) Real coordinates. (b) Natural coordinates.

In the general case, coordinates can be expressed in terms of interpolation functions as follows: ∑

#nodes

x=

Hi xi

(12.102)

i=1 #nodes

y=

∑

Hi yi

(12.103)

i=1

3. Write the strain-displacement equations. There are typically 9 equations: 3 force equilibrium equations and 6 constitutive equations linking the stresses to the strains. The other equations are the 3 moment equilibrium equations, but they simply lead to the fact that shear stresses on perpendicular planes are equal and in opposite directions so they have already been used up. However, there are 15 unknowns: 6 stresses, 6 strains, and 3 displacements. So we are short 6 equations. What saves the day is that the 6 strains are defined on the basis of the 3 displacements, so this adds 6 strain-displacement equations. In the end, we have 15 unknowns and 15 equations. Recalling Eq. 12.95, the normal strain in the x direction is 𝜀xx : 𝜕u (x, y) [ 𝜕H ] [uxi ] (12.104) 𝜀xx = x = 𝜕x 𝜕x The same equation holds true for 𝜀yy : [ ] 𝜕uy (x, y) 𝜕H 𝜀yy = (12.105) = [uyi ] 𝜕y 𝜕y For the shear strain 𝛾xy , the equation becomes: [ ] ] [ 𝜕uy 𝜕u 𝜕H 𝜕H [uyi ] 𝛾xy = x + = [uxi ] + 𝜕y 𝜕x 𝜕y 𝜕x (12.106) or, in matrix form: [𝜀] = [B][ui ]

(12.107)

where [𝜀] is the strain matrix (3 × 1 vector for a two-dimensional problem), [B] is the matrix containing the derivatives of the interpolation functions Hi (3 × 8 for a two-dimensional problem), and [ui ] is the matrix

12.5 NUMERICAL SIMULATION METHODS

of nodal displacements (8 × 1 for a two-dimensional problem). 𝜕H1 ⎡𝜖xx ⎤ ⎡ 𝜕x ⎢𝜖yy ⎥ = ⎢⎢ 0 ⎢ ⎥ ⎢ 𝜕H1 ⎣𝛾xy ⎦ ⎣ 𝜕y

0 𝜕H1 𝜕y 𝜕H1 𝜕x

𝜕H2 𝜕x

0 𝜕H2 𝜕y

0 𝜕H2 𝜕y 𝜕H2 𝜕x

𝜕H3 𝜕x

0 𝜕H3 𝜕y

⎡ux1 ⎤ ⎢u ⎥ ⎢ y1 ⎥ ⎢ux2 ⎥ ⎢u ⎥ × ⎢ y2 ⎥ u ⎢ x3 ⎥ ⎢uy3 ⎥ ⎢ux4 ⎥ ⎢u ⎥ ⎣ y4 ⎦

0 𝜕H3 𝜕y 𝜕H3 𝜕x

𝜕H4 𝜕x

0 𝜕H4 𝜕y

0 ⎤

𝜕H4 ⎥ 𝜕y ⎥ 𝜕H4 ⎥ 𝜕x ⎦

(12.108)

Because the interpolation functions H i are defined in natural coordinates, the derivatives 𝜕Hi ∕𝜕x and 𝜕Hi ∕𝜕y can be related to 𝜕Hi ∕𝜕r and 𝜕Hi ∕𝜕s through the Jacobian matrix [J] as follows: ⎡ 𝜕Hi ⎤ ⎡ 𝜕Hi ⎤ ⎢ 𝜕x ⎥ −1 ⎢ 𝜕r ⎥ (12.109) ⎥ ⎢ 𝜕H ⎥ = J ⎢ ⎢ 𝜕Hi ⎥ ⎢ i⎥ ⎣ 𝜕s ⎦ ⎣ 𝜕y ⎦ where the Jacobian matrix [J] is described as follows: ⎡ 𝜕x 𝜕y ⎤ ⎢ 𝜕r 𝜕r ⎥ (12.110) J=⎢ ⎥ ⎢ 𝜕x 𝜕y ⎥ ⎣ 𝜕s 𝜕s ⎦ Recalling Eq. 12.102 and Eq. 12.103, the components of this Jacobian matrix are written as follows: 𝜕x ∑ 𝜕Hi • x (12.111) = 𝜕r 𝜕r i 𝜕x ∑ 𝜕Hi • x (12.112) = 𝜕s 𝜕s i 𝜕y ∑ 𝜕Hi •y (12.113) = 𝜕r 𝜕r i 𝜕y ∑ 𝜕Hi •y (12.114) = 𝜕s 𝜕s i 4. Write the stress-strain equations for the soil. These are the constitutive equations, the ones that are specific to the soil involved. One of the simplest constitutive laws is the case where the stresses are linearly related to the strains (elasticity): [𝜎] = [C][𝜀]

(12.115)

where [𝜎] is the stress matrix, which is a 3 × 1 matrix for a two-dimensional problem and a 6 × 1 matrix for a three-dimensional problem; [𝜀] is the strain matrix, which is a 3 × 1 matrix for a two-dimensional problem and a 6 × 1 matrix for a three-dimensional problem; and [C] is the soil stiffness matrix, which is a 3 × 3 matrix for a two-dimensional problem and a 6 × 6 matrix for

349

a three-dimensional problem. In elasticity and for three dimensions, Eq. 12.115 is written as: ⎡𝜎xx ⎤ ⎢𝜎yy ⎥ ⎢ ⎥ E ⎢ 𝜎zz ⎥ = ⎢ 𝜏xy ⎥ (1 − 2v)(1 + v) ⎢ 𝜏yz ⎥ ⎢ ⎥ ⎣ 𝜏zx ⎦ v v ⎡1 − v v 1 − v v ⎢ ⎢ v v 1−v ⎢ 0 0 ⎢ 0 ⎢ 0 0 0 ⎢ ⎣ 0 0 0

1 2 1 2

0 0 0 −v −v 0

0 0 0 0 0 0

1 2

0 ⎤ ⎡𝜀 ⎤ xx 0 ⎥ ⎢𝜀 ⎥ yy 0 ⎥ ⎢𝜀 ⎥ ⎥ × ⎢ zz ⎥ 0 ⎥ ⎢𝜀xy ⎥ 0 ⎥ ⎢𝜀yz ⎥ ⎥ ⎢ ⎥ − v⎦ ⎣𝜀zx ⎦ (12.116)

In the case of two dimensions, the C matrix becomes: C=

v ⎡1 − v E ⎢ v 1−v (1 − 2v)(1 + v) ⎢ 0 0 ⎣

1 2

0 ⎤ 0 ⎥ (12.117) ⎥ − v⎦

5. Derive the equations governing the behavior of the soil element. You may have noticed that we have not yet written any fundamental equations such as the equilibrium equations. We do that in this step, but it is done by using another technique called a variational principle that includes the equilibrium equations. More specifically, we use the minimum total potential energy (MTPE) principle along with the virtual work technique. The MTPE principle states that the actual displacement solution of a deformable body is the ∏ solution that renders the TPE functional minimum, ∏ meaning that the derivative of is equal to zero. The two types of energies involved in the TPE are the work done by the external forces W and the internal strain energy U of the deformable soil mass. The TPE is minimum (system in equilibrium) when the change in work done by the external forces 𝛿W is equal to the change in internal strain energy 𝛿U: 𝛿 Π = 0 → 𝛿U = 𝛿W

(12.118)

The increment of virtual internal strain energy 𝛿U for a bar is: 𝛿U = 𝜎A 𝛿𝜀 dx = 𝛿𝜀 𝜎 dV

(12.119)

where 𝜎 is the axial stress, A is the cross-sectional area, 𝛿𝜀 is a virtual axial strain, and dx and dV are an infinitesimal length and volume of the bar. This is generalized for the three-dimensional soil element as: 𝛿U =

∫V

𝛿[𝜀]T [𝜎]dV

(12.120)

The change in work is calculated by assuming that the soil mass is subjected to virtual small displacements

350

12 PROBLEM-SOLVING METHODS

(virtual work). The increment in virtual external work 𝛿W for a bar is:

s

𝛿W = Fbody 𝛿u + Fboundary 𝛿u = bdV𝛿u + tdA𝛿u (12.121) where F body is the body force, 𝛿u is a virtual displacement, F is the boundary force, b is the body force density (e.g., unit weight), t is the boundary tractions (e.g., pressure), and dV and dA are an infinitesimal volume and area of the bar. This is generalized for the three-dimensional soil element as: 𝛿[u]T [b]dV +

𝛿W =

𝛿[u]T [t]dA

2(–1, 1) Point #2

𝛿[𝜀]T [𝜎] dV =

∫V

𝛿[u]T [b]dV +

∫V

Point #3

∫V

𝛿[u] [B] [C][B]𝛿 [u] dV = 𝛿[u] T

3(–1, –1)

𝛿[u]T [t]dA

4(1, –1)

Figure 12.25 Four integration points.

(12.123)

T

× ([Fbody ] + [Fboundary ]) = 𝛿[u]T [F]

Point #4

(12.122)

Using Eqs. 12.107 and 12.115, we get: T

Point #1

r

∫V ∫V Then the principle of virtual work states that the expressions in Eqs. 12.120 and 12.122 are equal: ∫V

1(1, 1)

(12.124)

e

The element stiffness matrix K is defined as: [K e ] =

[B]T [C][B]dV (12.125) ∫V To calculate the integral on the right side of Eq. 12.125, we select integration points where all the components of the B and C matrices are evaluated. In the special case of a plane strain problem, the components of the stiffness matrix can be reduced to the following expression: 2 2 ∑ ∑ t fmn (x, y)dxdy = t × fmn (ri , sj ). det J.wi .wj ∫ ∫Area i=1 j=1 (12.126) where t is the thickness of the element (1 in plane strain cases), fmn (x, y) is the function found at the intersection of the m row and n column of the Bt CB matrix of Eq. 12.125 expressed in real coordinates, fmn (r, s) is the same function but expressed in natural coordinates, i and j are the running indices identifying the location of the integration point, ri and sj are the natural coordinates of the chosen integration points on the element, wi and wj are the weighting factors that depend on the number and location of the integration points, and det J is the determinant of the Jacobian matrix. In the general case, the thickness is not a constant and must be calculated at each integration point by using the interpolation functions (see problem 12.7). Figure 12.25 shows an example of four integration points. Because Eq. 12.124 must be satisfied for any kinematically admissible virtual displacement field [u], we must have: [K e ][u] = [F] (12.127) In Eq. 12.127, most of the displacements u are unknown and most of the forces are either zero or known. This is

the equation governing the behavior of the element. If the element were a spring, K would be the spring constant, but in the case of the three-dimensional element, K is a square matrix. 6. Assemble the element equations into the global matrix equation. Equation 12.126 is the equation for one element. There are as many such matrix equations as there are elements in the mesh. They must be assembled to form the stiffness matrix for the entire soil mass. To do so, we specify that the body must remain continuous during the deformation. This means that each node can have only one displacement vector common to all elements containing this node. At each node, we also have only one body force and one external force value. The following example illustrates how the global matrix is assembled. Consider the two elements of Figure 12.26. The stiffness matrices for the 2 elements and their assembly Numbering of nodes and degrees of freedom 2 8 1 7 (4)

(1)

Positive direction of displacement at the nodes uy1 ux1

3

uy2 9

(5)

(2)

12

uy3

Element #2 6

11 (6) (a)

uy5

ux2

ux5 (5)

Element #2

5 (3)

(4) Element #1

10

(2)

ux4

(1)

Element #1 4

uy4

uy6 ux6 (6)

ux3

(3) (b)

Figure 12.26 Two 2D FEM elements and numbering the degrees of freedom: (a) Number of nodes and degrees of freedom. (b) Positive direction of displacements at nodes.

12.5 NUMERICAL SIMULATION METHODS

Stiffness Matrix of Element #1 1 K 77

1 K 78 1 K88

1 1 K 71 K 72 1 1 K81 K82 1 1 K11 K12 1 K 22

1 K 73 1 K 83 1 K13 1 K 23 1 K 33

SYM

1 K 74 1 K 84 1 K14 1 K 24 1 K 34 1 K 44

1 K 79 1 K 89 1 K19 1 K 29 1 K 39 1 K 49 1 K 99

351

Stiffness Matrix of Element #2 2 K 99

1 K 7,10 1 K 8,10 1 K1,10 1 K 2,10 1 K 3,10 1 K 4,10 1 K 9,10 1 K10,10

2 K 9,10 2 K10,10

2 K 93 2 K10,3 2 K 33

2 K 94 2 K10,4 2 K 34 2 K 44

2 K 95 2 K10,5 2 K 35 2 K 45 2 K 55

SYM

2 K 96 2 K10,6 2 K 36 2 K 46 2 K 56

2 2 K 9,11 K 9,12 2 2 K10,11 K 10,12 2 2 K 3,11 K 3,12 2 2 K 4,11 K 4,12 2 2 K 5,11 K 5,12

2 K 66

2 K 6,11 2 K11,11

2 K 6,12 2 K11,12 2 K12,12

(b)

(a) Assembled Global Stiffness Matrix 1 K111 2 3

K112

K113

K114

K

K

K

1 22

1 23

1 2 + K 33 K 33

4 5

0

K117

K118

K119

K11,10

0

0

0

K

K

K

K 21,10

0

0

1 2 + K 34 K 34

2 K 35

2 K 36

K

2 K 139 + K 39

2 K 13,10 + K 3,10

1 2 + K 44 K 44

2 K 45 K 525

2 K 46 K 526

K 0

1 2 + K 49 K 49 2 K 59

2 K 14,10 + K 4,1 0 2 K 5,10

2 K 3,11 2 K 4,11

K 23,12 2 K 4,12

K 52,11

K 52,12

K 0

K 62,12 0

1 24

0

K

6 7

2 66

1 27 1 37 1 47

0 K

1 77

8 9

SYM

K

1 28 11 38 1 48

K 0 0 K K

1 78 1 88

K K K

1 29

2 69 1 79 1 89

1 2 + K 99 K 99

K K K

2 6,10 1 7,10 1 8,10

1 2 + K 9,10 K 9,10 1 2 + K10,10 K10,10

10 11

2 6,11

0

0

0

2 K 9,11

2 K 9,12

2 2 K10,11 K10,12 2 K11,11

2 K11,12 2 K12,12

12

(c)

Figure 12.27 Assembling the global stiffness matrix: (a) Stiffness matrix of element #1. (b) Stiffness matrix of element #2. (c) Assembled global stiffness matrix.

into the global matrix of the soil mass of the two elements are shown in Figure 12.27. As can be seen, the coefficients of the individual element matrices are labeled Kjki . The index i designates the element number, j refers to the node number corresponding to the force F j , and k refers to the number of the node where a displacement uk contributes an additional displacement at node j. With these definitions for the indices, the stiffness coefficients for adjacent elements are simply added when they refer to the same j and k values while coming from different elements i. This simple example is extended to all nodes in the mesh to form the global stiffness matrix [K]. Then the global governing equation for the entire soil mass is: [K][u] = [F]

(12.128)

Figures. 12.26 and 12.27 show how to assemble the global matrix for two four-node elements. 7. Introduce the boundary conditions into the global matrix equation. Equation 12.128 describes how the soil mass will behave in general terms. The boundary conditions make the problem specific. These boundary conditions (also called constraints) are given in the way

of specified values of displacements, forces, temperatures, or any other parameters that affect the problem. In dynamics, these conditions involve the same types of parameters, but all of them are associated with a specific time. Examples of boundary(conditions include ) requiring no movement at a node uix = uiy = uiz = 0 , ) ( no external force at a node Fxi = Fyi = Fzi = 0 , or movement at a node allowed only in one direction, or a single force applied at a node. The specified values of displacement and forces go directly into the [u] and [F] matrices. Of course, for problems other than deformation problems, the boundary conditions are different and can be in terms of specified flow velocities, heat flux, and so on. 8. Solve the global matrix equation for the unknowns. The matrix equation to be solved is: [K][u] = [F]

(12.129)

In a three-dimensional problem, the [K] matrix is a 3n × 3n matrix where n is the number of nodes; the [u] matrix is a 3n × 1 matrix; and the [F] matrix is also a 3n × 1 matrix. The reason it is 3n is that there are 3 directions at each node with 3 associated displacements and

352

12 PROBLEM-SOLVING METHODS

3 associated forces. The displacement vectors and the force vectors will be: ⎡u1x ⎤ ⎡Fx1 ⎤ ⎢u1y ⎥ ⎢Fy1 ⎥ ⎢ 1⎥ ⎢ 1⎥ ⎢uz ⎥ ⎢Fz ⎥ ⎢ • ⎥⎢ • ⎥ ⎢ • ⎥⎢ • ⎥ ⎢ • ⎥⎢ • ⎥ ⎢ n⎥ ⎢ n⎥ ⎢ux ⎥ ⎢Fx ⎥ ⎢uny ⎥ ⎢Fyn ⎥ ⎢ n⎥ ⎢ n⎥ ⎣uz ⎦ ⎣Fz ⎦

[u] = [K ′ ]−1 [F ′ ]

In these vectors, most of the unknowns are the displacements at the nodes, except for the displacement boundary conditions. However, most of the forces at the nodes are known and are zero. Remember that we are talking about the external forces, not the internal forces. The soil experiences stresses all over its mass, but the external forces at the nodes are zero except at supports or at boundary conditions. This distinction between internal forces and external forces is critically important and can be illustrated as follows. Consider a simply supported beam resting on rigid supports at both ends. Place a heavy load in the center of the beam. If the beam is in equilibrium, the external moment is zero everywhere along the beam, but the internal moment (bending moment) is significant along most of the beam. You know the displacement at both ends (zero), but you do not know the force (support reaction). Along the rest of the beam, you do not know the displacement, but you know the force, which is zero except in the center where the force is equal to the applied external load. The same principle applies to the finite element method and Eq. 12.128. The displacement matrix [u] is largely unknown and the external force matrix [F] is largely known. Therefore, because we want to know [u], it will be necessary to invert the stiffness matrix [K] to get the displacements: [u] = [K]−1 [F]

to form the modified force matrix F ′ and the corresponding row (row i) in the K matrix is set to be all zeroes except for the diagonal value, which is 1. The same applies to column i, because the matrix is symmetrical. That way, the ith equation simply says that ui = 𝛿. This is repeated for all such cases and gives rise to a new matrix K ′ . The matrix K ′ is inverted and all displacements at all nodes are found by:

(12.130)

Because the global stiffness matrix is very large, this operation can require a lot of time when the mesh has many elements. Techniques for optimizing this operation have been developed in mathematics, including matrix banding. This banding is affected by the numbering of the nodes and it is always desirable to ensure that neighboring nodes do not have very different numbers. One issue arises with a boundary condition that specifies a displacement: say, ui = 𝛿. An example may be a support where no movement is allowed. In this case, the displacement is zero but the force is unknown. To solve the matrix problem (Eq. 12.130), all unknowns must be in the displacement matrix and all values in the force matrix must be known. To satisfy this mathematical need, the following trick is applied. The known displacement is entered in the displacement matrix as an unknown ui . The corresponding force is entered as the value of the known displacement 𝛿

(12.131)

Then the complete force matrix F is found as the matrix product Ku: [F] = [K][u]

(12.132)

Once the displacement matrix is obtained, the strains and stresses can be obtained by using the strain-displacements relationships (Eq. 12.107) and the stress-strain relationships (Eq. 12.115).

12.5.4

Example of Finite Element Solution

Use the FEM to solve the deformation field for a test performed on an elastic soil sample. The height of the sample is 0.1 m, the width is 0.05 m, and the length is infinite. The major principal stress is 300 kPa and the minor principal stress is 100 kPa. The modulus is 40 MPa and the Poisson’s ratio is 0.35. Consider a plane strain geometry and use two four-noded elements. Use numerical integration with four points to construct the stiffness matrix.

Step 1. Discretize the soil mass into finite elements connected by nodes The elements are shown in Figure 12.28. The element dimensions are a = 0.05 m and b = 0.05 m; the soil properties are E = 40,000 kPa and 𝜇 = 0.35.

a uy1 1

ux1

uy4

ux4

4

Element #1 b uy2 uy5 ux2 ux5 2 5 Element #2 uy3 uy6 ux3 ux6 3 6

Figure 12.28 Triaxial test in plane strain.

12.5 NUMERICAL SIMULATION METHODS

Step 2. Choose the interpolation functions in natural coordinates Recalling Eqs. considered:

12.98–12.101,

these

functions

1 (1 + r)(1 + s) 4 1 H2 = (1 − r)(1 + s) 4 1 H3 = (1 − r)(1 − s) 4 1 H4 = (1 + r)(1 − s) 4 H1 =

are

(12.133) (12.134) (12.135) (12.136)

⎡ 𝜕H ⎤ ⎡ b ⎢ 𝜕x ⎥ ⎢ 2 ⎢ 𝜕H ⎥ = ⎢ ⎥ ⎢0 ⎢ ⎣ 𝜕y ⎦ ⎣

𝜕H2 𝜕x 0 𝜕H2 𝜕y

0 𝜕H2 𝜕y 𝜕H2 𝜕x

𝜕H3 𝜕x 0 𝜕H3 𝜕y

0 𝜕H3 𝜕y 𝜕H3 𝜕x

√ 3 − √1 3

s= (12.137) 𝜕H4 𝜕x 0 𝜕H4 𝜕y

⎤ 0 ⎥ 𝜕H4 ⎥ ⎥ 𝜕y ⎥ 𝜕H4 ⎥ ⎥ 𝜕x ⎦

(12.138)

Constructing the [B] Matrix. a. Calculate the inverse of the Jacobian matrix used in the transformation from natural coordinates to real coordinates. ⎡ 𝜕x 𝜕y ⎤ ⎡ a ] ⎤ [ ⎢ 𝜕r 𝜕r ⎥ ⎢ 2 0 ⎥ 0.025 0 J=⎢ (12.139) ⎥= b = 0 0.025 ⎢ 𝜕x 𝜕y ⎥ ⎢⎣ 0 2 ⎥⎦ ⎣ 𝜕s 𝜕s ⎦ Therefore: det J = 6.25 ∗ 10−4 [ ] 0⎤ 40 0 −1 ⎥ J (12.140) a ⎥ = 0 40 ⎦ 2 b. Obtain the relation between the derivatives of the interpolation functions in real coordinates and in natural coordinates: ⎡ 𝜕Hi ⎤ ⎡ 𝜕Hi ⎤ ⎢ 𝜕x ⎥ −1 ⎢ 𝜕r ⎥ (12.141) ⎥ ⎢ 𝜕H ⎥ = J ⎢ ⎢ 𝜕Hi ⎥ ⎢ i⎥ ⎣ 𝜕s ⎦ ⎣ 𝜕y ⎦ 1 = det J

) ⎡b • ⎢2 ⎢0 ⎣

√ 3 − √1 3

(12.144)

d. Compute the components of the matrix [B] at the four integration points (Figure 12.29): Point #1. Recalling Eqs. 12.133–12.136, the derivatives of the interpolation function are: [ ] 𝜕H 1 1 1 1 = (1 + s) − (1 + s) − (1 + s) (1 + s) 𝜕r 4 4 4 4 (12.145) [ ] 1 1 1 𝜕H 1 = (1 + r) (1 + r) − (1 + r) − (1 + r) 𝜕s 4 4 4 4 (12.146) For integration at point #1. the natural coordinates are: 1 r= √ 3 1 s= √ 3 ) ) [ ( ( 1 1 1 𝜕H 1 − = 1+ √ 1+ √ 𝜕r 4 4 3 3 s 2

0.577 1 0.577

r

Element #1

and (

(12.142)

c. Select the natural coordinates of integration points r and s for a four-node element. This information is found in most FEM books (e.g., Zienkiewicz et al., 2005). ] [ 1 √ − √1 3 3 (12.143) r= 1 √ − √1 3 3 [ 1 1 ]

Step 3. Write the strain-displacement equations [𝜀] = [B][ui ] ⎡𝜕H1 0 ⎢ 𝜕x ⎡𝜖xx ⎤ ⎢ 𝜕H1 ⎢𝜖yy ⎥ = ⎢ 0 𝜕y ⎢ ⎥ ⎢ ⎣𝛾xy⎦ ⎢𝜕H 𝜕H 1 ⎢ 1 ⎣ 𝜕y 𝜕x ⎡ux1 ⎤ ⎢u ⎥ ⎢ y1 ⎥ ⎢ux2 ⎥ ⎢u ⎥ × ⎢ y2 ⎥ u ⎢ x3 ⎥ ⎢uy3 ⎥ ⎢ux4 ⎥ ⎢u ⎥ ⎣ y4 ⎦

⎤ ⎡ 𝜕H ⎤ ⎡ b • 𝜕H ⎤ 0⎥ ⎢ 𝜕r ⎥ ⎢ 2 𝜕r ⎥ • = a ⎥ ⎢ 𝜕H ⎥ ⎢ a 𝜕H ⎥ ⎥ ⎢ • ⎥ ⎥ ⎢ 2 ⎦ ⎣ 𝜕s ⎦ ⎣ 2 𝜕s ⎦

353

4

3 s

r Element #2

Figure 12.29 The integration points.

354

12 PROBLEM-SOLVING METHODS

(

( ) )] 1 1 1 1+ √ 1+ √ × 4 3 3 = [0.394 − 0.394 − 0.105 0.105] (12.147) 1 − 4

𝜕H = 𝜕s

) ( ) [ ( 1 1 1 1 1+ √ 1+ √ 4 4 3 3 ) )] ( ( 1 1 1 1 − − 1+ √ 1+ √ 4 4 3 3

= [0.394 0.105 − 0.105 − 0.394]

(12.148)

Point #2. 1 r = −√ 3 1 s= √ 3 ) [ ( 1 𝜕H 1 − = 1+ √ 𝜕r 4 3 ) ( 1 1 − × 1− √ 4 3

) [ ( 1 1 1− √ − 4 3 ) ( 1 1 − × 1+ √ 4 3

(

) 1 1− √ 3 )] ( 1 1 1+ √ 4 3

1 4

= [0.105 − 0.105 − 0.394 0.394] (12.151) ) ( ) [ ( 1 1 𝜕H 1 1 1− √ 1+ √ = 𝜕s 4 4 3 3 ) )] ( ( 1 1 1 1 − ×− 1+ √ 1− √ 4 4 3 3 = [0.105 0.394 − 0.394 − 0.105]

(12.152)

Point #4. (

)

1 4

1 1+ √ 3 )] ( 1 1 1− √ 4 3

= [0.394 − 0.394 − 0.105 0.105] (12.149) 𝜕H = 𝜕s

𝜕H = 𝜕r

) ( ) [ ( 1 1 1 1 1− √ 1+ √ 4 4 3 3 ) )] ( ( 1 1 1 1 − − 1+ √ 1− √ 4 4 3 3

= [0.105 0.394 − 0.394 − 0.105] Point #3. 1 r = −√ 3 1 s = −√ 3

(12.150)

1 r= √ 3 1 s = −√ 3 ) [ ( 𝜕H 1 1 − = 1− √ 𝜕r 4 3 ) ( 1 1 − × 1+ √ 4 3

) 1 1− √ 3 )] ( 1 1 1+ √ 4 3 (

1 4

= [0.105 − 0.105 − 0.394 0.394] (12.153) 𝜕H = 𝜕s

) ( ) [ ( 1 1 1 1 1+ √ 1− √ 4 4 3 3 ) )] ( ( 1 1 1 1 − − 1− √ 1+ √ 4 4 3 3

= [0.394 0.105 − 0.105 − 0.394]

(12.154)

Now Eqs. 12.138, 12.140, and Eqs. 12.145–12.154 are combined to create the B matrix. For example, the top right element of matrix B is 40 × 0.394 = 15.773.

0 −15.773 0 −4.226 0 4.226 0 ⎤ ⎡15.773 15.773 0 4.226 0 −4.226 0 −15.773⎥ B i=1 =⎢ 0 ⎢ ⎥ ⎣15.773 15.773 4.226 −15.773 −4.226 −4.226 −15.773 4.226 ⎦ j=2 0 −15.773 0 −4.226 0 4.226 0 ⎤ ⎡15.773 4.226 0 15.773 0 −15.773 0 −4.226⎥ B i=1 =⎢ 0 ⎢ ⎥ ⎣ 4.226 15.773 15.773 −15.773 −15.773 −4.226 −4.226 4.226 ⎦ j=2 0 −4.226 0 −15.773 0 15.773 0 ⎤ ⎡4.226 ⎥ ⎢ 0 4.226 0 15.773 0 −15.773 0 −4.226 B i=2 = ⎥ ⎢ 15.773 4.226 4.226 15.773 −4.226 −15.773 −15.773 −4.226 ⎦ ⎣ j=1 0 −4.226 0 −15.773 0 15.773 0 ⎤ ⎡ 4.226 15.773 0 4.226 0 −4.226 0 −15.773⎥ B i=2 =⎢ 0 ⎢ ⎥ ⎣15.773 4.226 4.226 −4.226 −4.226 15.773 −15.773 15.773 ⎦ j=2

(12.155)

12.5 NUMERICAL SIMULATION METHODS

Step 4. Write the stress-strain equations for the soil and obtain the constitutive matrix Recalling Eq. 12.116:

⎡ 1 ⎢ ⎢ 𝜇 E (1 − 𝜇) C= ⎢ (1 + 𝜇) (1 − 2𝜇) ⎢ (1 − 𝜇) ⎢ 0 ⎣

𝜇 (1 − 𝜇) 1

0 ⎤ ⎡6.419 3.457 0 ⎥ = 104 ∗ ⎢3.457 6.419 ⎢ ⎥ 0 0 1.481 ⎣ ⎦

0

Step 7. Introduce the boundary conditions into the global matrix equation

⎤ ⎥ ⎥ 0 ⎥ (1 − 2𝜇) ⎥ ⎥ 2 (1 − 𝜇 ) ⎦ 0

(12.156)

Step 5. Derive the equations governing the behavior of the soil element Recalling Eqs. 12.124 and 12.126: [K e ] =

[B]T [C][B]dV ∫V [K e ][u] = [F]

355

(12.157) (12.158)

and recalling the numerical integration from Eq. 12.125: 2 2 ∑ ∑ Ke = BT CBdv = BTij Cij Bij det J •wi •wj •t (12.159) ∫v i=1 j=1 For two-point Gauss integration, wi , and wj are equal to 1. In the case of plane strain, the thickness t of the elements is taken as 1. Therefore, the stiffness matrix for each element is as follows: K e = 104 ⎡2.63 1.23 −1.89 0.49 −1.32 −1.23 0.58 −0.49⎤ 2.63 −0.49 0.57 −1.23 −1.31 0.49 −1.89⎥ ⎢ ⎢ 2.63 −1.23 0.58 0.49 −1.31 1.23 ⎥ ⎢ ⎥ 2.63 0.49 −1.89 1.23 −1.31⎥ ×⎢ 2.63 1.23 −1.89 0.49 ⎥ ⎢ ⎢ SYM 2.63 −0.49 0.58 ⎥ ⎢ ⎥ 2.63 −1.23⎥ ⎢ ⎣ 2.63 ⎦ (12.160) Step 6. Assemble the element equations into the global matrix equation The global stiffness matrix equation j is based on the connected degrees of freedom shown in Figure 12.28, and is assembled as:

Referring to Figure 12.28, the degrees of freedom of the triaxial sample at nodes (3) and (6) should be constrained in both directions. Moreover, nodes (1) and (4) can only deform vertically. Thus, the rows and columns associated with those degrees of freedom should be zero. Step 8. Solve the global matrix equation for the unknowns The triaxial sample is subjected to a confining pressure 𝜎 3 and a vertical pressure 𝜎 1 . For this problem, 𝜎3 = 100 kPa, and 𝜎1 = 300 kPa. The force components applied at the nodes due to the confining pressure and the vertical stress are: At nodes 2 and 5 b 0.05 Phorizontal = 𝜎3 × × 2 = 100 × × 2 = 5kN∕m 2 2 (12.162) At nodes 1 and 4 a 0.05 = 7.5kN∕m (12.163) Pvertical = 𝜎1 × = 300 × 2 2 Now the force matrix is assembled as: ⎡F ′ x1 ⎤ ⎡ 0 ⎤ ⎢ Fy1 ⎥ ⎢−7.5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ Fx2 ⎥ ⎢ 5 ⎥ ⎢ Fy2 ⎥ ⎢ 0 ⎥ ⎢F ′ x3 ⎥ ⎢ 0 ⎥ ⎢ ′ ⎥ ⎢ ⎥ F 0 ⎥ F ′ = ⎢ ′ y3 ⎥ = ⎢ (kN) ⎢F x4 ⎥ ⎢ 0 ⎥ ⎢ Fy4 ⎥ ⎢−7.5⎥ ⎢ ⎥ ⎢ ⎥ ⎢ Fx5 ⎥ ⎢ −5 ⎥ ⎢ Fy5 ⎥ ⎢ 0 ⎥ ⎢F ′ ⎥ ⎢ 0 ⎥ ⎢ ′ x6 ⎥ ⎢ 0 ⎥ ⎣F y6 ⎦ ⎣ ⎦

(12.164)

Note that in fact the forces F x1 , F x3 , F y3 , F x4 , F x6 , and F y6 are actually unknown, but they are set equal to zero because of the mathematical trick mentioned at the end of Section 12.5.3 and because the corresponding displacements are zero. Note also that the matrix K ′ will have zeroes on rows corresponding to the displacement boundary conditions, except the diagonal, which will have a 1. The same applies to the corresponding columns. The 12 × 12 matrix K ′ is inverted by the computer and the displacement vector u is found as K ′ × F ′ :

⎡2.63 −1.23 0.58 0.49 0 0 −1.89 0.49 −1.32 1.23 0 0 ⎤ ⎢ 2.63 −0.49 −1.89 0 0 0.49 0.57 1.23 −1.32 0 0 ⎥ ⎢ ⎥ 5.27 0 0.58 0.49 −1.32 −1.23 0.39 0 −1.32 1.23 ⎥ ⎢ 5.27 −0.49 −1.89 −1.23 −1.32 0 1.15 1.23 −1.32⎥ ⎢ ⎢ 2.63 1.23 0 0 −1.32 −1.23 −1.89 0.49 ⎥ ⎢ ⎥ 2.63 0 0 −1.23 −1.32 −0.49 0.57 ⎥ Kg = 104 × ⎢ 2.63 1.23 0.57 −0.49 0 0 ⎥ ⎢ ⎢ 2.63 0.49 −1.89 0 0 ⎥ ⎢ ⎥ SYM 5.27 0 0.57 −0.49⎥ ⎢ 5.27 0.49 −1.89⎥ ⎢ ⎢ 2.63 −1.23⎥ ⎢ 2.63 ⎥⎦ ⎣

(12.161)

356

12 PROBLEM-SOLVING METHODS

⎡ux1 ⎤ ⎡ 0 ⎤ ⎢uy1 ⎥ ⎢−0.5134⎥ ⎢ ⎥ ⎢ ⎥ ⎢ux2 ⎥ ⎢−0.0428⎥ ⎢uy2 ⎥ ⎢−0.2567⎥ ⎢ux2 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ u 0 ⎥ u = ⎢ y3 ⎥ = ⎢ (mm) ⎢ux4 ⎥ ⎢ 0 ⎥ ⎢uy4 ⎥ ⎢−0.5134⎥ ⎢ ⎥ ⎢ ⎥ ⎢ux5 ⎥ ⎢ 0.0428 ⎥ ⎢uy5 ⎥ ⎢−0.2567⎥ ⎢ux6 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎣uy6 ⎦ ⎣ ⎦

Element

Node Boundary Γ

Domain Ω

(12.165) Figure 12.30 Discretization with the boundary element method.

Particle j

Then we can obtain the force vector by K × u: ⎡Fx1 ⎤ ⎡ 3.62 ⎤ ⎢Fy1 ⎥ ⎢ −7.5 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢Fx2 ⎥ ⎢ 5 ⎥ ⎢Fy2 ⎥ ⎢ 0 ⎥ ⎢Fx3 ⎥ ⎢ 3.62 ⎥ ⎥ ⎢ ⎥ ⎢ F 7.5 ⎥ F = ⎢ y3 ⎥ = ⎢ (kN) ⎢Fx4 ⎥ ⎢−3.62⎥ ⎢Fy4 ⎥ ⎢ −7.5 ⎥ ⎥ ⎢ ⎥ ⎢ ⎢Fx5 ⎥ ⎢ −5 ⎥ ⎢Fy5 ⎥ ⎢ 0 ⎥ ⎢Fx6 ⎥ ⎢−3.62⎥ ⎢ ⎥ ⎢ 7.5 ⎥ ⎦ ⎣Fy6 ⎦ ⎣ 12.5.5

vj Fn Fji

Fij

vi

(12.166)

Boundary Element Method

The FDM and the FEM model the continuum by discretizing the entire body of the soil mass. The boundary element method (BEM) (Crouch and Starfield, 1983; Brebbia et al., 1984) is different in that it models the continuum by discretizing only the boundaries of the continuum (Figure 12.30). The mathematical technique for the BEM consists of replacing the governing differential equations valid over the entire soil mass by integral equations that consider only the boundary values. If the soil mass extends to infinity, the FEM requires a boundary at some distance from the imposed loading or deformation. No artificial boundaries are needed in the BEM; this is an advantage of the BEM over the FEM and the FDM. Another advantage is that for a 3D problem, only the boundary surface need be discretized; this reduces the problem from a 3D problem (volume) to a 2D (surface) problem. This is attractive if the boundary surface is small compared to the volume of soil to be simulated. The BEM is particularly well suited to addressing static continuum problems with small boundary-to-volume ratios, with elastic behavior, and with stresses or displacements applied to the boundaries (Bobet, 2010). 12.5.6

Fn

Discrete Element Method

The discrete element method (DEM), also called the distinct element method (Cundall and Strack, 1979; Ghaboussi and Barbosa, 1990) differs from the finite element method in that

Particle i (a)

(b)

Figure 12.31 Distinct element method: (a) DEM domain. (b) Particle interaction. (Source: a: Courtesy of C. Couroyer, PhD thesis, 2000, University of Surrey, Guildford, Surrey, UK.)

Fn Fs Element (a) Contact forces

Cn

n s

Kn

Ks Cs

Element (b) Contact forces idealization

Figure 12.32 DEM element and idealized contact models.

it does not assume that the soil mass is a continuum; rather, it treats the soil mass as an assembly of particles of various sizes (Figure 12.31). Obviously, this is an improvement that gets us closer to reality for soils. The DEM addresses three issues during the calculations: the representation of the contacts, the representation of the solid material, and the detection and revisions of the contacts during deformation. Each soil particle is subjected to the forces transmitted at the contacts by adjacent particles and to its own body forces (gravity). The representation of the contact is usually handled through the use of spring and dashpot models (Figure 12.32). The springs have a stiffness kn for the normal force and ks for the shear force. The dashpots have damping factor cn for the normal force and cs for the shear force. The solution proceeds in small time steps and the finite difference method (FDM) is used in the solution (see Sections 12.5.1 and 12.5.2). The steps are:

12.6 PROBABILITY AND RISK ANALYSIS

where ks and kn are the stiffnesses in the normal and shear directions, respectively, cn and cs are the damping factors in the normal and shear directions, respectively, Δun and Δus are the incremental displacements in the normal and shear directions, respectively, and Δu̇ n and Δu̇ s are the incremental velocities in the normal and shear directions, respectively. The shear force F s cannot exceed the shear strength of the soil, so the following condition is checked at each increment:

1. The state of all the particles in the soil mass is known at time t. This includes contact forces, displacements, velocities, and accelerations. 2. A time increment Δt is considered. This time increment has to be small enough for the solution to be numerically stable. The following condition can be used (Hart et al., 1998): √ mmin (12.167) Δt < 0.1 2kmax where mmin is the smallest particle mass and kmax is the largest stiffness of all contacts. In the DEM, time comes into play for both dynamic and static problems. Even in a static problem, it takes time for the deformations to take place. 3. The differential equations of motion are then used to obtain the displacement and rotation of the particles at time t + Δt. The accelerations of the particles are calculated assuming that the forces and moments are constant over Δt: ∑ t Fi (12.168) ü ti = mi ∑ t Mi ̈𝜃 ti = (12.169) Ii where ü ti and 𝜃̈ i are the linear and angular acceleration of particle i at time t, respectively, Fit and Mit are the resultant force and resultant moment on particle i at time t, respectively, and mi and I i are the mass and the moment of inertia of particle i, respectively. Then the velocities of the particles are calculated assuming that the accelerations are constants over −Δt∕2 and +Δt∕2: t

t+ Δt2

u̇ i

t− Δt2

+ ü ti Δt

(12.170)

t+ t− t 𝜃̇ i 2 = 𝜃̇ i 2 + 𝜃̈ i Δt

(12.171)

Δt

= u̇ i

357

Fst+Δt ≤ c′ Ac + Fnt+Δt tan 𝜑′

(12.176)

′

where c is the effective stress cohesion intercept, Ac is the contact area, and 𝜑′ is the effective stress friction angle. 5. The cycle of calculations in 1 through 4 is repeated many times. The final solution is obtained when a chosen tolerance in the difference between two consecutive sets of calculations is achieved. The DEM is quite efficient with these calculations. The calculations are done through a straightforward process solving one equation at a time, and no large matrix has to be inverted. Where the computing power and storage capacity are required is in recognizing and keeping track of all the contacts between elements from one step to the next and by the extremely small time steps required by Eq. 12.167. The DEM is very useful for soils and fissured rock masses. One drawback of DEM is that the time step is inversely proportional to the stiffness of the particles. Since soil particles are typically very stiff (pieces of rock), the time step is extremely small, and many calculations must be performed for a few seconds of overall result. The other limitation is that the number of soil particle tends to be extremely large for a typical project and again much computing time is necessary for the DEM results to be useful.

Δt

where u̇ i and 𝜃̇ i are the linear and angular velocities respectively. Then the displacements and rotations of the particles are calculated assuming that the velocities are constant over Δt: t+ Δt2

= uti + u̇ i ut+Δt i

Δt

(12.172)

t+ 𝜃it+Δt = 𝜃it + 𝜃̇ i 2 Δt

(12.173)

Δt

where ui and 𝜃 i are the displacement and the rotation respectively. 4. The equations representing the behavior of the contacts are then used to update the forces and moments. Figure 12.32 gives a common model for the contact normal forces F n and the contact shear forces F s : Δt Fnt+Δt = kn ΔuΔt n + cn Δu̇ n

(12.174)

Δt Fst+Δt = ks ΔuΔt s + cs Δu̇ s

(12.175)

12.6

PROBABILITY AND RISK ANALYSIS

All the methods discussed so far are deterministic in nature, which means that they give one precise answer for one problem. Considering the fact that uncertainty exists in every step taken in arriving at a solution, it makes sense to calculate the uncertainty associated with the solution or predicted value. This is called the probabilistic approach. 12.6.1

Background

This subsection reviews some basic concepts of statistics because they are useful in the steps described for the general procedure. When many values of a certain variable are collected—such as the undrained shear strength su of a clay at a site and at a given depth, for example—they will vary and can be organized in a table from the lowest to the highest value (Table 12.2). These values sui can then be regrouped into sets of increments or ranges, as shown in Table 12.2. A histogram is a plot of the number of times the variable

358

12 PROBLEM-SOLVING METHODS

Table 12.2 Values of undrained shear strength and histogram input Undrained strength value (kPa) 49 62 67 73

Number of values (10 kPa increments)

Number of values (20 kPa increments)

1 value between 40 and 50 2 values between 60 and 70

1 value between 40 and 60 6 values between 60 and 80

4 values between 70 and 80

75 76 79 81

3 values between 80 and 90

85 86 93

1 value between 90 and 100 1 value between 100 and 110

105

4 values between 80 and 100

1 value between 100 and 120

is found in each increment as a function of the value of the variable (Figures 12.33a and 12.33b). Note that a different histogram is generated if a different increment magnitude is selected. A distinction is made between the variable X and the values of that variable xi . The mean 𝜇 of a set of values (x1 , x2 , x3 , … , xn ) is defined as follows and is called the expected value E(X) of X: n ∑

xi x1 + x2 + · · · + xn i=1 = E(X) (12.177) n n The standard deviation 𝜎 is a measure of the deviation of the values with respect to the mean. It is given by: √ (x1 − 𝜇)2 + (x2 − 𝜇)2 + · · · + (xn − 𝜇)2 𝜎= n−1 √ √∑ n √ √ (xi − 𝜇)2 √ i=1 = (12.178) n−1 The reason for using the squares is that the difference (xi − 𝜇) can be positive or negative and might cancel out during summation, thereby not giving a true rendition of the scatter around the mean. We could have used the absolute values of the difference, but that is not what was chosen in practice. Also, the reason for using (n − 1) rather than n is the fact that 𝜇=

only (n − 1) values of (xi − 𝜇) are independent, as the sum of the n values of (xi − 𝜇) is equal to zero. This is called the Bessel correction. The square of the standard deviation 𝜎 2 is called the variance v and the ratio of the standard deviation to the mean is the coefficient of variation CoV. The CoV is a measure of the scatter in the data. The CoV of structural dead loads may be around 0.05, whereas the CoV of soil data may be around 0.3: 𝜎 CoV = (12.179) 𝜇 For normal distributions, the inverse of the CoV is the reliability index 𝛽. The reliability index tells us how many standard deviations the mean is from the zero origin. It is very useful in reliability analysis and engineering code calibration. In this case, the variable is the difference between the resistance R and the load L and the reliability index 𝛽 tells us how many standard deviations 𝜎(R−L) the mean 𝜇(R−L) is from failure (R − L = 0). It serves as an indication of the safety level (reliability index). 𝜇 𝛽= (12.180) 𝜎 For distributions different from normal distributions, the generalized reliability index is still used, but is defined differently. If the number of values of xi increases, the histogram becomes smoother; if the number becomes infinity, a smooth function is obtained. This function is f (x) and is called the probability density function (PDF) (Figure 12.33c). It is defined as the function f (x) that satisfies: b

P(a < X < b) =

f (x)dx

∫a

(12.181)

where P(a < X < b) is the probability that X will be between a and b. The curves on Figure 12.34 are examples of the function f (x). The area under the curve between two values a and b is the probability that X will fall between those two values. The function also satisfies: +∞

P(−∞ < X < +∞) =

f (x)dx = 1

∫−∞

(12.182)

Recall that for the histogram, the distribution depended on the increment selected for the variable. The same happens for f(x): Different functions will be obtained depending on the units used for the variable axis. However, the integral in Eq. 12.181 will be the same because it is a relative measure. The cumulative distribution function (CDF) gives the value: x

P(X < x) =

∫−∞

f (x)dx

(12.183)

One of the most commonly used PDFs is the normal distribution. The normal distribution function is: 1 −1 f (x) = √ e 2 𝜎 2𝜋

(

x−𝜇 𝜎

)2

(12.184)

359

12.6 PROBABILITY AND RISK ANALYSIS

8 7

6 4 2 0

20

0

40 60 80 100 Undrained strength (kPa)

120

140

Number of values

Number of values

8

6 5 4 3 2 1 0

0

20

40 60 80 100 Undrained strength (kPa)

(a)

140

(b)

0.04

1 0.8

CDF (x)

0.03 PDF (x)

120

0.02

0.01

0.6 0.4 0.2

0 –50

0

50 100 150 Undrained strength (kPa) (c)

0 –50

200

0

50 100 150 Undrained strength (kPa) (d)

200

Figure 12.33 Histogram for two values of the variable increment. 1

μ = 0, σ = 0.45 μ = 0, σ = 1.0 μ = 0, σ = 2.24 μ = −2, σ = 0.71

0.9 0.8

Φ (u)

φ (X)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 –5

–4

–3

–2

–1

0

1

2

3

4

5

X

Figure 12.34 Examples of probability density function for normal distributions.

The corresponding CDF is: )) ( ( x−𝜇 1 1 + erf F(x) = √ 2 𝜎 2

(12.185)

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 –5

μ = 0, σ = 0.45 μ = 0, σ = 1.0 μ = 0, σ = 2.24 μ = −2, σ = 0.71

–4

–3

–2

–1

0 u

1

2

3

4

5

Figure 12.35 Examples of cumulative distribution function for normal distributions.

The “erf” function is called the error function; it does not have a closed-form expression, but can be tabulated. Figure 12.34 shows normal distributions and Figure 12.35 shows cumulative distributions.

360

12 PROBLEM-SOLVING METHODS

It is often advantageous to normalize the variable. The standard normal variable (SNV) is denoted u: x−𝜇 u= (12.186) 𝜎 Therefore, the mean and the standard deviation of the SNV are 0 and 1 respectively. The PDF and CDF for the SNV are: u2 1 PDF 𝜑(u) = √ e− 2 ( (2𝜋 )) 1 u 1 + erf √ CDF Φ(u) = 2 2

(12.187) (12.188)

Values of the CDF function Φ(u) for the standard normal variable u are presented in Table 12.3. So, if you wish to find out the probability P(X < x) that a normally distributed variable X is less than a chosen value x, the steps are: 1. Obtain the mean 𝜇 and standard deviation 𝜎 of X. 2. Calculate the value of the standard normal variable u = (x − 𝜇)∕𝜎. 3. Look in Table 12.3 to find Φ(u). 4. Then ( ) X−𝜇 x−𝜇 Φ(u) = P(U < u) = P < 𝜎 𝜎 = P(X − x) (12.189) 5. Remember that Φ(u) has the following properties: P(U < u) = 1 − P(U < −u) so

f (x) =

1 √

− 12

e

(

Lnx−𝜇Lnx 𝜎Lnx

)2

(12.192)

x𝜎Lnx 2𝜋 Note that the distribution differs slightly from the normal distribution equation. This is because the function is f (x) rather than f (Lnx). The function f (Lnx) would have the same expression as Eq. 12.184, but f (x) is equal to f (Lnx) times the derivative of Lnx with respect to x, which brings about the additional 1∕x. In this case, the mean and standard deviation of the lognormal distribution are: ) ( 𝜇2 (12.193) 𝜇Lnx = Ln √ 2x 2 𝜇x +𝜎x √ ( ) 𝜎2 (12.194) 𝜎Lnx = Ln 1 + 𝜇x 2 x

Table 12.3 can be used to obtain the probability P(X < x) that a lognormal distributed variable X is smaller than a chosen value x. The process takes place as follows: 1. Obtain the mean 𝜇x and the standard deviation 𝜎x of X. 2. Obtain the mean 𝜇Lnx and standard deviation 𝜎Lnx of LnX. This can be done by using Eqs. 12.193 and 12.194 once 𝜇x and 𝜎x are known. 3. Calculate the) value of the standard normal variable ( Lnx−𝜇Lnx . u= 𝜎 Lnx

Φ(u) = 1 − Φ(−u)

(12.190)

P(U < u) = P(U > −u)

(12.191)

Figure 12.36 shows some useful areas under the normal distribution. Another distribution that is very commonly used is the lognormal distribution (Figures 12.37 and 12.38). This distribution of a variable X is defined as a distribution

4. Look in Table 12.3 to find Φ(u) Then ( ) LnX − 𝜇Lnx Lnx − 𝜇Lnx Φ(u) = P(U < u) = P < 𝜎Lnx 𝜎Lnx = P(LnX < Lnx) = P(X < x) (12.195) 5. Remember that Φ(u) has the following properties: P(U < u) = 1 − P(U < −u) so Φ(u) = 1 − Φ(−u)

(12.196)

P(U < u) = P(U > −u)

(12.197)

and

0.5

Mean, μ = 0 Standard deviation, σ = 0.2

0.4 1σ

0.3

1σ

12.6.2

34.13% 34.13%

2σ 2.13% 0.14% 3σ

0 –0.8

–0.6

2σ 13.6 %

13.6%

–0.4

Procedure for Probability Approach

A method of calculating the uncertainty associated with a predicted value usually proceeds as follows:

0.2 0.1

such that the LnX (natural logarithm) is normally distributed. The probability density function of the lognormal distribution is therefore:

–0.2

0

0.2

2.13% 3σ 0.4

0.14%

0.6

Figure 12.36 Useful areas under the normal distribution.

0.8

1. First, the uncertainty associated with each variable involved in the solution is quantified. This quantification process often requires that the mean 𝜇 and standard deviation 𝜎 of each variable be determined, or that the

12.6 PROBABILITY AND RISK ANALYSIS

Table 12.3

361

Values of the areas under the distribution of the standard normal variable

Table 12.3 gives the cumulative probability up to the standardized normal value of x x

P[X < x] = ∫−∞

1 √ 2π

( exp

1 2 X 2

) dX

P[X < x]

0

x

x

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0 0.1 0.2 0.3 0.4

0.5000 0.5398 0.5793 0.6179 0.6554

0.5040 0.5438 0.5832 0.6217 0.6591

0.5080 0.5478 0.5871 0.6255 0.6628

0.5120 0.5517 0.5910 0.6293 0.6664

0.5159 0.5557 0.5948 0.6331 0.6700

0.5199 0.5596 0.5987 0.6368 0.6736

0.5239 0.5636 0.6026 0.6406 0.6772

0.5279 0.5675 0.6064 0.6443 0.6808

0.5319 0.5714 0.6103 0.6480 0.6844

0.5359 0.5753 0.6141 0.6517 0.6879

0.5 0.6 0.7 0.8 0.9

0.6915 0.7257 0.7580 0.7881 0.8159

0.6950 0.7291 0.7611 0.7910 0.8186

0.6985 0.7324 0.7642 0.7939 0.8212

0.7019 0.7357 0.7673 0.7967 0.8238

0.7054 0.7389 0.7704 0.7995 0.8264

0.7088 0.7422 0.7734 0.8023 0.8289

0.7123 0.7454 0.7764 0.8051 0.8315

0.7157 0.7486 0.7794 0.8078 0.8340

0.7190 0.7517 0.7823 0.8106 0.8365

0.7224 0.7549 0.7854 0.8133 0.8389

1.0 1.1 1.2 1.3 1.4

0.8413 0.8643 0.8849 0.9032 0.9192

0.8438 0.8665 0.8869 0.9049 0.9207

0.8461 0.8686 0.8888 0.9066 0.9222

0.8485 0.8708 0.8907 0.9082 0.9236

0.8508 0.8729 0.8925 0.9099 0.9251

0.8531 0.8749 0.8944 0.9115 0.9265

0.8554 0.8770 0.8962 0.9131 0.9279

0.8577 0.8790 0.8980 0.9147 0.9292

0.8599 0.8804 0.8997 0.9162 0.9306

0.8621 0.8830 0.9015 0.9177 0.9319

1.5 1.6 1.7 1.8 1.9

0.9332 0.9452 0.9554 0.9641 0.9713

0.9345 0.9463 0.9564 0.9649 0.9719

0.9357 0.9474 0.9573 0.9656 0.9726

0.9370 0.9484 0.9582 0.9664 0.9732

0.9382 0.9495 0.9591 0.9671 0.9738

0.9394 0.9505 0.9599 0.9678 0.9744

0.9406 0.9515 0.9608 0.9686 0.9750

0.9418 0.9525 0.9616 0.9693 0.9756

0.9429 0.9535 0.9625 0.9699 0.9761

0.9441 0.9545 0.9633 0.9706 0.9767

2.0 2.1 2.2 2.3 2.4

0.9773 0.9821 0.9861 0.9893 0.9918

0.9778 0.9826 0.9865 0.9896 0.9920

0.9783 0.9830 0.9868 0.9898 0.9922

0.9788 0.9834 0.9871 0.9901 0.9924

0.9793 0.9838 0.9874 0.9904 0.9927

0.9798 0.9842 0.9878 0.9906 0.9929

0.9803 0.9846 0.9881 0.9909 0.9931

0.9808 0.9850 0.9884 0.9911 0.9932

0.9812 0.9854 0.9887 0.9913 0.9934

0.9817 0.9857 0.9890 0.9916 0.9936

2.5 2.6 2.7 2.8 2.9

0.9938 0.9953 0.9965 0.9974 0.9981

0.9940 0.9955 0.9966 0.9975 0.9982

0.9941 0.9956 0.9967 0.9976 0.9982

0.9943 0.9957 0.9968 0.9977 0.9983

0.9945 0.9959 0.9969 0.9977 0.9984

0.9946 0.9960 0.9970 0.9978 0.9984

0.9948 0.9961 0.9971 0.9979 0.9985

0.9949 0.9962 0.9972 0.9980 0.9985

0.9951 0.9963 0.9973 0.9980 0.9986

0.9952 0.9964 0.9974 0.9981 0.9986

x P

3.00 0.9986

3.10 0.9990

3.20 0.9993

3.30 0.9995

3.40 0.9997

3.50 0.9998

3.60 0.9998

3.70 0.9999

3.80 0.9999

3.90 1.0000

362

12 PROBLEM-SOLVING METHODS

predicted function from which a mean and a standard deviation are calculated. 4. Once the standard deviation of the predicted function is known, one can find out the probability that the predicted function value will be higher or lower than a chosen target.

φ (X)

1.8 1.6

μ (LnX) = 0, σ (LnX) = 0.25

1.4

μ (LnX) = 0, σ (LnX) = 0.5

1.2

μ (LnX) = 0, σ (LnX) = 1.0

1 0.8 0.6 0.4

12.6.3

0.2 0

0

0.5

1

1.5 X

2

2.5

3

Figure 12.37 Examples of probability density function for lognormal distributions. 1

Φ (u)

There is a very important distinction to be made between probability of failure and risk. The probability of failure is simply the probability that something is going to collapse (e.g., a bridge, a slope, a building, a dam). Risk is defined as the probability of failure multiplied by the value of the consequence. It uses units of the value of the consequence, typically fatalities or dollars lost: R = P(F –5 –7.5 to –5 –10 to –7.5 –17 to –10

Severity

0–1 1–5 5–10 10–20 > 20

Negligible Moderate trouble Trouble Severe trouble Very severe trouble

(Source: After Jennings and Knight, 1975.)

1. 2. 3. 4.

2.6LL(%) 100

wo − PL(%) So > 0.85 PI(%)

Feda, 1966

𝛾 d = dry unit weight, LL = liquid limit, eo = natural void ratio, wo = natural water content, So = natural degree of saturation, PL = plastic limit, PI = plasticity index.

Liquid limit below 45. Plasticity index below 25. Dry unit weight between 10 and 17 kN/m3 . Porosity between 40 and 60%.

Table 15.20 gives additional criteria to determine if a soil can be collapsible or not. Once it is recognized that a soil may be collapsible using these criteria, the extent of the collapse and its severity can be gauged by the scale proposed by Jennings and Knight (1975). Their scale is based on the collapse potential index CP: e − ec × 100 (15.50) CP = o 1 + eo where eo is the void ratio of the soil at its natural water content under 200 kPa of vertical pressure in the consolidation test before wetting, and ec is the void ratio after soaking under 200 kPa of vertical pressure. Table 15.21 gives a severity scale for the collapse. The best way to determine the amount of collapse that may occur is to perform a consolidation test and simulate what would happen to the soil in the field. For this, the sample at its natural water content is placed in the consolidometer, the sample height h is recorded, and the vertical pressure p is increased in steps. For each step, the change in height Δh/h of the sample is recorded every 30 minutes and the curve of stress p vs. strain Δh/h is plotted (Figure 15.23).

Source 25.5 1 + 0.026LL(%)

Severity of collapsible soils scale

CP, collapse potential (%)

1,000 Applied stress, σ (kPa)

15.19

(

Table 15.21

D 100

pf

B C

10 A 1

0

5

10

15

Strain, ε (%)

Figure 15.23 Stress-strain curve for a collapsible soil.

20

15.19 COLLAPSE DEFORMATION BEHAVIOR

Each pressure is kept on the sample until the rate of strain is less than 0.1%/hour. When the vertical pressure p reaches the pressure pf anticipated in the field (under the foundation, for example) and at the end of that load step, the sample is inundated and the readings of strain continue during the collapse as a function of time. The end of the collapse step is when the strain has become less than 0.1%/hour. The next pressure step is applied, and so on, until the curve is completed (Figure 15.23). For collapse pressures less than pf , a line is drawn from A to C on the curve and used to estimate the collapse strain for intermediate values of pf . In this consolidation test, the sample is inundated and the collapse occurs under the extreme situation in which the sample can absorb all the water it is able to absorb. In reality, there could be a limit to the amount of water available to the soil (the rainstorm stops, for example) and the collapse

log uw

W

Δh/h

might be partial. A model to predict the collapse strain under partial wetting links the water content to the strain or the water tension to the strain (Figure 15.24) (Pereira and Fredlund, 2000).

Problem 15.1 Consider the stress-strain curve from a triaxial test shown in Figure 15.1s: Why is 𝜀r < 0 when 𝜀z > 0? Calculate the Poisson’s ratio. Calculate the soil modulus between 0 and A. Calculate the ratio between this soil modulus and the modulus of concrete.

150 σ′3 = 40 kPa 100

A

σ′1 50

0 0

O 0

0.01

0.02

0.03

0.04

–0.01 εr

Δh/h

Figure 15.24 Water content and water tension vs. collapse strain.

Problems and Solutions

a. b. c. d.

471

–0.02 –0.03

Figure 15.1s Triaxial test results.

0.05

εz εz

472

15 DEFORMATION PROPERTIES

Solution 15.1 The Poisson’s ratio is obtained from the elasticity equations (Eq. 15.3). Using the principal directions, the Poisson’s ratio is given by: −𝜀3 𝜎1 + 𝜀1 𝜎3 𝜈= (15.1s) 𝜀1 𝜎1 + 𝜀1 𝜎3 − 2𝜀3 𝜎3 Considering the geometry of the triaxial test: 𝜀z = 𝜀 1

and

𝜀r = 𝜀 3

(15.2s)

At point A, the values of the strains are: 𝜀z = 𝜀1 = 0.01

and

𝜀r = 𝜀3 = −0.004

(15.3s)

At point A, the values of the stresses are: 𝜎z = 𝜎1 = 100 kPa

and

𝜎r = 𝜎3 = 40 kPa

(15.4s)

Then the Poisson’s ratio is calculated as: v=

(−0.004) × 100 + 0.01 × 40 = 0.47 0.01 × 100 + 0.01 × 40 − 2 × (−0.004) × 40

(15.5s)

Note that V is different from the ratio of −𝜀r ∕𝜀z , which would be 0.4. The soil modulus E (Eq. 15.1) is given by: E=

𝜎1 ⋅ 2v𝜎3 100 − 2 × 0.47 × 40 = = 6240 kPa 𝜀1 0.01

(15.6s)

Note that E is different from the ratio 𝜎1 ∕𝜎3 , which would be 10,000 kPa. The ratio between this soil modulus and the modulus of concrete (20,000 MPa) is: EConcrete 20000 × 103 = = 3205 (15.7s) ESoil 6240 Problem 15.2 Given that Figure 15.2s is the result of an unconfined compression test, calculate the secant modulus (OA), the unload modulus (AB), the unload-reload modulus (BC), the reload modulus (BD), and the tangent modulus at A. Which is the smallest modulus? Which is the largest modulus? Which is the right modulus? σ Su

200

160

Sc

Ss

Sr

St

(120, 1.28) A

120

D (112, 1.55) 80

C (80, 1.23)

40 B (32, 0.82) 0

0

0.5

1

1.5

2

2.5

Figure 15.2s Modulus values.

ε (%)

15.19 COLLAPSE DEFORMATION BEHAVIOR

473

Solution 15.2 Because the test is an unconfined compression test, each modulus can be calculated directly from the slopes in Figure 15.2s. From the secant slope OA (Ss ), the secant modulus is: Δ𝜎 120 − 0 = = 9375 kPa Δ𝜀 0.0128 − 0 From the unload slope AB (Su ), the unload modulus is: Es =

Δ𝜎 120 − 0 = = 19130 kPa Δ𝜀 0.0128 − 0.0082 From the cyclic slope BC (Sc ), the unload-reload modulus is: Eu =

Δ𝜎 80 − 32 = = 10707 kPa Δ𝜀 0.0123 − 0.0082 From the reloading slope BD (Sr ), the reload modulus is: Ec =

Δ𝜎 112 − 32 = = 10960 kPa Δ𝜀 0.0155 − 0.0082 From the tangent slope (St ), the tangent modulus is: Er =

Et =

𝜎 167 − 80 = = 2900 kPa 𝜀 0.03 − 0

(15.8s)

(15.9s)

(15.10s)

(15.11s)

(15.12s)

• The smallest modulus is the tangent modulus. • The largest modulus is the unload modulus. • There is no “right” modulus. Each slope or modulus is used for different situations or applications. Problem 15.3 Explain the difference between the modulus, the stiffness, and the modulus of subgrade reaction. Comment on which one is a true soil property and say why. Solution 15.3 The modulus of deformation (kN/m2 ) is defined by the equations of elasticity and as the slope of the line of a stress-strain curve of a material in the case of an unconfined compression test. Stiffness (kN/m) is the ratio of a force Q applied on a boundary through a loading area divided by the displacement s experienced by the loaded area (square or circular shape). The modulus of subgrade reaction (kN/m3 ) is the ratio of pressure p applied to the boundary through a loading area divided by the displacement s experienced by the loaded area. Only the modulus of deformation is a true soil property, because stiffness and modulus of subgrade reaction depend on the size of the loaded area. The results of stiffness and modulus of subgrade reaction in one test will be different from the results of other tests with different areas. The modulus of deformation for the same material is not affected by the size of the loaded area. Problem 15.4 Equation 15.13s gives the secant modulus for any confinement level, any strain level, any time of loading, and any number of cycles. If Eai is equal to 1000 MPa, n is 0.5, 𝜎 ult is 100 kPa, to is 1 minute, m is 0.03, and p is 0.1: a. Plot the initial tangent modulus Ei as a function of the confinement level 𝜎 M for the reference loading time to and for monotonic loading (N = 1). b. Plot the secant modulus Es as a function of the strain level ε for a confinement of 50 kPa, for the reference loading time to , and for monotonic loading (N = 1). c. Plot the secant modulus Es as a function of the time of loading t for a confinement stress of 50 kPa, a strain of 0.5%, and for monotonic loading (N = 1). d. Plot the secant modulus Es as a function of the number of cycles N for a confinement of 50 kPa, an initial strain of 0.5%, and the reference time to . −1

E𝜎

𝜀tN

⎞ ⎛ ⎟ ( )−m ⎜ 1 𝜀 ⎟ t = ⎜ ( )n + N −p ⎜ 𝜎ult ⎟ to 𝜎M ⎟ ⎜ Eai pa ⎠ ⎝

(15.13s)

474

15 DEFORMATION PROPERTIES

Solution 15.4 a. For the initial tangent modulus, for the reference time to , and for monotonic loading (N = 1), the general equation becomes as shown in Eq. 15.14s, and the results are plotted in Figure 15.3s. ( ( )n ) 𝜎 (kPa) 0.5 𝜎 E𝜎 𝜀tN (MPa) = Eai M = 1000 M (15.14s) pa 100

Initial tangent modulus, Et (MPa)

1,500

1,200

900

600

300

0

0

30

60 90 120 Confining stress, σM (kPa)

150

Figure 15.3s Initial tangent modulus vs confining stress.

b. For the secant modulus Es as a function of the strain level ε for a confinement of 50 kPa, for the reference loading time to , and for monotonic loading (N = 1), the equation becomes as shown in Eq. 15.15s and the results are plotted in Figure 15.4s: −1 ⎞ ⎛ ⎟ ⎜ 1 𝜀 ⎟ = (1.414 × 10−3 + 10𝜀)−1 (15.15s) E𝜎 𝜀tN (MPa) = ⎜ ( )n + ⎜ 𝜎ult ⎟ 𝜎M E ⎟ ⎜ ai pa ⎠ ⎝ 800

700

Secant modulus, ES (MPa)

Secant modulus, ES (MPa)

800

600 500 400 300 200 100 0

0

2

4

6

8

700 600 500 400 300 200 100 0 0.0001

10

0.0010

Axial strain, ε (%)

0.0100 0.1000 Axial strain, ε (%)

1.0000

10.0000

Figure 15.4s Secant modulus vs. axial strain.

c. For the secant modulus Es as a function of the time of loading t for a confinement stress of 50 kPa, a strain of 0.5%, and for monotonic loading (N = 1), the equation becomes as shown in Eq. 15.16s and the results are plotted in Figure 15.5s: −1

⎞ ⎛ ⎟ ( )−m ⎜ 𝜀 ⎟ t 1 ⎜ = 19.45 × (t(min ))−0.03 E𝜎 𝜀tN (MPa) = ( )n + ⎜ 𝜎ult ⎟ to 𝜎M ⎟ ⎜ Eai pa ⎠ ⎝

(15.16s)

15.19 COLLAPSE DEFORMATION BEHAVIOR

475

Secant modulus, ES (MPa) 

20 16 12 8 4 0

0

40

80 120 Time, t (min)

160

200

Figure 15.5s Secant modulus vs. time.

d. For the secant modulus Es as a function of the number of cycles N for a confinement of 50 kPa, an initial strain of 0.5%, and the reference time to , the equation becomes as shown in Eq. 15.17s and the results are plotted in Figure 15.6s: −1 ⎞ ⎛ ⎟ ⎜ 1 𝜀 ⎟ E𝜎 𝜀tN = ⎜ ( )n + N −p = 19.45 × N −0.1 (15.17s) ⎜ 𝜎ult ⎟ 𝜎M ⎟ ⎜ Eai pa ⎠ ⎝

Secant modulus, ES (MPa) 

20 16 12 8 4 0

0

50

100 150 Number of cycles, N

200

Figure 15.6s Secant modulus vs. number of cycles.

Problem 15.5 A soil sample has a void ratio e = 0.6, an OCR = 2, a PI = 20%, and a shear strength of 40 kPa at a confining pressure of 70 kPa. Use an equation similar to Eq. 15.20 for the shear modulus G and prepare two plots of G/Gmax versus 𝛾. The first one is G/Gmax versus 𝛾 on natural scales and the second one is G/Gmax on the vertical natural scale and 𝛾 on the horizontal decimal log scale. What other influencing factors are missing from this classical G-𝛾 curve? Solution 15.5 We select an equation of the form

( G=

1 Gmax

+

𝛾 s

)−1 (15.18s)

We are given e = 0.6, OCR = 2, PI = 20%, pa = 101.325 kPa, s = 40 kPa, and 𝜎M = 70 kPa. The equation proposed by Jamiolkowski et al. (1991) can be used to estimate the maximum shear modulus (Gmax ). The overconsolidation exponent k is 0.18 for PI = 20% (Table 15.11). ( ′ )0.5 𝜎M Gmax 625 k = 1.3 (OCR) (15.19s) Pa Pa e

476

15 DEFORMATION PROPERTIES

(

Gmax = 101.325 Then the equation for G/Gmax is: 1 G = Gmax Gmax

(

)0.5 ) ( 625 70 0.18 (2) = 115844 kPa 101.325 (0.6)1.3

1 Gmax

+

𝛾 s

)−1 =

(15.20s)

( 𝛾 )−1 1 1 + 115844 115844 40

(15.21s)

1

Shear modulus ratio G/Gmax

Shear modulus ratio G/Gmax

The plots of G/Gmax versus 𝛾 are shown in Figure 15.7s. This plot includes the effect of strain level and confinement level, but not rate effect or the influence of cycles.

0.8 0.6 0.4 0.2 0 0.0001

2.0001

4.0001 6.0001 Shear strain, γ (%)

8.0001

10.000

1 0.8 0.6 0.4 0.2 0 0.0001

0.001

0.01 0.1 Shear strain, γ (%)

1

10

Figure 15.7s G/Gmax vs. shear strain.

Problem 15.6 Given the log of vertical effective stress vs. vertical strain curve (Figure 15.8s), find the preconsolidation pressure 𝜎p′ and calculate the compression index Cc . 0.6

0.55

Void ratio, e

0.5

0.45

0.4

0.35

0.3

1

10

100 1000 Effective vertical stress, σ′ (kPa)

10000

Figure 15.8s Strain vs. log of stress consolidation curve.

Solution 15.6 • Based on Cassagrande’s method shown in Figure 15.10 (Figure 15.9s): 𝜎p′ = 1200 kPa • The compression index Cc is: Cc =

Δe 0.52 − 0.30 = ) = 0.22 ( 10000 log 𝜎1 − log 𝜎2 log 1000

(15.22s)

15.19 COLLAPSE DEFORMATION BEHAVIOR

477

0.6

0.55

σ′p

Void ratio, e

0.5

α α

0.45

1 0.4

Cc

0.35

0.3

σ′p 1

10

100

1000

10000

Effective vertical stress, σ′ (kPa)

Figure 15.9s Void ratio vs. effective vertical stress.

Problem 15.7 Given a straight-line relationship (Figure 15.10s) between the vertical effective stress and the vertical strain (𝜎 ′ = ′ + 40, 000 𝜀), a vertical effective stress at rest of 100 kPa, and an initial void ratio eo of 1, draw the log of vertical 𝜎ov stress versus void ratio curve, and find the preconsolidation pressure 𝜎p′ and the compression index Cc from that curve. Discuss. 4500

Effective vertical stress, σ′ (kPa)

4000 3500 3000 2500 2000 1500 1000 500 0

0

2

4

6

8

10

Strain, ε (%)

Figure 15.10s Effective vertical stress vs. strain.

Solution 15.7 The relationship between the strain and the void ratio is 𝜀 = (e − eo )∕(1 + eo ). By substituting in the stress-strain equation, we get: ( ) e − eo ′ ′ 𝜎 = 𝜎ov + 40,000 = 100 + 20,000(e − 1) (15.23s) 1 + eo If we plot this equation as log 𝜎’ versus e, we get Figure 15.11s. Using this curve and Cassagrande’s construction, we find a preconsolidation pressure of the order of 1600 kPa (Figure 15.11s). Then we can calculate the compression index as: Cc =

Δe 0.965 − 0.80 = ) = 0.548 ( 4000 log 𝜎1 − log 𝜎2 log 2000

478

15 DEFORMATION PROPERTIES

1.05

1

Void ratio, e

0.95

0.9

0.85

0.8

σ′p = 1600 kPa

0.75 10

100

1000

10000

Effective vertical stress, σ′ (kPa)

Figure 15.11s Preconsolidation pressure.

As can be seen from Figure 15.11s, a preconsolidation pressure can be found for a soil that obviously does not have one (Figure 15.12s), as it is linear. The distortion created by the semi-log plot leads to the apparent preconsolidation pressure in this case. 4500

Effective vertical stress, σ′ (kPa)

4000 3500 3000 2500 2000 1500 1000 500 0

0

2

4

6

8

10

Strain, ε (%)

Figure 15.12s Effective vertical stress vs. strain.

Problem 15.8 Given the three vertical strain versus time curves of Figure 15.13s from a consolidation test with drainage top and bottom, and the original height of the sample of 14.2 mm, calculate the coefficient of consolidation cv by the t50 method and by the log time method.

15.19 COLLAPSE DEFORMATION BEHAVIOR

479

Time, t (min)

0.01

Strain, ε

0.02

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 Stress, σ′

0

0

0.03

Strain, ε

0.04 0.05 0.06

Strain, ε

0.06

Stress, σ′

0.05

0.07

Strain, ε

0.08

0.09

0.086

Strain, ε

0.088 0.09

Stress, σ′

0.084

Strain, ε

0.092 0.094 0.096

Figure 15.13s Strain vs. time consolidation curves.

Solution 15.8 √ a. t Method (Figures 15.14s–15.16s)√ 1. Plot the sample height H versus t. 2. Draw the tangent to the initial part of the curve. √ 3. Choose a point M at an arbitrary but convenient t1 value and a height H 1 on that tangent. √ √ 4. Plot a point N with coordinates t2 = 1.15 t1 and H 1 . 5. Connect N to the start of the curve. √ 6. The intersection with the curve gives t90 7. Calculate Cv from the equation: 0.848 d2 Cv = t90

(15.24s)

15 DEFORMATION PROPERTIES

14.1 14

H (mm)

13.9 13.8 13.7 13.6 13.5 13.4

t90 = 6.5

5

0

10

15

20

25

30

35

40

45

t

Figure 15.14s Square root of time method.

Cv1 =

0.848∗ (14.2∕2)2 0.848d2 = = 1.0118(mm2 ∕ min ) t90 6.52

13.6 13.5

H (mm)

13.4 13.3 13.2 13.1 13 12.9

t90 = 7

0

5

10

15

20

25

30

35

40

45

t

Figure 15.15s Square root of time method.

Cv2 =

0.848∗ (14.2∕2)2 0.848d2 = = 0.8724(mm2 ∕ min ) t90 72

12.98 12.96 12.94 12.92 H (mm)

480

12.90 12.88 12.86 12.84 12.82 12.80

t90 = 8

0

5

10

15

20

25

30

t

Figure 15.16s Square root of time method.

35

40

45

15.19 COLLAPSE DEFORMATION BEHAVIOR

Cv1 =

481

0.848∗ (14.2∕2)2 0.848d2 = = 0.6679(mm2 ∕ min ) t90 82

b. Log time method (Figure 15.17s). 1. Plot H vs. log time. 2. Find H 100 and H 0 (as shown in Figure 15.18s), then calculate H 50 from the equation: H50 =

H0 + H100 2

(15.25s)

3. Find t50 from the plot. 4. Calculate Cv as follows: 0.197d2 t50 H0 = 14.08 (mm) Cv =

H100 = 13.55 (mm) H50 = 13.815 (mm) → t50 = 20 min Cv =

) ( 0.197(14.2∕2)2 0.197d2 = = 0.50 mm2 ∕ min t50 20

(15.26s)

14.1 14 13.9

H50

13.8 13.7 13.6 H 100 13.5

t50 = 20 min

13.4 0.01

0.1

1

10

1000

100

Figure 15.17s Log time method. 0

0.01

ε0

Strain, ε

0.02

x x t1

t2 = 4t1

ε50

t50

0.03

0.04 ε100 0.05

0.06

End of primary consolidation

0.1

1

10

100

Log time (min)

Figure 15.18s Strain vs. log time consolidation curve.

1000

482

15 DEFORMATION PROPERTIES

Problem 15.9 Given the strain versus time curve of Figure 15.18s, and knowing that the initial void ratio eo is 0.7, calculate the secondary compression index C𝛼 . Solution 15.9 C𝛼 =

Δe Δ log t

(15.27s)

0

0.01

ε0

Strain, ε

0.02

x x t1

t2 = 4t1

ε50

t50

0.03

0.04 ε100 0.05

0.06

End of primary consolidation

0.1

1

10

100

1000

Log time (min)

Figure 15.19s Strain vs. log time consolidation curve.

From Figure 15.19s: 𝜀1 = 0.04696 t1 = 377(min ) 𝜀2 = 0.0499 t2 = 1400(min ) ΔH Δe ΔH = → ΔH = 𝜀1 H0 − 𝜀2 H0 → = Δ𝜀 H0 1 + e0 H0 Δe 1 + e0 Δe (0.0499 − 0.04696) = → Δe = 0.004998 1 + 0.7

(15.28s)

Δ𝜀 =

(15.29s)

Recalling Eq. 15.26s: C𝛼 =

0.004998 = 0.008772 log(1400) − log(377)

(in 1∕ log(minutes))

Problem 15.10 Devise a pressuremeter test procedure that allows you to measure as many parameters as possible for Eq. 15.20. Solution 15.10 Strain level influence: A PMT can be run by performing unload-reload cycles around a chosen mean borehole pressure. The cycles would be of increasing amplitude to vary the associated strain, but always around the same mean pressure so that the mean stress level would not change and the influence of the strain amplitude would be isolated. This would quantify the influence of the strain level.

15.19 COLLAPSE DEFORMATION BEHAVIOR

483

Stress level influence: A PMT can be run by performing unload-reload cycles with the same strain amplitude but at different stress levels. The loops of the cycles would have the same amplitude of Δ(ΔR∕Ro ), but would take place at increasingly higher pressure over limit pressure ratio. This would isolate the influence of the stress level. One must be cautious here and realize that because the pressuremeter test is primarily a shear test, the influence of the stress level would not be the influence of the confinement level. Time influence: A PMT can be run by holding a chosen pressure p and recording the relative increase in probe radius ΔR/Ro as a function of time t. This ΔR∕Ro vs. t curve will give information on the time dependency of the soil deformation. Cycle influence: A PMT can be run by performing cycles between two chosen pressure levels. The evolution of the relative increase in radius ΔR∕Ro with the number of cycles N will give a quantification of the sensitivity of the soil to cyclic loading. It is very important, when running these kinds of PMTs, to keep in mind the analogy or difference between the stress path and deformation process around the pressuremeter and in the geotechnical project. The closer the analogy, the more useful the information. Problem 15.11 Give the range of shrink-swell modulus that can be expected for soils. Use that range and the range of shrink-swell indices in Table 15.19 to give the range of expected relative volume change in shrink-swell soils. Solution 15.11 The shrink-swell modulus (Ess ) is a constant for a given soil and does not vary much from soil to soil, with values in the range of 0.5 to 1. Table 15.1s shows the expected range of relative volume change, Δ(ΔV∕V), for the corresponding range of shrink-swell indices (I ss ): ) ( I Δw ΔV = Δ = ss (15.30s) V Ess Ess Table 15.1s Relative change in volume of a soil in percent for various values of shrink-swell modulus and shrink-swell index

Shrink-swell modulus

Shrink-swell index

0.5 0.6 0.7 0.8 0.9 1.0

0–15

15–30

30–45

45–60

>60

0–30 0–25 0–21.4 0–18.8 0–16.7 0–15

30–60 25–50 21.4–42.9 18.8–37.5 16.7–33.3 15–30

60–90 50–75 42.9–64.3 37.5–56.3 33.3–50 30–45

90–120 75–100 64.3–85.7 56.3–75 50–66.7 45–60

>120 >100 >85.7 >75 >66.7 >60

Problem 15.12 Which of the following two soils is the most likely to collapse upon wetting? a. Silt with a dry unit weight of 14 kN/m3 , a liquid limit of 40%, a plastic limit of 20%, a porosity of 50%, and a natural water content of 10%. b. Clay with a dry unit weight of 16 kN/m3 , a liquid limit of 55%, a plastic limit of 20%, a porosity of 35%, and natural water content of 20%. Solution 15.12 Use the USACE (1990) indicators to determine if a soil is likely to collapse. a. Silt: 1. LL = 40% (less than 45%)

484

15 DEFORMATION PROPERTIES

2. PI = 40%–20% = 20% (less than 25%) 3. 𝛾d = 14kN∕m3 (between 10 and 17) 4. n = 50% (between 40% and 60%) b. Clay 1. LL = 55% (not less than 45%) 2. PI = 55%–20% = 25% (equals 25%) 3. 𝛾d = 16kN∕m3 (between 10 and 17) 4. n = 35% (not between 40% and 60%) According to these guidelines, the silt is more likely to collapse than the clay.

CHAPTER 16

Shear Strength Properties

16.1

GENERAL

Three strengths are usually considered for a material: compressive strength, tensile strength, and shear strength. Compressive strength is tested by applying an all-around pressure (hydrostatic loading) on a sample and recording the pressure at which the sample fails. In general, soils are very strong in all-around compression. Exceptions include soils with a very loose structure and a slight cementation, such as calcareous sands; under such loading, these soils can collapse on themselves and crush with a drastic reduction in volume. (For comparison purposes, other materials that are weak in compression are puffed rice and marshmallow.) Tension strength is tested by pulling on a sample at both ends. In general, soils are very weak in tension. This mode of failure does not often control the behavior of soils, however, because tensile stresses between the grains are rare, due in part to gravity stresses that impose a natural prestressing in the deposit. If tensile stresses develop between the grains, they first correspond to a decrease in compression rather than true tension. Tensile cracks do develop at the top of failing slopes or in shrinking soils near the ground surface. Shear strength can be tested by moving the top part of a sample with respect to the bottom part of a sample in the direction of the plane separating the top from the bottom. Most often, the shear strength is what controls the ultimate loads in geotechnical engineering projects. This is why it is so important to the geotechnical engineer. As an added example to convince you, think of the unconfined compression test. The soil sample is loaded vertically and has no lateral pressure applied. When the vertical stress becomes too high, the sample fails along a diagonal where the shear stress reaches the shear strength. So, even though the loading is compression, the failure is in shear. By comparison with concrete and steel, the strength of soil is very small (Table 16.1). Where does the shear resistance come from in a soil mass? It cannot be from the shearing resistance of the air or the water, because these shear resistances are negligible. In fact, it comes from the shearing resistance at the particle-to-particle contacts. The particles are pressed against each other by

Table 16.1

Strength of soils, concrete, and steel (in kPa)

Material

Shear strength

Unconfined compression strength

Tensile strength

Soil Concrete Steel

5–500 750* –1,100* 230,000

10–1,000 20,000–40,000 250,000

0–100 2,000–4,000 400,000

*This low value is explained in Section 16.4.

normal forces and the shear resistance is due in large part to the friction at the contacts. The normal stress between particles is quantified by the effective stress, and therefore one component of the shear strength is the product of the effective stress on the plane of failure times the coefficient of friction of the interface. The second component of the shearing resistance at the contacts is the glue that may exist at the contacts. This glue may be real, as in the case of calcium cementation, or apparent, as in the case of water tension between the particles that pulls the particles together when the soil dries. The apparent cohesion is in fact a part of the friction resistance, as the effective stress is enhanced by the tension in the water. As will be seen, many factors can affect the shear strength of a soil. The best way to obtain the shear strength of a soil is to measure it directly by a laboratory test or an in situ test and by reproducing in the test the same stress conditions as those anticipated in the field. Any shear strength parameter should be quoted by explaining how it was measured and over what stress range the soil was tested.

16.2 16.2.1

BASIC EXPERIMENTS Experiment 1

If a block of concrete with a weight N is placed on a concrete floor (Figure 16.1), the force F necessary to initiate motion by

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

485

486

16 SHEAR STRENGTH PROPERTIES N

N

N F

F

Glue

μN Experiment 1

F

Glue

μN

C

If I divide both sides of Eq. 16.5 by the total area A, I get:

Water

C uw > 0

Experiment 2

μN uwA

C (N − uw A) F = + tan 𝜑 or A A A 16.2.4

Experiment 3

𝜏f = c + (𝜎 − uw ) tan 𝜑 (16.6)

Experiment 4

dragging the concrete block on the concrete floor is given by:

Let’s repeat that last experiment, but this time, before I place the concrete block, I dry off some of the water in the holes such that the little amount of water that is left is held in the holes by tension in the water (uw < 0) (Figure 16.1). This creates a suction force between the block and the concrete floor that increases the force F necessary to move the block. This force is equal to the water tension times the area over which the water exists. This area is a fraction 𝛼 of the total area A and is represented by 𝛼A where 𝛼 is less than one. Equation 16.1 becomes:

F = 𝜇N

F = C + 𝜇(N − 𝛼uw A)

N

N

Water

N F

F

F

Glue uw < 0

αuwA

Experiment 4

μN

μdN

Experiment 5

μN

μcN

Experiment 6

Figure 16.1 Basic experiments.

(16.1)

where 𝜇 is the coefficient of friction of the concrete-toconcrete interface, F is the shear force, and N is the normal force. By dividing both sides of Eq. 16.1 by the interface contact area A, and replacing 𝜇 by tan 𝜑, the equation becomes: N F (16.2) = tan 𝜑 or 𝜏f = 𝜎 tan 𝜑 A A 16.2.2

Experiment 2

Imagine that before I place the concrete block on the concrete floor, I paint a layer of glue on the concrete floor (Figure 16.1) and then I place the concrete block on the glue and I let it set. Now I repeat the experiment and exert a force F, higher than in the first case because of the glue, to drag the block. Then Eq. (16.1) becomes: F = C + 𝜇N

(16.3)

where C is the force required to break the glue. If I divide both sides of Eq. 16.3 again by the total area A and use tan 𝜑 instead of 𝜇, I get: C N F (16.4) = + tan 𝜑 or 𝜏f = c + 𝜎 tan 𝜑 A A A 16.2.3

Experiment 3

Imagine now that I make some small holes on the concrete floor, that I paint the glue only on the top of the bumps between holes, and that I flood the holes with water before I place the block (Figure 16.1). When I place the concrete block on top of the concrete floor, two things happen: the glue sets and the water is squeezed between the two surfaces. If the water saturates the holes and if the water cannot escape, there will be a water compression stress uw (uw > 0) under the block and an associated uplift force uw × A, which decreases the normal force on the sliding plane. Equation 16.1 then becomes: (16.5) F = C + 𝜇(N − uw A)

(16.7)

If I divide both sides of Eq. 16.7 by the total contact area A, I get F C (N − 𝛼uw A) = + tan 𝜑 or A A A 16.2.5

𝜏f = c + (𝜎 − 𝛼uw ) tan 𝜑 (16.8)

Experiment 5

Let’s go back to experiment 1, but this time we design some special grooves in the concrete floor and matching grooves at the bottom of the concrete block. These grooves are inclined as shown in Figure 16.1, such that to move the concrete block, the block has to be pushed sideways and upward. This type of interface increases the shear force F necessary to move the block as follows. The friction force 𝜇N is still necessary, but a force 𝜇d N has to be added to overcome the roughness of the upward grooves. The subscript d is for dilation. The coefficient 𝜇d is equal to tan 𝜓 where 𝜓 is the angle of the groove with the horizontal. Here is why. Referring to Figure 16.2, the friction force T is equal to the normal force N cos 𝜓 times the coefficient of friction tan 𝜑; this is the constitutive law: T = N cos 𝜓 tan 𝜑

(16.9)

Then equilibrium in the direction of the force T can be written; this is the fundamental law: T + N sin 𝜓 = F cos 𝜓

(16.10)

By combining Eqs. 16.9 and 16.10, we get: F = N tan 𝜑 + N tan 𝜓 N

Ψ

Ψ

Ψ N

Ψ F T V

(16.11)

F

T V

Figure 16.2 Concrete block on slopes.

Ψ

487

16.3 STRESS-STRAIN CURVE, WATER STRESS RESPONSE, AND STRESS PATH

If we divide both sides of Eq. 16.11 by the total contact area, we get:

𝜏f = c + 𝜎 tan 𝜑 − 𝜎 tan 𝜓 = c + 𝜎 tan(𝜑 − 𝜓)(1 + tan 𝜑 tan 𝜓)

𝜏f = c + 𝜎 tan 𝜑 + 𝜎 tan 𝜓 = c + 𝜎 tan(𝜑 + 𝜓)(1 − tan 𝜑 tan 𝜓)

Again, if 𝜓 is relatively small, the term (1 + tan 𝜑 tan 𝜓) is close to 1.

(16.12)

Note that if 𝜓 is relatively small, the term (1 − tan 𝜑 tan 𝜓) is close to 1. 16.2.6

16.3 STRESS-STRAIN CURVE, WATER STRESS RESPONSE, AND STRESS PATH

Experiment 6

The stress-strain curve of a soil depends on a number of factors, including the soil stress history, the current stress level, the structure of the soil, and others. Two types of curves are usually encountered. The first exhibits a peak followed by a strain softening region; the second does not exhibit a peak but simply an increase toward a plateau at large strains (Figure 16.3). Overconsolidated soils, hard soils, and dense soils have curves exhibiting peaks (brittle), whereas normally

Let’s repeat experiment 5, but now with the grooves slanted in the other direction (Figure 16.1). This time the downward slope creates a force that decreases the value of F. Equation 16.11 becomes: F = N tan 𝜑 − N tan 𝜓

(16.13)

If we divide both sides of Eq. 16.13 by the total contact area, we get:

(a)

(b) 200

200

Normally consolidated or loose

Deviator stress: σ1–σ3 (kPa)

Overconsolidated or dense

150

150

100

100

Maximum or peak strength

50 0

0

2

4

Residual strength

6

8

50

10

75

0

0

2

4

Water stress (pore pressure), uw (kPa)

50

25

25

0

2

4

6

8

10

–25

10

0

0

2

4

6

8

10

–25

4

4 Normally consolidated or loose

Overconsolidated or dense

Volumetric strain, εν (%)

8

Normally consolidated or loose

50

2 0

6

75 Overconsolidated or dense

0

(16.14)

2

0

2

4

6

8

10

0

–2

–2

–4

–4

0

2

4

6

8

Figure 16.3 Stress-strain curves in soils: (a) Overconsolidated or dense. (b) Normally consolidated or loose.

10

16 SHEAR STRENGTH PROPERTIES

consolidated soils, soft soils, and loose soils have curves exhibiting no peak (ductile). For the same soil, under the same confinement, but for an overconsolidated and normally consolidated case, both curves tend to reach a common strength at large strain (Figure 16.3). This point is called the critical state. At that point the soil does not change volume while shearing. The water stress exhibits two different types of behavior for these two distinct types of curves. In the case of the curve with no peak, the soil compresses throughout the shearing process and the water goes into compression, thereby reducing the effective stress. In the case of the curve with a peak, the water goes into compression initially (reduction in effective stress) and then the soil starts to dilate; the associated increase in volume creates a decrease in water stress that ends up as tension. As a result, the effective stress increases. Note that water stress is not always measured during such tests. Nevertheless, the water stress is necessary for proper reduction of the data in terms of effective stress. The stress path in two dimensions is the path described by the top of the Mohr circle. It describes the evolution of certain stresses throughout the loading of the sample. Specifically, it tracks the path described by the points with p, q stress coordinates where p and q are defined as: 𝜎 + 𝜎h 𝜎 + 𝜎3 or p = v (16.15) p= 1 2 2 𝜎 − 𝜎h 𝜎 − 𝜎3 q= 1 or q = v (16.16) 2 2 where 𝜎 v and 𝜎 h are the vertical and horizontal total stresses in a triaxial test, for example. The most useful stress paths are plotted in terms of effective stresses (p′ and q′ ): p′ = q′ =

𝜎1′ + 𝜎3′ 2 𝜎1′ − 𝜎3′

or

p′ =

𝜎v′ + 𝜎h′ 2 𝜎v′ − 𝜎h′

(16.17)

or q′ = (16.18) 2 2 where 𝜎v′ and 𝜎h′ are the vertical and horizontal effective stresses in a triaxial test, for example. Examples of effective stress paths for different types of tests are shown in Figure 16.4. In any laboratory test, it is most desirable to match the effective stress path followed by the soil in the field during the project construction and the project life. Examples of field stress paths are shown in Figure 16.5. Stress path OA would be the case of the wetting of an unsaturated soil or the filling of an earth dam reservoir. Stress path OB might be associated with a slow excavation process. Stress path OC would correspond to a rapid embankment construction. Stress path OD would be the case of a slow embankment construction. As can be seen from Figure 16.5, stress paths OC and OB are those that will approach the failure envelope the fastest, because they go toward the strength envelope with the shortest distance. The shear strength 𝜏 f of a soil is defined as the highest shear stress the soil can resist. For the curve with a peak, it will

Half deviator stress q = qʹ= ( σ1– σ3) /2 (kPa)

488

Effective stress path (lightly overconsolidated) Effective stress path (normally consolidated)

Effective stress path (heavily overconsolidated) Total stress path

Mean effective stress pʹ= (σʹ1 + σʹ3) /2 (kPa)

Figure 16.4 Stress paths for an overconsolidated and a normally consolidated soil.

nve th E

e lop

ng

Stre

1 qʹ = (σʹ1 – σʹ3) 2

B A Koσ’0v

C D O σ’0v

pʹ =

1 ʹ (σ 1 + σʹ3) 2

Figure 16.5 Example of field stress paths.

be the shear stress corresponding to the peak of the curve, known as the peak shear strength. For the curve with no peak, it is the shear stress at large strain; a value of 10% strain is often used when no obvious plateau is reached. The residual shear strength of a soil is defined only when the curve has a peak. In this case, the value of the shear stress corresponding to the post-peak plateau is the residual shear strength. The remolded shear strength is the shear strength of the remolded soil. The remolded shear strength can be equal to the residual shear strength, but more often it is less than the residual shear strength.

16.4 16.4.1

SHEAR STRENGTH ENVELOPE General Case

Each one of the experiments described in Section 16.2 has a parallel for soils. Imagine now that the interface, instead of concrete on concrete, is a plane in a soil with no water. The shear strength that can be generated by the soil will definitely have a component due to friction, as explained in experiment 1: (16.19) 𝜏f = 𝜎 tan 𝜑 where 𝜏 f is the soil shear strength, 𝜎 is the normal stress on the plane of failure, and 𝜑 is the friction angle. Recall that tan 𝜑 is a coefficient of friction and as such is often between 0 and 1, although we will see later that it could actually be higher than 1. The glue added in experiment 2 refers to

489

16.4 SHEAR STRENGTH ENVELOPE

τ

any cohesion that may exist between the soil particles. This cohesion is relatively rare, and when it is not zero, it is quite small (5–20 kPa). The cohesion plus the friction give: 𝜏f = c + 𝜎 tan 𝜑

(16.20)

The water added in experiment 3 refers to the case where the voids between the soil particles are full of water or 100% saturation. In this case, the water is under a certain amount of pressure uw (compression below the groundwater level, GWL, or tension above the GWL) that changes the effect of the normal stress. The normal stress 𝜎 becomes the effective normal stress 𝜎 ′ : the difference between the total normal stress 𝜎 and the water stress uw . Also, the cohesion becomes the effective stress cohesion c′ and the friction angle becomes the effective stress friction angle 𝜑′ : 𝜏f = c′ + (𝜎 − uw ) tan 𝜑′ = c′ + 𝜎 ′ tan 𝜑′

(16.21)

In experiment 4, the water no longer filled the voids and covered a fraction 𝛼 of the total area. As a result, Eq. 16.21 is modified because the expression of the effective stress 𝜎 ′ has changed: 𝜏f = c + (𝜎 − 𝛼uw ) tan 𝜑 = c + 𝜎 tan 𝜑 ′

′

′

′

′

(16.22)

Experiment 5 conveys an important message regarding soil shear strength: the concept of dilatancy. When a very dense soil is sheared, it tends to increase in volume or dilate. This is due to each particle having to climb over the one in front of it during shearing. This increase in volume is associated with a lifting effect similar to that of the concrete block and increases the shear strength compared to a no-volume-change situation. The shear strength equation becomes: 𝜏f = c′ + 𝜎 ′ tan 𝜑′ + 𝜎 ′ tan 𝜓 ′

(16.23)

where 𝜓 ′ is the effective stress dilatancy angle. If 𝜓 ′ is small, Eq. 16.23 can be rewritten as: 𝜏f = c′ + 𝜎 ′ tan(𝜑′ + 𝜓 ′ )

(16.24)

In experiment 6, the problem of dilatancy became a problem of compression and the term 𝜎 ′ tan 𝜓 ′ had to be subtracted rather than added. In geotechnical engineering, it is common instead to use a negative value of 𝜓 ′ and keep Eq. 16.24 the same. In the general case, the shear strength of soils is measured and the effects of dilatancy or compression are absorbed in the value of 𝜑′ . The general equation for the shear strength of soils is therefore: 𝜏f = c′ + 𝜎 ′ tan 𝜑′

φʹ

s= cʹ + σʹ tan φʹ

(16.25)

where 𝜏 f is the shear strength, c′ is the effective stress cohesion intercept, 𝜎 ′ is the effective stress normal to the plane of failure (𝜎 − 𝛼uw ), and 𝜑′ is the effective stress friction angle. This equation works for all soils in all situations, including saturated or unsaturated, drained or undrained, dilative or compressive. If c′ and 𝜑′ are considered to be constants, then Eq. 16.25 is a straight line on the 𝜏 vs. 𝜎 ′ set of axes (Figure 16.6) and is referred to as the strength envelope. Any

τf

cʹ σʹ3

σʹ

σʹ1

σʹ = σ – αuw

Figure 16.6 Shear strength envelope for soils.

stress point below or on the envelope is possible, but it is not possible for any stress point to plot above that line. Thus, any failure Mohr circle will have to be tangent to the failure envelope. 16.4.2

The Case of Concrete

Concrete has a very large cohesion intercept compared to that of soils. Figure 16.7 shows the Mohr circle for an unconfined compression test on concrete (e.g.: fc = 35000 kPa) on one side and for an unconfined tension test on concrete (e.g.: ft = 3500 kPa) on the other. The shear strength envelope for concrete is also shown conceptually. The value of the shear strength used in the code is given by the equation s(kPa) = 5.25(fc (kPa)0.5 ) = 982 kPa for the values in the example. This value is shown on the strength envelope (Figure 16.7) and is associated with a significant tension. The reason is that in concrete beam design, shear typically occurs in sections near the supports where tension is large. It would be like using Eq. 16.25 with a large tension for the normal stress; this would decrease the shear strength significantly. This is why the shear strength of concrete in Table 16.1 τ (kPa)

20000

10000 A 10000

τF 10000

20000

30000

40000

Figure 16.7 Strength envelope for concrete.

σ(kPa)

490

16 SHEAR STRENGTH PROPERTIES

is quite small—much less than one-half the unconfined compression strength. 16.4.3

Overconsolidated Fine-Grained Soils

A special case occurs with overconsolidated soils where the shear strength envelope does not quite follow the straight line of Eq. 16.25. These soils exhibit a preconsolidation pressure 𝜎p′ , as measured in the consolidation test. For stresses less than 𝜎p′ , deformations are small; for stresses higher than 𝜎p′ , deformations are much larger for the same increase in effective stress. The preconsolidation pressure can be thought of as a yield stress on the consolidation stress-strain curve. This yield stress also affects the shear strength envelope. Indeed, when the effective stress 𝜎 ′ on the plane of failure is less than 𝜎p′ , the cohesion intercept found in many overconsolidated clays exists. However, when the effective stress 𝜎 ′ on the plane of failure is larger than 𝜎p′ , the cohesion intercept is destroyed by the stress level that destructures the soil, and the envelope goes through the origin (Figure 16.8). Others have proposed that the envelope be curved as shown in Figure 16.8. Mesri and Abdelghafar (1993) proposed an empirical equation, for stresses less than 𝜎p′ , that takes into account the overconsolidation ratio on the drained shear strength, as follows: ( ′ )1−m 𝜎p ′ ′ 𝜏f = 𝜎 tan 𝜑 (16.26) 𝜎′ where m is a shear strength coefficient given in Table 16.2. τ

16.4.4

Coarse-Grained Soils

A special case also arises for coarse-grained soils where the shear strength envelope does not quite follow the straight line of Eq. 16.25. These soils tend to dilate during shear at a low confinement level and compress at higher confinement levels. The stress level at which the change between dilation and compression occurs depends on the relative density of the coarse-grained soil: the higher the density, the larger the stress range over which the soil dilates. All soils end up compressing during shear at some level of stress. Recall the simplified equation including the dilation angle 𝜓 ′ : 𝜏f = c′ + 𝜎 ′ tan(𝜑′ + 𝜓 ′ )

(16.27)

Because 𝜓 ′ is positive at smaller stresses (dilation) and becomes negative at higher stresses (compression), the sum 𝜑′ + 𝜓 ′ is larger at smaller effective stresses than it is at higher effective stresses, and the shear strength envelope is curved (Figure 16.9). When the soil dilates, a distinction is made between the friction angle 𝜑′peak associated with the peak of the stress-strain curve and the post-peak large strain friction angle 𝜑′cv at which the soil reaches a point where shearing takes place at constant volume. The difference between the two is the dilation angle 𝜓 ′ : 𝜑′peak = 𝜑′cv + 𝜓 ′

(16.28)

For dense soils, 𝜑′peak is larger than 𝜑′cv and 𝜓 ′ is positive; for loose soils, 𝜑′peak is smaller than or equal to 𝜑′cv and 𝜓 ′ can be negative. In most tests, the angle 𝜑′peak is the one measured. The angle 𝜓 ′ can be inferred from the post-peak large strain shear strength that gives 𝜑′cv and then using Eq. 16.28.

φʹ2

16.5 φʹ1

UNSATURATED SOILS

For unsaturated soils, the effective stress can be calculated as explained in Section 11.13. The most general expression for

cʹ

σʹp

σʹ = σ – αuw

τ

Figure 16.8 Strength envelope for overconsolidated fine-grained soils.

Table 16.2

Parameter m for Equation 16.26

Soil

m for intact soil

m for destructured soil

Cemented soft clays Stiff clays and shales Soft clays

0.4–0.5 0.5–0.6 0.6–0.7

0.5–0.7 0.6–0.8 0.7–0.9

(Source: After Terzaghi et al.. 1996.)

Ψʹ > 0 Ψʹ < 0 τf

φʹcv

Dilation σʹ

Compression σʹ = σ – αuw

Figure 16.9 Strength envelope for coarse-grained soil.

16.6 EXPERIMENTAL DETERMINATION OF SHEAR STRENGTH (LAB TESTS, IN SITU TESTS)

the effective stress 𝜎 ′ is: 𝜎 = 𝜎 − 𝛼uw − 𝛽ua ′

(16.29)

Therefore, the general equation for the shear strength of a soil (unsaturated or saturated) is: 𝜏f = c′ + 𝜎 ′ tan 𝜑′ = c′ + (𝜎 − 𝛼uw − 𝛽ua ) tan 𝜑′

(16.30)

where 𝜎 is the normal effective stress on the plane of failure, 𝜎 is the normal total stress on the plane of failure, 𝛼 is the fraction of the total area of the failure plane covered by the water, 𝛽 is the fraction of the plane covered by the air, uw is the water stress, ua is the air stress, and 𝜑′ is the effective stress friction angle. As explained in Section 11.13, when the soil is saturated or when the air is occluded, Eq. 16.29 becomes: ′

𝜎 ′ = 𝜎 − uw

(16.31)

If the air is not occluded, there is a path for the air to be connected directly to the atmosphere and the air stress is atmospheric or zero gauge pressure. Then the most general expression of the effective stress in soils covering all real cases is: (16.32) 𝜎 ′ = 𝜎 − 𝛼uw Therefore, in all real cases for unsaturated soils (ua = 0 or ua = uw ) and saturated soils, the equation for the shear strength 𝜏 f is: 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′

(16.33)

The parameter 𝛼 can be estimated by taking it equal to the degree of saturation S (see Figure 11.16) or by using a slightly modified version of the Khalili and Khabbaz (1998) equation (see Figure 11.17): 𝛼=S (16.34) √ uwae 𝛼= (16.35) uw where uwae is the air entry value of the water tension and uw is the water tension. Shear strength equations other than Eq. 16.30 have been proposed, such as the one of Fredlund and Rahardjo (1993): 𝜏f = c′ + (𝜎 − ua ) tan 𝜑′ + (ua − uw ) tan 𝜑b

(16.36)

where 𝜑 is an angle indicating the rate of increase in shear strength relative to the matric suction ua − uw . Equation 16.36 can be reorganized as follows: ( ( ) ) tan 𝜑b tan 𝜑b ′ u − 1− ua tan 𝜑′ 𝜏f = c + 𝜎 − tan 𝜑′ w tan 𝜑′ (16.37) Comparison of Eq. 16.37 with Eq. 16.30 shows that the two equations are identical if: b

tan 𝜑b tan 𝜑b and 𝛽 = 1 − (16.38) tan 𝜑′ tan 𝜑′ You may recall from Section 11.13, Eq. 11.52, that 𝛼 + 𝛽 = 1; therefore, both conditions are satisfied automatically and the ratio tan 𝜑b ∕ tan 𝜑′ can be estimated through Eqs. 16.34 and 16.35. 𝛼=

491

16.6 EXPERIMENTAL DETERMINATION OF SHEAR STRENGTH (LAB TESTS, IN SITU TESTS) There are many ways to determine the effective stress shear strength parameters of soils. Because many factors influence the shear strength, it is best to aim at reproducing the initial stress conditions and the stress path during loading, while matching the drainage conditions to be encountered in the field. In the laboratory, the most common tests are the unconsolidated undrained triaxial test (UUT), the consolidated undrained triaxial test (CUT), the unconsolidated undrained direct shear test (UUDS), the consolidated undrained direct shear test (CUDS), the unconsolidated undrained simple shear test (UUSS), and the consolidated undrained simple shear test (CUSS). Unconsolidated means that no drainage is allowed when the confining pressure is applied; consolidated means that drainage is allowed during application of the confining pressure until the excess water stress has come back down to zero. The second letter in the acronym refers to the loading process; for example, a consolidated undrained test means that the loading process is done while allowing no drainage. Table 16.3 shows which test and test requirements are applicable to determining which shear strength parameters for saturated and unsaturated soils. Note that if the water in the soil voids is in tension (saturated or unsaturated), additional measurements are necessary to obtain the effective stress shear strength parameters. These additional measurements include the measurement of the water tension uw , the air entry water tension uwae , and the water content w or degree of saturation S. The reason is that the equation for the shear strength is: (16.39) 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ √ which requires estimating 𝛼 as S or uwae ∕uw The undrained shear strength su is simply read as the peak shear stress reached during an undrained test. The effective stress shear strength parameters (c′ , 𝜑′ ) require plotting the results on the shear stress 𝜏 vs. effective normal stress 𝜎 ′ , as shown in Sections 10.9, 10.10, and 10.12. In situ tests (see Chapter 8) can also be used to obtain the shear strength of soils. The most direct tests are the vane shear test (VST) and the borehole shear test (BST). The VST is simple and can be used to obtain the undrained shear strength of fine-grained soils. The BST is a bit more complicated but can be used to obtain the effective stress friction angle of coarse-grained soils. The BST can also be used for the undrained shear strength of saturated fine-grained soils by conducting a rapid test, and the effective stress shear strength parameters of saturated soils by conducting a test slow enough not to generate water stress. Water stress is not typically measured during the BST or the VST. Other tests such as the standard penetration test (SPT) and the cone penetration test (CPT) can be used to obtain shear strength parameters through correlations. For example, the blow count N of the SPT and the point resistance qc of the CPT have

492

16 SHEAR STRENGTH PROPERTIES

Table 16.3

Laboratory tests for shear strength determination of saturated and unsaturated soils

Test

Measurements

Shear strength

Comments

Direct shear test, unconsolidated undrained Direct shear test, consolidated undrained Direct shear test, consolidated drained

Normal stress, shear stress

su

Effective stress 𝜎 ′ = existing 𝜎 ′ in sample

Normal stress, shear stress

su

Effective stress 𝜎 ′ = chosen 𝜎 ′ for confinement

Normal stress, shear stress

c′ , 𝜑′

Simple shear test, unconsolidated undrained Simple shear test, consolidated undrained Simple shear test, consolidated drained

Normal stress, shear stress, displacement Normal stress, shear stress, displacement Normal stress, shear stress, displacement

su and complete stress-strain curve

Estimate of dilatancy angle 𝜓 ′ if horizontal and vertical displacements measured. If water is in tension, measurements of water tension, air entry water tension, and water content are also necessary. Effective stress 𝜎 ′ = existing 𝜎 ′ in sample

Triaxial test, unconsolidated undrained

Vertical stress, confinement stress, displacement Vertical stress, confinement stress, displacement Vertical stress, shear stress, displacement

su , complete stress-strain curve, and c′ , 𝜑′ if water stress measured su , complete stress-strain curve, and c′ , 𝜑′ if water stress measured ′ c , 𝜑′ and complete stress-strain curve

Triaxial test, consolidated undrained Triaxial test, consolidated drained

su and complete stress-strain curve

Effective stress 𝜎 ′ = chosen 𝜎 ′ for confinement

c′ , 𝜑′ , and complete stress-strain curve

Estimate of dilatancy angle 𝜓 ′ if horizontal and vertical displacements measured. If water is in tension, measurements of water tension, air entry water tension, and water content are also necessary. Effective stress 𝜎 ′ = existing 𝜎 ′ in sample

been used to estimate the friction angle of coarse-grained soils, as well as the undrained shear strength of fine-grained soils.

16.7 ESTIMATING EFFECTIVE STRESS SHEAR STRENGTH PARAMETERS The parameters referred to in this section are the effective stress cohesion intercept c′ , the effective stress friction angle 𝜑′ , and the effective stress dilation/compression angle 𝜓 ′ . 16.7.1

Coarse-Grained Soils

For coarse-grained soils, the effective stress cohesion intercept c′ is considered to be equal to zero, which often leads

Effective stress 𝜎 ′ = chosen 𝜎 ′ for confinement

Estimate of dilatancy angle 𝜓 ′ if volume change measured. If water is in tension, measurements of water tension, air entry water tension, and water content are also necessary.

to coarse-grained soils being called cohesionless soils. The parameter 𝜑′ controls the shear strength of these soils, along with the normal effective stress on the plane of failure. The friction angle 𝜑′ for coarse-grained soils varies between 25 and 50 degrees. Recall that tan 𝜑′ is the coefficient of friction 𝜇, which varies correspondingly between 0.5 and 1.2. A coefficient of friction higher than 1 is possible in soils because of the dilatancy effect, which combines friction and lifting. Tables 16.4 and 16.5 as well as Figure 16.10 give suggested values of the friction angle for coarse-grained soils. These are values of 𝜑′ typically obtained in a triaxial test or a direct shear test. Note that the value of 𝜑′ obtained in a plane strain test is about 10% higher than the one obtained in a triaxial compression test. The reason is that in the plane strain deformation process, the particles are forced to move in a

493

16.7 ESTIMATING EFFECTIVE STRESS SHEAR STRENGTH PARAMETERS

Range of values for 𝝋′

Table 16.4 Soil

Friction angle in degrees

Gravel with sand Sand, angular grains, well graded Sand, round grain, uniform Silty sand Inorganic silt

Coefficient of friction

Loose

Dense

Loose

Dense

35 33 27.5 27–33 27–30

50 45 34 30–34 30–35

0.7 0.65 0.52 0.51–0.65 0.51–0.58

1.2 1.0 0.67 0.58–0.67 0.58–0.7

(Source: After Terzaghi and Peck, 1967.)

Table 16.5

Guide for values for 𝝋′ Gravels and sands strength

Description

𝜑′ ∘ includes 𝜓 ′ (∘ )

𝜓 ′ (∘ )

N(bpf)

Simple field test*

Very loose

50

12 mm diameter rebar pushed in 0.3 m by hand. Shows definite marks of footsteps; hard to walk on. 12 mm diameter rebar pushed in 0.1 m by hand. Shows footsteps. 12 mm diameter rebar driven 0.3 m with carpenter hammer. Footsteps barely noticeable. 12 mm diameter rebar driven 0.1 m with carpenter hammer. No marks of footsteps. 12 mm diameter rebar driven 0.03 m with carpenter hammer. No marks of footsteps.

*Note that these tests are performed at the ground surface of the gravel-sand deposit, not on a sample.

for a strip footing and for a circular or square footing:

45°

ilt d-s san l velnd a e a r v s g gra ded um form gra edi Uni Well- nd -m a aded s d n rse l-gr sa coa Wel nd finea edrm s o f i m d Un diu -gra me ell rm Wand o f i es Un fin m or f i Un

40° Friction angle, 35° φ°

30° 0

20 40 60 80 100 Relative density, Dr (%)

Figure 16.10 Friction angle vs. relative density. (Source: From Schmertmann, 1975.)

restricted two-dimensional domain and cannot find the path of least resistance in three dimensions. Thus, the resistance is slightly higher and so is the friction angle. An application of this observation is in the difference between the friction angle

𝜑′plane

strain

= 1.1 × 𝜑′triaxial

compression

(16.40)

The dilation/compression angle 𝜓 is typically included in the measurement of the friction angle 𝜑′ . Therefore, it should not be added to the measured value of 𝜑′ . The following relationship between the two angles has been used: ′

𝜓 ′ = 𝜑′ − 30

(16.41)

Houlsby (1991) presents a plot (Figure 16.11) indicating that Eq. 16.41 should be modified to: 𝜓 ′ = 𝜑′ − 34

(16.42)

In any case, the angle 𝜓 ′ varies between -5 for very loose soils to +15 degrees for very dense soils. The parameter 𝜑′ can be measured directly in situ by using the BST. The BST may be the only tool that can give a direct measure of 𝜑′ for coarse-grained soils (see Section 8.6). The parameter 𝜑′ has also been correlated with in situ test results

494

16 SHEAR STRENGTH PROPERTIES

44

50

40 Friction 36 angle, φ' 32

45

φ’=28+0.27N

28 0 20 40 60 SPT blow count (b/0.3 m)

φʹ

40

Figure 16.12 Correlation between the SPT blow count N and the friction angle 𝜑′ for coarse-grained soils. (Source: After Terzaghi and Peck, 1967.)

35

30

50

Note: φʹ includes Ψʹ

0

5 Ψʹ

10

15

Figure 16.11 Peak friction angle vs. dilation/compression angle. (Source: After Houlsby, 1991.)

including the SPT blow count N and the CPT point resistance qc . It is not recommended to use the PMT limit pressure pL to obtain the friction angle. Using N or qc to obtain 𝜑′ requires understanding the following. The shear strength of a coarse-grained soil is expressed as: s = 𝜎 ′ tan 𝜑′

(16.43)

Therefore, there are two components involved in the soil response to the SPT or CPT: the effective stress level 𝜎 ′ at the depth of the test and the frictional characteristics of the soil tan 𝜑′ . Hence, it is important to extract the influence of 𝜎 ′ from N and qc before correlating them with 𝜑′ . The corrections for the influence of 𝜎 ′ on N were discussed in Section 8.1: ( ′ )−0.5 𝜎ov N1 = Nmeasured × (16.44) pa where N 1 and N measured are the corrected and uncorrected ′ values of the SPT blow count respectively, 𝜎ov is the vertical effective stress at the depth of the test, and pa is the atmospheric pressure used for normalization. There are other ways to include the influence of the stress level in the correlation. The following is a correlation between N and 𝜑′ that incorporates the stress level influence separately; it was proposed by Schmertmann (1975) and formulated into an equation by Kulhawy and Mayne (1990): 0.34

⎞ ⎛ ⎟ ⎜ N ⎟ tan 𝜑′ = ⎜ ′ ⎟ ⎜ 𝜎ov ⎟ ⎜ 12.2 + 20.3 pa ⎠ ⎝

(16.45)

Terzaghi and Peck (1967) proposed the simple correlation shown in Figure 16.12.

Triaxial φʹ (°)

46

25 –5

42 38 34 30

0

50

100

150

200

250

300

Normalized tip stress, qc1 = qc/ (σv0ʹ)0.5

Figure 16.13 Correlation between the CPT point resistance qc and the friction angle 𝜑′ for coarse-grained soils. (Source: After Mayne, 2007.)

The cone penetrometer point resistance qc should also be corrected for the stress level (Section 8.2) before attempting correlation with the friction angle 𝜑′ (Figure 16.13). Much like the correction for N, the correction for qc is: ( ′ )−0.5 𝜎ov qc1 = qc measured × (16.46) pa Then the following correlations exist between qc1 and 𝜑′ (Mayne, 2007): ( ) qc1 𝜑′deg = 17.6 + 11 × log p ) ( a qc measured (16.47) = 17.6 + 11 × log √ ′ 𝜎ov pa 16.7.2

Fine-Grained Soils

Normally consolidated fine-grained soils have no cohesion, but some overconsolidated fine-grained soils do exhibit true cohesion c′ . It is obtained by drawing a straight line (the shear strength envelope) through the failure points from shear strength tests and determining the intercept at 𝜎 ′ = 0. Sometimes fine-grained soils are called cohesive soils, but this is misleading, as the friction component of the shear strength still dominates. In fact, it is safe to ignore the cohesion c′ for most geotechnical problems. What creates the c′

495

16.8 UNDRAINED SHEAR STRENGTH OF SATURATED FINE-GRAINED SOILS

Table 16.6 Range of possible values for the effective stress cohesion c′ of fine-grained soils

Soil

Cohesion c′ in kPa

Coarse-grained soils Silts, low plasticity Silts, high plasticity, overconsolidated Clays, normally consolidated Clays, overconsolidated, low plasticity Clays, overconsolidated, high plasticity

0 0 5–10 0 10–15 15–20

Table 16.7 Range of possible values for the effective stress friction angle 𝝋′ of fine-grained soils Soil

Friction angle φ′ in degrees

Silts, low plasticity Silts, high plasticity Clays, low plasticity Clays, high plasticity

30–38 18–30 23–31 16–26

40

value? The phenomenological reason for any “glue” between particles can be attributed to electrical forces between fine particles and to cementation that may develop through chemical reaction. These bonds are sometimes called diagenetic bonds. This parameter c′ is called true cohesion and is not to be confused with the apparent cohesion capp that comes from water tension in the voids. In fact, capp is part of the friction term in Eq. 16.33: capp = −𝛼uw tan 𝜑′

Friction angle ϕʹ, (degrees)

ϕʹ 30

20 ϕʹres 10

(16.48)

where 𝛼 is the water area ratio, uw is the water tension, and 𝜑′ is the effective stress friction angle. Because uw has a negative value, capp is positive and can be large if the soil dries enough to generate significant water tension. This water tension can reach 10,000 kPa; therefore, capp can reach hundreds of kPa. The value of c′ , in comparison, is rarely higher than 25 kPa. Table 16.6 gives some possible values for different soils. The friction angle 𝜑′ corresponding to the peak shear strength for overconsolidated fine-grained soils and to the large strain strength for normally consolidated fine-grained soils is lower than the one for coarse-grained soils and varies from 20 to 35 degrees. Table 16.7 shows some possible values of 𝜑′ for various fine-grained soils. In general, the friction angle 𝜑′ decreases when the plasticity index increases. You will realize this if you wash your hands after handling a kaolinite clay (baby powder) and then after handling a bentonite clay. The bentonite will feel a lot more slippery than the kaolinite. Figure 16.14 shows general trends of 𝜑′ with the plasticity index I p . The effective stress parameters for fine-grained soils are not obtained from in situ tests because it is difficult to ensure that the test is a drained test. One exception to this statement is the use of the borehole shear test, which is essentially a direct shear test on the wall of the borehole; in this case the test must be performed slowly enough during the consolidation phase and the shearing phase that the assumption of no excess water stress can be made.

0 0

20

40

60

80

100

Plasticity index

Figure 16.14 Effective stress friction angle 𝜑′ versus plasticity index for fine-grained soils.

16.8 UNDRAINED SHEAR STRENGTH OF SATURATED FINE-GRAINED SOILS A particular case arises when a soil is loaded fast enough that the water does not have time to drain during the loading time or if drainage is prevented in a laboratory test. In this case the shear strength is called the undrained shear strength and designated as su . This undrained case occurs rarely for most construction problems concerning coarse-grained soils, but it is often encountered with construction problems involving fine-grained soils. For example, if it takes a month to build an embankment, a clean sand would have time to fully drain, but a high-plasticity clay would not. As will be shown next, during the undrained loading of a fine-grained soil, the effective stress does not increase significantly and therefore the shear strength does not increase significantly either. Instead, the water stress increases significantly. So the controlling design case for loading on a fine-grained soil is often the undrained case, also called the short-term case. Indeed, at that time the load is maximum and the shear strength is minimum. As time goes by, the water stress decreases due to water drainage, the shear strength increases accordingly, and the factor of safety against failure increases. The critical time

496

16 SHEAR STRENGTH PROPERTIES

τ

(a) Weak soil skeleton

Stiff water

Stiff soil skeleton

Stiff water

s=

su

cʹ

cʹ

an

ʹt +σ

φʹ

φʹ

Result of an undrained test

1

σʹ = σ – αuw σʹ1

σʹ3

uw + Δσ

uw + Δσ

Figure 16.15 Model of saturated soil skeleton and water.

in the case of a fine-grained soil is immediately after loading. This is why the undrained shear strength of fine-grained soils is so important: It controls the stability design of many geotechnical structures. As pointed out before, the general equation (Eq. 16.25) applies in all cases, including the undrained case, and su can be expressed as: (16.49) su = c′ + 𝜎 ′ tan 𝜑′ The problem is that it is often difficult to obtain the effective stress on the plane of failure 𝜎 ′ . One of the important factors in this case is how compressible the soil skeleton is compared to water (Figure 16.15). 16.8.1 Weak Soil Skeleton: Soft, Normally Consolidated Soils When a load is applied rapidly to a soft, normally consolidated soil, the water picks up the entire load because the soil skeleton is too weak to contribute. Therefore, the increase in normal stress Δ𝜎 on the soil due to loading is equal to the increase in water stress Δuw . The effective stress before loading 𝜎b′ is equal to: (16.50) 𝜎b′ = 𝜎b − uwb where 𝜎 b is the total stress before loading and uwb is the water stress before loading. The effective stress immediately after loading σ′a is equal to: 𝜎a′ = 𝜎b + Δ𝜎 − (uwb + Δuw ) = 𝜎b + Δ𝜎 − (uwb + Δ𝜎) = 𝜎b − uwb = 𝜎b′

(16.51)

As can be seen, the effective stress has not increased and therefore the shear strength has not increased. The undrained shear strength of a saturated, fine-grained soil with a weak skeleton is a constant su . This statement must be qualified by adding the following: provided that the stress level (confinement) is the same, the stress history is the same (OCR), and the stress path followed to go from the initial state to failure is the same. Indeed, all three factors can influence su and selecting the correct su is more complex than often thought (Ladd, 1991).

τ

uw

uw

(b) “φ = 0” concept

s = su

su

3

2

σ σ3

σ1 σ3 + Δσ

σ1 + Δσ

Figure 16.16 Undrained shear strength for weak soil skeleton.

When a fine-grained soil with a weak skeleton is loaded in an undrained test, the Mohr circle is as shown in Figure 16.16a in the effective stress set of axes (Mohr circle 1) and as shown in Figure 16.16b in the total stress set of axes (Mohr circle 2). The difference between the total stress and the effective stress is the water stress uw . If a second undrained test is performed on the same soil but after increasing the confining pressure by Δ𝜎 (Mohr circle 3), then the water stress also increases by Δ𝜎, the effective stresses do not change, and the effective stress Mohr circle does not change (still Mohr circle 1 on Figure 16.16a). The reason why the undrained shear strength is a constant independent of the total stress is because the effective stress Mohr circle remains the same regardless of the total stress. This is often called the 𝜑 = 0 concept because the envelope on the shear strength vs. total stress set of axes is horizontal. This is not to say that such a soil is frictionless (𝜇 = tan 𝜑 = 0). It simply means that the envelope is horizontal. In reality, the soil always has a nonzero friction angle, but it shows up in the shear strength vs. effective stress set of axes only (𝜇′ = tan 𝜑′ ≠ 0). Recall that the intergranular stress is represented by the effective stress, and in that set of axes, the soil friction is identified. 16.8.2

Strong Soil Skeleton: Overconsolidated Soils

In the case of an overconsolidated soil with a strong skeleton, when the load is applied rapidly, the soil skeleton is able to resist part of the load and the water picks up the rest of the load. The increase in water stress Δuw is not as large as the

16.8 UNDRAINED SHEAR STRENGTH OF SATURATED FINE-GRAINED SOILS

increase in normal stress Δ𝜎 on the soil, and is equal to f × Δ𝜎 where f is smaller than 1. The effective stress before loading 𝜎b′ is equal to: 𝜎b′ = 𝜎b − uwb (16.52) where 𝜎 b is the total stress before loading and uwb is the water stress before loading. The effective stress immediately after loading 𝜎a′ is equal to: 𝜎a′ = 𝜎b + Δ𝜎 − (uwb + Δuw ) = 𝜎b + Δ𝜎 − (uwb + f Δ𝜎) = 𝜎b − uwb + (1 − f )Δ𝜎 > 𝜎b′

(16.53)

As can be seen, the effective stress has increased and therefore the shear strength has increased. The undrained shear strength su of a saturated, fine-grained soil with a strong skeleton increases somewhat with the total stress because the effective stress increases somewhat. Again, factors like stress level reached under drained conditions (confinement), the stress history (OCR), and the stress path followed to go from the initial state to failure influence the value of su (Ladd, 1991). When a fine-grained soil with a strong skeleton is loaded in an undrained test, the Mohr circle is as shown in Figure 16.17a in the effective stress set of axes (Mohr circle 1) and as shown in Figure 16.17b in the total stress set of axes (Mohr circle 2). The difference between the total stress and the effective stress is the water stress uw . If a second undrained test is performed on the same soil, but after increasing the confining pressure by Δ𝜎 (Mohr circle 4 in the total stress set of axes), then the water stress increases by a fraction f × Δ𝜎 of Δ𝜎, the effective stress increases somewhat, and the effective stress Mohr circle moves (Mohr circle 3 on Figure 16.17a). The reason why the undrained shear strength increases slightly with an increase in total stress is that the effective stress increases slightly. τ φʹ (a)

su2

3

su1

Undrained tests

1

τ

σʹ1 = σ1 + Δσ – uw2

σʹ3

c 0 'σʹ31

σʹ11 = σ1 – Δw1 uw1

uw2

σʹ = σ – αuw uw2

uw1

(b) 4 2

σ σ3

σ3 + Δσ σ1

σ1 + Δs

Figure 16.17 Undrained shear strength for strong soil skeleton.

497

16.8.3 Rate of Loading Effect on the Undrained Strength Soils, like many other materials, are viscous: They increase in strength when the loading rate increases. The reason is attributed to the difference in water stress being developed at slower rates and at higher rates. At higher rates, the soil grains do not have time to move by finding the path of least resistance and more dilation is generated, thereby inducing higher effective stresses and shear strength. Also, the water in the voids has viscosity of its own. Indeed, water and air are viscous as well. They are Newtonian fluids and therefore are linearly viscous. They obey the following law: •

𝜏 = 𝜂𝛾

(16.54)

where 𝜏 is the shear stress, 𝜂 is the dynamic viscosity of the • material, and 𝛾 is the strain rate. The dynamic viscosity of water at 20∘ C is 10–6 kPa.s and the dynamic viscosity of air at 20∘ C is 1.8 × 10−8 kPa.s. The kinematic viscosity v takes units of m2 / s and is defined as: 𝜂 v= (16.55) 𝜌 where 𝜌 is the mass density of the material. For water, 𝜌 is 1000 kg/ m3 and for air, it is 1.2 kg/ m3 at the Earth’s surface. Soils are much less viscous than water and air. Equation 16.54 states that if the strain rate is doubled, the shear stress resistance will also double. In soils, if the strain rate is doubled, the shear stress resistance will be increased by a few percentage points. You might think: “Then why worry about it?” The issue is that sometimes the strain rate can be multiplied by factors of 1000 or more, and in such cases the increase or decrease can be significant. Briaud and Garland (1985) proposed the following model for the undrained shear strength of fine-grained soils: ( )−n su1 t = 1 (16.56) su2 t2 where su1 and su2 are the undrained shear strengths measured in time to failure t1 and t2 respectively and n is the viscous exponent for the fine-grained soil. This exponent was correlated to the reference undrained shear strength suref (Figure 16.18). The exponent n was also correlated with other soil parameters as follows: ( ) su ref −0.22 (16.57) n = 0.044 pa n = 0.028 + 0.00060 w (16.58) n = 0.035 + 0.00066 PI

(16.59)

n = 0.036 + 0.046 LI

(16.60)

where n is the soil viscous exponent in Eq. 16.56, suref is the reference undrained shear strength taken as the one obtained with a time to failure equal to one hour, pa is the atmospheric pressure, w is the natural water content in percent, PI is the plasticity index in percent, and LI is the liquidity index

498

16 SHEAR STRENGTH PROPERTIES

(a) 0.16

(b)

0.14

0.14

0.12

n = 0.044

0.10

Viscous exponent n

0.16

su ref pa

−0.22

0.12

0.08

0.06

0.06

0.04

0.04

0.02

0.02

0.00

0.00

0

50 100 150 Reference undrained shear strength, su (kPa)

n = 0.035 + 0.00066 PI

0.10 Viscous exponent n

0.08

200

0

20

40

60

80

100

Plasticity index, PI (%)

Figure 16.18 Viscous exponent n for fine-grained soils: (a) Influence of strength. (b) Influence of plasticity.

(as a fraction, not a percent). The scatter in those correlations is significant, as shown in Figure 16.18. All in all, the most common values of the exponent n vary from 0.03 to 0.06, with 0.03 occurring for a high-strength, low-plasticity clay and 0.06 for a low-strength, high-PI clay. ′ AND THE SHANSEP 16.9 THE RATIO SU ∕𝝈OV METHOD

The undrained shear strength, like any soil shear strength, depends on the effective stress on the failure plane at the time of failure. A measure of this effective stress level is the ver′ in the field at the depth z contical effective stress at rest 𝜎ov ′ sidered. The ratio su ∕𝜎ov has been used to try to normalize the variation of su with depth and with effective stress level. For normally consolidated, saturated, fine-grained soils, the ratio has been found to vary between 0.2 and 0.35, increasing slightly with the plasticity index. Holtz et al. (2011) propose that for normally consolidated, saturated, fine-grained soils: ( ) su = 0.23 ± 0.04 (16.61) ′ 𝜎ov NC

its disturbance. One is to do a drained recompression of the ′ . This approach tends sample to the in situ effective stress 𝜎ov to give too high an su value, as the recompression decreases the sample volume and water content below its natural state. Another is the stress history and normalized soil engineering properties (SHANSEP) method developed by Ladd and Foott (1974). The method consists of four steps: 1. Determine the preconsolidation pressure 𝜎p′ from consolidation tests. 2. Test samples of the soil in consolidated undrained tests (preferably under K o consolidation) at confining pressures well above 𝜎p′ to destructure the clay and obtain the normally consolidated behavior. These tests give the value of S in Eq. 16.62. 3. Obtain the influence of OCR by overconsolidating the sample, reducing the vertical stress, and measuring su at that point. These tests give the value of m in Eq. 16.62 and both S and m are therefore known. 4. Use Eq. 16.62 to develop the su profile for the consolida′ to be encountered under the structure tion pressure 𝜎vc (e.g., foundation or embankment): ( ′ )m 𝜎p su =S ′ (16.64) ′ 𝜎vc 𝜎vc

When the overconsolidation ratio (OCR) increases above 1, ′ of the overconsolidated soil becomes higher the ratio su ∕𝜎ov ′ of the normally consolidated soil. This than the ratio su ∕𝜎ov increase is not linear, and the following relationship has been proposed (Ladd et al., 1977): For overconsolidated, saturated, fine-grained soils: ( ) su = S(OCR)m (16.62) ′ 𝜎ov OC

As was shown, many factors affect the undrained shear strength of a soil. Therefore, any undrained shear strength value should be quoted by explaining how it was measured and over what stress range the soil was tested.

where S is the ratio for a normally consolidated soil (Eq. 16.61) and m is estimated to be 0.8. Thus, for overconsolidated saturated fine grained soils: ( ) su = 0.23 × (OCR)0.8 (16.63) ′ 𝜎ov OC

The undrained shear strength for unsaturated soils is obtained by shearing the soil while preventing any drainage of air or water during the test. A distinction must be made among four categories of soils:

Several factors influence the value of the undrained shear strength, one of which is the disturbance of the sample. Several methods have been proposed for “healing” a sample from

16.10 UNDRAINED SHEAR STRENGTH FOR UNSATURATED SOILS

1. Soils where the water is in tension and the air has a continuous path to the ground surface or the boundary (typical degree of saturation S < 0.85).

499

16.11 PORE-PRESSURE PARAMETERS A AND B

τ

Stiff soil skeleton spring

τ

Very weak air spring

Very stiff water spring

φʹ

φʹ

cʹ

cʹ σʹ = σ – αuw

σʹ = σ – αuw τ

τ

Figure 16.19 Model of unsaturated soil skeleton, air, and water.

2. Soils where the water is in tension and the air is occluded (typical degree of saturation 0.85 < S < 1). 3. Soils where the water is in tension and the soil is saturated. 4. Soils where the water is in compression and the soil is saturated. Sections 16.8 and 16.9 discussed results applicable to categories 2, 3, and 4 in the preceding list. This section discusses the undrained shear strength of soils in category 1: soils that are unsaturated and where the air has a continuous path to the boundary. In this case, the model in Figure 16.19 shows that part of the total stress applied to the soil will be transferred to the soil skeleton (effective stress) because the air spring is very compressible and must be compressed before stress is transferred to the water spring. The amount of total stress transferred to the water depends on the degree of saturation of the soil. For soils with very low degrees of saturation, most of the total stress will be transferred to the soil skeleton, whereas for soils with degrees of saturation close to about 0.85, most of the total stress will be transferred to the water. This has a big impact on the undrained shear strength. Indeed, if most of the total stress imposed is carried by the soil skeleton (low degree of saturation S), then the effective stress increases nearly as much as the total stress imposed and the shear strength increases with the total stress (Figure 16.20a). If, in contrast, most of the total stress imposed is carried by the water (S approaching 0.85), then the effective stress does not increase much and the shear strength is nearly independent of the total stress (Figure 16.20b). If the degree of saturation is low but the confining stress is high enough to compress the air, including bringing it into solution, then the initially low-saturation soil will start behaving more like a saturated soil (Figure 16.20a). As a result, the undrained shear strength can be highly variable for unsaturated soils, depending on the degree of saturation and the total stress level. Note that as the fine-grained soil becomes drier, the water tension that is generated increases the effective stress between particles and therefore the undrained shear strength. In the field, cases in which an unsaturated soil would be loaded in an undrained fashion are rare and are limited to

Undrained tests

Undrained tests σ

(a)

Unsaturated soil with low degree of saturation

σ (b)

Unsaturated soil with degree of saturation approaching 0.85

Figure 16.20 Strength envelopes for unsaturated soils.

high-rate dynamic loading. The concept of undrained shear strength should be handled with care for unsaturated soils, as the total stress level influences the value, especially for soils with a low degree of saturation. For these soils, the undrained shear strength case does not support the simplifying assumption that it offers for soft saturated soils, where the undrained shear strength can be considered independent of the total stress.

16.11

PORE-PRESSURE PARAMETERS A AND B

Pore-pressure parameters have been found convenient to quantify the variation of the water stress in response to undrained loading. Skempton (1954) and Bishop and Henkel (1962) proposed the following equation linking the change in water stress Δuw due to a variation in the major principal stress Δ𝜎 1 and a variation in the minor principal stress Δ𝜎 3 : Δuw = B[Δ𝜎3 + A(Δ𝜎1 − Δ𝜎3 )]

(16.65)

where B is the pore-pressure parameter associated with an increase in confining stress Δ𝜎 3 and A is the pore-pressure parameter associated with an increase in deviator stress Δ𝜎1 − Δ𝜎3 . For saturated soils, B is close to one and A depends on the overconsolidation ratio. At failure, Af is about 1 for normally consolidated soils, decreases with OCR, and can be negative for heavily overconsolidated soils. In practice, the coefficient B is sometimes used: Δuw = B Δ𝜎v

(16.66)

where Δ𝜎v is the increase in vertical stress. The coefficient B can be assumed in the design calculations, say, 0.5, and then construction is monitored with piezometers to ensure that the water stress does not rise above B Δ𝜎v . If it does, construction

16 SHEAR STRENGTH PROPERTIES

is halted until the water stress recedes sufficiently below that value. For unsaturated soils, there is a need to distinguish between the response of the water and that of the air. Fredlund and Rahardjo (1993) propose: duw = Bw (d𝜎3 + Aw d(𝜎1 − 𝜎3 ))

(16.67)

dua = Ba (d𝜎3 + Aa d(𝜎1 − 𝜎3 ))

(16.68)

Note that Bw , Aw , Ba , and Aa all depend on the degree of saturation of the soil. Also note that all pore-pressure parameters are like moduli, in that they depend on the strain level and strain rate at which they are defined. 16.12 ESTIMATING UNDRAINED SHEAR STRENGTH VALUES There are many ways to estimate the undrained shear strength of fine-grained soils. The problem is that the value of su is not unique and depends on many factors. Nevertheless, fine-grained soils are often categorized by their undrained strength, as shown in Table 16.8. The best way to determine su is to test the fine-grained soil in the laboratory using high-quality samples and to reproduce during the tests the initial stress conditions and the stress path during loading, while assuring no drainage. As discussed earlier, common laboratory tests available to obtain su include the: 1. 2. 3. 4. 5. 6.

unconsolidated undrained triaxial test (UUT) consolidated undrained triaxial test (CUT) unconsolidated undrained direct shear test (UUDS) consolidated undrained direct shear test (CUDS) unconsolidated undrained simple shear test (UUSS) consolidated undrained simple shear test (CUSS).

In situ tests can also be used, including the vane shear test (VST), the borehole shear test (BST), the cone penetrometer test (CPT), the pressuremeter test (PMT), and the standard Table 16.8

penetration test (SPT). The VST (see Section 8.5) is the best in situ test to obtain su and is particularly useful offshore, where sample decompression upon retrieval from deep-water boreholes can decrease the undrained shear strength by up to 40% (Denk et al., 1981). Bjerrum (1972) used 14 case histories to back-calculate the full-scale undrained shear strength su (field) from embankment failures and compare it to su (VST) obtained from the VST performed at the sites. Because the values did not correspond, Bjerrum proposed a correction factor 𝜇 (Figure 16.21) as a function of the plasticity index I p : su(Field) = 𝜇su(VST)

(16.69)

Ladd et al. (1977) collected additional failure case histories and confirmed the trend. As can be seen, the correction factor indicates that su (VST) is larger than su (field); this is attributed to the facts that the rate of shearing is much higher in the VST than in the failure of the embankment and that this rate effect is more prominent in high-plasticity clays than in low-plasticity clays. Differences in the influence of anisotropy and plane strain conditions between the VST and the embankment are also contributing factors. 1.50 Correction factor, μ

500

1.25 1.00 0.75 0.50 Bjerrum’s (1972) Recommended curve

0.25 0.00

0

20

40 60 80 Plasticity index, PI (%)

100

120

Figure 16.21 Bjerrum correction factor for undrained shear strength from vane test. (Source: After Bjerrum, 1972; Ladd et al., 1977.)

Classification of fine-grained soils by undrained shear strength Silts and clays strength

Description

su (kPa)

N(bpf)

Simple field test

Very soft Soft Medium or firm Stiff Very stiff Hard Very hard

400

50

Squeezes between your fingers Easily penetrated by light thumb pressure Penetrated by strong thumb pressure Indented by strong thumb pressure Slightly indented by strong thumb pressure Slightly indented by thumbnail Not indented by thumbnail

16.12 ESTIMATING UNDRAINED SHEAR STRENGTH VALUES

25 Cone factor, Nk

The borehole shear test (see Section 8.6) can be used to obtain a value of su in situ by direct measurement as long as the shearing is performed rapidly to ensure undrained behavior. Because the normal total stress (horizontal) on the plane of failure (vertical) can be varied in the BST, the influence of the total normal stress on su (as discussed in Sections 16.8–16.10) can be evaluated. The cone penetrometer test (see Section 8.2) has also been used to obtain su . The equation used is: q − 𝜎vo (16.70) su(CPT) = c Nk

where N c is a bearing capacity factor usually taken equal to 9 for deep localized failure. One would therefore expect that N k would be 9. However, many differences between a pile point and the CPT lead N k to be quite different from 9 and quite variable. The differences include the rate of loading effect, the scale effect, and the installation procedure. The penetration of the CPT goes much faster than the pile penetration during a typical load test (Nk > 9). The cone is much smaller in size than the pile; as a result, the cone detects thinner layers than the pile, which averages the soil resistance over a larger zone (again Nk > 9); also, the cone is pushed in, whereas the pile is either driven or drilled in place. All in all, the value of N k seems to average 14 ± 5 for su being determined from Eq. 16.70 (Figure 16.22), but correlations have led to values varying from 5 to 70. The main problem is that, as discussed earlier, su is not unique, so no general correlation can be proposed. The best way to approach the problem is to run a few lab tests to obtain the right su value needed for the project, run parallel CPT soundings, generate a local correlation to obtain a site value of N k from su and qc , and then extend the results by running additional CPTs. The undrained shear strength su can also be obtained from a pressuremeter test (PMT; see Section 8.3). In this case, the limit pressure pL is used as follows: p su(PMT) = L (16.72) Np where N p is the pressuremeter factor. This factor can be taken as 7.5 in first approximation, but the relationship is nonlinear and Briaud (1992) proposed: su(PMT) (kPa) = 0.67(pL (kPa))0.75

(16.73)

Figure 16.23 shows this relationship compared to two databases. The likely reason for this nonlinearity is that for lower values of su , the fine-grained soils tend to have stress-strain curves exhibiting no peak (strain hardening behavior), whereas at higher su values the fine-grained soils tend to exhibit peak strength and post-peak softening down

15 10 Nk =

5 0

10

qc – σv0 corr.su(FV)

20

30 40 50 60 Plasticity index, PI (%)

70

80

(a) 40

Cone factor, Nk

(16.71)

20

0

where qc is the point resistance of the CPT, 𝜎 vo is the vertical total stress at the depth where qc is measured, and N k is the cone factor. This equation comes from the ultimate bearing pressure pu under a pile point: pu = Nc su + 𝜎vo

501

30

20

10

0

0

10

20

30

40

50

60

Plasticity index, PI (%) (b)

Figure 16.22 N k factor for obtaining su from CPT qc value: (a) based on data from Baligh et al., 1980; Lunne and Kleven, 1981; su mostly from vane shear tests. (b) Based on data from Aas et al., 1986; su mostly from vane shear tests.

to a residual strength. Because the limit pressure involves large strains near the cavity wall and smaller strain at some distance from the cavity wall, the strength mobilized is an average between the two. This average will tend to be higher for strain hardening soils (low su ) than for strain softening soils (high su ). The advantage of using pL to get su is that the PMT involves a larger mass of soil than most other tests in the response to the expansion; as such, it can bridge over microfissures and other small-scale features and is more representative of the mass strength. The drawback is that the test is typically more expensive than the vane test, for example. The standard penetration test (SPT; see Section 8.1) and its blow count N have also been used to obtain the undrained shear strength su . Such correlations should be used as a last resort, however. Terzaghi et al. (1996) propose the following relationship to obtain a relatively conservative value of su : su(SPT) (kPa) = 4.4N60

(16.74)

where N 60 is the blow count (blows per foot) corrected to 60% of maximum energy (see Section 8.1). Terzaghi et al. (1996) point out that for low-plasticity, fine-grained soils, the factor

502

16 SHEAR STRENGTH PROPERTIES

400

su = 0.67 pL0.75

200 150

7N

300

CH

CL

200

Su

=

7N 6.

SC-ML 100

Su =

2.4N

Terzaghi and Peck (1967)

100

0

10

20

30

40

50

SPT N-value (blows/300 m)

50 0

Sowers, 1979

=1

250

Su

Undrained shear strength, su (kPa)

Undrained shear strength, su (kPa)

300

0

500

1000

1500

2000

Figure 16.24 Correlation between su and the SPT blow count N. (Source: After Sowers, 1979; Terzaghi and Peck, 1967.)

Limit pressure, pL (kPa)

where 𝜏f res is the residual shear strength, 𝜎 ′ is the effective normal stress on the plane of failure, and 𝜑′res is the residual friction angle. The amount of reduction from 𝜑′ to 𝜑′res depends on the soil type. Loose coarse-grained soils and normally consolidated, saturated, low-plasticity, fine-grained soils do not exhibit much reduction between the friction angle and the residual friction angle. The reduction for soils with higher plasticity is more significant, as exemplified by Figure 16.25 based on data from Stark and Eid (1994) and Lupini et al. (1981).

(a)

350 300 250

su = 0.67 pL0.75

200 150 100

30

0

0

500 1000 1500 Limit pressure, pL (kPa)

2000

(b)

Figure 16.23 Correlation between su and the pressuremeter limit pressure: (a) based on data from Briaud,1992; su mostly from unconfined compression tests. (b) Based on data from Baguelin et al., 1978; su from laboratory tests and vane tests.

4.4 in Eq. 16.74 can go up to 7. Terzaghi and Peck (1967) proposed: (16.75) su(SPT) (kPa) = 6.7N Sowers (1979) presents his experience in a figure relating N and su (Figure 16.24). 16.13 RESIDUAL STRENGTH PARAMETERS AND SENSITIVITY The residual strength of a soil is the strength at very large strains long after the peak strength. It exists for the effective stress shear strength and for the undrained shear strength. The residual effective stress cohesion can be taken as zero and the residual effective stress friction angle is reduced: 𝜏f res = 𝜎 ′ tan 𝜑′res

(16.76)

Residual Friction angle, φʹres (°)

50

20

10 Data from Stark and Eid, 1994 0

0

25

50

75

100

25 50 75 Clay fraction (% < 0.002 mm)

100

Liquid limit, wL (%) 30 Residual Friction angle, φʹres (°)

Undrained shear strength, su (kPa)

400

20

10 Data from Lupini et al.,1981 0

0

Figure 16.25 Correlation between effective stress residual friction angle and soil properties. (Source: After Stark and Eid, 1994; Lupini et al., 1981.)

16.15 TYPES OF ANALYSES

where su peak is the peak undrained shear strength and su rem is the remolded undrained shear strength. Some clays are not sensitive and some are very sensitive. For example, a low-plasticity, soft kaolinite clay is unlikely to be very sensitive (St < 2), but a quick clay may have a sensitivity in excess of 20. These quick clays do have some strength when undisturbed, say su = 25 kPa, but become a thick liquid when disturbed (see Section 14.2.9). A soil with a sensitivity of less than 4 would be qualified as a low-sensitivity soil; from 4 to 10 would be medium sensitivity; 10–20 would be highly sensitive; and above 20 would be quick.

16.14

STRENGTH PROFILES

The strength profile of a soil deposit can give a lot of information about the deposit. This strength can be measured by the CPT point resistance qc , or by the SPT blow count N, or by the PMT limit pressure pL , or by the undrained shear strength su for a fine-grained soil. If the profile shows a linear increase with depth with a zero value at the surface, the deposit could be a normally consolidated, soft, fine-grained soil, as would be expected in a city like New Orleans (Figure 16.26a). If the profile goes through zero at the surface but increases nonlinearly with depth with a downward curvature, then the deposit could be a dry sand deposit (Figure 16.26b). If the profile increased linearly with depth but had a definite nonzero value at the surface, the deposit could be a fine-grained soil overconsolidated by overburden removal through erosion or through the melting of a glacier (Figure 16.26c). If the profile indicated a constant strength with depth (Figure 16.26d), the deposit could be an unsaturated silty sand where the total stress increased with depth but the water tension decreased with depth, thereby maintaining the effective stress constant and the shear strength constant with depth. It could also be

Strength

Strength

Strength

(a)

(b)

(c)

Strength

Strength

Strength

(d)

(e)

(f)

Depth

Depth

The residual undrained shear strength su res is best measured directly, either in the laboratory or in the field. In the laboratory, the best apparatus is the ring shear apparatus, which consists of a split-donut type of device. In this apparatus the top half of the donut is rotated one way while the bottom half is held in place. In this fashion, very large strains can be reached until the shear strength reaches the residual strength plateau. In the field, the vane shear test can be used. The vane is rotated until the peak shear strength su peak is obtained and then rotation continues while recording the torque. When the torque stabilizes, the residual undrained shear strength su res is reached. ASTM recommends that after reaching the residual shear strength, the remolded shear strength be obtained by rapidly rotating the vane 5 to 10 times. The remolded undrained shear strength su rem is then obtained by repeating the vane test immediately after the rapid rotations. The sensitivity St of a fine-grained soil is defined as: su peak (16.77) St = su rem

503

Figure 16.26 Soil strength profiles.

an underconsolidated soft clay. If the profile showed a curved decrease with depth near the surface followed by an increase at larger depths, the deposit could be a high-plasticity, fine-grained soil overconsolidated by desiccation near the surface but becoming normally consolidated at depth where the seasonal shrink-swell cycles no longer have an influence (Figure 16.26e). If the profile shows a strong layer near the surface and a softer layer at some depth, it could indicate the presence of a crust of the softer layer below (Figure 16.26f ). Note that these strength profiles, if rotated 90 degrees counterclockwise, represent shear strength envelopes in the case of uniform soil. Indeed, after rotation, the vertical axis represents a measure of the strength and the horizontal axis a measure of the total stress. The horizontal axis can be transformed in a measure of the effective stress if the water stresses are known (such as a hydrostatic condition, for example).

16.15

TYPES OF ANALYSES

In this chapter we have talked about effective stress, total stress, undrained strength, and drained strength. Each strength is associated with a type of analysis in design, and it is important to understand which analysis is used for what strength. The types of strength analyses typically performed in geotechnical engineering include effective stress analysis, total stress analysis, undrained analysis, drained analysis, short-term analysis, and long-term analysis. Effective stress analysis: an analysis in which the soil is considered to be made of particles, water, and air. It is the most theoretically sound analysis, but it is also the most

504

16 SHEAR STRENGTH PROPERTIES

complicated analysis, because it requires knowledge of the total stress, the water stress, and the air stress (unless it can be assumed to be zero). It is applicable to all design cases. Total stress analysis: an analysis in which the soil is considered to be made of one material, without distinguishing between particles, water, and air. It is the easiest of the analyses because the number of variables is decreased significantly. It is also the most likely to be erroneous, because the fundamental principles are not respected, except in a few cases like the undrained behavior of fine-grained soils where the undrained strength can be considered constant. Undrained analysis: an analysis in which the water and air are not allowed to drain during loading. This analysis can be performed by using effective stress models and in a few specific cases total stress models. One of the difficulties in using this analysis together with an effective stress model is prediction of the water stress and possibly the air stress for unsaturated soils. Drained analysis: an analysis in which the water and air are allowed to drain until any excess water stress and any excess air stress have gone back to zero. It is one of the simplest of all effective stress analyses, but its usefulness is limited because it only applies to long loading times. Short-term analysis: an analysis of the behavior of the soil in the short term. A short-term analysis can be a drained analysis for a clean, coarse-grained soil and an undrained analysis for a fine-grained soil. It tends to control the design of structures that will load fine-grained soils. Long-term analysis: an analysis of the behavior of the soil in the long term. A long-term analysis is similar to a drained analysis because in the long term—sometimes in the very long term—the soil will drain and excess water stress and excess air stresses will vanish. This analysis tends to control the design of excavations.

16.16 TRANSFORMATION FROM EFFECTIVE STRESS SOLUTION TO UNDRAINED STRENGTH SOLUTION The results of an effective stress analysis can be transformed into the results of an undrained analysis when the undrained strength is constant and the 𝜑 = 0 concept applies. In this case the transformation consists of using the following correspondence principles: 1. Effective unit weight becomes total unit weight 𝛾eff → 𝛾t

(16.78)

2. Effective stress becomes total stress 𝜎′ → 𝜎

(16.79)

3. Effective stress cohesion becomes undrained shear strength c′ → su (16.80) 4. Effective stress friction angle becomes zero 𝜑′ → 0

(16.81)

In this fashion, for example, the shear strength changes as follows: (16.82) s = c′ + 𝜎 ′ tan 𝜑′ → s = su The ultimate bearing pressure changes as follows (see Chapter 18): 1 pu = c′ Nc + 𝛾eff BN𝛾 + 𝛾DNq → pu = Nc su + 𝛾D (16.83) 2 because for 𝜑 = 0, N𝛾 = 0, and Nq = 1. The passive earth pressure equation changes as follows (see Chapter 22): √ ′ ′ 𝜎ph = Kp 𝜎ov + 2c′ Kp → 𝜎ph = 𝜎ov + 2su (16.84) because for 𝜑 = 0, Kp = 1.

Problems and Solutions Problem 16.1 It is well known that a car with wider tires can take corners faster than the same car with narrower tires. That is to say, the shearing resistance of the car with wider tires is larger than the shearing resistance of the car with narrower tires. This seems counterintuitive when one considers that in both cases the weight of the car is the same, and therefore the friction should be the same regardless of the width of the tires. Explain why the car with wider tires develops more resistance to shear in the corners than the car with narrower tires. Solution 16.1 The weight of the car is the same in both cases, so the friction should be the same (in theory). However, we need to consider the force generated by the cohesion or “glue” between the tire and the asphalt. In equation form, F = 𝜇N + C, where 𝜇 is the friction coefficient, N is the normal force, and C is the cohesion force. The cohesion force C depends on the contact area, whereas the normal force N does not. The area of a wide tire is larger than the area of a narrow tire. A wide tire will thus

16.16 TRANSFORMATION FROM EFFECTIVE STRESS SOLUTION TO UNDRAINED STRENGTH SOLUTION

505

provide more area to resist the force between the tire and the pavement to turn around a corner. Direct shear tests between a piece of pavement and a piece of rubber from a tire would be required to demonstrate this possible explanation. Problem 16.2 A medium dense sand deposit has a dry unit weight of 17kN∕m3 , a saturated unit weight of 20kN∕m3 , and a friction angle of 32 degrees. Calculate the shear strength on a horizontal plane at a depth of 10 m if: a. The groundwater level is much deeper than 10 m and the sand has no water. b. The groundwater level is at the ground surface. c. The groundwater level is at 12 m and the sand is saturated by capillary action. Solution 16.2 a. The groundwater level is much deeper than 10 m and the sand has no water: 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ 𝜏f = 0 + (10 × 17 − 0 × 0) tan 32 = 106.2 kPa b. The groundwater level is at the ground surface: 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ 𝜏f = 0 + (10 × 20 − 1 × 10 × 9.81) tan 32 = 63.7 kPa c. The groundwater level is at 12 m and the sand is saturated by capillary action. In this case, there is suction in the soil. Equation 16.8 is used and the pore water pressure is negative: 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ 𝜏f = 0 + (10 × 20 − 1(−2 × 9.81)) tan 32 = 137.2 kPa Problem 16.3 In a simple shear test on a dense sand with no water, the normal stress is 100 kPa and the shear stress at failure is 80 kPa. At failure also, the vertical displacement is 0.5 mm upward and the horizontal displacement is 5 mm. a. Calculate the friction angle 𝜑′ and the dilation angle 𝜓 ′ . b. Calculate the shear strength of the sand if the normal stress increases to 200 kPa and the angles 𝜑′ and 𝜓 ′ remain the same. Solution 16.3 a. Using the shear strength equation and knowing that the effective stress cohesion of the dense sand is zero: 𝜏f = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ 80 = 0 + (100 − 0 × 0) tan 𝜑′

or

𝜑′ = 38.66 degrees

The tangent of the dilation angle is given by the ratio of vertical to horizontal displacement: tan 𝜓 ′ =

0.5 5

or

𝜓 ′ = 5.71 degrees

b. Again, using the shear strength equation: 𝜏f = 0 + (200 − 0 × 0) tan 38.66 = 160 kPa The dilation angle is not used because it is included in the friction angle 𝜑′ . Problem 16.4 A soft clay has formed a crust near the ground surface due to drying under the sun. At the ground surface the relative humidity has been 40% for a long time. A sample of the surface clay gives a unit weight of 17.5 kN/ m3 and a water content of 10%. Estimate the shear strength of the clay at the ground surface if the effective stress friction angle is 27 degrees and Gs is 2.7. What is the apparent cohesion of that clay?

506

16 SHEAR STRENGTH PROPERTIES

Solution 16.4 Based on the Kelvin equation (Eq. 11.71; see Chapter 11), we can calculate the water tension at the ground surface: u(kPa) = 135000 × ln(RH) = 135022 × ln(0.4) = −123719 kPa

(16.1s)

Based on the three-phase soil relationships, the void ratio is linked to the unit weight of solids, the water content, and the soil unit weight by: 𝛾 (1 + 𝜔) e= s −1 𝛾 Given Gs = 2.7 we can obtain: e=

𝛾s (1 + 𝜔) 2.7 × 9.81 × (1 + 0.1) −1= − 1 = 0.66 𝛾 17.5

Another useful equation links the degree of saturation to Gs , w, and e: S= We can obtain: S=

𝜔Gs e

𝜔𝛾s 0.1 × 2.7 = = 0.41 e𝛾w 0.66

Therefore, 𝛼 can be estimated as S (i.e., 0.41). At the ground surface, the shear strength of the clay can now be calculated as: 𝜏 = c′ + (𝜎 − 𝛼uw ) tan 𝜑′ = 0 + (0 − 0.41 × (−123719)) × tan 27∘ + 0 = 25846 kPa The apparent cohesion of the clay is also 25,846 kPa, because both c′ and 𝜎 are zero. Problem 16.5 A medium-stiff clay is tested in an undrained triaxial test. At failure, the effective stress on the failure plane is 230 kPa and the shear stress on the failure plane is 122 kPa. Calculate the undrained shear strength of this clay. Solution 16.5 su = 122 kPa Problem 16.6 A soft, saturated clay is tested in an unconsolidated undrained direct shear test with a normal stress of 50 kPa; the shear strength obtained is 20 kPa. An identical sample is tested, also in an unconsolidated undrained direct shear test, but this time the normal stress is 100 kPa. What would you expect the shear strength to be? Solution 16.6 Because the soil sample is being tested in an undrained condition, and because the soil skeleton is weak (soft clay), the increase in normal stress is taken up by the water and there is no increase in effective stress. Therefore, the expected undrained shear strength is the same as in the first test: 20 kPa. Problem 16.7 A sand layer has an SPT blow count of 27 bpf and a CPT point resistance of 13.5 MPa. Both measurements come from a depth of 12 m. The groundwater level is at a depth of 5 m. What is your best estimate of the friction angle for this sand at that depth? Solution 16.7 The vertical effective stress at the point of measurement of the SPT and the CPT is computed as (assuming a soil unit weight of 20 kN/m3 ): ′ 𝜎ov = 20 × 12 − 9.81 × 7 = 171.3 kPa The SPT blow count is corrected for stress level: ( ′ )−0.5 ) ( 𝜎ov 171.3 −0.5 = 27 × = 20.8 blows∕0.3 m N1 = Nmeasured × pa 101.3

16.16 TRANSFORMATION FROM EFFECTIVE STRESS SOLUTION TO UNDRAINED STRENGTH SOLUTION

507

Then the CPT point resistance is corrected for stress level: ( ′ )−0.5 ) ( 𝜎ov 171.3 −0.5 qc1 = qc measured × = 13.5 × = 10.4 MPa pa 101.3 The friction angle 𝜑′ can be evaluated in a number of ways. Using Mayne’s recommendation: ( ) ( ) qc1 10400 𝜑′ = 17.6 + 11 × log = 17.6 + 11 × log = 39.7 degrees pa 101.3 Using an equation from Schmertmann (1975) and Kulhawy and Mayne (1990): 0.34

⎞ ⎛ ⎟ ⎜ N ⎟ tan 𝜑′ = ⎜ ′ ⎟ ⎜ 𝜎ov ⎟ ⎜ 12.2 + 20.3 pa ⎠ ⎝

0.34

⎞ ⎛ ⎟ ⎜ 27 =⎜ ⎟ ⎜ 12.2 + 20.3 × 171.3 ⎟ ⎝ 101.3 ⎠

and

𝜑′ = 39.7 degrees

Using the Terzaghi and Peck (1967) figure relating the friction angle to the blow count, we get: 𝜑′ = 35 degrees Considering all the values collected, a cautiously conservative estimate of the friction angle might be 36 degrees. Problem 16.8 The undrained shear strength of a medium-stiff clay is 46 kPa when sheared in a time to failure equal to 3 minutes in a vane shear test. The medium-stiff clay has a water content of 35% and a plasticity index of 30%. Solve the following two problems: a. A guardrail post is placed in this clay on the side of the road to arrest cars upon impact. The rise time of the force during the impact is anticipated to be 20 milliseconds. What shear strength value should you use? b. An embankment is placed on that clay. In the design process it is assumed that if a failure occurs, the failure of the embankment would be slow and take place in about 6 hours. What undrained shear strength should be used in calculating the factor of safety against embankment failure? Solution 16.8 The rate effect equation for the undrained shear strength of a clay is: ( )−n su1 t = 1 su2 t2 The viscous exponent n is related to the water content by: n = 0.028 + 0.0006 w% so n = 0.049 The viscous exponent n is related to the plasticity index by: n = 0.035 + 0.00066 PI% so n = 0.0548 Use an average n value of nAvg = 0.0519 a. In this case su1 = 46 kPa, t1 = 180 sec, t2 = 0.02 sec, su2 =? ( )n t su2 = su1 1 t2 ) ( 180 0.0519 su2 = 46 = 73.8 kPa 0.02 b. In that case su1 = 46 kPa, t1 = 180 sec, t2 = 21600 sec, su2 =? ) ( 180 0.0519 = 35.9 kPa su2 = 46 21600 Problem 16.9 Use average and associated ranges of rate effect viscous exponent to generate a curve similar to the Bjerrum correction factor for the vane shear test, undrained strength. Assume that the vane reaches the peak undrained shear strength in 3 minutes and that the embankment reaches failure in half a day.

508

16 SHEAR STRENGTH PROPERTIES

Solution 16.9 Use is made of the rate effect equation and of the correlation between the rate exponent n and the plasticity index PI in percent: ( )−n su1 t = 1 su2 t2 n = 0.035 + 0.00066 × PI% The strength su1 is the field value of su , the strength su2 is the vane test value of su , the time t1 is the time for the embankment failure or half a day (720 min), the time t2 is the time for the vane test or 3 min. Therefore, the equation for the Bjerrum factor is: su,field = 240−(0.035 + 0.00066 PI%) 𝜇= su,VST To take into account the scatter in the n vs. PI correlation, the calculated 𝜇 value is bracketed between the following two expressions. su,field = 240−(0.01 + 0.00066 PI%) 𝜇= su,VST su, field 𝜇= = 240−(0.065 + 0.00066 PI%) su,VST Figure 16.1s shows the range of the function 𝜇 vs. PI based on the rate effect model. By comparing Figures 16.1s and 16.21, it appears that the rate effect model explains much of the correction factor except at low PI values.

Correction factor, μ

1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0

0

20

40 60 80 Plasicity index, PI %

100

120

Figure 16.1s Correction factor vs. plasticity index.

Problem 16.10 A clay has an overconsolidation ratio equal to 2.5. Use the SHANSEP method and reasonable values of the parameters to estimate the undrained shear strength of that clay at a depth of 20 m. The clay is offshore at the bottom of the North Sea in 300 m of water. Solution 16.10 𝛾sat = 19 kN∕m3 At depth 20 m below the sea floor in 300 m of water: ′ 𝜎ov = 300 × 9.81 + 20 × 19 − 320 × 9.81 = 183.8 kN∕m2

For normally consolidated saturated fine-grained soil: (

Su ′ 𝜎ov

) = 0.23 NC

and for overconsolidated fine-grained soil with OCR = 2.5 ( ) ( ) Su Su = (OCR)0.8 = 0.23 × 2.50.8 = 0.48 ′ ′ 𝜎ov OC 𝜎ov NC Therefore, Su = 0.48 × 183.8 = 88.2kN∕m2

16.16 TRANSFORMATION FROM EFFECTIVE STRESS SOLUTION TO UNDRAINED STRENGTH SOLUTION

509

Problem 16.11 An unsaturated sample of clay is tested in a simple shear test. At failure the total normal stress on the failure plane is 70 kPa and the shear stress is 175 kPa. a. Is that possible? b. After testing, the water content on the plane of failure is measured and the soil water retention curve gives a water tension of 1450 kPa. The water content coupled with the measurement of the unit weight and the assumption that Gs is 2.7 leads to a degree of saturation of 20%. If the clay has no effective stress cohesion, calculate the effective stress friction angle. Solution 16.11 a. Yes, it is possible. In unsaturated soils, the shear strength can be higher than the total normal stress because the water tension increases the effective stress. b. 𝜎 = 70 kPa 𝜏f = 175 kPa uw = −1450 kPa S = 20% ⇒ 𝛼 = 0.2 c′ = 0 𝜏f = c + (𝜎 − 𝛼uw ) tan 𝜑′ = 0 + (70 − 0.2 × (−1450)) tan 𝜑′ = 175 kPa tan 𝜑′ = 0.486 ⇒ 𝜑′ = 25.9∘ Problem 16.12 A lightly overconsolidated clay has a CPT point resistance of 1100 kPa, an OCR of 1.7, a PMT limit pressure of 590 kPa, an SPT blow count N of 13 bpf, and a unit weight of 18kN∕m3 . Estimate the undrained shear strength of that clay if the data comes from a depth of 6 m with the groundwater level being at a depth of 2 m. Solution 16.12 𝛾T = 18 kN∕m3 𝜎ov = 18 × 6 = 108 kPa ′ = 𝜎ov − 𝛼uw = 108 − 1 × (6 − 2) × 9.81 = 69 kPa 𝜎ov

The CPT data equation gives: su(CPT) =

qc − 𝜎ov Nk

Assuming an average value of N k equal to 14, the equation becomes: su(CPT) =

1100 − 108 = 70.9 kPa 14

The PMT data equation gives: su(PMT) =

pL NP

Assuming an average value of N p equal to 7.5: su(PMT) =

590 = 79 kPa 7.5

The recommendations also give: su(PMT) pa

( = 0.21 ×

pL pa

)0.75

510

16 SHEAR STRENGTH PROPERTIES

Using 101.3 kPa for the atmospheric pressure: ( )0.75 ) ( pL 590 0.75 su(PMT) = 0.21 × × pa = 0.21 × × 101.3 = 79.8 kPa pa 101.3 The SPT data equation gives: su(SPT) = 4.4N60 Assuming that N = N60 , then: su(SPT) = 4.4 × 13 = 57.2 kPa Given all the data, a cautiously conservative estimate of the undrained shear strength is 65 kPa. Problem 16.13 You are at the beach lying on dry, uniform sand. You take a handful of sand and let it fall from your hand onto a 0.3 m by 0.3 m wide plate. The sand pile on the plate has the shape of a pyramid and the angle of the pyramid with the horizontal is 𝛽. Demonstrate that 𝛽 is equal to the friction angle 𝜑′ . You then take that same pile of sand and add a bit of water. Now you are able to mold the sand pile into a cylinder standing vertically. Where does the sand strength come from? Is it cohesion or friction? Solution 16.13 The angle of the pyramid with the horizontal, 𝛽, is known as the angle of repose. When the sand falls, it comes to rest at the maximum angle possible; therefore, the slope of the sand pile is at impending failure. You can check that by tilting the plate slightly to one side: The side slope of the dry sand pyramid will fail to retain the same angle with the horizontal. If the slope is at impending failure, an element of soil as shown in Figure 16.2s is subjected to shear strength 𝜏 f . The equilibrium of the element leads to the following equations, which show that the slope angle is the friction angle of the sand: The shear force on the failure plane is: T = W sin 𝛽 The normal force on the failure plane is: N = W cos 𝛽 Therefore, T = N tan 𝛽 The maximum resisting force on the failure plane is: S = N tan 𝜑′ At failure: T=S

𝛽 = 𝜑′

therefore

T W

N

S Dry sand

β

Figure 16.2s Dry sand pile and angle of repose.

If we add a drop of water to the sand, water tension develops in the voids of the fine sand. This water tension pulls the particles against each other and creates an effective stress equal and opposite to the water tension. The shear strength of the sand is due to the friction related to the effective stress created by the water tension. This shear strength is often called apparent cohesion because the sand “sticks” together—yet the real mechanism is friction.

CHAPTER 17

Thermodynamics for Soil Problems

17.1

GENERAL

Heat flow in soils involves several different phenomena: convection, radiation, and conduction. Convection takes place when a fluid flows over a solid that is at a different temperature than the fluid. When you set up a fan to blow air toward your body and cool yourself down in the summer, you use convection heat transfer. Radiation refers to the fact that all bodies continuously emit energy because of their temperature. This energy propagates to other nearby fluids or bodies through electromagnetic waves. Hot radiators that you may use in the winter to warm yourself up operate by radiation heat transfer. Conduction is a heat transfer mechanism whereby energy moves from a region of high temperature to a region of lower temperature. The phenomenon is due to the motion and impact of molecules, which increase as the temperature rises. Conduction of heat in soil is very similar to the flow of water through soil and is the most important mechanism of heat transfer through soils.

17.2

DEFINITIONS

Because of the analogy between temperature propagation and water flow, it is useful to draw a parallel between the parameters used in both fields of geotechnical engineering. Heat, Q, is a quantity of energy measured in joules (N × m). It is named after James Prescott Joule (1818–1889), an English physicist. The heat Q is equivalent to the volume of water V (m3 ) in flow problems. Temperature, T, is a measure of how hot a material is; it is sometimes measured in degrees Kelvin (K), but more commonly in degrees Celsius (C). The Kelvin is named after the British engineer and physicist William Thomson, First Baron Kelvin (1824–1907). The Celsius is named after the Swedish astronomer Anders Celsius (1701–1744). The Kelvin scale starts at absolute zero temperature, which is −273∘ C. There is in fact a lower bound to the temperature scale: It corresponds

to the point where none of the molecules are moving. There is no known upper bound to the temperature scale. The temperature T is equivalent to the total head ht (m) in flow problems. The temperature gradient it is defined between two points in the soil mass; it is the ratio between the change in temperature dT over the distance dx separating the two points and is expressed in K/m. It corresponds to the hydraulic gradient i for the flow problem: dT in K∕m (17.1) dx The heat transfer rate H is the amount of heat transferred per amount of time and is expressed in joules per second or watts, named after the Scottish engineer James Watt (1736–1819) (J/s or W). The heat transfer rate is equivalent to the flow rate Q (m3 /s) in flow problems: it =

dQ in J∕s (17.2) dt The heat flow q is the amount of heat dQ per unit time dt and per unit area A or the heat transfer rate H per unit area A. It is expressed in watts per meter square (W/m2 ) or in joules per second and per meter square (J/s.m2 ). It is equivalent to the velocity v (m/s) in the flow problem: H=

dQ 1 H (17.3) × = in J∕s.m2 dt A A The thermal conductivity kt is a property of the soil. It takes units of J/s.K.m and is defined through Fourier’s law (Section 17.3) as the ratio between the heat flow and the thermal gradient: q in J∕s.K.m (17.4) kt = dT dx The thermal conductivity is an indication of the speed with which the heat flows through the soil under a given temperature gradient. It is equivalent to the hydraulic conductivity for the flow problem.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

q=

511

512

17 THERMODYNAMICS FOR SOIL PROBLEMS

The specific heat c is a property of the soil and takes units of J/kg.K. It is defined as: 1 dQ c= in J∕kg.K (17.5) m dT where m is the mass of the soil element considered, and dQ is the increase in heat stored in the element when the temperature is raised by dT. In the flow problem, the compressibility of the soil skeleton plays the role of the inverse of the specific heat. The inverse of the specific heat tells you how much heat you can squeeze out of the soil for a given change in temperature, much like the compressibility tells you how much water you can squeeze out of the soil if you apply a change in effective stress. The diffusivity 𝛼 appears in the governing differential equation. It is in m2 /s and is defined as: 𝛼=

k 𝜌c

in m2 ∕s

(17.6)

where 𝜌 is the mass density of the soil. The diffusivity gives the speed with which the temperature will decay in a soil. It is closely linked to the thermal conductivity kt , but is also influenced by the specific heat, which indicates how much heat can be squeezed out of the soil for a given change in temperature. In other words, you could have two soils with the same thermal conductivity but different specific heats. In this instance, the heat would flow at the same speed in both soils for the same thermal gradient, but if the heat source stopped, the one with the highest specific heat would cool down the slowest because it would be harder to squeeze the heat out of the soil. Table 17.1 summarizes the equivalency between soil thermal flow problems and soil hydraulic flow problems. Table 17.1 Equivalency between thermal conductivity and hydraulic conductivity Parameter

Flow of water

Flow of heat

Quantity Potential Gradient

Volume V (m3 ) Head ht (m) Hydraulic gradient ih (unitless) Flow rate Q (m3 /s)

Heat Q (J) Temperature T (K) Temperature gradient it (K/m) Heat transfer rate H (J/s) Heat flow q (J/s.m2 ) Thermal conductivity kt (J/s.K.m) Fourier Specific heat c (J/kg.K) Thermal diffusivity α (m2 /s)

Flux Flux density Conductivity

Law Storage Decay coefficient

Velocity v (m/s) Hydraulic conductivity kh (m/s) Darcy Compressibility Coefficient of consolidation cv (m2 /s)

17.3

CONSTITUTIVE AND FUNDAMENTAL LAWS

Fourier’s law is the constitutive law for heat flow. It is named after Joseph Fourier (1768–1830), a French mathematician and physicist. Fourier’s law (Fourier, 1822) states that the heat flow q is linearly related to the temperature gradient through the thermal conductivity kt : dT (17.7) dx where q is the heat flow, kt is the thermal conductivity, it is the temperature gradient, T is the temperature, and x is the length in the direction of the heat flow. Therefore, the units of thermal conductivity are J/s.K.m. The minus sign indicates that heat flows in the direction of decreasing temperatures. Fourier’s law is equivalent to Darcy’s law in the hydraulic flow problem. By the way, the R rating of house insulation comes from Eq. 17.7 and is based on very much the same concept as the resistance of an electrical conductor: dT dx or dT = R q (17.8) = R= q kt q = −kt it = −kt

The fundamental law is the conservation of energy. For the purpose of this chapter, this law states that during an amount of time dt, the amount of heat dQin flowing into an element of soil is equal to the amount of heat dQout flowing out of the element plus the heat stored or extracted dQstored from the element. dQ dQ dQ = + (17.9) dt in dt out dt stored 17.4

HEAT CONDUCTION THEORY

Let’s first address the problem of one-dimensional heat conduction. An example is the penetration of frost into a surface layer of soil due to low air temperature in the winter months. To solve this problem, we follow the normal steps (see Section 12.4): 1. Consider an element of soil dx wide, dy long, and with a unit length perpendicular to the page (Figure 17.1). 2. The heat flows through the element volume, which has a cross-sectional area A(dy × 1)and a length dx. During a time dt, the quantity of heat entering the element is Aqx dt, whereas the quantity of heat leaving the element is Aqx+dx dt. Area A

AqxΔt

0

Aqx + dx Δt

dy dx 0

Figure 17.1 Element of soil.

513

17.5 AXISYMMETRIC HEAT PROPAGATION

3. Conservation of energy allows us to state that the difference (Aqx dt − Aqx+dx dt) is equal to the stored heat in the element. dQ = A qx dt − A qx+dx dt 4. The constitutive law is Fourier’s law: dT(x, t) q(x, t) = −k dx

(17.10)

(17.11)

where q is the heat flow (J/s.m2 ), k is the thermal conductivity (J/s.K.m), T is the temperature (K), and x is the length (m) in the direction of the heat flow. 5. The second constitutive law is associated with the definition of specific heat. The amount of heat dQ will generate an increase in temperature dT in the element of mass m such that: dQ = m c dT = A dx 𝜌 c dT

(17.12)

where 𝜌 is the mass density of the material (kg/m3 ) and c is the specific heat of the material (J/kg.K). 6. Regrouping Eqs. 17.10 and 17.12 gives: Aqx dt − Aqx+dx dt = A dx 𝜌 c dT

(17.13)

Or, in partial derivative form: 𝜕q 𝜕T =𝜌c 𝜕x 𝜕t Combining Eqs. 17.11 and 17.14, we get: −

𝜕T 𝜕2T =𝜌c 𝜕t 𝜕x2 If we define the thermal diffusivity 𝛼 as: k

𝛼=

k 𝜌c

(17.14)

(17.15)

17.5

AXISYMMETRIC HEAT PROPAGATION

In the case of an axisymmetric geometry, Eq. 17.18 becomes: 1 𝜕T 1 𝜕T 𝜕 2 T + 2 = (17.19) r 𝜕r 𝛼 𝜕t 𝜕r where r is the radial distance from the axis, t is time, T is temperature, and 𝛼 is the thermal diffusivity of the soil. Carslaw and Jaeger (1947) solved this problem in the case of an infinitely long cylindrical heat source of radius Ro maintained at a temperature T o at the center of a full space, which was initially at a temperature equal to zero. In this case, the time t required for a given temperature T m to reach a certain distance R into the soil is given by: t = TF

R20

(17.20) 𝛼 where T F is the time factor (Figure 17.2), and is a function of the ratio R∕Ro and Tm ∕To . This equation is very similar to the consolidation equation, which yields the time for excess water stress dissipation around a pile. At first glance, Eq. 17.20 seems to indicate that t increases with R2o . But in fact, t decreases as Ro increases, because T F decreases with Ro faster than R2o increases. The following reasoning √ illustrates this point. In Eq. 17.20, if Ro is multiplied by 10, the time t is not multiplied by 10 because the time factor T F is not the same in both cases. If t was multiplied by 10, it would mean that it would take 10 times longer for the temperature to reach a value T m at a distance R − Ro from √ the boundary in the case of the largeradius heat source ( 10Ro ) than for the same temperature T m to be reached at the same distance R − Ro in the case of the

(17.16)

1.0

where 𝛼 is the diffusivity in m /s, then the governing differential equation for one-dimensional conduction heat is: 𝜕T 𝜕2T (17.17) =𝛼 2 𝜕t 𝜕x In three dimensions, it becomes: 2

0.8 0.7 Time factor Tf

0.6

(17.18)

7. Now the boundary and initial conditions have to be expressed. This depends on the problem at hand. The complexity of the solution depends on the complexity of the boundary conditions, but numerical methods can always be used to solve such problems. Note that Eq. 17.17 is similar to Eq. 12.56 for the consolidation theory, where the temperature T is replaced by the excess water stress ue . Therefore, the solutions are identical for identical boundary conditions. Jumikis (1977) presents the solution for a sinusoidal temperature fluctuation input at the ground surface to replicate seasonal variations.

Tm / T 0

𝜕2T 𝜕2T 𝜕2T 1 𝜕T + 2 + 2 = 2 𝛼 𝜕t 𝜕x 𝜕y 𝜕z

0.9

0.5 1000

0.4

300

0.3 3

100

0.2 0.1

0.1

1 0.3

30 10

0.03

0.0 0.0

0.5

1.0 Log10 R/R0

1.5

2.0

Figure 17.2 Time factor. (Source: After Carslaw and Jaeger, 1947.)

17 THERMODYNAMICS FOR SOIL PROBLEMS

smaller-radius heat source (Ro ). This does not make sense: Because the heat source is larger, it should take less time—and indeed it does, because the time factor T F decreases more than by a ratio of 10 in this case (Figure 17.2). Therefore, as Ro increases, t in fact decreases nonlinearly. For example, consider a hot cone penetrometer with a radius Ro of 20 mm that is kept at a temperature T o of 100∘ C in a soil with an initial temperature of 20∘ C and a diffusivity of 1 mm2 /s. Let’s calculate the time it will take for the temperature to reach 40∘ C at a distance of R equal to Ro + 100 mm = 120 mm. Considering that the temperature of the soil is at 20∘ C initially, the ratio of net temperature increase is (40 − 20)∕(100 − 20) = 0.25. For this temperature ratio and a radius ratio of 120∕20 = 6, Figure 17.2 yields a time factor T F equal to 32, and the time for the temperature to reach 40∘ C at R = 120 mm is: 202 t = 32 = 12800 s = 3.55 hours (17.21) 1 Now consider a hot oil conductor in the bottom of the Gulf of Mexico with a radius Ro of 500 mm, that is kept at a temperature T o of 100∘ C. Let’s calculate the time it will take for the temperature to reach a temperature of 40∘ C at a distance of R equal to Ro + 100 mm = 600 mm. Considering that the temperature of the soil is at 20∘ C initially, the ratio of net temperature increase is (40 − 20)∕(100 − 20) = 0.25. For this temperature ratio and a radius ratio of 600∕500 = 1.2, Figure 17.2 yields a time factor T equal to 0.02, and the time for the temperature to reach 40∘ C at R = 600 mm is:

1000 Liquid water 100

Offshore foundation

10 Pressure (Atm)

514

Onshore foundation

1 0.1 Permafrost 0.01

0.0001 –100

0 100 Temperature (°C)

200

Figure 17.3 Temperature phases for water. 40 m Platform

0

Temperature (°C) 10 20

Tendons

Temperature line

2000 m Template

(17.22)

100 m 15,000 m?

17.6

Water vapor

0.001

2

500 = 5000 s = 1.39 hours t = 0.02 1

Ice water

10

20

200

Foundations

T (°C)

Oil conductor

Geothermal gradient

THERMAL PROPERTIES OF SOILS

Any material can be found in solid, liquid, or gas form. For water, the transition from solid to liquid is at 0∘ C and the transition from liquid to gas is at 100∘ C. These temperatures correspond to 1 atmosphere of pressure but would be different at different pressure levels. Figure 17.3 shows the pressuretemperature phase diagram for water and its triple point. By the way, the latent heat of a material is the heat necessary to change the phase of the material (solid to gas, for example). The temperature on the Earth surface varies from about −50∘ C to about +50∘ C. The temperature in the Earth varies from −50∘ C on the surface to 5500∘ C at the center of the Earth. Rocks and soil particles melt at a temperature varying between 600∘ C and 1200∘ C. The temperature gradient in the Earth varies and may be taken as 15∘ C per km over the first 100 km of depth. The deepest types of projects involving the geotechnical engineer may be offshore platforms and the associated retrieval of oil. The water depth in which the largest platforms are constructed reaches several kilometers. At the bottom of such oceans, the temperature is only a few degrees Celsius. The oil reservoir may be at a depth of 15 km; thus, the temperature of the oil can easily be 100∘ C when it comes back up to the surface (Figure 17.4). So, for

Oil reservoir

Depth or pressure

Figure 17.4 Temperature gradient for an offshore platform. (Source: After Briaud and Chaouch, 1997.)

the geotechnical engineer, soil particles and rocks remain in solid form. However, within the range of Earth surface temperatures, water can be in liquid or in solid form (frozen), and the thermal properties of the soil may differ depending on whether the soil is frozen or not. The thermal properties of interest are the thermal conductivity k (J/s.m∘ C), the specific heat C (J/kg∘ C), and the diffusivity 𝛼 (m2 /s). A high value of thermal conductivity means that heat travels easily through the material; a high value of specific heat means that it takes a lot of heat to raise the temperature of the material; and a high value of diffusivity means that it will take little time for the temperature to rise in the material. These thermal properties depend on a number of factors, among which are the temperature level T, the pressure level p, the moisture content w, and the density 𝜌. Table 17.2

515

17.7 MULTILAYER SYSTEMS

Table 17.2

Thermal properties for various earth materials at standard conditions of temperature and pressure

Material Air Water Ice Clay (unfrozen) Clay (frozen) Sand (unfrozen) Sand (frozen) Rock

Density 𝜌 (kg/m3 )

Specific heat c (J/kg.∘ C)

Thermal conductivity k (J/s.m.o C)

Thermal diffusivity 𝛼 (mm2 /s)

1–1.4 960–1000 917–920 1400–1800 1400–1800 1500–2200 1500–2200 2200–3000

1000–1050 4190–4220 1960–2110 750–920 650–800 630–1460 500–1200 710–920

0.02–0.03 0.5–0.8 2.0–2.6 0.8–2.8 1.0–3.6 2.3–3.8 2.9–4.7 2–6

13–30 0.13–0.17 1.24–1.52 0.1–1.66 0.15–2.3 0.87–3.0 1.2–4.2 1.1–3.0

shows an estimate of the range of values one can expect for those thermal properties at ordinary temperature and pressure levels. The range of values in Table 17.2 helps one to understand the factors affecting the thermal properties. For example, a dry soil will have a thermal conductivity lower than the same soil once saturated, because air has a lower thermal conductivity than water. Also, sand in a very dense state will have a higher thermal conductivity than the same sand in a very loose state, because soil particles have a higher thermal conductivity than air or water. 17.7

where qe is the total heat flow, q1 is the heat flow in layer 1, and q2 is the heat flow in layer 2. Using Fourier’s law gives: ke ie (h1 + h2 ) × 1 = k1 i1 h1 × 1 + k2 i2 h2 × 1

where ke is the equivalent thermal conductivity, k1 is the thermal conductivity of layer 1, k2 is the thermal conductivity of layer 2, ie is the equivalent gradient, i1 is the gradient in layer 1, and i2 is the gradient in layer 2. Therefore: k h + k2 h2 ke = 1 1 (17.26) h1 + h2 This result can be generalized for n layers: n ∑

MULTILAYER SYSTEMS

Heat can flow through a layered system, such as an asphaltconcrete pavement over soil in the heat of the summer or a layer of snow covering the soil surface in the winter. Consider the case in which heat flows parallel to the interface of the two layers (Figure 17.5), where the starting temperature and the ending temperature are maintained at TA and TB . These temperatures exist at two points separated by a horizontal distance L. Layer 1 is h1 thick and layer 2 is h2 thick. The thermal gradient is the same in both layers: T − TA (17.23) = ie i1 = i2 = B L However, the total heat transfer rate H is the sum of the heat transfer rate H 1 in layer 1 plus the heat transfer rate H 2 in layer 2:

ke =

TA Heat flow

TB B

A

Layer 2 TA

L

∑

H = H1 = H2

and

H = ke ie L × 1 = k1 i1 L × 1 = k2 i2 L × 1

(17.28)

The change in temperature, however, is additive: ΔT = ΔT1 + ΔT2 = TA − TB

(17.29)

Heat flow TA Layer 1

k2, h2

Layer 2

Figure 17.5 Horizontal heat flow through two layers.

(17.27) hi

Now consider the case in which heat flows perpendicular to two layers (Figure 17.6), where the starting temperature and the ending temperature are maintained at T A and T B . Layer 1 is h1 thick and layer 2 is h2 thick. The temperatures T A and T B exist at two points separated by a distance (h1 + h2 ). The heat transfer rate is the same in both layers:

k1, h1

TB

ki hi

i=1 n

i=1

H = qe (h1 + h2 ) × 1 = H1 + H2 = q1 h1 × 1 + q2 h2 × 1 (17.24)

Layer 1

(17.25)

TA k1, h1

L

k2, h2 TB

TB

Figure 17.6 Vertical heat flow through two horizontal layers.

516

17 THERMODYNAMICS FOR SOIL PROBLEMS

But ie =

Air

Ta

ΔT h1 + h2

and

i1 =

ΔT1 h1

and

i2 =

ΔT2 h2

Ts

(17.30)

isn

Therefore, h1 + h2 h h H = 1 H1 + 2 H2 ke L k1 L k2 L

hsn

1

T1

if

zf

1

h1 + h2 h1 h2 + k1 k2

(17.32)

0 − Ts = 0 − T1 + T1 − Ts (17.33)

i=1 ki

17.8

Figure 17.8 Frost penetration depth for a two-layer system.

The difference in temperature between the bottom of the frozen soil layer at 0∘ C and the surface of the snow layer at T s can be written as:

n ∑

ke =

Frost penetration depth Td

This result can be generalized for n layers: hi i=1 n h ∑ i

Ground surface

Tf = 0

(17.31)

and ke =

T (°C) Snow surface

0

APPLICATIONS

Let’s consider a soil deposit in a cold country (Figure 17.7). The question is: How deep will the frost penetrate during a very cold period? At depth, where the soil is not frozen, the temperature is T d . The air is at temperature T a , much lower than 0∘ Celsius. It is assumed that the temperature of the soil surface T s is the same as the air temperature T a . The temperature at the bottom of the frozen soil is assumed to be 0∘ Celsius (C). The gradient of temperature in the frozen layer is i and is associated with a heat flow q and a thermal conductivity kf . Therefore, the depth of the frozen soil is:

(17.35)

where T 1 is the temperature at the interface between the bottom of the snow layer and the soil surface (Figure 17.8). By using the definition of the temperature gradient and then Fourier’s law, Eq. 17.35 can be rewritten as: qf q (17.36) 0 − Ts = isn hsn + if zf = sn hsn + zf ksn kf For continuity purposes, though, the heat flow has to be the same in the snow and the frozen part of the soil layer. Then: qsn = qf = q

(17.37)

and the frost penetration depth zf is: zf =

−Ts kf q

− hsn

kf ksn

(17.38)

As can be seen, the snow cover reduces the frost penetration k depth by hsn k f . sn

Tkf 0 − Ts =− zf = i q

(17.34)

Now let’s consider that a layer of snow covers the ground surface (Figure 17.8). The question here is: Would the depth of the frozen soil zf be the same? The thickness of the snow cover is hs and the thermal conductivity of the snow is ks . The air temperature is T a , and the temperature of the snow surface is T s and is assumed equal to T a . The thermal conductivity of the frozen soil is kf . Air

Ta Ts kf ku

zf

0 i

T (°C) Ground surface q

1

Frost penetration depth

Tf = 0 Td

Figure 17.7 Frost penetration depth for a uniform soil.

17.9

THERMAL CONE PENETROMETER TEST

The thermal cone penetrometer test (TCPT) can be used to obtain the thermal conductivity λ of the soil. Akrouch et al. (2016) developed that test which consists of pushing the CPT at the standard rate of 20 mm/s and stopping the test while recording the temperature on the friction sleeve. Indeed, the penetration of the CPT by itself creates enough increase in the temperature of the CPT metal shell that it is convenient to record its decay with time (Figure 17.9). Akrouch et al. (2016) performed experiments in the laboratory, in situ tests with the TCPT, and numerical simulations. Based on all the results, they came up with a recommended equation to obtain the thermal conductivity of the soil from the 50% time of the temperature decay curve (Figure 17.10): 𝜆(W∕m.K) =

77 (t50 (s))0.968

(17.39)

where λ is the thermal conductivity, in Watt per meter and per degree Kelvin, of the soil near the TCPT test and t50 is the time

17.10 FROZEN SOILS

Temperature, T (°C)

26

TCT - 4 m. TCT - 10 m. TCT - 13 m.

25

26 25

24

24

23

23

22 0

300

600 900 1200 1500 1800 Time, t (sec) (a)

517

22 1

TCT - 4 m. TCT - 10 m. TCT - 13 m. 10 100 Time, t (sec) (b)

1000

Figure 17.9 Temperature dissipation curve for the thermal cone penetrometer test (Source: Akrouch et al., 2016). (a) temperature vs time. (b) temperature vs log time.

Thermal conductivity, λ [W/m.K]

3

Numerical Analysis Results Average Calibrated Curve from Numerical Analysis

2.5

Liberal Arts Building Site Riverside National Sand Site

2

Fugro Backyard Site Adjusted Curve

1.5 1 0.5 0

0

50

100

150

200

250

300

350

400

Measured time, t50 [sec]

Figure 17.10 Thermal conductivity versus time to 50% decay. (Source: Akrouch et al., 2016.)

in seconds necessary to obtain 50% dissipation of the initial temperature. 17.10

FROZEN SOILS

The general term frozen soils covers problems of freezing soils, frozen soils, and thawing soils. Frozen soils are usually classified into three categories: (1) soils with nonvisible ice (N); (2) soils with visible ice and ice lenses less than 25 mm thick (V); and (3) soils with visible ice with ice lenses larger than 25 mm thick (ICE). Permafrost is a term indicating that the ground, including soil and rock, has been at or below 0∘ Celsius for more than two consecutive years. The temperature at which the water in the voids will freeze depends on many factors, including the salt content. The more salt there is, the lower the temperature has to be before the water will freeze. Generally, freezing starts at around −1∘ C, and at −20∘ C most soils are completely frozen. Figure 17.11 shows typical temperature profiles in frozen soils. It indicates

that close to the surface there is usually a zone that freezes and thaws each year, called the active zone. The water very close to the mineral surface of a particle can be tightly bound to the particle, especially for very small particles. This adsorbed water layer practically never freezes. Therefore, clays tend to resist freezing more than sands. The water film the furthest away from that boundary is the first one to freeze. Figure 17.12 shows conceptually the evolution of the water content of a soil as the temperature plunges below zero. As can be seen, the equilibrium frozen water content is higher for clayey soils than for sandy soils. Frost susceptibility is smallest for clean gravels and clean sands, on the one hand, and for high-plasticity clays, on the other. The most frost-susceptible soils are silts, as shown in Table 17.3. The reason is that frost heave requires the soil to have the ability to lift water by capillary action and let the water flow through its voids. Clean gravels and clean sands have high hydraulic conductivity but little ability for capillary action; in other words, it is easy for the water to move, but the water has no energy to go anywhere. High-plasticity clays, in contrast, have a very high ability for capillary action but a very low hydraulic conductivity; in other words, the water has plenty of energy, but it is very hard to move through the clay. Silts optimize the two requirements of capillary potential and water flow and are therefore some of the most frost-susceptible soils. Note that the unit weight of ice is about 10% less than the unit weight of water. Therefore, if a certain weight of water becomes ice, it will occupy about 10% more volume. This is why icebergs float with only one-tenth of the iceberg mass showing up above the water level and 90% below it (hence the expression “this is only the tip of the iceberg”). If a soil becomes frozen, it will expand according to the increase in volume of the water becoming ice. These ice lenses, once started, continue to attract water and become thicker by something called the cryosuction process. Such ice lenses have significant lifting potential; the uplift pressures can be several hundreds of kPa and can reach 2000 kPa if the heave is

518

17 THERMODYNAMICS FOR SOIL PROBLEMS

Temperature (°C) Ground surface 0m

Temperature (°C) Ground surface 0m

Frost depth Freeze-thaw layer

Permafrost No influence of weather

No influence of weather Depth

Depth

Water content unfrozen, (%)

Figure 17.11 Typical temperature profiles in frozen soils.

must be made between the unfrozen water content and the frozen water or ice content. They are defined as:

30

Total water content Clayey

20

Unfrozen water content Frozen water (ice) content

10

Sandy 0 10

–10

0

–20

Temperature, T (°C)

Figure 17.12 Evolution of water content with temperature.

Table 17.3

Frost susceptibility and soils

Soil type

Frost susceptibility

High-plasticity clays Low-plasticity clays, clays with sand and gravel Silty clays Silts, silty sands, very fine sands Gravels and sands with fines Clean sands and gravels

Negligible Moderate

(17.40) (17.41) (17.42) (17.43)

where W w is the weight of water, W i is the weight of ice, and W s is the weight of solids (Figure 17.13). In all other index parameters, it is necessary to state what is included and what is not. For example, the degree of saturation, the void ratio, and the porosity can be defined by including or not including the ice. The thermal properties of a frozen soil are the combination of the properties of the water, the ice, the air, and the soil skeleton. Table 17.2 shows these properties for each material individually and the impact they have on the soil. As can be seen, the frozen soil will have a higher thermal conductivity, a lower specific heat, and a higher thermal diffusivity than the unfrozen soil. In other words, the heat will flow faster in

Moderate to severe Severe Moderate Negligible

VV VT

confined. Heave magnitudes of 50–75 mm are common. The frozen soil can also develop an “adfreeze” bond with neighboring objects such as foundation piles. This bond can generate shear stresses from 50 to 150 kPa. Frozen soils have four phases instead of three. Note that nearly all frozen soils contain liquid water. The phase diagram is shown in Figure 17.13. For the water content, a distinction

Ww + Wi Ws Ww wu = Ws Wi wi = Ws w = wu + wi w=

Va

Air

Wa = 0

VwF

Water frozen

WwF

VwU

Water unfrozen

WwU

VS

Solids

WS

Figure 17.13 Phase diagram for a frozen soil.

WT

17.10 FROZEN SOILS

the frozen soil, and it will be easier to squeeze the heat out of the frozen soil than out of the unfrozen soil.The mechanical properties will also be affected. The shear strength will increase significantly, as the ice will contribute to increasing the cohesion intercept. The stiffness will also increase, as the ice essentially increases the amount of solids in the soil. However, the creep component of the settlement will be increased as the ice content increases. Indeed, ice exhibits creep properties that depend on the ice temperature; the lower the ice temperature, the less it will creep. The viscous exponent mentioned in Eq. 16.56 varies in the range of 0.1–0.5 for ice. Recall that the same exponent for unfrozen clays was

519

0.02–0.05 (see Figure 16.18). As a result, a frozen soil will creep more than the unfrozen soil under constant load, but the initial movement will be less. The hydraulic conductivity will decrease, as there is less area for the water to flow through. In that sense, frozen soils act according to the same principles as unsaturated soils. The best way to obtain the mechanical properties of frozen soils is to perform a laboratory or in situ test that duplicates the conditions under which the soil will be stressed in the project at hand. There is a close analogy between frozen soils and unsaturated soils, and more interaction between these two fields is likely to be very rewarding.

Problems and Solutions Problem 17.1 A house is built on a frozen soil layer. The house generates heat such that it maintains a temperature of 20∘ C in the house. If the thermal conductivity of the frozen soil is kfrozen = 1.3W∕m.K, if the thermal conductivity of thawed-out soil is kunfrozen = 1.1W∕m.K, and the temperature gradient in the frozen soil is ifrozen = −15∘ C∕m, what thickness of soil will thaw out? Solution 17.1 Based on the principle of continuity of heat flow, and assuming that the surface temperature is T s , the freezing temperature is T f (equal to 0∘ C), and the thawing depth is x, we can write: qunfrozen = qfrozen Tf − Ts kunfrozen × A × = kfrozen × A × ifrozen x Tf − Ts kunfrozen × A × = kfrozen × A × ifrozen x kunfrozen Tf − Ts x= × kfrozen ifrozen The thawing depth is thus: x=

1.1 0 − 20 × = 1.13 m 1.3 (−15)

Problem 17.2 A building is to be built with a geothermal foundation in a soil with a thermal diffusivity 𝛼 equal to 5 × 10−7 m2 ∕s. The energy piles are 0.4 m in diameter and water circulates up and down the piles to take advantage of the beneficial effect of the soil temperature (hotter in the winter and cooler in the summer). The energy piles operate 8 months out of the year, and, for optimum operation performance, the increase in temperature in adjacent energy piles due to the operation of one energy pile must not exceed 10% of the initial temperature difference between the pile and the soil. Calculate the minimum spacing between the energy piles. Solution 17.2 From the problem data and using Figure 17.2 above, Tm ∕To = 0.1. The time factor T F is calculated using Eq. 17.20: TF =

t×𝛼 8 × 30 × 24 × 3600 × 5 × 10−7 = = 259.2 R2o 0.22

From Figure 17.2, log10 R∕R0 = 1.3; therefore, R∕R0 = 19.95 and the minimum distance between energy piles should be R = Ro × 19.95 = 4 m.

520

17 THERMODYNAMICS FOR SOIL PROBLEMS

Problem 17.3 A cylindrical soil sample (D = 0.075 m, L = 0.150 m) is put in an oven where the temperature is kept at Tf = 45o C. The initial temperature of the soil sample is Ti = 25o C. The soil sample thermal conductivity k is 1.2 W/m.K and the volumetric heat capacity C is 1.2 × 106 J∕m3 .K. Using the literature, find the solution that gives the increase in temperature at the center of the cylindrical soil sample as a function of time and calculate how long it will take for the center of the sample to reach 30∘ C, 35∘ C, and 40∘ C. Solution 17.3 Carslaw and Jaeger (1947) developed the solution for the temperature increase at the center of a cylindrical sample as a function of time. The percentage increase or decrease in soil sample temperature U can be plotted versus the normalized time factor T as shown in Figure 17.1s, for both a finite-length sample and an infinite-length sample. T − Tmin U= Tmax − Tmin T=

𝛼(m2 ∕s) × t(s) D2 (m2 ) Time factor, T 0.10

0.01 0

1.00

Percent temperature change, U (%)

Soil sample

20

L D 40

60

80

Time factor curve for temperature change at the center of the sample of diameter D and infinite length L Time factor curve for temperature change at the center of the sample of diameter D and length L = 2D

100

Figure 17.1s

Percent temperature change U versus time factor T.

The thermal diffusivity 𝛼 of the soil sample is: k 1.2 = 10−6 = C 1.2 × 106 When the temperature reaches 30∘ C, U = 25%; from Figure 17.1s, T = 0.03: 𝛼 (m2 ∕s) =

T × D2 0.03 × 0.0752 = 168 sec = 𝛼 10−6 When the temperature reaches 35∘ C, U = 50%; from Figure 17.1s, T = 0.04: t (s) =

t (s) =

T × D2 0.04 × 0.0752 = 225 sec = 𝛼 10−6

17.10 FROZEN SOILS

521

When the temperature reaches 40∘ C, U = 75%; from Figure 17.1s, T = 0.07: t (s) =

T × D2 0.07 × 0.0752 = 393 sec = 𝛼 10−6

Problem 17.4 A two-layer system is made of a concrete pavement overlaying a sandy subgrade. The thermal properties of the two layers are shown in Figure 17.2s. a. What is the equivalent thermal conductivity of the system if the heat flows horizontally? b. What is the equivalent thermal conductivity of the system if the heat flows vertically?

Concrete k = 0.9 W/m.K

Sand k = 1.6 W/m.K

Figure 17.2s

0.2 m

0.6 m

Two-layer system.

Solution 17.4 a. The equivalent thermal conductivity of the system if the heat flows horizontally can be calculated using Eq. 17.26: ke =

k1 h1 + k2 h2 0.9 × 0.2 + 1.6 × 0.6 = = 1.42 W∕m.K h1 + h2 0.2 + 0.6

The equivalent thermal conductivity of the system if the heat flows horizontally can be calculated using Eq. 17.32: ke =

h1 + h2 0.2 + 0.6 = = 1.34 W∕m.K h1 h2 0.2 0.6 + + 0.9 1.6 k1 k2

Problem 17.5 Calculate the change in volume of a saturated soil with a water content of 30% and unit weight of 18kN∕m3 if 90% of the water by weight becomes frozen. Solution 17.5 First, we have to calculate the weight of the water in unfrozen conditions. The total unit weight γ is: 𝛾t =

Ww + Ws = 18 kN∕m3 Vt

The water content is 30%; therefore, Ww = 0.3Ws . Assuming a soil unit volume of 1 m3 , the water weight W w is 4.15 kN and the solid weight W s is 13.85 kN. The volume of water in the unfrozen condition is: W 𝛾w = w = 10 kN∕m3 or Vw = 0.415 m3 Vw If 90% of the water weight becomes frozen, then the weight of ice W i is 3.735 kN. The unit weight of ice 𝛾i is 10% less than the unit weight of water; therefore, the volume of ice is: Vi =

Wi 3.735 = = 0.415 m3 𝛾i 9

522

17 THERMODYNAMICS FOR SOIL PROBLEMS

The volume of remaining unfrozen water is: Vw =

Ww (4.15 − 3.735) = = 0.0415 m3 𝛾w 10

The total volume of water and ice when 90% of the water mass becomes frozen is 0.4565 m3 ; therefore, the change in volume of the soil is 0.0415 m3 or 4.15% of the original volume. Problem 17.6 Add a column to Table 17.1 dealing with electricity. Write Ohm’s law and compare it to Darcy’s and Fourier’s laws. Solution 17.6 Parameter

Flow of water

Flow of heat

Flow of current

Quantity Potential Gradient Flux Flux density Conductivity Law Storage Decay coefficient

Volume V (m3 ) Head ht (m) Hydraulic gradient ih (unitless) Flow rate Q (m3 /s) Velocity v (m/s) Hydraulic conductivity kh (m/s) Darcy Compressibility Coefficient of consolidation cv (m2 /s)

Heat Q (J) Temperature T (K) Temperature gradient it (K/m) Heat transfer rate H (J/s) Heat flow q (J/s.m2 ) Thermal conductivity kt (J/s.K.m) Fourier Specific heat c (J/kg.K) Thermal diffusivity 𝛼 (m2 /s)

Electric charge (C) Voltage (V) Electric field gradient E (V/m) Electric current flow (C/s) Electrical flux density (C/m2 ) Electric conductivity, 𝜎 (S/m) Ohm Capacitance Electrical diffusivity D (m2 /s)

Problem 17.7 A 0.3 m in diameter, 10 m-long probe is pushed into a clay. The clay has a thermal conductivity equal to 1.2 W/m.K and a thermal diffusivity equal to 2 × 10−6 m2 ∕s. The probe is at 2500∘ C and the intent is to bake the clay in place to create a baked-in-place pile with a wall thickness equal to 0.1 m. If the clay becomes permanently solidified at 1700∘ C and the initial clay temperature is 0∘ C, how long will it take before the pile is “cooked” and the probe can be removed to bake the next pile? Solution 17.7 The probe is considered to be an infinite cylindrical heat source, because the length-to-diameter ratio is very large. The increase in temperature that must be achieved at a radial distance R = Ro + 0.1 = 0.25 m is 1700∘ C. Using Figure 17.2: Tm ∕To = 1700∕2500 = 0.68 R∕Ro = 0.25∕0.15 = 1.67 log(R∕Ro ) = 0.222 From Figure 17.2, Tf = 1. Therefore: Tf = 1 =

R2o t×𝛼 0.152 ⇒ t = = 11250 sec . = 3.125 hours = 𝛼 R2o 2 × 10−6

Problem 17.8 Calculate the thermal conductivity for the soil near the thermal cone penetrometer test from Figure 17.9 at a depth of 13 m. Solution 17.8 It is assumed by extrapolation of the measured data that the equilibrium temperature at infinite time for that test is 22.5∘ C. Since the initial temperature is 25.1∘ C, the 50% dissipation temperature is 23.8∘ C. For that temperature the 50% time is read on the curve of Figure 17.9 as 400 seconds. Using Eq. 17.39 gives 𝜆=

77 = 0.233 W∕m.K (400)0.968

CHAPTER 18

Shallow Foundations

18.1

DEFINITIONS

Shallow foundations (Figure 18.1) are those placed close to the ground surface, typically at a depth less than one times the width of the foundation. A 1 m thick, 3 m by 3 m foundation under a column, placed at a depth of 1.5 m, would be a shallow foundation called a spread footing. Spread footings can be square, circular, or very long compared to their width, in which case they are called strip footings. A 3 m thick, 40 m by 40 m square foundation, placed at a depth of 10 m, would be considered a particular type of shallow foundation called a mat foundation. A 0.1 m thick, 15 m by 15 m foundation stiffened with 1 m deep beams 3 m apart in both directions would be a shallow foundation called a stiffened slab on grade.

18.2

CASE HISTORY

This case history illustrates the behavior of shallow foundations. Five tests of spread footings were performed at the National Geotechnical Experimentation Site at Texas A&M University. The soil at the site is a medium-dense, fairly uniform, silty fine silica sand with the following average properties near the footings and within the top 5 meters: mean grain size D50 = 0.2 mm, SPT (standard penetration test) blow count 18 blows per 0.3 m, CPT (cone penetrometer test) point resistance 6 MPa, PMT (pressuremeter test) limit pressure 800 kPa, PMT modulus 8.5 MPa, DMT (dilatometer test) modulus 30 MPa, borehole shear test friction angle 32o , estimated total unit weight 15.5 kN/m3 , and cross-hole shear wave velocity 240 m/s. The water table is 4.9 m deep. Additional data can be found in Briaud and Gibbens (1994; 1999). Geologically, the top layer of sand is a flood plain deposit of Pleistocene age about 3 m thick with a high fine content. The next layer of sand is a river channel deposit of Pleistocene age about 3 m thick, clean and uniform. The third layer is a mixed unit with an increasing amount of clay seams and gravel layers; it is also of Pleistocene age and was deposited by a stream of fluctuating energy. Below these 200,000-year-old sand layers and about 10 m below the ground surface is the 45-million-year-old Eocene bedrock;

this bedrock is a dark gray clay shale that was deposited in a series of marine transgressions and regressions. Erosion of the Eocene marine clay took place before the Pleistocene river sediments were deposited. The test setup is shown in Figure 18.2. The five footings were square with a side dimension equal to 1 m, 1.5 m, 2.5 m, 3 m, and 3 m. They were embedded 0.75 m into the sand and were 1.2 m thick. They were loaded in load step increments, each one lasting 30 minutes, while settlement was recorded every minute during the load step. All footings were pushed downward until the settlement reached 0.15 m. Figure 18.3 shows an example of the load settlement curve obtained for the 3 m by 3 m north footing, as well as the log of the settlement vs. the log of time for several load steps. The pressure vs. settlement curves for all footings are shown in Figure 18.4. These curves were normalized by dividing the pressure by the limit pressure of the pressuremeter and the settlement by the width of the footing. Figure 18.4 indicates that this normalization makes the footing size disappear: The p/pL vs. s/B curve becomes a property of the soil, much like a stress-strain curve. Telltales and inclinometers were placed below and on the side of the footing, respectively. They indicated the depth to which the soil was compressed and the lateral movement of the soil during the load application. Figure 18.5 shows the soil movement as a function of depth for four of the footings and the lateral movement for the 3 m north footing. The data indicate that most of the settlement and lateral movement occurs within one footing width below the footing. 18.3

DEFINITIONS AND DESIGN STRATEGY

The most important considerations in foundation design are to ensure: 1. The safety of the foundation against soil failure (ultimate limit state). 2. The functionality of the foundation and the structure above by minimizing the foundation movement and distortion (serviceability limit state). 3. The safety of the foundation against structural failure.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

523

524

18 SHALLOW FOUNDATIONS

1.5 m 3m Spread footing (square or circular)

1.5

20

Strip footing

m

m

Stiffened slab on grade Building

m

3m

12

30 m Mat foundation

Figure 18.1 Types of shallow foundations.

Item 3 is handled primarily by the structural engineer and is not covered in this book. Items 1 and 2 in the preceding list are primarily geotechnical engineering considerations involving soil shear strength and the soil increase and decrease in volume when loaded. They are the topic of this chapter for shallow foundations and of Chapter 19 for deep foundations. The geotechnical design of a shallow foundation consists of estimating the size and depth of the foundation. The depth is chosen on the basis of several factors, including profile of soil strength and compressibility, depth of the zone that shrinks and swells, depth of frost penetration, groundwater level, and ease of construction. The size is typically chosen once the depth is chosen. No foundation can be designed to ensure zero probability of failure. This is because any calculation is associated with some uncertainty; because the engineering profession’s knowledge, while having made great strides, is still incomplete in many respects; because human beings are not error-free; because budgets are limited; and because the engineer designs the bridge or building for conditions that do not include extremely unlikely events, such as a big airplane hitting the bridge at the same time as an earthquake, a hurricane, and a 500-year-flood during rush hour. The engineer and the public must accept a certain level of probability of failure. This acceptable level of probability of failure is tied to the number of deaths that the public accepts on a daily basis (fatalities) and to the amount of money that it can afford

8.5 m Settlement beam

LVDT

B

Sand

Jack

Load cell

0.5B

Telltales 2B

Inclinometer casings

10.7 m B Dywidag bars only no concrete

15 m

Clay shale 7.6 m

Drilled shaft (concrete + bars) Steel plates

2.7 m

2.7 m

Figure 18.2 Load test setup

18.3 DEFINITIONS AND DESIGN STRATEGY

Load settlement curve with 30-minute load steps 2

4

Load (MN) 6 8

10

12

3.0 m footing north

–20 –40 Settlement (mm)

Settlement time curves for each load step

0.06 Log displacement log10 (s/s1)

0

0

–60 –80 –100 –120 –140

525

–160

0.05 0.04

6.23 MN 7.12 MN 8.01 MN 8.9 MN 9.79 MN 10.24 MN

3.0 m footing north

0.03 0.02 0.01 0.00 0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

Log time, log10 (t/t1) (b)

(a)

Figure 18.3 Result for the 3 m by 3 m north footing: (a) Pressure-settlement curve. (b) Normalized curves. Pressure (MPa) 0.0 0

0.5

Pressure /PMT limit pressure 1.5

1.0

2.0

0.0 0.00

1m

20

2.5 m 80

Embedment = 0.75 m

0.08 0.10 0.12 0.14 0.16

140

2.0

0.06

3 m South

120

1.5

0.04

Settlement / width

Settlement (mm)

3 m North

60

100

1.0

0.02

1.5 m 40

0.5

B = 1.0 m B = 1.5 m B = 2.5 m B = 3.0 m (N) B = 3.0 m (S)

0.18

(a)

(b)

Figure 18.4 Pressure vs. settlement curve for all footings and normalized curves: (a) Pressure-settlement curve. (b) Normalized curves.

to spend (economy). In geotechnical engineering and in structural engineering, this acceptable probability of failure is typically less than 1 chance in 1000 (10–3 ). Design procedures have been developed to calculate a foundation size that meets these low probabilities of failure. These procedures involve: 1. Selecting the design issues (limit states). 2. Selecting load factors and resistance factors that are consistent with the low target probability of not meeting the design criterion.

3. Determining the minimum size of the foundation that satisfies the low probability of not meeting the design criterion. For example, let’s go back in time to the year 1100 and design the foundation of the Leaning Tower of Pisa, but with today’s knowledge. The load is calculated to be 150,000 kN. The uncertainty about this load is small because the dimensions of the structure are on the plans. Nevertheless, a load factor of 1.2 is used to obtain the factored load of 1.2 × 150,000 = 180,000 kN, which lowers the probability of

526

18 SHALLOW FOUNDATIONS

Pressure-settlement curve

Normalized curves 0.0

0.0

–2.0 –4.0

–1.0

–1.5

S(top) = 0.05 m 1.0 m

–6.0 –8.0

3.0 m footing north. load 8.9 MN

h = 5.3 m

–10.0

1.5 m

–2.0

Depth (m)

Relative depth, z/b

–0.5

3.0 m (n)

h = 2.7 m

–12.0

h = 0.2 m

3.0 m (S) –2.5 0.00

0.25 0.50 0.75 Relative soil movement, s/stop

1.00

–14.0 –20

–10

0

10

20

Deflection (mm)

Figure 18.5 Vertical and horizontal movement vs. depth.

exceeding the load. The resistance is the ultimate bearing pressure of the soil below the tower. It is calculated as 6 times the undrained shear strength su of the soil within the depth of influence of the foundation (Section 18.6.1). From the borings, in situ tests, and laboratory tests, a value of 80 kPa is selected for su . This leads to an ultimate bearing pressure of 480 kPa. The uncertainty associated with the undrained shear strength and the calculation model is not negligible, so a resistance factor of 0.6 is selected. The factored resistance is 0.6 × 480 = 288 kPa, which lowers the probability of not having the necessary resistance. The load factor 1.2 and the resistance factor 0.6 are based on the probability distribution of the load and of the resistance, and on ensuring that the probability that the difference between the factored load and the factored resistance is negative (failure) is less than approximately 10–3 . The difference between the load and the resistance is called the limit state function. We decide to place the 15 m-diameter Tower of Pisa on a circular mat foundation 1 m thick with a diameter B. Now the ultimate limit state equation is written as: 1.2 × 150000 < 0.6 × 480 × 𝜋 B2 ∕4

(18.1)

which leads to B > 28 m. The actual, as-built foundation was less than 15 m in diameter and the soil below the foundation failed. The design should also include other considerations such as the serviceability limit state, but this simple example illustrates the design process and the concept of load and resistance factors. More specifically, the design process proceeds as follows: 1. Decide on the foundation depth. 2. Make a reasonable estimate of the foundation size. 3. Calculate the ultimate bearing pressure of the foundation, pu .

4. Check if the ultimate bearing pressure satisfies the safety criterion under the given load (ultimate limit state). 5. Repeat steps 1 through 3 until the safety criterion is satisfied and obtain the safe foundation pressure ps , which is the unfactored load divided by the foundation area. 6. Under the safe foundation pressure ps , check that the foundation satisfies the serviceability limit state by calculating the movement of the foundation and ensuring that it is less than the allowable movement. 7. If the calculated movement is larger than the acceptable movement sa , increase the foundation size and/or the foundation stiffness and repeat step 6. 8. If the movement is acceptable, the design is complete, as the pressure applied is safe and allows only acceptable movement. In addition to the preceding steps concerning soil strength and compressibility, the foundation must be well designed structurally. For example, one must ensure that the column will not punch through the spread footing, or that the mat foundation will not bend excessively. The structural aspect of foundation design is not covered in this book. Shallow foundations are typically less expensive than deep foundations. Therefore, it is economically prudent, in most cases, to start with a shallow foundation solution. Only if it is shown to be insufficient or inappropriate should the design proceed with deep foundations. 18.4 LIMIT STATES, LOAD AND RESISTANCE FACTORS, AND FACTOR OF SAFETY Limit states are the loading situations and the associated equations that are considered during the design of a foundation. They must be satisfied to yield a proper design. There

527

18.4 LIMIT STATES, LOAD AND RESISTANCE FACTORS, AND FACTOR OF SAFETY

Table 18.1

Typical load factors for ultimate limit state

Type of loading

Load factor 𝜸 (AASHTO) for bridges

Load factor 𝜸 ′ (ASCE 7) for buildings

Load factor 𝜸 E (Eurocode 7) for buildings

Dead load and permanent live load Other live load Extreme events (earthquake, hurricane, etc.)

1.25 1.75 1

1.2 1.6 1

1.35 1.5 1

Table 18.2

Typical resistance factors for ultimate limit state and shallow foundations (Eurocode 7) Resistance factor 𝜑 (AASHTO)

Material factor 𝛾 M = 1/𝜑

Resistance factor 𝛾 R = 1/𝜑

Many high-quality tests

0.5–0.6

1.25–1.4

1.1–1.7 (footings)

Ordinary quantity and quality of tests

0.4–0.5

(may be reduced for extreme events)

1.1–1.6 (piles)

Type of soil testing

1

Extreme events (earthquake, hurricane, etc.)

are two major limit states: the ultimate limit state and the service limit state. In foundation engineering, the ultimate limit state involves calculations of ultimate capacity using primarily the shear strength of the soil. Satisfying the ultimate limit state ensures that the foundation will meet a chosen level of safety against failure. The service limit state involves calculations of movements using deformation parameters. Satisfying the service limit state ensures that the foundation will meet a chosen degree of confidence against excessive movement or distortion of the structure. The ultimate limit state refers to satisfaction of equations ensuring that the foundation will function far enough away from failure of the soil. This requires the choice of load factors 𝛾 and resistance factors 𝜑 that will achieve the chosen level of probability of success. These equations are of the form: 𝛾L < 𝜑R (18.2)

(may be reduced for extreme events)

An example of an ultimate limit state equation is: 1.25 DL + 1.75 LL < 0.5 Ru

(18.4)

where 𝛾 is the load factor, L is the load, 𝜑 is the resistance factor, and R is the resistance. The resistance here is meant to be the ultimate resistance of the soil. In the case of complex loading and multiple resistances, Eq. 18.2 becomes: ∑ ∑ 𝜑i Ri (18.3) 𝛾i Li
2 < 0.2 0.2 to 0.5 0.5 to 1 1 to 2 >2

4 < 1.5 1.5 to 4 4 to 10 10 to 20 > 20

Sand, gravel

(18.9)

Clay and silt—strip footing: )( ) ( D D 1 − e−1.3 B kp = 0.8 + 0.2 + 0.02 B

6.0 5.0

Nc(square) = 1.2Nc(strip)

The PMT bearing capacity factor kp is given in two steps (Frank, 1999/2013; Norme Française (2013), AFNOR P94–261), first, a soil classification step (Table 18.4) and then an equation for each soil category (Eqs. 18.13–18.18).

8.0

7.0

directions. The N c values for the square footing and the strip footing are related by:

(Source: After Norme Française (2013), AFNOR P94-261 and Frank (1999/2013).)

SPT N (bpf)

Shear strength su( ( kPa) < 75 75 to 150 150 to 300 > 300

10 4

Figure 18.27 Strain influence factor. (Source: After Schmertmann, 1970.)

in pressure expressed as the difference between the footing ′ pressure p (load over area) minus 𝜎ov . The coefficient C2 is: ( ) t(years) C2 = 1 + 0.2 log (18.73) 0.1 where t is the time in years. The strain influence factor I zi is such that Izi × Δp represents Δ𝜎i in Eq. 18.69. It is shown in Figure 18.27. In Figure 18.27, I z increases first and then decreases. The peak value of I z is called I zp . It is shown as 0.5 on Figure 18.27 but in fact it is given by: )0.5 ( Δp (18.74) Izp = 0.5 + 0.1 ′ 𝜎Izp ′ is the vertical effective stress at the depth of I zp . The where 𝜎Izp soil modulus Ei is recommended by Schmertmann as follows:

For circular or square footings

E = 2.5qc

(18.75)

For strip footings (L∕B > 10)

E = 3.5qc

(18.76)

where qc is the CPT point resistance. Schmertmann adds the conversion values of Table 18.7 between qc and N. 18.8.4

Chart Approach

The chart approach consists of simplifying the problem sufficiently so that the calculations are minimized and a

Table 18.7 sands

Conversion from CPT to SPT values for

Soil Silts, sandy silts, slightly cohesive silt-sand Clean, fine to medium sands and slightly silty sands Coarse sands and sands with little gravel Sandy gravel and gravel

qc (kPa)/ N (bpf) 200 350 500 600

(Source: After Schmertmann, 1970.)

chart can be read for the answer. Such a chart approach was developed by Terzaghi and Peck (1967) for footings on sand (Figure 18.28). This chart is only for footings on sands, and it gives the pressure that satisfies both the ultimate bearing pressure criterion and the settlement criterion of 25 mm. This chart was developed before LRFD was developed and as such is based on the following. The safe pressure criterion ensures that a reasonable factor of safety is applied to the ultimate bearing pressure: p psafe = u (18.77) F

544

18 SHALLOW FOUNDATIONS

γ

Settlement controls

0.5 γB N

Pressure (kPa)

11.1NSPT

Ultimate pressure controls Foundation width (m)

(a) Df /B = 1

Pressure for 25 mm settlement pall (kPa )

600 500

Df /B = 0.5

Df /B = 0.25

N = 50

N = 50

N = 50

N = 40

N = 40

N = 40

N = 30

N = 30

N = 30

N = 20

N = 20

N = 15 N = 10

N = 15 N = 10

N = 20 N = 15 N = 10

N=5

N=5

N=5

400 300 200 100 0

0 0.25 0.5 0.75 1.0 1.25

0 0.25 0.5 0.75 1.0 1.25

0

0.5

1.0

1.5

2.0

Width of footing, B (m) (b)

Figure 18.28 Chart for pressure leading to 25 mm settlement of footings on sand. (Source: Terzaghi and Peck, 1967.)

where psafe is the safe bearing pressure, pu is the ultimate bearing pressure, and F is the factor of safety. The allowable pressure criterion ensures that the settlement will be less than 25 mm in this case: pallowable = p for 25 mm settlement

(18.78)

The chart of Figure 18.28 gives the minimum of psafe and Pallowable . The first part of the design curves on the chart increases linearly with the width B of the footing for the following reason. For small values of B, it turns out that the ultimate bearing pressure criterion controls the design, and since there is no cohesion for sands, it is expressed as: 1 𝛾BN𝛾 (18.79) 2 As a result, psafe increases linearly with B. The influence of the depth of embedment D is included by having several charts for different relative depths of embedment D/B. For the settlement s of the footing, Terzaghi and Peck found that s was pu =

proportional to the SPT blow count as follows: pallowable (kPa) for 25 mm settlement = 11.1N(blows∕0.30 m) (18.80) This indicates that Pallowable is not a function of B and therefore it shows up as a horizontal line on Figure 18.28. As a result, the ultimate pressure criterion controls for small footings and the settlement criterion controls for larger footings. If Eq. 18.80 is extended to other settlement values, and assuming linear behavior, the equation becomes: p (kPa) (18.81) N (bpf) A database called TAMU-SHAL-SAND was collected to evaluate the precision of the Terzaghi and Peck’s chart (Bahmani and Briaud, 2020). An example result of this study is shown in Figure 18.29 indicating that the chart is conservative for shallow foundations up to 7 m in size and more true beyond that size. It was found that the settlement given by Eq. 18.81 needs to be doubled to predict s (mm) = 2.3

18.8 SETTLEMENT

545

1400 Pressure for 25 mm settlement Pall (kPa)

Burland 20 −u) W = 3200 kN/m

9m

Layer 1

1 2

Layer 2

4m

L=2

Su (μ = 70 kPa, σ = 20 kPa)

We are looking for the probability: ) ( F − 𝜇F 1 − 1.527 < P(F < 1) = P 𝜎F 0.436 = P(U < −1.21) = 1 − P(U < 1.21)

(20.112)

(20.113)

Table 12.3 gives: Figure 20.32 Probabilistic slope calculations example.

turns out to be 0.001, you are very comfortable, as such a low probability of failure is well within the acceptable range for common civil engineering projects. A background on probability is presented in Section 12.6.1. The procedure for obtaining the probability of failure is outlined in Section 12.6.2. A sample calculation of the probability of failure for a slope is given in Section 12.6.4. The following simple examples illustrate the calculations to obtain the probability of failure. 20.15.1

Example 1

A slope exists as shown in Figure 20.32. It is made of clay with a normally distributed undrained shear strength su having a mean of 70 kPa and a standard deviation of 20 kPa. The failure circle has a radius of 16 m and the length of the arc is 24 m. The weight of the soil mass within the circle is 3200 kN per meter of length perpendicular to the page on Figure 20.32 and the horizontal distance between the center of the circle and the center of gravity of the soil mass is 5.5 m. The deterministic value of the factor of safety is: 16 × 24 RL s = 70 = 1.527 Wa u 3200 × 5.5 Note that if: F=

Y = aX

then

𝜇Y = a𝜇X

(20.107)

(20.108)

Therefore, the value of the factor of safety in Eq. (20.107) is also the mean of F, 𝜇F . Let’s calculate the standard deviation of F. Again we note that if: Y = aX then

𝜎Y = a𝜎X

(20.109)

Therefore, the standard deviation of F is: RL 16 × 24 𝜎su = × 20 = 0.436 (20.110) Wa 3200 × 5.5 Failure occurs when F < 1 and the probability that F < 1 is P(F < 1), which can be evaluated as follows. The first step is to transform F into the standard normal variable U: F − 𝜇F F − 1.527 = (20.111) U= 𝜎F 0.436 𝜎F =

P(U < 1.21) = 0.8869

(20.114)

and the probability of failure for this case is: P(F < 1) = 1 − 0.8869 = 0.1131

(20.115)

For most civil engineering works, this would not be an acceptable probability of failure but the deterministic value of the factor of safety (1.527) would be. 20.15.2

Example 2

Consider the same slope geometry and the same circle, but now with the soil made of two layers. The top layer (crust) has a mean su value of 150 kPa, a standard deviation of 30 kPa, and a failure circle arc length of 6.5 m. The bottom layer has a mean su value of 70 kPa, a standard deviation of 20 kPa, and a failure circle arc length of 17.5 m. The weight and center of gravity of the soil mass are unchanged, with a weight of 3200 kN/m and a moment arm of 5.5 m. The deterministic value of the factor of safety is: R(L1 su1 + L2 su2 ) 16(6.5 × 150 + 17.5 × 70) = = 2.0 Wa 3200 × 5.5 (20.116) Note that if:

F=

Y = aX + bZ

then

𝜇Y = a𝜇X + b𝜇Z

(20.117)

Therefore, the value of the factor of safety in Eq. 20.116 is also the mean of F, 𝜇F . It makes sense that the factor of safety would be higher than in Example 1, because we have replaced part of the soil with a stronger layer. Let’s calculate the standard deviation of F. Again we note that if: √ (20.118) Y = aX + bZ then 𝜎Y = a2 𝜎X2 + b2 𝜎Z2 Therefore, the standard deviation of F is: √ ) ) ( ( 16 × 6.5 2 2 16 × 17.5 2 2 𝜎F = 30 + 20 = 0.364 3200 × 5.5 3200 × 5.5 (20.119) We follow the same process as in Example 1: U=

F − 𝜇F F − 2.0 = 𝜎F 0.364

(20.120)

20.16 THREE-DIMENSIONAL CIRCULAR FAILURE ANALYSIS

Then, we are looking for the probability: ( ) F − 𝜇F 1 − 2.0 P(F < 1) = P < = P(U < −2.747) 𝜎F 0.364 = 1 − P(U < 2.747) (20.121) Table 12.3 gives: P(U < 2.747) = 0.997

701

you are very likely to make errors that tend to balance each other or reduce the error. The reason for this balancing error is that if you randomly pick a value that is too high for the first layer, you are more likely to pick a value that is too low for the second layer, as there are more values lower than your first guess. If there is only one layer, you have only your first guess for the calculations.

(20.122)

and the probability of failure for this case is: P(F < 1) = 1 − 0.997 = 0.003

(20.123)

This is an acceptable probability of failure in civil engineering. The main reason why the probability of failure has dramatically decreased (from 0.113 to 0.003) is that the mean factor of safety (2 instead of 1.527) is now further away from the failure value (F = 1). 20.15.3

Example 3

Let’s repeat Example 1, but with two layers as in Example 2, except that these two layers are now identical and made of the Example 1 soil: mean su = 70 kPa and standard deviation of su = 20 kPa. The new calculations are as follows. The deterministic value of the factor of safety is: R(L1 su1 + L2 su1 ) 16(6.5 × 70 + 17.5 × 70) = Wa 3200 × 5.5 = 1.527 (20.124)

F=

It makes sense that we find the same factor of safety as in Example 1. The standard deviation of F has changed, however; it is now: √ ) ) ( ( 16 × 6.5 2 2 16 × 17.5 2 2 20 + 20 = 0.339 𝜎F = 3200 × 5.5 3200 × 5.5 (20.125) We follow the same process as in Example 1: U=

F − 𝜇F F − 1.527 = 𝜎F 0.339

20.16 THREE-DIMENSIONAL CIRCULAR FAILURE ANALYSIS The analysis of the circular failure in the preceding sections has assumed a plane strain condition. This means that the failing soil body has the shape of a cylindrical sector. Although in most cases this is a reasonable approximation, slope failures are always three-dimensional (Figure 20.33). Three-dimensional or 3D slope failure analyses can be performed, but are not done as commonly as plane strain analyses. One reason is that, most of the time, the assumption of a plane strain condition leads to a conservative factor of safety. One way to perform a 3D slope stability analysis is to decompose the soil volume into a series of slices, each of which is considered to be a plane strain case (Figure 20.34). Many different assumptions can be made for such a mechanism, as was done for the 2D case. Some of them include a common axis of rotation for all circles and no forces between the circle slices. It can be seen in Figure 20.34 that if the deepest circle in the center of the volume is the critical circle for the 2D case, then all other circles will have a factor of safety higher than the 2D critical circle. From this observation, it follows that the global factor of safety for the 3D volume will be higher than for the 2D case.

(20.126)

Then, we are looking for the probability: ) ( F − 𝜇F 1 − 1.527 < = P(U < −1.555) P(F < 1) = P 𝜎F 0.339 = 1 − P(U < 1.555) (20.127) Table 12.3 gives: P(U < 1.555) = 0.9400

(20.128)

and the probability of failure for this case is: P(F < 1) = 1 − 0.940 = 0.060

(20.129)

This is about half the probability of failure calculated in Example 1, yet the soil conditions are the same except that we divided the soil into two identical layers. The reason for the decrease in the probability of failure is that if you randomly select a shear strength value from two identical distributions,

Figure 20.33 Three-dimensional slope failure. (Source: Courtesy of Gordon W. Hunter, British Columbia Ministry of Transportation and Infrastructure.)

702

20 SLOPE STABILITY

the 3D volume becomes: m n ( ) ∑ ∑ 1 c′ij bij + (Wij − 𝛼ij uwij bij ) tan 𝜑′ij Rj wj m j=1 i=1 𝜃ij F3D = m n ∑ ∑ Rj wj Wij sin 𝜃ij

Axis of rotation W wj

j=1

Rj

H Plane strain circle

Shallow edge circle

Figure 20.34 Decomposition of a 3D soil body into 2D soil slices.

Ratio of 3D/2D factor of safety

3.5 1h.1v 3h.1v 5h.1v

3

2 1.5 1 2

0

4

6

8

10

12

Width/height

Figure 20.35 Comparison of 3D to 2D factor of safety. (Source: After Stark, 2003.)

Though there are some exceptions to this statement, it is the general trend and has been documented by Stark (2003), for example (Figure 20.35). In Figure 20.35, H is the height of the slope and W is the width of slope considered in the analysis (sum of wj on Figure 20.34). Three curves are shown for three different slope angles (1h to 1v, 3h to 1v, and 5h to 1v). For the 2D case, the factor of safety of the jth circle (Bishop simplified) is modified after Eq. 20.91: ) ∑ 1 ( c′ij bij + (Wij − 𝛼ij uwij bij ) tan 𝜑′ij m i=1 𝜃ij n

F2D, j =

∑ Wij sin 𝜃ij n

(20.131) where Rj and wj are the radius and width of the circle slice j respectively. The drastic assumptions associated with this equation limit its applicability to a first estimate. A number of computer programs are available to perform more sophisticated 3D analyses. The goal of the assumptions, as in the 2D case, is to make the problem a statically determinate problem and to satisfy equilibrium equations in all directions. In the end, the finite element method is again the best way to solve the 3D problem, because with this method all equilibrium equations will automatically be satisfied for all elements of soil.

20.17

2.5

(20.130)

i=1

where all parameters are defined in Table 20.1 and j is the number of the circle slice as shown in Figure 20.34. Then, if the axis of rotation is the same for all slices and if the forces between the circle slices are neglected, the factor of safety for

i=1

FINITE ELEMENT ANALYSIS

The finite element method (FEM) can be used to analyze the stability of slopes. The mesh should be large enough that the boundaries have only a small and tolerable influence on the stability calculations. If the height of the slope is H, the mesh should be at least 3 H high. If the horizontal distance between the toe and the crest of the slope is L, the mesh should be at least 5 L long. The advantages of the FEM over the limit equilibrium method (LEM) are that (Griffiths and Lane, 1999): 1. No assumptions need be made about the failure surface; the weakest surface will automatically be found through the stress field calculated as part of the solution. 2. All equilibrium equations are satisfied. 3. In addition to a factor of safety, information is obtained on the displacements of the slope. This information, of course, is only as good as the soil model and soil properties used to obtain it. 4. The FEM includes complex issues such as progressive failure up to complete failure. The factor of safety is determined through the use of a strength reduction factor (SRF), which is applied to the strength parameters of the soil: c′r =

c′ SRF

(20.132)

tan 𝜑′ (20.133) SRF where c′ and 𝜑′ are the effective stress cohesion and friction angle of the soil, respectively, and c′r 𝜑′r are the reduced effective stress cohesion and friction angle of the soil, respectively. tan 𝜑′r =

20.19 SLOPE STABILITY DATABASES

703

that convex corners deform more than the plane strain case and that concave corners deform less than the plane strain case. The radius r of the apex angle at the top of the slope also has an impact on the difference between corner and plane strain behavior; Figure 20.37 is for a corner radius over slope height, r/H, equal to 0.33. Figure 20.36 Failed slope in the finite element method. (Source: Courtesy of Griffiths and Lane, 1999.)

20.19

Databases of slope stability case histories have been organized over the years. They fall into two categories: natural landslides databases and human-made slopes databases either excavated or built. Timchenko (2021) reviewed some of those databases and organized a comprehensive database of open-pit mine slope failures called TAMU-Mine Slope. Using this database, Timchenko made recommendations for the setback distance LSB and for the travel distance LTRAV . The setback distance LSB is the horizontal distance between the crest of the slope before failure and the top of the failure surface while the travel distance LTRAV is the horizontal distance between the crest of the slope before failure and the end of the failed slope mass after coming to rest (Figure 20.38). LSB is useful to ensure the safety of people and equipment away from the slope crest and LTRAV is useful for managing slope failure and its impact on mining operations. Timchenko’s recommendations, based on regression analysis, are (Figures 20.39 and 20.40):

The FEM is performed repeatedly as the values of c′r and 𝜑′r are gradually decreased by using an increasing SRF. The failure criterion can be defined in various ways (Abramson et al., 2002): 1. Bulging of the slope surface. 2. Limiting shear stress reached on the failure surface. 3. Nonconvergence of the solution. When an agreed-upon failure criterion is reached, the SRF is equal to the safety factor. Figure 20.36 shows an FEM output of a failed slope.

20.18

STABILITY OF SLOPE CORNERS

Most slope stability analyses consider the plane strain case as the base case. Slope corners do not satisfy the plane strain condition and as a result the stability of slope corners can be significantly different from the plane strain case. Timchenko (2021) studied this issue and quantified the differences. She defined convex corners as corners with a plan view angle less than 180 degrees and concave corners as corners with a plan view angle larger than 180 degrees. Figure 20.37 indicates 3

SLOPE STABILITY DATABASES

LSB (m) = 0.41(HFAIL (m))0.82 LTRAV =

(20.135)

where H FAIL and H FALL are defined in Figure 20.38. Timchenko (2021) further recommends a factor of safety of ut(3D FEM corner)

ut(3D FEM plane strain)

2.5

HFALL tan 𝜑

=

. e4.57

43.0 +0.57 α (°) 180° +24.8

Slope angle, β=15° Slope angle, β=25°

1.5

Slope angle, β=35°

α r

Slope angle, β=45°

w

ut(3D FEM corner)

ut(3D FEM plane strain)

R2=0.94 2

1

0.5 Slope height H=15m 0 0.0

0.2

0.4

Convex 0.6

(20.134)

0.8

Concave 1.0

1.2

1.4

1.6

1.8

2.0

α (°)/180° Figure 20.37 Normalized displacement of slope corners vs. corner plan view angle. (Source: Courtesy of Timchenko, 2021.)

704

20 SLOPE STABILITY LMAX

Open Pit Top

LSB LTRAV

HFAIL

βFAIL

HFALL HMAX β

LR

HBE

Open Pit Bottom

βOP

Figure 20.38 Parameter definition for slope distances. (Source: Courtesy of Timchenko, 2021.) 1000

10,000 Case history and its number according to TAMU-MineSlope Best fit regression line

α=2.5

100 α=1.0

10

LSB = (HFAIL)0.82

1

100

. LTRAV = 0.98 (HFALL/tanφ) 2 R = 0.54

10 100 Measured HFAIL (m)

1000

Figure 20.39 Setback distance away from slope crest versus failing height. (Source: Courtesy of Timchenko, 2021.)

2.5 for the setback distance considering the scatter in the data. (20.136) LSB SAFE (m) = (HFAIL (m))0.82 SEISMIC SLOPE ANALYSIS

An earthquake can induce failure of a slope that is statically safe. The reason is twofold: The earthquake increases the driving moment, mostly through horizontal shaking; and in some soils it can decrease the shear strength of the soil by increasing the water compression stress through cyclic loading, possibly leading to liquefaction. There are several ways to include earthquake loading in slope stability analysis: 1. 2. 3. 4.

1,000

LSB = 0.41(HFAIL)0.82 R2 = 0.30

1

20.20

Measured LTRAV (m)

Measured LSB (m)

Case history and its number according to TAMU-MineSlope Best fit regression line

Pseudostatic method Newmark’s displacement method Post-earthquake stability method Dynamic finite element method

10 10

100

1,000

10,000

Measured HFALL /tanφ (m)

Figure 20.40 Slope failure mass travel distance versus falling height. (Source: Courtesy of Timchenko, 2021.)

20.20.1

Pseudostatic Method

The pseudostatic method is the most common and the simplest. It consists of adding a horizontal and vertical static force in the limit equilibrium analysis. These two forces are chosen to be equivalent to the effects of the inertia dynamic forces generated during shaking of the soil mass. They are assumed to be proportional to the weight W of the failing soil mass. The coefficients of proportionality are kh and kv for the horizontal and vertical direction, respectively (Figure 20.41). Most commonly, the vertical seismic coefficient kv is assumed to be zero and the horizontal seismic coefficient kh depends on the severity of the shaking. Table 20.6 (Abramson et al., 2002) is a summary of some common values for the seismic coefficient kh . The seismic force is usually placed at the center of gravity of the slice. Seismic analyses indicate that most of the time,

20.20 SEISMIC SLOPE ANALYSIS

Center of gravity of slice i

705

Table 20.7 Likelihood of failure for different values of ky

Wi

khWi kvWi

Figure 20.41 Pseudostatic seismic forces.

Relative position of ky and PGA

Observation

ky > PGA 0.5PGA < ky < PGA ky < 0.5 PGA

No failure likely Minor damage possible Failure likely

(Source: After Abramson et al.. 2002.)

Table 20.6 Values of the seismic horizontal coefficient kh Seismic coefficient kh

Comment

0.10

Major earthquake, U.S. Army Corps of Engineers, 1970, 1982 Great earthquake, U.S. Army Corps of Engineers, 1970, 1982 State of California Japan Marcuson and Franklin, 1983

0.15 0.05–0.15 0.15–0.25 1/3–1/2 of peak ground acceleration (PGA)

(Source: After Abramson et al., 2002.)

the peak acceleration increases as the wave propagates from the bottom to the top of the slope. This would mean that the point of application of the seismic force should be above the center of gravity of the slice (CG); this would generate smaller overturning moments than if that force were placed at the CG. Therefore, placing the seismic force at the CG of the slices, as is usual practice, is conservative in most cases. Another way to approach the problem is to find the horizontal seismic coefficient kh that would lead to failure of the slope. This value of kh , called the yield horizontal seismic coefficient ky , corresponds to a factor of safety of 1 (Figure 20.42). Then the value of ky can be compared

Factor of safety, f

2

Yield coefficient

1.5

kY

1

0.5

kY 0

0.1

0.4 0.2 0.3 Horizontal seismic coef, kh

Figure 20.42 Yield horizontal seismic coefficient.

0.5

to the peak ground acceleration (PGA) of the earthquake at the slope location. Abramson et al. (2002) suggest the observations in Table 20.7. Note that a very important part of the pseudostatic analysis, as for any slope stability analysis, is to select the correct shear strength. The issue here is that the shear strength during shaking is likely to be reduced compared to the static case. 20.20.2

Newmark’s Displacement Method

Newmark’s displacement method is credited to Newmark (1965). Whereas most other slope stability methods aim at predicting the factor of safety, this method aims at predicting the accumulation of displacement of the slope during a series of acceleration cycles, as in an earthquake, for example. The first step is to develop an acceleration history a(t) for the earthquake at the location of the slope (Figure 20.43). Then the yield acceleration ay is found by using the pseudostatic method. The relationship between ay and ky is: ay = ky g

(20.137)

where g is the acceleration due to gravity. Any acceleration above ay will lead to movement as the slope fails during a short time increment (points B to D in Figure 20.43). By integration of the net acceleration (a(t)–ay ) from B to D, the velocity of the soil mass is found (points B1 to D1 in Figure 20.43). Then the velocity decreases as the acceleration recedes below ay and the shear strength slows the soil mass down (points D1 to M1 in Figure 20.43). By integrating the velocity from B1 to M1 , the displacement of the slope mass is obtained (points B2 to M2 in Figure 20.43). At M2 , a permanent displacement has been accumulated; the displacement increases again when the acceleration exceeds the yield acceleration (point H in Figure 20.43). The process repeats itself until the earthquake is over. One of the key parts of this method is developing the acceleration history for the slope. This is discussed in Chapter 23. Makdisi and Seed (1978) performed a parametric analysis using actual and hypothetical dams and embankments. With the results, they simplified Newmark’s method and presented it in the form of a chart (Figure 20.44). The acceleration ratio ky ∕kmax is on the horizontal axis, where ky is the yield horizontal seismic coefficient corresponding to failure of the slope and kmax is the maximum acceleration

706

20 SLOPE STABILITY

this case is the proper selection of the shear strength existing after the earthquake. The steps for such an analysis (Duncan and Wright, 2005) include:

C

Acceleration a(t) B

A

D

H

E

G

ay

J K

Time F

J1

D1

L

Velocity v(t) B1

M1 H1

N1

20.20.4 Time

J2 N2 Displacement d(t)

M2

D2

H2

B2 Time

Figure 20.43 Newmark’s displacement method (1965). 1000 M = 81/4 (synthetic) M = 71/2 (taft)

Displacement (cm)

100

10 M = 61/2 (pasadena)

1

MONITORING

Monitoring consists of making observations or measurements on a slope in order to evaluate its stability. Among the most useful parameters to observe or measure are:

0

0.2

0.4

0.6

0.8

1.0

Acceleration - ky /kmax

Figure 20.44 Makdisi and Seed’s (1978) chart. (Source: This material is reproduced with permission of John Wiley & Sons, Inc.)

horizontal seismic coefficient corresponding to the maximum acceleration in the slope and defined as: amax = kmax g

(20.138)

The earthquake magnitude M (see Chapter 23) is selected for the site and the range of displacement is read on the vertical axis using the acceleration ratio and the magnitude M. 20.20.3

Dynamic Finite Element Analysis

A dynamic finite element analysis is a 2D or 3D dynamic analysis of the slope and its surroundings. The finite element mesh should be large enough that the effect of the boundaries does not significantly affect the stability calculations for the slope. The recommendations of Section 20.17 for static analysis may not be sufficient, as earthquake shaking creates waves that propagate against the boundaries and are reflected toward the slope. The soil model should incorporate the evolution of the strength and deformation properties as a function of cycles. The earthquake motion is usually induced at the bottom of the mesh and propagates upward through the slope mesh. The fact that the soil model can more closely follow the soil behavior and the fact that the dynamic equilibrium of the elements is satisfied at all times are two major advantages of this approach. The complexity of the approach is its drawback.

20.21

0.1

0.01

1. Study whether the soil has liquefied or not (see Chapter 23). 2. Determine the reduced shear strength due to cyclic loading (see Chapter 23). 3. Use that shear strength for a conventional stability analysis.

Post-Earthquake Stability Analysis

Post-earthquake stability analysis is a static analysis that considers the situation right after an earthquake. The main issue in

1. Cracks, particularly on top of the slope 2. Movements of the slope surface or at depth 3. Groundwater and water stress conditions Crack openings are indications that a slope is stressed (Figure 20.45). Cracks associated with instability on top of slopes are parallel to the slope crest and are several meters long. If the cracks are less than 25 mm wide, and if there is no difference in elevation between the two sides, the probability of failure is low. If the cracks are between 25 and 75 mm wide, with some difference in elevation between the two sides, the probability of failure is much higher. If the cracks are much larger than 100 mm with similar difference in elevation between the two sides, it is likely time to run. Instruments to measure cracks can be as simple as a tape measure or as sophisticated as an extensometer that monitors the distance between the two sides and sends readings to a remote monitoring station. It is very useful to monitor crack width b as a function of time t and plot the curve b vs. t.

20.21 MONITORING

(a)

(c)

707

(b)

(d)

Figure 20.45 Example of cracks on top of slopes. (Source: a: From Bray et al., 2001. Used by permission. b: Courtesy of Jonathan Wilkins. c: Courtesy of Dr. Ian West. d: British Geological Survey image P757993 © UKRI 1999. Source: BGS GeoScenic.)

If the growth rate of the crack opening decreases with time, it indicates a trend toward stability, but if the growth rate increases steadily with time, failure may be imminent. There are cases in which the growth rate decreases but then reverses to an increase with time (Figure 20.46). Movements of the slope surface can be tracked with useful and simple measurements. Tools can be as simple as surveying stakes driven in the ground and as advanced as GPS monitoring. Movement of the crest is a good indication, but swelling or heaving at the base of the slope is also an early sign. As in the case of cracks, the shape of the curve of the movement as a function of time is the best indication of potential failure.

Movement at depth in the slope is a very useful but more complicated measure, and is expensive to obtain. The most common method is to place a vertical inclinometer casing through the surface of the slope to a depth well below the potential failure surface (Figure 20.47). The bottom of the slope indicator casing should be in a soil zone that is not influenced by the slope movement. The reason is that the bottom readings will represent the zero position for the casing. The slope inclinometer casing is grooved and the probe has wheels (Figure 20.48) that fit in the grooves to keep the direction of the probe constant during the readings. The probe is lowered to the bottom of the casing and then pulled up the casing while readings are taken at regular intervals.

708

20 SLOPE STABILITY

Crack width, b

Impending failure

Unstable

Tending toward stability Time, t

Figure 20.46 Monitoring crack width over time. Recording device

Inclinometer casing Δ

α Inclinometer probe

Figure 20.47 Inclinometer in a slope.

This interval is usually the length L of the probe between wheels (e.g., L = 0.5 m). The first reading R1 is taken at the bottom of the casing at a depth z1 . The probe is pulled up an amount equal to L and the second reading R2 at depth z2 is obtained, and so on all the way to the top of the casing. The first set of readings in the casing is taken right after installation and provides a set of zero readings Roi . It is assumed that the bottom of the casing is low enough below

(a)

the slope potential failure surface that no movement takes place at the bottom of the casing; therefore, the readings at the bottom provide a reference for all the others. If there is doubt about whether the bottom is moving, then the top of the casing should be surveyed each time a set of inclinometer readings is taken. Each reading represents the angle 𝛼 between the probe direction and the vertical at that location. The instruments to measure this angle are servo-accelerometers located in the probe. A servo-accelerometer is essentially a mass placed at the end of a pendulum between two magnetic coils. When the pendulum begins to swing to a new position, a magnetic force is applied to keep the pendulum in the zero position. The current necessary for the magnetic force to keep the pendulum in the zero position is proportional to the angle that the pendulum would have taken had it not been restrained. The reading Ri at depth zi gives the angle of the inclinometer as: R sin 𝜃i = i (20.139) C where C is a constant specific to each inclinometer and 𝜃i is the angle of the casing with the vertical at depth zi . Often readings will be taken in two opposite directions of the probe (0 and 180∘ in a horizontal plane) and the average of the two readings will be used. The horizontal distance di between the two points separating two consecutive readings (often the length between probe wheels) is given by (Figure 20.49): di = L sin 𝜃i (20.140) where L is the increment of depth between readings. If the set of zero readings gave a value of di equal to doi , then the net horizontal distance is (di − doi ). Because we want the overall position of the casing after deformation, it is necessary to add all net horizontal distances between all consecutive points from the bottom of the casing to the depth where the

(b)

Figure 20.48 Inclinometer instrument and casing installation. (a) Inclinometer. (b) Casing installation. (Source: Courtesy of Landslide Technology.)

20.22 REPAIR METHODS

Casing position before movement

L L Casing position after movement

L L θn

zn

Dn

dn

θno L dno zi

di = Lsinθi dio = Lsinθio m Dn = Σ(di – dio) i=1

θi

L di

piezometer. If the water is in tension, the water stress can be measured with a field tensiometer up to a water tension of −90 kPa. Above that value, a soil sample can be taken, the water content determined, and the soil water retention curve used to go from the water content to the water tension (see Section 10.15).

20.22

REPAIR METHODS

There are essentially two ways to repair a slope that is getting close to failure (Figure 20.50):

dio

1. Increase the resisting moment (e.g., soil improvement, inclusions). 2. Decrease the driving moment (e.g., shallower slope). 20.22.1 z1

Increase the Resisting Moment

The shear strength s of the soil is:

z0

s = c′ + (𝜎 − 𝛼uw ) tan 𝜑′

horizontal movement is required. If it is assumed that the L∑ (di − dio ) = (R − Rio ) Dn = C i=1 i i=1 n

(20.141)

Groundwater and water stress conditions are very important aspects of slope stability. The groundwater level can be measured by simply measuring the equilibrium water level in an open standpipe or by using a piezometer. If the water is in compression, the water stress can be measured with a

Mechanical strengthening

Chemical strengthening

Nail

s

Adding a berm or retaining wall

Benching

(20.142)

where c is the effective stress cohesion intercept, 𝜎 is the normal total stress on the plane of failure, 𝛼 is the area ratio for the water phase, uw is the water stress, and 𝜑′ is the effective stress friction angle. Therefore, increasing s may consist of increasing c′ , or 𝜎 or tan 𝜑′ , or decreasing 𝛼uw . Increasing c′ can be done by chemical injection of cementing agents such as lime or cement. Increasing 𝜎 is usually not a good idea, as it also increases the driving moment. Increasing tan 𝜑′ is difficult but can be done through densification by compaction or vibration. If the soil is saturated and if the water is in compression, the term 𝛼uw becomes uw and decreasing uw consists of decreasing the water stress (through drainage, for example) if ′

Figure 20.49 Inclinometer data reduction.

n ∑

709

Crest removal

Lightweight material

Foam

Figure 20.50 Slope repair methods.

710

20 SLOPE STABILITY

there are excess water stresses or by lowering the water level (by pumping, for example) if the water stress is hydrostatic. If the soil is saturated with water in tension or if the soil is unsaturated, decreasing uw consists of drying or evaporation, for example. In this case uw becomes more negative, but 𝛼 also decreases, so the net result is not as efficient as a decrease in uw alone. In some instances, the water tension is naturally decreased (less negative) during the life of the slope because of the weather. This may lead to failure, and one way to prevent such failures is to keep the water tension from changing by isolating the soil from the weather. Geosynthetic covers can achieve this goal. The other way to increase the resisting moment is to insert inclusions in the slope and across the failure plane (Section 20.14). For existing slopes, soil nails or piles can be used. Soil nails are small-diameter inclusions that are placed most often by drilling and sometimes by driving. The drilling process consists of drilling a hole, removing the drilling tool, inserting a steel bar or cable with centralizers in the hole, and grouting the annulus between the bar and the soil. Soil nails have the advantage that they are relatively easy to place at any inclination, although they are most often placed nearly horizontally. Piles are placed vertically or near vertically and have a

larger diameter than nails. For new slopes like embankments, geosynthetic layers or reinforcing steel strips can be placed as reinforcement. 20.22.2

Decrease the Driving Moment

The driving moment M d for the slope is: Md = Wa

where W is the weight of the failing soil mass and a is the horizontal distance from the center of the failure circle for a circular failure surface and the center of gravity of the failing soil mass. Therefore, decreasing M d consists of decreasing W or a or both. To decrease W, lightweight material such as foam can be used for embankments. Also, the slope angle can be reduced by removing part of the crest, adding a berm at the bottom of the slope, or simply grading the slope to a flatter angle (Figure 20.50). In the end, the choice of one method or another is based on effectiveness of the method, feasibility, and cost. For landfill slopes, see Section 25.7.6. For slopes involving geosynthetics, see Section 26.6.3.

Problems and Solutions Problem 20.1 Calculate the factor of safety for the slope shown in Figure 20.1s in the following cases: a. b. c. d.

Slope alone. Slope plus building. Slope plus building and earthquake. What is the yield coefficient ky for that slope?

7m

2m

R= 10

m

100 kN/m 6m W=1400 kN/m kh=0.1 L=15 m Su=100 kPa

Figure 20.1s Slope with building and earthquake.

Solution 20.1 The factor of safety is the maximum resisting moment divided by the driving moment: FS =

(20.143)

Maximum resisting moment Driving moment

20.22 REPAIR METHODS

711

a. Slope alone: FS =

100 × 15 × 10 = 1.53 1400 × 7

b. Slope and building: FS =

100 × 15 × 10 = 1.40 1400 × 7 + 100 × (7 + 2)

c. Slope plus building and earthquake: FS =

100 × 15 × 10 = 1.30 1400 × 7 + 100 × (7 + 2) + 0.1 × 1400 × 6

d. The earthquake yield coefficient ky for that slope: FS =

100 × 15 × 10 = 1.00 ⇒ kY = 0.51 1400 × 7 + 100 × (7 + 2) + kY × 1400 × 6

Problem 20.2 An infinite slope is made of sand with a friction angle of 32∘ and a unit weight of 20 kN/m3 . The slope angle is 2.5 horizontal to 1 vertical. Calculate the factor of safety in the summer when the slope has no water, then calculate the factor of safety in the spring when the slope is filled with water flowing down the slope. Solution 20.2 For the case of sand with no water during the summer: FS =

tan 𝜑′ tan(32) = = 1.56 tan 𝛽 1∕2.5

For the case of the sand filled with water during the spring with no cohesion, and assuming a saturated unit weight of 20 kN/m3 : (𝛾 − 𝛾w ) tan 𝜑′ (20 − 9.81) tan(32) FS = sat = = 0.80 𝛾sat tan 𝛽 20 1∕2.5 The presence of water significantly reduces the factor of safety of the slope. Problem 20.3 Derive the expression for the factor of safety of an infinite slope with a failure plane parallel to the ground surface at a depth h and with a groundwater level at a height mh above the failure plane (m < 1). Solution 20.3 Let’s call 𝛾m the soil unit weight above the groundwater level and 𝛾sat the soil unit weight below the groundwater level. Referring to Figure 20.6 and the case of the infinite slope with seepage, the shear strength on the failure plane is: 𝜏f = c′ + (𝛾m (1 − m)h cos2 𝛽 + 𝛾sat mh cos2 𝛽 − 𝛾w mh cos2 𝛽) tan 𝜑′ 𝜏f = c′ + ((1 − m)𝛾m + m(𝛾sat − 𝛾w ))h cos2 𝛽 tan 𝜑′ The shear stress 𝜏 on the plane of failure is: 𝜏 = (𝛾m (1 − m)h cos 𝛽 + 𝛾sat mh cos 𝛽) sin 𝛽 𝜏 = ((1 − m)𝛾m + m𝛾sat )h cos 𝛽 sin 𝛽 The factor of safety is: c′ + ((1 − m)𝛾m + m(𝛾sat − 𝛾w ))h cos2 𝛽 tan 𝜑′ ((1 − m)𝛾m + m𝛾sat )h cos 𝛽 sin 𝛽 ((1 − m)𝛾m + m(𝛾sat − 𝛾w )) c′ FS = + ((1 − m)𝛾m + m𝛾sat )h cos 𝛽 sin 𝛽 ((1 − m)𝛾m + m𝛾sat ) tan 𝜑′ × tan 𝛽 FS =

712

20 SLOPE STABILITY

Problem 20.4 Design a safe slope angle for an excavated slope in a stiff clay to reach a 20 m deep deposit of lignite. The stiff clay has effective shear strength parameters of c′ = 10 kPa and friction angle 𝜑′ = 25∘ . Consider the case where the water level is not within the slope failure zone and then the one where the water level follows the slope contour. Use the chart method. In practice, it is not uncommon to see such excavations with much steeper slopes than the answer you will get in this problem; although failures do occur, they do not occur too often. Why do you think that is? Solution 20.4 If the soil is uniform and a circular failure surface is assumed, chart methods can be used. a. Case of No Water a-1. Taylor (1948) (Figure 20.2s)

H β

nH

DH Hard layer

Figure 20.2s Sketch of problem 20.4.

Let’s assume toe circles on Figure 20.13 with c′ = 10 kPa, 𝜑′ = 25∘ , and 𝛾 = 20kN∕m3 . We start with an assumed factor of safety Fc′ equal to 1.5. The developed friction angle is calculated using: ) ( tan 𝜑 ) ( tan 25 = tan−1 = 17.26∘ 𝜑d = tan−1 FS 1.5 tan 𝜑′ c′ Fc′ = ′ and F𝜑′ = cd tan 𝜑′d ′ c 10 Fc′ = ′ = 1.5 → c′d = 6.67 kPa 1.5 cd The stability number N =

c′d 𝛾H

=

6.67 20×20

= 0.0167 → Figure 20.13 → 𝛽 = 22∘

a-2. Spencer (1973) Use is made of Figure 20.15. uw = 0 because there is no water 𝜎0v tan 𝜑′ c′ Fc′ = ′ and F𝜑′ = cd tan 𝜑′d ru =

c′ 10 = 1.5 → c′d = = 6.67 kPa 1.5 c′d Figure 20.15 → ru = 0 → 𝛽 = 23∘ Fc′ =

and

N=

c′d 𝛾H

=

6.67 = 0.0167 20 × 20

b. Water Case b-1. Taylor (1948), Undrained: su = 70 kPa (𝝋′ = 0) Assuming a toe circle, a factor of safety of 1.5, and using su = 70 kPa, Figure 20.12 gives: FS =

Su S 70 → cd = u = = 46.67 kPa cd FS 1.5

20.22 REPAIR METHODS

N=

713

cd 46.67 = = 0.12 → Fig. 20.12 for n = 0 → 𝛽 = 11∘ 𝛾H 20 × 20

b-2. Spencer (1973), Drained Behavior: c′ ,𝝋′ ’ ru =

uw = 0.5 𝜎0v

Figure 20.15 → ru = 0.5 → 𝛽 = 11∘ Slope angle from Taylor and Spencer methods Slope angle 𝛽

Taylor (1948)

Spencer (1973)

22∘ 11∘

23∘ 11∘

Dry case Water case

Several factors come into play. First, the water in the soil is likely in tension, which increases the soil strength, thus the safe slope angle. Second, the excavation remains open for a limited amount of time and the soil behavior may be time-dependent. The slopes may be well drained. The open-pit mine slope industry seems to accept a higher probability of slope failure as part of the economical optimization of the open pit mining process. Problem 20.5 Calculate the factor of safety by using Janbu’s charts for the slope shown in Figure 20.3s in the following cases: a. Case 1: Hw = Hw′ = 10 m ′ = 10 m b. Case 2: Hw = 0, HW ′ c. Case 3: Hw = Hw = 10 xo

yo 10 m Hw

35°

Hʹw

c = 5 kPa φ = 30° γ = 20 kPa

Figure 20.3s Slope with different water levels.

Solution 20.5 Janbu Chart a. Case 1: Hw = H′w = 10 m 𝜇w = 𝜇w′ = 1, all other 𝜇 = 1 Pd =

20 × 10 − 10 × 10 + = 100 1×1×1

Pe =

20 × 10 − 10 × 10 + = 100 1×1

λc𝜙 =

100 tan(30) = 11.55 5

714

20 SLOPE STABILITY

For λc𝜙 = 11.55 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → Ncf = 27 27 × 5 = 1.35 100 For λc𝜙 = 11.55 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → xo = 0 and yo = 1.82 Therefore, Xo = 0 and Yo = 18.2 m. F=

b. Case 2: Hw = 0, H′w = 10 m 𝜇w = 𝜇w′ = 1, all other

𝜇=1

20 × 10 = 200 1×1×1 20 × 10 − 10 × 10 Pe = = 100 1×1 100 tan(30) = 11.55 𝜆c𝜙 = 5 = 11.55 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → Ncf = 27 Pd =

For 𝜆c𝜙

27 × 5 = 0.675 200 For 𝜆c𝜙 = 11.55 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → xo = 0 and yo = 1.82 Therefore, Xo = 0 and Yo = 18.2 m. F=

c. Case 3: Hw = 0, H′w = 10 𝜇w = 𝜇w′ = 1, all other

𝜇=1

20 × 10 = 200 1×1×1 20 × 10 Pe = = 200 1×1 200 tan(30) = 23.1 𝜆c𝜙 = 5 = 23.1 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → Ncf = 47 Pd =

For 𝜆c𝜙

47 × 5 = 1.175 200 For 𝜆c𝜙 = 23.1 and 𝛽 = 35∘ → cot 𝛽 = 1.43 → xo = −0.2 and yo = 2 Therefore, Xo = −2 m and Yo = 20 m. F=

Problem 20.6 A 3 horizontal to 1 vertical slope is cut in a clay that has an effective stress cohesion of 5 kPa and an effective stress friction angle of 30∘ . The slope is 6 m high. Calculate: a. The factor of safety of the slope against long-term failure if the water table is below the critical circle. b. The factor of safety of the slope against long-term failure if the water table coincides with the slope contour. Solution 20.6 The slope angle is: tan 𝛽 =

1 3

then 𝛽 = 18.42∘

a. Dry Case, No Water a-1. Taylor (1948) Assume the depth factor as D = 0.5, 𝛽 = 18.5∘ , c′ = 5 kPa, 𝜑′ = 30∘ , and 𝛾 = 20 kN∕m3 .

20.22 REPAIR METHODS

Fc′ = Iteration #1, FS = 1.5: Fc′ = The stability number N =

c′d 𝛾H

3.33 20 × 6

=

Fc′ = c′d

2.27 20 × 6

=

F𝜑′ =

tan 𝜑′ tan 𝜑′d

= 0.0277 → Figure 20.13 → 𝜑′d = 11∘

Iteration #2, FS = 2.2:

𝛾H

and

c′ 5 → 1.5 = ′ → c′d = 3.33 kPa ′ cd cd

F𝜑′ =

The stability number N =

c′ c′d

tan 𝜑′ tan 30 = = 2.97 tan 11 tan 𝜑′d

c′ 5 → 2.2 = ′ → c′d = 2.27 kPa c′d cd

= 0.019 → Figure 20.13 → 𝜑′d = 14.5∘ F𝜑′ =

tan 𝜑′ tan 30 = = 2.23 ′ tan 14.5 tan 𝜑d

The safety factor would be 2.21. a-2. Spencer (1967) ru =

uw assumed equal to 0 𝜎0v

Iteration #1, FS = 1.5: Fc′ =

c′ c′d

and

F𝜑′ =

tan 𝜑′ tan 𝜑′d

c′d c′ 5 3.33 ′ → 1.5 = → c = 3.33 kPa and N = = = 0.0277 d ′ ′ 𝛾H 20 × 6 cd cd Figure 20.15 → ru = 0 → 𝜑′d = 12∘ tan 𝜑′ tan 30 Fc′ = = = 2.7 tan 12 tan 𝜑′d Fc′ =

Iteration #2, FS = 2.5: Fc′ =

c′ 5 → 2.5 = ′ → c′d = 2 kPa c′d cd

The stability number N=

c′d 𝛾H

=

tan 𝜑′ tan 30 2 = = 0.017 → Figure 20.15 → 𝜑′d = 14∘ F𝜑′ = = 2.3 20 × 6 tan 14 tan 𝜑′d

The average safety factor would be 2.4. b. Water Case, Water Level at Ground Surface b-1. Spencer (1967), Drained Behavior:c′ , 𝝋′ ru = Iteration #1, FS = 1.5: Figure 20.15 → ru = 0.5, N =

c′d 𝛾H

=

3.33 20 × 6

= 0.0277 → 𝜑′d = 23∘ F𝜑′ =

Factor of safety is about 1.43.

uw = 0.5 𝜎0v

tan 𝜑′ tan 30 = = 1.36 ′ tan 23 tan 𝜑d

715

716

20 SLOPE STABILITY

Problem 20.7 Calculate the probability of failure of a slope that has a factor of safety with a mean of 1.5 and a coefficient of variation of 0.2 (assume that the factor of safety is normally distributed). If the acceptable probability of failure is 0.001, what must be the mean value of the factor of safety if the coefficient of variation remains equal to 0.2? Solution 20.7 Probability of Failure for a Given Factor of Safety Mean, 𝜇 = 1.5 Coefficient of variation, CoV = 0.2 Acceptable probability of failure, PoFac = 0.001 Standard deviation σ:

𝜎 𝜇 𝜎 = 𝜇 ⋅ CoV = 1.5 × 0.2 = 0.3

CoV =

Standard normal variable U of F = 1: F − 𝜇F 1 − 1.5 = = −1.67 𝜎 0.3 ) (F 1 − 𝜇F F − 𝜇F < = P(U < −1.67) P(F < 1) = P 𝜎F 𝜎F u=

Using Table 12.3, P(U < 1.67) = 0.9525 P(U < u) = 1 − P(U < −u) P(U < −1.67) = 1 − 0.9525 = 0.0475 or 4.75% probability of failure Factor of Safety for a Given Probability of Failure P(F < 1) = 0.001 ( ) ( ) ( ) F − 𝜇F F − 𝜇F 1 − 𝜇F 1 − 𝜇F 1 − 𝜇F P(F < 1) = P < < =P =P U< = 0.001 𝜎F 𝜎F 0.2𝜇F 0.2𝜇F 0.2𝜇F Using Table 12.3, we get: P(U < 3.1) = 0.999 or P(U < −3.1) = 0.001 Therefore, we must have:

1 − 𝜇F = −3.1 or 𝜇F = 2.63 0.2𝜇F

Problem 20.8 Which situation is more desirable? Explain and demonstrate. a. Mean factor of safety F = 1.5 and coefficient of variation of F = 0.2. b. Mean factor of safety F = 1.3 and coefficient of variation of F = 0.1. Solution 20.8 CoV =

𝜎 𝜇

𝜎a = CoV.𝜇 = 0.2 × 1.5 = 0.3 𝜎b = CoV.𝜇 = 0.1 × 1.3 = 0.13

( P (F < 1) = P

1 − 𝜇F F − 𝜇F < 𝜎F 𝜎F

20.22 REPAIR METHODS

)

( =P

F − 1.5 1 − 1.5 < 0.3 0.3

717

)

= P (U < −1.67) = 1 − P (U < 1.67) = 1 − 0.9525 = 0.0475 ( ) ) ( F − 𝜇F 1 − 𝜇F F − 1.3 1 − 1.3 P (F < 1) = P < =P < 𝜎F 𝜎F 0.13 0.13 = P (U < −2.31) = 1 − P (U < 2.31) = 1 − 0.9896 = 0.0104

Therefore, a mean factor of safety of 1.3 with a coefficient of variation of 0.1 (case b) is more desirable than a mean factor of safety of 1.5 and a coefficient of variation of 0.2 (case a). The reason is that the probability of failure is 1.04% in case b and 4.75% in case a. Problem 20.9 Calculate the factor of safety for the slope shown in Figure 20.4s. Select your best estimate of the critical circle and calculate the factor of safety by the Bishop modified method of slices.

0m

10 m

Range of possible locations for embankment crest (connection with bridge pavement) 15

15 10

Fill

5

γmat = 20.4 kN/m3 γsubm = 11 kN/m3 cʹ = 0 kPa ϕʹ = 32°

1

Range of possible locations for embankment toe

10

2

5

0

0

–5

–5 Sand γmat = 19.1 kN/m3 γsubm = 950 kg/m3 = 9.3 kN/m3 cʹ = 0 kPa ϕʹ = 30°

–10 –15

–10

Silty clay γmat = 19.1 kN/m3 γsubm = 9.3 kN/m3 c = 15 kPa ϕʹ = 20° suC0 = 100 kPa

–20

–15 –20

–25

–25 Bedrock

–30 0

5

10

15

20

25

30

35

–30 40

45

50

55

60

65

70

Figure 20.4s Slope of Fredericton embankment.

Solution 20.9 The sketch for the simplified method of slices is shown in Figure 20.5s. The results are shown in Table 20.1s.

Table 20.1s

Summary of the results of the Bishop simplified method of slices

Slice# 𝜑′

b (m)

Area Unit Total total weight Weight weight Area fill (m22 ) (%) (m2 ) (kN/m3 ) (kN/m) (kN/m)

1

32

3.2

Soil-1 (100)

9.0

9.0

20.40

183.6

183.6

60.0 0.50

159.00 3.20

0.00

0.00

0.00

114.73

0.00

0.00

0.87

131.38

2

32

5.0

Soil-1 (100) 43.5

43.5

20.40

887.4

887.4

51.1 0.63

690.61 5.00

0.00

0.00

0.00

554.51

0.00

0.00

0.96

575.61

3a

32

3.3

Soil-1 (95)

66.3

66.3

20.40

1352.5

1352.5

42.7 0.73

917.22 3.25

0.90

8.83

28.69

827.22

0.00

0.00

1.03

805.34

3b

30

1.8

Soil-2 (5)

3.5

3.5

19.10

66.9

66.9

36.6 0.80

39.86 1.75

2.60 25.51

44.64

12.83

0.00

0.00

1.04

12.33

5.0

Soil-1 (90) Soil-2 (10)

8.0 74.8

82.8

20.40 19.10

163.2 1428.7

1591.9

30.1 0.87

798.34 5.00

5.00 49.05 245.25

777.48

0.00

0.00

1.06

730.14

5.0

Soil-1 (60) Soil-2 (40)

26.6 17.8

44.4

20.40 19.10

543.5 339.2

882.7

21.8 0.93

327.80 5.00

7.22 70.83 354.14

305.15

0.00

0.00

1.08

283.50

51.7 21.3 3.0

76.0

20.40 19.10 19.10

1054.7 406.8 57.3

1518.8

11.9 0.98

313.18 5.00

8.90 87.31 436.55

393.91

15.00

75.00

1.03

455.14

4 5

30 30

𝜃

Wsin𝜃 cos 𝜃i (kN)

b (m)

hp (m)

uw (kPa)

uw bi (W-uw b) tan 𝜑′ c′ c′ b∕cos𝜃 (kN/m) (kN) (kN/m) (kN/m) m𝜃

[cb + (W-uw b) tan 𝜑′ ]∕m𝜃

6

20

5.0

Soil-1 (68) Soil-2 (28) Soil-3 (4)

7

20

5.0

Soil-1 (57) Soil-2 (33) Soil-3 (10)

40.6 23.3 7.3

71.2

20.40 19.10 19.10

828.2 445.0 138.5

1411.7

7.3 0.99

179.38 5.00

9.80 96.14 480.69

338.88

15.00

75.00

1.02

404.26

8

20

5.0

Soil-1 (49) Soil-2 (36) Soil-3 (15)

29.1 22.0 8.7

59.8

20.40 19.10 19.10

593.6 420.2 166.2

1180.0

0.0 1.00

0.00 5.00

10.14 99.47 497.37

248.46

15.00

75.00

1.00

323.46

9

20

5.0

Soil-1 (36) Soil-2 (43) Soil-3 (14) Water (7)

17.5 20.9 6.8 3.5

48.7

20.40 19.10 19.10 9.81

357.0 399.2 129.9 34.3

920.4

−9.1 0.99

−145.57 5.00

9.70

95.16 475.79

161.83

15.00

75.00

0.95

249.89

10a

20

3.5

Soil-1 (18) Soil-2 (44) Soil-3 (5) Water (33)

5.4 13.1 1.6 9.7

29.8

20.40 19.10 19.10 9.81

110.8 250.2 30.6 95.3

486.8

−15.1 0.97

−126.81 3.50

8.90

87.31 305.58

65.96

15.00

52.50

0.90

131.61

10b

30

1.5

Soil-1 (5) Soil-2 (43) Water (53)

0.5 5.0 6.2

11.8

20.40 19.10 9.81

10.8 96.1 60.8

167.7

−20.1 0.94

−57.63 1.50

8.00

78.48 117.72

28.86

0.00

0.00

0.80

35.97

11

30

5.0

Soil-2 (32) Water (68)

10.5 23.0

33.5

19.10 9.81

200.6 225.6

426.2

−23.7 0.92

−171.30 5.00

6.70

65.73 328.64

56.32

0.00

0.00

0.76

74.53

12a

30

5.0

Soil-2 (17) Water (83)

0.7 3.5

4.2

19.10 9.81

13.4 34.6

48.0

−31.2 0.86

−24.86 5.00

5.23

51.31 256.53

0.00

0.00

0.00

0.65

0.00

12b

0

3.5

Water (100)

12.7

12.7

9.81

124.6

124.6

−60.0 0.50

−107.90 3.50

3.86

37.87 132.53

0.00

0.00

0.00

0.50

0.00

13

0

3.2

Water (100)

4.1

4.1

9.81

40.2

40.2

−68.0 0.37

−37.29 3.20

1.50

14.72

0.00

0.00

0.00

0.37

0.00

𝚺

47.09

2754.0

Note: the percentage number in the section of the area stands for the corresponding area for each of the slices under consideration.

𝚺

352.50

4213.17 F.S-

1.53

20.22 REPAIR METHODS

719

O

5m 1

15

15 3b

1

2

15

15

2 3a

5

4

10b

5

12a 12b 13

5

0

0 11

6

–5

7

Soil-1 (fill) γmat = 20.4 kN/m3 γsubmerg = 11 kN/m3 cʹ = 0 kPa φ = 32°

–10 –15

8

9

Soil-2 (sand) γmat = 19.1 kN/m3 γsubmerg = 9.3 kN/m3 cʹ = 0 kPa φ = 30°

–20

10a

–5 –10

Soil-3 (silty clay) γmat = 19.1 kN/m3 γsubmerg = 9.3 kN/m3 cʹ = 15 kPa φ = 20° su = 100 kPa

–25

–15 –20 –25

Bedrock

–30 0

5

10

15

20

25

30

35

–30

40

45

50

55

60

65

70

Figure 20.5s Bishop simplified method of slices.

Problem 20.10 A slope is subjected to an acceleration history as shown in Figure 20.6s. The yield acceleration for that slope is 1.5 m/s2 . Calculate the displacement history of the slope according to Newmark’s method.

a (m/s2)

3

A

2

ay

1 0

0.5

1

1.5

2

t

Figure 20.6s Acceleration history.

Solution 20.10 From Figure 20.6s: single rectangular acceleration, A = 3 m∕s2 , yield acceleration, ay = 1.5 m∕s2 , relative acceleration arel (t). relative velocity vrel (t), and relative displacement drel (t). At t0 ≤ t ≤ t0 + Δt or 1 < t < 1.5 seconds, relative acceleration, relative velocity, and relative displacement are: arel (t) = A − ay = 3 − 1.5 = 1.5 m∕s2 t ]( ) [ vrel (t) = a (t) dt = A − ay t − t0 = [3 − 1.5] (t − 1) = 1.5t − 1.5 m∕s2 ∫t0 rel

720

20 SLOPE STABILITY t

drel (t) =

∫ t0

vrel (t) dt =

]( ) 1 1[ A − ay t − t0 = [3 − 1.5] (t − 1) = 0.75(t − 1)2 2 2

At t0 = t0 + Δt or 1.5 seconds: vrel (t + Δt) = [A − ay ]Δt = 3[3 − 1.5] × 0.5 = 0.75 m∕s 1 1 drel (t + Δt) = [A − ay ]Δt2 = [3 − 1.5]0.52 = 0.1875 m 2 2 At t0 + Δt ≤ t ≤ t1 or 1.5 < t < 2 seconds: arel (t) = 0 − ay = −1.5 m∕s2 t

vrel (t) = 0.75 +

t

arel (t)dt = 0.75 +

∫1.5

∫1.5

t

drel (t) = 0.1875 +

∫1.5

(−1.5)dt = −1.5t + 3 m∕s t

vrel (t)dt = 0.1875

∫1.5

(−1.5t + 3)dt = 0.75t2 + 3t − 2.625

All results are plotted in Figure 20.7s.

a (m/s2)

3

A

2

ay

1 0

0.5

1

1.5

2

t

2

t

v (m/s)

1.5 1

0.75

0.5 0

0.5

1

1.5

d (m)

0.4

0.375

0.3 0.1875

0.2 0.1 0

0.1875 0.5

1 t0

1.5 t0 + Δt

2 t1

t

Figure 20.7s Acceleration, velocity, and displacement history of a slope.

Problem 20.11 Consider the case of a c′ = 0, 𝜑′ > 0 soil and demonstrate that for one slice, the factor of safety is the same for the ordinary method of slices (OMS) and for the Bishop simplified method of slices (BSMS). If it is true for one slice, why is it not true for n slices (n > 1)?

20.22 REPAIR METHODS

721

Solution 20.11 OMS F= BSMS

W cos 𝜃 tan 𝜑 tan 𝜑 = W sin 𝜃 tan 𝜃

( ) tan 𝜃 tan 𝜑 m = cos 𝜃 1 + F 1 (W tan 𝜑) tan 𝜑 F= m = ( ) W sin 𝜃 tan 𝜃 tan 𝜑 cos 𝜃 sin 𝜃 1 + F ( ) tan 𝜑 tan 𝜃 tan 𝜑 = cos 𝜃 sin 𝜃 1 + F F tan 𝜑 (1 − cos 𝜃 sin 𝜃 tan 𝜃) = cos 𝜃 sin 𝜃 F tan 𝜑(1 − sin2 𝜃) tan 𝜑 cos2 𝜃 tan 𝜑 F= = = cos 𝜃 sin 𝜃 cos 𝜃 sin 𝜃 tan 𝜃

The assumption in both methods is with respect to side forces. Because we have only one slice, there are no side forces between slices. Most importantly, there are as many unknowns as there are equations, so equilibrium equations can be written in any direction and will lead to the same answer. The three unknowns are Sm , N ′ , and F; the three equations are vertical equilibrium, horizontal equilibrium, and the shear strength equation. Moment equilibrium is automatically satisfied because all forces go through the middle of the base of the slice. Thus, both methods should give an identical safety factor. This is no longer true when we have more than one slice, because the assumptions are different for the side forces. Problem 20.12 A 3D slope failure has a failure surface in the form of a sphere and a factor of safety F 3D . This sphere is sliced in a direction perpendicular to the crest. The slices have the same width b. The deepest slice in the center of the sphere has a factor of safety F min . Each slice has a 2D factor of safety equal to F t . What other assumptions must be made for the following equation to be true? n 1∑ F F3D = n i=1 i Solution 20.12 As explained in Section 20.16, in order to use this equation for the safety factor F 3D : 1. The axis of rotation must be the same for all slices (see Figure 20.34). 2. The forces between circle slices must be negligible. Problem 20.13 A dry, fine sand slope has a factor of safety of 1.5 on the Earth. a. Calculate and discuss the factor of safety for the same slope on the moon. b. Assume that there could be water on the moon. Would the result of the dry case still hold? Solution 20.13 𝜑 a. The factor of safety for dry sand is F = tan . It is independent of gravity acceleration (g), so it would be the same on tan 𝛽 the moon. b. The factor of safety on the moon would be different from the one on the Earth if the water was at the surface of the slope, but it would not be if the slope was above the groundwater level and water tension developed in the slope. Because water tension is a chemically based phenomenon and not a gravity-based phenomenon, it would be the same on the Earth and ′

722

20 SLOPE STABILITY

on the moon, but its ratio to gravity forces would be very different. Therefore, it would lead to different factors of safety. The same slope would be safer on the moon if water tension existed in both cases. Problem 20.14 Define the seepage force and discuss when the seepage force should be considered in a slope stability analysis. Why is it not usually considered? Solution 20.14 The seepage force is the force exerted in friction by water flowing around soil particles and trying to drag them away. The forces shown on a free-body diagram are the external forces. The internal forces are resolved internally. The seepage force is an external force when the soil skeleton is considered as the free body, but it is an internal force when the soil skeleton plus the water is considered as the free body. Most slope stability analyses consider the soil skeleton plus the water as the free body. In those instances, the seepage force must not be included in any slope stability calculations. Problem 20.15 Explain the difference between the following analyses, including what shear strength you would use: total stress analysis, effective stress analysis, undrained analysis, drained analysis, short-term analysis, long-term analysis. Solution 20.15 A total stress analysis considers that the soil is made of one material. During the analysis, the three components (particles, water, and air) are not recognized. This analysis can be used in the case of a soil with no water and in the case of a soil where the shear strength is independent of rapid variations in total stress. An effective analysis can be used in all cases. It makes no particular assumption regarding drainage and is based on sound fundamental principles. It makes use of the effective stress equation to obtain the shear strength of the soil based on effective stress cohesion c′ and the effective stress friction angle 𝜑′ (𝜏 = c′ + 𝜎 ′ tan 𝜑′ ). It can be used for an undrained analysis, a drained analysis, a short-term analysis, or a long-term analysis. The difficulty with this method is that the water stress in the mass must be known. An undrained analysis is used in the case where the water is not allowed or does not have time to drain away. The soil strength parameter used in this case is the undrained shear strength (su ). In a drained analysis, the water stress is considered to be hydrostatic throughout the mass. The soil strength parameters used are the drained strength parameters or effective stress parameters (𝜏 = c′ + 𝜎 ′ tan 𝜑′ ). A short-term analysis considers a time shortly after loading. It is often a drained analysis for fast-draining soils like free-draining sands and gravels, and an undrained analysis for slow-draining soils like silts and clays. In the case of free-draining sands and gravels, the drained strength parameters are used. In the case of silts and clays, the undrained shear strength (su ) is used. A long-term analysis considers that all water stresses induced by loading have had time to dissipate and are back to hydrostatic condition. In this regard a long-term analysis is similar to a drained analysis. The soil strength parameters used are the drained strength parameters or effective stress parameters (𝜏 = c′ + 𝜎 ′ tan 𝜑′ ). Problem 20.16 An inclinometer casing is attached to a 10 m high retaining wall. Zero readings taken before the wall is backfilled indicate that the wall is perfectly vertical. The backfill is placed and compacted. At the end of construction, the inclinometer readings are taken again. Find what the readings are if: a. The displacement y (m) of the wall obeys the equation y = 0.01(zmax − z) where zmax is the maximum depth the inclinometer probe can reach in the casing (10 m) and z is the depth at which the reading is taken. b. The displacement y (m) of the wall obeys the equation y = 0.01(zmax − z)2 where zmax is the maximum depth the inclinometer probe can reach in the casing (10 m) and z is the depth at which the reading is taken. The inclinometer has a calibration constant C = 20,000 and a wheel spacing of 0.5 m.

20.22 REPAIR METHODS

723

Solution 20.16 The constant for the inclinometer is C = 20,000, and the length of the probe between wheels is L = 0.5 m. Figure 20.8s shows the wall and inclinometer. y

Compacted soil

10 m

z

Figure 20.8s Illustration of the inclinometer.

The equation for the displacement Dn of the inclinometer casing is: Dn =

n n ∑ L∑ (di − dio ) = (R − Rio ) C i=1 i i=1

where Dn is the displacement at a depth zn from the surface, zn is the depth to the first (deepest) reading in the casing minus n times the distance L between readings, di is the difference in horizontal displacement between the i and i – 1 reading points, dio is the initial value of di , L is the length between readings, C is the inclinometer calibration constant, Ri is the reading at depth zi , and Rio is the initial value of Ri . a. In this case the wall deforms by simple rotation and the displacement y (m) of the wall is linear. The equation for Dn becomes: n n ∑ L∑ Dn = di = R C i=1 i i=1 But Dn is also given in the problem as: Dn = 0.01(10 − z) Therefore, the difference between two consecutive readings is: Dn − Dn−1 = dn = 0.01(10 − z) − 0.01(10 − z − 0.5) = 0.005 m The increment of displacement is constant and the angle of the wall is also constant: 𝜃n = sin−1

dn 0.005 = sin−1 = 0.01 rd L 0.5

Furthermore, Do is equal to zero, because the bottom of the wall does not move. Now the reading is equal to: Rn =

C 20000 d = × 0.005 = 200 L n 0.5

So the reading of the inclinometer is constant equal to 200; Table 20.2s summarizes the results.

724

20 SLOPE STABILITY

Table 20.2s Depth z (m) 10 9.5 9 8.5 8 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

Inclinometer readings for linear displacement of wall

Displacement y (m)

Inclined angle (radians)

Inclined angle (degree)

Inclinometer reading (R)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09 0.095 0.1

0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667 0.009999667

0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698 0.572938698

0 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200

b. In this case, the displacement y(m) of the wall is nonlinear. The equation for Dn is still: n n ∑ L∑ di = R Dn = C i=1 i i=1

But Dn is also given in the problem as:

Dn = 0.001(10 − z)2

Therefore, the difference between two consecutive readings is: Dn − Dn−1 = dn = 0.001(10 − z)2 − 0.001(10 − z − 0.5)2 = 0.001 (9.75 − z) The increment of displacement increases linearly with z and so does the angle of the wall: 𝜃n = sin−1

d 0.001(9.75 − z) = sin−1 n L L

Furthermore, Do is equal to zero, because the bottom of the wall does not move. Now the reading is equal to: Rn = Table 20.3s summarizes the results.

C 20000 d = × 0.001(9.75 − z) = 390 − 40z L n 0.5

20.22 REPAIR METHODS

Table 20.3s Depth z (m) 10 9.5 9 8.5 8 7.5 7 6.5 6 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

725

Inclinometer readings for linear displacement of wall

Displacement y (m)

Inclined angle (radians)

Inclined angle (degree)

Inclinometer reading (R)

0 0.00025 0.001 0.00225 0.004 0.00625 0.009 0.01225 0.016 0.02025 0.025 0.03025 0.036 0.04225 0.049 0.05625 0.064 0.07225 0.081 0.09025 0.1

0 0.0005 0.0014997 0.0024995 0.0034993 0.0044992 0.0054990 0.0064988 0.0074986 0.0084985 0.0094983 0.0104982 0.0114981 0.0124980 0.0134978 0.0144978 0.0154977 0.0164976 0.0174976 0.0184976 0.0194975

0 0.028648 0.085943 0.143239 0.200535 0.257832 0.315128 0.372425 0.429722 0.487020 0.544318 0.601617 0.658916 0.716216 0.773516 0.830818 0.888120 0.945423 1.002727 1.060032 1.117338

0 10 30 50 70 90 110 130 150 170 190 210 230 250 270 290 310 330 350 370 390

CHAPTER 21

Compaction

21.1

GENERAL

Compaction refers to the densification of shallow soil layers by rollers. These rollers may be static cylindrical rollers, smooth or with protrusions; vibratory cylindrical rollers; or impact noncylindrical rollers. Conventional compaction refers to the use of noninstrumented rollers, whereas intelligent compaction refers to the use of instrumented rollers with feedback loops. Dynamic compaction, also discussed in this chapter, refers to dropping large weights from a given height onto the ground surface; this process creates a crater and compacts the material under the crater. Compaction is required in many instances; examples include compaction for the base layer of pavements, for embankment fills, for retaining wall backfills, for fill around pipes, and for landfills. Depending on the soil type and the size of the project, different compactors are used (Figure 21.1). For example, hand tampers (also called jumping jacks) are used in small areas around pipes, rollers are used for roadway compaction, and drop-weight compactors are used for dynamic compaction of large areas at larger depth. The rollers are typically 50–150 kN in weight; the drums are 1–2 m in diameter and 2–3 m wide. The frequency of vibration for vibrating rollers is from 30–70 Hz. For dynamic compaction, the drop weight commonly varies from 50 to 250 kN and the drop height from 5–25 m. The depth over which the soil is compacted is up to 1 m for rollers and up to 10 m for dynamic compaction. The compaction process typically takes the following steps: 1. Perform laboratory tests on the material to be compacted (Proctor test, for example) and establish the value of the soil property to be reached in the field work. These properties are most commonly the dry density and the water content. The modulus of deformation can also be used. 2. Write the field specifications, including the target dry density or target modulus within a chosen range of water content.

(a)

(b)

(c)

(d)

(e)

Figure 21.1 Compaction equipment: (a) Hand tamper. (b) Sheep-foot roller. (c) Smooth cylindrical roller. (d) Impact noncylindrical roller. (e) Drop-weight compactor. (Source: a: Courtesy of Multiquip. b, c: Images courtesy of Caterpillar. d: Courtesy of LANDPAC. e: Courtesy of Serge Varaksin).

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

727

728

21 COMPACTION

3. In the field, use compacting equipment to compact the soil in 0.15 m lifts after it is brought to the chosen water content. 4. In the field also carry out field tests to verify that the target values listed in the specifications have been reached. COMPACTION LABORATORY TESTS

Laboratory tests are used to establish the characteristics of the soil to be compacted, to establish the target values to be achieved in the field, and to write the specifications for field work. The compaction process and the compaction curve associated with laboratory tests are described in detail in Chapter 10, Section 10.3; this section gives a brief summary. The compaction curve links the dry density or a soil modulus to the water content. The dry density vs. water content curve is relatively flat, as the dry density is not very sensitive to the water content. Within the range of dry density variation, the curve has a bell shape (Figure 21.2). The reason for this bell shape is that at point A, in Figure 21.2, the soil is relatively dry and it is difficult for a given compaction energy to bring the particles closer together. At point B, the water content is such that water tension exists between the particles and hinders the effectiveness of the compaction process. At point C, the water tension loses its effect and the primary role of the water becomes to lubricate the contacts between particles, thereby allowing the given compaction effort to reach a lower void ratio and a higher dry density. At point D, the soil is nearing saturation and the added water simply increases the volume of the voids, which negates the benefit of the compaction. The maximum dry density 𝛾 dmax and the optimum water content wopt are two important parameters obtained from the curve (Figure 21.2). This curve is obtained in the laboratory with the Standard Proctor or Modified Proctor Compaction Test (Figure 21.3). These tests are described in detail in Chapter 10, Section 10.3. Ralph Proctor was an American civil engineer who developed the Standard Proctor Test in 1933 and developed a new version of the test in 1955 known as the Modified Proctor Test.

Figure 21.3 Proctor compaction laboratory test.

21 S = 0.6

Dry unit weight, γd (kN/m3)

21.2

S = 0.8

S=1

Lines of equal degrees of saturation

20 γ dmax γd =

SGs S + Gsw

γw

19 Modified Proctor, 2700 kN*m/m3 18 γ dmax

17 Standard Proctor, 600 kN*m/m3 16

0

4

Wopt

8 12 Water content, w (%)

Wopt 16

20

Dry unit weight γd (kN/m3)

Figure 21.4 Compaction curves and saturation lines. 25

20 γdmax C 19

0

10

wopt 4

8

The lines of equal degree of saturation can be presented on the same diagram as the dry density vs. water content curve (Figure 21.4). The equation for the saturation lines is:

15 D

B A

18

γdmax

20

12

16

Water content w (%)

20

0

𝛾d =

wopt

5 0

4

8

12

16

Water content w (%)

Figure 21.2 Compaction curve: dry density.

20

SGs 𝛾 S + Gs w w

(21.1)

where 𝛾 d is the dry density, S is the degree of saturation, Gs is the specific gravity of the solids, w is the water content, and 𝛾 w is the unit weight of water. The derivation of this equation is shown in Chapter 10, Section 10.3.

729

21.3 COMPACTION FIELD TESTS 60

20

Modulus, E (MPa)

16

40 12 30 8 20 BCD modulus (MPa)

10 0

4

Dry unit weight (kN/m3) 0

2

4

6

8

10

12

Dry unit weight, γd (kN/m3)

50

0 14

Water content, w (%)

Figure 21.5 Compaction curve: modulus.

In compaction control, the dry density can be replaced by the soil modulus as a governing parameter for specifications and quality control. The advantage of using the modulus

is that the modulus is directly involved in the design calculations, whereas the dry density is not. The drawback is that the modulus depends on many factors (see Chapter 15, Section 15.2) and is not a single parameter for a given soil, whereas the dry density is. In the case of the modulus, the curve has the same bell shape as the dry density vs. water content curve, but is much more sensitive to the water content, especially on the wet side of the optimum water content (Figure 21.5). At low water contents, the modulus is influenced by the water tension that develops in the soil, whereas the dry density is not. As a result, the modulus curve can go back up at low water contents. The laboratory test to obtain the modulus vs. water content curve is the BCD test (Figure 21.6 and Chapter 10, Section 10.4), which is performed on the Proctor test sample. This is convenient because a dry density curve and a modulus curve can be obtained at the same time. 21.3

COMPACTION FIELD TESTS

The specifications indicate that the compacted soil must reach a dry density equal to a percentage of the maximum dry density measured in the laboratory (typically 95–100%) within a range of water content around the optimum water content. The specifications may also indicate that the compacted soil must reach a soil modulus equal to a percentage of the maximum soil modulus measured in the laboratory (typically 75% or so) within a range of water content around the optimum water content. Table 21.1 shows some possible target modulus values for pavement applications. Field tests are used to verify that the compaction work has been done according to specifications. The field tests are divided into classic tests and new tests. The classic tests have been used for a long time and are relatively slow (15–30 minutes per test). They include the sand cone test for dry density, the rubber balloon test for dry density, and the nuclear density gage for dry density and water content. The new tests take only a few minutes to perform. They include the lightweight deflectometer, the BCD, and the field oven (Figure 21.7). All these tests are described in detail in Chapter 8, Section 8.11.

Table 21.1

Figure 21.6 BCD-Proctor laboratory compaction test for modulus determination.

Modulus target values for pavements

Soil layer

Plate test reload modulus (MPa)

Lightweight deflectometer modulus (MPa)

BCD reload modulus (MPa)

Base course Subgrade soil

100–150 45–80

100–150 45–80

55–85 25–45

730

21 COMPACTION

(a)

(b)

(c)

Figure 21.7 Compaction control tests in the field: (a) Nuclear gage. (b) Lightweight deflectometer. (c) BCD. (Source: a: Photo by Lindsey D. Fields, Envirotech Engineering & Consulting, Inc. b: Courtesy of Minnesota Department of Transportation.) 20

19

n tio

ra e,

rv

cu S

18

=

0

5

10

0%

Coarsegrained to finegrained

17

16

Increase in compaction effort

10

Table 21.2

tu

Different soils react differently to different compaction equipment. Coarse-grained soils are most effectively compacted through vibration combined with pressure. Pressure alone increases the effective stress and therefore the friction between particles, thereby preventing their sliding into a more compact position. Vibration breaks the friction bonds and lets the particles settle into a tighter arrangement. Fine-grained soils are most effectively compacted through kneading and pressure. Vibration may simply increase the water stress if the soil is saturated. Also, coarse-grained soils tend to reach optimum compaction at water contents lower than fine-grained soils. However, coarse-grained soils tend to reach maximum dry densities that are higher than those of fine-grained soils (Figure 21.8). Table 21.2 shows a rating of applications for various pieces of compaction equipment.

Dry unit weight, γd (kN/m3)

COMPACTION AND SOIL TYPE

Sa

21.4

15

20

Water content, w (%)

Figure 21.8 Influence of soil type and compaction effort on dry density curve.

Applicability of compaction equipment for various soils

Soil type

Static sheep foot roller

Vibrating cylindrical roller

Impact noncylindrical roller

Gravel Sand Silt Clay

Poor Poor Good Very good

Good Very good Poor Poor

Good Good Poor Medium

Domestic waste

Good

Poor

Good

Dynamic compaction Good Good Medium Poor if saturated, good if unsaturated. Very good

21.5 INTELLIGENT ROLLER COMPACTION

21.5

feature that the roller is able to change its settings nearly instantaneously when it comes to a soft spot and to optimize the compaction process (Figure 21.10) while keeping track of the global position through GPS (Figure 21.11). In CCC and IC, the soil parameter most often measured is a soil modulus E.

INTELLIGENT ROLLER COMPACTION

Continuous control compaction (CCC) refers to compaction with rollers that are instrumented, make measurements on the fly, and give an image of the complete compacted area with the values of the soil parameter measured (Figure 21.9). Intelligent compaction (IC) refers to CCC with the added

(a)

(b)

Figure 21.9 Continuous coverage in CCC and IC: (a) Continuous mapping of soil stiffness. (b) Screen display. (Source: Courtesy of HAMM AG.) Amplitude = min

s es ss Pa 4 3 2 5

Soil modulus EVIB (MPa)

Amplitude = max

100 90 80 70 60 50 40 30 20 10 0

Soft spot 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

Distance (m)

Figure 21.10 Intelligent compaction roller adjustment settings. (Source: Courtesy of BOMAG.)

(a)

731

(b)

Figure 21.11 Intelligent rollers and readout equipment: (a) Roller. (b) Readout and control. (Source: a: Courtesy of BOMAG; b: Courtesy of Ammann.)

732

21 COMPACTION

21.5.1

Soil Modulus from Vibratory Rollers

In the case of vibratory rollers, the modulus E is obtained from measurements of the acceleration of the roller (Anderegg, 1997). A single degree of freedom model is used to represent the roller-soil interaction (Figure 21.12). Vertical equilibrium of the drum gives: F = −md ẍ d + me re 𝜔2 cos(𝜔t) + (mf + md )g

(21.2)

where F is the vertical force at the bottom of the drum, md is the mass of the drum, ẍ d is the linear vertical acceleration of the drum, me is the eccentric mass creating the vibration, re is the radial distance at which me is attached, 𝜔 is the circular frequency of the rotating shaft, t is the elapsed time, mf is the mass of the frame, and g is the acceleration due to gravity. All quantities on the right side of Eq. 21.2 come from the roller specifications and are known except for ẍ d , which is measured with an accelerometer on the drum axis. The vibration frequency of most rollers ranges from 30–70 Hz, their weight from 50–150 kN, their diameters from 1–2 meters, and their width from 2–3 meters. The soil resistance F can also be obtained as follows if the soil is represented by a spring-and-dashpot model: F = kS xd + cS ẋ d ACTION

(21.3) REACTION

mf g

.. md x d xd

xd

me reΩ2cos(Ωt) md g

F

F

ksxd . csxd

Contact force, Fc (kN)

Figure 21.12 Forces acting on the drum of a vibrating roller. (Source: Anderegg et al., ETH Zurich, 1997.)

where ks is the soil spring stiffness, xd is the vertical displacement of the drum, cs is the soil damping coefficient, and ẋ d is the vertical velocity of the soil boundary. In Eq. 21.3, ks is the parameter to be solved for, xd and ẋ d are obtained by integration of the acceleration signal, and cs is typically assumed to be about 20% of critical damping. Numbers in the range of 50–100 MN/m for ks and 150–250 kN s/m for cs have been measured (Van Susante and Mooney, 2008). An example of contact force vs. time and contact force vs. displacement is shown in Figure 21.13. Equations 21.2 and 21.3 are combined and the soil stiffness ks can be obtained from the combined equation: kS =

me re 𝜔2 cos(𝜔t) + (mf + md )g − md ẍ d − cS ẋ d xd

As explained in Chapter 15, Section 15.6, the soil stiffness ks is not an independent soil parameter because it depends on the size of the loaded area. The soil modulus E, in contrast, is an independent soil parameter; therefore, it is desirable to know how to obtain E from ks . This problem was solved by Hertz in 1895 and further developed by Lundberg (1939): F ks = xd 𝜋LE = ( [ ]) 1 𝜋L3 H 2(1 − v2 ) 2.14 + Ln 2 16(1 − v2 )(mf + md )Rg (21.5) where ks is the soil stiffness, F is the force applied, xd is the settlement of the drum, L is the drum width, 𝜐 is Poisson’s ratio, Ln is the natural logarithm, mf and md are the masses contributed by the frame and the drum of the roller respectively, R is the radius of the drum, and g is the acceleration due to gravity. Lundberg also gave the width b of the contact area between the drum and the soil: √ 16 R(1 − 𝜐2 ) F b= (21.6) 𝜋 E L

25

Contact force (kN) 200

50

Ks Pass 3

75

Pass 2

100 125

Ks Ks

Pass 1 0

20

40 60 Time, t (ms) (a)

80

100

(21.4)

4

2

0 –2 –4 Drum movement (mm) (b)

Figure 21.13 Contact force under the vibrating roller. (a) Force-time. (Source: Van Susante and Mooney, 2008/American Society of Engineers.) (b) Force-displacement. (Source: W. Fellin, D. Kolymbas/Taylor & Francis.)

21.6 IMPACT ROLLER COMPACTION

L F

2R d E, V

b

Figure 21.14 Drum on elastic soil problem. (Source: Adapted from Lundberg, 1939.)

As can be seen and as could be anticipated, the width b is inversely proportional to the soil modulus E. The roller soil contact area has a length L and a width b. Because the ratio L∕b is very large, the loading is similar to a strip footing. Under static conditions, the depth of influence under a strip footing is 4b. Thus, under the first pass, the width b is large because the soil is not very stiff and, as a result, the depth of influence is larger. Under subsequent passes, the soil stiffens and b decreases, and so does the depth of influence d (Figure 21.14). The width b varies commonly between 200 mm on soft soils to 20 mm on very stiff soils. 21.5.2

Roller Measurements as Compaction Indices

The machine drive power (MDP) (White et al., 2005) is a roller index that can be used to evaluate the degree of compaction generated by any roller. The principle is that if the soil is soft, it will take more power for the roller to roll forward and compact the soil; if the soil is stiff, it will take less power for the roller to roll forward. The difference, as illustrated in Figure 21.15, is that the roller on soft soil has to overcome a lot more soil deformation energy than the roller on stiff soil. It is similar to the difference you feel when you run on loose sand compared to running on pavement; it is a lot more difficult (and takes more energy) to run on loose sand than on pavement. The MDP is calculated as follows: [ ] a − (mv + b) (21.7) MDP = Pg − Wv sin 𝛼 + g where MDP is the machine drive power (kJ∕s), Pg is the gross power needed to move the machine (kJ∕s), W is the roller weight (kN), v is the roller velocity (m∕s), 𝛼 is the slope angle

(roller pitch from a sensor), a is the machine acceleration (m∕s2 ), g is the acceleration due to gravity (m∕s2 ), and m and b are the machine internal loss coefficient specific to a particular machine (kJ/m and kJ/s, respectively). The second and third terms in Eq. 21.7 represent the machine power associated with a sloping grade and the internal machine loss, respectively. The MDP represents only the machine power associated with the soil properties (White and Thompson, 2008) and decreases as the soil becomes more compact. For vibrating rollers, the compaction meter value (CMV) can be used. Some of the early work on continuous compaction control demonstrated that various indices incorporating drum acceleration amplitude and the amplitude of its harmonics could be linked to the stiffness of the underlying soil. Based on this early research, the CMV was proposed (Thurner and Sandström, 1980). The CMV is a dimensionless compaction parameter that depends on roller dimensions (drum diameter and weight), roller operation parameters (frequency, amplitude, speed), soil mechanical properties (strength and stiffness), and soil stratigraphy. It is determined using the roller acceleration signal and calculated as: a (21.8) CMV = C 2Ω aΩ where C is a constant (300), a2Ω is the acceleration of the first harmonic component of the vibration, aΩ is the acceleration of the fundamental component of the vibration, and Ω is the vibrating frequency of the roller. If the soil is soft, the roller stays in contact with the soil and the roller and the soil move together; therefore the signal is sinusoidal, there is no other frequency content in the signal except for Ω, and CMV is thus zero. As the soil becomes stiffer, the roller starts to jump and knock; this increases the frequency content of the signal, which becomes more complicated than just a sinusoidal signal, and the value of a2Ω increases. A Fourier transform analysis of the time domain signal gives the frequency content and therefore the value of a2Ω . The CMV at a given point indicates an average value over an area with a width equal to the width of the drum and a length equal to the distance the roller travels in a set time period (0.5 seconds, for example). The soil modulus E, or the MDP, or the CMV can be used to evaluate the degree of compaction achieved by the roller and a degree-of-compaction map of the area covered by the roller can be generated and located according to the GPS (see Figure 21.9).

21.6

(a)

(b)

Figure 21.15 Principle of machine drive power: (a) Soft soil = hard to push. (b) Hard flat soil = easy to push. (Source: Thompson, 2012.)

733

IMPACT ROLLER COMPACTION

Traditionally, compaction rollers have been cylindrical and have used their static weight, kneading action, or vibratory force to achieve the specific soil stiffness and soil strength. However, traditional rollers may have an energy capacity that is too low compared to the need. This might be the case for breaking the interparticle bonds of collapsible sands,

734

21 COMPACTION

(a)

(b)

Figure 21.16 Impact rollers. (Source: a: Courtesy of LAND-PAC; b: Courtesy of Brooms.)

Cylindrical roller

Triangular roller upon impact

3 Depth to ΔσTOP /10 (m)

for example. Impact compaction rollers were developed to alleviate this type of problem. They have noncircular drums (Figure 21.16) that rotate and fall to impact the ground surface. Such rollers tend to provide deeper compaction because the impact generates a wave that propagates at depth. Figure 21.17 demonstrates this point. It shows a freeze-frame picture of a numerical simulation movie describing the stress

Triangular drum

Pentagonal drum

Landpac drum

Octagonal drum

Cylindrical drum

2.5 2.3 2

1.9

1.9

1.6

1.5

1.4 1.25

1

1.6 1.4 1.05

1.5

0.95

0.93 0.9

1.2

0.5

0

5

10

15

20 25 30 35 40 Young’s modulus (MPa)

45

50

55

Figure 21.18 Depth of influence of rollers. (Source: Kim, 2010/Kukjoo Kim.)

Δσ/Δσmax 0.0 0.0

0.2

0.4

0.6

0.8

1.0

Depth (m)

0.5

1.0

1.5 Cylindrical 2.0

Triangular Pentagonal

2.5

Figure 21.17 Stress field under rollers. (Source: Kim, 2010/Kukjoo Kim.)

field in the soil as the roller passes over that spot (Kim, 2010). The simulation compares the case of a cylindrical roller with one of a triangular impact roller. These types of simulations were used to generate the depth chart of Figure 21.18, which indicates that the depth of influence decreases as the soil becomes stiffer and as the roller becomes closer to a cylindrical roller. Here the depth of influence is defined as the depth at which the stress becomes equal to one-tenth of the stress under the roller at the ground surface. In that sense, impact rollers are more efficient; however, the biggest drawback is that they do not provide evenly compacted surfaces (Figure 21.16). The following comments summarize the situation with cylindrical and impact rollers: 1. The width of the contact area between the drum and the soil controls the depth of compaction. The softer the soil is, the deeper the roller sinks into the soil, the wider the contact area is, and the deeper the compaction is. Therefore, the depth of compaction depends on the stiffness of the soil and the depth of compaction decreases with the number of passes.

21.7 DYNAMIC OR DROP-WEIGHT COMPACTION

2. The surface pressure controls the degree of compaction. This pressure is higher for impact rollers than for cylindrical rollers due to the dynamic effect. However, the distribution of the pressure is much more uneven for impact rollers than for cylindrical rollers. 3. The depth of compaction is larger for impact rollers because they impart higher stresses that increase the penetration of the roller drum into the soil, thereby increasing both the width and depth of influence. The increased depth of influence is also due to wave propagation during the impact. These waves can propagate much deeper than the typical depth of influence for static loading. 4. If time and equipment allow it, it makes sense to compact first with an impact roller and use several passes to minimize the extent of the areas between impacts. Then, finish by using a cylindrical roller to provide a more evenly compacted surface and optimize the compaction of the shallow layers. 5. The process described in item 4 above combines the benefits of both types of rollers: compaction of the deep layers (0.5–1.5 m) with the impact roller followed by evening out of the compaction of the shallow layers (0–0.5 m) with the cylindrical roller without disturbing the deep layers.

21.7

DYNAMIC OR DROP-WEIGHT COMPACTION

Dynamic compaction is often credited to Louis Menard (Menard and Broise, 1975). It consists of lifting a heavy weight of mass M and dropping it from a preset height H so as to pound the soil and compact it in the process (Figure 21.19). The pounding is repeated at the same spot for a number of drops (say, 6 times) and thus creates a crater; then the crane moves to another location and repeats the process in a grid pattern. The center to center spacing between impact points is about 2 times the diameter of the tamper. The crater should not be any deeper than 1.5–2 times the height of the

tamper, to avoid the collapse of the walls of the crater and associated difficulties in pulling the tamper out of the crater. The craters are typically backfilled with coarse-grained soil. After completing the first grid, the crane does a second pass by dropping the weight on the intermediate spots to complete the surface treatment. The drop weights commonly weigh 50–300 kN and drop from heights of up to 30 m, reaching velocities of 10–20 m/s at impact. During the final pass, called ironing, a flatter weight is dropped to smooth out the bumps. Upon each drop, the energy generated by the impact propagates to the deeper layers by compression and shear wave propagation. Thus, the effectiveness of this compaction process depends on the dynamic response characteristics of the soil being compacted. Trials are usually run ahead of time to evaluate the potential results, but dynamic compaction works best for unsaturated coarse-grained soils and is not applicable to saturated fine-grained soils. The maximum depth D that can be compacted by dynamic compaction is influenced by many factors, including the soil properties, the groundwater level, the number of drops at each location, and the amount of time elapsed between the grids. The following equation is recommended by Lukas (1995): √ D = n MH (21.9) where n is a site factor less than 1 (Table 21.3), M is the mass of the tamper in tonnes (1000 kg), and H is the average drop height in meters. For an applied energy of 1 to 3MJ∕m2 and for a temper drop using a single cable with a free spool drum, Eq. 21.9 was modified by Varaksin as follows (Chu et al., 2009): √ (21.10) D = C𝛿 MH where C is an equipment factor given in Table 21.4, and 𝛿 is a soil factor equal to 0.9 for metastable soils, young fills, or very recent hydraulic fills, and equal to 0.4–0.6 for sands. Compaction depths of 10 m can be achieved with the heavier tampers (e.g., 20 tonnes dropping 20 m). The improvement ratio f , defined as the ratio of the strength after dynamic

W H

D = n WH 0.2 < n < 0.9 (a)

735

(b)

Figure 21.19 Dynamic compaction. (Source: b: Courtesy of Menard Bachy Pty Ltd.)

736

21 COMPACTION

Table 21.3

Recommended values of n for different soils

compaction over the strength before dynamic compaction, varies with depth and is typically measured by in situ testing (PMT, CPT, SPT). Varaksin proposes the following variation of f with depth below the tamper: ( )2 z f = f1 + (f2 − f1 ) (21.11) D where f 1 and f 2 are the improvement ratios at the ground surface and at the depth D, respectively, and z is the depth at which f is evaluated. The energy E input in the soil for each drop by dynamic compaction can be presented per unit of surface area compacted (E2 in kJ∕m2 ) or per unit of soil volume compacted (E3 in kJ∕m3 ). The energy per unit surface area compacted E2 is: W ×H×N×P (21.12) E2 = s2 where W is the weight of the tamper in kN, H is the height of drop in meters, N is the number of drops, P is the number of passes, and s is the grid spacing in meters for the pounding pattern. The energy per unit volume of soil compacted E3 is:

Degree of Recommended saturation n value

Soil type Pervious soil deposits, granular soils Semi-pervious soil deposits, primarily silts with plasticity index 8

High

0.5

Low High

0.5–0.6 0.35–0.4

Low

0.4–0.5

High

Not recommended 0.35–0.40 Soil should be at water content less than the plastic limit

Low

(Source: Lukas, 1995/U.S. Department of Transportation.)

Table 21.4 Drop method Equipment factor C

Values of the equipment factor C

Double Free Rig Mechanical Hydraulic hydraulic drop drop winch winch winch 1.0

0.89

0.75

0.64

E2 (21.13) D where D is the depth of soil compacted. Lukas (1995) gives a list of typical energies used for different soil types (Table 21.5). E3 =

0.5

(Source: Chu et al., 2009/IOS Press.)

Table 21.5

Applied energy guidelines for E3

Type of deposit Pervious coarse-grained soil, Zone 1 Semi-pervious fine-grained soils, Zone 2; and clay fills above the water table, Zone 3 Landfills

Unit applied energy (kJ∕m3 )

Percent standard proctor energy

200–250 250–350

33–41 41–60

600–1100

100–180

U.S Standard sieve number 100

4

40

200

0

Percent finer by weight

80 70 60 50 40

10 20 30 40

Zone Pervious soils Plasticity index (PI) = 0 Permeability greater than 1×10–5m/s

50 60

30

70

20 10 0 5 mm

Zone 2 80 Semi-Pervious Plasticity index 0 < PI < 8 90 Permeability in the range of 1×10–5m/s to 1×10–8m/s

0.5 mm 0.74 mm Sand Coarse Medium Fine

Note: Standard Proctor energy equals 600 kJ/m3 . (Source: Lukas, 1995/U.S. Department of Transportation.)

100 0.005 mm 0.001 mm

Silt or clay

Percent coarser by weight

Zone 3 Impervious soils Plasticity index PI > 8 Permeability less than 1×10–8m/s

90

737

21.7 DYNAMIC OR DROP-WEIGHT COMPACTION

100

2

Depth (m)

ta

Before

8

10

Tolerable for extremely fragile old buildings 1 0.1

After

D=m M = tonnes H=m

da

6

Tolerable for modern buildings

1.7

d

4

MH D

PPV = 75 re su ea m 5) of 198 ge e, an n R ay (M

Peak particle velocity (mm/s)

0

1

10

Normalized distance ( D / MH ) 10

Figure 21.21 Peak particle velocity due to dynamic compaction. (Source: Lukas, 1995/American Society of Civil Engineers.) 12

0

0.5

1

1.5

2

Limit pressure (pL, MPa)

Figure 21.20 Improvement of soil strength due to dynamic compaction.

The degree of efficiency of dynamic compaction is measured by comparing the results of soil tests performed before and after the compaction process. The preferred test is the pressuremeter test (Figure 21.20), but the cone penetrometer test and the standard penetration test are also used. The depth of compaction is defined here as the depth to which the soil strength has increased compared to the initial state. This increase is not constant with depth, as seen in Figure 21.20. A typical use of dynamic compaction is to dynamically compact the soil deposit so that shallow foundations can be used instead of more expensive pile foundations. The decision is based on comparing the cost of a shallow foundation plus dynamic compaction to the cost of a deep foundation.

Dynamic compaction induces soil vibrations. These vibrations are typically measured in terms of the peak velocity of the soil particles or PPV. The PPV depends on a number of factors, primarily the energy of the impact and the distance from the impact. Mayne (1985) assembled a database giving the range of values shown in Figure 21.21. The following equation gives an upper bound of the PPV generated by a dynamic compaction impact according to the data in Figure 21.21: )1.7 (√ MH (21.14) PPV = 75 d where PPV is the peak particle velocity in mm/s, M is the mass of the tamper in tonnes (1000 kg), H is the drop height in m, and d is the horizontal distance from the impact location in m. This value of the PPV can be compared with what is tolerable. Typical values of PPV for damage threshold vary from 1 to 3 mm/s for very old and fragile buildings to 20–50 mm/s for modern buildings (Figure 21.21).

Problems and Solutions Problem 21.1 Referring to Figure 21.2, give the maximum dry density and the optimum water content. What is the degree of saturation of the soil at that point if the specific gravity of solids is 2.7? Solution 21.1 The maximum dry unit weight and the optimum water content are obtained from the compaction curve. The values from Figure 21.2 are: • Maximum dry unit weight, 𝛾dmax = 19.3 kN∕m3 • Optimum water content, wopt = 10% • Degree of saturation, S

738

21 COMPACTION

Using the equation that links these quantities: Gs 𝛾w 𝛾d = Gw 1+ s S Gs w 2.7 × 0.10 S= = = 0.725 = 72.5% Gs 𝛾w 2.7 × 9.81 − 1 −1 19.3 𝛾d Problem 21.2 Use the data points from Figure 21.5 to draw a correlation between dry density and modulus. Find the R square value for that correlation. Discuss whether there should be or should not be a correlation between dry density and modulus. Solution 21.2 The values of the modulus and dry density are as plotted in Figure 21.1s, and the R square value shows that there is not a good correlation between dry unit weight and soil modulus. The modulus depends on many other factors besides the amount of solids per unit volume. Factors such as structure, cementation, and stress history also affect the modulus. It is not surprising that there is no good correlation between dry density and modulus.

Modulus, E (MPa)

50 40 30 R2 = 0.3487

20 10 0 16

16.5

17 17.5 18 Dry unit weight, γd (kN/m3)

18.5

19

Figure 21.1s The correlation between dry unit weight and soil modulus.

Problem 21.3 Correlate the depth of the imprint that you can make with your thumb as a function of the soil modulus being compacted. Solution 21.3 (Figure 21.2s) 𝜐 = Poisson’s ratio of the soil (assumed to equal 0.35) S = settlement (m) E = soil modulus 𝜋 B s = (1 − 𝜐2 ) ⋅ p ⋅ 4 E

Depth of imprint (mm)

2.5 2 1.5 1 0.5 00

10

20

30

40

Soil modulus (MPa)

Figure 21.2s Depth of finger imprint vs. soil modulus.

50

21.7 DYNAMIC OR DROP-WEIGHT COMPACTION

739

Assuming that you can generate 0.1 kN with your thumb and that the area of contact between your thumb and the surface is 30 by 20 mm, the pressure under your thumb is: p=

F 100 × 10−3 = 166.67 (kN∕m2 ) = A 20 × 30 × 10−6

The settlement can be calculated as: B 𝜋 0.02 2.297 𝜋 = s(m) = (1 − 𝜐2 ) ⋅ p ⋅ = (1 − 0.352 ) × 166.67 × 4 E 4 E E or, with E in kPa and s in mm: E(kPa) = 2300∕s(mm) Problem 21.4 A vibratory intelligent roller weighs 140 kN; it has a drum diameter of 1.4 m and a drum length of 2.1 m. The eccentric weight generates a moment (me re in Eq. 21.2) equal to 1.5 kg.m at an angular frequency of 200 rd∕s. The drum weighs 30 kN and the added weight from the frame above the drum is 20 kN. The measured peak acceleration of the drum is + or −3g. Assume that the inertia force generated by the vibration of the frame is negligible compared to the one generated by the drum. Draw the acceleration signal, the velocity signal, and the displacement signal at the drum-soil contact point. Solution 21.4 (Figures 21.3s, 21.4s, 21.5s) ẍ = amax sin(𝜔t) = 30 sin(200t) 40 Acceleration (m/s2)

30 20 10 0 –10 0

0.01

0.02

0.03

0.04

0.05

0.06

–20 –30 –40 Time, t (sec)

Figure 21.3s Acceleration signal.

ẋ = −

amax cos(𝜔t) = −0.15 cos(200t) 𝜔

Vertical velocity (m/s)

0.02 0.15 0.01 0.05 0 0

0.01

0.02

0.03

0.04

–0.1 –0.15 –0.2

Time, t (sec)

Figure 21.4s Velocity signal.

0.05

0.06

0.07

740

21 COMPACTION

x=−

amax sin(𝜔t) = −0.75 × 10−3 sin(200t) 𝜔2

Vertical displacement (mm)

1 0.8 0.6 0.4 0.2 0 –0.2 0 –0.4 –0.6

0.01

0.02

0.03

0.04

0.05

0.06

0.07

–0.8 –1 Time (sec)

Figure 21.5s Vertical displacement signal.

Problem 21.5 The vibratory roller from problem 21.4 rests on a soil that has a stiffness ks to be determined. The damping coefficient of the soil is 200 kN∕m. Calculate the stiffness of the soil ks , the modulus of the soil E, and the width b of the contact area. Solution 21.5 (Figures 21.6s, 21.7s) 200 Contact force, F (kN)

160 120 80 40 0 –40 0.00

0.02

0.04

0.06

0.08

0.10

–80 –120

Time, t (sec)

Figure 21.6s Contact force versus time.

Contact force F (kN)

160 120 80 Ks 40 0

–1.0

–0.8

–0.6

–0.4

–0.2

0.0

0.2

0.4

0.6

–40 –80 Drum movement, xd (mm)

Figure 21.7s Contact force versus movement and soil stiffness.

0.8

1.0

21.7 DYNAMIC OR DROP-WEIGHT COMPACTION

741

Wroller = 140 kN Drum radius, R = 0.7 m Drum length, L = 2.1 m me re = 1.5 kg.m 𝜔 = 200 rad∕ sec Wdrum = 30 kN cs = 200 kN s∕m Poisson’s ratio, v = 0.35 (assumed) The equations are: ẍ d = ±3g sin(𝜔t) 3g cos(𝜔t) 𝜔 3g xd = ∓ 2 sin(𝜔t) 𝜔

ẍ d = ∓

F = −md ẍ d + me re 𝜔2 cos(𝜔t) + (mf + md )g The stiffness can be taken as the slope as shown in Figure 21.7s, or as the slope of the loop: 160 kN ks = = 150.9 kN∕mm 1.06 mm ks = 150900 kN∕m Then, the soil modulus is: ks =

ks =

Fs = xd

𝜋LE ]) ( [ 1 𝜋L3 E 2 2(1 − 𝜐 ) 2.14 + Ln 2 16(1 − 𝜐2 )(mf + md )Rg 𝜋 × 2.1 × E

⎤⎞ ⎛ ⎡ ⎥⎟ ⎜ ⎢ 3 1 × E 𝜋 × 2.1 ⎥⎟ 2(1 − 0.352 ) ⎜2.14 + Ln ⎢ ( ) ⎥⎟ ⎜ 2 ⎢ 20 30 2 + 0.7 × g ⎥⎟ ⎜ ⎢ 16(1 − 0.35 ) × g g ⎦⎠ ⎣ ⎝

6.6 × E ]) [ 1 29.1 × E 1.755 × 2.14 + Ln 2 491.4 E 40125.7 = ( ]) [ 1 29.1 × E 2.14 + Ln 2 491.4 and solving for E gives E = 281 MPa. To compute the contact width b between the roller and the soil at the time of the highest force, F = 160 kN, we use the following equation: √ 16 R(1 − 𝜐) F b= 𝜋 E L √ 16 0.7(1 − 0.35) 160 b= × × = 0.025 m 𝜋 280900 2.1 b = 25 mm ks =

(

742

21 COMPACTION

Problem 21.6 A landfill must be compacted by dynamic compaction to improve its bearing capacity. The required depth of compaction is 10 m. Determine the weight of the tamper to be used and the drop height required to achieve the 10 m depth of compaction. Solution 21.6 The following equation is used to evaluate the depth of compaction, D: √ D = 𝛼 MH From the problem statement, D = 10 m. Alpha is typically between 0.3 and 0.8; let’s assume 0.5: D∕𝛼 = 10∕0.5 = 20 √ 20 = MH M = 20 tonnes H = 20 m Problem 21.7 Regarding the landfill in problem 21.6, the closest building is located 100 m from the edge of the compaction zone. Calculate the peak particle velocity that can be expected. Would this be normally tolerable for a recently constructed building? Solution 21.7 The peak velocity of the soil particles (PPV) (in mm/s) caused by the dynamic vibration is calculated as: (√ )1.7 MH PPV = 75 d where M is the mass of the tamper in tonnes, H is the drop height in meters, and d is the distance from the impact zone in meters. From problem 21.6, M = 20 tonnes and H = 20 m. Then, with d = 100 m, the PPV is: )1.7 (√ ) ( MH 20 1.7 = 75 × = 4.86 mm∕s PPV = 75 d 100 The damage threshold for modern buildings is set to be 20 ∼ 50mm∕s, therefore, this PPV is tolerable for a recent building located 100 m away from the closest edge of the compaction zone.

CHAPTER 22

Retaining Walls

22.1 DIFFERENT TYPES (TOP-DOWN, BOTTOM-UP) There are many different types of retaining walls, but they are generally classified into two main categories: bottom-up walls and top-down walls. Bottom-up walls are walls that are built before the soil is placed behind the wall. In this case, the backfill is compacted in lifts from the bottom of the wall to the top of the wall, often with inclusions (e.g., metal strips, geosynthetics) being installed on the way up. Top-down walls are walls that are built in the ground; then the excavation in front of the wall takes place in stages, most often with inclusions (e.g., anchors, tiebacks, nails) being installed through the wall as excavation proceeds. Examples of bottom-up walls are gravity walls and mechanically stabilized earth (MSE) walls (Figure 22.1). Examples of top-down walls are cantilever walls, soil-nailed walls, and anchored walls (also known as tieback walls). The design of retaining walls requires calculations regarding: 1. Earth pressure distribution behind the wall 2. Deflection of the wall 3. Drainage issues. Bottom-up walls

Top-down walls

Gravity wall

MSE wall

Cantilever wall

Anchored wall

Figure 22.1 Types of retaining walls.

Soil nail wall

The body of knowledge regarding the issue of earth pressure is much more developed than that on the issue of deflection. One of the reasons is that historically, earth pressure theories came first. 22.2 ACTIVE, AT REST, PASSIVE EARTH PRESSURE, AND ASSOCIATED DISPLACEMENT Consider an imaginary wall in a lake. The water pressure uw on both sides of the wall would be hydrostatic and equal to 𝛾w z where uw is the water pressure against the wall at depth z below the water surface, and 𝛾w is the unit weight of water. As a result, the pressure diagram is triangular and the resultant is located at two-thirds of the wall height from the top of the wall. Note that the water pressure is the same in all directions, including horizontal and vertical, because water has a negligible resistance to shear (the Mohr circle for water is a point). Now consider an imaginary wall in the ground (Figure 22.2). The at-rest earth pressure 𝜎oh exists on both sides of the wall. If you push the wall horizontally, the pressure will increase on the side that penetrates into the soil up to soil failure (passive pressure 𝜎ph ) and decrease on the other side where the wall is moving away from the soil down to soil failure (active pressure 𝜎ah ). Note that if you push the wall far enough and if the soil is strong enough because of true or apparent cohesion, the pressure may become zero on the side where the wall is moving away from the soil and a gap opens up. No movement

At-rest pressure

Movement

Active pressure

Passive pressure

Movement

Passive pressure

Active pressure

Figure 22.2 Imaginary wall and earth pressures.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

743

744

22 RETAINING WALLS

H ya

H

yp

fc1 Passive wedge

Active wedge

Figure 22.3 Earth pressures wedges.

Wall

fc2

y>0

z

Wall Air Water

fc3

Grains

fc1 Air

fw1

fw2

On the passive side, the soil is pushed away and upward as a wedge of failing soil forms in front of the wall (Figure 22.3); as a result, the soil imposes an upward friction force on the wall. On the active side, the soil falls against the wall and downward as a wedge of failing soil forms behind the wall; as a result, the soil imposes a downward friction force on the wall. The passive wedge is much larger than the active wedge and requires more displacement to be mobilized. This is why the displacement required to mobilize the passive earth pressure is larger than the displacement required to mobilize the active earth pressure. The relationship between the soil pressure against the wall and the horizontal displacement of the wall is shown in Figure 22.4. Now let’s zoom in at the interface between the soil and the wall as shown in Figure 22.5. The soil particles contact the wall at several points where forces are transmitted between the soil and the wall. Between the particle contacts are the voids in the soil. These voids can be either completely filled with water (saturated soil) or filled with air and water (unsaturated soil). In the case of the saturated soil, the water will exert a pressure uw against the wall. This water stress can be compression below the groundwater level (GWL) or tension within the capillary zone above the GWL. The water stress times the area of wall over which the water acts is the force transmitted by the water on the wall. The horizontal force on the wall is the sum of the forces at the particle contacts and the force contributed by the water stress uw . Then we divide by the

Grains fw1 fa1 fw2

fa2 fc2

fc3 fa2 fw3 fc4

Water

Figure 22.5 View of the soil-wall contact.

total area and, as in the case of vertical stress (see Chapter 11, Section 11.13), the total horizontal stress 𝜎 h is 𝜎h′ + uw where 𝜎h′ is the effective horizontal stress and uw is the water stress (compression or tension). If the soil is unsaturated, the horizontal force on the wall is the sum of the forces at the particle contacts, the forces transmitted through the water, and the forces transmitted through the air. If the air is occluded in the water phase and does not contact the wall, then 𝜎 h is still equal to 𝜎h′ + uw , but the water phase is more compressible. Air tends to be occluded when the degree of saturation S is above 85%. If the air is not occluded (S < 85%), there is a continuous air path to the ground surface and the air stress is atmospheric or zero gage pressure. In this case (see Chapter 11, Section 11.13), the horizontal stress 𝜎 h is 𝜎h′ + 𝛼 uw where 𝜎h′ is the effective horizontal stress, 𝛼 is the ratio of the water area in contact with the wall over the total area, and uw is the water stress (which is in tension in this case). As pointed out in Chapter 11, Section 11.13, 𝛼 can be estimated as the degree of saturation with a ±30% precision or by the Khalili rule. The effect of the water tension in unsaturated soil will be to decrease the active horizontal pressure and increase the passive horizontal pressure compared to the case of the saturated soil with water in compression. Note that the active earth pressure and the passive earth pressure correspond to soil failure. Therefore, they should be thought of as strength rather than stress.

σh

22.3 σh At-rest pressure At-rest σoh pressure σah ya

22.3.1 σph Passive pressure yp

y

Figure 22.4 Earth pressure versus wall displacement.

EARTH PRESSURE THEORIES Coulomb Earth Pressure Theory

The earth pressure theories make the general assumption that the soil is at failure. In that sense, the earth pressures obtained by using these theories are similar to the concept of ultimate bearing capacity in foundation engineering; they represent strengths at failure rather than stresses at working loads. Coulomb, in 1776, was the first person to work on earth pressures. Charles Augustin de Coulomb was a French physicist who worked on this topic just before the French

22.3 EARTH PRESSURE THEORIES

β>0

F

A α+ β

ρ–β

B

C H ρ

α E

D

Figure 22.6 General geometry of the Coulomb soil wedge.

Revolution in the late 1700s, although he is better known for his work on electromagnetism. To develop his earth pressure theory, Coulomb made the following assumptions (Figure 22.6): 1. 2. 3. 4. 5. 6.

The problem is a plane strain problem. The soil has friction (𝜑′ ) and cohesion (c′ ). The soil has no water. The failure wedge is a rigid body. The failure surface and the ground surface are planes. The friction coefficient between the wall and the soil wedge is tan 𝛿.

Let’s first calculate the weight of the wedge W per unit length of wall. The area A of the triangle ABD is: ( ) sin (𝛼 + 𝛽) 1 1 A = BD × AC = AD 2 2 sin (𝜌 − 𝛽) × AD sin(180 − 𝛼 − 𝜌) (22.1) Because AD = then 2

A=

H sin(𝛼 + 𝜌) 2sin2 𝛼

and W=

H sin 𝛼 (

(

𝛾H 2 2sin2 𝛼

sin(𝛼 + 𝜌)

B A

H Pa

(22.3) )

and 𝛾H 2

R

D

Figure 22.7 Free body of the active soil wedge.

)

(22.6)

𝜕Pa =0 𝜕𝜌

(22.7)

and solve for 𝜌 as was done in Chapter 12, Section 12.4.2. The final result for Pa is: 𝛾H 2 2

sin2 (𝛼 + 𝜑′ ) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ − 𝛽) ⎤ sin(𝜑 2 ⎥ sin 𝛼 sin(𝛼 − 𝛿) ⎢1 + ⎢ sin(𝛼 − 𝛿) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦

1 K 𝛾 H2 2 a

(22.8)

and the coefficient of active earth pressure K a giving the magnitude of the vector Pa is:

α-δ ρ-φʹ

(22.5)

Equation (22.6) shows that Pa is a function of a number of factors, including the angle 𝜌 which is an unknown variable. The active earth pressure force will correspond to the value of 𝜌 that leads to the lowest value of Pa , because that will be the first value reached as the wall is pulled away from the soil. Therefore, the 𝜌 value corresponding to the active force is the one that minimizes Pa . For this we set:

=

R

(

sin (𝛼 + 𝛽) Pa = sin(𝛼 + 𝜌) 2 sin (𝜌 − 𝛽) 2sin 𝛼 ′ sin(𝜌 − 𝜑 ) × sin(180 − 𝜌 + 𝜑′ − 𝛼 + 𝛿)

(22.4)

Pa

ρ

Pa W = sin(𝜌 − 𝜑′ ) sin(180 − 𝜌 + 𝜑′ − 𝛼 + 𝛿)

Pa =

β

C φʹ α

)

sin (𝛼 + 𝛽) sin (𝜌 − 𝛽)

w

+δ

In the case of the active earth pressure (Figure 22.7), the external forces acting on the wedge are the weight W, the active force Pa on the wall side AD, and the resultant force R on the soil side BD. The force Pa is inclined at an angle 𝛿 with the normal to the wall-soil interface. If the wall does not settle excessively, the wedge goes down with respect to the wall and the wall friction acts upward on the wedge (positive 𝛿 value for the active case). The resultant R is inclined at an angle 𝜑′ with the normal to the soil-soil failure plane at the back of the wedge. Because the wedge goes down with respect to the soil mass beyond the wedge, the friction force acts upward on the wedge. We will ignore the cohesion force at this time. Then the polygon of forces can be drawn (Figure 22.7) and the law of sines gives:

(22.2)

sin (𝛼 + 𝛽) sin (𝜌 − 𝛽)

745

W

Ka =

sin2 (𝛼 + 𝜑′ ) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ − 𝛽) ⎤ sin(𝜑 ⎥ sin2 𝛼 sin(𝛼 − 𝛿) ⎢1 + ⎢ sin(𝛼 − 𝛿) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦ (22.9)

746

22 RETAINING WALLS

Note that the direction of the force Pa is not horizontal, but rather acts at an angle 90 − 𝛼 + 𝛿 with the horizontal. The horizontal component Pah is: 1 1 Pah = Ka 𝛾 H 2 cos(90 − 𝛼 + 𝛿) = Ka 𝛾 H 2 sin(𝛼 − 𝛿) 2 2 1 2 = Kah 𝛾 H (22.10) 2 Therefore, the coefficient of active earth pressure K ah giving the horizontal component Pah of the active push Pa is:

The passive earth pressure coefficient giving the magnitude of the vector Pp is:

sin2 (𝛼 + 𝜑′ ) (22.11) 2 √ ⎡ ⎤ ′ ′ sin(𝜑 + 𝛿) sin(𝜑 − 𝛽) ⎥ sin2 𝛼 ⎢1 + ⎢ sin(𝛼 − 𝛿) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦ In the simpler case where the backfill is horizontal, the wall is vertical, and there is no soil-wall friction (conservative), then 𝛽 = 𝛿 = 0, 𝛼 = 90∘ , and K a becomes: 1 − sin 𝜑′ Ka = (22.12) 1 + sin 𝜑′ In the case of the passive earth pressure (Figure 22.8), the external forces acting on the wedge are the weight W, the passive force Pp on the wall side AD, and the resultant force R on the soil side BD. The force Pp is inclined at an angle 𝛿 with the normal to the wall-soil interface. As the wall pushes against the wedge, the wedge goes up with respect to the wall and the wall friction acts downward on the wedge (positive 𝛿 value for the passive case). The resultant R is inclined at an angle 𝜑′ with the normal to the soil-soil failure plane at the back of the wedge. Because the wedge goes up with respect to the soil mass beyond the wedge, the friction force acts downward on the wedge. We will ignore the cohesion force at this time. Then the polygon of forces can be drawn (Figure 22.8) and the derivation proceeds as for the active case. In the end, the equation for Pp is:

Note that the direction of the force Pp is not horizontal, but rather acts at an angle 𝛼 + 𝛿 − 90 with the horizontal. The horizontal component Pph is:

Kah =

1 K 𝛾 H2 2 p

Pp =

(22.13)

B β A

H

Kp =

sin2 (𝛼 − 𝜑′ ) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ + 𝛽) ⎤ sin(𝜑 ⎥ sin2 𝛼 sin(𝛼 + 𝛿) ⎢1 − ⎢ sin(𝛼 + 𝛿) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦ (22.14)

1 1 K 𝛾 H 2 cos(𝛼 + 𝛿 − 90) = Kp 𝛾 H 2 sin(𝛼 + 𝛿) 2 p 2 1 2 = Kph 𝛾 H (22.15) 2 Therefore, the coefficient of passive earth pressure K ph giving the horizontal component Pph of the passive push Pp is: Pph =

Kph =

sin2 (𝛼 − 𝜑′ ) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ + 𝛽) ⎤ sin(𝜑 ⎥ sin2 𝛼 ⎢1 − ⎢ sin(𝛼 + 𝛿) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦

(22.16)

In the simpler case where the backfill is horizontal, the wall is vertical, and there is no soil-wall friction (conservative), then 𝛽 = 𝛿 = 0, 𝛼 = 90∘ , and K p becomes: 1 + sin 𝜑′ 1 − sin 𝜑′ and the product Ka × Kp is equal to 1. Kp =

22.3.2

(22.17)

Rankine Earth Pressure Theory

In 1857, Rankine took a different approach to the same problem. William J. Rankine was a Scottish civil engineer, physicist, and mathematician. He made the following assumptions: 1. 2. 3. 4. 5. 6.

The problem is a plane strain problem. The soil has friction (𝜑′ ) but no cohesion (c′ = 0). The soil has no water. The soil mass is in a state of plastic failure. The failure surface and the ground surface are planes. There is no friction between the soil and the wall.

w

Pp

C

+δ α

φʹ

R

ρ D

Pp

R

α-δ ρ-φʹ w

Figure 22.8 Free body of the passive soil wedge.

Coulomb considered the equilibrium of a rigid body wedge and reasoned in terms of equilibrium of forces, whereas Rankine considered the equilibrium of stresses at the element level in a failing mass. Rankine theory predates the work of Otto Mohr and the Mohr circle around 1882, but it is easiest to explain Rankine theory through the use of the Mohr circle, which will be done in Section 22.3.3. The active and passive earth pressures are as follows (Figures 22.9 and 22.10): 𝜎a = Ka 𝜎v = Ka 𝛾z

(22.18)

𝜎p = Kp 𝜎v = Kp 𝛾z

(22.19)

22.3 EARTH PRESSURE THEORIES

where 𝜎 a and 𝜎 p are the active and passive earth stresses on the wall, K a and K p are the active and passive coefficients, 𝛾 is the soil unit weight, and z is the depth below the ground surface. Note that the stress vectors 𝜎 a and 𝜎 p are parallel to the ground surface and therefore inclined at an angle 𝛽 with the horizontal (Figures 22.9 and 22.10). Rankine obtained the following expressions for K a and K p : √ cos 𝛽 − cos2 𝛽 − cos2 𝜑′ Ka = cos 𝛽 (22.20) √ cos 𝛽 + √cos2 𝛽 − cos2 𝜑′ cos 𝛽 + cos2 𝛽 − cos2 𝜑′ Kp = cos 𝛽 (22.21) √ cos 𝛽 − cos2 𝛽 − cos2 𝜑′ As can be seen from Eqs. (22.18) and (22.19), the stresses on the wall increase linearly with z. By integration of these two equations between 0 and H, the height of the wall, the active force Pa and the passive force Pp can be obtained and are given by Eqs. (22.8) and (22.13), but with different expressions for K a and K p given in Eqs. (22.20) and (22.21). Note also that the forces Pa and Pp are not horizontal, but rather parallel to the ground surface, which is at an angle 𝛽 with the horizontal (Figures 22.9 and 22.10). The horizontal components Pah and Pph are: 1 1 (22.22) Ka 𝛾H 2 cos 𝛽 = Kah 𝛾H 2 2 2 1 1 Pph = Kp 𝛾H 2 cos 𝛽 = Kph 𝛾H 2 (22.23) 2 2 Therefore, the coefficient of active earth pressure K ah giving the horizontal component Pah of the active force Pa and the Pah =

β

σν Pah β Pa

45 +

σʹ 2

Figure 22.9 Active pressure mass (Rankine). β

σν

σp

Pph β Pp

45–

coefficient of passive earth pressure K ph giving the horizontal component Pph of the passive force Pp are: √ cos 𝛽 − cos2 𝛽 − cos2 𝜑′ 2 Kah = cos 𝛽 (22.24) √ cos 𝛽 + cos2 𝛽 − cos2 𝜑′ √ cos 𝛽 − cos2 𝛽 − cos2 𝜑′ Kph = cos2 𝛽 (22.25) √ cos 𝛽 − cos2 𝛽 − cos2 𝜑′ In the simple case where the backfill is horizontal, then 𝛽 = 0, and K a and K p become: 1 − sin 𝜑′ 1 + sin 𝜑′ 1 + sin 𝜑′ Kp = 1 − sin 𝜑′

Ka =

ϕʹ 2

Figure 22.10 Passive pressure mass (Rankine).

(22.26) (22.27)

So, should we use Coulomb or Rankine earth pressure coefficients? The Coulomb solution is a limit equilibrium solution giving upper-bound values because the chosen failure surface and mechanism are not necessarily the weakest one. In this context, Coulomb passive earth pressure coefficients tend to be very optimistic (too large). In contrast, the Rankine solution is an equilibrium of stresses solution that gives lower-bound values. Therefore, if a lower bound is conservative, one could choose Rankine; if an upper bound is conservative, one could choose Coulomb. Note that for extreme values of the geometry parameters, it is advisable to use engineering judgment, as the K a and K p values can become unreasonable. Note also that for the simple case of a vertical wall, no wall friction, and horizontal backfill, both theories give the same answers (Eqs. (22.12), (22.17), (22.26), and (22.27)). The most common values vary from 0.25 to 0.40 for K a and from 2.5 to 4 for K p . 22.3.3

σa

747

Earth Pressure Theory by Mohr Circle

Consider an element of soil behind a retaining wall (Figure 22.11). This element is in an at-rest state of stress to ′ start with. The vertical effective stress is 𝜎ov , the horizontal ′ effective stress is 𝜎oh , and the corresponding Mohr circle is shown in Figure 22.11. If the wall is pulled very slightly away from the soil, the horizontal effective stress will decrease until the Mohr circle touches the failure envelope. At that point the soil element will be in a state of failure. It will have mobilized all the shear strength it can offer to support itself, ′ but will still need 𝜎ah from the wall to avoid collapse. This ′ value 𝜎ah is the active earth pressure. From triangle ABD in Figure 22.11, we can write that: ( ′ ) ′ 0.5 𝜎ov − 𝜎ah BD ′ (22.28) sin 𝜑 = = c′ ( ′ ) ′ AO + OD + 0.5 𝜎ov − 𝜎ah tan 𝜑′ which reduces to: √ ( ) ′ 1 − sin 𝜑 1 − sin 𝜑′ ′ ′ 𝜎ah = 𝜎ov − 2c′ ′ 1 + sin 𝜑 1 + sin 𝜑′

(22.29)

748

22 RETAINING WALLS

τ

Failed soil

45 +

ϕʹ

ϕʹ

2

σʹov B

σʹah ϕʹ

45 +

2

cʹ A cʹ tanϕʹ

O P σʹah

D

σʹoh

E

σʹ

σʹov

Figure 22.11 Element of soil and Mohr circle (active case).

or:

The direction of the failure lines can be found by using the Pole method (see Chapter 11, Section 11.5). The stress point ′ corresponds to a stress acting on a on the Mohr circle at 𝜎ph vertical plane, so a vertical line will intersect the circle at two points: the stress point and the Pole. Because the vertical line is tangent to the circle, the two points are the same and the Pole is at point P on Figure 22.12. A line from the Pole to the failure point B gives the direction of the failure plane on Figure 22.12. From geometry considerations, the angle of this plane with the horizontal is equal to 45 − 𝜑′ ∕2. Because the entire mass is at failure, a set of parallel failure lines exists. The conjugate failure lines on Figure 22.12 come from the failure point on the bottom part of the Mohr circle at failure that is not shown on Figure 22.12.

√ Ka − 2c Ka

1 − sin 𝜑′ = with Ka = (22.30) 1 + sin 𝜑′ The direction of the failure lines can be found by using the Pole method (see Chapter 11, Section 11.5). The stress point on the Mohr circle at 𝜎 ′ ah corresponds to a stress acting on a vertical plane, so a vertical line will intersect the circle at two points: the stress point and the Pole. Because the vertical line is tangent to the circle, the two points are the same and the Pole is at point P on Figure 22.11. A line from the Pole to the failure point B gives the direction of the failure plane on the diagram. From geometry considerations, the angle of this plane with the horizontal is equal to 45 + 𝜑′ ∕2. Because the entire mass is at failure, a set of parallel failure lines exists. Now if the wall is pushed into the soil instead of pulled away (Figure 22.12), the horizontal effective stress will ′ , increase, pass the value of the vertical effective stress 𝜎ov and continue to increase until the Mohr circle touches the failure envelope. At that point the soil element will be in a state of failure: It will have mobilized all the shear strength it can offer to resist the wall push and 𝜎 ′ ph will be generated. This value 𝜎 ′ ph is the passive earth pressure. From triangle ABD in Figure 22.12, we can write that: ) ( ′ ′ − 𝜎ov 0.5 𝜎ph BD (22.31) sin 𝜑′ = = ′ ) ( c AO + OD ′ ′ + 0.5 𝜎 − 𝜎 ov ph tan 𝜑′ ′ 𝜎ah

′ 𝜎ov

′

which reduces to: √ ( ) ′ 1 + sin 𝜑′ 1 + sin 𝜑 ′ ′ ′ 𝜎ph = 𝜎ov − 2c 1 + sin 𝜑′ 1 − sin 𝜑′ or ′ ′ 𝜎ph = 𝜎ov Kp − 2c′

√ Kp

with Kp =

1 + sin 𝜑′ 1 − sin 𝜑′

22.3.4 Water in the Case of Compression Stress (Saturated) Up to this point we have calculated the effective horizontal stress for the active case and the passive case. The wall is subjected to the total horizontal stress. When the soil next to the wall is saturated and the water is in compression (below the water table), the total active and passive earth pressures become: √ ′ 𝜎ah = 𝜎ov Ka − 2c′ Ka + uw (22.34) √ ′ 𝜎ph = 𝜎ov Kp + 2c′ Kp + uw (22.35) The water stress uw is obtained as follows: uw = 𝛾w hp

(22.32)

where 𝛾 w is the unit weight of water, hp is the distance from the groundwater level to the point considered if there is no flow, and hp is the pressure head obtained from a flow net if there is flow.

(22.33)

Failed soil

τ

ϕʹ

σʹov

B σʹph 45 –

ϕʹ 2

(22.36)

A

cʹ

O cʹ σʹoh tanϕʹ

E

σʹov

ϕʹ D 45 – 2

σʹph

Figure 22.12 Element of soil and Mohr circle (passive case).

P

σʹ

749

22.3 EARTH PRESSURE THEORIES

Note that there is a big difference between the pressure against a wall that has to retain a soil without water and the pressure against a wall that has to retain a soil with a water level at the ground surface. For example, if a wall is 3 m high and retains a dry sand with a unit weight of 18 kN/m3 and a friction angle of 30∘ , the active earth pressure behind the bottom of the wall will be: 𝜎ah = 3 × 18 × 0.33 = 18 kN∕m2

(22.37)

However, if the water rises to the top of the wall, increasing the unit weight of the soil to 20 kN/m3 , and if the water stress is hydrostatic, the active pressure behind the bottom of the wall becomes: 𝜎ah = (3 × 20 − 3 × 10) × 0.33 + 3 × 10 = 40 kN∕m2 (22.38) As can be seen, the pressure doubles due to the presence of the water. If we had assumed that no water could be present and designed the wall for a factor of safety of 2, the wall would have been close to failure when the water accumulated behind it. It is extremely important to pay great attention to water when designing retaining walls. 22.3.5 Water in the Case of Tension Stress (Unsaturated or Saturated) If the soil behind the wall is above the groundwater level, the water is in tension and the soil is either saturated or unsaturated. In both cases, the water stress uw is negative (tension). This increases the shear strength of the soil because it increases the effective stress. Thus, one would expect the active earth pressure to decrease and the passive earth pressure to increase. Equations (22.34) and (22.35) become: √ ′ 𝜎ah = 𝜎ov Ka − 2c′ Ka + 𝛼uw (22.39) ′ 𝜎ph = 𝜎ov Kp − 2c′

√ Kp + 𝛼uw

(22.40)

where 𝛼 is the water area ratio, which can be estimated as the degree of saturation S or by using the Khalili rule (see Chapter 11, Section 11.15). Note that the term 𝛼uw is also embedded in 𝜎 ′ ov . Regrouping gives: √ 𝜎ah = 𝜎ov Ka − 2c′ Ka + (1 − Ka )𝛼uw (22.41) 𝜎ph = 𝜎ov Kp + 2c′

√ Kp + (1 − Kp )𝛼uw

(22.42)

Equations (22.41) and (22.42) show that water tension decreases the active earth pressure and increases the passive earth pressure. However, it is very important to consider if the water tension used in these equations will always be present or if it is a seasonal occurrence. Furthermore, it would be uncommon for the water tension to pull on the wall. In the case of unsaturated soils and for earth pressure calculations, it is therefore prudent to consider that the water stress is equal to zero.

22.3.6 Influence of Surface Loading (Line Load, Pressure) Load is often applied at the top of a retaining wall (Figure 22.13) either during construction (e.g., compaction rollers) or after construction (e.g., bridge abutment, additional fill). In the case of a pressure p that covers the entire surface area at the top of the retaining wall, the active and passive earth pressures have an added term K a p and K p p respectively. The reason is that the pressure p simply adds to the total stress 𝜎 ov . In the case of a line load Q (kN/m) parallel to the wall crest and located at a perpendicular distance x from the wall, the increase in horizontal stress against the wall at a depth z below the top of the wall can be calculated by: 4Q x2 z (22.43) Δ𝜎h = 2 𝜋 (z + x2 )2 If the load is a point load P (kN) applied at a perpendicular distance x from the wall, the maximum increase in horizontal pressure against the wall at a depth z below the top of the wall can be calculated by: ( ) (z2 + x2 )1∕2 (1 − 2v) 3x2 z P − Δ𝜎h = 𝜋(z2 + x2 ) (z2 + x2 )3∕2 (z2 + x2 )1∕2 + z (22.44) The values obtained from Eqs. (22.43) and (22.44) are added to both the active earth pressure and the passive earth pressure. Solutions for other surface loading can be found in the Canadian Foundation Engineering Manual (Canadian Geotechnical Society, 2007) and the AASHTO LRFD Bridge Design Specifications (AASHTO, 2007). 22.3.7

General Case and Earth Pressure Profiles

In the general case, the total active and passive earth pressures are given by: √ ′ Ka − 2c′ Ka + Δ𝜎h + 𝛼uw (22.45) 𝜎ah = 𝜎ov √ ′ Kp − 2c′ Kp + Δ𝜎h + 𝛼uw (22.46) 𝜎ph = 𝜎ov where 𝜎 ah is the total active earth pressure on the wall at a ′ is the vertical effective depth z below the top of the wall, 𝜎ov stress at depth z, K a is the coefficient of active earth pressure, Δ𝜎ℏ is the earth pressure due to surface loading, c′ is the effective stress cohesion of the retained soil, 𝛼 is the water area ratio, uw is the water stress (tension or compression), 𝜎 ph is p (kN/m2)

Q (kN/m)

P (kN)

∆σh

∆σh

Ka.p or Kp.p

Pressure

Line load

Point load

Figure 22.13 Horizontal pressures due to surface loading.

750

22 RETAINING WALLS

the total passive earth pressure on the wall at a depth z below the top of the wall, and K p is the coefficient of passive earth pressure. These are the equations to use when calculating the active or passive earth pressure against the wall at a chosen depth z ′ and Δ𝜎h exist. Keep in mind that these pressures where 𝜎ov or stresses may not be horizontal if the ground surface is not horizontal, the back of the wall is not vertical, or the wall friction is not assumed to be zero. A distinction is made in this respect between K a and K ah , on the one hand, and K p and K ph , on the other (Sections 22.3.1 and 22.3.2). The next problem is to generate the complete profile of pressure against the wall versus depth. This is done by preparing a series of profiles using the following steps: Profile of total vertical stress 𝜎 ov versus depth. Profile of water stress uw versus depth. Profile of water area ratio 𝛼 versus depth. ( ′ ) = 𝜎ov − 𝛼 uw Profile of effective vertical stress 𝜎ov versus depth. ′ ′ = 𝜎ov Ka − 5. Profile of effective horizontal stress (𝜎ah 0.5 ′ 2c Ka ) versus depth. 6. Profile of horizontal stress due to surface loads (Δ𝜎ah ) versus depth. ′ 7. Profile of total horizontal stress (𝜎ah = 𝜎ov Ka − 2c′ Ka0.5 +Δ𝜎ah + 𝛼 uw ) versus depth.

1. 2. 3. 4.

Figure 22.14 shows an example of the series of profile steps. The same sequence is followed for the passive earth pressure profiles. If the soil is layered, the earth pressure has to be calculated twice at the depth of the layer boundary: once with the upper-layer soil parameters and once with the lower-layer soil parameters. As a result, there is typically a discontinuity in the earth pressure profile at the boundary between two soil layers (Figure 22.15). Q

σʹov

uw

α

σʹov

σʹah

Δσah

σah

Figure 22.14 Series of profiles to generate earth pressure profile versus depth.

22.4 SPECIAL CASE: UNDRAINED BEHAVIOR OF FINE-GRAINED SOILS As discussed in Chapter 16, Section 16.16, the equations for the undrained behavior of a fine-grained soil can be obtained from the effective stress equations by a simple transformation or correspondence principle: 1. Effective unit weight becomes total unit weight (22.47) 𝛾eff → 𝛾t 2. Effective stress becomes total stress 𝜎′ → 𝜎

(22.48)

3. Effective stress cohesion becomes undrained shear strength (22.49) c′ → su 4. Effective stress friction angle becomes zero 𝜑′ → 0

(22.50)

Using this transformation on Eqs. (22.45) and (22.46), the following equations are obtained for the undrained behavior active and passive earth pressures: (22.51) 𝜎ah = 𝜎ov − 2su + Δ𝜎h 𝜎ph = 𝜎ov + 2su + Δ𝜎h

(22.52)

where su is the undrained shear strength of the soil. These equations tend to give active earth pressures that are too low and passive pressures that are too high. One reason is that they assume that the soil is uniform with no fissures. These equations should be used with great caution and proper judgment. For example, imagine that you have to design a wall for a clay that has an undrained shear strength of 100 kPa and a unit weight of 20 kN∕m3 . Equation (22.51) says that no wall is needed until a depth of 10 m, as the active earth pressure is negative down to that depth. Now imagine that this clay has many fissures that are about 0.3 meters apart. The sample you tested was taken from one of the blocks between fissures and gave 100 kPa for su , but the soil mass is actually much weaker because of the fissures; the sample strength is not representative of the mass strength. If you dug a trench in such a material, it would be very surprising if you could dig down to 10 meters without a major collapse before that point. In contrast, if the material is truly uniform with no fissures (very rare), the theory says that you could dig to 10 m without support.

σaH Layer 1 sand Layer 2 sand

σah1 = Ka1σʹov –2c1 Ka1+αuw σah2 = Ka2σʹv –2c2 Ka2+αuw

22.5

AT-REST EARTH PRESSURE

The at-rest earth pressure is the horizontal stress that exists in the soil under geostatic stresses and without displacement. The coefficient of at-rest earth pressure K o is defined as: Ko =

Figure 22.15 Active pressures at a soil layer boundary.

′ 𝜎oh ′ 𝜎ov

(22.53)

22.5 AT-REST EARTH PRESSURE ′ ′ and 𝜎ov are the horizontal and vertical effective where 𝜎oh stresses, respectively. Note that K o is the ratio of the effective stresses, not the total stresses; also, K o does not involve the cohesion c′ , whereas the ratios K a and K p incorporate c′ in their definition: √ ′ 𝜎ah 2c′ Ka Ka = ′ + (22.54) ′ 𝜎ov 𝜎ov √ ′ 𝜎ph 2c′ Kp Kp = ′ − (22.55) ′ 𝜎ov 𝜎ov

Thus, it is theoretically possible for K o to have values higher than K p and lower than K a . For example, if ′ ′ = 300 kPa, 𝜎ov = 100 kPa, and c′ = 20 kPa, and if a high 𝜎ph horizontal stress at rest is locked up tectonically at the value ′ , then K p is 2.4 and K o is 3. of 𝜎ph In elasticity, the ratio of the horizontal stress to the vertical stress for a condition with no lateral movement (at-rest condition) is obtained in cylindrical coordinates from: )) ( ′ 1( ′ ′ =0 (22.56) + 𝜎oh 𝜎 − v 𝜎ov 𝜀h = E oh where 𝜀h is the horizontal strain, E is a modulus of deformation of the soil, and v is Poisson’s ratio. Therefore v Ko = (22.57) 1−v A commonly used value of Poisson’s ratio for a drained case is 0.33; then K o is equal to 0.5. However, measured K o values have been reported in the range of 0.4 to more than 2. A value of 2 would require a Poisson’s ratio equal to 0.67, which is possible for soils that dilate during compression, a well-known phenomenon. Such high K o values are found in cases where high horizontal stresses have developed during geological events that densify or overconsolidate the soil. They may also be generated during compaction of shallow layers. The coefficient of at-rest earth pressure is very difficult to measure, essentially because any instrument placed in the ground to measure K o will create disturbance and change the at-rest state of stress. The best measurements are thought to be possible with a self-boring pressuremeter. However, even the self-boring pressuremeter creates significant disturbance due to shearing and side friction upon descent of the probe. Furthermore, the choice of zero volume of the probe can significantly affect the value of K o obtained. The early part of the preboring pressuremeter test offers another way to obtain an estimate of the horizontal stress. As the horizontal pressure applied by the pressuremeter probe on the borehole wall is increased, it goes through the threshold of pressure corresponding to the at-rest horizontal pressure. The curved line that describes the horizontal pressure versus increase in radius until the elastic portion of the curve is reached could be used. A construction much like the Cassagrande construction for the preconsolidation pressure in the consolidation test would be needed, but calibration of such an idea has not been performed.

751

The step blade test consists of pushing a series of flat blades of increasing thickness into the soil while recording the horizontal stress on each blade. The idea was to extrapolate the horizontal stresses obtained on each blade back to a blade with zero thickness so as to find the at-rest horizontal stress. Although this idea was very clever, unfortunately the superposition of a penetration event and a lateral expansion event made the extrapolation unreliable. One method consists of measuring the water tension developing in fine-grained soils upon extrusion of saturated samples. When the saturated sample comes out of the sampling tube, it decompresses and the total stress suddenly becomes zero—but the sample cannot readily expand because of the low hydraulic conductivity, and the water goes into tension to prevent any increase in volume. This results in a transfer from the mean effective stress to the water tension. Sample at depth z: ) 1( ′ ′ (22.58) + uw 𝜎ov + 2𝜎oh 𝜎mean = 3 Sample extruded: ) 1( ′ 1 ′ ′ 0= (1 + 2Ko ) + uw or − uw = 𝜎ov 𝜎ov + 2𝜎oh 3 3 (22.59) Equation (22.59) shows that the water tension in the sample is a function of the horizontal effective stress. K o can then be calculated knowing the vertical effective stress. A K o triaxial test can be used to obtain a value of K o . This test consists of loading the sample vertically while increasing the horizontal stress (cell pressure) independently and in such a way that no lateral deformation will take place. During the test, the water stress is measured and the ratio between the horizontal effective stress (cell pressure minus water stress) and the vertical effective stress gives the K o value. Alternatively, consolidometer tests with an instrumented ring can be used to obtain a value of K o . The metal ring in which the sample is placed is instrumented with strain gages to measure the hoop strain in the metal, thereby giving the hoop stress that prevents lateral expansion. The radial stress is then obtained as: 𝜎 t (22.60) 𝜎oh = 𝜃 r where 𝜎oh is the radial or horizontal stress exerted by the soil on the metal ring that prevents expansion, 𝜎 𝜃 is the hoop stress in the metal obtained from the hoop strain measurements, t is the thickness of the metal ring, and r is the radius of the consolidometer. Knowing the vertical stress 𝜎ov imposed on the sample, and assuming that zero water stress is in the sample at the end of consolidation, gives data to calculate K o . One of the difficulties with this approach is to ensure that the strain gages are sensitive enough to detect the strain in the metal ring under the relatively small radial stresses imposed by the soil. Many correlations have also been proposed. The first one may be attributed to Jaky (1944), expressed as: Ko = 1 − sin 𝜑′

(22.61)

752

22 RETAINING WALLS

This equation was later revised to include the effect of the overconsolidation ratio (OCR) for uncemented sands and clays of low to medium sensitivity: sin 𝜑

Ko = (1 − sin 𝜑 )OCR ′

′

(22.62)

where 𝜑 is the effective stress friction angle of the soil, and OCR is the overconsolidation ratio, defined as the ratio of the effective preconsolidation stress 𝜎p′ over the current effective vertical stress. For clean quartz sand in chamber tests, Mayne (2007) proposed: ( )0.22 ( )0.31 q qc Ko = 0.1921 c (OCR)0.27 (22.63) ′ 𝜎a 𝜎ov

a

P (kN/m)- Roller

z

Passive d

σh

′

where qc is the CPT point resistance, 𝜎 a is the atmospheric ′ pressure, 𝜎ov is the vertical effective stress, and OCR is the overconsolidation ratio.

22.6

EARTH PRESSURE DUE TO COMPACTION

When soil is compacted behind bottom-up walls, the compaction process induces horizontal stresses that are higher than active earth pressures. This has been clearly documented (Duncan and Seed, 1986; Chen and Fang, 2008). The compaction roller creates high vertical stresses, which in turn create high horizontal stresses during compaction. Because the soil does not return to an undeformed state after unloading (not elastic), and because the soil locks in plastic strains after unloading, high horizontal stresses remain after the roller moves on. This horizontal prestressing is actually very beneficial for improving the behavior of pavement base courses. For retaining walls, this means that designing for the active earth pressure case may not be prudent. At the same time, the depth of influence of the roller is limited and after several lifts of compaction have been completed, the high stresses at depth (Figure 22.16) become smaller than the at-rest stresses at that depth. The U.S. Navy (1982) made some recommendations for earth pressures due to compaction, which, considering more recent data, lead to the profile shown in Figure 22.17. The pressure diagram starts at a slope equal to the passive earth pressure coefficient. From the surface to a depth where the horizontal pressure reaches the value 𝜎 h , the passive earth Roller

At rest

Figure 22.17 Wall pressure diagram including compaction stresses. (Source: Adapted from U.S. Government Printing Office.)

pressure profile, Kp 𝛾z, is used. Then the pressure remains constant at a value of 𝜎 h equal to: √ 2P 𝛾 L 𝜎h = (22.64) a+L 𝜋 where L is the length of the roller, a is the distance between the edge of the wall and the closest roller position, P is the line load imposed by the roller (weight of the roller plus the centrifugal force for vibratory rollers divided by the length of the roller), and 𝛾 is the unit weight of the soil being compacted. At a depth d, the pressure diagram joins the at-rest earth pressure profile, Ko 𝛾z, which is used beyond that point. That depth d is therefore equal to (Figure 22.17): √ 2P L (22.65) d= Ko (a + L) 𝜋 𝛾 where K o is the at-rest earth pressure coefficient. 22.7 EARTH PRESSURES IN SHRINK-SWELL SOILS When the backfill of a bottom-up wall or the soil behind a topdown wall has a high plasticity index (Ip) or a high swell index (Is), it is necessary to consider the soil shrink-swell behavior in calculating the pressure diagram. Hong et al. (2010) studied this issue and made the following recommendation (Figure 22.18). Water content Passive pressure

Roller Roller

Swell pressure At-rest pressure

Figure 22.16 Compaction earth pressure during backfilling.

Figure 22.18 Wall pressure diagram including swelling pressure. (Source: Adapted from Hong et al., 2010.)

22.9 GRAVITY WALLS

Three diagrams come into play in the resultant pressure diagram (Figure 22.18): the passive earth pressure diagram, the swell pressure diagram, and the at-rest earth pressure diagram. Both the passive and at-rest diagrams increase with depth according to K p and K o respectively. The swell pressure diagram, however, typically decreases with depth because the overburden pressure increases with depth and limits the swell pressure.The pressure diagram starts at a slope equal to the passive earth pressure coefficient. Although the swell pressure is higher than the passive pressure within that zone, the soil fails in shear before it can reach the swell pressure. When the passive pressure profile reaches the swell pressure profile, the swell pressure limits the earth pressure against the wall and the pressure diagram follows the swell pressure profile. When the swell pressure profile reaches the at-rest pressure profile, the at-rest pressure is maintained against the wall because the swell pressure is smaller than that. As a result, the pressure diagram switches to the at-rest pressure profile. The coefficients K p and K o have been discussed in previous sections. The swell pressure profile can be obtained by performing swell tests on samples from the retained soil. Note that the swell pressure decreases as movement is allowed.

22.8

DISPLACEMENTS

1.0 0.8 0.6 0.4 0.2 0 –0.03

Table 22.1 Possible range of displacement to generate active and passive earth pressures Soil type

Active, ya /H

Passive, yp /H

Loose sand Dense sand Soft clay Stiff clay

0.003–0.005 0.001–0.002 0.01–0.02 0.005–0.01

0.03–0.05 0.01–0.03 0.03–0.05 0.01–0.03

wall is high, it will take more movement to mobilize the earth pressure than if the wall is low. The argument in favor of this concept is that if the wall is high, the earth pressure wedge will be large and it will take more movement to completely fail the wedge of soil behind a high wall compared to a low wall. While it is clear that earth pressures depend on movement, and while it is also clear that predicting movements is important, our ability to make such predictions is not as good as our ability to calculate foundation settlement. Often the design of walls takes place solely on the basis of earth pressure distributions (ultimate limit state) rather than a combination of earth pressures and movements. Nevertheless, the trend in practice is toward increased inclusion of movement calculations in retaining wall design. Because the intact mass is the one deforming during such earth pressure problems, and because the overall strain level is quite small for well-designed systems, small strain moduli are most useful and can be obtained from cross hole sonic tests.

22.9

–0.02

–0.01

0

Deflection at top of the wall, utop/H

0.01

GRAVITY WALLS

Gravity walls are bottom-up walls usually made of reinforced concrete (Figure 22.20). In the early days they were heavy, massive blocks (concrete gravity wall), but such systems were soon replaced by walls that use less concrete weight and more backfill weight as dead weight to resist the soil push (cantilever gravity walls). In cantilever gravity walls, the slab under the retained portion of the backfill is subjected to the Earth pressure coefficient K

Earth pressure coefficient K

Figure 22.4 showed the general form of the earth pressure 𝜎 h or p vs. displacement y curve. This curve, sometimes called a P-y curve, represents the plane strain behavior of the wall at a depth z. Figure 22.19 shows some values coming from measurement and numerical simulations (Briaud and Kim, 1998). The vertical axis is a generalized earth pressure coefficient K, which is discussed further in Section 22.12, and the horizontal axis is the horizontal displacement normalized by the height of the wall. The amount of movement necessary to generate the active earth pressure 𝜎 ah is ya and the amount of movement necessary to generate the passive earth pressure 𝜎 ph is yp . Table 22.1 shows some possible values of ya /H and yp /H (H is the height of the wall) for different soil types. This means that if the

753

1.0 0.8 0.6 0.4 0.2 0 –0.015

–0.01

–0.005

0

0.005

Mean wall deflection, umean/H

Figure 22.19 Measured earth pressure coefficient versus normalized displacement of a wall. (Source: Adapted from Briaud and Kim, 1998.)

0.01

754

22 RETAINING WALLS

Concrete gravity

Cantilever gravity

Figure 22.20 Types of gravity walls.

backfill weight, which increases the sliding resistance and the resistance to overturning. Cantilever gravity walls have to be heavily reinforced, as a high bending moment develops at the connection between the slab and the stem. The word cantilever is also used for a type of top-down wall; this is why the word gravity is added to cantilever to designate the wall shown in Figure 22.20. The geotechnical design of gravity walls consists of a number of steps aimed at ensuring the safety (low probability of failure) and functionality (low probability of intolerable movements) of the wall. The purpose of the design is to satisfy the ultimate limit state and the serviceability limit state of the wall as it is subjected to the earth pressures behind the wall and in front of the wall. For gravity walls, however, the serviceability limit state is rarely addressed, as movements are difficult to estimate and often small. The design steps include estimating the pressure distribution behind the wall (active pressure), the pressure distribution in front of the wall (passive pressure), the resultant active force and its location, the resultant passive force and its location, the sliding ultimate limit state, the overturning ultimate limit state, the bearing capacity ultimate limit state, the slope stability ultimate limit state, and the settlement serviceability limit state (rare). Figure 22.21 identifies the possible failure modes for a gravity wall. 1. Active pressure behind the wall 𝜎 ah . For this, the steps in Section 22.3.7 are followed and the profile of total active earth pressure is prepared. Special earth pressure conditions, such as compaction stresses,

stresses due to shrink-swell soils, and stresses due to surface loading, are considered in arriving at the design active pressure diagram. 2. Passive pressure in front of the wall 𝜎 ph . This refers to any embedded portion of the wall that could generate a passive resistance. Here again, the steps of Section 22.3.7 are followed and the profile of passive earth pressure is prepared. 3. The resultant active push Pa (kN/m) is calculated as the area under the active earth pressure diagram (Figure 22.22): z=H+D

Pa =

∫z=0

𝜎ah

n ∑ dz = Aai

(22.66)

i=1

where 𝜎 ah is the horizontal active earth pressure at depth z below the top of the wall, H is the height of the wall, D is the embedded depth, and Aa is the area under the active earth pressure diagram. If there is more than one soil layer, Pa is given by the sum of the areas Aai (corresponding to layer i) under the active pressure diagram. 4. The resultant passive push Pp (kN/m) is calculated as the area under the passive earth pressure diagram (Figure 22.22): z=D

Pp =

∫z=0

𝜎ph dz =

m ∑ Api

(22.67)

i=1

where 𝜎 ph is the horizontal passive earth pressure at depth z below the bottom of the wall, H is the height of the wall, D is the embedded depth, and Ap is the area under the passive earth pressure diagram. If there is more than one soil layer, Pp is given by the sum of the areas Api (corresponding to layer i) under the passive earth pressure diagram. 5. The point of application of Pa is found by writing that the moment around a chosen point (often the bottom of the wall, O in Figure 22.22) created by the active B Z

Overturning

W

σahdz

H Xw

Slope Sliding

Z

Bearing capacity

Figure 22.21 Failure modes.

Pa Xa

D

Pp Xp M

O

Figure 22.22 Forces acting on a gravity wall.

22.9 GRAVITY WALLS

earth pressure diagram is the same as the moment created by the resultant Pa : z=H+D

Pa xa =

∫z=0

𝜎ah (H + D − z)dz =

n ∑ Aai aai i=1

(22.68) where xa is the moment arm of Pa , and aai is the moment arm of the individual areas under the pressure diagram corresponding to Aai . Of course, if the active earth pressure diagram is a simple triangle, then xa is 0.33 (H + D). 6. The point of application of Pp is found by writing that the moment around a chosen point (often the bottom of the wall, O in Figure 22.22) created by the passive earth pressure diagram is the same as the moment created by the resultant Pp : z=D

Pp xp =

∫z=0

n ∑ 𝜎ph (D − z)dz = Api api

(22.69)

i=1

where xp is the moment arm of Pp , and api is the moment arm of the individual areas under the pressure diagram corresponding to Api . Of course, if the passive earth pressure diagram is a simple triangle, then xp is 0.33D. 7. Sliding ultimate limit state is checked by evaluating the following equation: 𝛾1 Pa1 + 𝛾2 Pa2 ≤ 𝜑1 W tan 𝛿 + 𝜑2 Pp

(22.70)

where 𝛾 1 is the load factor associated with the active push Pa1 due to soil weight, 𝛾 2 is the load factor associated with the active push Pa2 due to surcharge, 𝜑1 is the resistance factor for the resistance to sliding due to soil weight, 𝜑2 is the resistance factor for the resistance to sliding due to the passive earth pressure in front of the wall, and 𝛿 is the friction angle for the interface between the bottom of the wall and the soil on which it rests. The angle 𝛿 is usually taken as equal to the friction angle 𝜑′ of the soil for rough interfaces. The load factor 𝛾 1 is typically taken as 1.5, and 𝛾 2 as 1.75. The resistance factor for the sliding resistance due to the weight of the wall is in the range of 0.8 to 0.9, whereas the resistance factor for the sliding resistance due to the passive earth pressure is usually around 0.5. 8. Overturning ultimate limit state is checked by evaluating the following equation related to the moment around the front of the wall (point M in Figure 22.22): 𝛾1 Pa1 xa1 + 𝛾2 Pa2 xa2 ≤ 𝜑1 W xw + 𝜑2 Pp xp (22.71) where 𝛾 a1 , 𝛾 a2 , Pa1 , and Pa2 are as defined in step 7; xa1 , xa2 , and xp are the moment arms of the forces Pa1 , Pa2 and Pp respectively; 𝜑1 and 𝜑2 are the same resistance factors as in step 7; W is the weight of the wall, and xw is the corresponding moment arm. The values

755

of the load and resistance factors for this ultimate limit state are the same as the values for step 7. 9. Bearing capacity ultimate limit state is checked as a shallow foundation subjected to the combination of W, Pa , and Pp (see Chapter 18, Section 18.4). This combination leads to the case of an inclined, eccentric load. 10. Slope ultimate limit state is checked in the same way as a slope with a wall loading the soil surface (see Chapter 20). The load and resistance factors were presented in Chapter 20. Section 20.2. 11. Serviceability limit state is usually not addressed in current practice. The following comments may be made on the movement of gravity walls. For gravity walls founded on competent soil, the movement usually takes place by rotation around the bottom of the wall (point O in Figure 22.22). Most of the horizontal movement tends to occur during construction and corresponds to the order of magnitude given in Table 22.1 for the active case. Note that the main source of horizontal movement comes from rotation of the base under the overturning moment. Indeed, if the sliding ultimate limit state is satisfied, the sliding movement should be very small. Vertical settlement of the wall will occur if the downdrag from the backfill and the high stresses under the front edge of the wall due to the applied moment compress the soil under the wall. Because this compression is uneven, with more settlement under the front edge, the wall will rotate with more horizontal movement at the top. To this extent, the settlement factors giving the settlement at the center and at the edge of the foundation (see Chapter 18, Section 18.7 on the load settlement curve approach) can be used to infer the rotation and movement of the wall. In that respect it is useful to study the case of a foundation subjected to a line load Q (kN/m) and an overturning moment M (kN.m/m) (Figure 22.23). The eccentricity e of the load Q is given by: M (22.72) e= Q The pressure diagram under the foundation is shown in Figure 22.23. The maximum pressure pmax and minimum pressure pmin under the foundation with a width B are given by: ) Q( 6e 1+ (22.73) Pmax = B B ) Q( 6e Pmin = 1− (22.74) B B Equation (22.74) indicates that pmin becomes zero when the eccentricity becomes equal to B/6. If pmin becomes zero, the instability of the wall is more likely, as the foundation cannot develop tensile resistance to overcome further increase in eccentricity. As long as the point of application of the resultant stays within a distance of B/6 from the axis of symmetry, the wall is more likely to be stable and experience limited movement. This is called the rule of the middle third as e can

756

22 RETAINING WALLS

Δ

Mass movement W

22.10 MECHANICALLY STABILIZED EARTH WALLS

Fd

Pa

Fs S

O

Pmin

Pmax

be ± B/6. Note that the wall can also move in the other direction (more horizontal movement at the bottom of the wall) if a slope stability problem exists.

e R

Figure 22.23 Pressure under a gravity wall.

A

B

D

C

Mechanically stabilized earth (MSE) walls are bottom-up walls made mostly of soil with some reinforcement. Henri Vidal, a French engineer, is credited with inventing reinforced earth in 1957. This technology is to geotechnical engineering what reinforced concrete is to structural engineering. It consists of placing inclusions in the soil to give it significant tensile strength. These walls were called reinforced earth walls in the beginning and are now called mechanically stabilized earth walls (MSE walls). An MSE wall is built by placing a layer of soil (say, 0.7 m thick), then a layer of reinforcement, then a layer of soil, then a layer of reinforcement, and so on until the desired wall height is reached (Figure 22.24). Panels are placed in front of and attached to the reinforcement for esthetic purposes and to retain any

Figure 22.24 MSE wall. (Source: Courtesy of The Reinforced Earth Company.)

757

22.10 MECHANICALLY STABILIZED EARTH WALLS

Cost of wall ($/m2)

800

Gr

600

0.3H1

l

L max

MSE (Metal)

400

s n wall

in

Crib/B 200

0

al yw avit

H1

L max

tics)

osynthe

e MSE (G

L max

3

5

7

9

11

H1 2

45 +

ϕʹ 2

FLEXIBLE

13

Figure 22.25 Cost of bottom-up walls. (Source: Adapted from Koerner and Soong, 2001.)

soil that might fall between reinforcement layers close to the front. The pressure on the panels is very small, as most of the earth pressure is taken up by the reinforcement. The reinforcement can be galvanized steel strips, steel grids, or geosynthetics. The success of MSE walls is due to their lower cost compared to cantilever gravity walls, particularly for very high walls (Figure 22.25). Indeed, MSE walls built to 50 meters in height have performed very well. The design of MSE walls includes an external stability design and an internal stability design. External Stability

For this case, the MSE wall is considered to be a gravity wall consisting of the front panels, the reinforcement, and the soil between the reinforcement. This reinforced soil mass (ABCD in Figure 22.24) is the gravity wall and has to satisfy the design criteria of a gravity wall outlined in Section 22.9. These include the sliding ultimate limit state, the overturning ultimate limit state, the bearing capacity ultimate limit state, the slope stability ultimate limit state, and the settlement serviceability limit state (rare). The design steps are identical to the steps detailed in Section 22.9. 22.10.2

L max

RIGID 0

Height of wall (m)

22.10.1

T max

H1 2

Internal Stability

Pull-out capacity and yield of the reinforcement are the two aspects of internal stability of an MSE wall. Let’s address pull-out capacity first. Pull-Out Design This design consideration ensures that the load in the reinforcement will not be high enough to pull the reinforcement out of the soil. An understanding of the load distribution in the reinforcement is necessary. Figure 22.26 shows the variation of the tension load T (kN) in the reinforcement as a function of the distance from the front of the wall. At the wall facing, the load T in the reinforcement is very small, and then it increases as the instability of the wedge of soil near the wall is transferred to the tension T (kN) in the reinforcement. At a distance Lmax from the front, the tension T

Figure 22.26 Load in the reinforcement.

reaches a maximum T max . Beyond T max , the tension decreases as the load is transferred to the stable soil mass and reaches zero at a certain distance from the front. The true embedment or anchoring length La available to resist the active pressure force against the wall is L – Lmax where L is the total length of the reinforcement. The design requires knowledge of Lmax , which is to be ignored in the length required to resist T max . Lmax is given in Figure 22.26; as can be seen for rigid inclusions, it is equal to 0.3 H in the top half of the wall and decreases to zero at the bottom of the wall. For flexible inclusions (geosynthetics), it is taken as the width of the active wedge. These recommendations are partially based on measurement and simulation data. The force T max is calculated as follows: Tmax = sv sh 𝜎h

(22.75)

where T max is the maximum line load (kN) to be resisted by the layer of reinforcement at depth z, sv is the vertical spacing between reinforcement layers at depth z, sh is the horizontal spacing between reinforcement inclusions at depth z, and 𝜎 h is the total horizontal stress at depth z. The stress 𝜎 h is calculated as: 𝜎h = kr 𝜎ov + Δ𝜎h (22.76) where kr is a coefficient of earth pressure defined in Figure 22.27 as a function of K a, 𝜎 ov is the vertical total stress at rest at depth z, and 𝛥𝜎 h is the increase in horizontal stress at depth z due to surface loading. The reason that kr is higher than K a for rigid inclusions is that during compaction of the backfill, the rigid inclusions (e.g., steel strips) can lock in higher horizontal stresses. Flexible inclusions (e.g., geosynthetics) do not lock in additional compaction stresses. As a result, kr is equal to K a for flexible inclusions. Now that we have calculated the load T max , we need to find the length of reinforcement that will safely carry this load without pulling out of the soil. The pull-out capacity T pull-out (kN) of the reinforcement inclusion is given by: Tpull out = 2fmax bLa

(22.77)

where fmax is the maximum shear stress that can be developed on both sides of the interface between the reinforcement and

22 RETAINING WALLS

2.5

the soil, b is the width of the inclusion, and La is the anchoring length beyond Lmax , the width of the active failure zone. The shear stress fmax is evaluated as follows: (22.78)

where F* is the friction factor given in Figure 22.28; 𝜎v′ is the vertical effective stress on the reinforcement; and 𝛼 is a scale factor taken as 1 for steel reinforcement, 0.8 for geogrids, and 0.6 for geotextiles. Note that although the recommended F* values are limited to 2 at the ground surface, values of F* much higher than 2 have been measured. The reason the coefficient of friction may be higher than 1 is that a combination of friction and bearing capacity is involved in the sliding-out of the reinforcement. The bearing capacity component comes from the protruding ribs for strips and from the transverse bars for grids. Then the ultimate limit state for pull-out must be satisfied: 𝛾1 Tmax1 + 𝛾2 Tmax2 = 𝜙Tpull out

(22.79)

where 𝛾 1 is the load factor for the active earth pressure due to soil weight (𝛾1 = 1.35), 𝛾 2 is the load factor for the active earth pressure due to any surcharge on top of the wall (𝛾2 = 1.50), 𝜑 is the resistance factor (𝜑 = 0.9), Tmax1 is the part of the load in the reinforcement due to the soil weight, T max2 is the part of the load in the reinforcement due to any surcharge on top of the wall, and T pull-out is the pull-out resistance calculated in Eq. (22.77). The required safe length La of the reinforcement is given by: ( ) ′ 𝛾1 kr 𝜎ov + 𝛾2 Δ𝜎h sv sh (22.80) La = ′ 2𝜑F∗ 𝜎ov 𝛼b The total length L of the reinforcement is largest at the top of the wall, but because it is common practice to keep the

0

1

t D

LO St St

Pullout resistance factor F* 2 3 4 5 6

7

8

0 2 Depth of fill (m)

fmax = F ∗ 𝜎v′ 𝛼

Ribbed steel strips F* = TANΦf

6000 mm Geotextiles F* = 0.67 tanΦf Geogrids F* = 0.67 tanΦf

Figure 22.27 Earth pressure coefficient for load in the reinforcement.

Grid-bearing member

Smooth steel strips F* = 0.4

1 1.2

*Does not apply to polymer strip reinforcement

Not to scale

Steel grids F* = 10 (t/St)

6m

1.2 + logCu < 2.0

20 (t/St)

0

Depth below top of wall Z or Z0 (mm)

Depth below top of wall, Z

0

AASHTO recommendations Default values for pullout friction factor, F*

Meta l Strip s Me W ta l B eld a ed r M W a ts ire & Gr ids

0

Coefficient of lateral stress ratio = kr/ka 1.7 1 1.2

*Geosynthetics

758

4 6 8

95% Predictive Lower Bound AASHTO (2010) F* for Ribbed Strips

10

Ribbed Strips - TxDOT 0-6493 Data Ribbed Stips - RECo Data

12 14

Ribbed Strips - RECo Data - Rupture (12.5 mm, 3000 Ohm.cm 0

p po yp

z>0

p = q-ky

p 0), the magnitude of the P value increases from the Po value corresponding to the at-rest earth pressure to the Pp value corresponding to the passive earth pressure. Because the soil pushes in a direction opposite to the chosen positive direction, P is negative. The net P-y curve for the zone below the excavation level is constructed by combining the two curves for Side 1 and Side 2 (Figure 22.35). Note that this P-y curve preparation is done for each node along the discretized wall. Then a finite difference program or spreadsheet is used, and the solution gives the following parameters as a function of depth: wall deflection y(z), slope of the wall y′ (z), bending moment in the wall M(z), shear force in the wall V(z), and pressure on the wall p(z). Sample outputs are shown in Figures 22.36 and 22.37. The deflection profile predicted by this method tends to underpredict

762

22 RETAINING WALLS

P (kPa)

Pressure (kPa) Depth (m) –100 –50

500

Moment (kN.m) 50

100

Deflection (mm) 4

2

0

4

2

0

400

3

Combined p-y curve

300

6 200 Side 2 p-y curve

20

Side 1 p-y curve

9

100

10

12

10

20

Y (mm)

15 18

–100

–100 –50

–200

Figure 22.35 Combining p-y curves below the excavation level. Depth (m)

p (kPa)

0

y (mm) 60

–60

Deflected shape of wall 50 60

y (mm) –60

60 50

9.5

60

–60 12.8

15

17.4

21

50 –60

60 50

–60

60 50

–60

60

y (mm)

y (mm) y (mm)

y (mm)

50 –60

Figure 22.37 Pressure, bending moment, and deflection. (Source: Adapted from Briaud et al., 1983).

1998). For improved prediction of deflections including mass movement, the finite element method should be used; nevertheless, a problem remains concerning the quality of the input parameters and the selection of the soil model.

22.12 ANCHORED WALLS AND STRUTTED WALLS

50 6.7

100 200

y (mm)

3.3 –60

50

60

y (mm)

Figure 22.36 P-y curves and deflected shape of cantilever wall. (Source: Adapted from Briaud et al., 1983).

the deflections observed in practice. The reason is that the mass movement of the retained soil is not included in the P-y curve in a theoretically sound manner. However, the bending moment profile predicted by this method and the maximum bending moment for design are much more consistently reliable than any hand calculation based on an assumed pressure distribution, such as shown in Figure 22.32 (Briaud and Kim,

Anchored walls (or tieback walls) and strutted walls are top-down walls (Figure 22.38). The wall portion may be a solid concrete wall built by the slurry wall method, a sheet pile wall, a bored pile wall, a deep soil mixing wall, or a soldier pile and lagging wall, to name but a few. Concrete slurry walls are built by excavating the soil one rectangular panel at a time with a clamshell rig and under slurry if necessary, lowering the reinforcing cage into the slurry-filled hole, and placing the concrete in liquid paste through a tremie (tube that goes to the bottom of the hole) from the bottom of the panel to the top while displacing the slurry out of the rectangular hole (Figure 22.38a). Bored pile walls and sheet pile walls were discussed in Section 22.11. In this section they are anchored to be able to retain a larger depth of soil. Deep soil mixing walls are like bored pile walls except that the bored piles are drilled by mixing the soil with about 20% cement; the resulting piles are not as strong, but they are less expensive. Soldier pile and lagging walls are constructed by driving or drilling piles in line on a 2–3 meter spacing and excavating in front of this line of piles while placing wood lagging to retain the soil between piles. Anchored and strutted walls are very convenient in tight settings like urban areas because they do not require much space for construction. On the one hand, struts clutter the excavation; on the other hand, anchors may hit underground utility lines. The design of anchored walls and strutted walls includes many parts, with the main ones being estimating the pressure

22.12 ANCHORED WALLS AND STRUTTED WALLS

(a) Slurry wall digging the wall

(d) Deep soil mixing wall

(b) Slurry wall lowering cage

(e) Soldier pile and lagging wall

763

(c) Slurry wall final with anchors

(f) Strutted wall

(g)

Figure 22.38 Various anchored and strutted wall techniques. (Source: a, b, c, d: Courtesy of Nicholson Construction; e, f, g: Courtesy of Schnabel Foundation Company.)

distribution behind the wall, calculating the anchor or strut loads, calculating the maximum bending moment in the wall, estimating the horizontal and vertical movements, and calculating the necessary length of anchors. 22.12.1

Pressure Distribution

Consider the pressure distribution behind a cantilever top-down retaining wall with the active pressure on the retained side (see Figure 22.32). If you install an anchor

within the excavated depth to hold the wall back, and if you stress that anchor in tension, the anchor head (plate) is going to press against the wall while you pull on the tendon, thereby increasing the local pressure (Figure 22.39). As a result, the pressure behind the anchor will be higher than the active pressure and will correspond to the prestressing load of the anchor or the strut. It is very common to stress all the anchors to the same load, so that the pressure behind the wall in the retained soil depth (above excavation level) is nearly constant and equal to

764

22 RETAINING WALLS

coefficient 0.2 would correspond to less fissured clays and 0.4 to more fissured clays. Also, if the fissures are large enough that water will exert pressure on the wall, the water pressure must be added.

F

A

G

D

F1

22.12.2 H

E

K L

I

Briaud and Kim (1998) collected a number of full-scale case histories on anchored walls and performed numerical simulations. For the case histories, the anchor loads were known, as were the horizontal deflections of the wall. The mean pressure 𝜎 h behind the wall was calculated as the ratio of the sum of the individual anchor loads F i divided by the total wall area A of soil retained by the anchors:

F2

B

J

n ∑

C

Figure 22.39 Influence of anchor stressing on pressure diagram. Sands

Soft to medium clays

Stiff fissured clays

0.25H H

σʹov

0.25H 0.50H

0.75H

0.25H 0.65Kaσʹov

uw

γH-4mSu

Pressure vs. Movement

0.2γH to 0.4γH

Figure 22.40 Pressure distribution for strutted walls. (Source: Adapted from Terzaghi et al., 1996.)

the sum of the anchor loads divided by the retained soil area. This is what led Terzaghi et al. (1996) to recommend a constant pressure diagram for strutted walls. Based on full-scale measurements, they recommended pressure diagrams for sand, for soft to medium clays, and for stiff fissured clays (Figure 22.40). The maximum total pressure 𝜎 h is as follows: For sands For soft to medium clays

′ 𝜎h = 0.65 K0 𝜎ov + uw 𝜎h = 𝛾H − 4 m su

For stiff fissured clay

𝜎h = 0.2𝛾H to 0.4𝛾H + uw (?)

(22.94) (22.95) (22.96)

′ is the where K a is the coefficient of earth pressure at rest, 𝜎ov effective vertical stress on the retained soil (sand) side at the bottom of the excavation, 𝛾 is the total unit weight of the clay, H is the height of the excavation, su is the clay undrained shear strength, and m is a parameter that depends on the depth of the soft to medium clay layer below the excavation. It is taken equal to 1 if the soft to medium clay layer stops at the bottom of the excavation, and equal to 0.4 if the clay layer goes much deeper than the bottom of the excavation. Note that, for sand, the analysis is an effective stress analysis and the water pressure must be added if water is present. For soft to medium clay, the analysis is an undrained analysis and the water pressure is included in 𝛾H. For stiff fissured clays, the

𝜎h =

Fi

i=1

A

(22.97)

The mean pressure 𝜎 h behind the wall was associated with the horizontal movement at the top of the wall utop and the mean horizontal deflection umean . Note that one case history led to more than one combination of pressure and displacement, as the construction sequence included several excavation levels and several anchor installations. The earth pressure coefficient K was calculated as the ratio of the mean pressure 𝜎 h over the vertical effective stress behind the wall at the bottom of the excavation: 𝜎h (22.98) K= ′ 𝜎ov (at z = H) Figure 22.41 shows the range of values of K versus utop ∕H and K versus umean ∕H. Terzaghi and Peck’s earth pressure value of 0.65 Ka 𝛾H for strutted excavations in sand leads to a K value of 0.21 if the friction angle is 30∘ (Ka = 0.33). For a K value of 0.21, Figure 22.41 gives a range of utop ∕H from 0.002 to 0.0045 and umean ∕H from 0.0015 to 0.0035. However, in the case of anchored walls, the engineer can choose the wall deflection by choosing the anchor loads. Indeed, if the anchor loads are very high, the wall could actually move back and go into passive resistance. In contrast, if the anchor loads are very low, there will be a lot of wall deflection toward the excavation. Figure 22.41 helps the engineer to select a K factor that will generate a targeted amount of wall movement. It appears that a K value of about 0.4 will lead to minimal displacements. For a given wall height H and for a chosen horizontal displacement utop or umean , the total earth pressure 𝜎 h at depth z is calculated according to: ′ 𝜎h = K𝜎ov (at z = H) + uw

(22.99)

where K is read on Figure 22.41 at the corresponding relative ′ displacement, 𝜎ov is the vertical effective stress at the bottom of the wall, and uw is the water pressure at depth z. Note that ′ the term K𝜎ov is a constant independent of depth, whereas uw increases with depth (Figure 22.42).

1.0

Earth pressure coefficient K

Earth pressure coefficient K

22.12 ANCHORED WALLS AND STRUTTED WALLS

0.8 0.6 0.4 0.2 0 –0.03

–0.02

–0.01

0

0.01

765

1.0 0.8 0.6 0.4 0.2 0 –0.015

–0.01

Deflection at top of the wall, utop/H

–0.005

0

0.005

0.01

Mean wall deflection, umean/H

Figure 22.41 Measured earth pressure coefficient versus normalized displacement of the wall. (Source: Adapted from Briaud and Kim, 1998.) A

22.12.4

J h1

Anchor 1 H

Anchor 2 Anchor 3 uw Kσʹov at z = H

D

K Chosen for target utop

B C D E F G H

h2/2 h2/2 h3/2 h3/2 h4/2 S

K

h3 M

h4

O N

R Q σph at z = D

The general shape of the deformed wall and adjacent ground surface has two components (Figure 22.44): a cantilever movement and a movement associated with deep deformations. The first one is associated with the lack of lateral support leading to the soil mass leaning into the excavation, and the second with the slope stability/bearing capacity type of deformation deeper in the soil mass. Predicting these displacements is not simple. Peck (1969) collected data on the movement of the ground surface near excavations and presented it in a very useful fashion (Figure 22.45). Peck divided the behavior according to soil type and to the value of the factor of safety F against base instability. He showed that the maximum settlement of the top of the wall can reach 0.01 H for excavations in sand or in soft to hard clay, it can reach 0.02 H for soft clays when F is larger than 1.3, and it can be larger than 0.02 H for soft clays when F is less than 1.3. Regarding the lateral extent over which the ground surface would be depressed, Peck found that it could be up to 2 H for excavations in sand or in soft to hard clay, it could be up to 4 H for soft clays when F is larger than 1.3, and larger than 4 H for soft clays when F is less than 1.3. Clough and O’Rourke (1990) collected additional data and revised Peck’s plots accordingly (Figure 22.46). In their work, Clough and O’Rourke also proposed a method to predict the maximum lateral movement of the wall depending on the

h2 L

P I σah at z = H + D

Figure 22.42 Pressure diagram for anchored walls.

22.12.3

Movement of Wall and Ground Surface

Base Instability

In the case of clays, one concern is an inverted bearing capacity failure. In the case of sands, the concern is a loss of effective stress and the development of a quick condition at the bottom of the excavation due to seepage. In clays, the bottom of the excavation may be unstable if the soil is not strong enough to sustain the lack of overburden on the excavated side. The factor of safety F against base instability is (Figure 22.43): Nc su F= (22.100) 𝜎ov (at z = H) where N c is a bearing capacity factor for a strip footing (see Figure 18.7), su is the undrained shear strength, and 𝜎ov(at z=H) is the vertical total stress behind the bottom of the wall.

F=

γT, su H

σz = H

B

PH

Ncsu γT H + q N s or c u σz = H

H1

Stability factor, Nc

10 Square or circle B/L = 1

9 8

0.5

7 0 = Infinite strip

6 5 4

0

1

2

3

Dimension, H/B

Figure 22.43 Base instability.

4

5

766

22 RETAINING WALLS

Horizontal displacement

Vertical displacement

(a) Cantilever movement curve 1

(b) Deep inward movement curve 2

(c) Cumulative movement curve 3

Figure 22.44 Components of excavation movements.

Settlement (%) Depth of excavation

0

0

1

2

3

relative stiffness L of the wall and the factor of safety F against base instability (Figure 22.47). The relative stiffness L(m) is defined as: EI L= 4 (22.101) 𝛾h

4

I II

1

III 2

3 Distance from excavation Depth of excavation Zone I

Sand and soft-to-hard clay

Zone II

Very soft to soft clay when the depth is limited, or when F > 1.3

Zone III

Very soft to soft clay when F < 1.3

Figure 22.45 Peck diagram for ground surface settlement. (Source: Adapted from Peck, 1969.)

where E is the modulus of elasticity of the wall material, I is its moment of inertia around the bending moment axis, 𝛾 is the unit weight of the soil, and h is the average vertical distance between anchors. For a given case, the relative stiffness L and the factor of safety for base instability are defined, the correct curve on Figure 22.47 is selected, and the corresponding ratio between the maximum horizontal deflection and the excavation height is read on the vertical axis. In general, the vertical and horizontal displacements of excavations are on the same order of magnitude unless the soil is very dilatant or collapsible. Some of the ways to decrease movements are to place the first anchor as shallow and as early as reasonably possible and to use high anchor loads (K = 0.4 in Figure 22.41).

Sheetpile walls, havg = 3.5 m

Distance from excavation Excavation depth 0

1

2

1 m-thick diaphragm walls, havg = 3.5 m

3

havg = (h1 + h2 + h3)/3

Settlement (%) Excavation depth

Sand

1

2

Soft to medium clay f* = 2

Stiff to very hard clay

Soft to medium clay f* = 1.2

Max. Lateral wall movement/ excavation depth (%)

0

2.5

h1 h2 h3

2.0 1.5

0.9 1.0

Base stability factor of safety

1.0 0.5

1.1 1.4

2.0 3.0

*F = factor of safety 5.1Su against base instability = σν (z = H)

Figure 22.46 Clough and O’Rourke’s diagram for ground surface settlement. (Source: Adapted from Clough and O’Rourke, 1990.)

0 10

20

40

100

200

400

1000

Relative stiffness (EI)/(γh4avg)

Figure 22.47 Clough and O’Rourke’s chart for maximum lateral movement. (Source: Adapted from Clough and O’Rourke, 1990.)

767

22.12 ANCHORED WALLS AND STRUTTED WALLS

Anchors

Ground surface

Lt = Lu + Lb = Ld + La

Active wedge

Sheath

Lock-off load

Tendon

Unb o leng nded th (L u)

Grout

Tend on b on leng th (L d b) Anc ho leng r bond th (L a)

Tota l

anch

or le

ngth

4.6 m

, (L

t)

Figure 22.48 Anchor or tieback.

9.2 m

Load (kN)

Distance (m) 1000

0

2

4

6

8

10

Load resisted by soil QS1 = 169 kN 0 1000

A 0

B 2

QS3 = 780 kN

QS2 = 260 kN

a

C 4

6

13.8

12

8

10

D 13.8

12

Load in the grout TENS.

QQ1 = 149 kN

QQ3

0 COMP. QQ2 = 520 kN 1000

(22.102)

The length Ld is taken as the length of the anchor within the active wedge behind the wall (Figure 22.48). An example of load distribution in the anchor under tension is shown in Figure 22.49. A long unbonded length is best for anchors in tension because it maximizes the length of grout in compression (Briaud et al., 1998). The design of anchors or tiebacks has two parts: calculating the anchor loads and calculating the required anchor capacity and associated length. The anchor load is determined by using the tributary method. Once the pressure diagram is obtained (Figure 22.42), the horizontal component Ahi of the load in anchor i is obtained by using the part of the pressure diagram D anch iscoun te or le ngth d (L ) d

780 kN

Load (kN)

Anchors can be constructed in different ways, but the most common way (Figure 22.48) is to drill a hole through the wall when the anchor depth is reached, insert a rod or multiple-strand cable in the open hole with centralizers, fill the hole with grout, wait for the grout to set, then tension the anchor, subject it to a proof test, and then lock the anchor at the design load. Sometimes a second injection of grout is performed through tubes left in place during the first injection to increase the anchor capacity. The rod or strand is in a bond-breaking sheath from the anchor head to a certain distance called the tendon unbonded length Lu . The sheath stops at Lu ; the rest of the rod or strand is barren and is called the tendon bond anchor length Lb . The length of the anchor in the active wedge is called the discounted anchor length Ld . The rest of the anchor is called the anchor bond length La . The total length of the anchor is Lt :

0.3 m

Load (kN)

22.12.5

1000

A 0

B 2

b

C 4

6

8

10

12

QT4 = 780 kN

TENS.

QT2 = 169 kN 0 A

Load in the steel tendon QT3 = 260 kN

QT1 = 20 kN B C

D 13.8

QT6

c

D

Figure 22.49 Example of load distribution in an anchor. (Source: Adapted from Briaud et al., 1998.)

tributary to anchor i. For example, the tributary area of the pressure diagram in Figure 22.42 for anchor 2 is CKLE. The expression is: ( ) hi hi+1 (22.103) + sh Ahi = Pi + 2 2 where Ahi is the horizontal component of the anchor load Ai , pi is the mean pressure behind the wall within the tributary depth, hi is the anchor spacing above anchor i, hi+1 is the anchor spacing below anchor i, and sh is the anchor spacing in the horizontal direction. Equation (22.103) applies to all anchor loads except the top anchor, where it becomes: ( ) h2 (22.104) sh Ah1 = p1 + h1 + 2 Often the anchor is not horizontal, but rather inclined at an angle 𝛼 to the horizontal (15∘ to 30∘ ). Thus, the anchor load Ai is (Figure 22.50): Ahi (22.105) cos 𝛼 Once the anchor load is determined, the anchor resistance and length can be calculated. The LRFD equation gives: Ai =

𝛾A1 = 𝜑R1

(22.106)

where 𝛾 is the load factor (𝛾 = 1.35), Ai is the anchor load, 𝜑 is the resistance factor, and Ri is the ultimate resistance of the

768

22 RETAINING WALLS

Table 22.3 gravel

Ground surface

Wall

Values of 𝛼 s anchorage factor for sand and

Soil type

Ahi α

Loose

Relative density Medium

Dense

Ai

anchor. If anchors are not proof tested, then 𝜑 is between 0.35 and 0.45. However, all anchors are usually proof tested and therefore there is little uncertainty as to the anchor capacity; in that case a resistance factor close to 1 can be used. Once Ri is obtained, the length of anchor necessary to obtain Ri is calculated. The design is very similar to the case of a pile in tension, and Ri is given by: Ri = 𝜋 DLa fmax = Fmax La

(22.107)

where D is the diameter of the anchor, La is the anchor bond length, fmax is the shear strength of the soil-grout interface, and F max is the maximum load that can be resisted per unit length of anchor. The parameter fmax is estimated as follows for various soils: For sand and gravel fmax =

′ 𝛼s 𝜎ov

with 𝛼s from Table 22.3 (22.108)

0.1 0.2 0.5 1.0

0.4 0.6 1.2 2.0

1.0 1.5 2.0 3.0

(Source: Adapted from Canadian Geotechnical Society, 2007.)

1 Kulhawy & Jackson (1989)

0.8 Factor α

Figure 22.50 Anchor load components.

Silt Fine sand Medium sand Coarse sand, gravel

Reese & O'Neill (1988)

Not applicable

Canadian Geotechnical Society (1985)

0.6 0.4 0.2 Range of Kulhawy & Jackson's data

0

0

40

80

120

160

200

240

280

Undrained shear strength, su (kPa)

For silts and clays fmax = 𝛼c su with 𝛼c from Figure 22.51 (22.109)

Figure 22.51 𝛼 factor for grouted anchors in clay. (Source: Adapted from Briaud et al., 1998.)

Tables 22.4, 22.5, and 22.6 present some presumptive values of fmax as recommended by AASHTO (2007). Furthermore, the values of F max , load per meter of anchor length, in Table 22.7 can be used for anchors satisfying the following criteria:

• Center-to-center spacing vertically and horizontally larger than 4 anchor diameter D. Design rules based on pressuremeter data for calculating the ultimate resistance of anchors also exist (Briaud, 1992). These rules, established by the LCPC in France, make a

• Diameter between 150 to 200 mm. • Grout pressure of about 1000 kPa. Table 22.4

Anchor type (grout pressure) Gravity grouted anchors (< 350 kPa) Pressure grouted anchors (350–2800 kPa)

Presumptive f max values for fine-grained soils

Soil type

Shear strength of soil su (kPa)

Shear strength of soil-grout interface fmax (kPa)

Silt-clay mixtures

Stiff to very stiff (50–200 kPa)

30–70

High-plasticity clay

Stiff (50–120 kPa)

30–100

Very stiff (120–200kPa)

70–170

Medium-plasticity clay

Stiff (50–120 kPa)

100–250

Medium-plasticity sandy silt

Very stiff (120–200kPa)

140–350

Very stiff (120–200kPa

280–380

(Source: Adapted from AASHTO, 2007.)

22.12 ANCHORED WALLS AND STRUTTED WALLS

Table 22.5

Anchor type (grout pressure)

Presumptive f max values for coarse-grained soils

Soil type

Gravity grouted anchors (< 350 kPa) Pressure grouted anchors (350–2800 kPa)

769

Sand or sand/gravel mixtures Fine to medium sand Medium to coarse sand with gravel

Silty sand Sandy gravel

Glacial till

Relative density and SPT N value N (blows/0.3 m)

Shear strength of soil-grout interface fmax (kPa)

Medium dense to dense (N = 10 to 50) Medium dense to dense (N = 10 to 50) Medium to dense (N = 10 to 30) Dense to very dense (N = 30 − 50+)

70–140

Medium dense to dense (N = 10 to 40) Dense to very dense (N = 40 to 50+) Dense (N = 30 to 50)

80–380 110–670 250–950

170–400 210–1400 280–1400

300–520

(Source: Adapted from AASHTO, 2007.)

Table 22.6

Presumptive f max values for rock

Rock type

Shear strength of soil-grout interface (kPa)

Soft shale Weathered sandstone Sandstone Slate and hard shale Soft limestone Dolomite limestone Granite or basalt

200–800 700–800 800–1700 800–1400 1000–1400 1400–2100 1700–3100

(Source: Adapted from AASHTO, 2007.)

distinction between several construction techniques for the anchors. Typically, all anchors are tested after installation and curing time. These tests include proof tests, performance tests, creep tests, and 70-day load-hold tests (Briaud et al., 1998). The proof test is the most common and consists of increasing the load in steps up to 1.33 times the design load. In the United States, anchors are accepted if the creep movement at that load is less than 2 mm per log cycle of time. This creep movement is due to the creep in the steel tendon, the progressive cracking of the grout in tension, and the creep of the soil

in shear. The loading history for the proof test and the result of a test are shown in Figure 22.52. 22.12.6

Embedment Depth and Downdrag

Another issue to be addressed is the embedment depth below the excavation level. You might think that the anchored wall would not need much embedment, since the anchors hold the soil back. The following reasoning shows the danger of having very little embedment (Briaud and Lim, 1999). When the excavation takes place, the soil mass behind the wall tends to move toward the excavation and downward (Figure 22.53). The downward movement drags the wall down, and if the embedment is insufficient, the downward movement can be significant. Even if the anchors are performing well, the wall can rotate; indeed, the anchors keep the soil from moving horizontally but not vertically. This rotation will generate horizontal movement as well. Therefore, to minimize horizontal movement, it is necessary to have well-designed anchors and a well-designed embedment depth to resist downdrag and the vertical component of the anchor loads. The embedment depth must also resist the unbalanced lateral load from the bottom of the pressure diagram. This is area GMNH in Figure 22.42. The depth of embedment for lateral resistance is obtained by designing the bottom part of the wall (GHI in Figure 22.42) to resist the pressure from areas GMNH and HOPI with the factored passive resistance SHIQR. This design follows the approach described for the cantilever top-down wall. As a guide, the depth of embedment

770

22 RETAINING WALLS

Table 22.7

Values of anchor load transfer capacity

Soil or rock type

Strength (SPT N values)

Estimated load transfer (kN/m)

Sand and gravel

Loose (N = 4 to 10) Medium (N = 10 to 30) Dense (N = 30 to 50) Loose (N = 4 to 10) Medium (N = 10 to 30) Dense (N = 30 to 50) Loose (N = 4 to 10) Medium (N = 10 to 30) Dense (N = 30 to 50) Stiff (N = 10 to 20) Hard (N = 20 to 40)

145 210 290 100 145 190 75 100 130 30 60 145 360 430 430 580 730

Sand

Sand and silt

Low-plasticity silt and clay Soft shale Slate and hard shale Soft limestone Sandstone Dolomite limestone Granite or basalt

(Source: Canadian Geotechnical Society, 2007; FHWA, 1984.)

Ground level

Fraction of design load

Proof test

Mass movement

0.8 0.4 20

40

60

Time, t (min)

Movement (mm)

0

Compression

1.2

0

0

Tension

200

Load (kN) 400 600

Excavation level 800

20

Anchor head movement

40 60 80

Residual Elastic Total

θ

θ

Vertical settlement

Horizontal movement due to settlement

100

Figure 22.52 Loading history for an anchor proof test. (Source: Adapted from Briaud et al., 1998.)

required for lateral resistance is on the order of 1.5 times the distance GH. The downdrag design requirement may be larger (see Chapter 19, Section 19.6). Note that two cases occur for the embedment depth: the case where the wall is continuous below the excavation level (e.g., slurry wall, sheet pile wall), and the case where only a row of piles exists below the

Figure 22.53 Downdrag creates horizontal movement. (Source: Adapted from Briaud et al., 1998.)

excavation level (e.g., soldier pile and lagging wall). The depth of embedment of a system using a row of piles will have to be larger than that of a continuous wall system.

22.13 SOIL NAIL WALLS

22.12.7 The P-y Curve Approach and the FEM Approach

P-y P

The P-y curve approach described in Section 22.11 is also applicable to anchored walls, and represents the best way to obtain the bending moment versus depth profile for the wall (Briaud and Kim, 1998). In the process of preparing the P-y curves, it is possible to follow the construction sequence as shown in Figure 22.54. The anchors must have their separate P-y curves, as shown in Figure 22.55. A sample result for the P-y curve approach is shown in Figure 22.56. This is a comparison between the P-y curve predictions and the actual measurement for a full-scale instrumented wall at Texas A&M University (Figure 22.57). The P-y curve approach is not as reliable for predicting movements as it is for predicting bending moments. For better movement predictions, the FEM is preferred, provided quality soil parameters are obtained and a realistic soil model is selected (Briaud and Lim, 1999). Figure 22.58 shows a sample result for the FEM approach. This is a comparison of the FEM predictions with the same full-scale wall at Texas A&M University (Figure 22.57).

Horizontal component of ultimate anchor load

1

4 3

6 5

P Low anchor loads 5

P-y Curve for stage 1

6 4

3

2 1

y P 3

High anchor loads

4 5

6

P = anchor design load y = wall deflection after stressing the anchor

y

Deflection of the wall between the original undeflected position of the wall and the position of the wall at the anchor location just before stressing the anchor

1

P-y Curve for stage 1 y

Figure 22.54 P-y curve path during construction sequence. (Source: Briaud et al., 1998/American Society of Engineers.)

Original undeflected position of the wall

Figure 22.55 Anchor P-y curve. et al.,1998/American Society of Engineers.)

(Source:

Briaud

The predictions from the P-y curve approach and the FEM approach can be compared (Figures 22.56 and 22.58).

SOIL NAIL WALLS

Soil nail walls are top-down walls reinforced with rigid inclusions. They are to the top-down walls what MSE walls are to the bottom-up walls. Soil nails are rigid inclusions that are built much like anchors, by drilling a 100–300 mm diameter hole, inserting a steel bar with centralizers in the open hole, and backfilling the hole with grout. Unlike anchors, however, they are not posttensioned. The load in the nail develops as the soil mass deforms. The spacing between soil nails is typically quite a bit smaller than the spacing between anchors. Whereas all anchors and tiebacks are load tested, only a small percentage of soil nails are load tested. Soil nail walls are particularly suited for cases where the soil can stand unsupported for a height of 1–2 m long enough to place a row of nails (a few hours) and where the drill hole can stand open long enough for nail insertion and grouting. The front of the wall is typically covered with shotcrete to a thickness of 100–200 mm projected over a reinforcement mesh. Figures 22.59 and 22.60 show the construction sequence. Much like the case of MSE walls, the design must consider external stability and internal stability as well as deformations. 22.13.1

2

Anchor

Deflection of the wall during stressing of the anchor Elastic elongation of the tendon during stressing of the anchor

22.13 2

Construction stages

771

External Stability

External stability includes global stability, sliding, and bearing capacity. Sliding and bearing capacity are handled in a fashion similar to the MSE wall design (Section 22.12). The global stability, however, is different from the MSE wall approach. It is a slope stability type of analysis that considers failure along a surface through the nails. This surface can

0

0

–1

–1

–2

–2

–3

–3

–4

–4

–5 –6 –7

Depth (m)

Depth (m)

22 RETAINING WALLS

Measured Sequence No sequence

–5 –6 –7

–8

–8

–9

–9

–10 –45

–25

–35

–15

–5

Measured Sequence No sequence

–10 –40

5

–20 0 20 40 Bending moment (kN•m)

Deflection (mm) 0

0

–1

–1

–2

–2

–3

–3

–4

–4

Depth (m)

Depth (m)

772

–5 –6

–5 –6

–7

–7

–8

–8

–9

–9

–10 –60 –40 –20 0 20 40 60 80 100 Horizontal line load (kN/m)

60

–10 –400

BMCOL prediction Measured –300

–200

–100

0

100

Axial load (kN)

Figure 22.56 Predicted and measured result for the P-y curve method. (Source: Briaud et al., 1998/American Society of Engineeers.) Post-construction ground level Wale 1.8 m

2.7 m

3m

4.8 m

2.7 m Final excavation level Soldier beam 12 @2.4 = 31.2 m

4 @2.4 = 9.6 m

4 @2.4 = 9.6 m

(a) Elevation view Ground surface 1.8 m

Load cell

5.05 m Tendon bonded length

Wale 3.0 m

Tube bracket

7.3 m Tendon bonded length 2.7 m

Final excavation level Pressure injected tieback

Soldier beam

Inclinometer casings

(b) Two-row anchored wall – section view

Figure 22.57 Full-scale instrumented wall at Texas A&M University. (Source: Briaud and Lim, 1999/American Society of Engineers.)

22.13 SOIL NAIL WALLS

40

30

20

10

0

–40 0

–20

0

20

40

60

–400 –300 –200 –100 0

–1

–1

–2

–2

–2

–3

–3

–3

–4 –5 –6

Depth (m)

–1

Depth (m)

Depth (m)

50 0

–4 –5 –6

–5 –6

–7

–7

–8

–8

–8

–9

–9

–9

–10

–10

Measured FEM Computed (a) Deflection (mm)

0

–4

–7

–10

773

Measured FEM Computed

Measured FEM Computed

(b) Bending moment (kN·m)

(c) Axial loads (kN)

Figure 22.58 Predicted and measured result for the FEM method. (Source: Briaud and Lim, 1999/American Society of Engineers.)

Excavate unsupported cut 1 to 2 m high

Step 1. Excavate small cut

Step 2. Drill hole for nail

Step 3. Install and grout nail

Step 4. Place drainage strips in initial shotcrete layer & install bearing plates/nuts

1

Figure 22.60 Soil nail wall construction. (Source: Courtesy of FHWA. www.fhwa.dot.gov/publications/publicroads/11septoct/ alongroad.cfm.)

2 3 4 Step 5. Repeat process to final grade

Step 6. Place final facing (on permanent wall)

Figure 22.59 Soil nail wall construction sequence. (Source: FHWA, 1998/U.S. Department of Transportation.)

be a circle, two straight lines, or one line (Figure 22.61). The one-line solution is the simplest and is discussed here. Computer programs such as SNAIL (CALTRANS, 1991) and GOLDNAIL (Golder Associates, 1993) can be used to solve the problem for more complex failure surfaces. Consider the soil nail wall of Figure 22.62. At equilibrium, the force resisted by the nails is T, the weight of the wedge is W, the surcharge force is Q, the shear force on the failure plane is S, and the normal force on the failure plane is N. The

Linear

Bilinear

Circular

Figure 22.61 Soil nail wall failure surfaces.

dimensions and angles involved are defined in Figure 22.62. The problem is to find the value of T to obtain a target factor of safety F (chosen value) on the ultimate shear resistance S.

774

22 RETAINING WALLS

controlling if threaded bars are used. Bending and shearing also do not appear to have a major influence on the behavior of the mass (Lazarte et al., 2003). Let’s look first at pull-out at the grout-soil interface.

Q

β

i W

H

Pull-out at Grout-Soil Interface

S

α

i φʹ Ψ

N

T

R = 𝜋 DLp fmax

L

Figure 22.62 Soil nail wall global stability analysis.

The factor of safety F is defined as: ∑ maximum resisting shear forces on failure plane F= ∑ driving shear forces on failure plane ∑ R (22.110) = ∑ L Alternatively, the LRFD expression would be: ∑ ∑ 𝛾 L=𝜑 R

(22.111)

Writing equilibrium equations normal and along the plane of failure gives: ∑ normal forces = (W + Q) cos 𝜓 + T sin(𝜓 + i) − N = 0 (22.112) ∑ tangent forces = (W + Q) sin 𝜓 − T cos(𝜓 + i) − S = 0 (22.113) The maximum value of the force S is Smax corresponding to the shear strength of the soil: Smax c′ L + N tan 𝜑′ = (22.114) F F where c′ and 𝜑′ are the effective stress cohesion and friction angle of the soil and F is the chosen factor of safety by design. The unknowns are N, S, and T and the three equations (Eqs. (22.112), (22.113), and (22.114)) give the three quantities. Actually, the angle 𝜓 corresponding to the lowest factor of safety is not known either, and must be found by trial and error. Once this is done, the load that must be safely carried by the nails is T. S=

22.13.2

The equation for the ultimate axial resistance R of a nail in tension is:

Internal Stability

Internal stability includes pull-out of the nails at the grout-soil interface, pull-out of the steel bar at the grout-bar interface, tensile yielding, and bending and shearing of the nails. The pull-out of the steel bar at the grout-bar interface is usually not

(22.115)

where D is the diameter of the nail (drill hole), Lp is the useful length of the nail beyond the failure zone, and fmax is the shear strength at the grout-soil interface. Table 22.8 gives some estimated values of fmax . Tensile Force Distribution in the Nail Figure 22.63 shows a simplified distribution of the tension in a nail within the reinforced soil mass. As in the case of the MSE wall, the tension load is lower at the wall face (T o ), then increases as the soil transfers the load to the nail until a maximum value is reached (T max ), and then decreases to zero as the load is transferred from the nail to the surrounding soil. At the wall face, the nail is usually connected to a plate pressed against the soil by a nut threaded on the nail steel bar. The load T o varies from 0.6 to 1 times the maximum load T max . The load T max starts to decrease at a distance Lp from the end of the nail. The measured locus of T max , the failure plane, and the distance Lp are shown in Figure 22.64. The load that must be globally carried by the nails is T, as calculated in the external stability analysis. The ultimate resistance that can be developed by individual nails over the length Lp is Ri , which must satisfy: n ∑ (22.116) 𝛾 T = 𝜑 Ri i=1

The distribution of Ri among the nails is not precisely defined and experience plays a role in that determination. In general, shorter nails are placed at the bottom of the wall and longer ones at the top. A pattern such as the one shown in Figure 22.65 is not uncommon. Length of Nails The required length of each nail Lpi to resist Ri is calculated by using Eq. (22.115). The total length for nail i is Lti ; it is obtained by adding the length Lpi required to safely carry the required load Ri plus the discounted length Ldt within the failure zone (Figure 22.64): Lti = Ldi + Lpi

(22.117)

22.13 SOIL NAIL WALLS

Table 22.8

775

Estimated ultimate grout-soil shear strength, fmax

Material

Construction method

Soil/rock type

Ultimate grout-soil shear strength, fmax (kPa)

Coarse-grained soils

Rotary drilling

Sand/gravel Silty sand Silt Piedmont residual Fine colluvium Sand/gravel low overburden high overburden Dense moraine Colluvium Silty sand fill Silty fine sand Silty clayey sand Sand Sand/gravel Silty clay Clayey silt Loess Soft clay Stiff clay Stiff clayey silt Calcareous sandy clay Marl/limestone Phyllite Chalk Soft dolomite Fissured dolomite Weathered sandstone Weathered shale Weathered schist Basalt Slate/Hard shale

100–180 100–150 60–75 40–120 75–150

Driven casing

Augered

Jet grouted Fine-grained soils

Rotary drilling Driven casing Augered

Rock

Rotary drilling

190–240 280–430 380–480 100–180 20–40 55–90 60–140 380 700 35–50 90–140 25–75 20–30 40–60 40–100 90–140 300–400 100–300 500–600 400–600 600–1000 200–300 100–150 100–175 500–600 300–400

(Source: Adapted from Elias and Juran, 1991.)

the nail, the resistance of the grout is ignored and only the steel is considered. The area of the steel bar must satisfy:

T (x)

To

Tmax

𝛾 Tmax = 𝜑At 𝜎y

(22.118)

X Lp

Figure 22.63 Tension load distribution in a soil nail.

Tensile Yielding of Nails The nails must be designed in such a way that the load applied does not break the nails. In calculating the tensile strength of

where 𝛾 and 𝜑 are the load and resistance factors respectively, T max is the highest load in the nail, At is the steel bar cross-section, and 𝜎 y is the yield strength of the steel. According to Briaud and Lim (1997) and Lazarte et al. (2003), the value of T max is given by: Tmax = 0.65 to 0.75 Ka 𝛾Hsv sh

(22.119)

776

22 RETAINING WALLS

(0.3 to 0.4) H

Locus of maximum nail axial force

1

Critical failure surface

T1 2

H

Lp

T2

Table 22.9

Estimates of soil nail wall movements

Variable

Weathered rock or stiff soil

Sandy soil

Fine-grained soil

0.001 1.25

0.002 0.8

0.003 0.7

𝛿h ∕H and 𝛿v ∕H C

(Source: Adapted from Lazarte et al., 2003.)

3

Lp

T3

Facing

Distribution of tension along nail

Lp

δH

Figure 22.64 Load in the nails and available resisting length.

D L

0.5 m 1.5 m

δV 15°

1.5 m 10 m

1.5 m

H

1.5 m 1.5 m

Deformed pattern

Initial configuration

α

1.5 m 0.5 m

Soil nail

Figure 22.65 Typical pattern of nail length distribution. Figure 22.66 Deformation of soil nail walls.

where K a is the active earth pressure coefficient, 𝛾 is the total unit weight of the soil, H is the height of soil retained, and sv and sh are the vertical and horizontal nail spacing respectively. Note that Eq. (22.119) assumes that there is no water within the retained depth of soil or rock. 22.13.3

Wall Movement

The movement of soil nail walls is similar to the movement of anchored and strutted walls. According to Lazarte et al. (2003), for soil nail walls with ratios of length of nails to height of wall between 0.7 and 1.0, negligible surcharge loading, and typical load and resistance factors (safety factors), empirical data show that the maximum long-term horizontal and vertical wall displacements at the top of the wall, 𝛿 h and 𝛿 v , can be estimated by the values in Table 22.9 where H is the wall height. The parameter C helps to estimate the extent of the movement behind the wall (Figure 22.66): D = CH(1 − tan 𝛼)

(22.120)

where D is the horizontal distance of influence of the excavation measured from the front of the wall where settlement of the ground surface takes place, C is the coefficient in

Table 22.9, H is the height of the wall, and 𝛼 is the batter of the wall (Figure 22.66). 22.13.4

Other Issues

Other issues include the details of the connection plates at the nail head, punching and bending of the wall cover at the front of the nail, corrosion resistance, and seismic loading. For more details on these matters, see Lazarte et al. (2003).

22.14

SPECIAL CASE: TRENCH

Trenches are narrow and fairly shallow excavations often used for placing utilities in congested areas. In the case of the undrained behavior of fine-grained soils, the relationship between the vertical total stress 𝜎ov and horizontal total stress 𝜎oh at failure of the trench is given by: 𝜎oh = 𝜎ov − 2su where su is the undrained shear strength of the soil.

(22.121)

22.14 SPECIAL CASE: TRENCH

777

σov σoh

Figure 22.67 Initiation of failure in a trench.

Figure 22.69 Much safer when protected by a trench box. (Source: Courtesy of www.cobletrenchsafety.com/jobprofile.php?id=95.)

Figure 22.68 Do not go in there! (Source: Courtesy of CDC’s Public Health Image Library.)

The most stressed element in the trench is the soil element at the bottom of the trench, as shown in Figure 22.67. For that element, the vertical total stress 𝜎ov is: 𝜎ov = 𝛾h

(22.122)

where 𝛾 is the total unit weight of the soil and h is the depth of the trench. Initiation of failure of the trench corresponds to failure of the element shown in Figure 22.67. For this element, 𝜎oh is zero and the depth hf at which initiation of failure starts is: hf =

2su 𝛾

(22.123)

So, for example, if su is 100 kPa and 𝛾 is 20 kN∕m3 , then hf is 10 m and a safe depth might be 5 m. Would you go and

work at the bottom of an open, unprotected trench 1 m wide and 5 m deep? You should not, and you should not allow anyone else to work in such a situation. The risk of collapse is too great, as evidenced by the number of related deaths every year. There is an average of 50 deaths per year due to trench accidents in the United States. Even going into a 1.2 m deep trench is not safe. You might think that as long as your head is above ground, you will be safe: Not true! If your head is above ground, you can open your mouth to take air in, but if your chest is below ground, you cannot expand your lungs, no air goes in, and you die. Do not go into an open, unsupported trench (Figure 22.68). You may go into a trench that is supported by what is called a trench box (Figure 22.69). The theory leading to Eq. (22.101) is correct, but the assumptions may not match the reality. It is assumed that the soil is uniform and that every part of it has a minimum undrained shear strength su . This may not be the case in the field, for many reasons: the soil may be fissured and you tested the soil between fissures rather than at the fissures, which may control the mass strength; you may have tested the soil in the summer when it is harder (e.g., water tension higher) and the trench is opened in the winter; the undrained shear strength may not be the appropriate strength if the soil drains during and after the trench is open. Additionally, if someone becomes partially buried in soil, do not try to pull that person out by rope and mechanical means. The tensile strength of the body is typically less than the pull-out capacity or force generated, and you can imagine the result! Excavate around the body to free the person.

778

22 RETAINING WALLS

Problems and Solutions Problem 22.1 Show the pressure diagram, calculate the resultant push, and give its location for a 10 m-high wall due to: a. Water only. b. Dry soil with unit weight of 20 kN/m3 and a friction angle of 30∘ (active and passive). c. The same soil but with water to the top of the wall (active and passive). Solution 22.1 (Figure 22.1s)

Uw

10 m Pw = 490.5 kN/m

3.33 m

98.1 kPa

Figure 22.1s Pressure diagram for water only behind the wall.

a. Water only uw = 𝛾w × H uw = 9.81 × 10 = 98.1 kPa The resultant push per unit length of wall is: uw × H 𝛾 × H2 = w 2 2 9.81 × (10)2 Pw = = 490.5 kN∕m 2

Pw =

Location from the bottom of the wall:

h = 3.33 m 3 b. Dry soil with unit weight of 20 kN/m3 and a friction angle of 30∘ (active and passive) The active force is: z=

′ 𝜎ov = 𝛾d × H = 20 × 10 = 200 kPa

Ka × 𝛾d × H 2 2 1 − sin 𝜑 1 − sin 30∘ Ka = = = 0.33 1 + sin 𝜑 1 + sin 30∘ Pa =

Pa =

0.33 × 20 × (10)2 = 333.3 kN∕m 2

22.14 SPECIAL CASE: TRENCH

The passive force is: Pp =

Kp × 𝛾d × H 2

2 1 − sin 𝜑 1 − sin 30∘ Kp = = = 0.33 1 + sin 𝜑 1 + sin 30∘ 0.33 × 20 × (10)2 Pp = = 3000 kN∕m 2 The location of the active and passive force measured from the bottom of the wall (shown in Figure 22.2s) is: z=

σovʹ = γd × H

h = 3.33 m 3

σʹa = Ka × σovʹ

σʹp = Kp × σovʹ

10 m Pa = 333.3 kN/m

Pp = 3000 kN/m

3.33 m

3.33 m

200 kPa

66.67 kPa

600 kPa

Figure 22.2s Pressure diagram for dry soil: Active and passive pressure profile.

a. Soil with water to the top of the wall (active and passive) The vertical effective stress at the bottom of the wall is: ′ 𝜎ov = 𝛾t × H − 𝛾w × H = 20 × 10 − 9.81 × 10 = 101.9 kPa

The active force is 1 ′ + Pw H Ka 𝜎ov 2 1 − sin 𝜑 1 − sin 30∘ Ka = = = 0.33 1 + sin 𝜑 1 + sin 30∘ 10 × 0.33 × 101.9 Pa = + 490.5 = 660.2 kN∕m 2 Pa =

The passive force is: 1 ′ + Pw H Kp 𝜎ov 2 1 + sin 𝜑 1 + sin 30∘ Kp = = = 0.33 1 − sin 𝜑 1 − sin 30∘ 10 × 3 × 101.9 Pp = + 490.5 = 2019 kN∕m 2 The location of the active and passive force measured from the bottom of the wall (shown in Figure 22.3s) is: Pp =

z=

h = 3.33 m 3

779

780

22 RETAINING WALLS

σt = γs × H

uw

σʹ = σt – uw

σʹa = Ka × σʹ

σa = σʹa + uw

σʹp = Kp × σʹ

σp = σʹp + uw

10 m Pp = 660 kN/m

Pp = 2019 kN/m

3.33 m 200 kPa

98.1 kPa

101.9 kPa

33.9 kPa

132 kPa

3.33 m 305.7 kPa

403.8 kPa

Figure 22.3s Pressure diagram for wall with water at ground surface.

Problem 22.2 Solve Coulomb’s wedge analysis for the passive case of a soil with friction and cohesion. Write vertical and horizontal equilibrium and demonstrate Eq. (22.17). Solution 22.2 (Figure 22.4s) Pp sin (𝜌 +

𝜑′ )

=

Pp =

W=

Pp = 𝜕 Pp 𝜕𝜌

W sin (180 − 𝜑′ − 𝜌 − 𝛼 − 𝛿) W sin (𝜌 + 𝜑′ ) sin (𝜑′ + 𝜌 + 𝛼 + 𝛿) 𝛾H 2 2sin2 (𝛼) 𝛾H 2 2

2sin (𝛼)

( sin (𝛼 + 𝜌) ( sin (𝛼 + 𝜌)

= 0 → Pp =

𝛾 H2 2

sin (𝛼 + 𝛽) sin (𝜌 − 𝛽) sin (𝛼 + 𝛽) sin (𝜌 − 𝛽)

)

)

sin (𝜌 + 𝜑′ ) sin (𝜑′ + 𝜌 + 𝛼 + 𝛿)

sin2 (𝛼 − 𝜑′ ) ]2 [ √ sin (𝜑′ + 𝛿) sin (𝜑′ + 𝛽) 2 sin (𝛼) sin (𝛼 + 𝛿) 1 − sin (𝛼 + 𝛿) sin (𝛼 + 𝛽)

(α + δ)

W

Pp (φʹ + ρ)

(180–φʹ–ρ–α–δ) R

Figure 22.4s Coulomb wedge analysis for the passive case.

Problem 22.3 Plot Coulomb and Rankine active and passive earth pressure coefficients for a vertical wall, no wall friction, as a function of the ground surface inclination 𝛽. Which one would you use?

22.14 SPECIAL CASE: TRENCH

781

Solution 22.3 Coulomb Theory a. Figure 22.5s shows the wedge analysis in Coulomb theory for this case. As stated, the wall is vertical, and there is no wall friction. Therefore, the active earth force is acting horizontally. Note that 𝛽 is the ground surface inclination, 𝜑 is the soil friction angle, and 𝛼 is the failure plane inclination. H is the height of the wall, Pa is the maximum active force acting on the wall, W is the weight of the wedge, and R is the resultant force.

β W

φ R

H Pa

α

Figure 22.5s Illustration of active wedge analysis in Coulomb theory.

In this case, the expression of Coulomb active earth pressure coefficient simplifies to:

To find the relationship between K ah shown in Figure 22.6s.

cos2 𝜑 Kah = [ √ ]2 sin 𝜑 sin(𝜑 − 𝛽) 1+ cos 𝛽 and 𝛽, a soil friction angle equal to 30∘ is assumed. The plot between K a and 𝛽 is

Horizontal earth pressure coefficient

10 For φ = 30 °

9

Kah - Coulomb

8

Kph - Coulomb

7

Kah - Rankine

6

Kph - Rankine

5

Kph - Average

4 3 2 1 0

0

5

10 15 20 Inclination β (degrees)

25

30

Figure 22.6s Plot of K a and K p vs. 𝛽 using both Coulomb theory and Rankine theory.

782

22 RETAINING WALLS

b. Figure 22.7s shows the wedge analysis in Coulomb theory for this case. As stated, the wall is vertical, and there is no wall friction. Therefore, the passive earth force is acting horizontally. Note that 𝛽 is the ground surface inclination, 𝜑 is the soil friction angle, and 𝛼 is the failure plane inclination. H is the height of the wall, Pp is the maximum passive force acting on the wall, W is the weight of the wedge, and R is the resultant force.

β W H

φ

R

Pp α

Figure 22.7s Illustration of passive wedge analysis in Coulomb theory.

In this case, the expression of Coulomb passive earth pressure coefficient simplifies to: cos2 𝜑 Kph = [ √ ]2 sin 𝜑 sin(𝜑 + 𝛽) 1− cos 𝛽 To find the relationship between K ph and 𝛽, a soil friction angle equal to 30∘ is assumed. The plot between K ph and 𝛽 is shown in Figure 22.6s. Rankine Theory Figure 22.8s shows an illustration of the retaining wall analyzed using Rankine theory.

β

Pa, Pp

H β

Figure 22.8s Retaining wall analyzed using Rankine theory.

Note that 𝛽 is the ground surface inclination, 𝜑 is the soil friction angle, H is the height of the wall, Pp is the passive earth force, and Pa is the active earth force. a. Based on Rankine theory, the active earth pressure coefficient that gives the horizontal component of Pa is: √ cos 𝛽 − cos2 𝛽 − cos2 𝜑 2 Kah = cos 𝛽 √ cos 𝛽 + cos2 𝛽 − cos2 𝜑 To find the relationship between K ah and 𝛽, a soil friction angle equal to 30∘ is assumed. The plot between K ah and 𝛽 is shown in Figure 22.6s.

22.14 SPECIAL CASE: TRENCH

783

a. Based on Rankine soil theory, the passive earth pressure coefficient that gives the horizontal component of Pa is: √ cos 𝛽 + cos2 𝛽 − cos2 𝜑 2 Kph = cos 𝛽 √ cos 𝛽 − cos2 𝛽 − cos2 𝜑 To find the relationship between K ph and 𝛽, a soil friction angle equal to 30∘ is assumed. The plot between K ph and 𝛽 is shown in Figure 22.6s. Discussion Figure 22.6s shows that for K ph the Coulomb solution is a limit equilibrium solution giving upper-bound values, whereas the Rankine solution is an equilibrium-of-stresses solution that gives lower-bound values. Therefore, if a lower bound is conservative, one should choose Rankine theory; if an upper bound is conservative, one should choose Coulomb theory. To that end, one would be tempted to use an average of the two values as a more reasonable estimate; however, such an average is not based on any theoretical reasoning. Note that in the case of extreme values of the geometry parameters, it is advisable to use engineering judgment, as the K a and K p values can become unreasonable. Problem 22.4 Evaluate the influence of wall friction on the active and passive earth pressure coefficients by comparing Rankine value (no friction) and Coulomb values (varying friction angle from 0 to 𝜑′ ) for a vertical wall and horizontal backfill. Which one would you use? Solution 22.4 Active Pressure For a vertical wall and horizontal backfill, the Coulomb value K ah that gives the horizontal component Pah of the active push Pa is: sin2 (90 + 𝜑′ ) Kah = [ ]2 √ sin(𝜑′ + 𝛽) sin 𝜑′ 1+ sin(90 − 𝛿) and the Rankine value is:

1 − sin 𝜑′ 1 + sin 𝜑 Figure 22.9s shows the K ah values versus friction angle for both the Coulomb and Rankine solutions. For Coulomb, different curves are presented for different values of the wall friction. As can be seen from Figure 22.9s, the Coulomb value of K ah decreases as the wall friction increases; the Rankine value does not change. The maximum value of Coulomb K ah is reached for zero wall friction, which is equal to the Rankine value. Active earth pressure coefficient Kah

Kah =

1

δ = 0 (Rankine and Coulomb) δ = 10 (Coulomb) δ = 20 (Coulomb) δ = 30 (Coulomb)

0.8 0.6 0.4 0.2 0

0

10

20 30 Friction angle of soil φʹ

40

50

Figure 22.9s K ah vs. soil friction angle for different wall friction angle.

Passive Pressure Coulomb: Kph = [

sin2 (90 − 𝜑′ ) ]2 √ sin(𝜑′ + 𝛿) sin 𝜑′ 1− sin(90 + 𝛿)

784

22 RETAINING WALLS

Rankine: Kph =

1 + sin 𝜑′ 1 − sin 𝜑

Passive earth pressure coefficient Kph

Figure 22.10s shows the K ph values versus friction angle for both the Coulomb and Rankine solutions. For Coulomb, different curves are presented for different values of the wall friction. 20

δ = 0 (Rankine and Coulomb) δ = 10 (Coulomb) δ = 20 (Coulomb) δ = 30 (Coulomb)

15 10 5 0

0

10

20

40

30

50

Friction angle of soil φʹ

Figure 22.10s K ph vs. soil friction angle for different wall friction angle.

Which one would you use? Rankine solution generally gives reasonable values. Coulomb theory is also reasonable in the case of the active earth pressure, but Coulomb’s passive earth pressure values are quite optimistic and should not be used. The problem is that the failure surface is optimistically chosen as a straight line instead of a curved surface which would offer less resistance. Problem 22.5 Demonstrate that the direction of the plane of failure for the active pressure case (PB in Figure 22.11s) is equal to 45 + 𝜑′ ∕2. Solution 22.5 First, we find the Pole on the failure circle in the active case (Figure 22.11s). The failure point is shown by point T on the failure circle. If we draw a line from the center M of the failure circle to the failure point T, the angle MTO will be 90∘ . In the triangle TOM: ∧

TMO = 180 − (90 + 𝜑′ ) = 90 − 𝜑′ τ φʹ Failure circle

T

φʹ O

Pole P 45 + φʹ/2

M

σʹ

90-φʹ

Figure 22.11s Pole method.

In the triangle TPM: ∧

∧

MPT = MTP ∧

∧

2 × MPT + (90 − 𝜑′ ) = 180 → MPT = 45 +

𝜑′ 2

22.14 SPECIAL CASE: TRENCH

785

Problem 22.6 A 6 m-high retaining wall has a backfill made of unsaturated sandy silt with a water tension equal to −1000 kPa and an area ratio (𝛼) equal to 0.3. The total unit weight is 20 kN∕m3 . The soil has no effective stress cohesion (c′ = 0), and an effective stress friction angle equal to 30∘ (𝜑′ = 30). The backfill is horizontal and the wall friction is neglected. Calculate the active and passive earth pressure diagram for this wall. Solution 22.6 Use Rankine theory to solve this problem. Active Earth Pressure The active earth pressure coefficient for this problem is: 1 − sin 𝜑′ 1 = 1 + sin 𝜑′ 3 Because the soil behind the wall is uniform, we only need to choose two calculation points: points a and b, shown in Figure 22.12s. Ka =

σv a

αuw –300 kPa

σvʹ

σahʹ

300 kPa

100 kPa

σah –200 kPa

H=6m

b

120 kPa –300 kPa

420 kPa

140 kPa

–160 kPa

Figure 22.12s Active earth pressure diagram.

Point a: total vertical stress 𝜎v = 0 Therefore, effective vertical stress 𝜎v′ = 𝜎v − 𝜎uw = 0 + 0.3 × (−1000) = 300 kPa ′ = Ka 𝜎v′ = Effective active horizontal stress is 𝜎ah ′ 𝜎ah

1 3

× 300 = 100 kPa

1 3

× 420 = 140 kPa

+ 𝛼uw = 100 + (−300) = 200 kPa Total active horizontal stress is 𝜎ah = Point b: total vertical stress 𝜎v = 𝛾t H = 20 × 6 = 120 kPa Therefore, effective vertical stress 𝜎v′ = 𝜎v − 𝛼uw = 120 − 0.3 × (−1000) = 420 kPa ′ Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ =

Total active horizontal stress is 𝜎ah =

′ 𝜎ah

+ 𝛼uw = 140 + (−300) = −160 kPa

The active earth pressure diagram is shown in Figure 22.12s. Practically, the suction should be ignored as it could disappear in the rainy season or cracks could develop in the backfill and the active earth pressure diagram would be the same as if the soil had no water. Passive Earth Pressure The passive earth pressure coefficient for this problem is Kp =

1 + sin 𝜑′ =3 1 − sin 𝜑′

′ = Kp 𝜎v′ = 3 × 300 = 900 kPa Point a: Effective passive horizontal stress is 𝜎ph ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + 𝛼uw = 900 + (−300) = 600 kPa

786

22 RETAINING WALLS

′ Point b: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ = 3 × 420 = 1260 kPa ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + 𝛼uw = 1260 + (−300) = 960 kPa

The passive earth pressure diagram is shown in Figure 22.13s. Practically, the suction would be ignored and the passive earth pressure diagram would be the same as if the soil had no water. σv a

σv ʹ

αuw –300 kPa

σphʹ

σph

300 kPa 900 kPa

600 kPa

H=6m

b

120 kPa –300 kPa 420 kPa

1260 kPa

960 kPa

Figure 22.13s Passive earth pressure diagram.

Problem 22.7 A wall is to be placed in a soil as described in Figure 22.14s. Prepare the active pressure diagram and the passive pressure diagram for that soil profile. σv

Sand, γ = 20 kN/m3 cʹ = 0 ϕʹ = 34°

uw

σvʹ

σahʹ

σah

3m

1m

Clay, γ = 18 2m cʹ = 10 kPa, ϕʹ = 28° kN/m3

Silt, γ = 19 kN/m3 cʹ = 0 ϕʹ = 30°

6m

Figure 22.14s Soil profile.

Solution 22.7 Use Rankine theory to solve this problem. For each layer, the active and passive earth pressure coefficients are calculated as follows: 1 − sin 𝜑′ 1 + sin 𝜑′ = 0.283, K = = 3.54 p 1 + sin 𝜑′ 1 − sin 𝜑′ 1 − sin 𝜑′ 1 + sin 𝜑′ Clay ∶ Ka = = 0.361, Kp = = 2.77 ′ 1 + sin 𝜑 1 − sin 𝜑′ 1 − sin 𝜑′ 1 + sin 𝜑′ = 0.333, K = =3 Silt ∶ Ka = p 1 + sin 𝜑′ 1 − sin 𝜑′

Sand ∶ Ka =

22.14 SPECIAL CASE: TRENCH

787

Active Earth Pressure Because several soil layers and a groundwater level are involved in this problem, five calculation points are chosen (Figure 22.15s). σv

σʹv

uw

σʹah

σah

a

Sand, γ = 20 kN/m3 cʹ = 0 ϕʹ = 34°

3m b 1m c

Clay, γ = 18 kN/m 2m cʹ = 10 kPa, ϕʹ = 28° d 3

60

60 80

10

17.0

70

19.8 13.2

116

30

86

29.8 23.2

28.6 19.0

Silt, γ = 19 kN/m3 cʹ = 0 ϕʹ = 30°

17.0

58.6 49.0

6m

e

230

90

140

46.6

136.6

(Unit: kPa)

Figure 22.15s Active earth pressure diagram.

Point a: Total vertical stress: 𝜎v = 0, uw = 0 Therefore, effective vertical stress: 𝜎v′ = 𝜎v − uw = 0 √ ′ Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ − 2c′ Ka = 0 ′ Total active horizontal stress is 𝜎ah = 𝜎ah + uw = 0 Point b: Total vertical stress: 𝜎v = 𝛾h = 20 × 3 = 60 kPa, uw = 0 Therefore, effective vertical stress: 𝜎v′ = 𝜎v − uw = 60 kPa √ ′ = Ka 𝜎v′ − 2c′ Ka = 0.283 × 60 − 2 × 0 = 17.0 kPa Effective active horizontal stress is 𝜎ah ′ Total active horizontal stress is 𝜎ah = 𝜎ah + uw = 17.0 kPa Point c: Total vertical stress 𝜎v = 𝛾h = 60 + 20 = 80 kPa, uw = 10 × 1 = 10 kPa Therefore, effective vertical stress: 𝜎v′ = 𝜎v − uw = 70 kPa Note that point c is on the interface between two different layers, so the effective active horizontal stress at that point should be calculated individually in each layer. √ ′ = Ka 𝜎v′ − 2c′ Ka = 0.283 × 70 − 2 × 0 = 19.8 kPa Point c, sand: Effective active horizontal stress is 𝜎ah ′ Total active horizontal stress is 𝜎ah + 𝜎ah + uw = 19.81 + 10 = 29.8 kPa √ √ ′ Point c, clay: Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ − 2c′ Ka = 0.361 × 70 − 2 × 10 × 0.361 = 13.2 kPa ′ Total active horizontal stress is 𝜎ah + 𝜎ah + uw = 13.2 + 10 = 23.2 kPa ∑ Point d: Total vertical stress: 𝜎v = 𝛾h = 80 + 18 × 2 = 116 kPa, uw = 10 × 3 = 30 kPa Therefore, effective vertical stress: 𝜎v′ = 𝜎v − uw = 116 − 30 = 86 kPa Note that point d is on the interface between two different layers, so the effective active horizontal stress at that point should be calculated individually in each layer. √ √ ′ Point d, clay: Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ − 2c′ Ka = 0.361 × 86 − 2 × 10 × 0.31 = 19.0 kPa ′ Total active horizontal stress is 𝜎ah = 𝜎ah + uw = 19.0 + 30 = 49.0 kPa

788

22 RETAINING WALLS

√ ′ Point d, silt: Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ − 2c′ Ka = 0.333 × 86 − 2 × 0 = 28.6 kPa ′ Total active horizontal stress is 𝜎ah = 𝜎ah + uw = 28.6 + 30 = 58.6 kPa ∑ Point e: Total vertical stress: 𝜎v = 𝛾h = 116 + 19 × 6 = 230 kPa, uw = 10 × 9 = 90 kPa Therefore, effective vertical stress: 𝜎v′ = 𝜎v − uw = 230 − 90 = 140 kPa √ ′ Effective active horizontal stress is 𝜎ah = Ka 𝜎v′ − 2c′ Ka = 0.333 × 140 − 2 × 0 = 46.6 kPa ′ Total active horizontal stress is 𝜎ah = 𝜎ah + uw = 46.6 + 90 = 136.6 kPa The active earth pressure diagram is shown in Figure 22.15s. Passive Earth Pressure Because several soil layers and a groundwater level are involved in this problem, five calculation points are chosen (Figure 22.16s.). σv

σʹv

uw

σph

σʹph

a Sand, γ = 20 kN/m3 cʹ = 0 ϕʹ = 34°

3m b 1m c

Clay, γ = 18 kN/m3 2m cʹ = 10 kPa, ϕʹ = 28° d

60

60

80

10

212.4

70

212.4 257.8

247.8

237.2

227.2 116

30

86

301.5 258 271.5

288

Silt, γ = 19 kN/m3 6 m cʹ = 0 ϕʹ = 30°

e

230

90

140

420

510

(Unit: kPa)

Figure 22.16s Passive earth pressure diagram.

For those calculation points, the vertical stresses, water stress, and effective vertical stresses are the same as for the active earth pressure. Here we only provide the calculation of effective passive horizontal stress and total passive horizontal stress at those five points. √ ′ = Kp 𝜎v′ + 2c′ Kp = 0 Point a: Effective passive horizontal stress is 𝜎ph ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + uw = 0 √ ′ Point b: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ + 2c′ Kp = 3.54 × 60 + 2 × 0 = 212.4 kPa ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + uw = 212.4 kPa Point c: Note that point c is on the interface between two different layers, so the effective passive horizontal stress at that point should be calculated individually in each layer. √ ′ = Kp 𝜎v′ + 2c′ Kp = 3.54 × 70 + 2 × 0 = 247.8 kPa Point c, sand: Effective passive horizontal stress is 𝜎ph ′ + uw = 247.8 + 10 = 257.8 kPa Total passive horizontal stress is 𝜎ph = 𝜎ph √ √ ′ Point c, clay: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ + 2c′ Kp = 2.77 × 70 + 2 × 10 = 2.77 = 227.2 kPa ′ + uw = 227.2 + 10 = 237.2 kPa Total passive horizontal stress is 𝜎ph = 𝜎ph

22.14 SPECIAL CASE: TRENCH

789

Point d: Note that point d is on the interface between two different layers, so the effective passive horizontal stress at that point should be calculated individually in each layer. √ √ ′ Point d, clay: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ + 2c′ Kp = 2.77 × 86 + 2 × 10 × 2.77 = 271.5 kPa ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + uw = 271.5 + 30 = 301.5 kPa √ ′ Point d, silt: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ + 2c′ Kp = 3 × 86 + 2 × 0 = 258 kPa ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + uw = 258 + 30 = 288 kPa √ ′ Point e: Effective passive horizontal stress is 𝜎ph = Kp 𝜎v′ + 2c′ Kp = 3 × 140 + 2 × 0 = 420 kPa ′ Total passive horizontal stress is 𝜎ph = 𝜎ph + uw = 420 + 90 = 510 kPa The passive earth pressure diagram is shown in Figure 22.16s. Problem 22.8 A 10 m-high retaining wall has a horizontal backfill made of soil without water. The soil properties are 𝛾 = 20 kN∕m3 , c′ = 0, 𝜑′ = 30∘ . Draw the active pressure diagram against the wall due to the following surcharges at the top of the wall: a. Uniform surcharge equal to 20 kPa. b. Line load of 20 kN/m at a distance of 1 m from the edge of the wall. c. A point load of 20 kN at a distance of 1 m from the edge of the wall. Solution 22.8 (Figure 22.17s)

0

0

10

Horizontal pressure (kPa) 20 30 40 50 60

70 a b c

1 2 3 Depth (m)

80

4 5 6 7 8 9 10

Figure 22.17s Horizontal pressure diagram.

Ka =

1 − sin 30 1 = 1 + sin 30 3

′ 𝜎ah = 𝜎ov Ka + Δ𝜎h

a. 𝜎ah = b. 𝜎ah =

1 × (20z + 20) = 6.67(z + 1) 3

25.46z 1 4 × 20 12 z = 6.67z + × 20z + 3 𝜋 (z2 + 1) (1 + z2 ) (

c. Assume that v = 0.35 𝜎ah

1 20 = z+ 3 𝜋(z2 + 1)

1

3z 3

(z2 + 1) 2

−

(z2 + 1) 2 (1 − 2v) 1

(z2 + 1) 2 + z

)

790

22 RETAINING WALLS

Problem 22.9 How deep would you dig an unsupported trench in a stiff clay with an undrained shear strength of 75 kPa and a unit weight of 18 kN/m3 ? The contract requires that you do the digging yourself while working at the bottom of the trench. Solution 22.9 Assuming that the soil is truly uniform, with no fissures: 𝜎ah = 𝜎0v − 2Su 𝜎ah = 𝛾H − 2Su = 18H − 2 × 75 = 0 → H = 4.16 m I would dig the trench to a depth of 1 m and stop there. Before going deeper, I would place a trench box to protect myself against trench collapse. Then I would dig further. Problem 22.10 Plot the coefficient of earth pressure at rest K o as a function of OCR for an overconsolidated clay with a friction angle 𝜑′ equal to 28∘ . On the same graph, plot K a and K p .

Earth pressure coefficient

Solution 22.10 (Figure 22.18s) 3 2.5

K0 Ka Kp

2 1.5 1 0.5 0

0

1

2 3 4 Over consolidation ratio, OCR

5

6

Figure 22.18s Earth pressure coefficients vs. OCR.

K0 = (1 − sin 𝜑′ )OCRsin 𝜑 1 − sin 𝜑 1 − sin 28∘ Ka = = = 0.36 1 + sin 𝜑 1 + sin 28∘ 1 + sin 𝜑 Kp = = 2.77 1 − sin 𝜑 ′

Problem 22.11 Draw the earth pressure diagram for a 7 m-high gravity retaining wall with a backfill compacted with a vibratory roller. The roller weighs 150 kN, has a centrifugal force amplitude of 50 kN, is 2 m wide, and gets as close as 1 m to the top edge of the wall. The soil has a unit weight of 19 kN/m3 , a passive earth pressure coefficient equal to 3, and an at-rest earth pressure coefficient equal to 0.6. √ ( √ 150+50 ) √2 × 19 2 2P𝛾 L 2 √ 𝜎h = = = 23.3 kN∕m2 a+L 𝜋 1+2 𝜋 √ ( √ 150+50 ) √2 √ √ 2 L 2 2P d= = = 2.0 m Ko (a + L) 𝜋𝛾 0.6(1 + 2) 𝜋 × 19 Ko 𝛾z = 0.6 × 19 × 7 = 79.8 kN∕m2

Solution 22.11 (Figure 22.19s)

√

22.14 SPECIAL CASE: TRENCH

791

P = (150 + 50)/2 kN/m 1m

0.41 m σh = 23.2 kN/m2

2m

K0γz = 79.8 kN/m2

Figure 22.19s Earth pressure diagram.

L: the length of the roller a: the distance between the edge of the wall and the closest roller position P: the line load imposed by the roller weight of the roller plus the centrifugal force for vibratory rollers 𝛾: the unit weight of the soil K 0 : the at-rest earth pressure coefficient d: depth to which the pressure diagram is modified due to the roller The depth z to reach the horizontal pressure equal to 23.2 kPa is such that Kp 𝛾z = 23.2 kPa, therefore, z = 0.41 m. Problem 22.12 An 8 m-high top-down wall is retaining a shrink-swell soil with a swell pressure profile decreasing with depth from 500 kPa at the ground surface down to 50 kPa at the bottom of the wall. The soil has a friction angle 𝜑′ equal to 28∘ and no cohesion c′ . K o is 0.6. Draw the pressure diagram for the wall. Solution 22.12 (Figure 22.20s)

0

100

200

300

400

500

σh (kPa)

Swell pressure 8m 235

80 50

At-rest pressure K0γh = 86 kPa

Passive pressure Kpγh = 399 kPa

Figure 22.20s Earth pressure diagram.

K0 = 0.6 1 + sin 𝜑′ 1 + sin 28∘ Kp = = = 2.77 ′ 1 − sin 𝜑 1 − sin 28∘

792

22 RETAINING WALLS

Assuming that the soil unit weight is 𝛾 = 18 kN∕m3 : Ko 𝛾h = 0.6 × 18 × 8 = 86.4 kPa Kp 𝛾h = 2.77 × 18 × 8 = 398.88 kPa Problem 22.13 Draw the displacement ya and yp necessary to mobilize the active and passive earth pressure as a function of the wall height H for a dense sand. Solution 22.13 (Figures 22.21s, 22.22s)

Displacement, ya (mm)

140 Loose sand Dense sand Soft clay Stiff clay

120 100 80 60 40 20 0

0

1

2

3 4 5 Wall height, H (m)

6

7

8

Figure 22.21s Active displacement vs. wall height.

Displacement, yp (mm)

350 Loose sand and soft clay Dense sand and stiff clay

300 250 200 150 100 50 0

0

1

2

3 4 5 Wall height, H (m)

6

7

8

Figure 22.22s Passive displacement vs. wall height.

From Table 22.1, the average displacements needed to generate active and passive earth pressures for different soil types are: Loose sand ya H

= 0.004,

Soft clay yp H

ya H

= 0.04

Dense sand ya H

= 0.0015,

= 0.015,

yp H

= 0.04

Stiff clay yp H

ya H

= 0.02

Problem 22.14 Derive Eqs. (22.73) and (22.74). Solution 22.14 The uniform soil pressure p1 due to the line load Q is: p1 =

Q B

= 0.0075,

yp H

= 0.02

22.14 SPECIAL CASE: TRENCH

793

The soil pressure p2 due to the overturning moment is the maximum pressure at the edge of the triangular distribution under the foundation. The pressure distribution under the foundation must resist the moment. Writing the moment equilibrium gives: 1 B 2B 6M p × × 2 = M ⇒ p2 = ± 2 2 22 3 2 B ⎧ ⎪pmax = Q 6M M p= ± 2 &e= ⇒⎨ B B Q ⎪ pmin = ⎩

(

Q B

6e B

1−

6e B

(

1+

Q B

) )

Problem 22.15 For the retaining wall shown in Figure 22.23s, calculate the pressure distribution against the wall, the resultant push, the factor of safety against sliding, and the factor of safety against overturning. 1m

Fill cʹ = 0, φʹ = 32° γt = 18 kN/m3

Reinforced concrete 2.4 m

No water 0.5 m

1.0 m

0.3 m 0.3 m 0.3 m

Clay cʹ = 5 kN/m3 φʹ = 28° γt = 20 kN/m3

Figure 22.23s Retaining wall.

Solution 22.15 (Figures 22.24s, 22.25s) Passive earth pressure, active earth pressure: ′ + 2c′ 𝜎ph = Kp 𝜎ov

Kp =

√ Kp + 𝛼u

1 + sin 𝜑′ 1 − sin 𝜑′

√ ′ 𝜎ah = Ka 𝜎ov − 2c′ Ka + 𝛼u Ka =

1 − sin 𝜑′ 1 + sin 𝜑′

a. Calculate the active earth pressure (active side): 1 − sin 32 = 0.307 1 + sin 32 1 − sin 28 = = 0.361 1 + sin 28 = 0.307 × 43.2 = 13.26 kPa at a depth of 2.4 m (in the fill) √ = 0.361 × 43.2 − 2 × 5 0.361 = 9.59 kPa at a depth of 2.4 m (in the clay) √ = 0.361 × 55.2 − 2 × 5 0.361 = 13.92 kPa at a depth of 3.0 m (in the clay)

Ka1 = Ka2 𝜎 ′ ah 𝜎 ′ ah 𝜎 ′ ah

′ = 𝜎ah Since there is no water 𝜎ah

794

22 RETAINING WALLS

b. Calculate the passive earth pressure (passive side): 1 + sin 𝜑′ 1 + sin 28 = = 2.77 1 − sin 𝜑′ 1 − sin 28 √ ′ = KP 𝜎ov + 2c′ Kp = 16.64 kPa at the ground level √ √ ′ = KP 𝜎ov + 2c′ Kp = 2.77 × 12 + 10 2.77 = 49.88 kPa at a depth of 0.6 m

Kp = ′ 𝜎ph ′ 𝜎ph

′ Since there is no water 𝜎ph = 𝜎ph

c. Draw the diagram (Figure 22.24s). σʹah (kPa)

σʹov (kPa)

σʹpH (kPa) 16.64

σʹov (kPa)

49.88

43.2

12

9.59 13.26

55.2

13.92

Figure 22.24s Earth pressure diagram.

Pah = Xa =

1 1 × 13.26 × 2.4 + 9.59 × 0.6 + × (13.92 − 9.59) × 0.6 = 22.96 kN 2 2 1 × 13.26 × 2.4 × 1.4 + 9.59 × 0.6 × 0.3 + 12 × (13.92 − 9.59) × 0.6 × 0.2 2

22.965 1 Pph = 16.64 × 0.6 + × (49.88 − 16.64) × 0.6 = 19.96 kN 2 16.64 × 0.6 × 0.3 + 12 × (49.88 − 16.64) × 0.6 × 0.2 4.99 Xp = = = 0.25 m 19.956 19.96

1.3

43.2 20.25

0.65 13.5

Pah = 22.97 kN

0.9 3 Pph = 19.96 kN Xp = 0.25 m

6

Xa = 1.06 m

C A

s

B

Figure 22.25s Forces diagram.

Wsoil1 = 2.4 m × 1 m × 18 kN∕m2 = 43.2 kN∕m Wsoil2 = 0.5 m × 0.3 m × 20 kN∕m2 = 3 kN∕m Wsoil3 = 1 m × 0.3 m × 20 kN∕m2 = 6 kN∕m

=

24.26 = 1.06 m 22.96

22.14 SPECIAL CASE: TRENCH

795

Wstem = (2.4 m + 0.3 m) × 0.3 m × 25 kN∕m2 = 20.25 kN∕mkN∕m Wbase = 1.8 m × 0.3 m × 25 kN∕m2 = 13.5 kN∕m Considering a sliding failure along AB: ∑ W tan 𝜑′ + Pp H (43.2 + 20.25 + 13.5 + 6 + 3) × tan 28∘ + 19.96 Fsliding = = Pa H 22.97 45.70 + 19.96 = 2.84 > 2 → OK = 22.97 Considering a rotation failure around point A: Mmax, resist

43.2 × 1.3 + 20.25 × 0.65 + 13.5 × 0.9 + 19.96 × 0.25 = Mdriving 22.97 × 1.06 86.46 = 3.55 > 2 → OK = 24.35

Foverturning =

It is also reasonable to consider rotation failure around point C: Foverturning =

Mmax, resist Mdriving

=

43.2 × 1.3 + 20.25 × 0.65 + 13.5 × 0.9 81.47 = = 4.67 > 2 → OK 22.97 × 1.06 17.46

Problem 22.16 Design the soil reinforcing strips required for a 20 m-high MSE wall. The precast concrete panels are 1.5 m by 1.5 m. The vertical and horizontal spacing between strips are 750 mm and 450 mm, respectively. The unit weight of the backfill material is 19 kN/m3 with an angle of internal friction of 34∘ and a coefficient of uniformity of 4.4. The location of the first layer of strips, measured from the finished grade, is 375 mm. Ignore the traffic surcharge. Solution 22.16 Panel section = 1.5 m × 1.5 m sv = 0.75 mm sh = 450 mm∕ 𝛾soil = 19 kN∕m3 Cu = 4.4 a. Design for Pull-Out The maximum line load (T max ) to be resisted by the reinforcement inclusions at depth z can be computed as: Tmax = sv sh 𝜎h The horizontal stress 𝜎 h can be calculated as:

𝜎h = kr 𝜎ov + Δ𝜎h 𝜎h = kr 𝜎ov

The coefficient of earth pressure kr is computed using Figure 22.26s (AASHTO). The ka value is computed as: ka = Then the kr value is computed as: a-1. If zi is less than 6 m, then:

a-2. If zi is larger than 6 m, then:

1 − sin 𝜑′ 1 − sin 34 = = 0.283 1 + sin 𝜑′ 1 + sin 34

z kr = 1.7 − i ka 12 kr = 1.2 ka

The calculation of T max for the different strips is summarized in Table 22.1s

22 RETAINING WALLS

Coefficient of lateral stress ratio = kr/ka 1.7

W eld

ed

tal

wi

ba

l strip

Me

Meta

2.5

rm a ts re & gr ids

s

1 1.2

0

*Geosynthetics

0 Depth below top of wall, Z

796

6000 mm

1 1.2 * Does not apply to polymer strip reinforcement

Figure 22.26s Coefficient of lateral stress ratio = kr /ka .

Table 22.1s Strip no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Summary of calculation of T max

Depth (m)

ka

kr ∕ka

kr

𝜎v (kPa)

𝜎h (kPa)

Tmax (kN)

0.375 1.125 1.875 2.625 3.375 4.125 4.875 5.625 6.375 7.125 7.875 8.625 9.375 10.125 10.875 11.625 12.375 13.125 13.875 14.625 15.375 16.125 16.875 17.625 18.375 19.125 19.875

0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283

1.67 1.61 1.54 1.48 1.42 1.36 1.29 1.23 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20

0.472 0.455 0.437 0.419 0.402 0.384 0.366 0.348 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340 0.340

7.1 21.4 35.6 49.9 64.1 78.4 92.6 106.9 121.1 135.4 149.6 163.9 178.1 192.4 206.6 220.9 235.1 249.4 263.6 277.9 292.1 306.4 320.6 334.9 349.1 363.4 377.6

3.365 9.716 15.564 20.907 25.747 30.082 33.913 37.240 41.134 45.973 50.813 55.652 60.491 65.331 70.170 75.009 79.848 84.688 89.527 94.366 99.206 104.045 108.884 113.724 118.563 123.402 128.241

1.14 3.28 5.25 7.06 8.69 10.15 11.45 12.57 13.88 15.52 17.15 18.78 20.42 22.05 23.68 25.32 26.95 28.58 30.22 31.85 33.48 35.12 36.75 38.38 40.01 41.65 43.28

22.14 SPECIAL CASE: TRENCH

797

Now that we have calculated the load T max , we need to find the length of reinforcement that will safely carry that load without pulling out of the soil. The pull-out capacity T pull-out (kN) of the reinforcement inclusion is given by: Tpullout = 2 × fmax × b × La ′ fmax = F ∗ × 𝜎ov ×𝛼

Using the ultimate limit state procedure, we have: 𝛾Tmax = 𝜙 Tpullout The active length of the reinforcement strip required to resist the pull-out load is: Tpullout = La = La = La =

𝛾Tmax 𝜑 Tpullout 2 × fmax × b ( ) 𝛾1 kr 𝜎 ′ov × sv × sh 2 × 𝜑 × F ∗ × 𝜎 ′ov × 𝛼 × b ( ) 𝛾1 kr × sv × sh

2 × 𝜑 × F∗ × 𝛼 × b ( ) 𝛾1 kr × sv × sh L = La + Lmax = + 0.3 H 2 × 𝜑 × F∗ × 𝛼 × b

The value of 𝛼 is taken as 1.0 for strip reinforcements (Section 22.10.2). The resistance and load factors are taken as 0.9 and 1.35, respectively. The coefficient of friction (F * ) is computed according to AASHTO LRFD using Figure 22.27s. Table 22.2s is a summary of calculation of total strip length (L). Default values for pull-out friction factor, F*

Not to scale

Grid-bearing member

Ribbed steel strips F* = TanΦf

Geotextiles F* = 0.67 TanΦf

Geogrids F* = 0.67 TanΦf

Smooth steel strips F* = 0.4

6000 mm

Steel grids F* = 10 (t/St)

Depth below top of wall Z or Z0 (mm)

1.2 + logCu ≤ 2.0

20 (t/St)

0

t D LO St St

Figure 22.27s Friction coefficient F* for MSE wall reinforcement.

If zi is less than 6 m, then: F ∗ = 1.2 + Log Cu = 1.8435 at z = 0 m F ∗ = 0.6745 at z = 6 m F ∗ = 1.8435 − 0.1948 × zi

798

22 RETAINING WALLS

If zi is larger than 6 m, then:

F ∗ = tan 𝜙 = 0.6745

Table 22.2s Strip no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Summary of calculation of total strip length (L)

Depth (m)

T max (kN)

F*

fmax (kPa)

La (m)

L (m)

0.375 1.125 1.875 2.625 3.375 4.125 4.875 5.625 6.375 7.125 7.875 8.625 9.375 10.125 10.875 11.625 12.375 13.125 13.875 14.625 15.375 16.125 16.875 17.625 18.375 19.125 19.875

1.14 3.28 5.25 7.06 8.69 10.15 11.45 12.57 13.88 15.52 17.15 18.78 20.42 22.05 23.68 25.32 26.95 28.58 30.22 31.85 33.48 35.12 36.75 38.38 40.01 41.65 43.28

1.770 1.624 1.478 1.332 1.186 1.040 0.894 0.748 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675 0.675

12.61 34.72 52.66 66.44 76.06 81.51 82.79 79.92 81.70 91.31 100.92 110.53 120.15 129.76 139.37 148.98 158.59 168.20 177.82 187.43 197.04 206.65 216.26 225.87 235.48 245.10 254.71

1.349 1.415 1.495 1.592 1.712 1.867 2.072 2.357 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546 2.546

7.349 7.415 7.495 7.592 7.712 7.867 8.072 8.357 8.546 8.546 8.546 8.546 8.546 8.471 8.021 7.571 7.121 6.671 6.221 5.771 5.321 4.871 4.421 3.971 3.521 3.071 2.621

b. Design for Yielding Using the ultimate limit state procedure, we have: 𝛾Tmax = 𝜑 Tyield The resistance and load factors are taken as 0.75 and 1.35, respectively. The T yield for steel reinforcement is given by: Tyield = 𝜎yield × A Tyield = 𝜎yield × b × Ec The value of A is the cross-sectional area of the strip after accounting for corrosion (AASHTO, 2007). The structural thickness of the strip at the end of the service life is computed according to AASHTO LRFD as: 0.086 − 2 × 0.015 Service Life of Zinc Coating (0.086 mm∕year) = 2 years + years 0.004 Service Life of Zinc Coating (0.086 mm∕year) = 16 years

22.14 SPECIAL CASE: TRENCH

Use a strip thickness 50 mm wide and 5 mm thick: Ec = 5 mm − 2Es Ec = 5 mm − 2 × (75 years − 16 years) × 0.12 mm∕year Ec = 3.58 mm Then: Tyield = 448159.2 kPa × (0.00358 mm × 0.05 mm) Tyield = 80.2 kN Then, using the result of T max at the bottom layer of strips where the maximum tension load is expected, we have: 0.75 × 80.2 kN > 1.35 × 43.3 kN 60.2 kN > 58.5 kNa

∴ OK

Detailed calculations for all the strips are shown in Table 22.3s. Table 22.3s Strip no. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Summary of calculations for strip resistance to yielding

Depth (m)

T max (kN)

R

φR

𝛾T max

Check

0.375 1.125 1.875 2.625 3.375 4.125 4.875 5.625 6.375 7.125 7.875 8.625 9.375 10.125 10.875 11.625 12.375 13.125 13.875 14.625 15.375 16.125 16.875 17.625 18.375 19.125 19.875

1.14 3.28 5.25 7.06 8.69 10.15 11.45 12.57 13.88 15.52 17.15 18.78 20.42 22.05 23.68 25.32 26.95 28.58 30.22 31.85 33.48 35.12 36.75 38.38 40.01 41.65 43.28

80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2 80.2

60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2 60.2

1.5 4.4 7.1 9.5 11.7 13.7 15.5 17.0 18.7 20.9 23.2 25.4 27.6 29.8 32.0 34.2 36.4 38.6 40.8 43.0 45.2 47.4 49.6 51.8 54.0 56.2 58.4

OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK

Note: Bearing capacity and slope stability failure were not checked as part of this problem. However, they must be checked to ensure that the system is safe against these failure modes.

799

800

22 RETAINING WALLS

Problem 22.17 A cantilever retaining wall is embedded 6 m below excavation level and retains 5 m of soil. An impervious layer exists 4 m below the bottom of the wall. The water level is at the ground surface on both sides of the wall and the soil deposit is uniform and deep. Draw the water pressure diagram against the wall on both sides of the wall, assuming that the water pressure is hydrostatic. Then draw a flow net and develop the water pressure diagram on both sides of the wall. Compare and comment. Solution 22.17 The hydrostatic water pressure diagram is shown in Figure 22.28s and the flow net in Figure 22.29s.

1 × 9.81 × 112 = 176.6 KN 2

1 × 9.81× 62 = 176.6 KN 2 6 × 9.81 = 58.9 kPa

11 × 9.81 = 107.9 kPa

Figure 22.28s Water pressure in hydrostatic conditions. A

G B F

C

E

D

Figure 22.29s Flow net.

Water pressures are calculated at points A, B, C, D, E, F, and G on each side of the wall to generate the water stress profile. The loss of total head through the flow net is 5 m. The loss of total head through each flow field is 5∕12 = 0.417 m. The total head ht(M) at any point M is calculated by: ht(M) = ht(beg) − nd Δht where ht(beg) is the total head at the beginning of the flow net (15 m), nd is the number of equipotential drops to go from the beginning of the flow net to point M, and Δht is the drop of total head across any flow field. Then the elevation head he(M) is measured on the scaled drawing and the pressure head hp(M) is obtained as the difference between the total head and the elevation head (Table 22.4s). Table 22.4s Point A B C D E F G

Calculation of heads and water stresses

Total head (m)

Elevation head (m)

Pressure head (m)

Water stress (kPa)

15 14.33 13.75 12.5 11.66 10.62 10

15 9.6 6.0 4 4 6.3 10

0 4.73 7.75 8.5 7.66 4.32 0

0 46.40 76.03 83.38 75.14 42.38 0

22.14 SPECIAL CASE: TRENCH

801

The water pressure diagram from the flow net is shown in Figure 22.30s together with the hydrostatic diagram. As can be seen, the hydrostatic diagram is more conservative. A

G

B

F C E D

Figure 22.30s Water pressure under flow conditions.

Problem 22.18 Demonstrate Eq. (22.91). Solution 22.18 Equation (22.90) expresses moment equilibrium at the bottom of a wall: Pa Xa − Ppm Xpm = 0 Using Eqs. (22.86)–(22.88), and (22.89) in Eq. (22.90), we get: ) ( )( )( ( ) 3 1 7 1 Ka 𝛾(H + D)2 Kp 𝛾D2 D =0 (H + D) − 2 3 8 18 ) ( )( )( ) ( 3 7 1 1 2 2 Kp D D = Ka (H + D) (H + D) 8 18 2 3 21 1 3 3 K D = Ka (H + D) 144 p 6 7 3 Kp D = Ka (H + D)3 8 √ 7 Kp 3 D =H+D 8 Ka ) (( )0.33 7 Kp −1 −D=H 8 Ka (( ) )0.33 7 Kp D −1 =H 8 Ka 3

D= (

7 Kp 8 Ka

H )0.33 −1

Problem 22.19 What is the depth of embedment d required for a cantilever wall retaining a height of sand H? Express the results as a function of H, Kp ∕Ka , and a factor of safety F applied to 𝜎 p , the passive pressure. (Note: There is no water.) Solution 22.19 1 Kp 𝛾D2 × 13 D 1 1 Ka 𝛾(H + D)2 × (H + D) − 2 =0 2 3 F.S 1 k D3 Ka (H + D)3 = F.S p

802

22 RETAINING WALLS

)0.33 1 Kp F.S Ka H D= ( )0.33 1 Kp −1 F.S K H+D = D

(

a

Problem 22.20 For the anchored slurry wall shown in Figure 22.31s, calculate the pressure distribution on both sides of the wall for a deflection of 25 mm at the top of the wall. Calculate the anchor forces. How important is the vertical capacity of the wall? Explain your answer. What would happen if the water level rose on both sides of the excavation to the top of the wall? What would happen if the water level rose to the top of the wall on the retained-soil side of the excavation and to 2 m below that on the excavated side? 1.9 m 30°

Sand γt = 18 kN/m3  ϕ = 32°  

3.0 m

Anchor horizontal spacing = 2.4 m 30°

2.7 m Sand  γt = 18 kN/m3 ϕ = 32°

0.4 m-thick reinforced  concrete slurry wall

1.55 m

Figure 22.31s Anchored slurry wall.

Solution 22.20 From Figure 22.19 and utop ∕H = 0.025∕7.6 = 0.003, K behind the wall is 0.2 (average). Using Eq. (22.99), the constant pressure from z = 0 to z = H is: ′ (at z = H) + uw 𝜎h = K𝜎ov

= 0.2(18)(7.6) = 27.4 kN∕m2 For z = H, just below the constant pressure, the K a active earth pressure is used: 𝜎ah = Ka 𝜎v 1 − sin(30) = (18)(7.6) 1 + sin(30) = 45.5 kN∕m2 For z = H + D, the active pressure is: 𝜎ah = Ka 𝜎v 1 = (18)(9.15) 3 = 54.8 kN∕m2 For z = H on the excavation side, the passive pressure is 0 and at z = H + D, the passive pressure is: 𝜎ah = Ka 𝜎v = 3(18)(1.15) = 83.7 kN∕m2

22.14 SPECIAL CASE: TRENCH

803

Using the tributary area for the top anchor, the horizontal component for that anchor is: F1h = 𝜎h A1 = 27.4 × (1.9 + 1.5) × 2.4 = 223.6 kN Using the tributary area for the bottom anchor, the horizontal component for that anchor is: F2h = 𝜎h A2 = 27.4 × (1.5 + 1.35) × 2.4 = 187.4 kN Because the anchors are inclined at 30∘ , the actual loads in the anchors are: F1h 223.6 = = 258.2 kN cos 𝛼 cos 30 F 187.4 F2 = 2h = = 216.4 kN cos 𝛼 cos 30 The vertical capacity is important because the soil mass tends to move toward the excavation and downward. The downward movement imposes downdrag on the wall. If the vertical capacity is insufficient, the wall will move downward and rotate around the anchor. This will cause horizontal movement as well. If water rises on both sides to the top of the wall, the water pressure on both sides will cancel out and the soil horizontal stress will decrease from the total stress (K𝛾H) to the effective stress (K𝛾 ′ H). This would lead to a pressure on the wall of about one-half of the pressure with no water on either side. If there was a difference in level of 2 m, there would be a net water pressure equal to 2 m of water on the wall in addition to the K𝛾 ′ H. F1 =

Problem 22.21 Explain Figure 22.49. Solution 22.21 Figure 22.49 shows an example of load distribution in an anchor in tension. The load resisted by the soil increases steadily from the back of the anchor to the front of the anchor. The load in the tendon is constant and equal to the anchor load along the tendon unbonded length because the greased sheath that covers the anchor does not permit any load transfer. Then the load in the tendon drops off as the grout contributes to the load being resisted. Within the zone where the grout is in tension, the tendon is the only one carrying load, because the grout cracks and contributes no load to the resistance. Within the tensile strains where the grout can resist tension, some of the load is carried by the tendon and some by the grout. The grout has zero load at the ground surface and the load increases in compression over the unbonded tendon length because the grout moves with respect to the soil and is loaded in compression. Beyond the tendon unbonded length, the grout is in tension to such a level that it cracks and cannot contribute to the resistance. Then, in the back of the anchor, the tension load decreases to the point where the strains are low enough and the grout can resist some tension. Problem 22.22 Use Tables 22.4 and 22.5 and add a column giving the back-calculated alpha values. Solution 22.22 (Table 22.5s) Table 22.5s Anchor type (grout pressure) Gravity grouted anchors (< 350 kPa) Pressure grouted anchors (350–2800 kPa)

Soil type

Shear strength of soil su (kPa)

Shear strength of soil-grout interface fmax (kPa)

𝛼s

Silt-clay mixtures

Stiff to very stiff (50–200

30–70

0.35–0.6

High-plasticity clay

Stiff (50–120)

30–100

0.6–0.83

Very stiff (120–200)

70–170

0.58–0.85

Stiff (50–120)

100–250

2–2.1 1.2–1.75

Very stiff (120–200) Very stiff (120–200)

140–350 280–380

2.3–1.9

Medium-plasticity clay Medium-plasticity sandy silt

804

22 RETAINING WALLS

fmax = 𝛼c su f 𝛼c = max su Sample calculation: Stiff silt-clay mixture su = 50 kPa, fmax = 30 kPa, 𝛼c =

30 50

= 0.6

Problem 22.23 For Figure 22.62, the height H is 9 m, 𝛼 is 17∘ , 𝛽 is 18∘ , 𝜓 is 44∘ , and i is 10∘ . The stiff clay weighs 20 kN/m3 with some cohesion c′ (to be ignored), and a friction angle 𝜑′ of 32∘ . A uniformly applied surcharge of 10 kPa is to be considered on top of the wall. Calculate the required nail force T for a factor of safety against shear failure along the chosen plane to be 1.5. Distribute that force among the four nails and find the required length for each nail. Solution 22.23 (Figure 22.32s) 10.2 m Q

β i W H=9m

α

S i

N Ψ = 44º

T L

Figure 22.32s Nailed wall.

Equations (22.112), (22.113), and (22.114) are used to find the three unknowns N, S, and T: (W + Q) cos 𝜓 + T sin(𝜓 + i) − N = 0 (W + Q) sin 𝜓 − T cos(𝜓 + i) − S = 0 S c′ L + N tan 𝜑′ S = max F F The weight of the soil is obtained by multiplying the unit weight of the soil by the area of the triangle. The area of the triangle is 41.4 m2 per meter perpendicular to the page and W is 828 kN/m. Substituting these values in the equations: (828 + 10.2 × 10) cos 44 + T sin(44 + 15) − N = 0 (828 + 10.2 × 10) sin 44 − T cos(44 + 15) − S = 0 N tan 30 −S=0 1.5 Then: 669 + 0.857T − N = 0 646 − 0.515T − S = 0 0.385N − S = 0 After solving the system of equations, N = 1065 kN∕m, S = 410 kN∕m, and T = 462 kN∕m. For simplification, if all the nails in the wall carry the same force, then the force T is divided by the four nails and the force at each nail is approximately 115.5 kN/m. The required length of the nails can be found using Eq. (22.115) plus a factor of safety F: Ra =

𝜋 DLp fmax T = n F

22.14 SPECIAL CASE: TRENCH

805

where Ra is the allowable load on each nail, T is the total nail load, n is the number of nails, D is the diameter of the nail (drill hole), Lp is the useful length of the nail, and fmax is the shear strength at the grout-soil interface. The shear strength fmax depends on the soil and the construction method and is estimated using Table 22.8. For a stiff clay, an fmax of 50 kPa can be used and a diameter of 200 mm. FT 1.5 × 462 Lp = = = 5.5 m n 𝜋 D fmax 4𝜋 × 0.2 × 50 The required length Lp of each nail is 4.4 m. The length of the nail inside the failure zone is the discounted length Ld . The total length of each nail is the sum of the required length and the discounted length. The total length of each nail is: Lti = Ldi + Lpi Lt1 = Ld1 + Lp1 = 5.6 + 5.5 = 11.1 m Lt2 = Ld2 + Lp2 = 4.3 + 5.5 = 9.8 m Lt3 = Ld3 + Lp3 = 3 + 5.5 = 8.5 m Lt4 = Ld4 + Lp4 = 1.8 + 5.5 = 7.3 m Problem 22.24 A 3 m-wide strutted excavation is planned in a clay with an undrained shear strength equal to 40 kPa and a total unit weight of 19 kN/m3 . What depth of excavation corresponds to a factor of safety against base failure equal to 1.5? Solution 22.24 The safety factor for the base failure can be calculated using the following equation: F=

Nc su 𝜎ov(z=H)

Assume that H = 3 m and H∕B = 1. Then, using the Skempton chart, the Nc = 6.4 6.4 × 40 and H = 8.9 m 19 × H Assume that H = 9 m and H∕B = 3. Then, using the Skempton chart, the Nc = 7.3 1.5 =

7.3 × 40 and H = 10.24 m 19 × H Assume that H = 10 m and H∕B = 3.33. Then, using the Skempton chart, the Nc = 7.3 1.5 =

7.3 × 40 and H = 10.24 m 19 × H H = 10 m

1.5 =

CHAPTER 23

Earthquake Geoengineering

T

his chapter serves as an introduction to the large and complex field of geotechnical earthquake engineering. The book by Kramer (1996), the book by Towhata (2008), and the FHWA manual by Kavazanjian et al. (2011) are excellent references for further study.

23.1

BACKGROUND

Plate tectonics is the main reason for earthquakes on our planet. The Earth’s crust is made up of six continental plates (Figure 23.1) that have traveled large distances over geologic times. The plates move because of the thermal difference between the earth surface and the deeper layers. This thermal difference creates convection currents in the molten core of the planet, which move the surface rock plates. The boundaries between plates are called faults. The problem is that the movement of the plates with respect to one another is not smooth. Indeed, the interface between plates or faults is rough and stresses accumulate along the fault over time. When the stress becomes equal to the strength of the fault surface, the fault shears in a dramatic motion known as an earthquake (Figure 23.2). A bit of lubricant would solve that problem! An earthquake originates at some depth below the ground surface; this point is called the hypocenter. The point on the ground surface directly above the hypocenter is the epicenter. The distance between a site on the ground surface and the epicenter is the epicentral distance. An earthquake starts at one location, propagates along a fault, then propagates up into the rock mass and then into the soil mass (earthquake), and then sometimes into an ocean (tsunami). This propagation is in the form of compression waves and shear waves (see Chapter 9, Section 9.2.1). Seismographs record the passage of the waves. A compression wave moves faster than the corresponding shear wave and will arrive first. The difference in time between the arrival of the compression wave and the

shear wave can be used to determine the distance d between the location of the seismograph and the epicenter: d=

Δtp−s 1 vs

−

(23.1)

1 vp

where Δtp-s is the difference in time between the arrival of the p wave and the s wave, vs is the shear wave velocity, and vp is the compression wave velocity. One seismograph can give the distance d but not the direction of the wave generating the signal; three seismographs are necessary to locate the epicenter (Foster, 1988) (Figure 23.3).

23.2

EARTHQUAKE MAGNITUDE

The size of an earthquake can be quantified in several ways. The first and oldest way is the earthquake intensity, which is a qualitative description of the effect of the earthquake. The Mercalli scale (1883) is the best known and goes from I (not felt) to XII (total destruction). Giuseppe Mercalli was an Italian seismologist and volcanologist who proposed this scale in the late 1800s. It was revised a few times after that. The problem with the Mercalli scale is that it relies on human reactions and structural damage observations, both of which depend on more than just the size of the earthquake. The Richter scale is the best-known of the magnitude scales (Richter, 1935). The Richter magnitude (M L , with the subscript L used to designate local magnitude) is defined as the logarithm base 10 of the magnitude trace amplitude in micrometers recorded on a Wood-Anderson seismometer located 100 km from the epicenter of the earthquake. Seismic instruments were developed and installed around 1930 and are used extensively today to quantify earthquake magnitude. The Richter scale has been modified over the years and led to the use of the body wave magnitude and surface wave magnitude scales.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

807

808

23 EARTHQUAKE GEOENGINEERING

Figure 23.1 Tectonic plates on Earth. (Source: United States Geologic Survey/Public Domain.)

(a)

(b)

Figure 23.2 Movement of tectonic plates.

is that they are unable to recognize large earthquakes; this is called saturation. Saturation occurs at a magnitude of about 6.2 for mb and 8 for M s . Saturation is due to the fact that very large earthquakes release more of their energy at longer periods; because the periods associated with the mb and M s calculations are fixed, they cannot acknowledge higher periods and therefore larger earthquakes. Another limitation of these magnitude scales is that they do not address the amount of time associated with the shaking. The moment magnitude (M w ) takes that aspect into account and is broadly used today. It is rooted in the seismic moment M o associated with the work done by the earthquake along the fault: Mo = GAD

B Distance from station B

A C Epicenter

Figure 23.3 Locating the epicenter with three seismographs.

The body wave magnitude (mb ) is calculated from the amplitude of compression waves with periods of about 1 sec toward the beginning of the record. The surface wave magnitude (M s ) is calculated from the amplitude of Rayleigh waves with periods of about 20 sec. One limitation with these scales

(23.2)

where G is the shear modulus of the rock, A is the area over which the slip occurs, and D is the amount of slip movement. Because M o is a very large number, and because Mw = 0.66 log M0 (N.m) − 6.05

(23.3)

As an example, let’s calculate the moment magnitude of the December 26, 2004, Sumatra-Andaman earthquake. As reported by Lay et al. (2005), the fault surface was 1300 km long and 220 km deep, and the slip distance was 5 m. For a typical value for G of 3 × 1010 N∕m2 , the seismic moment M o was 4.7 × 1022 N.m. Then the moment magnitude is very close to 9 or a huge earthquake. Though this is the moment magnitude, the news media would report that the earthquake registered as 9 on the Richter scale. Note that all these scales

23.5 GROUND MOTION

are log scales, so that, for example, a magnitude 9 earthquake is 10 times larger than a magnitude 8 earthquake. Yet another way to classify an earthquake is to calculate the energy released during the slip. Bath (1966) proposed to obtain the energy E from: log E = 5.24 + 1.44M

(23.4)

where E is the energy in N.m or Joules and M is the magnitude. So if M is 9, then E is 1.6 × 1018 Joules. To give you an idea of how much energy this represents, it is enough to cover the electrical consumption of the entire United States for one month. So if we could harness that energy and turn it to good use, it would be very valuable, and unfortunately it seems to be renewable energy!

• See Chapter 9, Sections 9.2.2, 9.2.3, 9.2.4, and 9.2.5 for dynamic in situ tests based on wave propagation. • See Chapter 10, Section 10.13 for the resonant column test, shear modulus, and damping ratio. • See Chapter 15, Section 15.10 for the initial tangent shear modulus Gmax . • See Chapter 15, Section 15.11 for the normalized shear modulus G∕Gmax and damping ratio vs. shear strain 𝛾 curves. • See Chapter 15, Sections 15.15 and 15.16 for the resilient modulus. • See Chapter 19, Sections 19.8.7 and 19.8.8 for the rate of loading and cyclic loading effects. 23.5

23.3

WAVE PROPAGATION

For specifics on wave propagation, see Chapter 9, Section 9.2.1. 23.4

DYNAMIC SOIL PROPERTIES

Dynamic soil properties have been discussed in previous chapters:

Acceleration (m/sec2)

• See Chapter 8, Section 8.2 for the seismic CPT. • See Chapter 8, Section 8.11.5 for the lightweight deflectometer test.

Velocity (cm/sec)

During an earthquake, the rock fault shears and sends shear waves and compression waves through to the ground surface. This shaking of the rock and soil mass can be recorded using instruments sensitive to motion. These are generally accelerometers that use the piezoelectric effect. They contain microscopic crystal structures (crystal quartz) that get stressed by inertia forces and react by creating a change in voltage. This voltage is measured and correlated by calibration to accelerations. While the soil motion created by an earthquake is in three directions, the horizontal motion is usually the one of most interest because it tends to cause the most damage. Figure 23.4 shows an acceleration record for

Peak ground acceleration - PGA 0

0

5

50

10

15

20

25

15

20

25

15

20

25

Peak ground velocity - PGV

0

–50

Displacement (cm)

GROUND MOTION

5

–5

0

5

10

20 Peak ground displacement - PGD 10 0 –10

0

5

809

10 Time (sec)

Figure 23.4 Ground motion for an earthquake. (Source: FHWA, 1998/U.S. Department of Transportation.)

810

23 EARTHQUAKE GEOENGINEERING

an earthquake along with the velocity and the displacement. The velocity and the displacement are obtained by integrating once and then twice the acceleration versus time. These time domain signals are quite complex, and there is a need to report simpler parameters to describe an earthquake. These parameters include information on the amplitude A, the frequency f , and the duration t of the acceleration a; velocity v; and displacement u. The amplitudes of a, v, and

u can be characterized by the peak values, which are the highest values in the signal. The PGA is the peak ground acceleration, the PGV is the peak ground velocity, and the PGD is the peak ground displacement. The PGA, PGV, and PGD are indicated in Figure 23.4. A huge earthquake can generate 10 m∕s2 or 1 g acceleration, whereas acceleration of 0.1m∕s2 or 0.01 g is associated with small earthquakes. Figure 23.5 shows a PGA map of the United States

PGA with 2% in 50 year PE. BC rock. 2008 USGS

50°

45°

40°

1.00 0.77 0.59 0.46 0.35 0.27 P 0.21 G 0.16 0.12 A 0.10 0.08 g 0.06 0.04 0.03 0.02 0.01

35°

30°

km 0

500

25°

–125

°

–120

°

–115°

–110°

–105°

–100°

–95°

–90°

–85°

–80°

–75°

–70°

–65°

(a) PGA with 10% in 50 year PE. BC rock. 2008 USGS

50°

45°

40°

1.00 0.77 0.59 0.46 0.35 0.27 0.21 S 0.16 A 0.12 0.10 g 0.08 0.06 0.04 0.03 0.02 0.01

35°

30°

km 500

0

25°

–125

°

–120

°

–115°

–110°

–105°

–100°

–95°

–90°

–85°

–80°

–75°

–70°

(b)

Figure 23.5 USGS maps of peak ground acceleration for 2% and 10% probability of exceedance over 50 years corresponding to 2275-year and 475-year return periods, respectively. (Source: United States Geologic Survey/Public Domain.)

–65°

23.5 GROUND MOTION

prepared by the United States Geologic Service (USGS) for two distinct return periods: 2275 years and 475 years. Also useful are the effective acceleration (acceleration closest to the structural response and damage of the structure), the sustained maximum acceleration (acceleration sustained for 3 or 5 cycles), and the effective design acceleration (peak acceleration after filtering accelerations above 8 Hz). A more detailed inspection of Figure 23.4 shows that the frequencies associated with the acceleration signal are higher than the frequencies associated with the velocity signal, which are themselves higher than the frequencies associated with the displacement signal. The frequency content is different and is best obtained by performing a Fourier transform analysis (Kramer, 1996). This transformation is a mathematical transformation named after the work of Jean Baptiste Joseph Fourier, a French mathematician and physicist, who developed it around 1800. It transforms the signal from a plot of amplitude vs. time into a plot of amplitude vs. frequency (Figure 23.6) or from the time domain to the frequency domain. This amplitude vs. frequency plot is called a Fourier spectrum, so one will have a Fourier acceleration spectrum, a Fourier velocity spectrum, and a Fourier displacement spectrum. The Fourier spectra describe the ground motion. Another spectrum is the response spectrum to a particular earthquake input motion. A response spectrum is a plot of the maximum response (a, v, or u) of a linear single degree of freedom (SDOF) system to an earthquake input motion versus the natural period T of the system for a given damping ratio 𝛽. The natural period T of an undamped SDOF

system is given by:

√ T = 2𝜋

m k

(23.5)

where m is the mass of the system (kg) and k is the spring stiffness (N/m). The damping ratio 𝛽 in percent is given by: 𝛽=

c ccrit

c × 100 × 100 = √ 2 mk

1. Choose the input motion signal for the earthquake. 2. Choose a value of the damping ratio 𝛽 and the mass m for the SDOF. 3. Select a value of the stiffness k of the SDOF and excite the system with the selected earthquake motion.

Sa (m/sec2)

10 5

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5 Period (sec)

2

2.5

3

Sv (cm/sec)

150 100 50 0

Sd (cm)

40

20

0

(23.6)

where c is the damping coefficient of the dashpot (N.s/m) and ccrit is the critical damping. The critical damping is what brings the system back to zero without oscillations. Nowadays many doors are equipped with critically damped pistons so they close back without oscillations. (Old saloon bar doors, for example, did not have critical dampers.) In earthquake engineering, a damping ratio equal to 5% is common. Three response spectra are typically created: one each for the acceleration (Sa ), the velocity (Sv ), and the displacement (Sd ). The process followed to obtain an acceleration response spectrum is illustrated in Figure 23.7 and an example is shown in Figure 23.8. The step-by-step procedure is as follows:

15

0

811

Figure 23.6 Fourier transforms of signals spectrum.

812

23 EARTHQUAKE GEOENGINEERING

t a2max

a a1max

m

k1

a

m

t

14

t

m

t m c

m

anmax

Fundamental Period (seconds)

a

aomax = amax

a

a a3max

c

c c

k2

k3

kn

aomax

Spectral acceleration N, Sa (g)

t

a2max

aomax = amax

T3

Tn

Period, T (sec)

Gilroy No. 2 (soil) Fourier spectrum amplitude (m/sec)

Empire State

6

0.05

0.04

0.03

0.02

0.01

2

3

Building

Taipei 101

4 Washington

Monument

2 Tower of Pisa

0

100

200

300

400

500

600

c1

0.06

1

828 m, 11.3 s

Figure 23.9 Relationship between fundamental period and height of a building.

anmax

0.07

0

8

c2 < c1

a1max

T1 T2

Petronas Towers

Building Height (m)

Figure 23.7 Development of an acceleration response spectrum. (Source: Adapted from Matasovic, 1993.)

0

10

0 a3max

Burj Khalifa

World Trade Center

12

4 5 Period, T (sec)

6

7

8

Figure 23.8 Examples of response spectrum.

4. Record the highest value (acceleration, velocity, displacement) of the output motion of the SDOF. 5. Repeat steps 3 and 4 for different values of k. 6. Plot the maximum values of step 4 (spectral value) versus the fundamental period of the SDOF. This is the response spectrum. Note that structures have fundamental periods that increase with their height (Figure 23.9); these are generally in the range of 0.1 seconds for very small buildings to 10 seconds for extremely tall and flexible buildings. The spectral acceleration on the ordinate of the spectrum depends on the rock motion, the soil properties, the damping ratio, and the ratio

of the SDOF stiffness over mass. It does not depend on k and m separately, because of the mathematics behind the problem. The spectrum itself—meaning the curve of a vs. T—is a function of the rock motion, the soil properties, and the damping ratio, but not of the stiffness and the mass of the SDOF. This unique property makes it possible to recommend a single design spectrum that can be used for any structure, given a damping ratio. The response spectrum is very useful because, for a building with a given fundamental period, this spectrum defines the highest spectral acceleration to which the structure is likely to be subjected. This acceleration times the mass of the building gives the inertia force to be resisted by the structure and the foundation. Table 23.1 shows orders of magnitudes of accelerations and velocities for the soil and the structures placed on it. The duration of the earthquake’s strong motion has a major influence on the amount of damage inflicted. The most common way to measure duration is to use the bracketed duration, which is defined as the time between the first and last Table 23.1 Order of magnitude of soil and structure horizontal acceleration and velocity for different earthquake magnitudes Magnitude

8 7.5 7 5.5

Accelerations (gravities)

Velocity (mm/second)

Ground motion

Structure

Ground motion

Structure

0.60 0.45 0.30 0.12

0.33 0.22 0.15 0.1

740 560 360 150

410 280 180 130

(Source: After Hall and Newmark, 1977.)

23.6 SEISMIC HAZARD ANALYSIS

813

0.4 Loma Prieta Station Gilroy #2 0.3

1

1

0.1

0.1

First exceedance

PGA (g)

0.2 Acceleration (g)

Last exceedance +0.05g

0.1

Mag = 6.0

0 –0.1

Mag = 7.5

0.01

–0.05g

0.01 1

10 Distance (km)

100

1

10 Distance (km)

100

–0.2

Figure 23.11 Attenuation of peak horizontal ground acceleration. (Source: Adapted from Boore et al., 1997.)

–0.3 –0.4

0

5

10

15

20 Time (s)

25

30

35

40

Figure 23.10 Example of bracketed duration.

exceedance of a chosen threshold of acceleration. This threshold is often taken as 0.05 g. Figure 23.10 shows an example in which the bracketed duration is 15 seconds.

23.6

SEISMIC HAZARD ANALYSIS

Now that we know how to characterize ground motion, we need to establish what parameters to consider for the site where the construction will take place, or where the stability must be evaluated, or where liquefaction is an issue. A distinction is made here between a deterministic analysis and a probabilistic analysis. In a seismic hazard deterministic analysis, the steps are as follows:

shows a correlation by Boore et al. (1997), including the data points used. In a seismic hazard probabilistic analysis, the steps are somewhat different and consist of the following: 1. Identify all earthquake sources capable of creating significant ground motion at the site. This is the same step as in a deterministic analysis, except that a probability distribution is associated with the location of the source to quantify that uncertainty. 2. Determine the magnitude and recurrence of earthquakes from each source. Small earthquakes occur more often than large earthquakes. The magnituderecurrence relationship gives the number N of earthquakes of a given magnitude or higher that may occur every year. Gutenberg and Richter (1944) proposed the following model: log N = a − bM

1. Identify all earthquake sources capable of creating significant ground motion at the site. 2. Determine the distance between the source and the site. 3. Select the controlling earthquake, that is, the earthquake most likely to produce the highest level of shaking at the site. 4. Determine the ground motion parameters at the site associated with the controlling earthquake: peak acceleration, peak velocity, response spectrum. Step 3 requires the use of an attenuation relationship, a relationship that gives the decrease in acceleration, for example, as a function of the distance from the source. Such attenuation relationships have been developed based mostly on experimental data. These empirical equations (e.g., Cornell et al., 1979) are typically of the form: log(PGA) = A + BM − C log(R + D)

(23.7)

where PGA is the peak ground acceleration, M is the earthquake magnitude, R is the distance from the source to the site, and A, B, C, and D are calibration constants. Figure 23.11

(23.8)

where N is the number of earthquakes per year of magnitude M or greater, and a and b are regional parameters. This model has been revised (Figure 23.12) based on further measurements over the years and also by including geologic and geodetic data. Note that the reciprocal of N is called the recurrence interval or return period. In the building industry, the recurrence interval of the earthquake to use for insuring collapse prevention is the 2500-year earthquake; in the bridge industry, it is the 1000-year earthquake. 3. Determine the ground motion at the site by using an appropriate attenuation relationship. This is the same step as in a deterministic seismic hazard analysis, except that the uncertainty regarding the attenuation is now included as shown in Figure 23.11. 4. The uncertainties in steps 1 through 3 are combined to obtain the probability that the ground motion parameters will be exceeded over a chosen period of time and ensure that this probability meets a target value chosen as acceptable by design.

814

23 EARTHQUAKE GEOENGINEERING

u (0, t)

0.1

10

Ground surface

Guttenberg and Richter model (1944)

1

Schwartz and Coppersmith (1984)

10–1

10

10–2

10

100 Seismicity data

–3

1000

Linear soil

Recurrence interval (yrs)

Annual number of earthquake

1

Up wave

Down wave

Rigid rock

Figure 23.13 Linear soil on rigid rock.

to this differential equation in complex notation reflects this decomposition and is of the form: u(z, t) = aei(𝜔t+kz) + bei(𝜔t−kz)

10–4

10–5

Geologic data

4

5

6 Magnitude, m

10000

7

8

(23.10)

where a and b are the amplitude of the wave going up and the wave going down respectively, 𝜔 is the circular frequency of the harmonic motion, and k is a wave number given by:

100000

𝜔 vs

(23.11)

1 2𝜋 = f 𝜔

(23.12)

k= Recall that:

Figure 23.12 Magnitude-recurrence relationship. (Source: Kavazanjian et al., 2011; Gutenberg and Richter, 1944; Schwartz and Coppersmith, 1984.)

23.7

GROUND RESPONSE ANALYSIS

When a rock fault slides, it shakes the adjacent rock into motion. This motion is propagated in all directions, including upward toward the ground surface. Although most of the travel is through rock, the last few 100 meters may be through soil. The propagation through soil may have a significant impact on the motion of the ground surface, and this ground response is addressed in this section. Propagation analysis can be done as a one-dimensional (1D), two-dimensional (2D), or three-dimensional (3D) analysis. The theory for such analyses can get quite complex. Two simple cases of one-dimensional analysis are presented here. More advanced coverage is given in Kramer (1996). 23.7.1 One-Dimensional Solution for Undamped Linear Soil on Rigid Rock In the case of undamped linear soil on rigid rock (Figure 23.13), the shaking of the rock generates, among other waves, a shear wave traveling in the soil at a shear wave velocity vs and generating a horizontal movement equal to u(t, z) where t is time and z is depth. The equation of motion for an element of soil (see Chapter 20, Section 19.3.4) is: 𝜕2u 𝜕2u = vs 2 2 (23.9) 2 𝜕t 𝜕z If the rock imposes a harmonic motion at the base of the soil layer, which is H thick, two waves will be generated in the soil layer: one going up and one going down. The solution

T=

where T is the period (s), f is the frequency (Hz), and 𝜔 is the circular frequency (rd∕s) of the harmonic motion. Because the ground surface is considered to be a free boundary, the shear stress 𝜏(z, t) and shear strain 𝛾(z, t) must be zero on that boundary (z = 0). The shear strain 𝛾 (0, t) must therefore satisfy: 𝜕u(0, t) 𝛾(0, t) = =0 (23.13) 𝜕z This leads to the condition that a = b and the final expression for u is: u(z, t) = 2a cos(kz)ei𝜔t (23.14) This represents a stationary wave (a wave that remains in a constant position) due to the superposition of the upward wave and the downward wave. Remember that we are interested in transforming the motion of the rock at the base of the soil layer into a motion at the ground surface. The transfer function F(𝜔) is therefore: F(𝜔) =

umax (0, t) 1 1 = = ( ) umax (H, t) cos(kH) cos 𝜔H v

(23.15)

s

While in the general case the transfer function will be a complex number, in this simple case it is a scalar. To obtain the horizontal displacement vs. time signal at the soil surface, the horizontal displacement vs. time at the rock boundary is simply multiplied by the transfer function for each time in the record. Note that if 𝜔H∕vs is equal to 𝜋∕2, Eq. (23.15) indicates that the transfer function becomes infinite and the soil is in resonance with the rock motion. Therefore, the natural

23.7 GROUND RESPONSE ANALYSIS

period T of a soil layer with a height H and a shear wave velocity vs , also called the characteristic site period, is:

14 12 10

(23.16)

Equation (23.15) shows that the important factors in the response of a soil layer to an earthquake are the frequency of the rock motion, the thickness of the layer, and its shear wave velocity or small strain shear modulus, which are closely related (see Chapter 9, Section 9.2.1). In the United States, the natural period of soil deposits is on the order of 0.4–2 seconds.

|F (ω)|

2𝜋H 𝜋 4H 𝜔H 𝜋 = or = or T = vs 2 Tvs 2 vs

8

(23.17)

where 𝜂 is the soil viscosity (N.s∕m2 ) and 𝜌 is the mass density of the soil (kg∕m3 ). The solution is similar to the preceding undamped case except that the wave number k is now a complex number k* with a real part k1 and a complex part k2 : ∗

∗

with: k∗ =

𝜔 (1 − i𝛽) vs

(23.18) (23.19)

where 𝛽 is the damping ratio, which is usually a small number between 0.05 and 0.1. The transfer function is: F(𝜔) =

1 cos(k∗ H)

(23.20)

and the modulus of that function or amplification function is the ratio of the movement at the ground surface over the movement at the rock level: |f (𝜔)| =

umax (0, t) 1 =√ ( ) ( ) umax (H, t) 𝜔H 2 + sinh 𝛽 cos2 𝜔H v v s

≈√ cos2

(

𝜔H vs

1 )

( )2 + 𝛽 𝜔H v

β = 10%

4

β = 20%

2 0

5

0

10

15

ω

Figure 23.14 Amplification function.

Let’s assume that the damping in a soil layer can be represented by a Kelvin-Voigt model (see Chapter 13, Section 13.2.1). In the case of a damped linear soil on rigid rock, the governing differential equation becomes a bit more complicated (Kramer, 1996):

u(z, t) = aei(𝜔t+k z) + bei(𝜔t−k z)

β = 5%

6

23.7.2 One-Dimensional Solution for Damped Linear Soil on Rigid Rock

𝜕2u 𝜕2u 𝜂 𝜕3u = vs 2 2 + 2 𝜌 𝜕z2 𝜕t 𝜕t 𝜕z

815

23.7.3

Layered Soils

One very useful solution is that for a layered system. This solution was coded by Schnabel et al. (1972) with the program SHAKE and modified by Idriss and Sun (1992) with the program SHAKE91. It solves the problem of a soil deposit made of n layers (i = 1 to n), Δz thick, with a shear modulus Gi for each layer i. A column of soil is considered, and each layer is represented by an element being deformed in simple shear (Figure 23.15). The horizontal cross-section of the column is 1 m × 1 m. The horizontal equilibrium of an element leads to: ) ( 𝜕𝜏 𝜏 + dz 1 × 1 − 𝜏 × 1 × 1 = ma 𝜕z 𝜕2u (23.22) = 1 × 1 × dz × 𝜌 × 2 𝜕t or: 𝜕𝜏 𝜕2u (23.23) =𝜌 2 𝜕z 𝜕t where 𝜏 is the shear stress on the horizontal plane, z is depth, 𝜌 is the soil mass density, u is the horizontal displacement, and t is time. This equation is solved by the finite difference method (see Chapter 12, Section 12.5.1) and ends up for a

2

s

i–1 i i+1

(23.21)

s

This amplification function |F(𝜔)| is shown in Figure 23.14 for several values of the damping ratio 𝛽. As can be seen, the amplification is maximum for the lower frequencies. Solutions can also be found for more complex situations, such as damped linear soil on elastic rock, layered damped soil on elastic rock, and nonlinear soil behavior (Kramer 1996).

Deflected shape AT t

Initial shape (t = 0) z1 = 0 1

u u

ui, t

τ

γ

n–1 n Rock

t

Figure 23.15 Column of soil elements deformed in simple shear during shear wave propagation.

816

23 EARTHQUAKE GEOENGINEERING

centered expression as: 𝜏i+1,t − 𝜏i−1,t

=𝜌

displacements and shear stress, which are usually zero for most nodes except for the boundary condition nodes. At the ground surface, the shear stress is always zero, whereas the displacement at the rock level is set equal to the displacement of the bottom element. The displacement at the rock level at the beginning of the earthquake provides the first value. More realistic analyses include the strain level dependency of the shear modulus and damping ratio and the influence of the confinement on the shear modulus. The process of deconvolution is the reverse process, where the ground surface motion is observed during an earthquake and the rock motion is back-calculated at the base of the soil column. The use of programs like SHAKE and other techniques has helped produce graphs like the one in Figure 23.16, which shows the acceleration at the ground surface for a given acceleration at the rock level.

ui,t+Δt − 2ui,t + ui,t−Δt

(23.24) Δz Δt2 where 𝜏 i,t is the shear stresses at time t on element i, Δz is the increment of depth, 𝜌 is the mass density of the soil, ui,t is the displacement of element i at time t, and Δt is the increment of time. Boundary conditions exist at the bottom and at the top of the soil column. At the bottom, the displacement is equal to the input rock displacement at any time t. At the top, the shear stress is zero. The shear strain 𝛾 i, t is linked to the horizontal displacements of the nodes of the soil column as follows: ui+1,t − ui,t 𝛾i,t = (23.25) Δz The shear stress 𝜏 i, t can then be calculated as: 𝜏i,t = Gi 𝛾i,t + 𝜂i

𝜕𝛾i,t

23.8

(23.26) 𝜕t where G is the shear modulus of the soil and 𝜂 is the viscosity of the soil. It can be shown (Kramer, 1996) that the viscosity 𝜂 of the soil is linked to the damping ratio 𝛽 by:

DESIGN PARAMETERS

The design approach often considers two levels of earthquakes: a rare earthquake and an expected earthquake. A rare earthquake may be defined as an earthquake with a 2% probability of exceedance in 50 years, whereas an expected earthquake would correspond to a 10% probability of exceedance in 50 years (Figure 23.5). These definitions correspond approximately to a return period of 2500 years and 500 years, respectively. The design parameters, including ground motion, are selected in one of two ways. The first way is to perform a site-specific analysis (as discussed in Section 23.7) or by using recommendations outlined in building codes. This next section addresses the code approach.

2G 𝛽 (23.27) 𝜔 where G is the shear modulus and 𝜔 is the circular frequency of the motion. The input of the problem consists of a shear modulus, a damping ratio, and a mass density for each element in the soil column. Then, the boundary conditions together with Eqs. (23.24), (23.25), (23.26), and (23.27) are written for all nodes in the soil column to solve for the unknown displacements in all elements. The solution consists of starting at t = 0 and stepping into time an amount Δt per step. The boundary conditions provide the first values of the 𝜂=

23.8.1

Site Classes A–E for Different Soil Stiffness

The design spectrum depends on the soil at the site and in particular its stiffness. This is why site classes have been defined,

Acceleration at soft soil sites (g)

0.6 Range based on analytical studies

0.5 1985 Mexico City

0.4

0.3 Recommended

0.2

1989 Loma Prieta 0.1 Earthquake magnitude ≈ 7 0

0

0.1

0.2

0.3

0.4

0.5

Acceleration at rock sites (g)

Figure 23.16 Amplification of rock motion at soil sites. (Source: From Idriss, 1990.)

0.6

23.8 DESIGN PARAMETERS

Table 23.2

817

Site class definitions Average properties in top 30 m

Site class Soil profile name

Shear wave velocity, vs , (m∕s)

Standard penetration resistance, N (blow/0.3m)

Undrained shear strength, su , (kPa)

A B C

vs > 1500 750 < vs ≤ 1500 360 < vs ≤ 750

N/A N/A N > 50

N/A N/A Su > 100

180 ≤ vs ≤ 360 vs < 180

15 ≤ N ≤ 50 N < 15

50 ≤ su ≤ 100 Su < 50

Hard rock Rock Very dense soil and soft rock Stiff soil profile Soft soil profile

D E

Any profile with more than 3 m of soil having the following characteristics: 1. Plasticity index PI > 20s 2. Moisture content 𝜔 ≥ 40%, and 3. Undrained shear strength su < 25 kPa

E

Any profile containing soils having one or more of the following characteristics: 1. Soils vulnerable to potential failure or collapse under seismic loading, such as liquefiable soils, quick and highly sensitive clays, collapsible weakly cemented soils. 2. Peats and/or highly organic clays (H > 3 m of peat and/or highly organic clay where H = thickness of soil) 3. Very high-plasticity clay (H > 7.5 m with plasticity index PI > 75) 4. Very thick soft/medium stiff clays (H > 36 m)

F

(Source: Adapted from Kavazanjian et al., 2011.)

ranging from site class A for hard rock to site class E for soft soil, as shown in Table 23.2. Site class F is a special category for which a site-specific dynamic site response analysis is recommended (see Section 23.7) instead of a code approach. The site classes use the average soil parameters within the top 30 m from the surface as a classification basis, because this depth is most influential in determining the dynamic response. The shear wave velocity is the parameter of choice, but the SPT blow count and the undrained shear strength are also helpful. Once the soil is classified according to the definitions listed in Table 23.2, the amplification factors can be selected to modify the acceleration spectrum. 23.8.2

Code-Based Spectrum

In the code approach, the acceleration response spectrum is constructed from the analysis of existing data, past experience, and engineering judgment. Such a spectrum is shown in Figure 23.17. First the reference spectrum is developed, assuming that the soil at the site is rock (site class B) and then the values obtained for the reference spectrum are modified for the proper site class. The reference spectrum parameters are: 1. The spectral acceleration at a period equal to 0 seconds taken as the peak ground acceleration PGA. The PGA is used here because for a fundamental period of

SDS = Fa × SS

Elastic seismic coefficient, Csm

AS = Fpga × PGA

0.2 T0 = 0.2Ts

Csm = SD1/Tm SD1 = Fv × S1

1.0 Ts = SD1/SDS Period, Tm (second)

Figure 23.17 Design code acceleration response spectrum. (Source: From Kavazanjian et al., 2011 / U.S. Department of Transportation.)

0 seconds, the structure is infinitely stiff and the maximum acceleration of the structure is the same as the maximum acceleration from the ground. 2. The spectral acceleration at a short period equal to 0.2 seconds, called Ss . 3. The spectral acceleration at a long period equal to 1 second, called S1 .

818

23 EARTHQUAKE GEOENGINEERING

These values come from the selection of the design earthquake (e.g., 1000-year or 2500-year return period) and the use of the dedicated USGS website, for example (Kavazanjian et al., 2011). The site-specific spectrum is obtained from the values of the reference spectrum. The site-specific spectrum parameters are: 1. The spectral acceleration at a period equal to 0 seconds, called As: (23.28) AS = FPGA × PGA where F PGA is the site factor for the PGA found in Table 23.3. 2. The spectral acceleration at a short period equal to 0.2 seconds, called SDS : SDS = FA × SS

(23.29)

where F A is the site factor for SS , found in Table 23.4. 3. The spectral acceleration at a long period equal to 1 second, called SD1 : SD1 = FV × S1

To = 0.2TS

(23.32)

In that fashion the design spectrum is completely defined (Figure 23.17) and can be used for structural or geotechnical analysis. In Figure 23.17, the elastic seismic coefficient Csm is the ratio between the design horizontal shear force due to inertia and the effective weight of the structure.

Mapped spectral response accelerations at short periods Site class A B C D E F

Site class A B C D E F

Ss ≤ 0.25 g

Ss = 0.50 g

Ss = 0.75 g

Ss = 1.00g

Ss ≥ 1.25 g

0.8 1.0 1.2 1.6 2.5 a

0.8 1.0 1.2 1.4 1.7 a

0.8 1.0 1.1 1.2 1.2 a

0.8 1.0 1.0 1.1 0.9 a

0.8 1.0 1.0 1.0 0.9 a

a: Site-specific geotechnical investigation and dynamic site response analysis are required in this case. (Source: Adapted from Kavazanjian et al., 2011.)

Table 23.5

PGA ≤ 0.1 g

PGA = 0.2 g

PGA = 0.3 g

PGA = 0.4 g

PGA ≥ 0.5 g

0.8 1.0 1.2 1.6 2.5 a

0.8 1.0 1.2 1.4 1.7 a

0.8 1.0 1.1 1.2 1.2 a

0.8 1.0 1.0 1.1 0.9 a

0.8 1.0 1.0 1.0 0.9 a

a: Site-specific geotechnical investigation and dynamic site response analysis are required in this case. (Source: Adapted from Kavazanjian et al., 2011.)

Site factor FV

Mapped spectral response accelerations at 1-second periods Site class A B C D E F

S1 ≤ 0.1 g

S1 = 0.2 g

S1 = 0.3 g

S1 = 0.4 g

S1 ≥ 0.5 g

0.8 1.0 1.7 2.4 3.5 a

0.8 1.0 1.6 2.0 3.2 a

0.8 1.0 1.5 1.8 2.8 a

0.8 1.0 1.4 1.6 2.4 a

0.8 1.0 1.3 1.5 2.4 a

a: Site-specific geotechnical investigation and dynamic site response analysis are required in this case. (Source: Adapted from Kavazanjian et al., 2011.)

23.8.3

Site factor FPGA

Site factor FA

Mapped spectral response accelerations at short periods

(23.30)

where F V is the site factor for S1 , found in Table 23.5. 4. The period T S corresponding to the end of the spectrum plateau is given by: S (23.31) TS = D1 SDS 5. The period T o corresponding to the beginning of the spectrum plateau:

Table 23.3

Table 23.4

Hazard Levels

The severity of an earthquake is described by the hazard level, which ranges from I to IV. Each level is tied to the SD1 or SDS value. Recall that SD1 is the long-period (1 second) spectral acceleration, adjusted for the site factor, and SDS is the short-period (0.2 second) spectral acceleration, also adjusted for the site factor. The hazard levels are defined in Table 23.6. 23.9 23.9.1

LIQUEFACTION Phenomenon

When a loose coarse-grained soil under the groundwater level is shaken rapidly enough, it tends to decrease in volume. The decrease in volume causes the water to be pushed out of the pores. If the water cannot escape fast enough, the water stress uw increases and can reach a value equal

23.9 LIQUEFACTION

Table 23.6

If the soil is resistant to liquefaction, a liquefaction study is not necessary even for hazard levels III and IV. Liquefaction-resistant soils include:

Seismic hazard levels

Hazard level

SD1 = FV S1

SDS = FA SS

I II III IV

SD1 < 0.15 0.15 < SD1 < 0.25 0.25 < SD1 < 0.40 0.40 < SD1

SD1 < 0.15 0.15 < SD1 < 0.35 0.35 < SD1 < 0.60 0.60 < SD1

Note: These hazard levels apply for site classifications A, B, C, and D. Further description and conditions apply for site classifications E and F (see Kavazanjian et al., 2011).

to the (vertical total stress )𝜎 ov . At that point, the effective ′ = 𝜎ov − uw = 0 becomes zero: The soil loses stress 𝜎ov its strength and behaves like a thick liquid. This is the phenomenon of liquefaction. Loose sands under the groundwater level are particularly sensitive to this condition. Dense coarse-grained soils and fine-grained soils are much less sensitive. Liquefaction of the soil leads to flow slides, lateral spreading, sand boils, loss of bearing capacity, increased earth pressures against retaining walls as the soil becomes a heavy liquid, and post-earthquake settlement as the water stress dissipates. 23.9.2

When to Do a Liquefaction Study?

The need for liquefaction studies is tied first to the severity of the earthquake. This severity is described by hazard levels ranging from I to IV, as described in Table 23.6. For hazard levels I and II, a liquefaction study is not required. For hazard level IV, a liquefaction study is always required. For hazard level III, a liquefaction study is required unless: 1. The mean earthquake magnitude for the design event is less than 6, or 2. The mean magnitude is between 6 and 6.4, and N1−60 > 20 (mean normalized SPT blow count; see Chapter 8, Section 8.1). 3. The mean magnitude is between 6 and 6.4, N1−60 > 15, and SDS < 0.35g. τ Dense sand DR = 70% +τ c

1. Bedrock. 2. Fine-grained soils with more than 15% clay, liquid limit wL higher than 35%, and water content lower than 0.9wL. 3. Sands with N1−60 > 30 bpf or qt1 > 160 (mean corrected and normalized cone penetrometer resistance; see Chapter 8, Section 8.2). 4. Soils where the water table is deeper than 15 m below the ground surface. Note that quick clays should be considered as potentially liquefiable; however, the liquefaction is not due to the same process as the liquefaction of fine sands discussed here. 23.9.3

When Can a Soil Liquefy?

To predict whether a soil can liquefy, cyclic tests can be performed in the laboratory by cyclic triaxial testing, or (better) by cyclic simple shear testing, or (even better) by cyclic torsional shear testing. During such tests, the sample is subjected to an initial effective stress and then a chosen value of shear stress or deviator stress is applied cyclically in two-way symmetrical shearing. This means that the shear stress varies between +𝜏 c and −𝜏 c . The frequency of the cycles is selected to be representative of earthquake frequencies (say, 1–10 Hz). Typical results for cyclic simple shear tests on saturated sand are shown in Figures 23.18 and 23.19. Liquefaction may or may not occur after a number of cycles or an amount of time consistent with typical earthquakes (less than 30 seconds for most cases and up to 2 minutes for a huge earthquake). The cyclic stress ratio (CSR) is the ratio of the horizontal shear stress 𝜏 c applied cyclically to a soil at a depth z over the ′ on the soil at the same depth. The vertical effective stress 𝜎vo lowest value of the CSR that triggers liquefaction is called the cyclic resistance ratio or CRR. Figure 23.20 shows the results of shaking table tests performed by De Alba et al. (1976); Figure 23.20 indicates how the CRR decreases as the number of cycles increases. The goal of liquefaction studies is to

Loose sand DR = 40%

τ

+τc

γ

–τc

819

γ

–τc

Figure 23.18 Undrained constant stress cyclic simple shear test results (stress-strain curves).

820

23 EARTHQUAKE GEOENGINEERING

Uw σ

+τc

σv′

–τc

Time

Figure 23.19 Undrained constant stress cyclic simple shear test results (stress path and water stress.)

0.4

Relative density = 54% Relative density = 68% Relative density = 82% Relative density = 90%

Cyclic stress ratio

0.3

Initial confining pressure = 55 kPa 0.2

0.1

0

1

10

100

1000

Number of cycles

Figure 23.20 Cyclic stress ratio to reach liquefaction. (Source: From De Alba et al., 1976. With permission from ASCE.)

calculate the CSR and the CRR within the depth of interest. Liquefaction is predicted if: (23.33)

The drawbacks of using laboratory tests to predict liquefaction include sample disturbance, difficulty in reproducing in situ stresses, and difficulties in reproducing a true earthquake cyclic shear loading. As a result, the preferred approach in design has been to use earthquake case histories at sites where liquefaction did or did not occur. The combination of the average shear stress 𝜏 av due to the earthquake shaking, a measure of the soil strength, and the knowledge of whether the soil liquefied are used to produce design charts. The results of in situ tests are preferred in this approach to quantify the soil strength. The first charts, such as the one shown in Figure 23.21, were based on the SPT blow count. Additional charts were then proposed, based on cone penetrometer data (Figure 23.22) and then shear wave velocity data (Figure 23.23). The chart based on the dilatometer is preliminary in nature (Figure 23.24) and so is the chart based on the pressuremeter limit pressure (Figure 23.25).

Percent fines = 35 Cyclic stress or resistance ratio CSR or CRR τav /σ′0v

CRR < CSR

0.6 15

≤5

0.5 Liquefaction

0.4

0.3 No liquefaction 0.2

0.1

0.0

No liquefaction Liquefaction 0

10

20

30

40

50

Corrected blow count, (N1)60 (blow/0.3m)

Figure 23.21 SPT-based liquefaction chart for magnitude 7.5. (Source: Adapted from Youd and Idriss, 1997.)

Cyclic stress or resistance ratio CSR or CRR τav /σ′0v

0.6

Cyclic stress or resistance ratio CSR or CRR τav /σ′av

23.9 LIQUEFACTION

0.25 < D50(mm) < 2.0 FC (%) < 5

MW = 7.5

γe = 20% = 10% = 3%

0.5

Liquefaction

CRR curve

0.4

0.3

0.2 No liquefaction 0.1

0

Liq.

0

50

100

150

200

No Liq.

250

300

Normalized corrected CPT TIP resistance, qc1N

Figure 23.22 CPT-based liquefaction chart for magnitude 7.5. (Source: From Robertson and Wride, 1998/Canadian Science Publishing.)

Cyclic stress or resistance ratio CSR or CRR τav /σ′av

0.6

Mw = 7.5

Fines content (%) ≥ 3520≤5

0.4 Liquefaction

No liquefaction

0.2

Liquefaction No liquefaction

0

0

100

200

300

Overburden stress-corrected shear wave Velocity, vs1, m/s

Figure 23.23 Shear wave velocity-based liquefaction chart for magnitude 7.5. (Source: Adapted from Andrus and Stokoe, 2000.)

In these charts, the vertical axis is the cyclic stress ratio ′ where 𝜏av is the average shear (CSR), defined as 𝜏av ∕𝜎ov ′ stress generated during the design earthquake and 𝜎ov is the vertical effective stress at the depth investigated and at the time of the in situ soil test. The shear stress 𝜏av is related to the maximum shear stress 𝜏max , which is obtained from a site response analysis (e.g., using the program SHAKE) for the design earthquake, or, more simply, by using the peak ground acceleration PGA obtained from maps such as the one shown in Figure 23.5. If the PGA is used to obtain 𝜏av for a

0.5

821

Mw = 7.5 Liquefaction

0.4 0.3

0.2 No liquefaction 0.1

0

0

2 4 6 8 Horizontal stress index, KD

10

Figure 23.24 DMT-based liquefaction chart for magnitude 7.5. (Source: Adapted from Monaco et al., 2005.)

7.5 magnitude, the expression is (Seed and Idriss, 1971): ( )( ) 𝜏av 𝜎vo amax CSR = ′ = 0.65 (23.34) rd ′ g 𝜎vo 𝜎vo where amax is the PGA for the design earthquake, g is the acceleration due to gravity, 𝜎vo is the total vertical stress at the ′ is the effective vertical stress at depth being investigated, 𝜎vo the depth being investigated, and rr is a flexibility factor. The flexibility factor depends on the depth at which the liquefaction is being evaluated. Such a factor is necessary because the PGA is acting at the ground surface while the possibility of liquefaction is evaluated at a depth z. Figure 23.26, after Seed and Idriss (1971), gives a range of values for rd . On the horizontal axis of the charts in Figures 23.21–23.25 is the in situ test parameter normalized and corrected for the effective stress level in the soil at the time of the test and for fine content. The SPT blow count is N 1–60 and the procedure to correct it for effective stress level is described in Chapter 8, Section 8.1. The correction for fine content is embedded in the chart of Figure 23.21. The CPT point resistance is qt1 and the procedure to correct for effective stress level and fine content is described in Chapter 8, Section 8.2. The shear wave velocity is vs1 and the procedure to correct for effective stress is: ( )0.25 𝜎a (23.35) vs1 = vs ′ 𝜎vo where vs is the shear wave velocity measured in the field, 𝜎 a ′ is the atmospheric pressure, and 𝜎vo is the vertical effective stress at the depth investigated. The pressuremeter limit pressure is pL1 and the procedure to correct for effective stress is ( )0.5 pa pL1 = pL (23.36) ′ 𝜎ov where pL is the pressuremeter limit pressure, pa is the atmo′ is the vertical effective spheric pressure (∼100 kPa) and 𝜎ov stress at the depth where pL is measured.

822

23 EARTHQUAKE GEOENGINEERING

0.6

0.6

Mw = 7.5

Percent fines = 35 15 ≤ 5

γe = 20% = 10% = 3%

0.5

0.5

0.25 < D50(mm) < 2.0 FC (%) < 5

Mw = 7.5 Liquefaction

Liquefaction

0.4

Cyclic stress or resistance 0.3 ratio CSR or CRR τav /σ′0v 0.2

CRR curve

0.4

Cyclic stress or resistance 0.3 ratio CSR or CRR τav /σ′av 0.2

No liquefaction

0.1

0.1

No liquefaction

0

0.0 0

1500 2000 500 1000 Normalized limit pressure, pL1 (kPa)

2500

0

500 1000 1500 2000 Normalized limit pressure, pL1 (kPa)

(a)

2500

(b)

Figure 23.25 PMT-based liquefaction charts for magnitude 7.5. (a) Using Youd and Idriss, 1997. (b) Using Robertson and Wride, 1998. (Source: Adapted from Briaud, 2013).

4.5

0

0.2

0.4

0.6

0.8

1.0

5 Range for different soil profile

Depth z (m)

10

Approximate average

15

20

Simplified procedure not verified with case history data in this region

25

Magnitude scaling factor, MSF

Soil flexibility factor (rd) 0

4 3.5 3 2.5 2 1.5 1 0.5 0 5.0

6.0

7.0

8.0

Figure 23.27 Magnitude scaling factor (MSF). (Source: After Kavazanjian et al., 2011; Youd and Idriss, 1997.)

of a different magnitude, a magnitude scaling factor (MSF) is applied as follows: CRRM = MSF × CRRM=7.5

30

9.0

Earthquake magnitude, Mw

(23.37)

Figure 23.26 Soil flexibility factor to modify the PGA for a depth z. (Source: Seed and Idriss, 1971/American Society of Civil Engineers.)

Kavazanjian et al. (2011), building on the work of Youd and Idriss (1997), suggested that the hatched area in Figure 23.27 be used for MSF. In summary, the way to use the charts is (Figure 23.28):

Once the soil parameter is corrected, it is entered on the horizontal axis of the chart and the CRR is read on the liquefaction design curve of the corresponding chart. Note that the charts in Figures 23.21–23.25 give the cyclic resistance ratio (CRR) for an earthquake of magnitude 7.5. For earthquakes

1. Obtain the soil parameter profile. 2. Correct the profile for stress level due to depth effects and fine content if necessary. Prepare a corrected soil parameter profile. 3. Enter the chart corresponding to the soil parameter and read the cyclic resistance ratio. Prepare a CRR profile.

23.10 SEISMIC SLOPE STABILITY

SPT N60 0

0

10 20 30 40

Fines (%) 0

10

20

30

CSR & CRR 40

FSeq

0.0 0.2 0.4 0.6 0.8 0.0 0.5 1.0 1.5 2.0 Sand

1 2

823

Earthquake CSR

Depth below ground surface (m)

3 4 5 6

8

Sand & sand with gravel

9

Clay CH

7

10 11

Silty Sand

12 13

Clay CH

14

Figure 23.28 Profiles of liquefaction analysis. (Source: Idriss and Boulanger, 2008/Earthquake Engineering Research.)

4. Modify the CRR profile for a magnitude different from 7.5. 5. Calculate the CSR generated by the design earthquake. Prepare the CSR profile. 6. Compare the CSR profile and the CRR profile. 7. The zone of potential liquefaction is the zone where CSR > CRR.

23.10

SEISMIC SLOPE STABILITY

Seismic slope stability was covered in Chapter 20, Section 20.18. The following summarizes the procedure to select the appropriate value of the horizontal seismic coefficient k for a pseudostatic analysis. The step-by-step procedure that follows is as recommended by Kavazanjian et al. (2011): 1. Perform a static slope stability analysis without any earthquake loading to ensure that the slope is stable and that the factor of safety F is sufficient in the case of no earthquake (say, 1.5). 2. Using maps (USGS map, such as the ones shown in Figure 23.5, for example), obtain the peak ground acceleration PGA and the spectral acceleration at one second S1 for site class B at the base of the slope for the design earthquake.

3. Select the site adjustment factor F PGA from Table 23.3 and the factor F V from Table 23.5 for the correct site class and the correct acceleration value. 4. Calculate the value of the maximum horizontal seismic inertia coefficient kmax as: kmax = FPGA × PGA

(23.38)

5. Calculate the value of the average horizontal seismic inertia coefficient kav as follows. The coefficient kav is lower than kmax because the average horizontal acceleration over the slope mass is less than the PGA due to wave scattering: ( ( )) F S kav = 𝛾 1 + 0.01H 0.5 V 1 − 1 kmax (23.39) kmax where 𝛾 is equal to 1 for all site classes except for site classes A and B, where it is taken as 1.2; H is the height of the slope; and F v is the site factor from Table 23.5. 6. If the slope can tolerate a movement of 25–50 mm, the value of kav can be further reduced by a factor of 2. In the end, the factor kh is given by: ( ( )) F S kh = 0.5𝛾 1 + 0.01H 0.5 V 1 − 1 kmax (23.40) kmax 7. Under the combined static and earthquake inertia loading, the target factor of safety should be at least 1.1.

824

23 EARTHQUAKE GEOENGINEERING

23.11

SEISMIC DESIGN OF RETAINING WALLS

The design of retaining walls under static conditions is described in Chapter 22. This section addresses what happens under earthquake conditions. Gravity walls are considered first, followed by MSE walls, cantilever walls, and tieback walls. 23.11.1

Seismic Design of Gravity Walls

The horizontal force Pa per unit length of wall due to the active earth pressure behind a retaining wall when there is no earthquake, and the water table is below the bottom of the wall (see Chapter 22, Sections 22.3.1 and 22.9) is given by: 1 (23.41) K 𝛾H 2 2 a where 𝛾 is the soil unit weight, H is the wall height, and K a is the active earth pressure coefficient expressed as (Figure 23.29): Pa =

Ka =

sin2 (𝛼 + 𝜑′ ) ]2 [ √ ′ +𝛿) sin(𝜑′ −𝛽) sin2 𝛼 sin(𝛼 − 𝛿) 1 + sin(𝜑 sin(𝛼−𝛿) sin(𝛼+𝛽)

1 (23.43) K 𝛾H 2 2 ae where Pae is the active force per unit length of wall due to the active earth pressure during an earthquake, K ae is the active earth pressure coefficient in the earthquake case, 𝛾 is the soil unit weight, and H is the wall height. The coefficient K ae is obtained in the same fashion as K a (see Chapter 21, Section 21.3) except that the earthquake force kh W is added to the equilibrium equations. The final expression of K ae after finding the most critical wedge angle is credited to Mononobe and Okabe (Okabe, 1926; Mononobe and Matsuo, 1929): Pae =

Kae =

(23.42)

where 𝛼 is the angle of the back of the wall with the horizontal, 𝜑′ is the effective stress friction angle of the soil behind the wall, 𝛿 is the angle of friction between the soil and the back of the wall, and 𝛽 is the angle of the ground surface behind the wall with the horizontal. In the case of earthquake loading on a gravity wall, the earth pressure is increased by the horizontal shaking of the soil and the associated horizontal inertia force. The vertical acceleration can also modify the weight of soil acting on the wall, but this vertical inertia force is usually ignored, mainly because it does not occur at the same time as the horizontal force; indeed, the horizontal and vertical accelerations are rarely in phase, so the peak horizontal and peak vertical accelerations do not occur simultaneously. The horizontal inertia force generated by the earthquake is written as kh W where kh is the seismic coefficient and W is the weight of the soil wedge. The coefficient kh is similar to the coefficient used for slope stability; it is taken as kav (Eq. (23.39)) if the wall cannot

sin2 (𝛼 + 𝜑′ − 𝜓) cos 𝜓sin2 𝛼 sin(𝛼 − 𝛿 − 𝜓) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ − 𝛽 − 𝜓) ⎤ sin(𝜑 ⎢1 + ⎥ ⎢ sin(𝛼 − 𝛿 + 𝜓) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦

(23.44)

where 𝛼 is the angle of the back of the wall with the horizontal, 𝛽 is the angle of the ground surface behind the wall with the horizontal, 𝛿 is the angle of friction between the back of the wall and the soil, 𝜑′ is the friction angle of the soil, and 𝜓 is the angle representing the earthquake inertia force through: ( ) kh −1 𝜓 = tan (23.45) 1 − kv where kh and kv are the horizontal and vertical seismic coefficients respectively. Note that kv is often ignored (taken as equal to zero). The angle of the critical surface with the horizontal is flatter in the active earthquake case than in the static case (Kramer, 1996). Note also that Pae includes the static component Pa and a dynamic component ΔPae of the active push (Figure 23.29) and can be rewritten as: Pae = Pa + ΔPae

(23.46)

For the passive earth pressure, the equations become: 1 (23.47) K 𝛾H 2 2 pe where Ppe is the passive force per unit length of wall due to the passive earth pressure during an earthquake, K pe is the passive earth pressure coefficient in the earthquake case, 𝛾 is the soil unit weight, and H is the wall height. The expression of K pe (Figure 23.30) is: Ppe =

Movement 1 ΔPae = — (Kae – Ka) γH 2 2 H

tolerate any movement and as kh (Eq. (23.40)) if a movement of 25–50 mm is tolerable. In the case of an earthquake, the force Pa becomes Pae , which is written as:

1 K γH 2 Pa = — 2 a

0.6H

0.33H KaγH

Figure 23.29 Gravity retaining wall with earthquake loading: active case.

Kpe =

sin2 (𝛼 − 𝜑′ + 𝜓) cos 𝜓sin2 𝛼 sin(𝛼 + 𝛿 + 𝜓) 2 √ ⎡ ′ + 𝛿) sin(𝜑′ + 𝛽 + 𝜓) ⎤ sin(𝜑 ⎢1 − ⎥ ⎢ sin(𝛼 + 𝛿 + 𝜓) sin(𝛼 + 𝛽) ⎥ ⎣ ⎦

(23.48)

23.11 SEISMIC DESIGN OF RETAINING WALLS

Movement 1 ΔPpe = — (Kpe – Kp) γH 2 2 H

1 Pp = — KpγH 2 2

0.6H

0.33H KaγH

Figure 23.30 Gravity-retaining wall with earthquake loading: passive case.

Note that Ppe includes the static component Pp and dynamic component ΔPpe of the passive push (Figure 23.30) and can be rewritten as: (23.49) Ppe = Pp + ΔPpe

Earthquake active earth pressure coefficient, Kae

The point of application of the static component of the active and passive forces, Pa and Pp , is located at 0.33 H (H is wall height) from the bottom of the wall in the simplest case of a uniform soil. Note that the static pressure distribution is triangular, but the pressure distribution associated with the earthquake inertia force is not triangular. Recall that K ae was obtained from a Coulomb wedge analysis, which gives a global force solution, and not a Rankine stress analysis, which gives a pressure distribution. In fact, the point of application of the seismic component is different from the point of application of the static component. The point of application of the dynamic components of the active and

passive forces, ΔPae and ΔPpe , is higher than the one for the static components, because the amplitude of the soil movement due to the earthquake generally increases as the shear wave propagates upward. As a result, the point of application of ΔPae and ΔPpe is located at 0.6 H from the bottom of the wall. The Mononobe-Okabe expressions of K ae and K pe in Eqs. (23.44) and (23.48) have the advantage of being simple to use. They also have their shortcomings. One of them is that the failure surface is assumed to be the same for the static case and the dynamic case, which is not true. In the active case, the slope of the failure surface is flatter for the dynamic case than for the static case. Second, the effect of cohesion is not included, although it can reduce the effect of the dynamic part of the active pressure. Figure 23.31 shows the influence of the cohesion c′ on K ae for a friction angle of 35∘ and for different values of the horizontal seismic coefficient kh . The cohesion c′ is normalized in the figure by 𝛾H where 𝛾 is the soil unit weight and H is the wall height. Another shortcoming is that the wedge approach assumes a straight-line failure surface, which is not necessarily the weakest surface. This difference is particularly severe for the passive resistance Ppe , which can be seriously overestimated and should be used with caution if at all. A log spiral failure surface gives more conservative values for K pe and should be preferred. Such values are shown in Figure 23.32 (after NCHRP, 2008). 23.11.2 Water Pressures on Walls during an Earthquake It is generally desirable to ensure that the groundwater table is below the bottom of the retaining wall, as the water

2 c/γH = 0 c/γH = 0.05

1.5

c/γH = 0.10 c/γH = 0.15 1

c/γH = 0.20 c/γH = 0.25

0.5

c/γH = 0.30

0 0

0.2

0.4

0.6

825

0.8

1

Horizontal seismic coefficient, kh (g)

Figure 23.31 Influence of cohesion c′ for 𝜑′ = 35∘ on the earthquake active earth pressure coefficient K ae for different values of the horizontal seismic coefficient kh . (Source: Adapted from NCHRP, 2008.)

826

23 EARTHQUAKE GEOENGINEERING

Earthquake passive earth pressure coefficient, Kpe

14

Φ = 40°

12

Φ = 37.5

10

Φ = 35° Φ = 32.5

8

Φ = 30°

6

Φ = 27.5

4

Φ = 25° Φ = 22.5

2 0

Φ = 20 0

0.2

0.4

0.6

0.8

1

Horizontal seismic coefficient, kh (g)

Active

Rankine

Passive

Coulomb Log spiral

Figure 23.32 Coefficient of passive earth pressure in the case of earthquake loading for a log spiral failure surface and a wall friction angle equal to 2/3 of the soil friction angle. (Source: NCHRP, 2008, Kavazanjian et al., 2011.)

pressure significantly increases the active force. This is also true for walls in earthquake-prone areas. However, this may not be possible; a high water level is often encountered for walls in harbors or near shore. In such instances it is necessary to account for the water behavior during an earthquake in addition to the hydrostatic pressure associated with the static case. Water on the Side That Has No Soil

fw =

If there is water on the side of the wall that has no soil (e.g., berthing wall in a harbor, earth dam), the pressure in the static case pwh is hydrostatic and given by: pwh = 𝛾w z

(23.50)

where 𝛾 w is the unit weight of water and z is the depth below the water level. The dynamic pressure during an earthquake Δpwe is given by Westergaard (1931): Δpwe =

√ 7 kh 𝛾w zHw 8

where kh is the horizontal seismic coefficient, z is the depth below the water level, and H w is the total height of water against the wall (Figure 23.33). The assumptions made by Westergaard to develop this solution limit the application of this formula to the case where the earthquake frequency is below the fundamental frequency f w of the water body. This frequency is given by:

(23.51)

vp 4Hw

(23.52)

where vp is the compression wave velocity. Note that the dynamic pressure works alternatively in both directions. The most detrimental condition for the retaining wall is likely to be when the dynamic pressure decreases the hydrostatic pressure, thereby decreasing the stabilization effect of the water. By integrating the expression in Eq. (23.51), we can obtain the resultant force ΔPwe : ΔPwe =

7 k 𝛾 H2 12 h w w

(23.53)

23.12 SEISMIC DESIGN OF FOUNDATIONS

827

Earthquake + static p′ae

pw

Δpwe

p′ae

Pw

H

0.67H

0.6Hw

Pwh

Δpwe 0.67Hw

Hw

pwh

0.67H

Static + earthquake

zw

Figure 23.33 Water pressure on wall due to earthquake.

The point of application of ΔPwe can be calculated by moment equilibrium and is found to be 0.6H w below the water surface (Figure 23.33). Water on the Retained-Soil Side If there is water in the backfill, the problem becomes a bit more complicated, as the inertia force is proportional to the total unit weight 𝛾 and the shear resistance is proportional to the effective unit weight 𝛾 ′ . Therefore, Eq. (23.43) must be altered to reflect this dual effect. In the case where the water level in the backfill is at the ground surface and no excess water stress is generated, Towhata (2008) recommends the following approach: 1. Use 𝛾 ′ in the active earth pressure equation. 2. Increase the horizontal seismic coefficient kh to reflect the increase in inertia force. 3. Add the hydrostatic pressure. The equation then becomes (Figure 23.33): 1 (23.54) K 𝛾 ′H2 2 ae However, the horizontal seismic coefficient kh is increased to kh′ : 𝛾 (23.55) kh′ = ′ kh 𝛾

Because of the triangular distribution of pressures, both P′ae and Pw act at 0.67 H from the top of the wall in the simplest case of a uniform soil. Kramer (1996) gives recommendations for the more complex case where the water level behind the wall is not at the ground surface. 23.11.3

Seismic Design of MSE Walls

MSE walls retain the soil through a reinforced soil mass. The earthquake design of these types of walls follows the same approach as the static design (see Chapter 22, Section 22.10), except that the coefficient K a is replaced by the coefficient K ae in the calculation. 23.11.4

Seismic Design of Cantilever Walls

Cantilever walls retain the soil without anchors or strut simply by the resistance of their embedment in the foundation soil. The earthquake design of these types of walls follows the same approach as the static design (see Chapter 22, Section 22.11), except that the coefficients used for the earth pressure are K ae and K pe instead of K a and K p .

P′ae =

This nearly doubles the value of kh . Then the angle 𝜓 used in the expression of K ae is: ( ′ ) ( ) kh kh 𝛾 −1 −1 × = tan (23.56) 𝜓 = tan 1 − kv 1 − kv 𝛾 ′ The hydrostatic thrust must then be added: 1 Pw = 𝛾w H 2 2 1 1 Pae = Pw + P′ae = 𝛾w H 2 + Kae 𝛾 ′ H 2 2 2

23.11.5

Anchored walls retain the soil through the use of anchors or struts and through their depth of embedment. The earthquake design of these types of walls follows the same approach as the static design (see Chapter 22, Section 22.12), except that the coefficient K used for the earth pressure above the excavation level is increased by the ratio Kae ∕Ka . Below the excavation level, the earth pressure coefficients are K ae and K pe .

23.12 (23.57) (23.58)

Seismic Design of Anchored Walls

SEISMIC DESIGN OF FOUNDATIONS

During an earthquake, a foundation and the soil around it will interact. Two kinds of interactions are identified: kinematic and inertial. Kinematic interaction refers to the interaction

828

23 EARTHQUAKE GEOENGINEERING

between the soil and the foundation as the foundation modifies the free field movement of the soil because of its presence. Inertial interaction refers to the interaction between the soil and the foundation as the foundation movement due to soil shaking generates accelerations throughout the building and associated inertia forces at the foundation level. In many instances, kinematic interaction can be ignored, and the foundation need only be designed to resist the inertia forces due to the inertial interaction. The approaches used for the design of foundations to resist earthquake loading are the same for shallow foundations and deep foundations. There are two main categories: the design code approach and the dynamic analysis approach. Both approaches aim at obtaining the inertia forces and moments on the foundation and then designing the foundation to handle these forces on a pseudostatic basis. In the design code approach, a response spectrum is specified, and then the fundamental period of the building is calculated. This fundamental period is entered on the horizontal axis of the spectrum and the corresponding spectral acceleration is obtained. The horizontal force H to be resisted by the foundation is the product of the spectral acceleration and the associated mass of the building. In this approach, the ductility of the structure and foundation are not considered. This ductility tends to reduce the inertia force and is included through the use of a reduction factor Rf . Table 23.7 shows such reduction factors for bridge substructures. The reduced force used for design purposes is H∕Rf . This force is applied to the foundation and the ultimate limit state is checked to ensure safety against failure. Because an earthquake is considered to be an extreme event, the load and resistance factors are close to 1 if not equal to 1.

Table 23.7

Force reduction factor Rf for bridges Importance category

Substructure Wall-type piers, larger dimension Reinforced concrete pile bents • Vertical piles only • With batter piles Single columns Steel or composite steel and concrete pile bents • Vertical pile only • With batter piles Multiple-column bents

Critical Essential Other 1.5

1.5

2.0

1.5 1.5 1.5

2.5 1.5 2.9

3.0 2.0 3.0

1.5 1.5 1.5

3.5 2.0 3.5

5.0 3.0 5.0

(Source: Adapted from Kavazanjian et al., 2011.)

In the dynamic analysis approach, the structure and foundation are simulated numerically. The foundation is usually simplified and represented by a system of translational and rotational springs and dashpots. The simulation gives the inertia forces and moments applied to the foundation. This approach has the advantage of including the ductility of the structure more directly. Again, this force is applied to the foundation and the ultimate limit state is checked to ensure safety against failure. Because an earthquake is considered to be an extreme event, the load and resistance factors are close to 1 if not equal to 1. There is a trend toward displacement base design (service limit state) rather than load-based design. In this approach the displacement due to the earthquake loads are calculated and allowance is made for what leads to no damage, medium damage, heavy damage but still standing, and total collapse. In the case of deep foundations, it is possible for the liquefied soil to load the piles by flowing past them. The load generated by liquefied soil must be added to the inertia load. This brings in the importance of the shear strength of liquefied soils. Seed and Harder (1990) proposed a correlation of the liquefied soil shear strength to the corrected standard penetration test (SPT) blow count (N1 )60 (see Chapter 8, Section 8.1). Further correction was added to the (N1 )60 value for the presence of fines, which can be approximated as follows: P (23.59) 10 where (N1 )60 is the SPT blow count corrected for stress and energy level, P is the percent finer than 0.075 mm (expressed in percent), and (N1 )60−cs is the SPT blow count further corrected for the fine content. The correction increases the value of N to bring it back to the value that would have been obtained had the sand not contained fines (cs means clean sand). Olson and Stark (2002) further developed the original work of Seed and Harder, added data, and proposed the following equation on the basis of the corrected SPT blow count and the corrected CPT point resistance as follows (Figures 23.34 and 23.35): su−liq = 0.03 + 0.0075 (N1 )60 for N1 ≤ 12 bpf (23.60) ′ 𝜎vo su−liq = 0.03 + 0.0143 × qc1 for qc1 ≤ 6.5 MPa (23.61) ′ 𝜎vo (N1 )60−cs = (N1 )60 +

′ is the where su-liq is the shear strength of the liquefied soil, 𝜎vo pre-failure vertical effective stress in the soil, and (N1 )60 qc1 are the prefailure corrected SPT blow count and CPT point resistance respectively (see Chapter 8, Sections 8.1 and 8.2). The pressure generated by the liquefied soil flowing past the pile can be estimated as 7 times the shear strength of the liquefied soil: (23.62) pu = 7su−liq

where pu is the pressure generated on the pile by the flowing soil and su-liq is the shear strength of the liquefied soil.

23.12 SEISMIC DESIGN OF FOUNDATIONS

829

0.3

Liquefied shear strength Prefailure vertical effective stress

Stark and Mesri (1992) Davies and Campanella (1994) 0.2

Olson and Stark (2002)

3 20 7

50 85

0

20

40

35+

32

100 55 7 13 7

?

50 7

7

1

74

0

56 3 30 15

13 ?

0

20 20

15

85

0.1

2

4

6

Stark and Mesri (1992) 8

10

12

14

16

18

Normalized SPT blow count, (N1)60

Figure 23.34 Shear strength of liquefied coarse-grained soils based on SPT blow count. (Source: Adapted from Olson and Stark, 2002.)

Liquefied shear strength Prefailure vertical effective stress

0.3

0.2

Olson and Stark (2002)

16 86

0.1

50 56

3

3

?

13

100

20

0 30

3 15 13 7

40

50 36+ 8

20

20 32

20

7 ?

7 7

1

Olson (1998)

74

0

0

1

2

3

4

5

6

7

8

9

10

Normalized CPT tip resistance, qc1 (MPa)

Figure 23.35 Shear strength of liquefied coarse-grained soils based on CPT point resistance. (Source: Adapted from Olson and Stark, 2002.)

Problems and Solutions Problem 23.1 After an earthquake, a seismograph installed in the bedrock records the arrival of a compression wave and 10 seconds later the arrival of a shear wave. The rock has a compression wave velocity equal to 3000 m/s and a shear wave velocity equal to 1500 m/s. How far is the earthquake epicenter from the seismograph? How would you find the exact location of the epicenter? Solution 23.1 The distance between the epicenter and seismograph is: d=

Δtp−s 1 vs

−

1 vp

=

1 1500

10 = 30,000 m 1 − 3000

830

23 EARTHQUAKE GEOENGINEERING

where Δtp–s is the arrival time difference of a shear wave and compression wave, vs is the shear wave velocity, and vp is the compression wave velocity. The earthquake epicenter is 30,000 m away from the seismograph. Three seismographs are needed to find the exact location of the epicenter: The intersection of the three circles gives the location. Problem 23.2 An earthquake takes place along a fault and creates 2 m of relative displacement between two tectonic plates. The area over which the slip takes place is 500 km by 100 km and the shear modulus of the rock is 20 GPa. Calculate the seismic moment M o , the moment magnitude M w , and the energy E of the earthquake. Solution 23.2 Seismic moment M o :

Mo = GAD = (20 × 109 ) × (5 × 1010 ) × (2) = 2 × 1021 N ⋅ m

where G is the shear modulus of the rock, A is the area over which the slip occurs, and D is the amount of slip movement. Moment magnitude M w : Mw = 0.66 log Mo (N ⋅ m) − 6.05 = 0.66 log(2 × 1021 ) − 6.05 = 8 Energy E: log E = 5.24 + 1.44M = 5.24 + 1.44 × 8 = 16.76 Therefore, the energy E is E = 10

16.76

= 5.8 × 1016 N.m = 5.8 × 1016 joules.

Problem 23.3 Search the Pacific Earthquake Engineering Research (PEER) Center website (http://peer.berkeley.edu/nga/) and select an earthquake ground acceleration vs. time record. From this record, determine the peak ground acceleration. Then integrate the acceleration record to generate the velocity vs. time record and find the peak ground velocity. Then integrate the velocity record to generate the displacement vs. time record and find the peak ground displacement.

Displacement (cm)

Velocity (cm/sec)

Acceleration (m/sec2)

Solution 23.3 A sample record chosen from the PEER website is the Loma Prieta Station Gilroy #2 record. From Figure 23.1s: Earthquake Loma Prieta Station: Gilory#2 5

0

–5

0

5

10

15

20

25

0

5

10

15

20

25

0

5

10

15

20

25

50

0

–50 20 10 0 –10

Time (sec)

Figure 23.1s Acceleration, velocity, and displacement of an earthquake record.

23.12 SEISMIC DESIGN OF FOUNDATIONS

831

Peak ground acceleration in gs (PGAg) = 0.322 g Peak ground acceleration (PGA) = 3.159 (m∕s2 ) Peak ground velocity (PGV) = 0.391m∕s Peak ground displacement (PGD) = 0.121(m)

Problem 23.4 From the acceleration record of problem 23.3, find the bracketed duration for a threshold acceleration of 0.05 g and the sustained maximum acceleration for 3 cycles and then for 5 cycles. Solution 23.4 The horizontal lines on Figure 23.2s show the threshold accelerations of ±0.05 g. The bracketed duration for this earthquake (the time between the first and last exceedance) is 15.26 seconds.

0.4 Loma Prieta Station Gilroy #2 0.3 First exceedance 0.2

Acceleration (g)

Last exceedance 0.1

+0.05g

0 –0.05g

–0.1 –0.2 –0.3 –0.4

0

5

10

15

20

25

30

35

40

Time (s)

Figure 23.2s Bracketed duration of the ground acceleration.

The maximum acceleration for three cycles is 0.145 g. The maximum acceleration for five cycles is 0.13 g.

Problem 23.5 An event with a return period T has a yearly probability of exceedance equal to 1/T. The equation linking the return period T of an event to the probability of exceedance P over a period of time L is: P = 1 − (1 − 1∕T)L Calculate (a) the return period for an earthquake that has a 2% probability of exceedance in 50 years, and (b) the return period for an earthquake that has a 10% probability of exceedance in 50 years.

832

23 EARTHQUAKE GEOENGINEERING

Solution 23.5 a. For an earthquake with a 2% probability of exceedance in 50 years, the return period is: P = 1 − (1 − 1∕T)L 1 − P = (1 − 1∕T)L (1 − P)1∕L = (1 − 1∕T) 1∕T = 1 − (1 − P)1∕L 1 T= 1 − (1 − P)1∕L 1 T= = 2476 years 1 − (1 − 0.02)1∕50 For an earthquake with a 10% probability of exceedance in 50 years, the return period is: 1 T= 1 − (1 − P)1∕L 1 T= = 475 years 1 − (1 − 0.10)1∕50 Problem 23.6 An 828 m-tall tower weighs 6000 MN and has an equivalent stiffness of 200 MN/m. Calculate the natural period of the tower. A one-story house weighs 1.4 MN and has a natural period of 0.15 seconds. What is the equivalent stiffness of the house? Solution 23.6 a. The natural period T of the tower is:

√ T = 2𝜋 √

m = 2𝜋 k

√

W gk

(6000∕9.81) = 10.98 sec 200 b. Rearranging the natural period equation, the stiffness k of the house is: 1.4 W k = 4𝜋 2 2 = 4 × 3.142 × gT 9.81 × 0.152 = 250.1MN∕m T = 2𝜋

Problem 23.7 Search the PEER Center website (http://peer.berkeley.edu/nga/) and select an earthquake ground acceleration vs. time record. For the acceleration record: a. Develop the Fourier acceleration spectrum. b. Develop the response spectrum, for a damping ratio of 5%, by choosing m and varying k. c. Choose a first set of values for k and m and find the spectral acceleration a1 , then find the spectral acceleration a2 for a second set of values equal to 2k and 2m. Compare a1 and a2 .

23.12 SEISMIC DESIGN OF FOUNDATIONS

833

Solution 23.7 The record selected for this example is the station Gilroy #2 on soil. The Fourier acceleration spectrum and the three response spectra (acceleration, velocity, and displacement) are given in Figures 23.3s and 23.4s. 0.07 Gilroy No. 2 (soil) Fourier spectrum amplitude (m/sec)

0.06

0.05

0.04

0.03

0.02

0.01

0

0

1

2

3

4

5

7

6

8

Period, T (sec)

Figure 23.3s Fourier acceleration spectrum.

Sa (m/sec2)

15 10 5 0

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5 Period (sec)

2

2.5

3

Sv (cm/sec)

150 100 50 0

Sd (cm)

40

20

0

Figure 23.4s Fourier response spectrum.

834

23 EARTHQUAKE GEOENGINEERING

Note: The response spectra are defined as the response of the SDOF with a natural period T. It is obtained by solving the equation of motion: m̈x + cẋ + kx = −ma(t) √ By setting 𝜔 = k∕m and c = 2m𝜔𝛽, and then dividing by m, the equation becomes: ẍ + 2𝛽𝜔ẋ + 𝜔2 x = −a(t) Thus, by multiplying k and m by the same value, 𝜔 will not change and the response spectrum will not change, including the spectral acceleration. However, a change in 𝛽 (damping) will change the response spectrum. Problem 23.8 The PGA for a magnitude 6 earthquake is 0.5 g. What is the most likely PGA 50 km away? Solution 23.8 The peak ground acceleration at a distance R(km) for a magnitude M earthquake can be estimated by using Figure 23.5s. A line is drawn parallel to the trend line starting at the PGA value of 0.5 g. Then the PGA value is read on that line at a distance of 50 km.

1

PGA (g)

0.5g

0.1 0.09g Mag = 6.0 50 km

0.01 1

10

100

Distance (km)

Figure 23.5s Attenuation of peak horizontal ground acceleration.

Problem 23.9 What is the likely return period or recurrence interval for a magnitude 6 earthquake? Solution 23.9 According to Figure 23.12, the recurrence interval of a magnitude 6 earthquake is about 200 years. Problem 23.10 Calculate the natural period of a 20 m-thick stiff soil layer if the soil shear wave velocity is 200 m/s. Then calculate the natural period of a 50 m-thick soft soil layer if the shear wave velocity is 100 m/s. Solution 23.10 Using Eq. (23.6), the natural period is: T=

4H vs

23.12 SEISMIC DESIGN OF FOUNDATIONS

Stiff soil: T= Soft soil: T=

835

4 × 20 = 0.4 sec 200 4 × 50 = 2 sec 100

Problem 23.11 What is the transfer function (amplification factor) for the displacement at the ground surface during an earthquake if the natural period of the deposit is 1 second and the depth of soil layer above rock level is 100 m? Assume an undamped linear soil on rigid rock. Redo the calculation for a damped linear soil on rigid rock if the damping ratio is 5%. The shear wave velocity of the soil is 250 m/s. Solution 23.11 The 1-second period is used to find 𝜔:

2𝜋 =1 𝜔 Therefore, 𝜔 = 2𝜋. The first calculation is when the soil is undamped, 𝛽 = 0. The transfer function is: 1 F(𝜔) = √ ( ) ( )2 𝜔H + 𝛽 cos2 𝜔H v v T

s

s

1

= 1.24 F(𝜔) = √ ( ) 2𝜋 × 100 +0 cos2 250 With damping, 𝛽 = 0.05, the transfer function is: F(𝜔) = √

(

cos2

1 = 1.22 ) ( )2 2𝜋 × 100 2𝜋 × 100 + 0.05 250 250

Problem 23.12 A soil has a shear wave velocity equal to 250 m/s and an SPT blow count equal to 30 bpf. The design earthquake corresponds to a PGA equal to 0.3 g. Develop the response spectrum according to Figure 23.17 if the reference spectrum has the following characteristics: spectral acceleration at 0.2 seconds = 0.5 g, spectral acceleration at 1 second = 0.2 g. Solution 23.12 The site-specific spectral parameters are found in Table 23.2. With the given soil parameters: vs = 250m∕s and NSPT = 30 ⇒ Soil classification is“ D” From Table 23.3, with a PGA = 0.3 g and a site classification of D, FPGA = 1.2. AS = FPGA × PGA As = 1.2 × 0.3g = 0.36 g SDS = 0.5 g and SDl = 0.2 g (from the problem), therefore: SDl SDS 0.2 g T= = 0.4 0.5 g To = 0.2TS

TS =

836

23 EARTHQUAKE GEOENGINEERING

To = 0.2 × 0.4 = 0.08

Elastic seismic coefficient

The constants found using the site classification are used to develop the site-specific acceleration response spectrum, shown in Figure 23.6s. 0.6

0.4 0.36 0.2

0

0 0.08

0.3 0.4 0.6 0.9 PTCL, Tm (second)

1.2

Figure 23.6s Design code acceleration response spectrum.

Problem 23.13 At a depth of 5 m below the ground surface, a saturated sand deposit has a corrected SPT blow count equal to 10 bpf, a CPT corrected and normalized point resistance of 90, and a corrected shear wave velocity of 170 m/s. The fine percentage is less than 5%. The groundwater level is at the ground surface and the soil has a total unit weight of 18 kN∕m3 . Will the soil liquefy in a magnitude 7.5 earthquake if the PGA is 0.6 g? What would be the highest magnitude for which the soil would not liquefy? Solution 23.13 a. The cyclic stress ratio CSR =

𝜏av ′ 𝜎vo

( = 0.65

amax g

)(

𝜎vo ′ 𝜎vo

) rd

( ( ) ) amax = 0.6g, 𝜎vo = 𝛾H = 18 × 5 = 90 kN∕m2 , 𝜎 ′ vo = 𝛾 ′ H = (18 − 9.8) × 5 = 41 kN∕m2 Fig. 23.25 → rd = 0.95. 𝜏 CSR = ′av = 0.65 (0.6) (2.195) 0.95 = 0.81 𝜎 vo NSPT = 10

Fig. 23.21−−−−−→Liquefy, q = 90

Fig. 23.22−−−−−→Liquefy, vs = 170(m∕s)

b. CSR =

𝜏av ′ 𝜎vo

( = 0.65

amax g

)(

𝜎vo ′ 𝜎vo

Fig. 23.23 −−−−−→ Liquefy

) rd

CSR=0.1

Fig.23.21−−−−−→amax = 0.07 g CSR=0.13

Fig.23.22−−−−−→amax = 0.095 g CSR=0.13

Fig.23.23−−−−−→amax = 0.095 g Problem 23.14 A slope is cut in a medium-stiff clay with an undrained shear strength su equal to 50 kPa. The height of the slope is 10 m. The site has a site class B, a PGA of 0.45 g, and a spectral acceleration at 1 second equal to 0.3 g. Calculate the horizontal seismic coefficient kh to be used in the slope earthquake stability analysis.

23.12 SEISMIC DESIGN OF FOUNDATIONS

837

Solution 23.14 Table 22.3

Class B, S1 = 0.3 g−−−−−→FPGA = 1.0 Table 22.5

Class B, PGA = 0.45 g−−−−−→FV = 1.0 kmax = FPGA × PGA = 0.45 g ( ( )) FV S1 −1 kmax kav = 𝛾 1 + 0.01H 0.5 kmax ( ( )) 1 × 0.3 g kh = kav = 1.2 1 + 0.01 × 10 0.5 −1 0.45 g = 0.504 g 0.45 g Problem 23.15 Write the expression of the earthquake active earth pressure coefficient and the corresponding static active earth pressure coefficient. Plot the ratio versus kh for kv = 0, vertical back wall, horizontal backfill, frictionless wall, and a 30∘ friction angle for the backfill. Solution 23.15 The expression for the active earth pressure coefficient in the earthquake case, K ae , is found after finding the most critical wedge angle. K ae is: sin2 (𝛼 + 𝜑′ − 𝜓) Kae = ]2 [ √ ′ +𝛿) sin(𝜑′ −𝛽−𝜓) cos 𝜓 sin2 𝛼 sin(𝛼 − 𝛿 − 𝜓) 1 + sin(𝜑 sin(𝛼−𝛿−𝜓) sin(𝛼+𝛽) where 𝛼 is the angle of the back of the wall with the horizontal, 𝛽 is the angle of the ground surface behind the wall with the horizontal, 𝛿 is the angle of friction between the back of the wall and the soil, 𝜑′ is the friction angle of the soil, and 𝜓 is the angle representing the earthquake inertia force as: ( ) kh 𝜓 = tan−1 1 − kv where kh and kv are the horizontal and vertical seismic coefficients, respectively. The expression for the static active earth pressure coefficient, K a , is: sin2 (𝛼 + 𝜑′ ) Ka = ]2 [ √ ′ +𝛿) sin(𝜑′ −𝛽) sin2 𝛼 sin(𝛼 − 𝛿) 1 + sin(𝜑 sin(𝛼−𝛿) sin(𝛼+𝛽) The ratio of Kae ∕Ka for a vertical wall (𝛼 = 90), no wall friction (𝛿 = 0), horizontal backfill (𝛽 = 0), and a 30∘ angle of friction for the backfill can be plotted as in Figure 23.7s. 4.5 4 3.5 Kae/Ka

3 2.5 2 1.5 1 0.5 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

kh

Figure 23.7s Ratio of the earthquake active earth pressure coefficient and the corresponding active earth pressure coefficient versus kh .

838

23 EARTHQUAKE GEOENGINEERING

Problem 23.16 A 3 m-high vertical gravity retaining wall has a dry horizontal backfill with a friction angle equal to 30∘ and a unit weight of 20 kN∕m3 . It must be designed for a horizontal seismic coefficient equal to 0.2. Calculate: a. Static coefficient of active and passive earth pressure, K a and K p . b. Seismic coefficient of active and passive earth pressure, K ae and K pe . c. The static component and dynamic component of the active push against the wall and their point of application, Pa , ΔPae , X a , and X ae . d. The static and dynamic components of the passive push against the wall if the wall was pushed into the soil backfill and their point of application, Pp , ΔPpe , X p , and X pe . Solution 23.16 a. Static coefficient of active and passive earth pressure, K a and K p : 1 − sin 𝜑′ 1 − sin(30) = = 0.333 1 + sin 𝜑′ 1 + sin(30) 1 + sin 𝜑′ 1 + sin(30) = Kp = =3 1 − sin 𝜑′ 1 − sin(30)

Ka =

b. Seismic coefficient of active and passive earth pressure, K ae and K pe : ( ) ) ( kh 0.2 −1 = 113∘ 𝜓 = tan = tan−1 1 − kv 1−0 The seismic coefficient of active earth pressure is: Kae =

sin2 (𝛼 + 𝜑′ − 𝜓) ]2 [ √ ′ +𝛿) sin(𝜑′ −𝛽−𝜓) cos 𝜓 sin2 𝛼 sin(𝛼 − 𝛿 − 𝜓) 1 + sin(𝜑 sin(𝛼−𝛿−𝜓) sin(𝛼+𝛽)

Kae =

sin2 (90 + 30 − 11.3) ]2 = 0.473 [ √ sin(30−0−11.3) cos(11.3) sin2 (90) sin(90 − 0 − 11.3) 1 + sin(30+0) sin(90−0−11.3) sin(90+0)

The seismic coefficient of passive earth pressure is: Kpe =

sin2 (𝛼 + 𝜑′ − 𝜓) ]2 [ √ ′ +𝛿) sin(𝜑′ +𝛽+𝜓) cos 𝜓 sin2 𝛼 sin(𝛼 + 𝛿 + 𝜓) 1 − sin(𝜑 sin(𝛼+𝛿+𝜓) sin(𝛼+𝛽)

Kpe =

sin2 (90 − 30 + 11.3) ]2 = 5.29 [ √ sin(30+0+11.3) cos(11.3) sin2 (90) sin(90 + 0 + 11.3) 1 − sin(30+0) sin(90+0+11.3) sin(90+0)

c. The static component of active push is: 1 K 𝛾H 2 2 a 1 kN Pa = × 0.33 × 20 × (3)2 = 30 2 m The dynamic component of active push is: 1 ΔPae = (Kae − Ka )𝛾H 2 2 1 kN ΔPae = (0.473 − 0.333)(20)(3)2 = 12.6 3 2 m The point of application of Pa from the bottom of the wall is: Pa =

1 1 Xa = H = (3) = 1 m 3 3 The point of application of Pae from the bottom of the wall is: Xae = 0.6H = 0.6(3) = 1.8 m

23.12 SEISMIC DESIGN OF FOUNDATIONS

839

d. The static component of passive push is: 1 K 𝛾H 2 2 p 1 kN Pp = (3)(20)(3)2 = 270 3 2 m

Pp =

The dynamic component of passive push is: 1 (K − Kp )𝛾H 2 2 pe 1 kN ΔPpe = (5.29 − 3)(20)(3)2 = 206.1 3 2 m The point of application of Pa from the bottom of the wall is: ΔPpe =

1 1 H = (3) = 1 m 3 3 The point of application of Pae from the bottom of the wall is: Xp =

Xpe = 0.6H = 0.6 × 3 = 1.8 m Problem 23.17 The wall of problem 23.16 has water on the no-soil side and water in the backfill up to the ground surface. The water depth on the no-soil side is 2 m. Calculate: a. The hydrostatic pressure and the resultant water push on both sides of the wall, pw1 , pw2 , Pw1 , and Pw2 . b. The earthquake pressure and the resultant push on both sides of the wall if the horizontal seismic coefficient is 0.2.

Solution 23.17 a. The hydrostatic pressure and the resultant water push on both sides of the wall, pw1 , pw2 , Pw1 , and Pw2 : The hydrostatic pressure on the no-soil side is: pw1 = 𝛾w z1 pw1 = 9.81(2) = 19.62

kN m2

The resultant push for the hydrostatic pressure on the no-soil side is: 1 1 kN p z = (19.62)(2) = 19.62 2 w1 1 2 m The hydrostatic pressure on the backfill side is: Pw1 =

pw2 = 𝛾w z2 kN m2 The resultant push for the hydrostatic pressure on the backfill side is: pw2 = 9.81(3) = 29.43

1 1 kN pw2 z2 = (29.43)(3) = 44.15 2 2 m b. The earthquake pressure and the resultant push on both sides of the wall if the horizontal seismic coefficient is 0.2: The earthquake water pressure on the no-soil side is: √ 7 Δpwe1 = kh 𝛾w zHw 8 √ 7 kN Δpwe1 = (0.2)(9.81) (2)(2) = 3.43 2 8 m Pw2 =

840

23 EARTHQUAKE GEOENGINEERING

The resultant push for the earthquake water pressure on the no-soil side is: 7 k𝛾 H 2 ΔPwe1 = 12 h w w 7 kN ΔPwe1 = (0.2)(9.81)(2)2 = 4.58 12 m The resultant earthquake water push on the no-soil side is: Pw1 + ΔPwe1 = 19.62 + 4.58 = 24.2

kN m

The earthquake push on the backfill side is obtained as follows: kN 𝛾 ′ = 𝛾 − 𝛾h = 20 − 9.81 = 10.19 3 m ( ) kh 𝛾 −1 𝜓 = tan × 1 − kv 𝛾 ′ ) ( 20 0.2 = 21.4∘ × 𝜓 = tan−1 1 − 0 10.19 sin2 (𝛼 + 𝜑′ − 𝜓) Kae = [ ]2 √ ′ +𝛿) sin(𝜑′ −𝛽−𝜓) cos 𝜓 sin2 𝛼 sin (𝛼 − 𝛿 − 𝜓) 1 + sin(𝜑 sin(𝛼−𝛿−𝜓) sin(𝛼+𝛽) Kae =

sin2 (90 + 30 − 21.4) [ ]2 = 0.685 √ sin(30+0) sin(30−0−21.4) 2 cos (21.4) sin (90) sin (90 − 0 − 21.4) 1 + sin(90−0−21.4) sin(90+0)

1 K 𝛾 ′H2 2 ae 1 kN P′ ae = (0.685) (10.19) (3)2 = 31.41 3 2 m The resultant push on the backfill side is: P′ ae =

Pae = Pw2 + P′ae Pae = 44.15 + 31.41 = 75.6

kN m

Problem 23.18 Demonstrate that the point of application of the dynamic water pressure in Eq. (23.51) is 0.6H w from the top of the water level. Solution 23.18 By writing the moment equation: z=HW

Z.ΔPwe =

∫z=0

(Δpwe × z)dz

From Eqs. (23.51) and (23.53), we have: √ 7 kh 𝛾w zHw 8 7 = k𝛾 H 2 12 h w w

Δpwe = ΔPwe By plugging Eqs. (23.51) and (23.53) into: Kae =

sin2 (90 + 30 − 11.3) ]2 = 0.473 [ √ sin(30+0) sin(30−0−11.3) 2 cos(11.3) sin (90) sin(90 − 0 − 11.3) 1 + sin(90−0−11.3) sin(90+0)

23.12 SEISMIC DESIGN OF FOUNDATIONS

we get: Hw

Z=

∫0

(

7 k𝛾 8 h w

) zHw z dz

7 k 𝛾 H2 12 h w w 3

H

Z=

√

841

12 ∫0 w z 2 dz 3

8Hw2 3 2 5 6 1 Z = × 3 × Hw2 = H 2 5 10 w Hw2 Problem 23.19 An anchored wall retains 10 m of sand with a blow count of 18 bpf and a unit weight of 20 kN∕m3 . The wall is vertical, the backfill is horizontal, and the wall friction is zero. The water level is deeper than the excavation level. The allowable movement at the top of the wall is 30 mm. The design earthquake will generate a horizontal seismic coefficient equal to 0.25. Calculate: a. The pressure p against the wall above the excavation in the case of no earthquake. b. The pressure pe against the wall above the excavation in the case of the design earthquake.

Solution 23.19 1. The constant pressure p against the wall above the excavation in the case of no earthquake utop Fig.22.19 Fig. 22.19 utop = 30 mm, H = 10000 mm → = 0.003−−−−−→K = 0.2 H p = K × 𝛾H = 0.2 × 20 × 10 = 40kN∕m2 2. The constant pressure pe against the wall above the excavation in the case of the earthquake. Calculate the coefficient of active earth pressure in the case of no earthquake 33∘ = 0.3 Based on Figure 16.12 → 𝜑′ = 35∘ Ka = 1−sin 1+sin 33∘ Calculate the coefficient of active earth pressure in the case of earthquake: sin2 (𝛼 + 𝜑′ − 𝜓) ]2 [ √ ′ +𝛿) sin(𝜑′ −𝛽−𝜓) cos 𝜓 sin2 𝛼 sin(𝛼 − 𝛿 − 𝜓) 1 + sin(𝜑 sin(𝛼−𝛿−𝜓) sin(𝛼+𝛽) ) ( ) = tan−1 0.25 = 14∘ Kae =

where: 𝜓 = tan−1

(

kh 1−kv

1−0

Kae =

sin2 (33 − 14) ]2 = 0.466 [ √ sin(33+0) sin(33−0−14) 2 cos(14) sin (90) sin(90 − 0 − 14) 1 + sin(90−0−14) sin(90+0)

Calculate the pressure against the wall in the case of the earthquake: pe =

Kae 0.466 × K𝛾H = × 0.2 × 20 × 10 = 62.1 kN∕m2 Ka 0.3

842

23 EARTHQUAKE GEOENGINEERING

Problem 23.20 A building is 60 m tall, weighs 500 MN, and has a horizontal stiffness of 400 MN/m. The design earthquake gives the response spectrum shown in Figure 23.8s. Calculate the horizontal force that must be resisted by the foundation.

0.4

a (g)

0.3

0.2

0.1

0

1

2

3

4

5

T (sec)

Figure 23.8s Response spectrum for problem 23.20.

Solution 23.20 Fundamental period: T = 2𝜋

√

M K

√ = 2𝜋

500∕g 400

= 2.24 sec

Spectrum

T −−−−−→a = 0.245 g F = Ma = (500∕9.81) × 0.245 ∗ 9.81 = 122.5 MN

CHAPTER 24

Erosion of Soils and Scour Problems

24.1

THE EROSION PHENOMENON

An erosion problem always has three components: the soil or rock, the water, and the geometry of the obstacle that the water is interacting with. The resistance of the soil or rock is characterized by its erodibility, the water action is quantified by its velocity, and the geometry of the obstacle is quantified by its dimensions. Background on erosion from Briaud (2008) appears in this chapter, including the associated case histories. Figure 24.1 shows a free-body diagram of a soil particle, a cluster of particles, or a rock block at the bottom of a lake. The water imposes a normal stress (hydrostatic pressure) around the soil particle or rock block. The normal stress is slightly higher at the bottom than at the top because the bottom is slightly deeper in the water column. This normal stress difference creates the buoyancy force, which reduces the weight of the soil particle or rock block. Figure 24.2 shows the same particle, cluster of particles, or rock block at the bottom of a flowing river. Three things

Vw ≠ Vy ≠ 0 uw

τ W

fci

C.G

foi

foi foi

C.G

water soil

fci

C.G

fci

fci

C.G

foi C.G

foi = electrical forces between particles fci = forces at contacts between particles C.G = center of gravity W = weight of particle uw = water pressure around particle τ = shear stress around particle

Figure 24.2 Free-body diagram of a soil particle or rock block when the water flows. (Source: Adapted from Briaud, 2008/American Society of Civil Engineers.)

V w = Vy = 0 uw W

fci

C.G

foi

foi foi

C.G

water soil

fci

C.G

fci

fci

C.G

foi C.G

foi = electrical forces between particles fci = forces at contacts between particles C.G = center of gravity W = weight of particle uw = water pressure around particle

Figure 24.1 Free-body diagram of a soil particle or rock block for a no-flow condition. (Source: Adapted from Briaud, 2008/American Society of Civil Engineers.)

happen when water starts flowing. First, a drag force and associated shear stresses develop at the interface between the soil particle or rock block and the water flowing over it. This drag force is very similar to the seepage force. Second, the normal stress on top of the soil particle or rock block decreases because of the water flow. Indeed, as the velocity increases around the particle or the obstacle, the pressure drops to maintain conservation of energy, according to Bernoulli’s principle. This phenomenon is similar to the air flow on top of an airplane wing where the pressure is lower than below the wing, thereby developing the uplift force necessary for the plane to fly. Third, the normal stresses and shear stresses applied at the boundaries fluctuate with time because of the turbulence in the water. These fluctuations find their roots in the appearance and disappearance of eddies, vortices, ejections, and sweeps in the flowing water; they can contribute significantly to the erosion process, especially at

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

843

844

24 EROSION OF SOILS AND SCOUR PROBLEMS

higher velocities. In some cases, they are the main reason for erosion. The contribution of turbulence fluctuations to the erosion process has been studied by several authors, including Croad (1981), Raudkivi (1998), Hoffmans and Verheij (1997), Bollaert (2002), and Hofland et al. (2005). The combination of the mean value and the fluctuations around the mean of the drag force and uplift force can become large enough to pluck and drag the soil particle, soil particle cluster, or rock block away and generate erosion. Note that in the case of unsaturated soils or saturated soils with water tension, the mechanical interparticle compressive forces (f ci in Figures 24.1 and 24.2) can be significantly larger than in the case where the water is in compression. This apparent cohesion may increase the resistance to erosion, at least until the flow and presence of water destroy the water tension.

24.2

EROSION MODELS

The erodibility of a soil or rock can be defined as the relationship between the erosion rate ż and the velocity of the water v near the soil-water interface. This definition is not very satisfactory because the velocity varies in direction and intensity in the flow field. In fact, strictly speaking, the water velocity is zero at the soil/rock-water interface. A more satisfactory definition is the relationship between the erosion rate ż and the shear stress 𝜏 at the soil/rock-water interface: ż = f (𝜏)

(24.1)

The erosion function described by Eq. (24.1) represents the constitutive law of the soil or rock for erosion problems, much like a stress-strain curve represents the constitutive law of the soil or rock for a settlement problem. Although a definition based on shear stress is an improvement over a velocity-based definition, it is still not completely satisfactory, as shear stress is not the only stress that contributes to the erosion rate. A more complete description of the erosion function is given by Eq. (24.2): ( ( ) )n )p ( 𝜏 − 𝜏c m Δ𝜏 Δ𝜎 ż +𝛽 +𝛾 (24.2) =𝛼 u 𝜌u2 𝜌u2 𝜌u2 where ż is the erosion rate (m/s), u is the water velocity (m/s), 𝜏 is the hydraulic shear stress, 𝜏 c is the threshold or critical shear stress below which no erosion occurs, 𝜌 is the mass density of water (kg/m3 ), Δ𝜏 is the turbulent fluctuation of the hydraulic shear stress, and Δ𝜎 is the turbulent fluctuation of the net uplift normal stress. All other quantities are parameters characterizing the soil being eroded. While this model is quite thorough, it is rather impractical at this time to determine the six parameters needed in Eq. (24.2) on a site-specific and routine basis. Today Eq. (24.3), which corresponds to the first term of Eq. (24.2), is widely accepted: ) ( 𝜏 − 𝜏c m ż (24.3) =𝛼 u 𝜌u2

As additional fundamental work is performed in erosion engineering, it is likely that Eq. (24.3) will evolve toward Eq. (24.2).

24.3

MEASURING THE EROSION FUNCTION

In the early 1990s, an apparatus was developed to measure the erosion function. This erosion function apparatus (EFA) was described in detail in Chapter 10, Section 10.20.1, including the data reduction (Briaud et al. 2001a). The principle is to go to the site where erosion is being investigated, collect samples within the depth of concern, bring them back to the laboratory, and test them in the EFA. A 75 mm outside diameter sampling tube containing the sample is placed through the bottom of the conduit where water flows at a constant velocity (Figure 24.3). The soil or rock is pushed out of the sampling tube only as fast as it is eroded by the water flowing over it. For each velocity, an erosion rate is measured, and a shear stress is calculated using Moody’s chart (Moody, 1944). Thus, the erosion function is obtained point by point. Examples of erosion functions are shown in Figure 24.4 for a fine sand and Figure 24.5 for a low-plasticity clay. Note that for the same average velocity of 1 m/s in the EFA test conduit, the rate of erosion for the sand is about 1000 times faster than for the clay. This indicates that the rate of erosion can be very different for different soils. Other devices have also been developed to evaluate how resistant earth materials are to water flow. These include the rotating cylinder test (RCT) to measure the erosion properties of stiff soils (e.g., Chapuis and Gatien, 1986), the jet erosion test (JET) to evaluate the erodibility of soils (e.g., Hanson, 1991), and the hole erosion test (HET) to measure the erosion properties of stiff soils (e.g., Wan and Fell, 2004). More recently, a simple and inexpensive tool for field use has been developed called the pocket erodometer test (PET) (Briaud et al., 2012). This tool is described in Chapter 8, Section 8.10. Tests with the pocket erodometer can be performed at the site on the end of a sample to get a first indication of the erodibility of the soil within minutes after sampling. Even more

Water flow V

τ Soil

· z (mm/hr)

Piston pushing · at rate z

τ (N/m2)

(a)

(b)

Figure 24.3 Erosion function apparatus to measure erodibility. (Source: Adapted from Briaud et al., 1999.)

24.4 SOIL EROSION CATEGORIES

10000 Erosion rate (mm/hr)

Erosion rate (mm/hr)

10000

845

1000 100 10 1

Sand D50 = 0.3 mm

0.1 0.01 0.1

1 10 Velocity (m/s)

1000 100 10 1

0.01 0.1

100

Sand D50 = 0.3 mm

0.1

1 10 Shear stress (N/m2)

100

Figure 24.4 Erosion function for a fine sand as measured in the EFA. 100

10 1 Porcelain clay PI = 16% Su = 23.3 kPa

0.1 0.01 0.1

1 10 Velocity (m/s)

100

Erosion rate (mm/hr)

Erosion rate (mm/hr)

100

10 1 Porcelain clay PI = 16% Su = 23.3 kPa

0.1 0.01 0.1

1

10

100

Shear stress (N/m2)

Figure 24.5 Erosion function for a low-plasticity clay as measured in the EFA.

recently, a new test has been developed called the borehole erosion test (BET) (Briaud et al., 2016). The BET is described in Chapter 8, Section 8.13.

24.4

SOIL EROSION CATEGORIES

Categories are used in many fields of engineering: soil classification categories, hurricane strength categories, and earthquake magnitude categories, among others. Such categories have the advantage of quoting one number to represent a more complex condition. Erosion categories are proposed (Figure 24.6) to bring erodibility down in complexity from an erosion rate vs. shear stress function to a category number. Such a classification system can be presented in terms of velocity (Figure 24.6) or shear stress (Figure 24.7). The categories proposed are based on 30 years of erosion testing experience and an erosion data base called TAMU-Erosion (Briaud et al., 2019). To classify a soil or rock, the erosion function is plotted on the category chart; the erodibility category number for the material tested is the number for the zone in which the erosion function fits. Note that, as discussed earlier, using the water velocity is less representative and leads to more uncertainty than using the shear stress; indeed, the velocity and the shear stress are not linked by a constant. The velocity chart has

the advantage that it is easier to gage a problem in terms of velocity. An erodibility classification chart developed on the basis of the pocket erodometer test is shown in Figure 8.30. One of the most important soil parameters in erosion studies is the threshold of erosion. Below this threshold, erosion does not occur; above this threshold, erosion occurs. In terms of shear stress, this threshold is the critical shear stress 𝜏 c ; in terms of velocity, it is the critical velocity vc . Figure 24.8 shows a plot of the critical velocity as a function of the mean grain size, and Figure 24.9 shows the same plot for the critical shear stress. The data come from measurements using the EFA as well as measurements published in the literature. As can be seen in Figures 24.8 and 24.9, the relationship between the critical value and the grain size has a V shape, indicating that the most erodible soils are fine sands with a mean grain size in the range of 0.1–0.5 mm. This V shape also points out that particle size controls the erosion threshold of coarse-grained soils, whereas particle size does not correlate with the erosion threshold of fine-grained soils. Note that Shields (1936) proposed a curve for coarse-grained soils in his doctoral work; his data are included in Figures 24.8 and 24.9. Shields’s recommendations do not extend to fine-grained soils. Note also that Hjulström (1935) proposed such a curve for both coarse-grained soils and fine-grained soils, but his recommendations for fine-grained soils turned out to be too simple.

846

24 EROSION OF SOILS AND SCOUR PROBLEMS

100000

-Fine Sand -Non-plastic Silt

10000

100 SW

10

SW-SM SM

SP-SM SP

1

-Jointed Rock (Spacing < 30 mm) -Fine Gravel -High Plasticity Silt

0.1

Very low erodibility V -Jointed Rock

CL SP-SC MH

ML

Low erodibility IV

-Low & High PI Clay -All fissured Clays -Jointed Rock GP (30–150 mm Spacing) GC -Cobbles -Coarse Gravel CH

SC-SM SC ML-CL

0.1

Medium erodibility III

-Medium Sand -Low Plasticity Silt

- Increase in Compaction (well-graded soils) - Increase in Density - Increase in Water Salinity (clay)

1000

Erosion rate (mm/hr)

High erodibility II

Very high erodibility I

(150–1500 mm Spacing) -Rip Rap Nonerosive

Rock

1.0

-Intact Rock -Jointed Rock (Spacing > 1500 mm)

10

VI

100

Velocity (m/s) (a)

100000 Very high erodibility I

10000 1000 Erosion rate

100

High erodibility II

Medium erodibility III

AmorphousFine Fibrous Granular Coarse Peats Peats Fibrous Peats

(mm/hr)

Very low erodibility V

10 1 0.1 0.1

Low erodibility IV

Nonerosive VI 1.0

10

100

Velocity (m/s)

(b) Figure 24.6 Proposed erosion categories for soils and rocks based on velocity: (a) mineral soils;, (b) organic soils.

24.4 SOIL EROSION CATEGORIES

-Fine Sand -Non-plastic Silt

10000

-Medium Sand -Low Plasticity Silt

- Increase in Compaction (well-graded soils) - Increase in Density - Increase in Water Salinity (clay)

1000 Erosion rate (mm/hr)

High erodibility II

Very high erodibility I

100000

100 SW-SM

SW

10

SM SP-SM SP

1

ML

-Jointed Rock Low (Spacing < 30 mm) -Fine Gravel erodibility -High Plasticity Silt IV -Low & High PI Clay -All Fissured -Jointed Rock Clays (30–150 mm Spacing) GP -Cobbles -Coarse Gravel

GC CH CL SP-SC MH

1

-Rip Rap

Rock

10

Very low erodibility V

-Jointed Rock (150–1500 mm Spacing)

SC-SM SC ML-CL

0.1 0.1

Medium erodibility III

100 1000 Shear stress (Pa)

Nonerosive

VI -Intact Rock -Jointed Rock (Spacing > 1500 mm) 10000

100000

(a)

100000

Very high erodibility I

10000

Erosion rate (mm/hr)

High erodibility II

Medium erodibility III

AmorphousGranular Fine Fibrous 1000 Peats Coarse Peats Fibrous Peats

Low erodibility IV

100

Very low erodibility V

10 1

Nonerosive VI

0.1 0

1

10 100 1000 Shear stress (Pa)

10000

100000

(b) Figure 24.7 Proposed erosion categories for soils and rocks based on shear stress: (a) mineral soils; (b) organic soils.

847

24 EROSION OF SOILS AND SCOUR PROBLEMS

Intact rock

Clay

Silt

Fine sand

Coarse sand

Rip-rap & jointed rock

Grave

Critical Velocity, Vc (m/s)

1000 100

US Army Corps of Engineers

D50 = 0.3 mm

Vc = 0.07 (D50)–1.45

10 1 0.1

Vc = 0.315 (D50)0.5

Vc = 0.1 (D50)–0.12

0.01

0.01 0.0001 0.001 0.1 1 10 Mean Grain Size, D50 (mm)

1000

100

10000

Joint Spacing for Jointed

Legend: TAMU Erosion EFA Data - Coarse TAMU Erosion EFA Data - Fine Data from Shields, Casey, US.WES, Gilbert, White as repoted by Vanoni, V.A., ed. (1975). “Sedimentation Engineering.” ASCE Manuals and Reports on Engineering Practice, ASCE, New York.

Figure 24.8 Critical velocity as a function of mean grain size.

Intact rock

Silt

Clay

Fine sand

Coarse sand

Rip-rap & jointed rock

Grave

100000

Critical Shear Stress, τc (Pa)

848

10000 1000

–2.3

τc = 0.06 (D50)

D50 = 0.3 mm

100 US Army Corps of Engineers

10

τc = D50

1 0.1

τc = 0.05 (D50)–0.25

Curve proposed by Shields (1936)

0.01 0.01 0.0001 0.001 0.1 1 10 Mean Grain Size, D50 (mm)

100

1000

10000

Joint Spacing for Jointed

Legend: TAMU Erosion EFA Data - Coarse TAMU Erosion EFA Data - Fine Data from Shields, Casey, US.WES, Gilbert, White as repoted by Vanoni, V.A., ed. (1975). “Sedimentation Engineering.” ASCE Manuals and Reports on Engineering Practice, ASCE, New York.

Figure 24.9 Critical shear stress as a function of mean grain size.

24.5 ROCK EROSION

Table 24.1 Soil and water properties influencing erodibility Soil water content Soil unit weight Soil plasticity index Soil undrained shear strength Soil void ratio Soil swell Soil mean grain size Soil percent passing #200

Soil clay minerals Soil dispersion ratio Soil cation exchange cap Soil sodium absorption rate Soil pH Soil temperature Water temperature Water salinity Water pH

The erodibility of soils varies significantly from one soil to the next; therefore, erodibility depends on the soil properties. It depends also on the properties of the water flowing over the soil. For clays, the higher the salt concentration in the water, the more erosion-resistant the clay is (Cao et al., 2002; Croad, 1981). The properties influencing erodibility are numerous; some of them are listed in Table 24.1. It appears reasonable to expect that a relationship would exist between common soil properties and erodibility—but erodibility is a function, not a number, so correlations can be made only with elements of that function, such as the critical shear stress or the initial slope of the erosion function. Such correlations have been attempted (Cao et al., 2002; Briaud et al., 2019) but did not lead to significant success and had marginal coefficients of correlation. On the one hand, there should be a correlation; on the other hand, the correlation is complex and requires multiple parameters, all involved in the resistance of the soil to erosion. All in all, it is preferable to measure the erosion function directly with an apparatus such as the EFA.

24.5

ROCK EROSION

Soil erosion is not very well known, but rock erosion is even less known, so the engineer must exercise a great deal of engineering judgment when it comes to rock erosion. Nevertheless, many engineers and researchers have contributed to the advancement of knowledge in this relatively new field. They include Temple and Moore (1994), Annandale (1995), Kirsten et al. (1996), van Schalkwyk et al. (1995), Bollaert (2002), and Manso (2006). Rock erodes through two main processes: rock substance erosion and rock mass erosion. Rock substance erosion refers to the erosion of the rock material itself, whereas rock mass erosion refers to the removal of rock blocks from the jointed rock mass. Rock substance erosion includes three sub-mechanisms: erosion

849

due to the hydraulic shear stress created by the water at the rock-water interface, erosion due to abrasion caused by sediments rubbing against the rock during the flow, and erosion from the impact of air bubbles that pit the rock surface due to cavitation at very high velocities. Rock mass erosion includes two sub-mechanisms: erosion due to slaking, and erosion due to block removal between joints. Slaking can occur when a rock, such as a high-plasticity shale in an ephemeral stream, dries out and cracks during summer months; these small blocks are then removed by the next big flood. Block removal can occur if, during high-turbulence events, the difference in pressure between the top and the bottom of a rock block becomes large enough to overcome the weight and side friction on the block. Bollaert (2002) points out that brittle fracture and fatigue failure can contribute to breaking the rock into smaller pieces which are then carried away by the water. Note that most of the time, rock mass erosion will be the dominant process in rock erosion, with rock substance erosion occurring only rarely. The critical velocity associated with rock erosion is much higher than the critical velocity associated with soil erosion in general. At the same time, the erosion rate for a given velocity is much lower for rock erosion than for soil erosion in general. Table 24.2 is an attempt at quantifying the critical velocity and the erosion rate of jointed rocks where rock mass erosion may control the process. This table is preliminary in nature and should be calibrated against field behavior. The critical velocities quoted in Table 24.2 refer to the mean depth velocity necessary to move a particle with a size equal to the spacing between joints; as such, they are likely lower bounds because they ignore any beneficial effect from the shear strength of the joints. Note that the orientation of the bedding of the rock mass is important, as shown in Figure 24.10. Engineering judgment must be used to increase or decrease the critical velocity when the bedding is favorable or unfavorable to the erosion resistance. In addition, it is highly recommended in all cases to measure the erosion function of the rock substance on core samples obtained from the site. Examples of rock erosion rates can be collected from geology. For example, the Niagara Falls started about 12,000 years ago on the shores of Lake Erie and have eroded back primarily through undercutting of the falls rock face to halfway between Lake Erie and Lake Ontario. This represents 11 km and an average rate of 0.1 mm/hr, through sandstones, shales, and limestone sedimentary rocks (http://en.wikipedia.org/ wiki/Niagara_Falls). Another example is the Grand Canyon, where the Colorado River has generated 1600 m of vertical erosion through complex rock layers over an estimated 10 million years for an average rate of 0.00002 mm/hr (http://en .wikipedia.org/wiki/Geology_of_the_Grand_Canyon_area) as the Colorado Plateau was upheaving. These rates appear

850

24 EROSION OF SOILS AND SCOUR PROBLEMS

Table 24.2

Rock mass erosion

Joint spacing (mm)

Critical velocity (m/s)

Erosion category

Orientation of joints

1500

0.5–1.35 1.35–3.5 3.5–10 >10

Category III Medium Category IV Low Category V Very Low Category VI Non-erosive

Not applicable Evaluation needed Evaluation needed Not applicable

Note: This table is preliminary in nature and should be calibrated against field behavior.

Favorable orientation

Unfavorable orientation

Flow direction

Flow direction

τ

Vx

τ=μ

dγ

=μ

dt τ du γ Water dz element τ

dVx dz

γ = shear strain

Figure 24.10 Effect of joint orientation on erosion resistance.

negligible at first glance yet neglecting them would be to neglect the Grand Canyon or the retreat of Niagara Falls. The lesson is clear: it is not only the rate of erosion that is important, but also the length of time over which that rate is applied.

24.6

WATER VELOCITY

Figure 24.11 shows the profile of water velocity as a function of flow depth. The water velocity is largest near the top of the water column and zero at the bottom. This has been measured repeatedly in hydraulic engineering. By comparison, the shear stress is highest at the bottom and near zero at the top of the water column. The relationship between the shear stress and the velocity can be established as follows. Because water is a Newtonian fluid, there is a linear relationship between the shear stress 𝜏 and the shear strain rate d𝛾/dt: ( 𝜏=𝜇

d𝛾 dt

) (24.4)

where 𝜇 is the dynamic viscosity of the water (10–3 Pa.s at 20∘ C). This viscosity is different from the kinematic viscosity v of water (10–6 m2 /s at 20∘ C) defined as v = 𝜇∕𝜌 where 𝜌 is the mass density of water (1000 kg/m3 ). Because, as shown in Figure 24.11, 𝛾 is du/dz, then d𝛾/dt is dv/dz where v and u are the water shear velocity and horizontal displacement at

The hydraulic shear stress is proportional to this slope

τ = shear stress

du dz

Figure 24.11 Velocity and shear stress profile versus flow depth.

a depth z respectively. Then the shear stress 𝜏 at depth z is given by: ( ) dv 𝜏=𝜇 (24.5) dz Therefore, the shear stress is proportional to the gradient of the shear velocity profile with flow depth, and the shear stress at the soil/rock-water interface is the slope of the profile at the interface. If the slope of the water velocity profile at the water-soil or water-rock interface (interface shear stress) is kept constant, and if the water depth is varied, it can be shown that the mean depth velocity will vary as well. This implies that there is no constant ratio between mean depth velocity and interface shear stress. This is one reason why velocity alone is not as good a predictor of erosion as shear stress. Thus, any erosion design tool presented in terms of velocity should be used with caution. Nevertheless, velocity is much easier for the engineer to gage than shear stress, and this is why both velocity and shear stress are used in practice. The magnitude of these shear stresses is very small and measured in N/m2 . They are much smaller than the shear stresses that the geotechnical engineer is used to calculating in foundation engineering, for example, which are in the range of kN/m2 . Figure 24.12 gives examples of the range of

24.6 WATER VELOCITY

Geotechnical shear strength of soils Soft

Van der Waals forces, weight of one soil grain

851

V.Hard

104 Oil wells

Depth involved (m)

102 Foundations Slopes Retaining walls 100 Pavements 10–2

10–4

10–6 0 10

Scour

101

102

103

104

Soil shear stress involved

105

106

(N/m2)

12000 10000 8000 6000 4000 2000 0 1960

1970

1980

1990

1998

Velocity (m/s)

Time (yrs)

Water depth (m)

shear stresses associated with various fields of engineering. If the undrained shear strength is a reasonable measure of the strength of a clay for foundation engineering design, the critical shear stress is the “shear strength” of the same clay for erosion studies. The difference in magnitude of the stresses and the strengths between foundation engineering and erosion is that in erosion studies one looks at the resistance of one particle, or a small cluster of particles, whereas in foundation engineering one looks at the resistance of the soil mass associated with the scale of the foundation. The water in a river does not flow at a constant velocity, so the velocity history over a period of time is a necessary input to many erosion problems. This velocity history or hydrograph is not usually readily available. Often, a discharge (m3 /s) hydrograph is available and must be transformed into a velocity (m/s) hydrograph and a water depth (m) hydrograph. This is commonly done by using software such as HEC-RAS (Brunner, 2002). An example of the results of this transformation is shown in Figure 24.13. HEC-RAS solves the one-dimensional energy equation for gradually varied flow in natural or constructed channels and adds the one-dimensional momentum equation around hydraulic structures such as bridges, culverts, and weirs where the energy equation is no longer applicable.

Discharge (m3/s)

Figure 24.12 Range of shear stresses encountered in different engineering fields.

3 2 1 0 1960

1970

1980 Time (yrs)

1990

1998

1970

1980 Time (yrs)

1990

1998

15 10 5 0 1960

Figure 24.13 Discharge, velocity, and water depth hydrographs.

852

24 EROSION OF SOILS AND SCOUR PROBLEMS

The hydrograph can be used to determine the 100-year flood or the 500-year flood. One simple graphical method (e.g., Chow et al., 1988) consists of obtaining the yearly maximum flows from the hydrograph, ranking them in descending order of intensity, calculating for each flow the probability of exceedance as the rank divided by the total number of observations + 1, then plotting the flow versus the probability of exceedance on a semilog paper, such as the one if Figure 24.14. Once the data are plotted, a linear regression is performed over, say, the first 20 to 30 years of data and extrapolated to the 0.01 probability of exceedance for the 100-year flood and to the 0.002 probability of exceedance for the 500-year flood. The return period is the inverse of the probability of exceedance. There are other and more refined ways of obtaining these design floods, but this simple graphical method helps one to understand the process and the meaning of a 100-year flood: a flood that has a 1% chance of being exceeded in any one year. Figure 24.14 shows the result of an analysis for the hydrograph at the Woodrow Wilson bridge. As can be seen from Figure 24.14, the 100-year flood has a discharge of 12,600 m3 /s and the 500-year flood has a value of 16,600 m3 /s. Note that the 500-year flood does not have a flow 5 times larger than the flow of the 100-year flood. The probability of exceedance R of the design flood with a given return period T r depends on the design life Lt of a structure: R = 1 − (1 − 1∕Tr )Lt

(24.6)

If the design life of the bridge is 75 years, the probability that a flood with a return period of 100 years will be exceeded

Flood-frequency curve based on original hydrograph (1931–1999) 20000 y = –2491.6Ln(x) + 12629 2

R = 0.9563

10000

100-year flood: 12629 m3/s 500-year flood: 16639 m3/s

100

10 1 Percent probability of exceedance in any one year (%)

Streamflow (m3/s)

15000

5000

0 0.1

Figure 24.14 Flood frequency curve obtained from measured discharge hydrograph.

during the 75-year design life is 53%, according to Eq. (24.6); that probability is 14% for the 500-year flood. Only when one gets to the 10,000-year flood does the probability get lower than 1% (0.75%). Therefore, looking at those numbers alone, it seems desirable to use the 10,000-year flood for design purposes. This flood is used in design in The Netherlands for regions of the country deemed critical. The United States uses the 100- and 500-year floods for design purposes in hydraulic engineering; this leads to probabilities of exceedance in the tens of percent. By comparison, structural engineers use a probability of exceedance of about 0.1% for the design of bridge beams (LRFD target) and, judging from measured vs. predicted pile capacity databases (Briaud and Tucker, 1988), geotechnical engineers use a probability of exceedance of the order of a few percent. While these numbers can be debated, it is relatively clear that these different fields of civil engineering operate at vastly different probability of exceedance levels. Note that risk is different from the probability of exceedance (see Chapter 12, Section 12.6.3), as it also involves the value of the consequence. Hence, the probability of exceedance target should vary with the consequence of the failure.

24.7

GEOMETRY OF THE OBSTACLE

The geometry of the obstacle encountered by the water influences the velocity of the water and the flow pattern, including turbulence intensity. When water approaches a pier in a river, it has to go around the pier. In doing so, it faces a restricted area and has to accelerate to maintain the flow rate. This acceleration results in a local mean depth velocity that can be 1.5 times higher than the approach mean depth velocity. If the approach velocity is lower than the critical velocity, but the local velocity around the pier reaches a value higher than the critical velocity, then scour occurs around the pier. This scour type is called clear water scour: that is, scour created by water that does not carry soil particles. In contrast, if the approach velocity and the velocity around the pier are both higher than critical, then the scour type is live bed scour. This means that the water is carrying a significant amount of soil particles. The scour depth reached under live bed scour conditions can be less than the scour depth reached under clear water scour conditions. The reason is that during live bed scour, some of the particles in suspension or rolling on the bottom fall down in the scour hole, thereby limiting the depth of the scour hole around the pier. Figures 24.15a and b show the results of numerical simulations of erosion created by water flow in a contracted channel. The CHEN 4D computer program (Chen et al., 1990; Chen, 2002) was the program used.

24.8 BRIDGE SCOUR

853

(a)

(b)

Figure 24.15 Predicted scour hole shape and streambed shear stresses around abutments and piers: (a) t = 2000 min . (b) t = 15,000 min . Scour depth and shear stress distributions at: (a) t = 2000 min and (b) t = 15,000 min . (Source: From Chen, 2002/International Society for Soil Mechanics and Geotechnical Engineering.)

24.8

BRIDGE SCOUR

Bridge scour refers to the erosion of the soil surrounding the foundation of bridge piers in rivers. The water flows at the approach velocity v, arrives at the bridge support, has to accelerate around that obstacle to maintain the flow rate, and thus has a higher potential to erode the river bottom around the foundation. Figure 24.16 shows the scour hole resulting from this erosion around a bridge pier. Bridge scour accounts for 60% of all bridge failures in the United States (Briaud, 2006a). Figure 24.17 shows the progression of the scour depth as a function of time as a response to the flow history (hydrograph) at the bridge. It is important to know how deep the hole is going to be so that this scour depth can be ignored in the resistance of the foundation. The prediction of that scour depth requires knowledge of the soil erosion

function, the water velocity, and the geometry of the obstacle. The obstacle can be a bridge pier, a bridge abutment, or the contraction of the river. As a result, we talk about pier scour, abutment scour, and contraction scour (Figure 24.18). The simplest problem is that of a constant water velocity v flowing for an infinite time around a cylindrical pier of diameter B. Figure 24.19 shows a typical curve giving the scour depth as a function of time in this case. Experiments have shown that the scour depth z vs. time t curve is well described by a hyperbola: t z= 1 (24.7) + zt ż i

max

where z is the scour depth, ż i is the initial erosion rate at a time equal to zero under a velocity v, zmax is the scour depth at a time equal to infinity (asymptotic value) under a velocity v,

854

24 EROSION OF SOILS AND SCOUR PROBLEMS

CL z(abut) applies

z(cont) applies Probable flood level

z(abut) z(cont)

z(pier)

Normal water level

Figure 24.18 Pier scour, abutment scour, and contraction scour.

Figure 24.16 Scour hole around bridge pier.

it is economical to perform the more complex zfinal analysis. This is often the case with fine-grained soils.

and t is the time during which the water has been flowing at the velocity v. The scour depth zmax is called the maximum scour depth under v and would occur if a flood-creating v lasted a long time. If instead the flood lasted a finite amount of time, say, 24 hours, then the scour depth is called the final scour depth zfinal at the end of the flood event, say, 24 hours. If ż i is large, then zfinal will quickly approach zmax , and one flood may be long enough to create zmax . This is the case with very erodible soils, such as sands, where a maximum scour depth analysis called zmax analysis is sufficient. If, however, ż i is small, then zfinal is likely to be much lower than zmax and

24.8.1

Maximum Scour Depth (zmax ) Analysis

Pier Scour The following equation gives the maximum scour depth for pier scour, that is to say the maximum depth of the hole that can form around the pier for a given set of parameters (Briaud, 2012, 2015a, Figure 24.18): zmax (pier) B′

= 2.2 ⋅ Kpw ⋅ Kpsh ⋅ Kpa ⋅ Kpsp ⋅ (2.6 ⋅ Fr(pier) − Frc(pier) )0.7 (24.8)

Discharge (m3/sec)

Woodrow Wilson Bridge hydrography 12000 10000 8000 6000 4000 2000 0 1960

1965

1970

1975

1980

1985

1990

1995

1990

1995

Time (year) Scour depth vs. time

Scour depth (mm)

7000 6000 5000 4000 3000 2000 1000 0 1960

1965

1970

1975

1980

1985

Time (year)

Figure 24.17 Increase in scour depth versus time as result of applied hydrograph.

24.8 BRIDGE SCOUR

Scour equation

Velocity V(m/s)

Hydrograph

z (t) =

t 1 t 1 zmax z∙ i Max. scour depth

Time t(hrs)

Scour curve Scour depth z(mm)

∙ Scour rate z(mm/hr)

Erosion function

zi τ0 τmax Shear stress τ(N/m2)

zmax zfinal

∙ zi Flood duration

Time t(hrs)

critical velocity vc . The projected width B′ (Figure 24.20) is given by: ( ) L B′ = B cos 𝜃 + ⋅ sin 𝜃 (24.9) B where B′ is the projected width, B is the pier width, L is the pier length, and 𝜃 is the attack angle, which is the angle between the flow direction and the main direction of the pier. The water depth influence factor K pw corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.8) was developed for a pier in deep water. Deep water is defined as a water depth hw larger than 1.43 B′ . If the water depth is shallower than 1.43 B′ , the scour depth is reduced. The equation for K pw is: { ( )0.33 hw h , for Bw′ < 1.43 0.89 B′ K = (24.10) pw

,

1.0

Figure 24.19 Scour depth vs. time curve for constant velocity.

B hw

Velocity V1

Velocity V1 B′

A. Water depth effect

B Velocity V1 θ B

L

S

else

The pier shape influence factor K psh is given in Table 24.3; it corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.8) was developed for a cylindrical pier. The aspect ratio influence factor K pa corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.8) was developed for a cylindrical pier. This influence factor is taken care of by use of the projected width B′ instead of B, so K L/B is always 1. The pier spacing influence factor K psp corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.8) was developed for a single pier. If another pier is placed within the influence zone of the first one, the scour depth will be larger. The equation for K psp is: { ( )−0.91 S , for BS′ < 3.42 K = 2.9 B′ (24.11) psp

B. Pier spacing effect

855

D. Attack angle effect

,

1.0

else ′

Velocity V1 B L C. Pier shape effect

Figure 24.20 Definition of pier parameters.

where zmax(pier) is the maximum depth of pier scour, B′ is the projected width of the pier perpendicular to the flow, K pw is the water depth influence factor for pier scour depth, K psh is the pier shape influence factor for pier scour depth, K pa , is the aspect ratio influence factor for pier scour depth, the aspect ratio is L/B ratio of pier length L over pier width B, K psp is the pier spacing influence factor for pier scour depth, Fr(pier) is the pier Froude Number (defined later) based on the water depth influence factor approach velocity v1 and pier width B′ , and Frc(pier) is the critical pier Froude Number based on

where S is the pier spacing and B is the projected width. Equation (24.11) indicates that piers spaced more than 3.42 times the projected pier width from each other do not increase the scour depth at the pier. The pier Froude Number Fr(pier) is given by: ( ) V1 (24.12) Fr(pier) = √ g ⋅ B′ where V 1 is the water velocity at the location of the pier if the pier were not there, g is the acceleration due to gravity, and Table 24.3

Correction factor for pier nose shape (K psh )

Shape of pier nose

K psh

Shape of pier nose

K psh

Square nose Round nose

1.1 1.0

Circular cylinder Sharp nose

1.0 0.9

(Source: Adapted from Richardson and Davis, 2001.)

856

24 EROSION OF SOILS AND SCOUR PROBLEMS

B′ is the projected width of the pier. The critical pier Froude Number Frc(pier) is given by: Frc(pier) = √

Vc

(24.13)

g ⋅ B′

where V c is the critical velocity for the soil. A comparison between predicted pier scour depth by Eq. (24.8) and measured pier scour depth in flume tests is shown in Figure 24.21 (Briaud, 2015a). The measured data are the data used to develop the equation. Comparison against other data is presented in Briaud (2015a).

location for a given set of parameters (Briaud 2012, 2015a, Figure 24.22). zmax (Cont) = 1.27(1.83Frm2 − Frmc ) (24.14) hwml where zmax(Cont) is the maximum depth of contraction scour, hwm1 is the water depth in the main channel at the approach section, Frm2 is the Froude Number for the main channel at the bridge in the contracted zone, and Frmc is the critical Froude Number for the main channel at the bridge. The Froude Number Frm2 is given by: V ∕C Frm2 = √1 R ghwml

Contraction Scour

Predicted(Zmax(pier)/B')

Contraction scour involves two regions of the river: the approach zone, called zone 1; and the contracted zone, called zone 2 (Figure 24.22). Equation (24.14) gives the maximum scour depth for contraction scour, that is to say the maximum depth of scour that can develop in the contracted channel at the bridge

(24.15)

where V 1 is the velocity in the approach section, g is the acceleration due to gravity, hwm1 is the water depth in the main channel at the approach section, and CR is the contraction ratio, defined as: Q − Qblock CR = (24.16) Q where Q is the total discharge, and Qbiock is the part of the discharge Q blocked by the approach embankments. The critical Froude Number Frmc is given by:

6

R2 = 0.8242

5

Line 1 to 1

4 3 2 1 0 0

1

2

3

4

5

6

Measured(Zmax(pier)/B')

Figure 24.21 Predicted vs. measured maximum pier scour depth. (Source: Briaud, 2015a).

B

θ

L′right

Initial ground level

hwm1 zmax (Cont)

L1

A

A’

hwm1

L2

L′left C’ Section B-B’

Wa

hwm1

(24.17)

where V mc is the critical velocity for the soil in the main channel, g is the acceleration due to gravity, hwm1 is the water depth in the main channel at the approach section, 𝜏 c is the critical shear stress for the soil in the main channel, 𝜌 is the mass density of the soil, and n is the Manning’s coefficient. Manning’s coefficient characterizes the roughness of the river bottom. Estimated values are given in Table 24.4. A comparison between predicted contraction scour depth by Eq. (24.14) and measured contraction scour depth in flume tests is shown in Figure 24.23 (Briaud, 2015a). The measured data are the data used to develop the equation. Comparison against other data is presented in Briaud (2015a).

C

Q

B’

V (𝜏 ∕𝜌)0.5 Frmc = √ mc = c 0.33 gnhwml ghwm1

V1

V2 = V1/CR

zmax (Cont)

Section C-C’

Initial ground level

ys (uni_Cont)

Section A-A’

Figure 24.22 Definition of contraction scour parameters.

Abutment Scour Equation (24.18) gives the maximum scour depth for abutment scour, that is to say the maximum depth of scour that can develop around an abutment in the contracted channel at the bridge location for a given set of parameters (Briaud, 2012, 2015a, Figure 24.18): zmax (Abut) = 243 × Kash Kask Kal Kag Re−0.28 × (1.65Frf 2 − Frfc ) f2 hwf1 (24.18) where zmax(abut) is the maximum depth of abutment scour, hwf 1 is the water depth in the flood plain in the approach flow next to the abutment, K ash is the shape factor for abutment scour, K ask is the skew angle influence factor for abutment scour, K al is the influence factor taking into account the proximity of

24.8 BRIDGE SCOUR 𝟎.𝟓 Manning coefficient n in V = n𝟏 R𝟎.𝟔𝟕 h Se *

Table 24.4 Roughness

n (s.m–0.33)

Smooth clay surface Sand (D50 = 0.2 mm) Sand (D50 = 0.4 mm) Sand (D50 = 1 mm)

Predicted(Zmax(cont)/hwm1)

∗

0.011 0.012 0.020 0.026

R2 = 0.8627

2 1.5

Line 1 to 1

1 0.5 0 0.5

1

1.5

2

2.5

Measured(Zmax(cont)/hwm1)

Figure 24.23 Predicted vs. measured maximum contraction scour depth. (Source: From Briaud, 2015a.)

(a)

(b)

(c)

Figure 24.24 Abutment shapes: (a) Wing-wall abutment. (b) Spillthrough abutment. (c) Vertical wall abutment.

the abutment from the main channel, K ag is the geometry of the channel influence factor for abutment scour, Ref 2 is Reynolds Number around the toe of the abutment, Frf 2 is the Froude Number around the toe of the abutment, and Frfc is the critical Froude Number for the soil near the toe of the abutment. The shape factor K ash corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.18) was developed for a wing-wall abutment (Figure 24.24). The values of K ash are:

Kash

Roughness

n (s.m–0.33)

Gravel (D50 = 2–64 mm) Cobbles (D50 = 64–230 mm) Boulder (D50 > 230 mm)

0.028–0.035 0.030–0.050 0.040–0.070

With V velocity in m/s, Rh hydraulic radius of channel in m, and Se slope of the energy line (m/m).

2.5

0

857

⎧1.0 for wing-wall abutment ⎪1.22 for vertical-wall abutment =⎨ 0.73 for spill-through abutment with 2 ∶ 1 Slope ⎪ ⎩0.59 for spill-through abutment with 3 ∶ 1 Slope (24.19)

The skew angle factor K ask corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.18) was developed for an approach embankment perpendicular to the river bank (Figure 24.25). If the embankment alignment is oblique to the river bank, the abutment scour depth is different. The equation for K ask is: { 1.0 − 0.005(|𝜃 − 90∘ |) for 60∘ ≤ 𝜃 ≤ 120∘ Kask = 0.85 for other 𝜃 values (24.20) where 𝜃 is the skew angle as shown in Figure 24.25. The influence factor for the proximity of the abutment to the main channel K al corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.18) was developed for an abutment far away from the bank of the main channel. When the abutment is close to the bank of the main channel, the abutment scour depth becomes larger. The equation for K al is: (Lf − Le ) ⎧ (Lf − Le ) + 1.35 for < 1.5 ⎪−0.23 h wf1 hwf1 Kal = ⎨ ⎪1.0 otherwise ⎩ (24.21) where Lf is the length of the flood plain, Le is the length of the embankment, and hwf 1 is the water depth in the approach channel near the abutment. The channel geometry influence factor K ag corrects for the fact that the parenthetical expression on the right-hand side of Eq. (24.18) was developed for a compound channel geometry. For a rectangular channel geometry, the abutment scour depth is smaller. The values for K ag are: { 1.0 for compound channel (24.22) Kag = 0.42 for rectangular channel The Reynolds Number Ref 2 is in the equation to respect the scaling laws and the influence of size. It is defined as: Ref 2 =

Vf 2 hwf1 v

(24.23)

where hwf 1 is the water depth in the approach channel near the abutment, v is the kinematic viscosity of water (10–6 m2 /s at

858

24 EROSION OF SOILS AND SCOUR PROBLEMS 3.5

Side view

βm

hwf1 Y Top view Le

C L

X

Flow

Θ

Predicted Zmax (Abut) /hwm1

ya

βa hwm1

R2 = 0.9005

Z

Lf

Lm

3 2.5 2 1.5 1 0.5 0 0

Wa

Vf2

V1 or Vm1

Shape of abutment Spill-through abutment

Wing-wall abutment

Vertical wall abutment

Top view Side view

βa

Right flood plain

Le Lf

A

m d1 β

Lm

Main channel

Flow

Left flood plain

Lf

d1 βm β

a

Le

wa θ A’

ddeck

Figure 24.25 Abutment parameter definitions.

20∘ C), and V f2 is the local velocity near the abutment in the flood plain, obtained as follows: ⎧ Q0.5 , for short setback ((L − L ) ≤ 5h ) f e wm1 ⎪ A2 ) ( ⎪ Qf 1 for long setback Le ≤ 0.25Lf ⎪ Af 2 , ⎪ Vf 2 = ⎨ otherwise use a linearly interpolated velocity between ) ( ⎪ Q0.5 for Lf − Le = 5hwml and ⎪ A ⎪ Q2 ⎪ f1 for Le = 0.25Lf ⎩ Af 2 (24.24) where Q0.5 is the flow in half the channel defined as the sum of half the upstream flow in the main channel, 0.5 Qm1 , plus the flow in the flood plain immediately upstream of

0.5

1

1.5

2

2.5

3

3.5

4

Measured Zmax (Abut) /hwm1

Figure 24.26 Predicted vs. measured maximum abutment scour depth. (Source: From Briaud, 2015a.)

the abutment where the abutment is situated, Qf 1 , hwm1 is the water depth in the main channel in the approach flow; A2 is the cross-sectional flow area in the contracted zone corresponding to the flow Q0.5 ; Af 2 is the cross-sectional flow area on the floodplain at the contracted section; Lf is the width of the floodplain in the approach zone; and Le is the length of embankment leading to the abutment. The Froude Number Frf 2 is calculated around the toe of the abutment and is given by: Vf 2 Frf 2 = √ (24.25) ghwf1 where V f 2 is defined in Eq. (24.24), g is the acceleration due to gravity, and hwf 1 is the water depth in the approach flow near the abutment. The critical Froude Number Frfc is calculated around the toe of the abutment and is given by: V (24.26) Frfc = √ c ghwf1 where V c is the critical velocity for the soil around the toe of the abutment, g is the acceleration due to gravity, and hwf 1 is the water depth in the approach flow near the abutment. A comparison between predicted abutment scour depth by Eq. (24.18) and measured contraction scour depth in flume tests is shown in Figure 24.26 (Briaud, 2015a). The measured data are the data used to develop the equation. Comparison against other data is presented in Briaud (2015a). 24.8.2 Maximum Shear Stress at Soil-Water Boundary When Scour Begins The maximum scour depth is the scour depth reached when the flood velocity v is applied long enough to reach zmax . If the flood velocity stops before zmax is reached, then only zfinal is reached (see Figure 24.19). To predict zfinal , it is necessary to develop the relationship between scour depth z and time t. It was found that a hyperbolic equation would fit well with measured curves of z vs. t: t (24.27) z= 1 + zt ż i

max

24.8 BRIDGE SCOUR

where z is the scour depth, ż i is the initial erosion rate at a time equal to zero under a velocity v, zmax is the scour depth at a time equal to infinity (asymptotic value) under a velocity v, and t is the time during which the water flows at the velocity v. The scour depth-time curve of Eq. (24.27) is defined once zmax and ż i are known. The maximum scour depth zmax is obtained as discussed in Section 24.8.1. The initial erosion rate ż i is obtained from the erosion rate vs. shear stress curve measured in the EFA test or deduced from the soil classification and Figure 24.6. Therefore, it is necessary to know the maximum shear stress 𝜏 max created by the water when it flows around the obstacle at the beginning of the scour process. The following equations were developed based on numerical simulations to calculate the shear stress 𝜏 max for pier scour, contraction scour, and abutment scour. Maximum Shear Stress for Pier Scour

kpw = 1 + 16e(−4hw ∕B)

(24.29)

where hw is the water depth and B is the width of the pier. The pier shape influence factor corrects for the fact that the expression on the right-hand side of Eq. (24.28) excluding the influence factors was developed for a circular pier. For square piers, the factor is 1.15; for rectangular piers, it depends on L/B where L is the pier length and B is the pier width. The equation for kpsh is: kpsh = 1.15 + 7e(−4L∕B)

where 𝜃 is the skew angle or attack angle, which is the angle between the flow direction and the main direction of the pier (see Figure 24.20). The pier spacing influence factor kpsp corrects for the fact that the expression on the right-hand side of Eq. (24.28), excluding the influence factors was developed for an isolated pier. For a line of piers, the pier spacing influence factor kpsp is given by: kpsp = 1 + 5e(−1.1S∕B) (24.32) where S is the center-to-center spacing of the piers and B is the width of the pier (see Figure 24.20). Maximum Shear Stress for Contraction Scour For contraction scour, the equation for 𝜏 max(Cont) (Nurtjahyo, 2003) is: 1

For pier scour, the equation (Nurtjahyo, 2003) is: [ ] 1 1 2 − 𝜏max (Pier) = kpw kpsh kpsk kpsp ⋅ 0.094𝜌V1 log Re 10 (24.28) where 𝜏 max(pier) is the maximum shear stress for pier scour, kpw is the water depth influence factor for pier scour shear stress, kpsh is the pier shape influence factor for pier scour shear stress, kpsk is the skew angle or angle of attack influence factor for pier scour shear stress, kpsp is the pier spacing influence factor for pier scour shear stress, 𝜌 is the mass density of water, V 1 is the mean depth velocity of the water at the location of the pier if the pier were not there (also called upstream velocity in line with the pier), and Re is the pier Reynolds Number. The water depth influence factor corrects for the fact that the expression on the right-hand side of Eq. (24.28) excluding the influence factors was developed for a pier in deep water. At very shallow water depths, the shear stress 𝜏 max increases significantly. The equation for kpw is:

(24.30)

where L is the length of the pier and B is the width of the pier. The skew angle or angle of attack influence factor kpsk corrects for the fact that the expression on the right-hand side of Eq. (24.28) excluding the influence factors was developed for a cylindrical pier. For square and rectangular piers with a length L and a width B, the factor kpsk is given by: ( )0.57 𝜃 (24.31) kpsk = 1 + 1.5 90

859

𝜏max (Cont) = kcr kcl kc𝜃 kcw 𝜌gn2 V12 Rh − 3

(24.33)

where 𝜏 max(Cont) is the maximum shear stress for contraction scour shear stress, kcr is the contraction ratio influence factor for contraction scour shear stress, kcl is the contraction length influence factor for contraction scour shear stress, kc𝜃 is the transition angle influence factor for contraction scour shear stress, kcw is the water depth influence factor for contraction scour shear stress, 𝜌 is the mass density of water, g is the acceleration due to gravity, n is Manning’s coefficient, V 1 is the mean depth velocity of the water in the approach zone, and Rh is the hydraulic radius of the contracted channel. The contraction ratio influence factor kcr corrects for the fact that the velocity V 1 in the equation is the approach velocity, not the velocity in the contracted zone. It is given by: ( )1.75 A (24.34) kcr = 0.62 + 0.38 1 A2 where A1 is the cross-sectional flow area in the approach zone and A2 is the cross-sectional flow area in the contracted zone. Because A2 is smaller than A1 , kcr increases the shear stress in the contracted zone. The contraction length influence factor kcl corrects for the fact that the main part of Eq. (24.33) (right-hand side without correction factors) was developed for abutment widths that were larger than 0.7 times the length of the approach embankments. For abutments narrower than that, the kcl factor is given by: ) ( )2 ( ⎡ Wa Wa , − 1.98 ⎢0.77 + 1.36 L1 − L2 L1 − L2 ⎢ Wa kcl = ⎢ ⎢ ≤ 0.35 for L1 − L2 ⎢ ⎢ otherwise ⎣1.0, (24.35) where W a is the width of the top of the abutment (see Figure 24.23), L1 is the width of the river in the approach zone, and L2 is the width of the river in the contracted zone (see Figure 24.21).

860

24 EROSION OF SOILS AND SCOUR PROBLEMS

The transition angle influence factor kc𝜃 corrects for the fact that the main part of Eq. (24.33) (right-hand side without correction factors) corresponds to no abutment (𝜃 = 0). If the abutment appears through a nonzero transition angle, then kcθ must be used; it is given by: ( )1.5 𝜃 kc𝜃 = 1.0 + 0.9 (24.36) 90 where 𝜃 is the transition angle (see Figure 24.22). The water depth influence factor for contraction scour shear stress kcw was found to be equal to 1 in all conditions. Maximum Shear Stress for Abutment Scour For abutment scour, the equation for 𝜏 max(Abut) is: 𝜏max (Abut) = 12.5kacr kaar kaw kash kask kal 𝜌V12 Re−0.45

(24.37)

where 𝜏 max(Abut) is the maximum shear stress for abutment scour shear stress, kacr is the contraction ratio influence factor for abutment scour shear stress, kaar is the influence factor for the aspect ratio of the approach embankment for abutment scour shear stress, kaw is the influence factor for Froude Number for abutment scour shear stress, kash is the influence factor for abutment shape for abutment scour shear stress, kask is the influence factor for the skew angle of the abutment for abutment scour shear stress, kal is the influence factor related to the location of the abutment in the flood plain for abutment scour shear stress, 𝜌 is the mass density of water, V 1 is the mean depth velocity of the water in the approach zone, and Re is the abutment Reynolds Number. The contraction ratio influence factor kacr corrects for the fact that the velocity V 1 in the equation is the approach velocity and not the local velocity around the abutment. It is given by: ( ) Qtot kacr = 3.65 − 2.91 (24.38) Qtot − Qblock where Qtot is the total discharge and Qblock is the part of the total discharge blocked by the approach embankments. The influence factor kaar takes into account the aspect ratio of the abutment. It is given by: ( )−0.24 Le (24.39) kaar = 0.85 Wa where Le is the length of the approach embankment and W a is the width of the top of the abutment (see Figure 24.25). The influence factor kaw takes into account the water depth. It is given by: { 2.07Fr + 0.8 for Fr > 0.1 (24.40) kaw = 1 for Fr ≤ 0.1 where Fr is the Froude Number, defined as: Vf 2 Fr = √ ghwf1

(24.41)

where V f 2 is the water velocity in the approach zone in line with the abutment and hwf 1 is the water depth in the approach zone in line with the abutment (see Figure 24.25). The influence factor kash takes into account the shape of the abutment. It is given by: kash

⎧1.0 vertical-wall abutment ⎪ = ⎨0.65 wing-wall abutment ⎪0.58 spill-through abutment ⎩

(24.42)

The influence factor kask takes into account the skew angle of the abutment. The reference case is the case when the embankment is perpendicular to the river bank with a skew angle equal to 90∘ . The skew angle can be smaller or larger than 90∘ , but was found to have little influence on the maximum bed shear stress and is conservatively taken as equal to 1: (24.43) kask = 1 The influence factor kal takes into account the location of the abutment in the flood plain. The factor kal is different from 1 only when the abutment is near the edge of the main channel. It is given by: ) ( ⎧1.0 for Lf − L(e ∕hwf 1 )≤ −2 ⎪ for − 2 ≤ Lf − Le ∕hwf 1 ≤ 0 ⎪(1.2 + 0.1) ⎪ Lf − Le ∕hwf 1 ( ) kal = ⎨ for 0 ≤ Lf − Le ∕hwf 1 ≤ 1 ( − 0.2) ⎪1.2 ⎪ Lf − Le ∕hwf 1 ) ( ⎪1.0 for Lf − Le ∕hwf 1 ≥ 1 ⎩ (24.44) 24.8.3 Final Scour Depth (zfinal ) Analysis for Constant Velocity Flow and Uniform Soil Once the maximum shear stress 𝜏 max is known, the erosion curve linking the erosion rate ż to the shear stress 𝜏 is used to find the erosion rate ż i corresponding to 𝜏 max . Equation (24.27) can then be used to find out what zfinal is, as both zmax and ż i are known. The following example illustrates these calculations. A round-nose pier, with a width of 2 m and a length of 6 m, is located in a river where the water depth is 7.89 m, the approach flow velocity is 1.4 m/s, and the attack angle is 0∘ (Figure 24.27a). EFA tests were conducted on soil samples in the vicinity of the pier, and gave the average erosion function shown in Figure 24.27b. The critical velocity of the soil V c is 1.57 m/s and the duration of the flood is 48 hours. Find the pier scour depth after 48 hours of flood. 1. The maximum scour depth zmax is calculated first. The correction factors for water depth K pw , pier shape K psh , pier aspect ratio K pa , and pier spacing K psp are all equal to 1.0. The pier Froude Number needed in Eq. (24.8) is: V 1.4 = 0.316 Fr(pier) = √ 1 = √ ′ g⋅B 9.81 × 2

(24.45)

861

24.8 BRIDGE SCOUR L=6m

Plan view

V1 = 1.4 m/s

V1 = 1.4 m/s Hw = 7.9 m

Erosion rate, (mm/hr)

7

B=2m

5 4 3 2 1 0

zmax (pier)

Vc = 1.6 m/s

6

0

5

Elevation view

(a)

10 15 Shear stress, τ (Pa) (b)

20

Figure 24.27 Data for example of bridge scour calculations: (a) Bridge pier geometry. (b) Erosion function of the soil.

× (2.6 × 0.316 − 0.356)0.7 = 2.58 m (24.47) 2. The maximum shear stress 𝜏 max around the pier at the beginning of the scour process is calculated next. The correction factor for water depth kpw is 1.0, for pier spacing kpsp is 1.0, for attack angle kpa is 1.0, and for pier shape kpsh is 1.15 (kpsh = 1.15 + ). The pier) Reynolds Num7e−4L∕B = 1.15 + 7e−12( ×2 6 . Therefore, the ber Re is 2.8 × 10 Re = 1.4 10−6 maximum shear stress around the pier is: 𝜏max (pier) = 1.0 × 1.15 × 1.0 × 1.0 × 0.094 × 1000 ( ) 1 1 2 × 1.4 − = 11.7 Pa log 2800000 10 (24.48) 3. The initial rate of scour ż i around the pier is read on the EFA curve (Figure 24.27) at 𝜏 = 𝜏max = 11.7 Pa, and gives 4.8 mm/hr. 4. The final depth of pier scour after 48 hours of flow can then be obtained from Eq. (24.27) as: 48 = 211 mm (24.49) zfinal (48h) = 48 1 + 4.8 2580 Therefore, the pier scour depth generated by the 48-hour flood is 8.2% of the maximum pier scour depth. Note that the erosion function used for this example corresponds to a soil with a medium resistance to erosion (Category 3) and that the flood is a relatively small flood (1.4 m/s). Major floods in rivers can reach 3 and 4 m/s. In very steep mountain torrents and at the bottom of levees during overtopping, the velocity can reach more than 10 m/s.

Section 24.8.3 dealt with a uniform soil subjected to a constant velocity. However, in reality, the flow velocity is not constant in a river, and the soil is likely to exhibit different layers versus depth. Let’s look first at the velocity varying over time. The graph presenting the velocity as a function of time over many years is called a velocity hydrograph (see Figure 24.17). This hydrograph represents an accumulation of events where the velocity vi can be considered constant for a short period of time Δti . The solution (Briaud et al., 2001b) progresses by stepping into time with a time increment equal to Δti for each iteration. For Δt1 , the velocity is V 1 and zfinal 1 can be calculated. When the second velocity V 2 appears, the question is to know how to accumulate the second scour depth to the first one. The accumulation principle is as follows (Figure 24.28). The two scour depth z vs. time t curves for the velocities V 1 and V 2 are drawn separately. The scour depth zfinal 1 is found on the V 2 curve and corresponds to the starting point for the scour depth increment for the second velocity. The time t* is the time required for velocity V 2 to create zfinal 1 . Then the scour depth due to V 2 applied for Δt2 can be calculated by using the z vs. t curve for the velocity V 2 starting at t*. More generally, the time t* is the time required for velocity V i to create the same scour depth as all the previous velocities. If that scour depth is larger than zmax for V i , the

V2

V1 Δt1

Δt2 t (hrs)

(a) Sequence of two velocities

z (t) (mm)

zmax (pier) = 2.2 × 1.0 × 1.0 × 1.0 × 1.0 × 2.0

24.8.4 Final Scour Depth (Zfinal ) Analysis for a Velocity Hydrograph and Layered Soil

V (m/s)

and the critical pier Froude Number is: V 1.58 Frc(pier) = √ c = √ = 0.356 (24.46) g⋅a 9.81 × 2 Therefore, the maximum pier scour depth zmax(pier) is:

z-t curve for V2

z-t curve for V1 Δt1

Δt2 t* t (hrs)

(b) Accumulation principle

Figure 24.28 Accumulation of scour depth for two consecutive floods.

862

24 EROSION OF SOILS AND SCOUR PROBLEMS Main Channel

z (t) (mm)

Bascule pier

V Layer 1 (Hard) Layer 2 (Soft) (a) Layered soil

H1

20.8 m

H1

1.2 m

t* t1

t (hrs)

(b) Accumulation principle

29.4 m End of pile tips

24.8.5

The Woodrow Wilson Bridge Case History

The following case history (Kwak et al., 2002) describes the process followed to evaluate the scour depth around the main piers of the Old Woodrow Wilson Bridge, which carried the I-95 across the Potomac River in Washington, D.C., from 1960, when it was built, until 2005 when it was replaced. Soil Erodibility The soil stratigraphy is presented in Figure 24.30. It shows that at the location of the main pier in the main channel,

Secondary Channel Su = 10–20 kPa Pier w = 45% 5.4 m PI = 41 1.2 m

Loose sand Loose medium dense brown silty sand with lenses of gravel. Dense sand Gray and brown dense to very dense sand with silt, gravel, and clay lenses, cobbles and boulders.

N = 10–25 b/0.3 m

25.6 m 15 m

N > 50 b/0.3 m

End of pile tips

Hard clay Su = 100–250 kPa Hard gray and low and high w = 24% plasticity clay with some lenses PI = 72 of sand and gravel.

Figure 24.29 Scour depth for a layered soil system.

velocity V i does not increase the scour depth. This accumulation principle is applied for the entire hydrograph by stepping into time over the design life of the bridge. For a layered soil system, the process is very similar (Figure 24.29). If the soil layer 1 is H 1 thick, the scour depth is predicted as a function of time by using the erosion function of soil 1 and the velocity accumulation principle. When the scour depth becomes equal to H 1 , the erosion function is switched to that of soil layer 2 and the time t* required for the first velocity impacting soil layer 2 to generate a scour depth equal to H 1 is found. After that, the calculations proceed using the erosion function of layer 2. These two algorithms have been automated in a program called SRICOS-EFA and are available at https://briaud.engr.tamu.edu/. The SRICOS-EFA method also allows the user to develop a probability of exceedance P vs. scour depth z curve so that the engineer can choose a scour depth corresponding to an acceptable probability of exceedance. The steps to develop the P-z curve are as follows (Brandimarte et al., 2006; Briaud et al., 2007a; Bolduc et al., 2008). First, the flow values in the hydrograph for the chosen period of time are organized in a log normal cumulative distribution function. Second, a random number generator is used to sample that distribution and create, say, 1000 equally likely future hydrographs. Third, for each of these 1000 future hydrographs, the final depth of scour, Z final , is obtained according to the SRICOS-EFA method. Fourth, the 1000 values of Z final are organized in a log normal distribution and presented as a cumulative density function referred to earlier as the P-z curve. This process is an integral part of the SRICOS-EFA computer program (Kwak et al., 2001; https:// briaud.engr.tamu.edu/). The following case history gives an example of the calculation of scour depth, including probabilistic results.

Soft clay Very soft to soft, gray to brown silt and clay, with some organic and lenses of sand and gravel.

Figure 24.30 Soil stratigraphy at the location of the New Woodrow Wilson Bridge.

the soil stratigraphy consists of a soft organic clay overlying a layer of hard plastic clay. Twelve ASTM Standard thin-wall steel tube samples were collected at the bottom of the Potomac River and sent to Texas A&M University for EFA testing. Examples of the erosion functions obtained for samples close to the main pier are shown in Figure 24.31. As can be seen, the soft layer has a much higher critical velocity than the hard clay below, demonstrating yet again that critical velocity does not necessarily increase with shear strength. Water Velocity The nearest gage station (Gage Station 01646500; www.usgs .gov) on the Potomac River is located approximately 13 km upstream of the Woodrow Wilson Bridge and has a drainage area of 29,965 km2 . The discharge hydrograph from this gage station was multiplied by the drainage area ratio between the bridge location and the gage location (30742/29965) to obtain the discharge hydrograph at the bridge (Figure 24.32). The program HEC-RAS (Hydrologic Engineering Center’s River Analysis System) (Brunner, 2002) is a commonly used 1D flow analysis program. It was used to develop the relationship between the discharge and the velocity on the one hand and the relationship between the discharge and the water depth on the other (Figure 24.33). Note that the velocity in Figure 24.33 is the mean depth velocity of the water at the main pier location if the bridge were not there. That is the velocity, also called approach velocity, used in pier scour depth calculations. Using these relationships, the discharge vs. time curve was transformed into the water depth hydrograph and into the velocity hydrograph or velocity vs. time curve (Figure 24.34). Geometry of the Obstacle The old Woodrow Wilson Bridge was a bascule bridge and the obstacle to the flow considered for this case history was the main bascule pier for the bridge. The pier is square and 9.75 m by 9.75 m in plan view. The attack angle is zero, as the pier is in line with the flow.

24.8 BRIDGE SCOUR

Hard clay Scour rate (mm/hr.)

Scour rate (mm/hr.)

Soft clay 150 100 50 τc = 5.09 (N/m2) 0

0

5

10

15

20

25

100 80 60 40 τc = 0.16 (N/m2)

20 0

30

0

10

150

vc = 1.35 (m/sec.) 100 50 0

1 2 3 Velocity (m/sec.)

0

20

30

40

Shear stress (N/m2)

4

Scour rate (mm/hr.)

Scour rate (mm/hr.)

Shear stress (N/m2) 100 80 60 40

vc = 0.2 (m/sec.)

20 0 0

1

2 3 4 Velocity (m/sec.)

5

Figure 24.31 Erosion functions for the two main soil layers at the main pier location.

12000

Discharge Q (m3/s)

10000 8000

6000 4000 2000

0 1960

1970

1980

1990

2000

2010

Time (years)

Figure 24.32 Discharge hydrograph (1960–2012) for the Potomac River at the Woodrow Wilson Bridge.

15 Water depth (m)

Velocity, v (m/s)

4 3 2 1 0

0

5000

10000

15000

20000

25000

12 9 6 3 0

0

5000

10000

15000

20000

Discharge, Q (m3/s)

Discharge, Q (m3/s)

(a)

(b)

25000

Figure 24.33 Calculated relationship between discharge, velocity, and water depth at the Woodrow Wilson Bridge if the bridge were not there.

863

864

24 EROSION OF SOILS AND SCOUR PROBLEMS

15 Water depth (m)

Velocity (m/s)

3

2

1

0 1960

1970

1980

1990

1998

10

5

0 1960

1970

1980

1990

Time (years)

Time (years)

(a) Velocity Hydrograph

(b) Water Depth Hydrograph

1998

Figure 24.34 Velocity and water depth hydrograph.

Scour Depth Calculations

Probabilistic Scour Calculations

The time step for the 38-year period of observation (1960 to 1998) was chosen as one day, for a total of 13,870 time steps. The scour depth calculations progressed one day at a time by following the accumulation principle detailed in Section 24.8.4. The program SRICOS-EFA (https://briaud.engr.tamu .edu/) was used with the soil erosion functions, the water velocity and water depth hydrographs, and the pier geometry as input. The resulting scour depth vs. time plot is shown in Figure 24.35. The same procedure was repeated to predict the scour depths at the other piers of the old Woodrow Wilson Bridge where measured values were available (Hunt, 2001, personal communication). The comparison between predicted and measured values for all the piers that did not have rip-rap protection and where scour depth measurements were collected as a function of time is shown in Figure 24.36 (Kwak et al., 2002).

Figure 24.37 is an example of a probability vs. scour depth P-z curve for values of the design life Lt of the bridge. With this graph, the engineer can decide at what probability of exceedance to operate and choose the corresponding scour depth.

Scour depth (mm)

8000

24.8.6

The Observation Method for Scour

The Observation Method for Scour (OMS, Briaud et al., 2018) relies significantly on past observations at the bridge. The OMS works in four steps. Step 1 consists of collecting the maximum observed scour depth at the bridge, Z mo . Step 2 consists of finding out what is the biggest flood velocity V mo that the bridge has been subjected to since its construction. Step 3 answers, by using an extrapolation function, the question: what will be the scour depth Z fut if the bridge is subjected to a major flood velocity V fut . Step 4 is a comparison between Z fut and the allowable scour depth Z all for the foundation. Figure 24.38a shows predicted scour depths by the OMS compared to measured scour depth for 11 bridges in Texas and Massachusetts. Figure 24.38b shows the comparison for the same 11 bridges where the prediction is made by using the current HEC-18 guidelines. Briaud et al. (2018) outline the advantages and drawbacks of the OMS method.

6000

24.9 24.9.1

4000

2000

0 1960

1970

1980

1990

1998

Time (yrs)

Figure 24.35 Predicted scour depth vs. time for pier 1E of the Old Woodrow Wilson Bridge.

RIVER MEANDERING Predicting River Meandering

Rivers are active system where meanders can move laterally several meters per year. This lateral migration of the main channel affects bridges, embankments, and other structures straddling the river. It is important to predict future meander movements to design remedial measures or move the structure. Many have contributed to the advancement of knowledge in this field, including Brice (1974), Hickin and Nanson (1984), Hooke (1980), Lagasse et al. (2001), and W. de Moor et al. (2007). Briaud et al. (2007a) developed the MEANDER method to predict the movement of a meander over time.

24.9 RIVER MEANDERING

865

4 Pier 3W Predicted local scour depth (m) by SRICOS-EFA method

Pier 4E 3

Pier 5E Pier 23E Pier 24E

2

Pier 25E Pier 26E 1

Pier 27E Pier 28E Pier 29E

0 0

1 2 3 Measured local scour depth (m)

4

Probability of exceedance (%)

Figure 24.36 Predicted vs. measured scour depths at the old Woodrow Wilson Bridge. 100

Lt = 50 year Lt = 75 year Lt = 100 year Lt = 150 year

10

1 0.1 0.01 5

7.5

10

12.5

15

Scour depth (m)

Figure 24.37 Probability of exceedance over the design life vs. scour depth curve for the bascule pier of the new Woodrow Wilson Bridge.

It proceeds along the same steps followed to predict scour depth. First, the initial geometry of the river is described by fitting circles to the meander bends and placing straight-line tangents to the circles between circles. Second, the erosion function of the riverbanks is input. This can be done by using the results of EFA tests or by using the erosion classification charts of Figures 24.6 and 24.7 adjusted for the presence of vegetation, trees, or other erosion-retarding layers. Third, the velocity hydrograph is input from measurements at a nearby gage station. Fourth, the circles describing the meanders are moved according to erosion rules developed through a series of very large-scale laboratory meander experiments (in sand and then in clay) as well as numerical simulations (Briaud et al., 2007a; Wang, 2006; Park, 2007; Yeh, 2008). This leads

to a prediction of the location of the river after the period of time corresponding to the hydrograph. The MEANDER method also allows the user to develop a map indicating the probability that the river will move a certain distance or more. The steps to develop that probabilistic river location are as follows (Briaud et al., 2007a). First, the flow values in the hydrograph for the chosen period of time T are organized in a log normal cumulative distribution function. Second, a random number generator is used to sample that distribution and create, say, 1000 equally likely future hydrographs. Third, for each of these 1000 future hydrographs lasting a time T, the final location of the river is obtained according to the MEANDER method. Fourth, the 1000 traces of the future river location are organized in a probabilistic map (Briaud et al., 2007b). This map gives

866

24 EROSION OF SOILS AND SCOUR PROBLEMS 8.00

8.00

0188-02-023

0188-02-023 170-0177-05-119

6.00 5.00 4.00

0382-05-021 B13001-1EA

y = 1.0669x R2 = 0.8612

B28032-0JC D06002-0U4

3.00

B28009-0JD D10005-367

2.00

D12026-1XX

1.00 0.00 0.00

7.00

0072-04-020

E01001-41Q Linear (Regression)

1.00

2.00

3.00

4.00

Zpredicted(HEC-18)(m)

Zpredicted(OMS) (m)

7.00

0072-04-020 170-0177-05-119

y = 1.7687x R2 = 0.3061

6.00

0382-05-021

5.00

B13001-1EA B28032-0JC

4.00

D06002-0U4

3.00

B28009-0JD D10005-367

2.00

D12026-1XX

1.00

E01001-41Q Linear (Regression)

0.00 0.00

1.00

2.00

3.00

4.00

Zfut(measured)(m)

Zmeasured (m) (a)

(b)

Figure 24.38 Predicted scour depth vs. measured scour depth: (a) Observation Method. (b) Hydraulic Engineering Circular No. 18. (Source: From Briaud et al., 2018.)

Legend River (today) 50% Probability that the river will reach here or further in 20 years 10% Probability that the river will reach here or further in 20 years 1% Probability that the river will reach here or further in 20 years 0.1% Probability that the river will reach here or further in 20 years

SH 10

5

Figure 24.39 Conceptual presentation of the meandering risk for a river.

the location of the river corresponding to the probability that the river will reach that location or go further after a time t. A conceptual example of this probabilistic map is shown in Figure 24.39 for a period of 20 years. This process is an integral part of the MEANDER computer program (https:// briaud.engr.tamu.edu/). The following case history illustrates the meander migration calculation process. 24.9.2 The Brazos River Meander Case History (Park, 2007) The river is the Brazos River in Texas, USA. The meander is located near Navasota, Texas (Figure 24.40) and the bridge carries highway SH105 over the Brazos River.

Observations Records indicate that the meander has migrated significantly and rather steadily over a long period of time. Figure 24.40 shows the migration rate, which averages 4 m/yr. Observations at the site and large-scale laboratory experiments at Texas A&M University (Wang, 2006; Park, 2007; Yeh, 2008) indicate that the process by which the meander progresses is erosion of the base of the exterior river bank, which undercuts the steep slopes and leads to overhang failures of the banks. The material that falls into the flow is then moved to the other side of the main channel and slightly downstream. This cross-channel movement is due to the helical flow of the

24.9 RIVER MEANDERING 1981 1995 2006

867

Channel movement (m)

North

FM 159

asota

Nav

05 SH 1

Brenham

0

150

100

50

0 1980

1985

1990

400 (m)

1995

2000

2005

2010

Time (year)

(a) River migration from 1981 to 2006

(b) Meander migration versus time

Figure 24.40 Measured migration of the meander over a 25-year period.

North

0 0 30 (m)

100

200

300

2006 profile

400

1951 profile

To Navasota (East) 0

West

100

500

600 (m) To scale

To Brenham (West) 200

300

400

600 (M)

500

0 Erosion

10

Deposition

20 2006 profile 30

1951 profile Not to scale

(M)

Figure 24.41 Lateral movement of the main channel between 1951 and 2006.

water in the meander. Such helical flow has been experimentally measured and numerically reproduced (Briaud et al., 2007a; Yeh, 2008). This process leads to the formation of sand bars on the inside of the meander and to steep banks on the outside of the channel (Figure 24.41).

erode faster than the clay layer above. This will undercut the overhanging clay and lead to sloughing, as observed in the field. The prediction of meander migration was made using the erosion function of the deeper sand layer, as it was the controlling layer in this case.

Soil Erodibility

Water Velocity

Borings were done at the site of the meander from the top of the bank. The stratigraphy according to boring B-2 (Figure 24.42) shows 8 m of clay underlain by 7 m of sand. Thin-wall steel tube samples were collected and tested in the EFA. The results are shown in Figure 24.43. As could be predicted, the deeper layers were more erodible than the shallow ones. This means that the sand layer below will

Gage Station ST #08110200 is located at the SH105 bridge over the Brazos River very close to the meander where the data were collected. This gage station was in operation from 1965 to 1987. To obtain the hydrograph over the prediction period 1958 to 2006, a process was developed (Park, 2007) to make use of other nearby stations that had longer records (ST #08110200, ST #08108700, and ST #08109000). Then

868

24 EROSION OF SOILS AND SCOUR PROBLEMS

To Navasota (East)

To Brenham (West)

Erodibility category Depth (m)

V IV III II I

0 Hard brown clay

Stiff brown sitty clay

6.0 8.0

Very dense brown silty sand (SM)

13.1 Very dense sand (SP)

15.0 2006 Profile

NORTH

Figure 24.42 Soil stratigraphy at boring B-2.

Erosion rate (mm/hr)

100000 10000

Very high erodibility I

High erodibility II

Meander Migration Calculations Medium erodibility III

Low erodibility IV

1000 100

Very Low erodibility V

10 1 0.1

1 10 Velocity (m/s)

Non-erosive VI 100

Figure 24.43 EFA test results on the soil from the meander bank.

the relationship between discharge, velocity, and water depth was obtained from the actual measurements made during the period of 1965 to 1987 at gage ST #08110200. The velocity hydrograph of Figure 24.44 was finally obtained. Geometry of the Obstacle In this case, the obstacle is the shape of the meander, which is characterized primarily by its radius of curvature R and the width of the river channel W. To obtain R, a circle is fitted to the meander and the radius of the best-fit circle is retained as the value of R. The bend angle Φ is the angle to the center of that circle bounded by the beginning and the end of the meander on that circle. Any point M on the meander is then identified by the angle 𝜃 between the beginning and point M. Migration of the meander at point M is predicted as the movement over a period of time in the direction of the circle radius.

The program MEANDER (https://briaud.engr.tamu.edu/) was used to predict the migration of the meander over the period of time 1981 to 2006. The measured river centerline and the predicted river centerline are shown in Figure 24.45. 24.9.3

Observation Method for Meanders

The Observation Method for Meanders (OMM, Briaud and Montalvo, 2017) relies on past observations at the meander and on erosion testing of the riverbank. The past movement and velocity history of the meander are used to back-calculate site specific erosion parameters. Those parameters serve as input to predict the meander migration for a chosen future velocity hydrograph. The steps are as follows 1. Obtain the flow history at the meander site: flow Q vs time t. 2. Convert the flow history into velocity history at the meander site: velocity v vs. time t. 3. Select a point M on the outside bank of the meander and a direction along which the future position of point M is to be predicted; obtain the meander migration history at the site along that direction from historical maps: migration distance M vs. time t. 4. Obtain the erosion function for the soil at the site: erosion rate vs. velocity v. 5. Transform the soil erosion function of step 4 into the meander erosion function by matching the meander migration rate vs. velocity v of step 3. This is automated by using the TAMU-OMM spread sheet.

24.10 LEVEE OVERTOPPING

869

Daily mean velocity (m/s)

3.0 2.5 2.0 1.5 1.0 0.5 0.0 1957

1962

1967

1972

1977 1982 1987 Time (year)

1992

1997

2002

2007

Figure 24.44 Velocity hydrograph for the Brazos River meander.

1600 1400

Y (m)

1200

2 Velocity (m/s)

Brazos River at SH105 from 1981 to 2006

Velocity Hydrograph Critical Velocity

1.6 1.2 0.8 0.4 0 1965

1000

(a)

1975

1985 1995 Time (years)

2005

2015

800 600 400 200 0 –400 –200

0

200

400

600

800 1000 1200

X (m)

Figure 24.45 Predicted and measured migration of the Brazos River from 1981 to 2006.

6. Consider a future flood with a velocity vfut sustained over a period of time tfut and calculate the movement of point M called meander migration M fut during the future flood by using the meander erosion function of step 5. Briaud and Montalvo (2017) show examples of the OMM application (Figure 24.46).

24.10 24.10.1

LEVEE OVERTOPPING General Methodology

Levees or dikes are small dams built along a river or an ocean to prevent the water from inundating the land in case of flood. The top of the levee is set at a predetermined height corresponding to the water level for a chosen design flood. This flood corresponds to a certain return period, such as a 100-year flood. If the flood exceeds the design return period, water is likely to flow over the levee and generate potential erosion. One of the first observations is that if the water flows above a levee of height H, by the time the water reaches the

Migration (m)

160

1981 (Initial) 2006 (Measured) 2006 (Predicted)

120 80 40

M predicted M observed 0 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010 2015 (b) Time (years)

Figure 24.46 Velocity hydrograph and corresponding meander migration for the Nueces River. (Source: From Briaud and Montalvo, 2017.)

bottom of the dry side of the levee it will have a velocity V, which can be very high. One simple way to evaluate that velocity is to write conservation of energy: √ 1 mgH = mV2 or V = 2gH (24.50) 2 For example, if the levee is 5 m high, the velocity V will be approximately 10 m/s. Of course, Eq. (24.50) does not take into account the energy lost in friction between the water and the levee surface, but it does indicate that the velocity range is much higher than typically encountered in rivers, where water rarely flows faster than 3–4 m/s. Furthermore, a distinction should be made between events such as hurricanes, on one hand, and river floods, on the other. The major distinction is that hurricanes may overtop a levee for about 2 hours, whereas river floods may overtop a levee for 2 days. A levee-overtopping erosion chart developed for these two types of events is presented in Figure 24.47. It indicates which soil categories and associated erosion functions are likely to resist overtopping during a 2-hour and a 2-day overtopping. Recall that categories I to IV on the erosion

870

24 EROSION OF SOILS AND SCOUR PROBLEMS

Erosion rate (mm/hr)

100000 10000

Very high erodibility I

High erodibility II

Medium erodibility III

Failure 2 days

1000 100 10

S-11

10.0

S-14

S-2 S-16

S-15

S-3

Lake Borgne

S-4 Orieans East Bank

S-8 St. Bernard Parish

Non-erosive VI 1.0

S-13

New Orieans East

S-9

Resist 2 hrs

0.1 0.1

S-12

S-10

Very Low erodibility V Resist 2 days

Failure 2 hrs

1

S-1

Lake Pontchartrain

Low erodibility IV

Mi

ss

iss

S-17

ipp

i

River

100.0

Velocity (m/s)

S-10

S-5 S-6/Lock #2

Legend

S-7

- Sample #10 - Levee

St. Bernard Parish

Figure 24.47 Levee overtopping.

24.10.2 Hurricane Katrina Levee Case History: New Orleans On August 29, 2005, levee overtopping and associated erosion contributed significantly to the Hurricane Katrina disaster in New Orleans, where some places are 6 m below the tops of the levees. This case history (Briaud, 2006b) describes the process by which overtopped levees erode and discusses whether unprotected soils can resist overtopping erosion.

Figure 24.48 Location of shallow samples collected from the top of the levees.

100000 10000 Erosion rate (mm/hr)

chart are soils and categories V and VI are rocks. As can be seen, only the most erosion-resistant soils can resist 2 hours of overtopping without protection (Category IV), and no soil can sustain 2 days of overtopping without being totally eroded away. Armoring or vegetation satisfying strict criteria must be used to ensure that overtopping can be sustained for longer than 2 hours. The following case history illustrates how the levee overtopping chart was generated and how it can be used.

Very high erodibility I

High erodibility II

Medium erodibility III Low erodibility IV

1000 100

Very low erodibility V

10 1

Non-erosive VI

0.1 0.1

1.0

10.0

100.0

Velocity (m/s)

Figure 24.49 EFA test results in terms of velocity for some levee soils.

Soil Erodibility

Water Velocity Hurricanes are large rotating masses of moisture that can be 400 km in diameter. They travel relatively slowly at speeds

100000 10000 Erosion rate (mm/hr)

Thin-wall steel tube samples and bag samples were obtained from the top of the levees at shallow depth (0–1 m). These samples were collected from locations S1–S15 in Figure 24.48. The bag samples were reconstituted in a Shelby tube by recompacting the soil at a low and at a high compaction effort (Briaud, 2006b). The soil type varied widely, from loose, uniform fine sand to high-plasticity stiff clay. EFA tests were performed on the samples. The results of all the tests are shown in Figures 24.49 and 24.50. One of the first observations from those figures is that the erodibility of the soils obtained from the New Orleans levees varies widely, all the way from very high erodibility (Category 1) to low erodibility (Category 4). This explains in part why some of the overtopped levees failed while other overtopped levees did not.

Very high erodibility I

High erodibility II

Medium erodibility III

1000

Low erodibility IV

100

Very low erodibility V

10 1 0.1 0.1

Non-erosive VI 1

10

100

1000

10000

100000

Shear stress (Pa)

Figure 24.50 EFA test results in terms of shear stress for some levee soils.

24.11 COUNTERMEASURES FOR EROSION PROTECTION 100000

Very high erodibility I

10000 Erosion rate (mm/hr)

of about 40 km/hr. Therefore, a hurricane takes about 10 hours to go over a levee or a bridge. The worst part of the storm, however, is only a fraction of that time. The friction generated by the wind at the air-water interface drags the water into a storm surge that can reach several meters above the mean sea level and kilometers in length. The surge associated with Katrina was about 8.5 m at Bay St. Louis, 4.6 m at Lake Borgne, and 3 m at Lake Pontchartrain. The storm surge was high enough to overtop some of the levees. As discussed earlier, the water velocity at the bottom of such levees can reach 10m/s.

Low erodibility IV

100

Very low erodibility V

10 1

Non-erosive VI 10.0

24.11 COUNTERMEASURES FOR EROSION PROTECTION Rip Rap

Countermeasures for erosion protection include a number of solutions, the most prevalent of which is the use of rip rap (Figure 24.52). Rip rap can be sized by the following equation (U.S. Army Corps of Engineers, 1991): ( )2.5 Vdes d30 = Hw FCst Cv Ct √ (24.51) Csl (Gs − 1)gHw where d30 is the particle size of the rip-rap grain size distribution curve corresponding to 30% fines, H w is the water depth, F is the factor of safety, Cst is the stability coefficient, Cv is the velocity distribution coefficient, Ct is the blanket thickness coefficient, V des is the mean depth water velocity, Csl is the side slope correction factor, Gs is the specific gravity of the rip rap, and g is the acceleration due to gravity (9.81 m/s2 ).

Waves Design still water level

M

ed 1V ium :1 sl .5 op H e

Normal water level

Geotextile or granular filter (a)

100.0

Velocity (m/s)

Figure 24.51 EFA test results for the soils of levees that failed and for the soils of levees that did not fail by overtopping erosion.

24.11.1 There was overwhelming evidence that the water overtopped the levees in many places; such evidence consisted mostly of ships being trapped on top of the levees when the water receded, but also included debris stuck in trees at levels higher than the top of the levees. Some levees resisted the overtopping well, whereas some levees were completely eroded. In Figure 24.51, the erosion functions for the samples taken from levees that were overtopped and resisted well are plotted as open circles; the solid dots are for the samples of levees that were completely eroded. As can be seen, the eroded levees were made of soils in erodibility categories 1 and 2, whereas the levees that resisted well were made of soils in erodibility categories 3 and 4. This led to the levee overtopping chart shown in Figure 24.47.

1.0

Legend: Levee breaches No levee damage

Geometry of the Obstacle

Predicting Levee-Overtopping Erosion

Medium erodibility III

1000

0.1 0.1

Most levees around New Orleans are between 3 and 6 m high. They have two main shapes. The first one consists of a flat top that is some 4 m wide with side slopes at about 5 horizontal to 1 vertical. Because the width of such a levee configuration takes a lot of space, the second shape consists of the same shape as the first, but at a reduced scale with a vertical wall extending from the top of the levee. The problem addressed here is limited to the first shape (no wall).

High erodibility II

871

(b)

Figure 24.52 Rip rap with geosynthetic filter installation: (a) Design plan. (b) Field installation. (Source: Right picture: Courtesy of FHWA.)

872

24 EROSION OF SOILS AND SCOUR PROBLEMS

The stability coefficient Cst takes into account the roughness of the rip-rap blocks; it is 0.3 for angular rock and 0.375 for round rocks. The velocity distribution coefficient Cv takes into account the fact that water tends to accelerate on the outside of river bends; it is 1 for straight channels and inside of bends, and 1.23 in most other cases. The blanket thickness coefficient Ct is a function of the rip-rap gradation, with a default value of 1 in the absence of additional data. The velocity V des is the mean depth velocity for straight channels. For river bends, it is given by: ( ) Rc Vdes = Vave 1.74 − 0.52 log (24.52) W where V ave is the mean depth velocity upstream of the bend, Rc is the centerline radius of curvature of the river bend, and W is the river width at the water level. The side slope coefficient Csl is given by: √ ( )1.6 sin(𝜃 − 14∘ ) (24.53) Csl = 1 − sin 32∘ where 𝜃 is the slope angle of the riverbank in degrees. The specific gravity of solids Gs is usually taken as 2.65. It is very important to place a filter between the soil to be protected and the rip-rap layer. Without a filter, the soil under the rip rap may continue to erode through the large voids in the rip rap. In the end, the rip rap may not move away, but may simply go down significantly as the underlying soil erodes away. The filter may be a sand filter or a geosynthetic filter (see Chapter 27). Design guidelines can be found in Heibaum (2004) for sand filters and in Koerner (2012) for geosynthetic filters.

Erosion rate (mm/hr)

100000

10000

Very High Very high Erodibility erodibility I I

Other countermeasures to prevent erosion include (Lagasse et al., 2009): 1. Flow deflectors such as spurs, jetties, dikes, and guide banks. 2. Rigid armoring of the soil surface, such as soil-cement mixing and grouted mattresses. 3. Flexible armoring, such as rip rap, gabions, and articulated blocks. 4. Pier geometry modification, such as slender pier shape and debris deflectors. 5. Vegetation, such as woody mats and root wads. 6. Fixed and portable instrumentation, such as sonars and float-out devices. 7. Periodic inspection. 24.11.2

Grass

Vegetation can help significantly in retarding erosion. To be effective, though, this vegetation has to satisfy the following minimum requirements: It should have a mat-like appearance, have a sod-forming root system, be made of perennial grasses, have a dense consistent coverage, and have a minimum height of 0.3 m during flood season. Tree roots can be considered to help reinforce the levee slope; however, a tree on a levee that is uprooted by a storm will create a major hole in the levee. Also, if the tree dies, the disappearance of the roots will leave channels for the water to seep through the levee. On the whole, trees on levees or near levees are not a good idea. Erosion tests on grass were conducted in the EFA (Shidlovskaya, 2021) and lead to the erosion functions shown in Figures 24.53 and 24.54. As can be seen, grass generally fits in erosion category III. For grass, the critical

High High Erodibility erodibility II II

Medium Medium erodibility Erodibility III III

1000

Low Very Low erodibility Erodibility V IV

100

10

Very low Non-Erosive VI erodibility V

1

Low Nonerosive Erodibility IV

VI

0.1 0.1 Z2 (TRF) B2 (AA) B3 (AA) BR6 (TRF)

1 Z3 (TRF) StA2 (TAMU) B4 (AA) BR7 (TRF)

Velocity (m/s) B1 (AA) StA3 (TAMU) BR1 (JLB)

10 Z1 (TRF) BR2 (JLB) StA4 (TRF)

100 BR5 1 year old (TRF) BR3 (JLB) StA5 (TRF)

Figure 24.53 Erosion tests results for grass in the velocity space. (Source: From Shidlovskaya, 2021.)

24.12 INTERNAL EROSION OF EARTH DAMS

Erosion rate (mm/hr)

100000

Very High Very high Erodibility erodibility I I

10000

High High Erodibility erodibility II II

873

Medium Medium erodibility Erodibility III III Low Low erodibility Erodibility IVIV

1000 100

Very Very low Low erodibility Erodibility V V

10 1

Non-Erosive Nonerosive VIVI

0.1 0.1

1

Z2 (TRF) StA2 (TAMU) BR1 (JLB)

Z3 (TRF) StA3 (TAMU) StA4 (TRF)

10

100 Shear stress (Pa)

B1 (AA) BR2 (JLB) StA5 (TRF)

1000

Z1 (TRF) BR3 (JLB) BR6 (TRF)

10000 BR5 1 year old (TRF) B3 (AA) BR7 (TRF)

100000 B2 (AA) B4 (AA)

Figure 24.54 Erosion tests results for grass in the shear stress space. (Source: From Shidlovskaya, 2021.)

(a)

(b)

Figure 24.55 Internal erosion of an earth dam. (a) Blackman Creek Dam. (b) Teton Dam. (Source: a: Photograph by Mark S. Harrison, Oklahoma Conservation Commission. Used by permission. b: Courtesy of Eunice Olson.)

velocity or critical shear stress are defined as the values corresponding to the detachment of the grass cover with its root system (ellipse in Figures 24.53 and 24.54). These critical velocities varied from 1 m/s to 6 m/s with the most erosion-resisting grass being St Augustine grass. The following factors have an influence on the erosion resistance of grass covers: 1. Coverage of the grass cover. The grass cover must be very dense as any exposed soil will be eroded by the water leading to unrooting of the grass. 2. Health of the grass. If the grass is not healthy, the blades of grass and the roots will not be as resistant.

3. Age of the grass. Older, well established grass covers are more erosion resistant 4. Strong and deep root system. Shallow roots will be eroded away more easily.

24.12 24.12.1

INTERNAL EROSION OF EARTH DAMS The Phenomenon

It is estimated that 46% of earth dam failures occur due to internal erosion, and half of those failures occur during the first filling of the reservoir (Fell et al., 2005; Figure 24.55). Yet, the handling of internal erosion of earth dams is still

874

24 EROSION OF SOILS AND SCOUR PROBLEMS

based primarily on engineering judgment and experience. Although guidelines and publications exist, much remains to be studied and researched in this field. For internal erosion of an earth dam to take place, the following are required: 1. A source of water and a seepage flow path. 2. Erodible material that can be carried by the seepage flow within the flow path. 3. An unprotected exit from which the eroded material may escape. 4. For a pipe to form, the material must be able to form and support the roof of the pipe. Four different phenomena can lead to internal erosion of an earth dam (Figure 24.56): 1. 2. 3. 4.

Backward erosion. Concentrated leak. Suffusion. Soil contact erosion.

Backward erosion is initiated at the exit point of the seepage path when the hydraulic gradient is too high and the erosion gradually progresses backward, forming a pipe. A concentrated leak is internal to the soil mass; it initiates a crack or a soft zone emanating from the source of water and may or may not progress to an exit point. Erosion gradually continues and can create a pipe or a sinkhole. Suffusion develops when the fine particles of the soil wash out or erode through the voids formed by the coarser particles. This occurs when the amount of fine particles is smaller than the void space between the coarse particles. If, in contrast, the soil has a well-graded particle size distribution with sufficiently small voids, suffusion is unlikely. Soils are called internally unstable if suffusion takes place and internally stable if particles are not eroding under seepage flow. Soil contact erosion refers to sheet flow at interfaces between soil types. It may occur, for example, when water seeps down the back face of the core at the interface with the filter and then the stabilizing mass.

Earth dams deform during and after construction. This movement can be compression, extension, and/or shear distortion. Because typical dams are made of different zones playing different roles, they exhibit different deformation characteristics. This can lead to differential movement, resulting in cracks or soft zones where internal erosion can be initiated. Shrinkage can also create cracks that are prone to erosion if water comes to flow through them. Fell and Fry (2005) summarize the most likely locations where internal erosion can start in an earth dam (Figure 24.57). 24.12.2

Most Susceptible Soils

Coarse silt and fine sand are among the most erodible soils. Therefore, earth dams containing significant amounts of such materials will be more prone to internal erosion. Clays in general, and high-plasticity clays in particular, are more resistant to erosion as long as the electrical bonds between particles are not destroyed by chemicals. It seems that some core materials of glacial origin, such as glacial tills, can be particularly susceptible to internal erosion. Sherard (1979) gives a range of gradation of soils that can lead to internal erosion problems (Figure 24.58). The soils that are most susceptible to suffusion are those where the volume of fines is less than the volume of the voids between coarse particles. In this case, the fines can move easily between the coarse particles and erode away to an exit face. After suffusion, such soils are devoid of fines and become very pervious clean gravel, for example. Fell and Fry (2005) indicate that gap-graded soils and coarsely graded soils with a flat tail of fines (Figure 24.59) are most susceptible to suffusion. 24.12.3 Criterion to Evaluate Internal Erosion Potential One of the important criteria for evaluating erosion is to calculate the hydraulic gradient and compare it to the critical gradient. The critical gradient is given by: icr =

water level

Suffusion Contact erosion Concentrated leak Filter

Fill

Backward erosion

Core

Figure 24.56 Mechanisms of internal erosion failures. (Source: Adapted from Perzlmaier, 2005.)

𝛾sat − 𝛾w 𝛾w

(24.54)

Values of icr typically vary in the range of 0.85–1.2. The hydraulic gradient in dams depends on many factors, including the difference in water level between the upstream and the downstream, the length of the drainage path, and the relative hydraulic conductivity of the various zones. The target maximum gradient in the flow must be kept much lower than the critical value, especially in areas where internal erosion is possible. Figure 24.60 shows ranges of hydraulic gradient values that are associated with initiation of internal erosion, on the one hand, and full development of piping, on the other, for unfiltered exit faces. Generally speaking, there is a

24.12 INTERNAL EROSION OF EARTH DAMS Downstream

B

Conduit

Plan view

Spillway Dam axis 1

3

A

A 2

Left abutment

Right abutment

Upstream

B

Section A-A Spillway

Left abutment

1

Right abutment

3

Conduit

2 Section B-B

Upstream

Downstream

4

Reservoir

7 Core 5

Foundation

6

1. Spillway wall interface 2. Adjacent to conduit 3. Crack associated with steep abutment profile 4. Desiccation on top of core

5. Embankment to foundation 6. Foundation (if the foundation is soil or erodible rock) 7. Embankment through poorly compacted layer, crack (or by backward erosion if the core is cohesionless)

Figure 24.57 Possible locations of initiation of internal erosion. (Source: Adapted from Fell and Fry, 2005.) 100

Percent finer (%)

80

Range of broadly graded problem soils reported by Sherard, 1979

60

40 When coarser, usually impossible to use as core material

20

0 0.001

0.01

0.1

1

10

100

1000

Grain size, mm

Figure 24.58 Range of problem soils for internal erosion. (Source: After Sherard, 1979.)

875

876

24 EROSION OF SOILS AND SCOUR PROBLEMS

developed from sheet flow tests, and the critical velocity may be different from those initiating internal erosion. Several methods, based in part on the analysis of the grain size curve, have been developed to evaluate the instability of soils in dams and their sensitivity to the suffusion phenomenon. They include Sherard (1979), Kenney and Lau (1986), Burenkova (1993), and Fell and Wan (2005).

100 Clay to silt

Sand

Gravel

Percent passing

80 60 Gap graded soil 40

Coarsely graded soil with a flat tail of fines

20 0 0.001

0.01

0.1 1 10 Particle size (mm)

100

24.12.4 1000

Figure 24.59 Range of problem soils for suffusion. (Source: Adapted from Fell and Fry, 2005.)

1.2

Hydraulic gradient i

1.0

Complete piping erosion

0.8

0.6

Initiation of backward erosion

0.4

0.2

0 1

2

3

4

5

6

Coef. of uniformity CU = D60/D10

Figure 24.60 Range of hydraulic gradient values associated with internal erosion. (Source: Adapted from Perzlmaier, 2005.)

trend toward higher-porosity soils beginning to erode at lower hydraulic gradients, even lower than 0.3. Yet, soils with plastic fines erode at higher gradients, and gap-graded soils begin to erode at lower gradients than non gap-graded soils with the same fine content. The U.S. Army Corps of Engineers uses a lower-bound value of the critical hydraulic gradient equal to 0.8 and allows a hydraulic gradient of up to 0.5 at the toe of levees, provided a number of conditions are met (U.S. Army Corps of Engineers, 2003). Another way to address the incipient motion of soil particles in internal erosion problems is to use the concept of critical velocity and charts such as Figures 24.8 and 24.9. However, these critical velocities were

Remedial Measures

Internal erosion of earth dams often occurs very quickly, leaving limited time for remedial action (Foster et al., 2000a, 2000b). Most of the time, complete breach occurs within 12 hours of first visual detection of internal erosion and sometimes in less than 6 hours. The majority of failures occur during the first filling or within 5 years after first filling. The process of suffusion tends to develop more slowly than the back erosion and piping processes. One solution to many internal erosion problems is the use of quality filters. A filter is a layer of soil placed between a fine-grained soil and a coarse-grained soil to transition the flow without having the fines of the fine-grained soil erode through the voids of the coarse-grained soil. The grain size distribution curve of the soil filter layer is designed to provide this transition in a gradual fine-to-coarse fashion. Terzaghi and Peck (1967) proposed filter criteria which have evolved over the years. If a fine layer needs to be protected from erosion by a coarser layer, the following criteria are relevant. The piping criterion aims to minimize the amount of fines which flow through to the coarser material. The permeability criterion aims to ensure sufficient permeability for water to flow through without creating excess pore water pressures. The gradation criterion aims to provide some compatibility of gradation between the two layers. Note that the subscripts F and C will be used for the fine and coarse layers respectively. For example, if the core material of an earth dam needs to be protected and since the core material is usually a fine-grained soil, the filter would be the coarse layer. If, on the other hand, rip rap was used to protect a bridge abutment from scour, the rip rap would be the coarse layer and the filter placed below the rip rap and used to protect the native soil would be the fine layer. 1. Piping criterion

D15 C ≤ 5D85 F

2. Permeability criterion D15 C ≥ 5D15 F 3. Gradation criterion

D50 C ≤ 25D50 F

(24.55) (24.56) (24.57)

where D15C , and D50C are the grain size of the coarse layer corresponding to 15%, and 50% finer by weight while D15F , D50F , and D85F are the grain size of the fine layer corresponding to 15%, 50%, and 85% finer by weight. More details can be found in U.S. Bureau of Reclamation (2011).

24.12 INTERNAL EROSION OF EARTH DAMS

877

Problems and Solutions Problem 24.1 If a faucet drips on a pebble for 20 million years, will there be a hole in the pebble? Solution 24.1 Common sense might lead you to say yes. Then the question might become: How is it possible for a stress level as small as the one created by a drop of water to destroy the bonds of the rock? The answer may be that any stress, no matter how small, can defeat any strength, no matter how large, provided the number of cycles is high enough. Experiments to check such a statement would be very valuable. Problem 24.2 Water flows in a river at a mean depth shear velocity of 2 m/s. The gradient of the shear velocity at the bottom of the river is 7000 m/s per m of depth. Calculate the shear stress applied by the water to the bottom of the river. The soil particles at the bottom of the river are cubes 1 mm in size. They have a unit weight of 26.5 kN/m3 and a friction angle equal to 35∘ . Calculate the shear stress necessary to move the soil grains. Compare this shear stress to the shear stress applied by the water; will there be erosion? Solution 24.2 From the problem statement: v = 2m∕s dv 7000m∕s = dz m Particle size: 1 mm cube 𝛾s = 26.5kN∕m3 𝜇 = 1 × 10−3 Pa ⋅ s a. Shear stress applied by the water at the bottom of the river: ( ) ( ) m∕s dv −3 = (1 × 10 Pa ⋅ s) × 7000 = 7 Pa 𝜏w = 𝜇 dz m b. Shear stress necessary to move the grains: ( ( ) ) 𝛾s × V 26500 × 10−9 𝜏s = 𝜎N × tan 𝜑′ = × tan 35 = 18.56 Pa × tan 𝜑′ = A 10−6 c. Comparison: Stress necessary to move the grains (𝜏 s = 18.56 Pa) is larger than the shear stress generated by the water (𝜏 w = 7 Pa); therefore, there will be no erosion. Problem 24.3 The particle of problem 24.2 is now a 1 mm diameter sphere that rests between two other spherical particles (Figure 24.1s). The particle is subjected to the same shear stress as in problem 24.2. Will the particle be able to roll over its neighbors and erode away?

878

24 EROSION OF SOILS AND SCOUR PROBLEMS

R

R/2 R 3/2 O

60° 60°

R = 1 mm 60°

Figure 24.1s Soil particle.

Solution 24.3 The driving moment M D and resisting moment M R around point O in Figure 24.1s are calculated. For the driving moment, it is assumed that the drag force exerted by the shear stress acts on the projected surface of the spherical particle and that the normal stress applied by the water due to the flow is negligible: ( √ ) 3R 𝜋 𝜋D2 MD = 𝜏w R+ = 7 ∗ 10−6 ∗ ∗ 12 (1 + 0.87) = 10.25 ∗ 10−6 N.mm 4 2 4 R R 26500 4 3 0.5 ∗ 𝜋D ∗ = 𝛾s V = = 27.74 ∗ 10−6 N.mm 2 2 3 2 109 MR > MD → The particle won′t be able to roll over its neighbors and erode away MR = W

Problem 24.4 The straight part of a river is at flood stage and experiences a 160-year flood. During the flood, the water depth is 6 m and the mean depth water velocity is 3 m/s. The bottom of the river is made of sand and the banks have a bank angle of 30∘ . Would you expect the sand to erode? If yes, what size rip rap would you recommend to place on top of the sand to prevent erosion? Would you place a geosynthetic filter between the sand and the rip rap? Explain. Solution 24.4 Yes, one would expect the sand to erode. Indeed, the water velocity is 3 m/s and the critical velocity of the sand will be at most 1 m/s (Figure 24.8). Rip rap can be sized as follows: ( )2.5 Vdes d30 = Hw FCst Cv Ct √ Csl (Gs − 1)gHw The height of water is 6 m and we choose a factor of safety equal to 2. Also, we assume that the rip-rap blocks are angular, so Cst is equal to 0.3. The magnitude of Csl is calculated with 𝜃 = 30∘ : √ ( )1.6 √ ( ) sin(𝜃 − 14∘ ) sin 16∘ 1.6 1 − = = 0.8 Csl = 1 − sin 32∘ sin 32∘ ( )2.5 3 d30 = 6 × 2 × 0.3 × 1.23 × 1 √ = 0.3 m 0.8(2.65 − 1)9.81 × 6 It is very important to place a filter between the soil to be protected and the rip-rap layer. Without a filter, the soil under the rip rap may continue to erode through the large voids in the rip rap; in the end, the rip rap may not move away, but may simply go down significantly as the underlying soil erodes away.

24.12 INTERNAL EROSION OF EARTH DAMS

879

Problem 24.5 A bridge is designed for a life of 50 years and you wish to design the bridge for a flood that has a probability of occurring or being exceeded of 0.001. What should the recurrence interval of the design flood be? Solution 24.5

( )Lt 1 R=1− 1− TR

where T R is the return period and R is the probability of exceedance of the flood. ( )50 )50 ( 1 1 → 1− = 0.999 0.001 = 1 − 1 − TR TR ( ) 1 50 × log 1 − = log 0.999 TR ( ) 1 −4.34512 × 10−4 log 1 − = = −8.690 × 10−6 TR 50 1 1− = 0.99997999 → TR = 50000 yrs. TR The 50,000-year flood is the one to be considered. Problem 24.6 A round-nose pier is 3 m wide and 6 m long. The center-to-center spacing of the piers is 50 m. The water depth at the site is 10 m and the approach flow velocity of 3 m/s has an attack angle equal to 10∘ (Figure 24.2s). EFA tests were conducted; the average erosion function representing the soil is given in Figure 24.3s. The critical velocity of the soil is 1.6 m/s. The duration of the flood is 48 hours. Find the pier scour depth after 48 hours. L = 6m

u = 108 B 5 3m

V1 = 3 m/s Plan view

V1 = 3 m/s Hw = 10 m

zmax (pier) Elevation view

Figure 24.2s Pier scour problem.

880

24 EROSION OF SOILS AND SCOUR PROBLEMS

Vc = 1.6 m/s

Erosion rate, (mm/hr)

12

8

4

0

0

20

40

60

80

Shear stress, τ(Pa)

Figure 24.3s Erosion function.

Solution 24.6 The maximum scour depth and the maximum shear stress around the pier can be calculated as follows. Maximum Scour Depth The correction factors for water depth (K pw ), pier shape (K psh ), pier aspect ratio (K pa ), and pier spacing (K psp ) must be calculated first: { ( )0.33 hw h , for Bw′ < 1.43 0.89 B′ K = pw

, else ) ( ) L 6 B′ = B cos 𝜃 + ⋅ sin 𝜃 = 3 cos 10 + sin 10 = 4 m B 3 hw 10 = = 2.5 > 1.43 so Kpw = 1 B′ 4 Kpsh = 1 because the pier has a round nose (

1.0

K1∕B = 1 because the projected width has been used. { ( )−0.91 S , for BS′ < 3.42 K = 2.9 B′ psp

1.0

, else

50 S = = 12.5 > 3.42 so Kpsp = 1 B′ 4 All correction factors are equal to 1.0. The Froude Number is calculated with the approach velocity and pier width: V 3 = 0.48 Fr(pier) = √ 1 = √ ′ g⋅B 9.81 × 4 The critical pier Froude Number is calculated: V 1.6 Frc(pier) = √ c = √ = 0.255 g ⋅ B′ 9.81 × 4 Therefore, the maximum pier scour depth in given condition is: zmax (pier) = 2.2 ⋅ Kpw ⋅ Kpsh ⋅ Kpa ⋅ Kpsp ⋅ a′ ⋅ (2.6Fr(pier) − Frc(pier) )0.7 = 2.2 × 1 × 1 × 1 × 1 × 4 × (2.6 × 0.480 − 0.255)0.7 = 87.57 m = 8757 mm Maximum Shear Stress around Pier The correction factors for water depth (kpw ) and for pier spacing (kpsp ) are calculated and found equal to 1. For pier shape, kpsh = 1.15 + 7e(−4L∕a) = 1.15 + 7e(−4×6∕2) = 1.15

24.12 INTERNAL EROSION OF EARTH DAMS

881

The angle of attack factor is:

( )0.57 ( )0.57 𝜃 10 = 1 + 1.5 = 1.429 kpsk = 1 + 1.5 90 90 The Reynolds Number based on pier width is: VD 3 × 4 = 12000000 = v 10−6 Therefore, the maximum shear stress around the pier in the given condition is: ( ) 1 1 𝜏max (pier) = kpw × kpsh × kpsp × kpsk × 0.094𝜌V12 − log Re 10 ) ( 1 1 2 = 1 × 1.15 × 1 × 1.429 × 0.094 × 1000 × 3 − log 1200000 10 = 57.36 Pa Re =

Erosion rate, z (mm/hr)

Initial Rate of Scour ż i(pier) is read on the EFA curve at 𝜏 = 𝜏max , and is 10.25 mm/hr (Figure 24.4s).

Vc = 1.6 m/s

10.25mm / hr

12

8

4 57.36 kPa 0

0

20

40

60

80

Shear stress, τ(pa)

Figure 24.4s Erosion function and initial erosion rate for pier scour.

The depth of pier scour after the 48-hour duration of the 160-year flood can be calculated as: zfinal (t) =

t(hrs) = t(hrs) 1 + ż i ys(max )

1 10.2

48 = 463.7 mm 48 + 8757

Problem 24.7 Calculate the abutment and contraction scour depth after 48 hours of flood for the following case. The geometry of the channel and the bridge are given in Figure 24.5s. The compound channel is symmetrical, and the discharge during the flood is Q = 2000 m3 ∕s. The critical velocity of the soil in the main channel and flood plain is 1.2 m/s. The erosion function of the soil from an EFA test is given in Figure 24.6s. The duration of flood is 48 hours, and the hydraulic data are as follows: Mean velocity in the general approach cross section: V1 = 1.13 m∕s Mean velocity in the approach floodplain: Vf1 = 0.78 m∕s Mean velocity in the approach main channel: Vm1 = 1.4 m∕s Water depth in the approach flood plain: Hwf1 = 2.55 m Water depth in the approach main channel: Hwm1 = 7.9 m Mean velocity in the general contracted cross-section: V2 = 1.75 m∕s

882

24 EROSION OF SOILS AND SCOUR PROBLEMS

A

Vm1 = 1.4 m/s

Vf1 = 0.78 m/s

0.5Q total = 1000 cms

Mean velocity in the contracted main channel: Vm2 = 1.83 m∕s Hydraulic radius in the approach main channel: Rh1 = 3.65 m Find the abutment scour depth and the contraction scour depth after 48 hours of flood. CL

A’

Wa = 6 m Le = 124 m Lm = 77 m

Lf = 154 m

Plan view CL Hwf1 = 2.55 m V2 = 1.75 m/s zmax (Abut)

Hwm1 = 7.9 m

zmax (Cont)

Section A-A’

Figure 24.5s Channel geometry.

Erosion rate, z (mm/hr)

7 Vc = 1.6 m/s

6 5 4 3 2 1 0

0

5

10 Shear stress, τ(pa)

15

20

Figure 24.6s Erosion function.

Solution 24.7 Step 1. Calculate the maximum shear stress in the middle of the channel and around the abutment. Contraction Scour The maximum shear stress in the middle of the channel can be calculated as: 𝜏max (Cont) = kcr kcl kc𝜃 kcw 𝛾n2 V1 2 Rh −1∕3 where kcr is the correction factor for the contraction ratio, kcl is the correction factor for the contraction length, kc𝜃 is the correction factor for the transition angle, and kcw is the correction factor for the water depth.

24.12 INTERNAL EROSION OF EARTH DAMS

For this case,

( kcr = 0.62 + 0.38

A1 A2

(

)1.75 = 0.62 + 0.38 ×

V2 V1

)1.75

( = 0.62 + 0.38 ×

1.75 1.13

883

)1.75 = 1.44

Width of the channel at approach section: L1 = (154 + 77) × 2 = 462 m Width of the channel at contraction section: L2 = (30 + 77) × 2 = 214 m ) ( )2 ( 6 6 − 1.98 kcl = 0.77 + 1.36 = 0.80 462 − 214 462 − 214 ( )1.5 90 kc𝜃 = 1 + 0.9 = 1.9 90 kcw = 1 Therefore, 𝜏max (Cont) = kcr kcl kc𝜃 kcw 𝛾n2 V1 2 Rh −1∕3 = 1.44 × 0.80 × 1.9 × 1 × 9810 × 0.0182 × 1.132 × 3.65−1∕3 = 5.77 Pa Abutment Scour The maximum shear stress around the abutment can be calculated as: 𝜏max (Abut) = 12.45 × kacr × kash × kaw × kas × kask × kal × 𝜌 × V1 2 × Re−0.45 where kacr is the contraction ratio influence factor for abutment scour shear stress, kash is the correction factor for aspect ratio of the approach embankment, kaw is the correction factor for Froude Number, kas is the correction factor for abutment shape, kask is the correction factor for the skew angle of the abutment, and kal is the correction factor for abutment location in the flood plain. For this case: q 1.75 − 2.91 = 2.74 kacr = 3.65 2 − 2.91 = 3.65 × q1 1.13 ( )−0.24 ) ( Le 124 −0.24 kash = 0.85 = 0.85 × = 0.41 Wa 6 V 1.13 Fr = √ 1 = 0.23 > 0.1 =√ gHwf1 9.81 × 2.55 Therefore, kaw = 2.07Fr + 0.8 = 2.07 × 0.23 + 0.8 = 1.27 Because this is a spill-through abutment, kas = 0.58 kask = 1 Because

Lf − Le Hwf1

=

154 − 124 = 11.76 > 1, kal = 1 2.55

Therefore, 𝜏max (Abut) = 12.45 × kacr × kash × kaw × kas × kask × kal × 𝜌 × V1 2 × Re−0.45 ( )−0.45 1.13 × 6 2 = 12.45 × 2.74 × 0.41 × 1.27 × 0.58 × 1 × 1 × 1000 × 1.13 × 10−6 = 11.09 Pa

884

24 EROSION OF SOILS AND SCOUR PROBLEMS

The initial rate of scour zi for contraction scour and abutment scour are read on the EFA curve at 𝜏 = 𝜏max : it is 2.01 mm/hr and 4.7 mm/hr respectively as shown in Figure 24.7s.

Erosion rate, z (mm/hr)

7 6 4.7 mm/hr

5 4 3

2.01 mm/hr

2 1 0

5.77 Pa 0

5

11.09 Pa

10 Shear stress, τ(pa)

15

20

Figure 24.7s Erosion function and initial erosion rate for abutment scour and contraction scour.

Step 2. Calculate the maximum contraction scour depth and the maximum abutment scour depth. Contraction Scour The maximum contraction scour depth can be calculated as: zmax (Cont) = 1.27(1.83Frm2 − Frmc ) ⋅ Hwm1 where Frm2 is the Froude Number for the main channel at the bridge in the contracted zone, Frmc is the critical Froude Number for the main channel at the bridge, and H wm1 is the water depth in the main channel at the approach section. For this case, V ∕CR V 1.75 = 0.199 =√ 2 =√ Frm2 = √1 gHwm1 gHwm1 9.81 × 7.9 V 1.2 Frmc = √ mc = √ = 0.136 gHwm1 9.81 × 7.9 Therefore, ys(Cont) = 1.27(1.83Frm2 − Frmc ) ⋅ Hwm1 = 1.27 × (1.83 × 0.199 − 0.136) × 7.9 = 2.29 m Abutment Scour The maximum abutment scour depth can be calculated as: ys(Abut) = Hwf1 ⋅ Kash ⋅ Kask ⋅ Kal ⋅ Kag ⋅ 243 ⋅ Ref 2 −0.28 (1.65Frf 2 − Frfc ) where K ash is the correction factor for the abutment shape, K ask is the correction factor for the abutment skew, K al is the influence factor that takes into account the proximity of the abutment to the main channel, K ag is the geometry of the channel influence factor for abutment scour, Ref 2 is the Reynolds Number around the toe of the abutment, Frf 2 is the Froude Number around the toe of the abutment, and Frfc is the critical Froude Number for soil near the toe of the abutment. For this case, assume that it is a spill-through abutment with 2:1 slope, Kash = 0.73. Kask = 1 − 0.005|𝜃 − 90| = 1 For this compound channel, Kag = 1 Because

Lf − Le Hwf1

Therefore, Kal = 1 Because

=

Lf − Le Hwm1

154 − 124 = 11.8 > 1.5 2.55

=

154 − 124 = 3.8 < 5 7.9

24.12 INTERNAL EROSION OF EARTH DAMS

885

it is a short setback condition. Therefore, 0.5Q = 0.5 × V2 = 0.5 × 1.75 = 0.875 m∕s A2 Vf 2 ⋅ Hwf 1 0.875 × 2.55 Ref 2 = = 2.23 × 106 = v 10−6 Vf 2 0.875 = 0.175 Frf 2 = √ =√ gHwf 1 9.81 × 2.55 Vfc 1.2 Frfc = √ = 0.24 =√ gHwf 1 9.81 × 2.55 Vf 2 =

and thus zmax (Abut) = Hwf1 ⋅ Kash ⋅ Kask ⋅ Kal ⋅ Kag ⋅ 243 ⋅ Ref 2 −0.28 (1.65Frf 2 − Frfc ) = 2.55 × 0.73 × 1 × 1 × 1 × 243 × (2.23 × 106 )−0.28 (1.65 × 0.175 − 0.24) = 0.368 m Step 3. Calculate the depth of contraction scour and abutment scour after 48 hours. z(Cont) (t) = zmax (Abut) (t) =

t(hrs) 1 ż i

+

zmax (Cont)

t(hrs) 1 ż i

+

=

t(hrs)

t(hrs) ys(Abut)

=

1 2.01

1 4.7

48 = 93 mm 48 + 2290

48 = 140 mm 48 + 368

Therefore, the contraction scour depth generated by the 48-hour flood is 4.1% of the maximum contraction scour depth, whereas the abutment scour depth generated by the same flood is 38% of the maximum abutment scour depth. Problem 24.8 Download the SRICOS-EFA program from the website https://briaud.engr.tamu.edu/ and run Example 1 from the list of examples. Solution 24.8 See the website. Problem 24.9 Download the MEANDER program from the website https://briaud.engr.tamu.edu/ and run Example 1 from the list of examples. Solution 24.9 1. Install MCRInstaller and run MEANDER. 2. Open the first example, Brazos1958C_const_NGP.meander, in the Data folder included with the program. This example is for migration with a constant discharge. 3. Choose between SI units or English units in Input > Units . . . . Choose the SI Units. 4. Open the Geometry window. This window lets you open the file with the initial coordinates of the river and fit circles that represent the meanders. Browse the geometry file Brazos_1958C_2006.dat included in the Data folder, which has the initial coordinates of the river. 5. The numbers in the Geometry window (Figure 24.8s) must be: Average River Width is 110 m, the Tick Spacing is 200, and the Criterion Lines 1, 2, and 3 are 10, 0, and 0 respectively. Click Fit Circles. Click Return after the circles are fitted. 6. The next window (Figure 24.9s) lets you input the data soil. Input the EFA curve on the Soil Data window and choose the sand option for the type of the soil.

886

24 EROSION OF SOILS AND SCOUR PROBLEMS

Figure 24.8s Input the geometry data.

Figure 24.9s Soil data window.

24.12 INTERNAL EROSION OF EARTH DAMS

887

7. Open the Water Data window (Figure 24.10s). The Critical Froude Number is 0.17 and the Time Step is 240. The speed of the program depends on this increment. The discharge, in the case of this example, is constant. The discharge units are in cubic meters per second. The time period is one year or 365 days. The discharge versus velocity and the discharge versus water depth have to be obtained from software such as HEC-RAS or TAMU-FLOW. These programs perform their analyses based on the cross-section of the river. Click OK after you are done. 8. Before running the program, you may want to check the data again by clicking Input Tables and Input Plots. These two options let you review your data. 9. Once all the data are in, you can click the Run button. After the program finishes the calculations, click the Output Plots icon (Figure 24.11s). Click Center Line or One Bank to see the results of the meander migration (Figure 24.12s).

Figure 24.10s Water data input.

888

24 EROSION OF SOILS AND SCOUR PROBLEMS

Figure 24.11s Output plots windows.

Figure 24.12s Results of the meander migration.

24.12 INTERNAL EROSION OF EARTH DAMS

889

Problem 24.10 A 5 m-high levee is overtopped for 2 hours during a hurricane. The levee material and the soil below the levee are borderline between a high-plasticity clay CH and a low-plasticity clay CL. Draw a contour of the levee after 2 hours of overtopping. Solution 24.10 To draw the contour of the levee after 2 hours of overtopping erosion, we must first find where the erosion will start. When the water overtops the levee, it accelerates and reaches the critical velocity of the soil Vc = 1.1 m∕s after some distance from the levee crest. This distance is such that: √ V = 2gH √ 1.1 = 2 × 9.81 × H → H = 0.06 m So, erosion will start once the water has reached a levee height equal to 5 − 0.06 m = 4.94 m. Then we select five points along the levee and compute the erosion after 2 hours (problem 24.10) and after 72 hours (problem 24.11). To calculate the erosion depth, we first calculate the water velocity, then find the corresponding erosion rate from the erosion function, and then multiply the erosion rate by the overtopping duration (2 or 72 hours). The levee soil is borderline between a high-plasticity clay CH and a low-plasticity clay CL, so the erosion function is selected as the boundary line on the erosion chart of Figure 24.13s. Example calculations are shown for a depth of 4 m below the crest of the levee. H=4m √ √ V = 2gH = 2 × 9.81 × 4 = 8.86 m∕s → erosion rate = 58 mm∕hr Erosion depth 100000

Erosion rate (mm/hr)

10000

1000

Very high erodibility I

High erodibility II

Medium erodibility III

-Jointed rock Low (spacing < 30 mm) -Fine gravel erodibility -Coarse sand IV -High plasticity silt -Jointed rock -Low plasticity clay (30–150 mm spacing) -All fissured -Cobbles clays -Coarse gravel - Increase in compaction Very low -High plasticity clay erodibility (well-graded soils) V - Increase in density -Jointed rock - Increase in water salinity (150–1500 mm spacing) (clay) -Riprap

-Fine sand -Non-plastic silt

100

10 SM 1

z = 58 × 2 = 116 mm

SP

ML

-Medium sand -Low plasticity silt

MH

CH

CL

0.1 0.1

1.0

Non-erosive VI

Rock

-Intact rock -Jointed rock (Spacing > 1500 mm)

10

100

Velocity (m/s)

Figure 24.13s Proposed erosion categories for soils and rocks based on velocity.

Table 24.1s shows all the calculations, and the contours of erosion after 2 hours and 72 hours of overtopping erosion are shown in Figure 24.14s. Table 24.1s

Erosion calculations and contours

Drop height m

Velocity m/s

Erosion rate mm/hr

Erosion depth after 2 hr m

Erosion depth after 72 hr m

0 0.06 1 2 3 4 5

0 1.1 4.43 6.26 7.67 8.86 9.9

0 0.1 6 18 35 58 91

0 0.0002 0.0012 0.036 0.07 0.116 0.182

0 0.0072 0.432 1.296 2.52 4.176 6.552

890

24 EROSION OF SOILS AND SCOUR PROBLEMS

v = 1 m/s

5

1 5.00

0 2 4

x

6

Erosion contour after 2 hours

8 10 12

Z

Erosion contour after 72 hours

Figure 24.14s Contours of erosion after 2 and 72 hours.

Problem 24.11 Repeat problem 24.10 for a flood that lasts 72 hours. Solution 24.11 Same approach but multiply by 72 hours instead of 2 hours. Problem 24.12 The core of an earth dam is made of clay with the grain size distribution curve shown in Figure 24.15s. Design the first filter layer to be placed next to the core of the dam as part of the transition to the rockfill stabilization layer.

100

Clay

Sand

Silt

Gravel

Percent finer (%)

80

60

40

20

0 10–4

10–3

10–2

10–1

100

101

102

Particle size (mm) Figure 24.15s Grain size distribution curve for core of dam.

Solution 24.12 The grain size distribution curve of the granular filter must satisfy the following requirements where subscript F refers to the fine core material and subscript C refers to the coarser filter material. 1. Piping criterion D15 C ≤ 5D85 F ; D15 C ≤ 5 × 2 × 10−2 = 0.1 mm 2. Permeability criterion D15 C ≥ 5D15 F ; D15 C ≥ 5 × 10−3 = 0.005 mm 3. Gradation criterion D50 C ≤ 25D50 F ; D50 C ≤ 25 × 2.4 × 10−3 = 0.06 mm

24.12 INTERNAL EROSION OF EARTH DAMS

A possible filter gradation is shown on Figure 24.16s.

100

Silt

Clay

Sand

Gravel

D85F

Percent finer (%)

80

60 D50F

25D50F

40

20

D15F

5D15F 5D85F

0 10–4

10–3

10–2

10–1

100

101

102

Particle size (mm) Figure 24.16s Grain size distribution curve for core of dam and possible filter.

891

CHAPTER 25

Geoenvironmental Engineering

T

his chapter introduces a relatively recent yet very important field: geoenvironmental engineering. The book by Sharma and Reddy (2004) is an excellent reference for further study. 25.1

INTRODUCTION

After World War II, there was a remarkable development of industries aimed at improving the quality of life. These chemical and manufacturing industries created a significant amount of waste in an unregulated environment. The advent of nuclear power plants in the late 1950s brought the problem of nuclear wastes, which can remain deadly for thousands of years. Disposal of such deadly wastes caught the attention of the public and emphasized the need for a more organized approach. The Love Canal disaster occurred when an old canal that had been filled with toxic chemicals for many years started to leak and seriously affect the health of local residents in New York State in the 1970s. This broadly publicized disaster contributed to the development of laws and regulations aimed at ensuring the health and safety of the public by avoiding any environmental contamination. 25.2

TYPES OF WASTES AND CONTAMINANTS

Wastes are unwanted or useless materials. They come from many different human activities and take many different

forms. Sources of wastes include dredging, mining, and farming: they can also be generated by the residential, commercial, institutional, industrial, nuclear power, and defense sectors. The waste generated by those human activities amounts to something like 100 billion kN or 10 billion m3 per year worldwide. Wastes can be in gas form, liquid form, or solid form. In geotechnical engineering, the liquid and gas forms are of concern because they propagate through the soil that must be cleaned up; this falls under the topics of contamination and remediation. The solid wastes, hazardous or not, end up being stored in landfills or other facilities designed for that purpose. In the United States alone, the amount of waste generated each year is staggering, as shown in Figure 25.1 and Table 25.1. Wastes can be categorized into the following different types (Sharma and Reddy, 2004): 1. Solid wastes. The term solid waste is misleading, as a solid waste can be a solid, a liquid, or a gas. These wastes include municipal solid waste (MSW), industrial solid wastes, and construction and demolition wastes. MSW includes household solid wastes and commercial solid wastes. Figure 25.2 shows the distribution of MSW in percent of the total quantity and its evolution over the past 50 years in the United States. Industrial solid wastes are nonhazardous wastes generated by manufacturing and industrial processes, including the production of furniture, Table 25.1

Type and quantity of waste in the USA

Amount of waste in the USA in meganewtons per year (mn/yr) Municipal solid waste

92.3% 46.5%

0.005%

0.006%

Meganewtons per year (MN/YR)

Industrial solid waste Hazardous waste

3.0%

Waste

Radioactive waste Medical waste

Figure 25.1 Amount of waste per year in the United States. (Source: Sharma and Reddy, 2004/John Wiley & Sons.)

Municipal solid waste Industrial solid waste Hazardous waste Radioactive waste Medical waste

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

2,500,000 76,000,000 3,800,000 3,750 4,650 893

894

25 GEOENVIRONMENTAL ENGINEERING

40

Percent of total

35

Total municipal solid waste in us = 2.5 millions mn/yr

35

Paper

30 25

30

Yard trimmings

25

20 15

40

20

Food scraps

15

Metals

10

Wood

5 0 1960

10 5

Glass

Plastics 1970

1980

1990 Years

2000

2010

0 2020

Figure 25.2 Distribution of MSW and its evolution.

apparel, machinery, buses, trucks, cars, airplanes, jewelry, shoes, and so on. Construction and demolition wastes include wood, concrete, bricks, and plumbing materials, among others. 2. Hazardous wastes. These are wastes that can cause death or serious illness or pose a substantial hazard to human health or the environment. In the United States, they are defined in subtitle C of the Resource Conservation and Recovery Act (RCRA). The Environmental Protection Agency (EPA) distinguishes three categories of hazardous wastes: nonspecific source wastes (e.g., solvents, dioxins), specific source wastes (e.g., sludge from petroleum refining and organic chemical manufacturing), and commercial chemical product source waste (e.g., creosote, acids, pesticides). Another important factor is the concentration level of the chemical in the liquid (water); the acceptable level is defined in the regulations. For example, 0.05 milligram of mercury per liter would be acceptable but 0.3 mg/L would not; 1 mg of lead per liter would be acceptable but 7 mg/L would not. 3. Radioactive waste. These wastes are classified into four categories of their own: a. High-level wastes (HLW). These are liquid or solid wastes that are extremely dangerous and must not come into contact with humans. They come from defense or nuclear power plant activities. They require permanent isolation, as the radiation penetration remains lethal for 10,000 years. b. Transuranic wastes (TRU). These come from manufacture of nuclear weapons and processing of nuclear fuels. The radiation penetration remains lethal for 20 years. TRUs are relatively rare. c. Low-level wastes (LLW). These are the lowest-level radioactive wastes, which can be disposed of as regular waste after sufficient isotope decay. d. Mill tailings. These are the ore residues from mining uranium. There is a significant amount of it and it is typically stored, sometimes buried in large remote areas of a country.

4. Medical waste. These wastes come from hospitals and other health agencies. They include microbiological wastes (e.g., infectious cultures), human blood, pathological wastes (e.g., organs, body parts), contaminated animal wastes, isolation wastes (e.g., waste contaminated with highly communicable diseases), contaminated sharps (e.g., needles, scalpels), and uncontaminated sharps. Some MSW, such as disposable diapers and sanitary napkins, do contain pathogens, but usually not as much as medical wastes. One common solution for medical wastes is incineration to kill the disease-causing pathogens. 25.3

LAWS AND REGULATIONS

In the United States, once a bill is proposed, passes both Houses of Congress, and is signed by the president, it becomes a law (also known as an act or statute). To enforce the new act, the responsible agency—the Environmental Protection Agency in most cases for geoenvironmental engineering—develops regulations to implement the act. All federal government regulations are collected in enormous books of the Code of Federal Regulations; this series of books is divided into volumes called titles (Title 40 covers topics relating to geoengineering), which are further divided into parts and then sections. The EPA also develops Guidance Documents for technical issues and Policies for decision management to help industry comply with the regulations. Each state can use the federal regulations as is or enforce stricter, state-specific versions. The EPA office in each state is in charge of enforcing the regulations. Laws have been passed on a wide variety of environmental topics. Some of the most important and with the widest range include: • • • • • • •

Solid Waste Disposal Act (SWDA; 1965, 1970) National Environmental Policy Act (NEPA; 1969) Occupational Safety and Health Act (OSHA; 1970) Clean Air Act (CAA; 1970, 1977, 1990) Clean Water Act (CWA; 1977, 1981, 1987) Safe Drinking Water Act (SDWA; 1974, 1977, 1986) Resource Conservation and Recovery Act (RCRA; 1976, 1980) • Comprehensive Environmental Response, Compensation, and Liability Act (CERCLA; 1980) These laws have positively impacted geoenvironmental engineering practice. In particular, CERCLA, also known as Superfund, generated a tremendous amount of work for geotechnical engineers regarding the cleaning up of soil deposits. As society evolves and humankind creates new products, new laws are enacted and amendments are made to existing acts. This is the case of the Hazardous and Solid Waste Amendments to RCRA (HSWA; 1984) and the Superfund Amendments and Reauthorization Amendments

25.4 GEOCHEMISTRY BACKGROUND

Table 25.2 (OSHA)

Levels of safety and health protection

Protection

Level A

Level B

Level C

Level D

Respiratory Maximum Maximum Moderate Minimum Skin Maximum Very high Moderate Minimum Eye Maximum Very high Moderate Minimum to CERCLA (SARA; 1986). CERCLA/Superfund addressed the issue of cleaning up contaminated sites, whereas RCRA addressed the issue of landfill design. Safety and health are the top priority regarding investigation and remediation at contaminated sites. The following levels of protection for humans working on contaminated sites have been formulated by the Occupational Safety and Health Administration (OSHA) in the United States. They address the level of protection required for eye, skin, and respiratory safety (Table 25.2). 1. Level D. Minimum protection including coveralls, gloves, chemical-resistant steel-toed boots, safety glasses, hard hat, escape mask, and face shield. 2. Level C. Moderate protection, including full-face or half-mask, air-purifying respirator, hooded chemicalresistant clothing, inner and outer chemical-resistant gloves, chemical-resistant boots and boot covers, hard hat, escape mask, and face shield. 3. Level B. Very high protection, including positivepressure, full-face, self-contained breathing apparatus (SCBA), hooded chemical-resistant clothing, chemical-resistant inner and outer gloves, chemicalresistant steel-toed boots and boot covers, hard hat, and face shield. 4. Level A. Maximum protection, including positivepressure, full-face, self-contained breathing apparatus (SCBA), totally encapsulating chemical-protective suit, chemical-resistant inner and outer gloves, chemicalresistant steel-toed boots and boot covers, and hard hat (under suit).

25.4 25.4.1

GEOCHEMISTRY BACKGROUND Chemistry Background

A discussion of geochemistry starts with a background lesson on chemistry. Electrons, protons, and neutrons make up the universe. They are extremely small subatomic particles with masses of 1.66 × 10−27 kg for neutrons and protons and 9.1 × 10−31 kg for the much lighter electron. Electrons are negatively charged, protons are positively charged, and neutrons are electrically neutral. Atoms, which consist of electrons, protons, and neutrons in various combinations, are the basic building blocks of matter. Hydrogen, oxygen, and

895

carbon, for example, are atoms. You can break them down into smaller pieces, but you will no longer have hydrogen, oxygen, or carbon. The nucleus of the atom is made of neutrons and protons, while electrons gravitate around the nucleus. An element is made of only one kind of atom. The periodic table (Figure 25.3) in chemistry gives the list of the different elements, 92 of which occur naturally (e.g., hydrogen, oxygen, carbon) and some two dozen others of which have been created by scientists. By changing the number of protons and electrons in the atom, you create different elements. For example, oxygen has 8 protons, but hydrogen has only 1. This is an element’s atomic number. The periodic table gives the atomic number (number of protons in the nucleus) and the atomic mass, which is the sum of the weight of the protons and the neutrons times the Avogadro number. Amadeo Avogadro was an Italian scientist who contributed significantly to molecular theory in the early 1800s. Note that the weight of the electrons is negligible compared to the weight of the protons and neutrons. Stable atoms have the same number of electrons and protons. The Avogadro number (6.022 × 1023 ) is the number of atoms in 12 grams of carbon12. This number helps define the mole. In biology, a mole is an underground rodent, but in chemistry it is the mass corresponding to 6.022 × 1023 (Avogadro number) molecules, atoms, or some other pure chemical substance unit. The periodic table gives the mass of one mole of each of the elements. For example, a mole of water (H2 O) has a mass of two moles of hydrogen (2 × 1.00794 g) plus one mole of oxygen (15.9994 g) or 18.01528 g. Note that, in chemistry, mass is calculated with a much larger number of significant figures than in geotechnical engineering. The reason is the reproducibility of the results and the need to be that accurate. Molecules are combinations of atoms. Two atoms of oxygen (O), for example, form the molecule O2 ; two hydrogen atoms and an oxygen atom form a molecule of water (H2 O); a molecule of sugar is C12 H22 O11 . If an atom or a molecule loses or gains an electron on its outer orbital shell, it becomes an ion, either a cation or an anion. Cations have lost electron(s), are positively charged, and move toward the cathode because a cathode is negatively charged in an electrical circuit. Examples of cations are Na+ , Ca++ , and Al+++ . Anions have gained electron(s), are negatively charged, and move toward the anode because an anode is positively charged in an electrical circuit. Examples of anions are Cl– , O– , and N– . Radicals are groups of atoms that are common to many molecules. Such radicals include hydroxyls (OH– ), carbonates (CO3 – ), sulfates (SO4 – ), and nitrates (NO3 – ). Molecules are bound by ionic bonds (attraction between ions having opposite charges) or covalent bonds (bound by shared electrons). The valence or oxidation number of an element in a molecule is the number of electrons transferred to or from an atom of the element with atoms of other elements in the molecule (e.g., O2– , H1+ , C4+ ). The molecular mass is the sum of the atomic masses of the elements in the molecule.

896

25 GEOENVIRONMENTAL ENGINEERING

Figure 25.3 Periodic table of elements. (Source: Courtesy of National Institute of Standards and Technology, U.S. Department of Commerce.)

The concentration of a chemical in solution can be defined in one of three ways: 1. Molar concentration: number of moles per liter of solution. 2. Mole fraction: number of moles of a substance divided by the total number of moles in the solution. 3. Mass concentration (most commonly used in geoenvironmental engineering): mass of the element or substance in milligrams per liter (mg/L) of solution. If the solution is water (mass density = 1000 g/L), then the mass concentration is mg per 1,000,000 mg of solution or part per million (ppm). The unit of ppm is commonly used in geoenvironmental engineering. So, for the mass concentration of a chemical in water: 1 ppm = 1 mg∕liter

(25.1)

The pH, a measure of acidity, gives the concentration of hydrogen ions by: pH = − log(H+ )

(25.2)

where log is the logarithm base 10 and (H+ ) is the concentration of hydrogen ions in moles per liter of solution—or, more accurately, the hydrogen ions’ activity, which can be slightly different. The pH scale varies from 0 to 14, with a neutral solution (distilled water) having a pH of 7. A pH less than 7 is acidic and more than 7 is basic or alkaline. Example of pH values include battery acids pH = 1, vinegar and lemon juice 2.5, wine 3.2, beer 4, human blood 7, baking soda 8.5, soap 10, and ammonia 12. The p in pH is said to stand for potential and pH for potential of hydrogen. Chemical reactions take place when reactants are transformed into products. For example: NaOH + HCl → Nacl + H2 O Sodium Hydroxide + Hydrochloric Acid → Salt + Water (25.3) These chemical reactions must satisfy conservation of mass on both sides of the reaction equation. There are at least four types of chemical reactions:

25.4 GEOCHEMISTRY BACKGROUND

1. Acid-base reactions. These reactions affect the pH of the soil and groundwater. Equation (25.3) is an example. 2. Precipitation-dissolution reactions. Some chemicals are more soluble (sugar and salt) in water than others (oil). This is important in geoenvironmental engineering, as it can affect a remedial operation. 3. Oxidation-reduction reactions (redox reactions). Oxidation is the loss of electrons by a molecule, atom, or ion. Reduction is a gain of electrons. For example, oxidation of carbon can produce carbon dioxide, CO2 , but reduction of carbon can produce methane, CH4 . Both gases are produced by the long-term degradation of landfills. 4. Complexation reactions. These are chemical reactions that take place between a metal ion and a molecular or ionic entity known as a ligand. The properties of these complexes, including solubility, can be quite different from the properties of the metal itself, and such transformations can help in cleanup strategies. Inorganic chemistry and organic chemistry are two very important branches of chemistry. The difference between organic and inorganic compounds is that most organic compounds contain carbon, whereas most inorganic compounds do not. Carbon is the fourth most abundant element on and in our planet, and organic compounds vastly outweigh (100 to 1?) inorganic compounds. Living organisms, petroleum and its derivatives, plastics, rubber, fat, sugar, proteins, and enzymes are examples of organic compounds. Metals and salt are examples of inorganic compounds. Nuclear chemistry is another branch of chemistry. Elements with high atomic numbers, like uranium, tend to be unstable and break down. During this process, these elements, called radionuclides, emit radiation (𝛼 rays, 𝛽 rays, 𝛾 rays) with an intensity that decays very slowly. A radionuclide is characterized by its half-life, which is the time required for 50% of the atoms in a substance to decay into more stable substances. All these aspects of chemistry are important to the geoenvironmental engineer who wishes to select the best response possible to contamination and disposal problems. 25.4.2

Geochemistry Background

Geochemistry is the application of chemistry to the field of geoenvironmental engineering. It is concerned with the interaction between chemicals and soils at temperatures and pressures associated with soil deposits. From the point of view of inorganic chemistry, contamination of soils by toxic metals is the main issue. These toxic metals include, for example, lead (Pb), mercury (Hg), and arsenic (As). They can be dissolved in the pore water (aqueous phase), attached to the particle surface (adsorbed phase), or stuck in the pores as separate solids (solid phase). The geochemical processes controlling the distribution of metals among the three phases include the four chemical reactions

897

mentioned in Section 25.4.1, plus adsorption and desorption. Adsorption is the accumulations of ions on the charged surface of soil particles; desorption is the decrease of ions on the particle surface. The impact of each of these geochemical processes on the contaminant and the soil should be carefully evaluated before any remediation decision is made. This can be done on samples in the laboratory or by computation and modeling. The total concentration of metal in soil is obtained by washing the soil with an acid and using atomic absorption spectrophotometry (AAS), for example, to study the leachate. The toxicity characteristics leaching procedure (TCLP) simulates the leaching that a waste might undergo when disposed of in a landfill. The TCLP is used to determine if a waste is hazardous or not and to determine the necessary level of treatment if it is hazardous. The different types of metals present in the soil can be identified by sequential extraction, which consists of using solutions of increasing strength and analyzing the leachate. The metal concentration is the mass of metal divided by the mass of dry soil expressed in mg/kg or ppm (parts per million) or in μg/kg or ppb (parts per billion). From the point of view of organic chemistry, contamination of soils by hydrocarbons is the main issue. These hydrocarbons include, for example, benzene, toluene, and xylene. Organic contaminants also include polychlorinated biphenyls (PCBs), and pesticides (aldrin, endrin). A commonly encountered group of hydrocarbon contaminants is the nonaqueous phase liquids (NAPLs), which do not mix with water (e.g., oil). NAPLs are further separated into those that float on water, called light NAPLs or LNAPLs; and those that sink through water, called dense NAPLs or DNAPLs. LNAPLs typically come from spills of fuels like gasoline, kerosene, or diesel, whereas DNAPLs come from degreasing, metal stripping, and pesticide manufacturing. NAPLs are found in soils in the gas phase, in the liquid phase, and attached to the surface of the particle. Transformation from one phase to another involves volatilization (liquid to gas), dissolution (mixing in water), adsorption (attachment to particle surface), and biodegradation. Biodegradation is a redox reaction that is particularly suited to the action of microbes on NAPLs dissolved in water. The properties of NAPLs are studied in the laboratory and include density, viscosity, solubility, volatility, and surface tension. These properties all affect the optimization of the remedial measure. One way to quantify the amount of NAPLs in soils is to measure the NAPLs’ degree of saturation SNAPL , defined as: V SNAPL = NAPL (25.4) Vv where V NAPL is the volume of NAPLs in the voids and V v is the total volume of voids. Another way to find out how many NAPLs are in the soil is to wash the soil with a solvent and then analyze the solution obtained by a process such as gas chromatography-mass spectrometry (GC-MS).

898 25.5 25.5.1

25 GEOENVIRONMENTAL ENGINEERING

CONTAMINATION Contamination Sources

Contamination can be due to sources on the ground surface, in the zone above the water table (the vadose zone; vadosus means shallow in Latin), or in the zone below the water table. On the ground surface, sources include infiltration of contaminated surface water, land disposal of liquid or solid wastes, accidental spills, fertilizers, pesticides, disposal of sewage, wastewater treatment plant sludge, salt used on roads in icy conditions, animal feedlots, and fallout from automobile emissions. In the vadose zone, sources include landfills, surface impoundments, leakage from underground storage tanks (e.g., service station tanks and septic tanks), leakage from underground pipelines, and disposal at the bottom of shallow excavations. Below the groundwater level, sources include deep well injections, mines, abandoned oil wells, and disposal in deep excavations. Both the soil particles and the groundwater can potentially be contaminated, and the most serious contaminants are heavy metals, hydrocarbons, and radionuclides. Although the total contaminated land area may be a fraction of a percent of a country’s total surface, it is important to remediate all sites. The cost of remediation can be very high, with an estimated average in the range of $1 million per site. As there are some hundreds of thousands of such sites, the cost could potentially reach hundreds of billions of dollars. Whatever the cost, and wherever the site, the cleanup or remediation process starts with detection of the contamination. 25.5.2 Contamination Detection and Site Characterization The following are the steps in the remediation process: 1. Detection of the contamination. 2. Establishment of the vertical and horizontal extent of the contamination. 3. Identification of the contaminants. 4. Assessment of the risk and impact. 5. Choice and design of the remediation scheme. 6. Execution of the remediation work. 7. Verification of the solution. Environmental site assessments (ESAs) are part of the detection process. An ESA may be required when purchasing a piece of property in the United States. There are three levels: ESA I. This phase consists of collecting information regarding previous ownership and prior use, using records of contaminated sites in the area, aerial photos, geologic and topographic maps, visit(s) to the site, and talking to neighbors. An ESA I indicates whether there are reasons to believe the site may be contaminated. If so, ESA II comes into play.

ESA II. This phase consists of testing the soil and the groundwater to find out if there is contamination and, if so, to what extent and to what level of severity (type of contaminants). If it is found that there is contamination requiring cleanup, ESA III comes into play. ESA III. This phase consists of designing and implementing the remediation scheme, including verification that a satisfactory level of cleanup has been achieved. The plan for an environmental site characterization always includes a Safety and Health plan (S&H) and a Quality Assurance-Quality Control plan (QA-QC). The site characterization can make use of drilling and sampling methods, geophysical methods, or in situ testing methods. Drilling and sampling are described in Chapter 7 for uncontaminated sites. Drilling can be done by hollow stem auger drilling, wet rotary drilling, or air pressure rotary drilling, but the hollow stem auger is usually favored for contaminated sites. The reason is that it can be used dry, and minimizes the amount of contaminated fluid generated and the associated disposal cost. For most levels of contamination, the drill rig must be decontaminated after each boring. This is accomplished by pressure steam-washing the rig and washing the drilling and sampling tools with a strong detergent solution and rinsing with clean water. The purpose of this cleaning process is to avoid cross-contamination between borings. Another difference in the case of contaminated sites is that the drillers have to wear the appropriate level of protection (see Section 25.3). The soil samples are the same as in the case of uncontaminated sites, but a much stricter chain of custody is followed for the samples. The chain of custody is a documentary trail that follows the sample through its entire life, including when and where it was taken, who was responsible for it, what happened to it, and everything else, all the way to final disposal. When two people are involved in the transfer of a sample, they both sign and date the chain of custody document. Geophysical methods are described in Chapter 9 for uncontaminated sites. They are useful for determining the large-scale stratigraphy of the site and therefore the boundaries of the potential contamination. Surface geophysical methods are particularly convenient for contaminated sites because they are nonintrusive. In situ testing methods are described in Chapter 8 for uncontaminated sites. In the case of contaminated sites, the cone penetrometer is particularly useful because it limits the amount of contaminated cuttings and contaminated water generated during testing. As a result, a number of techniques have been developed for using the CPT at such sites. One of them is the characterization of petroleum-contaminated sites with laser-induced fluorescence or LIF (ASTM D6187). In this test, the CPT probe is equipped with a side window (Figure 25.4) and a laser beam shines on the soil as the cone is penetrating at 20 mm/s. The laser beam causes the soil and the hydrocarbon to generate fluorescence, which is measured.

25.5 CONTAMINATION

899

Table 25.3 Some values of fluorescence for hydrocarbons Aromatic Molecular Fluorescence Fluorescence hydrocarbon weight (g/mole) range (nm) color Toluene Naphthalene Anthracene Benzo(a) Pyrene Perylene Figure 25.4 LED-induced fluorescence cone penetrometer probe. (Source: Courtesy of Vertek, a Division of Applied Research Associates, Inc.) Gasoline

92 128 178 252 252 228

270–310 310–370 370–470 400–500 440–530 470–580

Faint purple Blue Blue-green Green Green Green-yellow

(Source: Adapted from https://www.vertekcpt.com/wp-content/ uploads/2020/09/Vertek_Geotechnical_and_Environmental_ Catalog.pdf.)

Signal intensity

Signal intensity

Jet fuel

Wave length (nm)

Diesel fuel

Creosote/coal tar

Signal intensity

Signal intensity

Wave length (nm)

Wave length (nm)

Wave length (nm)

Figure 25.5 Fluorescence response of several hydrocarbons. (Source: Courtesy of Fugro.)

Depth (m)

Each type of hydrocarbon has a “fingerprint” or “signature” signal in terms of intensity and wave length, as shown in Figure 25.5 and Table 25.3. Side-by-side LIF soundings give the extent of the contaminated plume (Figure 25.6).

Another CPT technique adapted for contaminated sites is the BAT water sampler (Figure 25.7). BAT is the name of a company. In this case, the CPT probe is equipped with a porous filter that is obstructed until the CPT probe is pushed to the required depth. Then the filter is exposed and water is allowed to penetrate through the filter into a water sampling tube that can be removed through the CPT rods when full. Drawing the water into the sampling tube is accomplished by using the suction of a hypodermic needle. Yet another CPT technique developed for contaminated sites is the MIP gas sampler (Figure 25.8). MIP stands for membrane interface probe. The cone penetrometer probe is fitted with a hydrophobic, semipermeable membrane and a heater. The heater is kept at a temperature higher than 100∘ C and vaporizes any volatile organic compound (VOC). The natural pressure gradient created by the heat source forces the gas to penetrate through the semipermeable membrane. Once through the membrane, the gas is swept by an inert carrier gas to the surface, where it is analyzed by a series of detectors.

0

0

0

0.5

0.5

0.5

1

1

1

1.5

1.5

1.5

2

2

2

2.5

2.5

2.5

3

3

3

3.5

3.5

3.5

4

4

4

4.5

4.5

4.5

5 –3

0 3 6 9 12 15

5 –3

0 3 6 9 12 15 Output (V)

5 –3

0 3 6 9 12 15

Figure 25.6 LIF CPT soundings and plume identification.

900

25 GEOENVIRONMENTAL ENGINEERING

Extension pipe

Sampling barrel

Evacuated sample tube

Retractable tip

Septum Double-ended injection needle Septum BAT MKIII filter tip

Figure 25.7 BAT CPT water sampler.

Carrier gas supply

Gas return tube (to GC)

Permeable membrane

Volatile organic contaminants Fugro's cone penetrometer

Figure 25.8 MIP CPT gas sampler.

The CPT soil sampler is also convenient for contaminated sites (Figure 25.9). The test consists of pushing a cone penetrometer to the depth where a sample is needed. At the required depth, the cone tip insert is disengaged, the CPT probe is advanced to collect the soil sample while the cone tip stays stationary as in a piston sampler, and then the CPT probe is pulled back to the surface. Such CPT samples are about 40 mm in diameter and up to 400 mm in length.

Figure 25.9 CPT soil sampler.

Hydrogeologic data are also very important, as contamination of the groundwater is of great concern to human life. There are two components to these studies. One deals with determining where the groundwater table is and which way the water is flowing (if at all). The other is a determination of the hydraulic conductivity of the soil. Determining the groundwater level is done through the use of monitoring wells and piezometers, as described in Chapter 7, Section 7.6; in situ hydraulic conductivity measurements are covered in Chapter 8, Section 8.12. Finally, the chemical analysis of any soil samples and water samples collected has to be conducted under controlled conditions. This chemical analysis aims at identifying the type and concentration of the chemicals in the ground. The final report should include the geologic data, the soil data, the hydrogeologic data, and the chemical data. It should identify the type of contamination, the extent of the contamination, and the future movement of the contaminants. 25.5.3

Contaminant Transport and Fate

The word transport refers to the flow of the contaminant and the word fate to the change in form and concentration of the contaminant through chemical reactions. For uncontaminated sites, the steady-state (no influence of time) flow of water through soil is governed by Darcy’s law (constitutive law) and conservation of mass (fundamental law). These two equations, combined with boundary conditions and soil properties (hydraulic conductivity), give the solution to steady-state flow problems. This topic was covered in Chapter 14. For transient (influence of time) flow of water through soil, additional equations must be used. Consolidation of saturated layers is an example of transient flow; in that case, the additional equation is the one linking the stress applied to the change in volume of the element. This change of volume (settlement) varies with time and adds to the volume of water flowing through the soil. This problem is presented in Chapter 12, Section 12.4.6 and Chapter 15, Section 15.14. For contaminant flow through soil, an additional issue is the movement of the contaminant in terms of concentration.

25.5 CONTAMINATION

901

The contaminant may be found in the gas phase, in the liquid phase, or in the solid phase. This movement can take place through transport processes, chemical reaction processes, and biological processes. Transport Processes Transport processes include advection, diffusion, and dispersion. Advection. Advection is the movement of the contaminant liquid under a hydraulic gradient. It is the same case as water flow, with Darcy’s law and conservation of mass as the equations. One difference is that the seepage velocity vs is used rather than the discharge velocity v (see Chapter 14, Section 14.2.1) because vs represents the actual velocity of the contaminant movement: Q = vA = vs Av or

v = nvs

(25.5)

where Q is the flow, A is the total cross-sectional area, Av is the area of the voids, and n is the soil porosity. The concentration C of the contaminant is defined as: C=

M V

(25.6)

where M is the mass of contaminant (solute) within the volume V of the liquid carrying the contaminant (solvent). Associated with the concentration C is the contaminant mass flux F, defined as the mass of contaminant passing through a unit area of soil per unit time. The area is the area perpendicular to the flow direction: Fadv =

M Mx = = Cv = nvs C At V t

(25.7)

where F adv is the mass flux of contaminant due to advection (subscript adv), M is the mass of the contaminant, A is the total cross-sectional area perpendicular to the flow, t is the increment of time considered, and x is the distance traveled by the contaminant during t; other parameters are defined in Eqs. (25.5) and (25.6). Then the governing differential equation is: dC dC = −nvs (25.8) dt dx Diffusion. Diffusion refers to the change in concentration of a contaminant due to a chemical gradient. If you put a drop of colored food dye in a glass of water, you will observe diffusion of the dye in the water. This is due to the initial difference in concentration (gradient) of dye between locations in the glass. Another example is the intrusion of seawater into freshwater aquifers (Figure 25.10). This diffusion process is described by Fick’s law, which states that the contaminant mass flux is linearly proportional to the change in mass concentration between two points: Fdif = −nD∗

dC dx

(25.9)

Sea level

H Fresh Water

ce) erfa (int n sio iffu of d Salt water one

40H

Z

Figure 25.10 Example of diffusion: Seawater intrusion in aquifer. (Source: https://www.lenntech.com/groundwater/seawaterintrusions.htm.)

where F dif is the mass flux of contaminant due to diffusion (subscript dif ), n is the soil porosity, D* is the diffusion coefficient, and dC/dx is the concentration gradient. Adolph Fick was a German-born physician who proposed this law in 1855. Remarkably, the value of D* does not vary much, with a range of 1 × 10−9 to 2 × 10−9 m2 ∕s (Mitchell, 1976). Using Fick’s law and the equation of continuity gives the governing differential equation: d2 C dC (25.10) = nD∗ 2 dt dx Note that this equation is the same as the governing differential equation for the consolidation theory (Chapter 12, Eq. 12.56), except that C is replaced by the excess water stress uwe in the consolidating layer and nD* is replaced by the coefficient of consolidation cv . The solution to this equation for simple boundary conditions was proposed by Crank (1956). Dispersion. Dispersion refers to the fact that the velocity is not the same at all points in the flow field (Figure 25.11). This creates a problem similar to the diffusion component where the concentration C varies with the distance (longitudinal or transversal), and is written as: dC dC = −n𝛼vs (25.11) Fdsp = −nD dx dx

C B

C B A

A

A arrives first, then b, then c Mechanical dispersion

Figure 25.11 Mechanical dispersion.

902

25 GEOENVIRONMENTAL ENGINEERING

where F dsp is the mass flux of contaminant due to dispersion (subscript dsp), n is the soil porosity, D is the dispersion coefficient, 𝛼 is the dispersivity, vs is the seepage velocity, and dC/dx is the concentration gradient along the flow. Then the change in concentration with time is given by: dC d2 C = n𝛼vs 2 dt dx

(25.12)

The total mass flux at one point is the sum of the three components. The soil parameters entering into the equations (n, D*, 𝛼) can be measured through laboratory tests. One such test is the column test, in which the contaminant is injected at the top of the column and concentration measurements are made on the effluent at the bottom of the column. The soil parameters can also be obtained from in situ tests where a dye is injected at one location and the dye concentration is checked as a function of time in adjacent monitoring wells. Chemical Reaction Processes Chemical reaction processes include many different types of chemical reactions. Sorption and Desorption. Sorption is the process by which the contaminant becomes attached to the surface of the soil particle. Desorption is the process by which the contaminant gets detached from the surface of the particle. Empirical equations link the mass of contaminant sorbed per unit dry mass of soil S (e.g., mg/kg or ppm) to the concentration of contaminant in solution at equilibrium C (e.g., mg/L). Due to sorption, the contaminant velocity vc is slower than the seepage velocity vs ; in other words, the contaminant movement is slowed down and the water goes faster than the contaminant. A retardation coefficient R (larger than 1) then links the seepage velocity vs to the true contaminant velocity vc : v vc = s (25.13) R The retardation coefficient is estimated by laboratory testing or correlation with the ratio S/C. Precipitation and Dissolution. Precipitation and dissolution reactions involve the level of solubility of a contaminant in the carrier liquid (most often water). The degree of solubility varies from one contaminant to the next and is characterized by an equilibrium constant K, which is high for high solubility and vice versa. Sulfate salts and chlorides tend to be highly soluble, sulfide and hydroxides tend to be least soluble, and carbonates and silicates have intermediate solubility. Oxidation and Reduction. These reactions are often called redox reactions. Oxidation is a loss of electrons. Reduction is a gain of electrons. Redox reactions are characterized by the redox potential Eh expressed in volts. The redox potential is a measure of the affinity of a substance for electrons (its electronegativity). The reference potential is that of hydrogen, which is set at 0 volt. Substances more strongly electronegative than hydrogen have positive redox

potentials. These are substances that are capable of oxidizing; for example, oxygen has a redox potential of 1.23 V. In contrast, substances less strongly electronegative than hydrogen have negative redox potentials and are capable of reducing; for example, calcium has a redox potential of −2.87 V. Hydrogen peroxide is a powerful oxidant that is used in low concentrations to treat wounds because it releases oxygen, which kills bacteria. Acid-base reactions. An acid-base reaction is a gain or loss of a proton (H+ ) or the gain or loss of a hydroxyl group (OH– ). An acid is a proton donor, and a base is a proton taker. These reactions affect the pH of the soil and the groundwater and therefore the type of remediation strategy. Other reactions. Other reactions include complexation, ion exchange, hydrolysis, and volatilization. Complexation takes place when organic or inorganic ions or molecules combine in the dissolved phase. Ion exchange occurs when an ion is replaced by another one. This type of reaction, called isomorphous substitution, can take place at the surface of clay particles, with, for example, a Ca++ taking the place of an Na+ . Hydrolysis is the reaction between an organic molecule and water. Volatilization refers to the transformation of a liquid or solid into gas; it applies to volatile organics, for example, and is governed by Henry’s law: Ca = KH Cw

(25.14)

where Ca is the concentration of the contaminant in the gas phase, K H is Henry’s constant for that contaminant, and Cw is the concentration of the contaminant in the liquid phase. William Henry was a British chemist who proposed this law in 1803. Henry’s constant and many other constants associated with chemical reactions for typical contaminants in soils can be found in EPA publications such as Subsurface Contamination Reference Guide (U.S. Environmental Protection Agency, 1991) or in reference books, such as Sharma and Reddy (2004). Biodegradation Biodegradation is a redox reaction mediated by microorganisms; it can be aerobic or anaerobic. Aerobic biodegradation takes place in the presence of oxygen, which acts as an electron acceptor from the substance; anaerobic biodegradation takes place in the absence of oxygen. Some microorganisms naturally occurring in soil have the remarkable ability to degrade and transform many compounds, including hydrocarbons, polychlorinated biphenyls (PCBs), polyaromatic hydrocarbons (PAHs), pharmaceutical substances, radionuclides, and metals. However, for degradation to take place, the microorganism must be matched to the contaminant to be degraded. The microorganisms facilitate the transfer of electrons from a donor to an acceptor and in the process transform the substance to which it is attached. Typical outputs of biodegradation are methane, carbon dioxide, and water.

903

25.5 CONTAMINATION

Governing Differential Equation The governing differential equation (GDE) can be obtained by adding the contributions from the previously discussed processes. In the simple case of a one-dimensional flow, the total mass of contaminant per unit volume of soil CT is: CT = 𝜌d Cs + 𝜃w Cw + 𝜃g Cg

(25.15)

where 𝜌d is the dry density of the soil (mass of dry soil over volume of soil), Cs is the concentration of contaminant in the solid phase (mass of contaminant over mass of solids), 𝜃 w is the volumetric water content (volume of water over volume of soil), Cw is the concentration of contaminant in the liquid phase (mass of contaminant over volume of water), 𝜃 g is the volumetric gas content (volume of air over volume of soil), and Cg is the concentration of contaminant in the gas phase (mass of contaminant over volume of air). For a saturated flow, 𝜃 g is zero, 𝜃 w is equal to the porosity n, and Eq. (25.15) becomes: CT = 𝜌d Cs + nCw (25.16)

This leads to: 𝜕(𝜌d Cs ) 𝜕(nCw ) + 𝜕t ( 𝜕t ) 𝜕C 𝜕 =− nvs Cw − (D∗ n + 𝛼L vs n) w ± S 𝜕x 𝜕x

The term S is added to include any sources or sinks of contaminants, such as those due to chemical reactions. The partition coefficient K d is defined as: Kd =

(25.17)

where n is the soil porosity and vs is the seepage velocity. The contaminant mass flux F dif (mass of contaminant crossing a unit area of soil per unit time; see Eq. (25.9)) due to diffusion is given by: dC Fdif = −D∗ n w (25.18) dx where D* is the effective diffusion coefficient, n is the porosity, and x is the coordinate along the travel direction of the water. The contaminant mass flux F dsp (mass of contaminant crossing a unit area of soil per unit time; see Eq. (25.11)) due to dispersion is given by: dC dC (25.19) Fdsp = −DL n w = −𝛼L vs n w dx dx where DL is the dispersion coefficient in the longitudinal direction, n is the porosity, dx is the increment of x, and 𝛼 L is the longitudinal dispersivity. If we combine the transport from advection, diffusion, and dispersion, we obtain, for saturated flow: FT = Fadv + Fdif + fdsp dC dC = nvs Cw − D∗ n w − 𝛼L vs n w (25.20) dx dx Then we write that for conservation of mass, the change in concentration of mass of contaminant with time has to be equal to the slope of the curve describing the flux vs. distance along the flow path: dF dCT =− T dt dx

(25.21)

Cs Cw

(25.23)

If we ignore sources and sinks, we get the equation: 𝜕Cw 𝜕Cw vs (D∗ + 𝛼L vs ) 𝜕 2 Cw (25.24) = −( +( ) ) 𝜕t 𝜌d Kd + n 𝜕x 𝜌d Kd + n 𝜕x2 n n After setting Rd , the retardation factor, and DH , the hydrodynamic dispersion as: Rd =

The contaminant mass flux F adv (mass of contaminant crossing a unit area of soil per unit time; see Eq. (25.7)) due to water flow or advection is given by: Fadv = Cw nvs

(25.22)

𝜌d Kd + n 𝜌 K =1+ d d n n

(25.25)

DH = D∗ + 𝛼L vs

(25.26)

Then the GDE for contaminant transport for this simplified case is: v 𝜕Cw DH 𝜕 2 Cw 𝜕Cw (25.27) = s + 𝜕t Rd 𝜕x Rd 𝜕x2 Before we can solve this equation, we have to define the boundary conditions and the initial conditions. Let’s take the simple case in which the water is flowing horizontally in a confined aquifer with a point source of contamination that remains constant versus time (Figure 25.12). In this case the boundary conditions and initial conditions are: C0 = constant contaminant concentration at point x = 0 and t = 0

(25.28)

The solution was presented by Fetter (1992): ) ( Rd x − vs t C0 Cw (x, t) = erfc √ 2 4Rd DH t

(25.29)

where erfc is the complementary error function. Propagation of the contamination is shown as normalized concentration

Confined aquifer

v X X=0

Figure 25.12 One-dimensional transport in a confined aquifer.

904

25 GEOENVIRONMENTAL ENGINEERING

Relative concentration C/C0

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3

Mean

0.2 0.1 X +σ

–σ

X = Vs t / Rd The profile of a diffusing front as predicted by the complementary error function

Figure 25.13 Diffusing front for one-dimensional contaminant transport. (Source: After Fetter, 1992.)

(C∕C0 ) versus lateral extent x = vs t∕Rd in Figure 25.13. Solutions for one-dimensional flow and more complicated boundary conditions can be found in Hemond and Fechner-Levy (2000). For two-dimensional and threedimensional conditions, the GDE can be solved by the finite difference method or other numerical schemes.

source of the contamination, and the soil in which the contaminant may propagate. Exposure is defined as the contact of a chemical or biological agent with the outer boundary of a human being. The amount of exposure is measured by the extent of contact with that outer boundary during a specified amount of time. Exposure assessment includes determining the exposure setting (e.g., weather, topography, geology, hydrogeology), the exposure pathways (e.g., likely transport routes, speed of propagation, proximity of human activities), and the exposure concentrations (e.g., current and future chemical concentration). Toxicity is the adverse effect of a contaminant on human life (e.g., cancer, birth defect). Toxicity assessment includes determining if the chemical is a carcinogen or a noncarcinogen. The toxicity of noncarcinogenic chemicals is quantified by the reference dose (RfD in mg/kg/day), which is the daily dose that would not create an appreciable risk of deleterious health effects during a lifetime. The toxicity of carcinogenic chemicals is quantified by the slope factor (SF in (mg/kg/day)–1 ) that is the upper bound (95% confidence) on the increased cancer risk from a lifetime exposure to the chemical. The EPA publishes tables of RfDs and SFs for many chemicals (Sharma and Reddy, 2004). The risk quantification is done by using the hazard quotient HQ for noncarcinogens and the risk R for carcinogens, as follows: E (25.30) HQ = RfD R = CDI × SF

25.6

REMEDIATION

Remediation of contamination is the action of reducing the risk of detrimental effect on human life to an acceptable level. Once the type and extent of contamination have been identified, there are three common options: monitoring, containment, and remediation. Let’s look first at risk assessment, as it influences the choice of remedial measure. 25.6.1

Risk Assessment and Strategy

Risk is the product of the probability of an event happening or being exceeded times the value of the consequence. The value of the consequence and therefore the risk can be expressed in cost units, in human or animal fatalities, or in number of people sick (among others). Acceptable risk is then established by considering the risk associated with normal life activities and accepting such levels as targets for contamination. In the United States, there are two risk assessment techniques: one general technique from the EPA and one technique for leaking petroleum tanks from the ASTM. EPA Procedure The EPA procedure (U.S. Environmental Protection Agency, 2001) advances in four steps: (1) data collection and evaluation; (2) exposure assessment; (3) toxicity assessment; and (4) risk characterization. Data collection consists of identifying the contaminants, their concentration, the

(25.31)

where E is the chemical intake (mg/kg/day) and CDI is the chronic daily intake averaged over 70 years (mg/kg/day). The EPA considers a value of HQ higher than 1 and R larger than 10–6 to be unacceptable. ASTM Procedure The ASTM procedure (ASTM, 1995) is aimed at the remediation of sites with leaking petroleum tanks, where the chemical contaminants may include benzene, toluene, and xylene, for example. It proceeds in three “tiers” or steps. Tier 1, much like the EPA method, consists of collecting data, including the concentration of chemicals. These concentrations are then compared with risk-based screening levels (RBSLs) found in published tables. If the levels are below the RBSL for the most severe contamination propagation pathway, no action is required. If not, remediation can be planned to meet the tier 1 RBSL, or a more sophisticated Tier 2 analysis of the problem can be chosen. In Tier 2, the points of compliance are selected and the site-specific target levels (SSTLs) are determined. Additional soil and water data are collected and simple calculations are made to predict the transport and fate of the chemical over time. If the SSTL is met, no further action is necessary. If not, remediation to meet the SSTL is undertaken, or a more refined Tier 3 evaluation takes place. Tier 3 evaluation makes use of analyses more sophisticated than Tier 2 analyses and remediation takes place if the levels obtained after the Tier 3 analysis do not meet the target levels.

25.6 REMEDIATION

905

Remedial alternatives Monitored natural attenuation

Remediation Containment Ex-situ methods

Ex-situ containment Landfills

In-situ methods

In-situ containment

Impoundments Passive systems

Active systems

Horizontal barriers

Pumping Subsurface drains

Vertical barriers Caps

Figure 25.14 Remediation alternatives and decision tree. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

The general strategy, once the type and extent of contamination have been identified, is to choose among the following three common options: (1) monitoring; (2) containment; or (3) remediation (Figure 25.14). Monitoring is selected if the level of contamination is minor and the risk is low; containment is most often a temporary measure; and remediation of soil and water is the long-term alternative. 25.6.2

In Situ Waste Containment

In situ waste containment is typically used as a temporary measure to prevent further propagation of the contaminant while a more permanent remediation solution is set up. Containment is sometimes used as a permanent measure if the cost of cleanup technologies is prohibitive or the technologies are impractical. Two types of systems exist: passive systems and active systems. Passive systems are barriers to enclose the waste and minimize spreading (vertical barriers, bottom barriers, or surface covers), whereas active systems are generally pumping wells and drains.

with a liquid slurry made of water and bentonite clay (about 5% bentonite by weight). The role of the bentonite slurry is to seal the walls of the trench and provide a horizontal pressure that minimizes the chance of trench wall collapse. The slurry, which is at least 3% heavier than water, is kept level with the top of the trench, and is therefore higher than the water level in the soil. As a result, the slurry permeates out through the trench wall and deposits a film of very fine bentonite clay particles on the wall. This very-low-permeability film can be 10 mm thick or more and seals the wall against water penetration in the trench. When the trench is completed and full of slurry, the trench is backfilled with low-permeability backfill. This backfill

Up-gradient background monitoring well

Down-gradient monitoring well

Cover

Vertical Barriers Vertical barriers (Figure 25.15) are built to surround the waste. They are built from the surface down to a naturally impervious soil layer. They can be hanging barriers above an impervious layer if the contaminant will not propagate with depth, as in the case of LNAPLs that float on the water table. There are several types of vertical barriers: slurry trench barriers, grouted barriers, and steel sheet pile barriers. Slurry Trench Barriers. Slurry trench barriers are the most common type of vertical barrier. They are constructed by excavating a narrow trench about 0.5 to 1 m wide to the depth required. During the excavation, the trench is filled

Vertical barrier wall

Waste

Bottom seal or barrier Aquitard

Figure 25.15 Vertical, bottom, and surface containment barriers. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

906

25 GEOENVIRONMENTAL ENGINEERING

may be a soil-bentonite mix (SB), or a cement-bentonite mix (CB), or it can involve a geomembrane. Soil-bentonite barrier backfill is typically a mixture of sand (excavated soil if possible), dry bentonite, and bentonite slurry. It has the consistency of wet concrete with 25–50% fines but no gravel. This low-permeability backfill is placed at the end of the slurry trench and displaces the slurry forward toward the ongoing excavation. The target hydraulic conductivity of the trench is 10–9 to 10–10 m/s. Cement-bentonite barrier backfill is similar to SB backfill except that the soil is replaced by cement. A typical mix would be 5% bentonite, 15% cement, and 80% water. CB barriers are stronger than SB barriers but have lower permeability because the cement hinders the full expansion of the bentonite. Geomembrane barriers are installed by lowering a membrane into the open hole so that it seals the bottom and the side walls of the trench. Installation of a geomembrane can be a complex and difficult operation, but the permeability of the membrane is much lower than that of the other two systems. One important aspect of slurry trench barriers is the design for stability of the trench. An earth pressure analysis must be conducted to calculate the global factor of safety against collapse of the trench wall when the trench is fully excavated. Such design is rooted in the content discussed in Chapter 23. Grouted Barriers. Grouted barriers are prepared first by rotary drilling and then by grout injection. A column is constructed, then another one in line with the first one, and then another one in line with the first two, and so on in sequence so Table 25.4

that in the end a wall of columns forms the barrier. Typically, a first set of columns is built by skipping the intermediate columns, and then the intermediate columns are built when the first set of columns has hardened. Pressure grouting, jet grouting, and soil mixing are the techniques most commonly used. Pressure grouting consists of injecting grout under some pressure after a hole is drilled with diameters in the range of 1–1.5 m. The injection is done by the point injection technique or by the “tube a manchette” technique. Jet grouting consists of drilling a small hole and then rotating the drill rod upon withdrawal while jetting grout laterally under pressure to enlarge the hole. The pressure of the jet is in the range of 35–40 MPa. Grouted columns created with this technique can reach 3 m in diameter. Soil mixing consists of literally mixing the soil with grout (e.g., 20% by weight) as drilling takes place. In other words, the drilling mud is replaced by the soil-grout mixture. This minimizes the amount of cuttings generated. With grouted columns as barriers, it is very important to ensure overlapping of the columns to achieve a good seal against contaminant flow. Steel Sheet Pile Barriers. To create a steel sheet pile barrier, steel sheet piles are driven one beside the other to form a barrier in the soil. The advantages of this technique are that no excavation is necessary and the barrier installation is rapid. One problem is the corrosion issue and the lack of a good seal across the joints between sheet piles. Table 25.4 summarizes the advantages and drawbacks of the various barrier techniques.

Advantages and drawbacks of waste containment systems

System

Advantages

Drawbacks

Slurry barriers

Long-term, inexpensive, no maintenance required, well proven, available materials

Grout barriers

Geomembrane barriers

Injection of grout only requires small holes, can go to very large depth, can fill caverns, can vary setting time Effective, compatible with many contaminants

Steel sheet pile barriers

No excavation needed, no maintenance required

Pumping

Less costly than barrier, design flexibility, control of pumping rates, common technology, depth not a problem Economical to operate, drain location flexible, fairly reliable, simple and economical construction

Compatibility between slurry and contaminant; need for natural impervious layer; problems with boulders, caverns Difficult when soil is not pervious; holes and gaps more likely to jeopardize containment of liquids Sealing between sheets is complex process; keying membrane in bottom layer is complex as well; expensive Seal between sheet piles is not effective; corrosion problems; iron not compatible with many contaminants Requires frequent monitoring to limit propagation; maintenance required; capture zone limited Not for low-permeability soils; underdrains tough to place; monitoring required; not for deep contamination; potential clogging; excavation required

Subsurface drains

(Source: Sharma and Reddy, 2004/John Wiley & Sons.)

25.6 REMEDIATION

907

Bottom Barriers Bottom barriers (see Figure 25.15) are built to seal the bottom of a contaminated zone. This may be necessary when there is no natural low-permeability layer under the contaminated zone and the contaminant can propagate downward. A bottom barrier can be constructed by grouting or directional drilling. Grouting may be done by pressure grouting or jet grouting, but in both cases the injection pipe is driven or vibrodriven to the depth of the bottom barrier and a grout bulb is constructed. The operation is repeated until the overlapping bulbs form a bottom barrier. The drawback of this technique is that holes have to be punched through the waste or contaminated zone. Directional drilling can be used to reduce this problem: It consists of setting an inclined drill outside of the contaminated zone and reaching underneath that zone by drilling at an angle. Then the hole is grouted. Side-by-side holes are drilled and grouted to form the bottom barrier. Surface covers (see Figure 25.15) are built to cap the contaminated zone. They prevent the infiltration of running water and rain, thereby minimizing leaching; they prevent atmospheric contamination, reduce erosion, and improve aesthetics at the site. They are typically made of multiple layers, each with a specific purpose. The base layer or foundation layer creates a uniform surface on which to build other layers. Above that is the gas collection layer, made of coarse-grained soil and equipped with venting pipes. Above that is the barrier layer, made of compacted clay and a geomembrane or a geosynthetic clay liner (see Chapter 26) to prevent the surface water from entering the contaminated zone. This layer typically has a hydraulic conductivity of 10–9 m/s or less. Above that is the drainage layer, made of coarse-grained soil and/or geotextile to collect any water percolating down through the top layers. Above that is the surface and protection layer, made of topsoil and erosion control geosynthetic to prevent erosion and foster plant growth. Pumping wells (Figure 25.16) are built to pump contaminated water out of the soil or introduce a hydraulic gradient that will force the plume to move in the desired direction. These wells are used when the contaminant is mixed into the water (soluble). Injection wells can be used at some distance from pumping wells to force the liquid to go toward the pumping wells. This strategy can be used when the contaminant is a liquid not miscible with water. Pumping and injection are most effective when the soil is coarse-grained with high permeability. In the design of the well, the following issues must be addressed: depth, spacing, zone of influence, pumping rate, and number of wells. The treatment of the effluent must be addressed through an on-site or off-site treatment. Subsurface drains (Figure 25.17) play essentially the same role as pumping wells, but they drain the contaminated zone by gravity instead of pumps. Drains have the advantages of being a more economical solution than pumping, and they can be used in low-permeability soils where pumping is not

Domestic well

Extraction wells with radius of influence To treatment Domestic well

Contaminated Clean Impermea ble bedrock

Figure 25.16 Pumping wells. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

Discharge Lateral Collector

Main Collection sump

Figure 25.17 Subsurface drains. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

efficient. Drains can be placed horizontally as perforated pipes, or vertically as draining boreholes with a sump pump, or a combination of both. As in the case of pumping, the issues to be addressed are the location of the drains to take best advantage of gravity forces, depth, spacing, and zone of influence. Other issues specific to drains include pipe diameter, gradient of the slope, filters to prevent clogging, and size of the sump and pump.

908 25.6.3

25 GEOENVIRONMENTAL ENGINEERING

Soil Remediation

The most commonly used methods of soil remediation are soil vapor extraction, soil washing, and solidification. However, many other techniques are available. Soil Vapor Extraction Soil vapor extraction (SVE), also called soil venting, vacuum extraction, and aeration, consists of sucking the contaminated air out of the voids in the unsaturated soil zone above the water table. It is applied mostly to the volatile organic compounds (VOCs) that form in petroleum-contaminated soils. Extraction wells are installed on a grid, the vapor is removed by the vacuum gradient, and the collected vapor is treated using carbon filters, for example. The best candidates for such treatment are soils that are highly permeable and gasses that have a vapor pressure larger than 70 Pa and a Henry’s law constant higher than 0.01. Soil Washing Soil washing (SW) consists of excavating the soil from the site and treating it on site with a soil scrubbing system. A chemically suitable washing fluid is selected on the basis of the contaminants to be removed. After excavation, the soil is washed by forcing it through an energetic scrubbing system where it is mixed with the washing fluid. The coarse fraction is usually easier to clean than the fine fraction, because of the size and chemical complexity of clay particles. The clean soil is returned to its initial location; the remaining contaminated soil and the contaminated effluent are sent elsewhere for further treatment or disposal. This technique becomes economically attractive when the amount of soil to be treated is large (say, more than 50 MN). Stabilization and Solidification Stabilization and solidification (S/S), also called immobilization or fixation, consists of treating the contaminant in such a way that it is bound to the soil particles and remains trapped at the site. The method can be done ex situ or in situ. For the ex situ method, the contaminated soil is excavated and mixed with a stabilizing agent that fixes the contaminant to the soil particles. The mix is cured, and then the stabilized soil is replaced at the site or disposed of in a landfill. For the in situ method, the stabilizing agent is injected or mixed with the in situ contaminated soil to prevent the contaminants from moving away from the site. Portland cement is an example of a stabilizing agent. Electrokinetic Remediation Electrokinetic (EK) remediation consists of applying a potential difference between two electrodes (e.g., steel bars) driven into the contaminated soil. The potential difference (say, 40 V/m) drives the contaminated fluid to the electrodes where it is collected and removed. The anode is positively charged

and attracts negatively charged contaminants (anions); the cathode is negatively charged and attracts positively charged contaminants (cations). The flow rate (m3 /s) generated by EK remediation is proportional to the electrical potential between electrodes and the dielectric constant of the soil, but inversely proportional to the distance between electrodes and the viscosity of the fluid. EK remediation works well in fine-grained soils, which are otherwise difficult to clean. Thermal Desorption Thermal desorption (TD) consists of heating the contaminated soil to temperatures between 100 and 500∘ C. These high temperatures vaporize the contaminants, which are then removed by a vapor extraction system. This method works for volatile and semi-volatile organic compounds, but not for metals. Note that the contaminants are not destroyed as they would be during incineration (much higher temperature). TD can be performed in situ or ex situ. For ex situ treatment, the soil is excavated, brought to the treatment plant, and subjected to the heating process. In situ, heating blankets are placed on the surface for shallow treatment and heating wells are installed for deeper zones. If the water content is too high (e.g., more than 15%), dewatering may be necessary as a first step. Vitrification Vitrification (VT) consists of melting the contaminated soil into glass. This requires a lot of heat, with temperatures of about 1800∘ C. At such a temperature, organics are either destroyed or vaporized and stable inorganic compounds are surrounded by the molten soil. Upon cooling, the mass turns to glass and the inorganic contaminants are fixed in place. The method can be applied in situ or ex situ. In situ electrodes are placed and very high voltage and very high current are applied (e.g., 4000 V, 4000 A) with a power requirement of 3–4 MW. Gas collection hoods are placed on top of the electrodes to evacuate the gasses created to a treatment system. Bioremediation Bioremediation (BR) involves microorganisms or microbes eating the contaminant, transforming it into nontoxic byproducts through their digestive systems, and releasing those by-products to the atmosphere. The by-products are usually carbon dioxide and water or organic acids and methane. Appropriate microorganisms are found naturally in the soil and include yeast, fungi, and bacteria. The best contaminant food for them is petroleum hydrocarbons; other organic contaminants and inorganic contaminants are not as well suited to bioremediation. Microorganisms operate best in the presence of moisture, nutrients, and oxygen, so biostimulation consists of providing them with those three components to enhance the transformation process. Bioaugmentation consists of adding selected microorganisms to degrade a specific contaminant or supplement the work of the indigenous microorganisms.

25.6 REMEDIATION

Bioremediation may be aerobic (with oxygen) or anaerobic (without oxygen), but the aerobic process takes less time and is favored. Both in situ and ex situ treatments are possible. Either way, monitoring (e.g., for CO2 and O2 ) is necessary to adjust the stimulation process when appropriate. When stimulation involves providing more oxygen, it can be done in the form of bioventing (bringing air into the soil) or injection of hydrogen peroxide. Phytoremediation Phytoremediation (PR), from the word phyto in Greek, which means “plant,” is the natural soil remediation work done by plants through their root systems. The contaminant crosses the root membrane to enter the plant, which either degrades the contaminant or stores it in the plant tissue. Phytoremediation is best suited for sites with low levels of contamination at shallow depth (less than 3 m). It represents a final cleanup strategy rather than a main remediation method. Table 25.5 summarizes the advantages and drawbacks of soil remediation methods. 25.6.4

Groundwater Remediation

Groundwater is a very important resource for humankind. It represents 40% of our drinking water and must be kept free of contaminants. Groundwater remediation includes several Table 25.5 Method

different methods: pump and treat, in situ flushing, permeable reactive barriers, in situ air sparging, monitored natural attenuation, and bioremediation. Pump and Treat The pump and treat (PT) technique consists of installing a well from which the water is pumped out of the contaminated soil, treating the contaminated water, and pumping the water back into the soil or to another appropriate location (Figure 25.18). A typical configuration is to have a row of pumping wells downstream of the contaminant flow and a series of recharging wells upstream of the contaminant flow. That way, the pumping wells can also serve as monitoring wells for the efficiency of the treatment. The zone of influence and the depth of the wells, as well as the pumping rate, are part of the design. In Situ Flushing In situ flushing (ISF) consists of setting up the same kind of wells as in the pump and treat solution, but in this case the cleaning liquid is injected through the upstream wells, passes through the contaminated soil, cleans it, and is pumped out at the downstream wells. The cleaning liquid is carefully selected to remove the contaminant from the water without hurting the environment. The difference between PT and ISF

Advantages and drawbacks of soil remediation methods

Advantages

Soil vapor extraction

Easy installation, low disturbance, short time, economical Soil washing Reduces volume of contaminated soil, excavation and efficient treatment on site, few permits required Stabilization/solidification Low cost, widely applicable, simple, high throughput rates Electrokinetic remediation Applicable to fine-grained soils, wide range of contaminants, less expensive Thermal desorption Very rapid treatment, readily available equipment, very good for volatile organics Vitrification

Bioremediation Phytoremediation

909

Long-term durability, wide applicability, reduction of volume, public acceptance, cost-effective for difficult sites Very good for organic contaminants, minimum equipment required, no excavation, low cost Less expensive, safe, in-place treatment

(Source: Sharma and Reddy, 2004/John Wiley & Sons.)

Drawbacks Not possible in low permeability areas; need air emission permits; only for unsaturated soils Ineffective for soils with high fine content; relatively expensive; public exposure possible Contaminants remain; increased volume; volatiles created; limits future use Changes pH; buried metal is a problem; stagnant zones between electrodes Dewatering may be necessary; not good for fine-grained soils; not usable for heavy metals; large space required Difficult for very wet soils; limited depth ( 10% organics; high energy cost; dangerous in some cases Highly sensitive to local conditions; monitoring required; long treatment time Shallow treatment ( 10–6 m/s).

Contamination plume

25.7 Ground water flow Confining bed

Figure 25.18 Pump and treat setup. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

is that, with ISF, the cleaning is done in the soil rather than outside the soil as with PT. Permeable Reactive Barriers Permeable reactive barriers (PRBs) are treatment walls placed in the soil; they let the water go through but not the contaminants, which are immobilized or degraded as the groundwater flows through the barrier. Typically a trench is built and filled with a carefully selected reactive agent. As the contaminated water flows through the PRB, the clean water comes out and the contaminant is transformed into nontoxic by-products. In Situ Air Sparging In situ air sparging (ISAS) consists of drilling injections wells through the contaminated soil to reach underneath the plume and injecting compressed air under the contaminated plume. Because it is lighter, the air flows upward through the contaminated water and entrains contaminants vapors, which are then evacuated through soil vapor extraction. As a positive side effect and in the process the air brings oxygen, which enhances the activity of microorganisms and bioremediation. ISAS is best suited to high permeability soils (k > 10–5 m/s) and to the decontamination of volatile organic compounds, as in the case of leaking underground petroleum storage tanks. Monitored Natural Attenuation Monitored natural attenuation (MNA) is the “do nothing and monitor” or “watch and wait” solution. It consists of monitoring the process of natural remediation, including natural bioremediation, dilution, dispersion, and volatilization. There is no human intervention in this decontamination process. Bioremediation Bioremediation (BR) for contaminated groundwater works according to the same principles as bioremediation for contaminated soil. It is different from MNA in that there is

LANDFILLS

Most landfills (Figures 25.19 and 25.20) are used as permanent repositories of municipal solid waste, which is the main topic of this section. In the United States, each person generates about 20 N (1 N is the weight of a small apple) of MSW per day (20 N/person/day). This number used to be 12 N in 1960, reached 20 N in 1990, and has stabilized since then, but the population continues to grow, so landfills have to handle more and more MSW. The total amount of MSW per year in the United States is close to 2.5 million MN per year. The best ways to reduce waste, in order of preference, are: 1. Source reduction. 2. Recycling and/or composting. 3. Disposal in combustion facilities and landfills. Although the amount of waste being recycled has increased 10-fold over the past 40 years, today more than 50% of all MSW still ends up in a landfill. It is extremely important that these landfills be designed to keep the waste from contaminating soil and water and burdening future generations with unwanted problems. 25.7.1

Waste Properties

As mentioned in Section 25.2 and Figure 25.2, municipal solid waste in landfills consists primarily of paper, plastic, and food scraps. However, you can also find the odd rusted refrigerator and car tires. It is difficult to come up with the friction angle or modulus of elasticity of an old fridge or a car tire, yet these are the type of properties we are accustomed to using. To complicate matters further, the waste can be in various stages of decomposition, which affect its engineering properties. The only way to answer this problem is by testing the site-specific waste at a large-enough scale. This has been the effort of many researchers and engineers, including Landva and his colleagues (Landva and Clark, 1990). The following values are given to provide an order of magnitude of such properties, but the best approach consists of obtaining site-specific values of these parameters through testing at large scale—a scale large enough to be representative of the MSW behavior. The unit weight 𝛾 of MSW has been measured in large pits and reported by many authors. The first observation is that 𝛾 is highly variable depending on the type of waste, the degree of compaction, the state of decomposition, the proportion of daily soil cover, and the depth of the landfill.

25.7 LANDFILLS

911

Moisture barrier layer Granular drainage material Landfill gas to flare station or to energy utilization plant

Final clay and synthetic cap with vegetation

Gas monitoring probe

Gas collection well Rainwater retention pond Leachate to treatment plant

Ground water monitoring well

Stormwater outlet

Working face Storm water control berm

In place refuse Existing ground

Drainage liner Leachate collection sump with riser

Synthetic liner Compacted clay liner

Perforated leachate collection pipe

Ground water

Modern landfill design

Figure 25.19 Cross-section of a landfill.

Clark, 1990). Shear strength data collected by many authors was reviewed by Kavazanjian (1999), who proposed a bilinear lower-bound envelope. The first part applies to normal stresses lower than 30 kPa and gives c = 24 kPa and 𝜑 = 0. The second part applies to normal stresses higher than 30 kPa and gives c = 0 and 𝜑 = 33∘ : For 𝜎 < 30 kPa,

s = 24 kPa

(25.32)

For 𝜎 > 30 kPa,

s = 𝜎 tan 33 = 0.65 𝜎

(25.33)

Figure 25.20 Landfill under operation. (Source: Photo provided courtesy of the Texas Comptroller of Public Accounts from the Energy Report 2008, available at: http://www.window.state.tx.us/ specialrpt/energy/.)

Kavazanjian (1999) suggested a shear wave velocity for MSW, which varies from 150 m/s at the surface to 350 m/s at a depth of 60 m. For compressibility, most investigators favor the consolidation equation (see Chapter 18, Section 18.8.9): ( ′ ) 𝜎ov + Δ𝜎v Cc log (25.34) ΔH = Ho ′ 1 + eo 𝜎ov

Numbers ranging from 3–14 kN/m3 have been reported, with an average of 8 kN/m3 . The unit weight increases and the variability decreases as depth increases in the waste increases. Porosity is reported to vary between 0.4 and 0.6, void ratio between 0.67 and 1.5, and water content between 0.15 and 0.4 (Sharma and Reddy, 2004). Field permeability measured in MSW pits gave a range of 10–5 to 4 × 10−3 m∕s (Landva and

where ΔH is the settlement, H o is the initial thickness of the waste layer, Cc is the compression index, eo is the ′ is the effective vertical stress before initial void ratio, 𝜎ov ′ loading, and 𝜎ov + Δ𝜎ov is the effective stress long after loading. Values of Cc ∕(1 + eo ) between 0.1 to 0.4 have been suggested (NAVFAC, 1983), with the higher values corresponding to higher organic content. With MSW, a significant amount of delayed settlement (creep) can be expected over

912

25 GEOENVIRONMENTAL ENGINEERING

10–15 years, with the magnitude of ΔH/H o as much as 50% for new landfills and 15–20% for old landfills (Sharma and Reddy, 2004). The creep settlement equation is written as: ) ( t C𝛼 ΔH = Ho (25.35) log end 1 + eo tstart where ΔH is the creep settlement, H o is the layer thickness, eo is the initial void ratio, C𝛼 is the secondary compression index, tstart is the start time, and tend is the end time. Values of C𝛼 ∕(1 + eo ) have been reported (Sharma, 2000) as varying from 0.1 to 0.4, with the higher values for higher organic content and higher degree of decomposition of the waste. 25.7.2

Regulations

The U.S. Resource Conservation and Recovery Act (RCRA) was passed in 1970 and amended in 1980 and 1984. Subtitle D of RCRA applies to MSW landfills, whereas subtitle C of RCRA applies to hazardous solid waste landfills. The issues covered are location, operation, design, monitoring, closure, and post-closure. Restrictions exist when landfill locations are proposed near airports, wetlands, floodplains, and fault areas. The surface area A required for a landfill in a city is calculated by: WPt A= (25.36) D𝛾 where W is the weight of waste generated by a person per day, P is the total population of the city, t is the design period for the landfill, D is the depth of the landfill, and 𝛾 is the unit weight of the compacted landfill. The weight generated by one person per day is about 20 N. The unit weight of waste in a landfill varies widely, with an average of around 8 kN/m3 .

The period t varies from 10 to 30 years, and the depth D is between 10 and 30 m. There are many aspects to operating a landfill properly. First, a daily cover of about 0.3 m thick coarse-grained soil is required to cover the waste that was brought in that day. Other aspects include monitoring of the gas generated by the waste, control of public access, control of discharge and surface water, and record-keeping regarding compliance. One of the main components of the design of a landfill is the bottom composite liner, with a leachate collection system, a gas venting system, and a groundwater monitoring system; the top cover is another primary component. Closure takes place when the final cover is completed. Mandatory postclosure activities, including maintenance of the top cover and of the leachate collection system, as well as monitoring of the gas generated and the groundwater, must continue for 30 years. 25.7.3

Liners

Liners are barriers constructed at the bottom and on the side of landfills. Their purpose is to keep the waste and any by-product(s) out of the surrounding soil and groundwater. For municipal solid waste landfills, the liner composition is specified by RCRA Subtitle D (40 C.F.R. 258), and consists of a series of layers performing different functions. Going from the top to the bottom of the bottom liner, the following layers (2 through 6 for the liner) are encountered (Figure 25.21): 1. Waste. 2. Protective soil cover to minimize damage to the underlying geotextile.

Final cover Erosion cover (vegetative soil) Flexible membrane liner (fml) (minimum 20 mm thick)

Composite base liner and Leachate collection system Protective soil cover

Infiltration layer (hydraulic barrier) (k ≤ 1 × 0–5 cm/sec.)

Geotextile filter

Riser & cleanout pipes

Granular leachate collection layer Refuse fill

Flexible membrane liner (minimum 30 mm thick or 60 mm HDPE) Low permeability soil layer (k ≤ 1 × 0–7 cm/sec.) minimum 609 mm thick

Anchor trench for fml Composite side liner

Slope

To leachate collection drain and sump

Figure 25.21 Landfill cover and bottom liner composition for municipal solid wastes. (Source: After Sharma and Reddy, 2004. This material is reproduced with permission of John Wiley & Sons, Inc.)

25.7 LANDFILLS

913

4. Coarse-grained soil layer to serve as a leachate collection system. 5. Geomembrane layer to serve as a barrier preventing liquid penetration into the underlying layers. 6. Low-permeability soil layer (k < 10–9 m/s) with a minimum thickness of 0.9 m. 7. Natural soil.

Figure 25.22 Installing a geomembrane in a bottom liner. (Source: Courtesy of Layfield Environmental Systems, Layfield Group Limited, 11120 Silversmith Place, Richmond, British Columbia, Canada V7A 5E4.)

3. Geotextile layer that acts as a filter for any liquid coming down from the waste. 4. Coarse-grained soil layer to serve as a leachate collection system. 5. Geomembrane layer to prevent liquid penetration into the underlying layers; this geomembrane must be at least 75 mm thick for a flexible membrane liner (fml) and 1.5 mm thick for a high-density polyethylene (Figure 25.22). 6. Low-permeability soil layer (k < 10–9 m/s) with a minimum thickness of 0.6 m. 7. Natural soil. The liner should have a slope so that the leachate can drain naturally by gravity, be collected at a low point or sump, and be pumped and treated on a regular basis. The liner on the side slopes is the same as the bottom liner except that it does not typically have a leachate collection layer. The leachate naturally goes to the bottom of the landfill, where it is collected in the leachate collection layer. For hazardous solid waste landfills, the liner composition is specified by RCRA Subtitle C (40 C.F.R. 244) with a series of layers as follows. Going from the top to the bottom of the bottom liner, the following layers (2 through 6 for the liner) are encountered: 1. Waste. 2. Protective soil cover (optional) to minimize damage to the underlying geomembrane. 3. Geomembrane to act as a barrier for any liquid coming down from the waste. This geomembrane must be at least 0.76 mm thick if there is a protective soil layer above it or at least 1.14 mm thick if there is no protective layer above. For HDPE liners, the minimum required thickness is larger, varying from 1.5 to 2.5 mm.

As can be seen, a municipal solid waste liner is a single liner, whereas a hazardous solid waste liner is a double liner with the leachate collection system sandwiched between the two liners. The geomembranes and geotextiles used in landfill liners are discussed in Chapter 26. The hydraulic conductivity k of the liner must be less than 10–9 m/s. The hydraulic conductivity k of soils is discussed in Chapter 14, Sections 14.2.5 and 14.2.6. The measurement of k in the laboratory is discussed in Chapter 10. Sections 10.16–10.19 and in the field in Chapter 8, Section 8.12. The k values of clays permeated by contaminated liquids may differ significantly from the values obtained with water because of the chemistry of the permeating fluid. Various experiments starting in the late 1980s (e.g., Bowders and Daniel, 1987; Shackelford, 1994) indicated that when a clay is permeated with different chemicals, the hydraulic conductivity changes—sometimes dramatically. For example, a high concentration of methanol or heptane or trichloroethylene in the fluid will increase k; this is because such chemicals decrease the thickness of the clay particle double layer. In contrast, diluted acid in the permeating fluid will tend to decrease the value of k because the acid can create precipitates that clog the clay pores and render flow more difficult; however, k will likely increase in the long term. One first step in gaging whether a chemical will alter the hydraulic conductivity of a soil is to investigate the change in Atterberg limits when the soil is mixed with the chemical; note, though, that the link between the effect on the Atterberg limits and k is not always clear. Mitchell and Madsen (1987) concluded that permeation with hydrocarbons may affect k, but only if the concentration in the permeating fluid exceeds their solubility limit. Similar caution should be exercised for geosynthetic bentonite-clay liners (Shackelford et al., 2000). In all cases, it is best to run site-specific tests with the clay from the site and the anticipated fluid, including the appropriate chemical concentration. 25.7.4

Covers

Covers (Figure 25.21) are placed on top of landfills that are full and must be closed. A cover has many purposes, including minimizing the infiltration of rainwater, decreasing the hydraulic head on the bottom liner, resisting surface erosion, keeping out rodents, insects and birds, controlling gas emissions, and improving aesthetics. The typical cross-section of a cover consists of a series of layers (1 through 5) as follows:

914

25 GEOENVIRONMENTAL ENGINEERING

1. Vegetative layer for aesthetics and erosion protection. 2. Protective soil layer (optional). 3. Drainage layer to collect water, made of gravel and sand. 4. Barrier layer to stop water from penetrating into the waste. This layer may consist of a compacted clay layer, a geosynthetic clay liner (GCL), a geomembrane, or combinations thereof. 5. Drainage layer to collect gas generated by the waste, made of sand and gravel or geotextile. 6. Waste. The specifications for covers of hazardous solid waste landfills (RCRA subtitle C) are more stringent than for covers of municipal solid waste landfills (RCRA subtitle D). For hazardous wastes, the required thickness of the layers is larger than for municipal wastes. The final elevation of the top of a landfill is usually higher than the surrounding ground elevation (Figure 25.23). The side slopes of the final cover may be at 24∘ with the horizontal if the cover is made of soil layers, but it may be prudent to have the slopes at only 18∘ if a geomembrane is included in the cover, unless special measures to improve geomembrane roughness are taken. The top of the landfill is also sloped, but only at 2–5% on either side of the center to provide natural drainage. 25.7.5

Leachate Collection

The amount of leachate that would go through a single compacted clay liner is given by: q=k

Δh A L

(25.37)

where q is the flow in m3 /s, k is the soil hydraulic conductivity in m/s, Δh is the change in total head when crossing the compacted clay layer, L is the length of the flow path through the liner (thickness), and A is the plan view area of the liner. The hydraulic conductivity k is required by design to be less than 10–9 m/s, so this is the number used in Eq. (25.37). The change in total head Δh is usually taken as the sum of the height of liquid standing on top of the liner plus the thickness of the liner. This assumes that the total head under the liner is zero. A composite liner is made of a geomembrane underlain by a compacted clay liner. The amount of leachate that would go

Figure 25.23 General cross-section of a landfill.

through a composite liner was studied by Giroud and Bonaparte (1989), who recommended the following equation: q=

a0.1 k0.88 hw A 170000

(25.38)

where q is the flow in m3 /s, a is the cumulative area of holes in the geomembrane in m2 per acre (4047 m2 ), k is the soil hydraulic conductivity in m/s, hw is the height of liquid on top of the geomembrane in m, and A is the area of the bottom liner over which the flow of leachate q is calculated. Furthermore, Giroud and Bonaparte (1989) recommend assuming one hole of 3.2 mm2 per 4047 m2 of geomembrane under operating conditions, but a much larger hole for conservative sizing of the leachate collection system. For sizing purposes, they recommend a hole of 103 mm2 per 4047 m2 of geomembrane. A cover is exposed to rain, runoff, and evaporation. The amount of leachate through a cover’s top layer is calculated as follows: I = P − R − E ± ΔS (25.39) where I is the infiltration, P is the precipitation, R is the runoff, E is the evapotranspiration, and ΔS is the change in water volume per unit time of the soil cover. All terms in Eq. (25.39) take the same units (m3 /yr, for example). If there is no cover on the waste, as is the case during operation, the amount of leachate reaching the bottom liner should be reflected by adding another term to Eq. (25.39), to represent the amount of liquid generated by the waste itself by compression or by chemical reaction. The leachate collection system within covers and liners is built with a slope such that the leachate flows downward in the drainage layer toward a sump. At the sump, the leachate is collected and pumped to the surface, where it is analyzed and treated. 25.7.6

Landfill Slopes

The topic of slope stability is covered in Chapter 20. In the case of a landfill, slope stability comes into play in a number of instances (Figure 25.24), including the side slopes of the excavation, the stability of the side slope liner at the time of construction, the stability of the side slope liner when loaded unevenly by the waste pile, the stability of the waste when the landfilling operation advances through the landfill area, and the stability of the waste and cover upon closure of the landfill. The stability of the side slope of the excavation can be addressed by using the methods described in Chapter 20. The stability of the waste, for the case of a failure in the waste itself, can also be addressed using conventional methods, except that the shear strength of the waste may or may not follow soil mechanics principles (Section 25.7.1). The stability of the side slope liner is the case of a thin and long slope feature; it can be addressed by using the infinite slope method (see Chapter 20, Section 20.3). The stability

25.7 LANDFILLS

915

200

Not too steep Waste

Slope-liner

Excavation

Figure 25.24 Slope stability design issues in a landfill.

Shear stress τ (kPa)

Final cover

100

0

of the side slope liner when loaded by the waste is usually a controlling factor in design because it is more severe than the case of the liner by itself. In this case, the most likely failure mechanism is a block failure along the side and bottom liner, because the liner may be the weak link in the resistance to shear. Thus, it is best not to have the front face of the waste at a steep slope. One important issue is the shear strength of the interface between the various materials making up the liner. Each interface should be checked and the associated factor of safety calculated. Factors of safety between 1.3 and 1.5 are common. The interfaces involve the geomembrane, the geotextile, the geosynthetic clay liner (GCL), the drainage layer, the natural soil, and the waste. The geomembrane should be textured rather than smooth, to improve its interface shear strength, and the GCL should be stitched to dramatically increase the shear strength of the bentonite layer. The best way to obtain design values for the interface shear strength is to perform direct shear strength tests (ASTM D5321 and D 6143) on site-specific materials under simulated field conditions. Some aspects of the behavior are important to document: the peak shear strength, the postpeak residual shear strength, the influence of the normal stress level, and the nonlinearity of the strength envelope. It is useful to place all shear strength envelopes on the same graph when the tests are completed (Figure 25.25) to find out which of the interfaces is the weak link for a given normal stress. A seismic slope analysis is also necessary (see Chapter 20, Section 20.18). An additional problem may arise when during construction of the landfill, the excavation proceeds through a clay layer with an underlying sand layer under water pressure especially if it is an artesian pressure (Figure 25.26). Though this case is rare, it can be disastrous, because if the excavation is dug to a depth where the water pressure (𝛾 w hw ) at the top of the sand layer overcomes the downward pressure of the clay remaining on top of the sand (𝛾h), the bottom of the excavation will blow up and a mixture of sand and water will run into the excavation. A factor of safety must be applied to the maximum depth of excavation to guard against such an event. The safe remaining thickness h of the clay layer is: h=

F𝛾w hw 𝛾

(25.40)

Textured HDPE on non-woven geotextile

0

100

GCL internal residual strength

200 300 Normal stress σ (kPa)

400

Figure 25.25 Strength envelopes of various interfaces. (Source: After Sharma et al., 1997.)

h>

hw h

Fhwγw γ Clay unit weight, γ Sand under artesian pressure

Figure 25.26 Blowout problem at bottom of excavation.

25.7.7

Gas Generation and Management

Landfills generate gas, mostly carbon dioxide and methane, through biodegradation. These gases are flammable, are toxic to humans, can create excessive deformation of the liners, and smell bad. Furthermore, methane is a greenhouse gas. The gas generation process is due to the work of bacteria that transform some of the waste through digestion. The product is approximately 50% carbon dioxide (CO2 ) and 50% methane (CH4 ). The carbon dioxide is usually generated first, followed by the methane. The intensity of this process depends on a number of factors, including the availability of nutrients for the bacteria, temperature, humidity, pH, and age of the waste. Landfill temperatures vary from 20 to 60∘ C. Higher temperature and higher water content of the waste are more favorable to gas generation, which can reach 10,000 m3 per kN of waste over the life of the landfill. Gas generation in a landfill does have a finite life, which can vary from 20 years under favorable conditions where biodegradation is rapid (e.g., humid climates) to 100 years under unfavorable conditions where biodegradation is slow (e.g., arid climates). The gas generated must be disposed of and the disposal process monitored. There are essentially three ways to dispose of gasses: (1) vent to the atmosphere; (2) vent and burn with no energy recovery; and (3) vent and burn with energy recovery. The most common of the three is vent and burn without energy recovery, through the use of flares. Venting

916

25 GEOENVIRONMENTAL ENGINEERING

is achieved by placing gas wells into the waste, which facilitate gas migration to the surface where the gas is burned. Old landfills used open flame flares, which are the simplest kind, but modern landfills use enclosed flares because they allow for measurement of the gas coming out of the waste and yield better overall control. Wells typically consist of perforated pipes 50–300 mm in diameter that extend to 75% of the full depth of the landfill. The spacing varies from 15 to 100 m and averages 60 m. The energy recovery systems use the gas to power gas turbines or combustion engines to generate electricity, but the initial cost of such a system is worth the investment only for large landfills with more than 10 million kN of waste. 25.8

FUTURE CONSIDERATIONS

Lowering the generation of waste at the source is the first and best way to decrease the amount of waste generated by

humankind. Recycling is the second best option. Recycling of household waste or municipal solid waste has become part of everyday life, and over the past 20 years has reduced the amount of waste going to landfills to about 50% of the MSW generated (Figure 25.27). The most successful programs have been recycling of aluminum and paper, because in both cases the cost-benefit ratio is favorable. Recycling does not stop at household waste, but extends as well to the industrial sector, which is by far the largest generator of waste. Efforts at recycling fly ash, blast furnace slag, foundry sand, paper mill sludge, incinerator ash, glass, plastics, scrap tires, demolition and concrete debris, and wood waste are being made (Sharma and Reddy, 2004). Note that once landfills are closed, the area can be used for various activities including parks, golf courses, airports, and sports stadiums (Figure 25.28).

120

Recycling rate in percent

100

96.2

80 71.6 67.0 57.5

60

49.6 40

35.5

35.4

29.2

27.5

20

0

Auto Newspapers/ batteries mechanical papers

Steel cans

Yard trimmings

Aluminum beer & soda cans

Tires

Products

Glass containers

Pet HDPE (natural bottles white & jars translucent bottles)

Figure 25.27 Recycling rate for various waste products in 2011. (Source: EPA.)

Figure 25.28 Post-closure use of landfills.

25.8 FUTURE CONSIDERATIONS

917

Problems and Solutions Problem 25.1 How do you define waste? Solution 25.1 Waste is unwanted or useless material. Problem 25.2 What is the biggest generator of waste in the United States? Solution 25.2 The industrial sector is the largest generator of waste. Problem 25.3 Are solid wastes solids? Solution 25.3 No. The term solid waste is misleading, as a solid waste can be a solid, a liquid, or a gas. Problem 25.4 What are the four main categories of waste? Solution 25.4 The four main categories of waste are: solid wastes, hazardous wastes, radioactive wastes, and medical wastes. Problem 25.5 How long can a high-level radioactive waste continue to be deadly? Solution 25.5 The radiation penetration from high-level wastes, which are generated by defense or nuclear power plant activities, remains lethal for 10,000 years. Problem 25.6 What are RCRA and CERCLA and what do they regulate? Solution 25.6 RCRA is the Resource Conservation and Recovery Act; it addresses the issue of landfill design. CERCLA is the Comprehensive Environmental Response, Compensation, and Liabilities Act; it addresses the issue of cleaning up contaminated sites. Problem 25.7 What does an OSHA level C mean and what does it require? Solution 25.7 OSHA stands for Occupational Safety and Health Administration. OSHA Level C refers to moderate protection for humans working at contaminated sites. That is, it requires moderate protection including full-face or half-mask air-purifying respirator, hooded chemical-resistant clothing, inner and outer chemical-resistant gloves, chemical-resistant boots and boot covers, hard hat, escape mask, and face shield.

918

25 GEOENVIRONMENTAL ENGINEERING

Problem 25.8 What is the name of the smallest piece of matter and what are its components? Solution 25.8 An atom is the smallest piece of matter. Atoms are made of protons, electrons, and neutrons. Problem 25.9 What is the difference between an atom and a molecule? Solution 25.9 Atoms consist of a nucleus containing protons and neutrons with electrons surrounding the nucleus; they are the basic building blocks of matter (e.g., hydrogen atom). Molecules are combinations of atoms bonded together. For example, two hydrogen atoms and an oxygen atom form a molecule of water (H2 O). Problem 25.10 What is the difference between an ion, an anion, and a cation? Solution 25.10 Ion is the general term for an atom that has lost or gained an electron on its outer orbital. More specifically, an ion can be a cation or an anion. Cations are neutral atoms that have lost one or more electrons, making them positively charged, such as Na+, Ca++ , and Al+++ . Anions are the opposite: They have gained one or more electrons and thus have a net negative charge, such as Cl– , O– , and N– . Problem 25.11 In the case of concentration in water, how many parts per million (ppm) are in 1 mg/m3 ? Solution 25.11 For the mass concentration of a chemical in water, 1 ppm = 1 mg/liter and 1 liter = 0.001 m3 . Therefore, there are 0.001 ppm in 1 mg/m3 . Problem 25.12 What is the difference between organic and inorganic materials? Solution 25.12 The difference between organic and inorganic compounds is that most organic compounds contain carbon, whereas most inorganic compounds do not. Problem 25.13 What is the difference between atomic absorption spectrophotometry (AAS) and gas chromatography-mass spectrometry (GC-MS)? Solution 25.13 GC-MS is used to identify the components of a chemical mixture. AAS is used to measure the molar concentration of chemicals. Problem 25.14 What are ESA I, II, and III, and when are they used? Solution 25.14 Environmental site assessments or ESAs are part of the contamination detection process. They are often required when purchasing a piece of property in the United States. There are three levels: ESA I: This phase consists of collecting information regarding previous ownership and prior use through records of contaminated sites in the area, aerial photos, geologic and topographic maps, visits to the site, and talking to neighbors. An ESA I indicates whether there are reasons to believe the site is contaminated. If so, ESA II comes into play.

25.8 FUTURE CONSIDERATIONS

919

ESA II: This phase consists of testing the soil and the groundwater to find out if there is contamination and, if there is, to what extent and to what level of severity (type of contaminants). If contamination that requires cleanup is found, ESA III comes into play. ESA III: This phase consists of designing the remediation scheme and achieving it, including verification that a satisfactory level of cleanup has been realized. Problem 25.15 What is the difference between the LIF CPT, the MIP CPT, and the BAT CPT? Solution 25.15 LIF (laser-induced fluorescence) is a CPT technique used to determine the extent of plumes at petroleum-contaminated sites and the type of petroleum product contaminating the site. A laser beam is shone on the soil, which emits different fluorescence depending on the hydrocarbon present. MIP (membrane interface probe) is a CPT technique used to identify the type of volatile organic compound by heating the soil and letting the gas permeate through a membrane located on the side of the CPT. Once in the CPT housing, the gas is swept by an inert carrier gas to the surface where it is analyzed. BAT is a CPT technique used to collect groundwater. (BAT is the name of a company.) The CPT probe is equipped with a porous filter that is obstructed until the CPT probe is pushed to the required depth. Then the filter is exposed and water is allowed to penetrate through the filter into a water sampling tube, which can be removed through the CPT rods when full. Problem 25.16 In contaminant transport, what is the difference between concentration and flux? Solution 25.16 Concentration C is the mass of contaminant (solute) per volume of liquid carrying the contaminant (solvent); it is measured in kg/m3 . Flux F is the mass of contaminant flowing through a unit area of soil per unit of time; it is measured in kg/m2 s. They are related through F = Cv. Problem 25.17 Choose some reasonable values of the parameters in the solution to contamination propagation (Eq. (25.29)) and draw the propagation plot. Then vary each parameter to understand the influence each one has on the propagation. For help with the solution, go to www.lmnoeng.com/Groundwater/transportStep.htm Solution 25.17 C Cw (x, t) = 0 erfc 2

(

Rd x − vs t √ 4 Rd DH t

)

C0 = Constant contaminant concentration at point X = 0 and t = 0 X = Distance V s = Seepage velocity 𝜌 K Rd = Retardation factor Rd = 1 + dn d where 𝜌d is dry density and n is total porosity K d = Partition coefficient DH = Hydrodynamic dispersion DH = D∗ + 𝛼L vs where 𝛼 L dispersivity varies from 0.1 to 100 D* = Molecular diffusion coefficient; typical value 1 × 10−9 (m2 ∕s) Propagation of the contamination is plotted as normalized concentration Cw ∕C0 versus x for the following input parameters: 𝛼L = 100, 𝜌d = 1.6(g∕cm3 ), n = 35(%), D∗ = 1 × 10−5 (cm2 ∕s), vs = 1.92e − 5 (cm∕s), Kd = 0.1(cm3 g), t = 1000 (days)

25 GEOENVIRONMENTAL ENGINEERING

• Propagation plot (Figure 25.1s) 1 t = 1000 Days 0.8

Cw /C0

0.6 t = 100 Days

0.4

0.2 t = 10 Days 0

0

10

20

30

40

50

Distance, x (m)

Figure 25.1s Propagation plot.

• Varying parameter 𝛼 L (Figure 25.2s). 1 a = 100 a = 10 a=1

Cw /C0

0.8

0.6

0.4

0.2

0

0

10

20 30 Distance, x (m)

40

50

Figure 25.2s Propagation plot for different values of 𝛼 L .

• Varying parameter 𝜈 s (Figure 25.3s). 1 Vs = 1.92e-5 m/s Vs = 4e-5 m/s Vs = 1.5e-5 m/s

0.8

0.6 Cw /C0

920

0.4

0.2

0

0

10

20

30

40

Distance, x (m)

Figure 25.3s Propagation plot for different values of vs .

50

25.8 FUTURE CONSIDERATIONS

921

• Varying parameter 𝜌d (Figure 25.4s). 1 ρ = 1.6 g/cm3 ρ = 1.1 g/cm3 ρ = 1.8 g/cm3

0.8

Cw /C0

0.6

0.4

0.2

0

0

10

20 30 Distance, x (m)

40

50

Figure 25.4s Propagation plot for different values of 𝜌d .

Problem 25.18 How can you form a bottom barrier at depth for a waste containment system? Solution 25.18 A bottom barrier can be constructed by grouting or directional drilling. Grouting can be pressure grouting or jet grouting, but in both cases the injection pipe is driven or vibrodriven to the depth of the bottom barrier and a grout bulb is constructed. The operation is repeated until the overlapping bulbs form a bottom barrier. The drawback with this technique is that holes have to be punched through the waste or contaminated zone. Directional drilling consists of setting an inclined drill outside of the contaminated zone and drilling at an angle to reach underneath that zone. Then the hole is grouted. Side-by-side holes are drilled and grouted to form the bottom barrier. Problem 25.19 A landfill has been closed for one day and the long-term settlement must be evaluated. The waste is 15 m deep, has a unit weight of 8 kN/m3 , and has a C𝛼 ∕(1 + eo ) coefficient equal to 0.2. Calculate the creep settlement after 20 years. Solution 25.19 𝛥H = Ho

C𝛼 log 1 + eo

(

tend tstart

)

( = 15 m ∗ 0.06 ∗ log

20∗ 365 day 1 day

) = 3.48 m

Problem 25.20 Calculate the area of the landfill necessary to handle the municipal solid waste generated by a city of 1 million people over a period of 10 years. Each person in that city generates 20 N of MSW per day. The depth to the water table is 20 m and a high-plasticity clay layer exists at a depth of 15 m. Solution 25.20 The surface area A required for a landfill in a city is calculated by: A = WPtD𝛾 where W is the weight of waste generated by a person per day, P is the total population of the city, t is the design period for the landfill, D is the depth of the landfill, and 𝛾 is the unit weight of the compacted landfill (estimated at 8 kN/m3 ).

922

25 GEOENVIRONMENTAL ENGINEERING

The depth of the water table is 20 m, and there is high-plasticity clay at depth 15 m, so the depth of landfill is selected as 15 m: 20 N∕day × 106 × 3650 day 7.3 × 107 kN = = 6.08 × 105 m2 A= 3 15 m × 8 kN∕m 1.2 × 102 kN∕m2 Problem 25.21 Calculate the flow of leachate through a 0.6 m-thick clay liner covering a 200 m × 200 m area. The leachate level is 0.4 m above the top of the liner and the hydraulic conductivity of the clay meets the specification of 10–9 m/s. Solution 25.21 The amount of leachate that would go through a single compacted clay liner is calculated by q = K 𝛥h A, where q is the flow L in m3 /s, k is the soil hydraulic conductivity in m/s, Δh is the change in total head when crossing the compacted clay layer, L is the length of the flow path through the liner (thickness), and A is the plan view area of the liner: 𝛥h = 0.6 + 0.4 = 1 m A = 200 × 200 = 40000 m2 𝛥h 1 q = K A = 10−9 × × 40000 = 6.67 × 10−5 m3 ∕s L 0.6 Problem 25.22 Chloride dissolved in water is leaching through a liner and permeating into an aquifer-bearing 2 m-thick layer of silty sand. The concentration of the dissolved chloride is 1500 mg/liter, the discharge velocity is 3.7 × 10−7 m∕s, and the porosity of the silty sand is 0.25. Calculate the mass flux of chloride into the aquifer per unit area of landfill liner due to advection. Solution 25.22 Fadv = vC = nvs C = 3.7 × 10−7 × 1500 × 10−3 = 5.55 × 10−7 mg∕m2 s Problem 25.23 Calculate the flow of leachate through a composite liner with a 0.75 m-thick compacted clay layer over an HDPE geomembrane. The clay has a hydraulic conductivity of 10–9 m/s and the HDPE membrane is 1.5 mm thick. The height of liquid above the geomembrane is 0.2 m. Give the answer for expected operating conditions first and then give a more conservative estimate for sizing the leachate pumping system. Solution 25.23 q=

a0.1 k0.88 hw A 170000

Case 1: Operating Conditions )0.1 ( ( )0.88 3.2 × 10−6 × 10−9 × 0.2 × 1 4047 q= = 1.73 × 10−15 m3 ∕s = 0.55 cm3 ∕year 170000 Case 2: Pumping System Design )0.1 ( 103 × 10−6 × (10−9 )0.88 × 0.2 × 1 4047 = 2.64 × 10−15 m3 ∕s = 0.83 cm3 ∕year q= 170000 Problem 25.24 An excavation is dug for a landfill in a 20 m-thick stiff, high-plasticity clay layer underlain by a sand layer under water pressure. The unit weight of the clay is 19 kN/m3 and that of the sand is 20 kN/m3 . The water pressure is such that a casing through the clay into the sand has water rising 5 m below the top of the clay layer (ground surface). How deep can the excavation be dug into the clay to maintain a minimum factor of safety of 1.5 against bottom blowout failure?

25.8 FUTURE CONSIDERATIONS

Solution 25.24 The safe thickness h of the clay layer after excavation is: h=

F𝛾w hw 1.5 × 9.81 × 15 = = 11.6 m 𝛾 19

Therefore, the safe excavation depth is 20 - 11.6 = 8.4 m.

923

CHAPTER 26

Geosynthetics

26.1

GENERAL

Geosynthetics have been to geotechnical engineering what computers have been to humankind in general: a revolution. The use of these planar synthetic materials in soils to reinforce, to drain, and to separate has grown remarkably over the past 50 years to the point where it is a huge industry today. According to ASTM D4439, a geosynthetic is a planar product manufactured from polymeric material (plastics) to be used with soil, rock, earth, or other geotechnical engineering-related materials as an integral part of a humanmade project, structure, or system. There are many types of geosynthetics, including geotextiles, geomembranes, geogrids, geosynthetic clay liners, geofoam, geonets, geocells, geobags, and geocomposites. Geotextiles and geomembranes are the two largest groups of geosynthetics. In 2013, the cost of geosynthetics was between $1 and $7m2 . The book by Koerner (2012) is an excellent reference on geosynthetics.

26.2

TYPES OF GEOSYNTHETICS

Geotextiles (Figure 26.1a) are textiles made of synthetic fibers. The fibers are either woven together or tied together (non-woven). Weaving consists of standard interlacing with textile machinery. In non-woven fabrics, the fibers are tied together by heating, gluing, or needle-punching. In needle-punching, short needles with barbs are punched through the fabric to provide a mechanical interlocking. Geotextiles are flexible and porous to liquid flow. They are used mainly for separation, reinforcement, filtration, and drainage. Geomembranes (Figure 26.1b) are relatively thin, impervious sheets of plastic material. They are made by first preparing the polymer resin and its additives. The actual forming of the membrane takes place by extrusion through two parallel plates or rollers. The resulting sheet is between 1 and 3 mm thick and can be smooth or roughened. Geomembranes are used mostly as nearly impervious barriers to contain liquids or vapors.

Geogrids (Figure 26.1c) are plastic grids that have a very open configuration; they have large holes between ribs. They are formed by bonding rods together, by weaving and then coating, or by stretching. Their main use is reinforcement. Geosynthetic clay liners or GCLs (Figure 26.1d) are made of a thin layer of bentonite clay sandwiched between two layers of geotextiles or geomembranes. GCLs are manufactured by feeding the bentonite on top of a conveyor-belt-style geosynthetic and covering it after the feed point by a top geosynthetic. The two geosynthetic layers are kept together by needle punching, stitching, or gluing. The bentonite will expand dramatically when wetted. GCLs are about 4–6 mm thick when the bentonite is hydrated at water contents of 10–35%. They are used mostly as nearly impervious barriers to contain liquids or vapors. Geofoams (Figure 26.1e) are extremely light blocks made of polymer bubbles. They are fabricated by thermal expansion and stabilization of polystyrene bubbles. The density of the blocks is about 2% of the density of soils, but three to four times more expensive per unit volume. They are stacked together to form lightweight fills, and are used as compressible layers behind retaining walls, as vibration dampers for seismic protection, and as thermal insulation in foundations. Geonets (Figure 26.1f ), like geogrids, are open netting geosynthetics made of plastic. They are different from geogrids in that they are thicker; they are sometimes called spacers as they provide space for fluid to flow within the structure. Also, the openings are more like diamonds than the squares of geogrid openings. They are used primarily for drainage purposes. Geocells (Figure 26.1g) are a form of geogrid in the sense that they have a very open configuration, but their purpose is to reinforce by confining the soil within the cells. The cells may be 1 m × 1 m in plan view and 1 to 2 m high. The soil is placed within the cells, which provide lateral confinement and thereby significantly increase the bearing capacity of the soil layer. Geobags (Figure 26.1h) are literally bags made of geosynthetic material; they are usually filled with sand and used for

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

925

926

26 GEOSYNTHETICS

(a) Geotextiles

(d) Geosynthetic clay liners

(g) Geocells

(c) Geogrids

(b) Geomembranes

(e) Geofoam blocks

(f) Geonets

(h) Geobags

(i) Geocomposites

Figure 26.1 Examples of geosynthetics. (Source: Photographs compliments of the Geosynthetic Institute.)

erosion protection in lieu of rip rap. Their size is in the range of rip rap, and can be as large as 5 m3 . Geocomposites (Figure 26.1i) are combinations of the previous geosynthetics that are intended to maximize the usefulness of a geosynthetic layer. They are used as filter layers, for example. Geosynthetics are useful in a number of geotechnical applications, as shown in Table 26.1.

soils. The reason is that the material and the applications are quite different and more versatile than those associated with soils alone. Also, the field of geosynthetics is quite a bit younger than the field of geotechnical engineering. Although very significant progress has been made, some of the properties’ definitions, the tests used to determine their value, and the design guidelines are still evolving. 26.3.1

Properties of Geotextiles

Physical Properties 26.3

PROPERTIES OF GEOSYNTHETICS

The parameters used to characterize geosynthetics are much more numerous than and often different from those used for

The unit weight of typical plastics varies from 9 to 13 kN/m3 . The unit weight of dry, clean geotextiles is between 3 and 7kN/m3 —but that is not the way it is typically given. Instead,

26.3 PROPERTIES OF GEOSYNTHETICS

Table 26.1

Applications for some geosynthetics

Geotextiles

Geogrids

Geomembranes

Separation Roadway reinforcement Soil reinforcement Filtration Drainage Drainage

Reinforcement Roads Slopes Walls Foundations

Liners Ponds Canals Landfills, dry Landfills, wet Landfill covers

Steel plate Length L Geotextile G=M

L 2

927

3

41.5° Figure 26.2 Flexure stiffness test for geotextile.

it is quoted as mass per unit area (ASTM D5261) with values between 150 and 750g/m2 (Koerner, 2012). The thickness of commonly used geotextiles is between 0.5 and 4 mm (ASTMD5199). Mechanical Properties Stiffness is usually defined as the ratio between the force applied and the resulting displacement, as in stiffness of a spring. The stiffness of a geotextile is defined in a very different way; it is obtained from a laboratory test (ASTM D1388) in which a 25 mm-wide strip of geotextiles is gradually pushed over the edge of the crest of a slope under controlled conditions (Figure 26.2). The slope is 41.5∘ with the horizontal and when the strip touches the slope, the length L of the overhanging strip is recorded. The stiffness of the geotextile is defined as: ( )3 L (26.1) G=M 2 where G is the flexural stiffness (g.m), M is the mass per unit area (g∕m2 ), and L is the overhang length (m). The G values for geotextiles are in the range of 0.01 to 1 g.m. The average modulus of deformation of geotextile under tension stresses varies widely. It can be 60 MPa for some nonwoven, needle-punched geotextiles all the way to 400 MPa for some woven monofilament geotextiles (Koerner, 2012). Because the evolution of the thickness during the test is not certain, this modulus is not commonly quoted for these products. Instead, it is more commonly presented as the

Geosynthetic clay liners Liners

Geofoam Lightweight fill Compressible inclusions Thermal insulations Drainage

ratio of the force per unit length of fabric over the normal strain generated. Numbers in the range of 30–150 kN/m are common. The average tensile strength of geotextiles (ASTM D4632) is in the range of 50–100 MPa; again, however, that is not the way it is typically cited. Instead, the average tensile strength is quoted as the force St per unit length of fabric that creates rupture; average numbers are in the range of 25–60 kN/m. One of the problems is that the thickness varies during elongation of the geotextiles. Another problem is that the strain to failure is much larger than in soils, with values around 25% for some woven fabrics and up to 70% for some nonwoven fabrics. As a result, the tensile strength is usually quoted together with a value of the strain at failure. The tensile strength of the seams (ASTM D4884) is typically 50–75% of the tensile strength of the intact fabric. The compressibility of geotextiles is generally not a concern except when they are used to convey water or other liquids in the in-plane direction. In this case it is important to make sure, by testing, that the small conveyance tubes within the geotextiles will not collapse under the in situ compression. The puncture strength is important and may be quoted as an impact puncture strength or a static puncture strength. The impact puncture strength is tested by dynamically puncturing the fabric with a pendulum test (energy to puncture) (ASTM D256) or a drop cone (penetration distance). The impact puncture strength of a geotextile is quoted as the energy that leads to puncture. Common values of geotextile impact puncture strength vary from 25 Joules to 300 Joules. It is named after the English physicist James Prescott Joule who contributed in the mid to late nineteenth century. The static puncture strength (ASTM D6241) is determined by slowly pushing a 50 mm diameter beveled plunger into the fabric and recording the puncture failure load P. The following empirical relationship between the puncture load P (kN) and the tensile strength St (kN/m) has been proposed (Cazzuffi and Venezia, 1986): P = St 𝜋d

(26.2)

where d is the diameter of the punching plunger. The interface shear strength between a geotextile and a soil can be very important in design and should be measured using site-specific materials. The accepted test (ASTM D5321) is a

928

26 GEOSYNTHETICS

Normal force Block Geotextile Shear force Soil

Figure 26.3 Direct shear test for soil-geotextiles interface shear strength.

variant of the soil direct shear test in which the top part of the soil is replaced by a geotextile-covered block (Figure 26.3). The interface shear strength is typically in the range of 75% to 100% of the soil shear strength. Creep or deformation under constant stress is also important for geotextiles, and creep tests are necessary for long-term applications under load (ASTM D5262). The general model used for soils can be extended to geotextiles: ( )n 𝜀1 t = 1 (26.3) 𝜀2 t2 where 𝜀1 and 𝜀2 are the strains reached in a time t1 and t2 respectively and n is the viscous exponent. Table 26.2 gives estimates of the n values for several polymers based on the data from den Hoedt (1986). Note that the time-temperature equivalency, which has been used for a long time in the asphalt field, is also used for geotextiles for speeding up time in creep tests. Hydraulic Properties The percent open area of a geotextile can be measured by shining a light through the geotextile onto a poster board and measuring the illuminated area on the poster. Monofilament woven geotextiles have percent open areas in the range of 6–12%. The apparent opening size (AOS) of a geotextile is obtained through a test (ASTM D4751) in which glass beads of uniform diameter are placed on top of the geotextile and wet-sieved through the geotextile. The diameter for which Table 26.2

95% of the beads by weight are retained on the geotextile is the AOS, designated as O95 . Typical values range from 0.01–0.5 mm. A distinction is made between the cross-plane hydraulic conductivity kcp and the in-plane hydraulic conductivity kip . Cross-plane refers to the case where the liquid flows in a direction normal to the plane of the geotextile, this is called filtration. In-plane refers to the case where the liquid flows parallel to the plane or within the geotextile; this is called drainage. Typical values of the hydraulic conductivity of geotextiles (ASTM D4491) range from 10−3 m/s to 10−5 m/s for the stress-free product. This is in the range of gravel to coarse sand. Because the thickness of the geotextile can vary due to the in situ stress condition, the permittivity 𝜓(S−1 ) is used for the cross-plane flow and the transmissivity Θ (m2 /s) is used for in-plane flow instead of the hydraulic conductivity. They are defined as follows: kcp

q q q = Δh = iAt Δh A At t q q q t= Transmissivity Θ = kip t = t iA iWt iW Permittivity Ψ =

t

=

(26.4) (26.5)

where q is the flow, A is the flow area perpendicular to the flow, i is the hydraulic gradient, Δh is the loss of total head over the flow distance, t is the thickness, and W is the width of the geotextile involved in the flow. The hydraulic conductivity, the permittivity, and the transmissivity should be tested under the compressive stress likely to be experienced in the field (ASTM D5493). While it is always important to do this testing, the values under load do not appear to be very different from the values under no load. As mentioned earlier, the cross-plane hydraulic conductivity k of geotextiles for water flow typically ranges from 10−3 to 10−5 m/s. If the fluid is not water, then the hydraulic conductivity and the permittivity should be corrected as follows: kf kw

=

Ψf Ψw

=

Θf Θw

=

𝜌 f 𝜇w 𝜌 w 𝜇f

(26.6)

where kf and kw are the hydraulic conductivity for the fluid and for water respectively, 𝜓 f and 𝜓 w are the permittivity for the fluid and for water respectively, Θf and Θw are the transmissivity for the fluid and for water respectively, 𝜌f and 𝜌w are

Viscous exponent n for several polymers

Geotextile polymer

Viscous exponent n value at 20% of ultimate strength

Viscous exponent n value at 60% of ultimate strength

Polyester (PET) Polyamide (PA) (nylon) Polypropylene (PE) Polyethylene (PP)

0–0.01 0.02 0.07 0.08–0.14

0–0.01 — 0.19–0.2 0.12–0.19

(Source: Adapted from den Hoedt, 1986.)

26.3 PROPERTIES OF GEOSYNTHETICS

the density of the fluid and of water respectively, and 𝜇f and 𝜇w are the viscosity of the fluid and of water, respectively. Other properties of geotextiles include resistance to abrasion from repeated action of gravel impacting the geotextile, soil retention and clogging (as in filters and silt fences), sunlight degradation from long-term exposure to ultraviolet rays, and degradation due to temperature, oxidation, chemical action, and biological action. Designers use reduction factors to take into account these factors, all of which affect the long-term strength and function of the geotextile. 26.3.2

Properties of Geomembranes

Physical Properties Remember that geomembranes are solid and have no intended holes in them. The unit weight of dry, clean geomembranes is between 8.5 and 15 kN/m3 depending on the polymer used to make them (ASTM D792). The high-density polyethylene (HDPE) membranes have a unit weight of about 9.2 kN/m3 . However, the density of geomembranes is usually given in terms of mass per unit area (ASTM D1910), in g/m2 . The thickness of commonly used smooth geomembranes is between 0.5 and 3 mm (ASTMD5199). The height of asperities for textured geomembranes can be 0.25–0.75 mm; these asperities do increase the interface shear strength. Mechanical Properties The stress-strain curve of geomembrane specimens tested in tension exhibits the same two types of shapes as soils: some have a peak strength followed by a residual strength, like overconsolidated soils; and some have a gradual increase in strength with no strain softening, like normally consolidated soils. One major difference is that the range of strains is drastically larger for geomembranes. Strains to failure for soils are in the 2–10% range, whereas strains to failure for geomembranes are in the 20–100% range. Failure refers to no more increase in resistance, but rupture may take as much as 1000% strain. The initial tangent modulus of deformation in tension can vary from 30 MPa for polyvinyl chloride (PVC) to 250 MPa for HDPE geomembranes. The peak tensile strength shows values in the range of 10–50 MPa depending on the type of polymer. The high peak strengths tend to lead to lower residual strengths, which can be 50–70% of the peak value. The tensile strength of the seams can be less than that of the parent material and should be tested. Seams are manufactured by overlapping two sheets and fusing them together, or by pinching two sheets and fusing them together. The tensile test for the overlapped seam is a shear test and the tensile test for the pinched seam is a peel test (e.g., ASTM D6392 and D882). Peeling tends to offer less resistance than shearing. The interface shear strength between a geomembrane and the soil is important in many design aspects, especially for slope stability. It is measured with the same test as for

929

geotextiles (Figure 26.3). A major difference exists between smooth membranes and textured membranes. Koerner (2012) quotes friction angles of 17∘ and 18∘ for fine sand and smooth HDPE, and 22–30∘ for fine sand and textured HDPE. Sometimes geomembranes are placed against geotextiles. Again Koerner (2012) quotes friction angles of 6–11∘ for geotextiles and smooth HDPE, and 19–32∘ for geotextiles and textured HDPE. The worst combination seems to be a woven monofilament on top of a smooth HDPE (6∘ ); the best combination appears to be a nonwoven, needle-punched geotextile on top of a textured HDPE (32∘ ). The puncture strength of a geomembrane is important and can be quoted as impact puncture strength or static puncture strength. The impact puncture strength relates to the ability of the geomembrane to resist shocks from falling objects. It is tested by dropping a heavy object on the membrane (ASTM D3029) or through a pendulum test (energy to puncture) (ASTM D1822). The static puncture strength is related to the ability of the geomembrane to resist puncturing when the membrane is in contact with large aggregates under high pressures. Two alternatives exist to test static puncture strength: a small-scale test and a large-scale test. In the small-scale test (ASTM D4833), an 8 mm diameter beveled-edge piston is pushed through a geomembrane stretched over a 45 mm diameter empty mold. The puncture failure load P is expected to be in the range of 50–500 N for thin, nonreinforced geomembranes and 200–2000 N for reinforced geomembranes (Koerner, 2012). The large-scale test consists of pressing the geomembrane against a bed of cones simulating aggregates (ASTM D5514). Hydraulic Properties Geomembranes are often used to prevent a fluid from passing from one side of the membrane to the other: this is called separation. Such geomembranes are essentially impervious. Nevertheless, nothing is truly and completely impervious. When the minute amount of fluid passing through the geomembrane must be known, the hydraulic properties of the geomembrane become important. Conventional hydraulic conductivity of geomembranes is in the range of 10−13 to 10−15 m/s, but, in the language of geosynthetics, terms such as water-vapor transmission and permeance (which actually designate other parameters) are often used. If solvents are to be retained instead of water, the hydraulic conductivity can increase drastically, by a factor of 100 or even 1000, depending on the nature of the chemical. As in the case of geotextiles, other properties of geomembranes include sunlight degradation with long-term exposure to ultraviolet rays, radioactive degradation (limit of 106 –107 rads), and degradation due to hot and cold temperatures, chemical action, and biological action. Designers use reduction factors to take into account these issues, all of which affect the long-term strength and function of the geomembrane.

930 26.3.3

26 GEOSYNTHETICS

Properties of Geogrids

Top cover

Air bag

Physical Properties Geogrids are open-grid geosynthetics (see Figure 26.1c). The distance between ribs is in the range of 10–100 mm. They can be unidirectional (applied stress is in one direction) or bidirectional (direction of applied stress can be random). The percent open area (POA) is measured by shining a light through the geogrids onto a poster board where the illuminated area is measured. Most geogrids have POAs in the range of 40–95%. The mass per unit area varies quite a bit, from 200–1000 g/m2 . Mechanical Properties The flexural stiffness G was defined in Eq. (26.1). Stiff geogrids have G values above 10 g.m, whereas flexible geogrids have G values of less than 10 g.m. From the point of view of tensile strength, several strengths can be identified: the rib strength, the junction strength, and the wide width strength. The rib strength refers to the strength of the individual longitudinal elements. The junction strength refers to the strength of the connection between the longitudinal and transversal elements. The wide width strength F ug is the strength of the geogrid at the field scale where all element strengths are integrated (ASTM D6637). The wide width tensile strength F ug of many geogrids is in the range of 20–140 kN/m (force per unit length of fabric) reached at a wide range of strains from 5–30%. The tensile modulus of a geogrid is defined as the tensile load applied per unit length of geogrid (kN/m) divided by the corresponding strain of the geogrids. Numbers in the range 125–255 kN/m have been measured at small strains (1–5%) for stiff geogrids (Austin et al., 1993). The interface shear strength between soil and geogrid can be measured in a direct shear test, as shown in Figure 26.3. In this test the geogrid is glued to a solid block that fits in the upper part of the direct shear box, which is 0.3 m by 0.3 m minimum (ASTM D5321). Results of such tests indicate that the ratio of the soil-geogrid shear strength over the soil-soil shear strength (called efficiency) is close to 1 in bidirectional loading and somewhat less (e.g., 75%) in unidirectional loading (Koerner, 2012). Furthermore, Sarsby (1985) showed that if the geogrid aperture (distance between two ribs) is at least equal to 3.5 times the mean particle size d50 of the soil, it is likely that the efficiency of the geogrid will be 1, meaning that the shear strength of the soil-geogrid interface will be equal to the soil-soil shear strength. The pull-out strength of geogrids embedded in soils is very important, as geogrids are most often used as reinforcement. There are two pull-out strengths: you can break the geogrid (F ug given by the manufacturer), or you can break the soil (F us ). Breaking the soil means failure in shear at the interface between the soil and the geogrid. That ultimate pull-out load can be written as: Fus = 2Le K𝜎v′ tan 𝜑′

(26.7)

Load cell Geosynthetic

Jack

~1 m Soil

Clamp ~2 m

Figure 26.4 Laboratory container for pull-out tests on geogrid.

where F us is the ultimate pull-out load per unit width of geogrid, Le is the embedment length of the geogrid, K is a pull-out coefficient specific to the soil and geogrid involved, 𝜎v′ is the vertical effective stress at the depth of the geogrids, and 𝜑′ is the soil effective stress friction angle. The value of K should be obtained by testing. Pull-out tests can be carried out in the field on full-scale structures, or in the laboratory on large containers simulating the field conditions (Figure 26.4). Note that two phenomena contribute to the ultimate load F us and therefore the coefficient K: friction between the soil and the geogrid on one hand and penetration of the geogrid transversal elements into the soil. The first one can be calculated from friction laws while the second calls for bearing capacity estimates. Consider a 2.5 mm-thick geogrid made of 7 m-long, 5 mm-wide longitudinal ribs with a spacing of 100 mm and 3 mm-wide transverse elements with a spacing of 200 mm. It is located 2 m below the ground surface of a soil weighing 20 kN/m3 . The interface shear strength between the geogrid and the soil is 18 kPa and the bearing capacity of the transverse element is 900 kPa, then the pull out load per unit width of geogrid is given by: Fus = friction longitudinal + friction transverse + bearing capacity transverse (26.8) 1 + 2 × 0.003 Fus = 2 × 0.005 × 7 × 18 0.1 ) ( 7 1 × 0.005 × 18 × + 0.0025 × 1− 0.1 0.2 ) ( 7 1 × 0.005 × 900 × (26.9) × 1− 0.1 0.2 Fus = 12.6 + 3.59 + 74.81 = 91kN∕m

(26.10)

As can be seen from this example, the bearing capacity on the ribs is the major contribution, but the friction on the longitudinal ribs is not negligible. Recall that it takes a lot less displacement to mobilize the friction than the bearing capacity, so the friction will be mobilized first and the bearing capacity last. Now we can calculate the global friction factor K in Eq. (26.7) knowing that the friction angle is 30∘ and the vertical effective stress at the depth of the geogrid is 40 kPa: Fus 91 = = 0.281 (26.11) K= 2 × 7 × 40 × tan 30 2Le 𝜎v′ tan 𝜑′

26.3 PROPERTIES OF GEOSYNTHETICS

931

Tensile strain (%)

10

Bentonite clay

8 6

Geotextile or geomembrane

Needle punch or stitch bonds

(a) Non-reinforced GCL

(b) Reinforced GCL

Creep failure

4

Geotextile

2 0 –2

Creep 0

2

4

Figure 26.6 Geosynthetic clay liners’ cross-section.

6

Log time (hours)

26.3.4 Figure 26.5 Creep behavior of geogrids.

Physical Properties

The geogrid covers an area Ag per meter of geogrid width, which is a very small fraction of the total area Af : ( ) 1 1 + 0.003 × 1 − × 0.005 Ag = 0.005 × 7 × 0.1 0.1 7 2 (26.12) = 0.35 + 0.10 = 0.45 m × 0.2 The total area At is: At = 7 × 1 = 7 m2

Properties of Geosynthetics Clay Liners

(26.13)

Therefore, the area covered by the geogrid is 6.4% (0.45/7) of the total area, yet it develops 28.1% (Eq. (26.11)) of the total friction. Of course, in addition to the pull-out resistance of the geogrid itself, it is important to ensure that the connection can handle such a force. Creep tension properties are also very important for geogrids, as they are usually subjected to constant tension during their design life. At low stress levels (low fraction of the tensile strength F ug ), the geogrid will exhibit strain that increases linearly with the log of time. The slope of that line is the constant strain rate (e.g., 0.5% strain per log cycle of time). This creep strain rate depends on the type of polymer, the temperature, and the stress level. At intermediate stress levels, the geogrid may exhibit a delayed failure, in which the creep strain rate is constant for a while but increases dramatically after a certain time, leading to failure (Figure 26.5). Delayed creep failure typically occurs within the range of 25–50% of the ultimate tension F ug . Geogrids should be tested for creep response (ASTM D5292 and ASTM D6992). The time temperature superposition (TTS) principle can be used to shorten the testing time. In TTS, advantage is taken of the fact that a long time at low temperature is equivalent to a short time at high temperature. As in the case of geotextiles and geomembranes, other properties of geogrids include temperature effects, chemical effects, biological effects, radioactive effects, and sunlight effects. Designers use reduction factors to take into account these factors, all of which affect the long-term strength and function of the geogrid.

Geosynthetic clay liners (GCLs) are a recent innovation; they are made of a thin layer of bentonite clay sandwiched between two layers of geotextiles or geomembranes (Figure 26.6). They come in large flexible rolls and are used as containment barriers in the case of landfill liners and covers, for example. They are either non-reinforced or reinforced. The reinforcement solves the following problem. When hydrated, the bentonite clay is extremely slick and represents a weak shear plane that would initiate failure when placed in a slope. To remedy this situation, GCLs can be reinforced by fibers (needle-punched) or stitches (stitch bonds) that tie the two sides of the GCL together. Non-reinforced GCLs are used as barriers on flat ground, whereas the more common reinforced GCLs are used on sloping ground. The clay type can be sodium bentonite or calcium bentonite. Sodium bentonite has the lowest hydraulic conductivity, but its availability worldwide is limited. The thickness of a GCL varies significantly because of the difference in hydrated and dry thicknesses. Furthermore, it is difficult to isolate the thickness of the clay layer from its boundaries. The hydrated thickness is more important, as it affects the hydraulic properties. The total hydrated thickness of GCLs typically varies between 10 and 30 mm. The mass per unit area of GCL is in the range of 5–6 kg/m2 , with 4–5 kg/m2 of dry bentonite between the geosynthetic layers. Once hydrated, the GCL can easily become twice as heavy. The GCL is sold “dry,” which means that it has a low initial water content of around 10%. When the bentonite hydrates, it can reach water contents well over 100%. Hydraulic Properties The hydraulic properties of GCLs are very important, as GCLs are mostly used as barriers. The chemistry of the liquid hydrating the GCL can make a significant difference. Koerner (2012) reports on the difference between distilled water, tap water, mild landfill leachate, harsh landfill leachate, and diesel fuel. The results show that the swell movement is largest with distilled water and zero with diesel fuel; the other liquids lead to intermediate swell movement. The swell test consists of placing a “dry” sample of bentonite in a consolidometer, submerging it in water, and allowing it to

932

26 GEOSYNTHETICS

swell under light vertical pressure. The bentonite will swell, reaching the maximum swell movement in a time that can vary from 2 weeks to 2 months. The hydraulic conductivity k of a GCL can be measured in the laboratory using a flexible-wall triaxial permeameter (ASTM D5887). The in situ conditions should be reproduced as closely as possible, including the applied pressure and type of liquid. The value of k is obtained as follows: ( ) Δh A (26.14) q = kiA = k t where q is the flow rate, k is the hydraulic conductivity, I is the hydraulic gradient, A is the cross-sectional area through which the liquid flows, Δh is the loss of total head across the GCL, and t is the thickness to be permeated. This thickness is very difficult to measure accurately and leads to inaccuracies in quoting the k value. Daniel et al. (1997) reported that the same GCL tested by many different laboratories yielded values between 2 × 10−11 and 2 × 10−12 m∕s. This range can be expected for the k values of sodium bentonite GCLs and for water as permeate. Because of the difficulties associated with thickness measurements, the results of a GCL permeability test are usually given in terms of flow per unit area (q/A in m3 /s.m2 ). At the junction between contiguous sheets of GCL, there is an overlap. A minimum overlap of 150 mm is recommended to maintain a hydraulic conductivity equal to that of the GCL itself. Sometimes a layer of bentonite without GCL is added at the junction. From the long-term endurance point of view, freeze-thaw cycles and shrink-swell cycles do not seem to affect the hydraulic conductivity of GCLs significantly. Mechanical Properties The bentonite contributes very little to the wide width tension strength of a GCL. As a result, the tension strength of a GCL is estimated by using the values of the geotextile or geomembrane within which the bentonite is sandwiched. The shear strength of the GCL depends on the interface considered: upper geosynthetic and soil or waste, bentonite clay layer with or without reinforcement, lower geosynthetic and soil or waste. The upper and lower interface shear strengths between the materials above or below the GCL are addressed by considering the type of geosynthetic involved. For geotextiles, see Section 26.2.1; for geomembranes, see Section 26.2.2. The shear strength of the bentonite clay layer is measured by a direct shear test (ASTM D6243). Koerner (2012) reports on tests where the shear strength parameters c and 𝜑 decrease dramatically upon hydration of the bentonite clay layer with water when tested in a relatively rapid direct shear test. This decrease in shear strength parameters is not as severe when the hydrating liquid is leachate, and no decrease was found when the hydrating liquid was diesel fuel. The shear strength of reinforced GCL is much higher than that of unreinforced GCL and larger displacements are required to reach failure.

One concern is the long-term shear strength of reinforced GCLs. This is related to the long-term strength of needle-punched fibers or stitch bonds. The long-term (100-year) internal shear strength of reinforced GCL is up to 50% of the short-term shear strength (Koerner, 2012). The peel strength of reinforced GCLs refers to the maximum force per unit length that the upper and lower geosynthetic layers can resist when pulled away from each other at a 90∘ angle to the main direction of the GCL seam; it is measured in kN/m (ASTM D6496). Resistance to puncturing is also important and should be measured. The tests include ASTM D4883 and ASTM D6241. Squeezing of the bentonite layer away from a location by local pressure is avoided by placing a layer of sand, for example, above the GCL. 26.3.5

Properties of Geofoams

Physical Properties Geofoams are blocks made of light yet hard polystyrene materials. They are used as light fills, as thermal insulations, and as compressible inclusions. The width and the height of the blocks vary from 0.3 to 1.2 m, and the length from 1.2–5 m. A distinction is made between expanded polystyrene (EPS) and extruded polystyrene (XPS). EPS is made from solid beads of polystyrene expanded by blowing gas through them. XPS consists of melted polystyrene crystals mixed with additives and a blowing agent and shaped by extrusion through a die; the white Styrofoam coffee cups are made of extruded polystyrene. EPS geofoam blocks are typically larger than XPS geofoam blocks. The unit weight of geofoams ranges from 0.1–0.5 kN/m3 , which is much smaller than the average unit weight of soil (∼ 20 kN∕m3 ). Geofoams do not absorb much water, but are combustible and should not be exposed to temperatures in excess of 95∘ C. Mechanical Properties Because geofoams are often used as lightweight fill, the unconfined compressive strength is of interest. Figure 26.7 shows results from Negussey (1997), as presented by Koerner (2012). This unconfined compression stress-strain data indicates that the geofoam exhibits a linear behavior with a modulus E until a yield strength 𝜎 y and then strain hardens at a modest rate. The yield strength 𝜎 y is reached at around 2% compressive strain. The unit weight of the geofoam 𝛾 GF has a direct impact on its mechanical properties and the following equations can be derived from Negussey’s data: Modulus of EPS geofoam E(MPa) = 20𝛾GF (kN∕m3 ) (26.15) Modulus of XPS geofoam E(MPa) = 60𝛾GF (kN∕m3 ) (26.16) Yield strength of EPS geofoam 𝜎y (kPa) = 500𝛾GF (kN∕m3 ) (26.17)

26.4 DESIGN FOR SEPARATION

933

250

Stress, (kPa)

200 150 100

22.8 kg/m3

50

26.1 kg/m3 32.4 kg/m3

0

0

5

10

15

20

Strain (%)

Figure 26.7 Stress-strain curves for geofoam. (Source: After Negussey, 1997.)

Yield strength of XPS geofoam 𝜎y (kPa) = 800𝛾GF (kN∕m3 ) (26.18) The internal shear strength of geofoams can be tested by following ASTM C253, and the shear strength between geofoam blocks can be tested by direct shear testing (ASTM D5321). The tensile strength of geofoams, 𝜎 t , is much larger than that of soils. Using the data from Styropor (1993), the following equation can be proposed: Tensile strength of EPS geofoam 𝜎t (kPa) = 1250𝛾GF (kN∕m3 )

(26.19)

Creep properties are important because geofoams may be subjected to long-term loads (in lightweight embankments, for example). In creep testing of geofoams, the time temperature superposition can be used to shorten the time required to characterize the long-term behavior. Data from Negussey (1997) indicates that when the sustained compression stress is below 50% of the unconfined compression strength, the viscous exponent (Chapter 15, Eq. 15.9; Chapter 16, Section 16.8) of geofoam is within the range of values found in soils (n = 0.01 to 0.08). Thermal Properties Geofoams can be used as thermal insulation under buildings. Therefore, their thermal properties are important. The main property is the R value (see Chapter 17, Section 17.3). The R value is defined as: ΔT(∘ C) (26.20) R(∘ C.m2 ∕W) = q(J∕s.m2 ) where R is the R value or thermal resistance in ∘ Celsius.m2 / Watt, ΔT is the difference in temperature on either side of the geofoam in ∘ Celsius, and q is the heat flow in J/s.m2 . The higher the R value is, the more insulating the geofoam is. The R value per unit width of geofoam is R′ expressed in ∘ Celsius.m/Watt. The R′ value of geofoams varies from 20–40 and increases with unit weight. It is typically higher for XPS than for EPS. Other aspects to be addressed are the chemical resistance of geofoams, which are readily attacked by hydrocarbons, such

as gasoline; degradation due to long-term exposure to UV rays; and flammability. This is why it is best for geofoams to be covered by a soil backfill as soon as possible after installation. 26.3.6

Properties of Geonets

Geonets are open-grid geosynthetics very similar to geogrids; however, their purpose is to serve as spacers by providing flow conduits within their thickness. They are typically used in conjunction with a geotextile or geomembrane on top and bottom of the geonet. Whereas geogrids have a single layer of ribs typically perpendicular to each other, the ribs in geonets are stacked on top of each other (2 or 3 layers) and lined up in diagonals to facilitate flow. The mass per unit area varies from 0.8–1.6 kg/m2 and the thickness from 4–8 mm. The mechanical properties of geonets are similar to those of geogrids, but the hydraulic properties are most important, as geonets are used primarily for drainage purposes. The drainage capacity is quoted in flow per unit width of geonet. Values in the range of 10–3 to 10–4 m3 /s.m are common, but can decrease by 30% when the pressure increases to 1000 kPa. The drainage capacity of geonets may also be quoted in terms of transmissivity Θ (Eq. 26.5), which is related to the flow rate per unit width (q/W) by: q 1 × (26.21) i W where i is the hydraulic gradient. The EPA has regulations indicating that a geonet must have a transmissivity of at least 3 × 10−5 m2 ∕s for landfills and 3 × 10−4 m2 ∕s for surface impoundments. Θ=

26.4

DESIGN FOR SEPARATION

Separation means that the two materials on each side of the geosynthetic cannot penetrate it. This is associated with failure mechanisms by impact, punching, or tear (Figure 26.8). Impact refers to the case where a stone falls on top of the geosynthetic. Punching refers to the case where the geosynthetic is pushed through an opening between large aggregates.

934

26 GEOSYNTHETICS

Impact Stone

Punching W

Tire

b

H

is subjected to a tension force F tension , which is given by (Giroud, 1981): ( ) 2y b + (26.25) Ftension = 0.1pda 2 f (𝜀) = 0.025pda 2 b 2y

Tear P

y Stone

Figure 26.8 Modes of failure of geosynthetics in separation.

Tear refers to the case where the geosynthetic is pulled apart by stones that are moving away from each other in the deformation process of the geotechnical structure. Designing for impact first requires estimating the energy of the falling object. If the stone is represented by a sphere, the energy Estone to be dissipated at impact is: 𝜋d3 𝛾h (26.22) 6 where W is the weight of the stone, h is the height of drop, d is the stone diameter, and 𝛾 is the unit weight of the stone. This energy is absorbed in part by the geosynthetic layer and in part by the soil immediately below the geosynthetic. If the soil is soft, the stone has a soft landing and the peak force in the geosynthetic is lower than if the soil is stiff but the deformation is large. If the soil is extremely weak, only the geosynthetic resists the impact. To take the soil support contribution into account, the value of Estone is divided by a soil support factor F s varying between 5 and 25 (Koerner, 2012). This energy is then compared to the impact strength Egeosyn (Joules) of the geosynthetic (Section 26.3.1). The impact strength of the geosynthetic is divided by a cumulative reduction factor F, which accounts for installation damage, creep, and chemical/biological degradation, for example. F r varies from as low as 1.1 to as high as 9. The design ensures that: Estone = Wh =

Egeosyn Estone ≤ Fsoil Freduc

(26.23)

Designing against puncture requires estimating the force F stone generated by the stone protruding into the geosynthetic due to a pressure p applied. The pressure p may be applied by a rolling truck, for example. Koerner (2012) proposed: Fstone = pda 2 S1 S2 S3

(26.24)

where da is the diameter of the penetrating stone, and S1 , S2 , and S3 are the protrusion factor, the scale factor, and the shape factor, respectively. The product S1 S2 S3 varies from 0.65 in the most severe condition (angular large stone) to 0.01 in the most favorable condition (rounded small particles). The value of F stone is then compared to the strength of the geosynthetic (Eq. (26.2)) divided by the cumulative reduction factor. Designing against tear starts by calculating the tension force generated in the geosynthetic when squeezed between two layers of soil. When the upper and lower layers of soil are subjected to a rolling truck, for example, the layers deflect and bend locally under the wheel load. During this bending, the geosynthetic trapped between the two soil layers

where p is the pressure applied, da is the particle or stone diameter, f (𝜀) is a function of the strain in the geosynthetic, y is the displacement into the stone void, and b is the width of the stone void (Figure 26.8). Then the value of F tension is compared to the strength of the geosynthetic (Eq. (26.2)) divided by the cumulative reduction factor.

26.5

DESIGN OF LINERS AND COVERS

Liners are barriers placed at the bottom of landfills to prevent the waste and the liquid it generates from contaminating the soil and water below the waste. Covers are barriers placed on top of landfills to close them, prevent the waste from contaminating the surrounding environment, and prevent the gas it generates from escaping without control. Both liners and covers have evolved dramatically in the past 30 years, with most of the change taking place between 1980 and 1990. Before 1980, only a compacted clay liner was required at the bottom of landfills (Figure 26.9). The leachate collection system was a layer of sand and gravel with perforated pipes; there was no leachate detection system and no secondary liner to decrease the probability of leaks through the liner. Nowadays, liners are double composite systems (Figure 26.10) with a leachate collection system, a primary liner, a leak detection system, and a secondary liner. The liner involves many layers: geosynthetic for the purpose of separation (geomembrane), barrier (geosynthetic clay liner), and drainage and leachate detection-collection system (geocomposite with geonet-geotextile-geomembrane) in addition to layers of compacted clay. The change from compacted clay liners to geosynthetic base liners was prompted by the fact that the compacted clay layer had to be quite thick (up to 1.5 m thick), thereby taking up space that could otherwise have been used for waste; and by the discovery that certain chemicals, such as organic solvent leachate, dramatically increase the hydraulic conductivity of clays. Given this, a compacted clay liner alone could not ensure that no leakage would occur.

Waste Protective layer

Leachate collection system

Sand & gravel Compactive clay

1m

Figure 26.9 Early liner cross-section.

26.6 DESIGN FOR REINFORCEMENT

935

Waste Filter soil Gravel pipes Geosynthetic clay liner or compacted clay

Primary liner Secondary liner

Compacted clay Native soil

Leachate collection Geomembrane (1.5 mm) Geonet Geotextile Geomembrane (1.5 mm)

Figure 26.10 Example of a modern liner cross-section. 200

(30)

(21) GM

300

GM/CCL (31)

200

(32) (19)

100

(41)

GM/GCL 0

0

1

(8,15)

(19)

2 Life cycle stage

Leakage rate (lphad)

Leakage rate (lphd)

400

150 GM/CCL 100

(11)

(7) GM

(27)

(12)

50 (6)

(4) 3

1

GM/GCL

(4) 2 Life cycle stage

3

(b) Geonet leak detection system

(a) Sand leak detection system

Figure 26.11 Leakage rates in landfill liners. (Source: From Koerner, 2012/Xlibris.)

The leakage through a liner is very rarely zero. It is measured in liters per hectares per day (lphd). A liter is 10–3 m3 and a hectare is 10,000 m2 . The leakage varies with time as the landfill is being constructed and during its operation. Koerner (2012) defines stage 1 as the stage during construction, stage 2 as when considerable waste is placed, and stage 3 as when the final cover is placed. Furthermore, a distinction should be made between the different types of liners: geomembrane alone (GM), geomembrane over a compacted clay liner (GM/CCL), and geomembrane over a geosynthetic clay liners (GM/GCL). Based on the work of Othman et al. (1997) and Bonaparte et al. (2002), who gathered leakage rates for 289 landfills, Koerner gives the rates shown in Figure 26.11. The geomembranes used in liners are typically made of high-density polyethylene (HDPE). Some global minimum recommendations for the survivability of geomembranes used in liners are presented in Table 26.3. Low severity refers to a careful manual placement with light loads on smooth ground, for example; very high severity refers to machine handling on rough, stiff ground under heavy loads. Landfill covers are necessary so the waste does not contaminate the surrounding environment. These covers prevent rainwater from accessing the waste and keep the gas generated by the landfill from escaping into the atmosphere without control. An example of a cover cross-section is shown

in Figure 26.12. The layers involved include vegetation for a positive landscape impact, an erosion control geosynthetic, a top soil and cover soil layer for the plants to grow, a drainage layer (combination of geotextile/geonet/geomembrane), and then a second barrier (compacted clay or GCL), a gas collection layer, and the waste. The erosion control geosynthetic is often necessary because the top of the landfill is like a big hill, so the runoff water can erode the top soil in the cover. The drainage layer drains the rainwater away from the landfill. The barrier layer provides a second assurance that the water will not penetrate and also that the gas produced will not escape without control. The gas collection layer is necessary because most municipal landfills generate a lot of gas, mainly methane (CH4 ) and carbon dioxide (CO2 ). This layer can be made of sand with perforated pipes that collect the gas; the perforated pipes are connected to risers (vertical unperforated pipes) that bring the gas to a collection point; alternatively, the gas may be burned and the combustion products released into the atmosphere (flare). 26.6 26.6.1

DESIGN FOR REINFORCEMENT Road Reinforcement

The role of geosynthetics in road reinforcement is threefold: separation, reinforcement, and minimization of crack propagation. Geotextiles can be used for separation; geotextiles

936

26 GEOSYNTHETICS

Table 26.3 Minimum requirements for geomembrane survivability Required value considering the degree of severity Property

Low

Medium

High

Very high

Thickness (mm) Tensile strength (kN/m) Tear resistance (N) Puncture (N) Impact resistance (J)

0.63 7 33 110 10

0.75 9 45 140 12

0.88 11 67 170 15

1.00 13 90 200 20

(Source: After Koerner, 2012.)

Top soil Cover soil Landfill cover

Vegetation erosion protection geotextile Geotextile Geonet

Compacted clay or geosynthetic clay liner

Geomembrane (1 mm)

Gas collection layer Waste

Figure 26.12 Example of a landfill cover. (Source: From Koerner, 2012/Xlibris.)

and geogrids can be used for reinforcement and mitigation of crack propagation. A distinction is made here between applications for unpaved roads, paved roads, and overlay of asphalt flexible pavements. Separation has already been addressed in Section 26.4. It applies to the case of unpaved and paved roads, but rarely in the case of the overlay of asphalt flexible pavements. Prevention of crack propagation applies to the case of asphalt flexible pavements only. Indeed, for overlay of asphalt pavements, the rolling surface may contain vertical cracks. Minimizing the chances that the crack will propagate from the lower cracked asphalt layer vertically through the overlay to the rolling surface can be achieved by using a thicker overlay asphalt layer or the combination of a geotextile and a thinner asphalt overlay. It is important to keep moisture from rising through the overlay. For this, the geotextile is first rendered impervious by impregnating it with bitumen; then it is placed on top of the old pavement; then the overlay is constructed. The concept is that the geotextile will provide horizontal reinforcement with significant tensile strength and contain the future increase of the crack growth. The role of geosynthetics in road reinforcement is better suited to unpaved road than paved roads. The reason is that, on the one hand, geosynthetics tend to generate their resistance over a level of strain much larger than materials like asphalt

and concrete and, on the other hand, unpaved roads deflect more under traffic load than paved roads. Thus, geosynthetics will contribute more to the capacity of unpaved roads than paved roads. The design concept is to calculate the pavement thickness with and without the geosynthetics layer and perform an economic analysis on the two options. The benefit of using the geosynthetic layer is derived by assuming that the pressure level on the rolling surface (tire pressure) can be increased due to the presence of the geosynthetics. Without the geosynthetics, the stress on the subgrade must be kept within the elastic limit, whereas with the geosynthetics, the stress on the subgrade can reach the bearing capacity of the subgrade, as failure will be prevented by the geosynthetics. Both geotextiles and geogrids can be used for this application. Details of these designs can be found in Koerner (2012).

26.6.2 Walls

Mechanically Stabilized Earth Geosynthetic

Retaining walls (see Chapter 22) may be top-down walls, such as tieback walls; or bottom-up walls, such as gravity walls. Mechanically stabilized earth (MSE) walls are bottom-up walls, meaning that they are built starting at the bottom and going up until the top of the wall is completed. MSE walls are built by placing a layer of soil (say, 0.3 m thick), compacting it, then placing a layer of reinforcement, then a layer of soil and compacting it, placing a layer of reinforcement (geotextile or geogrids in this case), and so on to the top of the wall (Figure 26.13). These walls can be built in such a fashion to heights reaching tens of meters and are less expensive than conventional gravity or cantilever retaining walls (Figure 26.14). The reinforcement can be made of rigid inclusions such as steel strips and steel wire mesh or flexible inclusions such as geosynthetics (geotextiles and geogrids). The front of the wall is covered with panels that are tied to the reinforcement. The design of geosynthetic MSE walls includes internal stability and external stability. The minimum length of reinforcement is set at 0.7 H where H is the height of the wall.

937

26.6 DESIGN FOR REINFORCEMENT

2 Layers @650 mm

where W is the weight of the wall mass per unit length of wall, 𝜑′ is the friction angle of the interface at the bottom of the wall, and Pa is the horizontal force per unit length of wall due to the active earth pressure against the back of the wall (see Chapter 22). Alternatively, for the LRFD approach, the equation is:

4 Layers @500 mm

𝛾Pa = 𝜑W tan 𝜑′

Layer numbers 15 14 13 12 11

6.0 m

9

10 8

7 5

9 Layers @300 mm

6 4

3 1

O

2

(26.27)

where 𝛾 is the load factor and 𝜑 is the resistance factor. Overturning of the wall is addressed through a factor of safety expressed as a ratio of moments. The moments are taken around the bottom of the front of the wall (point O on Figure 26.13): ∑ Resisting moments Wd = Foverturing = ∑ (26.28) Pa xa Driving moments where d is the moment arm of the weight W of the wall and xa is the moment arm of the active earth pressure force Pa . Alternatively, for the LRFD approach the equation is:

Figure 26.13 MSE wall with geosynthetics reinforcement.

𝛾Pa xa = 𝜑 W d 900

(26.29)

where 𝛾 is the load factor and 𝜑 is the resistance factor.

800 Mean values Cost (dollars/m2)

700

ity av Gr

Internal Stability

lls wa

600 500 MSE (Metal)

400 alls

B/B CRI

M

200 100

) hetics osynt e G ( SE

IN w

300

1

3

5

7

9

11

13

Height of wall (m)

Figure 26.14 Cost of retaining walls. (Source: From Koerner, 2012/Xlibris.)

External Stability Bearing capacity at the base of the wall, general slope stability of the wall and the slope within which it rests, sliding of the wall mass, and overturning of the wall mass are all external stability issues. Commonly used factors of safety for a global factor of safety approach and for each one of those failure modes are presented in Table 26.4. Average load and resistance factors for an LRFD approach are also shown in Table 26.4. Bearing capacity and slope stability are dealt with in Chapters 18 and 20 respectively. Sliding of the wall mass can be addressed through a factor of safety expressed as: ∑ Resisting forces W tan 𝜑′ = (26.26) Fsliding = ∑ Pa Driving forces

Pull-out capacity and yield of the geosynthetic reinforcement are the two aspects of internal stability of an MSE wall with geosynthetic reinforcement. Commonly used factors of safety for a global factor of safety approach and for each one of those failure modes are presented in Table 26.4. Average load and resistance factors for an LRFD approach are also shown in Table 26.4. Pull-out capacity requires an understanding of the load distribution in the reinforcement. Figure 26.15 shows the variation of the tension load T (kN/m) in the reinforcement as a function of the distance from the front of the wall. At the wall facing, the load T in the reinforcement is very small, and then it increases as the instability of the wedge of soil near the wall is transferred into the geosynthetic as a tension force T (kN/m). At a distance Lmax from the front, the tension T reaches a maximum T max . Beyond T max , the tension decreases and reaches zero at a certain distance from the front. This distance must be less than the actual length L of the reinforcement, or significant deformations and possibly failure will occur. The true embedment or anchoring length La available to resist the active pressure force against the wall is L − Lmax . The design requires a knowledge of Lmax , which is to be ignored in the length required to resist T max . The force T max is calculated as: Tmax = sv 𝜎ah

(26.30)

where T max is the maximum line load (kN/m) to be resisted by the geosynthetic layer at a depth z, sv is the vertical spacing between reinforcement layers at the depth z, and 𝜎 ah is the total horizontal active stress at the depth z. The stress 𝜎 ah is discussed in Chapter 22.

938

26 GEOSYNTHETICS

Table 26.4

Some possible load and resistance factors for external and internal stability of geosynthetic MSE wall Load factor

Stability

Design issue

Factor of safety

Resistance factor

Dead

Live

External stability

Bearing capacity Slope stability Sliding Overturning Pull-out Breakage

2 1.5 1.5 2 1.5 1.5

0.5 0.7 0.85 0.7 0.9 0.9

1.25 1.25 1.25 1.25 1.5 1.5

1.75 1.75 1.75 1.75 1.75 1.75

Internal stability

T L

Active pressure σ′v

Z Lmax

H

Tmax Y

La Sv

Lmax

Sv

σah

La L

45+ϕ/2

Figure 26.15 Load in the reinforcement. (Source: Adapted from Theisen, 1992.)

Now that we have calculated the load, we need to find the length of reinforcement that will safely carry the load without pulling out of the soil. The pull-out line capacity T pullout (kN/m) of the geosynthetic layer is given by: Tpull out = 2fmax La

(26.31)

where f max is the maximum shear stress that can be developed on both sides of the interface between the geosynthetic and the soil, and La is the anchoring length. Recall that La is the length beyond the failure wedge. The shear stress f max is evaluated as follows: (26.32) fmax = 𝜎v′ tan 𝛿 where 𝜎v′ is the vertical effective stress on the geosynthetic layer at depth z (including any effect from surcharge or load at the ground surface), and tan 𝛿 is the coefficient of friction between the soil and the geosynthetic. Then the ratio between the load T max and the resistance T pullout must satisfy a factor of safety F (see Table 26.4): Tpullout = F × Tmax

(26.33)

and the required safe length La of the geosynthetic sheet is given by: Fs 𝜎 (26.34) La = ′ v ah 2𝜎v tan 𝛿 In the simple case where 𝜎ah = = 0.33, F = 2, and tan 𝛿 = 0.5, then La is equal to 0.66 sv , which is quite small for normal vertical spacing of 0.3–0.5 meters. Ka 𝜎v′ , where Ka

A load and resistance factor approach would consist of replacing Eq. (26.34) by: 𝛾2La 𝜎v′ tan 𝛿 = 𝜑sv 𝜎ah

(26.35)

where 𝛾 is the load factor and 𝜑 is the resistance factor. Note that the anchoring length La is constant with depth. The reason is that as the load increases with depth, so does the resistance. In practice, the minimum embedment length is set at 1 m. The distance Lmax required to develop the load in the reinforcement layer is taken as the width of the active wedge (see Chapter 22 and Figure 26.15): ( ) 𝜑′ Lmax = (H − z) tan 45 − (26.36) 2 Therefore, the final length of the geosynthetic layer L is: 𝛾 ) ( s𝜎 𝜑′ 𝜑 v ah (26.37) + ′ L = (H − z) tan 45 − 2 2𝜎v tan 𝛿 For construction simplicity, the length L is often kept constant for the entire wall. Because the length L is largest at the top of the wall, practically the length of reinforcement is taken as: 𝛾 ) ( s𝜎 𝜑′ 𝜑 v ah (26.38) + ′ L = H tan 45 − 2 2𝜎v tan 𝛿 where 𝜎 ah and 𝜎v′ are calculated at the depth of the first reinforcement layer. Yield of the geosynthetic layer is the next design issue. We must make sure that the geosynthetic can safely carry the

26.6 DESIGN FOR REINFORCEMENT

load T max without yielding or rupturing. For this, we need to find the allowable tensile resistance of the geosynthetic layer T allow . This allowable tensile resistance T allow is obtained from the measured ultimate tensile resistance T ult and given by: Tallow

Tult = RFID × RFCR × RFCBD

26.6.3

Reinforced Slopes

A distinction must be made here between humanmade bottom-up slopes (e.g., embankments and dams) and natural slopes or top-down slopes (e.g., hillside slopes and cuts). In the first case, it is possible to install geosynthetic reinforcement layers (geotextiles or geogrids) as the slope is built. In the second case, it is not possible to use geosynthetics as reinforcement; however, natural slopes and cuts can be reinforced with geosynthetics by covering the slope with a geosynthetic layer and anchoring the cover deeply beyond the failure plane. For bottom-up slopes, the factor of safety of the slope with reinforcement is calculated as presented in Section 20.14. 26.6.4

Reinforced Foundations and Embankments

Geosynthetics can be placed below a foundation or embankment to improve its carrying capacity and performance. Geotextiles and geogrids are most commonly used this way. The main design issues are bearing capacity, settlement, and anchoring length. Bearing capacity. Bearing capacity is improved because the failure plane has to pull on the geosynthetic layer. The degree of improvement is analyzed by the same method used for slope stability analysis once a circular surface has been chosen. For example, consider a strip footing of width B at

pu

Strip footing O B

B/2

A2

A1

(26.39)

where T allow and T ult are the allowable and ultimate resistance of the geosynthetic layer (kN/m); and RF ID , RF CR , and RF CBD are strength reduction factors that take into account installation damage, creep, and chemical and biological factors. These strength reduction factors vary between 1 and 2 depending on the application (Koerner, 2012), and average 1.55, 2.15, and 1.32, respectively. Note that once combined, these reduction factors lead to using an allowable tension that is about 20% of the measured ultimate tensile strength T ult of the geosynthetic. The required ultimate strength T ult of the geosynthetic is such that: Tult 𝛾 Tallow = = Tmax or RFID × RFCR × RFCBD 𝜑 𝛾 Tult = Tmax × RFID × RFCR × RFCBD (26.40) 𝜑 We must also make sure that there is no slip at the location of the overlap between geosynthetic layers. Thus, the overlap distance must satisfy Eq. (26.35). Because the overlap is located near the wall where the tension load is less than T max , a length equal to 1∕2La is typically used for the required overlap distance.

939

Geosynthetic layers su

Figure 26.16 Foundation reinforcement.

the surface of a soft clay with an undrained shear strength su (Figure 26.16). A geosynthetic layer is placed at a depth of B/2 with a tensile strength T kN/m. Let’s find the value of T required to increase the bearing capacity by a factor of 2. At failure and for a circular failure surface as shown on Figure 26.16, moment equilibrium around O gives: B T = su 𝜋BB + TB or pu = 2𝜋su + 2 (26.41) 2 B Note that it is assumed here that the geosynthetic is flexible and that it will deform at the intersection with the failure surface in such a way that it will become tangential to the failure circle. For the geosynthetic layer to double the bearing capacity, we must have: pu B

2𝜋su + 2 TB 2𝜋su

= 2 or T = 𝜋su B

(26.42)

For a 2m-wide strip footing and su = 120 kPa, the value of T is 240 kN/m. This is a very high allowable tensile strength for a geosynthetic and several layers would have to be used to safely achieve this level of tensile strength. However, for su = 40 kPa, the value of T is 80 kN/m, which can be achieved with one layer. Settlement. Settlement is also affected by the presence of a geosynthetic layer. At small displacements, the contribution is limited, as the geosynthetic typically has to deform enough to make a difference. The magnitude of settlement necessary for this contribution to be significant is more consistent with embankments than foundations. Indeed, larger settlements are more readily accommodated by flexible embankments than by rigid foundations. Figure 26.17 shows an embankment with a width L and a geosynthetic layer at the bottom of it. If it is assumed that the embankment settles s at its center, that the deflected shape of the bottom of the embankment is an arc of a circle, and that the circle passes through the ends of the embankment, then the relationship between the settlement s and the radius R of the circle is: L2 = R2 or (neglecting higher-order terms) 4 L2 R= (26.43) 8s where L is the width of the embankment. (R − s)2 +

940

26 GEOSYNTHETICS

R

total head through the flow net, and A is the cross-sectional area perpendicular to the flow. The steps include drawing a flow net, calculating the flow q through the flow net, determining the required permittivity from Eq. (26.47), and seeking the geotextile that satisfies this requirement. Note that the permittivity of the geotextile has to be corrected for reduction factors as follows:

R L

s

Geosynthetic layers

Exaggerated deflection profile

Figure 26.17 Embankment reinforcement.

Then the geosynthetic stretches from an initial length L to a deformed length L′ : L 2R which leads to a strain 𝜀 in the geosynthetic of: L′ = 2R Arc sin

𝜀=

Arc sin(4s∕L) −1 4s∕L

(26.44)

(26.45)

and a tension T equal to: T = E𝜀

(26.46)

where E is the geosynthetic modulus (Section 26.3.1). The geotextile or geogrid required needs to have a tensile strength much higher than T because of the reduction factors (Eq. (26.39)). Anchoring length. The geosynthetic layer must extend far enough beyond the edges of the embankment or the foundation to ensure that it will not pull out when loaded. The anchor length is calculated as presented in Eq. (26.7). The anchor length can be shortened by wrapping the geosynthetic around large embedded stones or timber cribbing.

26.7

DESIGN FOR FILTRATION AND DRAINAGE

Filtration refers to the case in which water is flowing perpendicular to the plane of the geosynthetic; drainage is the case in which water flows in the direction of the geosynthetic. The design of a geosynthetic filter or drain (mostly nonwoven, needle-punched geotextile) has two aspects: water passage and soil retention. The problem is that for water conveyance, the geotextile should have large openings, whereas for soil retention it should have small openings. A compromise must be found. For filtering, the required permittivity for water passage is calculated: q k Ψreq = = (26.47) t Δh A where 𝜓 req is the permittivity, k is the hydraulic conductivity, t is the thickness of the geotextile, q is the flow through the flow net to be handled by the geotextile, Δh is the drop of

Ψallow

Ψult (RFSCB × RFCR × RFIN × RFCC × RFBC )

(26.48)

where 𝜓 allow is the permittivity that can be used in design, 𝜓 ult is the permittivity quoted by the manufacturer, RF SCB is the reduction factor for soil clogging and blinding, RF CR is the reduction factor for creep reduction of void space, RF IN is the reduction factor for adjacent materials intruding into the geotextile void space, RF CC is the reduction factor for chemical clogging, and RF BC is the reduction factor for biological clogging. These reduction factors vary between 1 and 10 depending on the application, and average 4.41 (RFSCB ), 1.83 (RFCR ), 1.1(RFIN ), 1.25 (RFCC ), and 2.2 (RFBC ). As can be seen, multiplying all these factors leads to using an allowable permittivity that is a very small fraction of the ultimate value. In addition, a regular factor of safety is applied as follows: F=

Ψallow Ψreq

(26.49)

For critical applications, this factor of safety should be as high as 5–10. For filtering, the problem of soil retention occurs when the water flows from a soil with fines into a much coarser soil or an open space. In this case the fines may wash out through the coarser soil and follow the water flow. The filter ensures that the transition from fine-grained to coarse-grained is gradual and lets the water go through while retaining the fines of the soil. The design makes use of the geotextile AOS. (Recall from Section 26.3.1 that the AOS is the apparent opening size, defined as the diameter of glass beads corresponding to 95% retained by weight.) Typical AOS values range from 0.01 mm to 0.5 mm. A simple criterion for the opening of the geotextile is (Carroll, 1983): O95 < 2.5D85 (26.50) where O95 is the AOS of the geotextile and D85 is the particle size corresponding to 85% passing by weight of the soil to be protected. More detailed criteria for geosynthetic filters have been developed (e.g., Luettich et al., 1992; Koerner, 2012; Giroud, 2010). In particular, Giroud (2010) proposed two new criteria based on porosity and thickness in addition to water conveyance and soil retention. The porosity criterion ensures that the geotextile has enough openings per unit area and makes a clear distinction between woven and nonwoven geotextiles. The thickness criterion recognizes that, unlike granular filters, the opening size of a geotextile filter depends on its thickness.

26.8 DESIGN FOR EROSION CONTROL

26.8

DESIGN FOR EROSION CONTROL

Geosynthetics have been used for decades in the field of erosion control. There are several application domains, including geosynthetic filters under rip rap, geosynthetics to facilitate revegetation, and silt fences. Filters Rip rap is often placed to prevent erosion when high water velocities affect the ground surface. Sizing the rip rap consists of finding out the highest water velocity to be handled and choosing the rip-rap size accordingly. Figure 24.8 can be used for rip-rap size selection. In addition, one must check that the rock itself is not degradable over time when subjected to wet-dry cycles. Once the rip rap is chosen, it is very important to place a geosynthetic layer between the soil to be protected and the rip rap. If such a layer is not placed, the soil can erode from underneath the rip rap. The rip rap will not move downstream, but will sink into the soil below and not prevent erosion. The geosynthetic layer has two functions: soil retention to prevent the soil underneath from eroding away, and water conveyance to prevent compression water stresses from developing in the underlying soil. These water stresses would weaken the soil and lead to failure (e.g., slope instability). Therefore, the geosynthetic must be a filter, and geotextiles are best suited for this purpose. The design of the filter follows the same rules as those discussed in Section 26.7. Revegetation Erosion on the slopes of embankments, dams, levees, and river banks can be minimized by strong and thick vegetation. The problem is that it takes time for appropriate vegetation to grow and become dense and deeply rooted. To help fix the vegetation, geosynthetics can be used. A distinction is made between temporary erosion and revegetation materials (TERMS) and permanent erosion and revegetation materials (PERMS). TERMS are completely biodegradable

6 Long-term allowable velocity (m/s)

For drainage, the water flows in the direction of the geosynthetic. The design of a geosynthetic for drainage purposes follows an approach similar to that used in the design for filtering. Instead of permittivity, however, we use transmissivity in this case, defined as: q (26.51) Θreq = kt = iw where Θreq is the transmissivity required, k is the in-plane hydraulic conductivity, t is the thickness of the geosynthetic, q is the water flow to be handled, i is the hydraulic gradient, and w is the width of the geosynthetic. Although geosynthetics have very useful applications in filtering and drainage, they provide limited flow capacity for water conveyance (10–8 to 10–6 m3 /s per meter of geosynthetic).

941

Hard armor

5 4

Soft armor and vegetation

3 2

Vegetation

1

Barren soil

0 1

10 Flow duration (hr)

Figure 26.18 Allowable velocities for erosion control measures. (Source: Adapted from Theisen, 1992.)

(hay straws, mulches) or partially biodegradable (hydraulic mulch geofibers, erosion control blankets). PERMS include turf reinforcement mats (TRMs) and vegetated geocellular containment systems (GCSs). The geosynthetics used in PERMS have openings to let the vegetation and roots grow through and some filtering capability. After the seeds are sown, the vegetation grows and gets entangled with the geosynthetic, which provides reinforcement to the root system. Figure 26.18 gives a range of velocities that can be resisted by various forms of armoring, including the soft armor with vegetation discussed here. Silt Fences Water flowing on barren soil along roadways or construction sites erodes the soil. To prevent this erosion, silt fences are often placed to let the water go through but stop and collect the silt-size particles that would otherwise flow downstream. Silt fences consist of a geosynthetic (most often woven geotextile) placed above ground by attaching it to vertical posts driven in the soil (Figure 26.19). The silt fence catches and retains the fine soil particles yet lets the water flow through. The water flow is typically quite shallow compared to a river flow, but the velocity can be high on steep slopes. The following design issues must be addressed: maximum length of slope between fences, runoff flow rate, sediment flow rate, height of fence, spacing and strength of fence posts, and geotextile selection. The maximum length of slope Lmax that can be handled by one silt fence may be estimated by (Koerner, 2012): Lmax (m) = 36.2e−11.1 tan 𝛼

(26.52)

where 𝛼 is the slope angle. If the length of the slope to be protected is longer than that, a sequence of silt fences separated by a distance less than Lmax is used. The runoff flow rate Q is tied to the recurrence interval of the rainfall selected (often taken as the 10-year flow) and is given by: Q(m3 ∕hr) = C × I(m∕hr) × A(m2 )

(26.53)

942

26 GEOSYNTHETICS

H

Geotextile

Accumulated fine soil

Fence

Post

S

L/2

L/2

Water seeps through

R

R

α

Figure 26.19 Silt fences. (a: Courtesy of Robert Koerner, 2012.)

E(kN∕km2 ⋅ yr) = 10 × R × K × LS × C × P

(26.54)

where R is the dimensionless rainfall coefficient, K is the dimensionless soil erodibility factor, LS is the dimensionless length of slope or gradient factor, C is the dimensionless vegetation cover factor, and P is the dimensionless conservation practice factor. These factors are given in Wishmeier and Smith (1960). The USLE equation has shortcomings and does not apply to channel and gully flow. Nevertheless, it provides a first estimate. The height H of the silt fence can be calculated by finding out the volume V of water and soil that can be retained by the fence over a 1 m width of fence: ) ( H V(m3 ) = Qt = H × 1m (26.55) tan 𝛼 where Q is obtained from Eq. (26.53), t is the duration of the rainstorm, H is the height of the fence, and 𝛼 is the angle of the slope on which the water flows. Silt fences are usually 0.3–0.9 m high. Then the spacing of the posts retaining the fence is chosen. This spacing is usually between 1 and 3 m. The load on the fence due to the water pressure can then be calculated to obtain the lateral load and maximum bending moment on the posts. This bending moment is in the range of 5–30 kN.m. The last step is to calculate the tensile load in the fence material. If it is assumed that the fence deflects an amount s in an arc of circle under the average water pressure of 0.5𝛾 w H behind the fence, the tension in the geosynthetic is given by the following equation: 𝛾w HL2 (

T= 16s

Arc sin

) or T = 4s L

𝛾w HL2 if 4s∕L is small (26.56) 16s

4s L

where T is the tension per meter of geotextile, 𝛾 w is the unit weight of water, H is the height of the fence, L is the distance between posts, and s is the horizontal deflection of the fence at midspan (Figure 26.19). Typical values of T range from 5–30 kN/m.

26.9

OTHER DESIGN APPLICATIONS

26.9.1

Lightweight Fills

Lightweight fills are most commonly built of geofoam blocks (Horvath, 1994; Saye et al., 2000). The unit weight of the geofoam blocks is at most 10% of the unit weight of soil. Therefore, the pressure on the native soil and the associated settlement can be reduced significantly. Note that a pavement layer still has to be constructed with heavier materials (granular base course for drainage and asphalt rolling layer) on top of a geofoam embankment. Of course, the compression of the geofoam must be added to the settlement of the soil below, but considering the typical application (embankment on soft soils), that compression is most of the time negligible compared to the settlement of the soil below. Overall, 80% reduction in settlement is not uncommon. 26.9.2

Compressible Inclusions

Another notable application of compressible inclusions is the case of geofoam blocks behind retaining walls to decrease the earth pressure. This type of solution is particularly useful for walls that cannot tolerate much lateral deflection without damaging the structure. This is the case of basement walls and bridge abutments in shrink-swell soil areas. The pressure-absorbing layer may be 50–600 mm thick and tends to decrease the pressure as shown in Figure 26.20. The thicker the geofoam layer is, the lower the pressure Depth over wall height (z/h)

where C is a dimensionless coefficient taken as 0.5 for barren soil, I is the rainfall intensity, and A is the drainage area. The weight of soil accumulated per unit area of soil drained and per unit time behind the silt fence can be estimated by using the Uniform Soil Loss Equation (USLE; Wishmeier and Smith, 1960):

0 0.2

50 mm Geofoam 600 mm Geofoam

0.4 0.6

At rest

0.8 Active 1

0

5

10 15 Earth pressure

20

25

Figure 26.20 Decrease in earth pressure by compressible inclusions. (Source: From Horvath, 1997.)

26.9 OTHER DESIGN APPLICATIONS

Table 26.5

R factor for various materials

Material

R factor (M2 .K/W or M2 .C/W)

Steel Ice Concrete Glass Water (25∘ C) Glass wool Air (25∘ C) Geofoam blocks

0.022 0.45 0.95 1.25 1.64 23.8 38.5 25 to 40

(Source: From Koerner, 2012/Xlibris.).

is likely to be. Note also that the pressure distribution is altered toward the bottom of the wall where the geofoam is most effective. Another application is to mitigate seismically induced pressures (Athanasopoulos, 2007).

in watts per degree Kelvin per meter (W/K.m). Therefore, the R rating is expressed in m2 .K/W. Because the degree Kelvin is equal to the degree Celsius, the R rating has the same value in m2 .K/W and in m2 .C/W. Table 26.5 shows that geofoam blocks have some of the highest R ratings of any materials. Within the geofoam range of R values, extruded polystyrene (XPS) has higher R values than expanded polystyrene (EPS). Styrofoam coffee cups are made of XPS. Once the R factor is known, the heat flow can be calculated: dQ dT A = kt A = dT (26.58) dt dx R where dQ is the amount of heat (J) flowing in a time dt (s), kt is the thermal conductivity (J/s.K.m or W/K.m), A is the area perpendicular to the heat flow (m2 ), dT is the change in temperature (K), dx is the length over which the change of temperature is occurring (m), and R is the thermal resistance or R factor (m2 .K/W). The applications include insulation under a house on permafrost to avoid ground thawing, or under a refrigerated building to avoid ground freezing. 26.9.4

26.9.3

Thermal Insulation

Geofoams are among the best temperature insulators. Recall from Chapter 17, Eq. 17.8 that the R factor of an insulator is a measure of the resistance to temperature propagation and is given by: dx (26.57) R= kt where dx is the thickness of the insulating layer in meters and kt is the thermal conductivity of the insulating material

943

Geosynthetics and Landfill Slopes

Modern landfill liners are made of many different layers, including geosynthetics. These geosynthetic layers, particularly GCLs, can represent planes of lower shear strength where slope failure can develop. This issue should be addressed at the time the liner is designed, together with a plan and possible restrictions on where and how high the waste can be piled up at one location. This topic is addressed in Chapter 27.

Problems and Solutions Problem 26.1 A geosynthetic is placed on the ground surface and stones are to be placed on top of it. The maximum diameter of the stones is 60 mm and the drop height from the truck is 1.5 m. The soil below the geosynthetic is medium stiff with a soil support reduction factor of Fs = 15; the cumulative reduction factor for the geosynthetic is Fr = 5. What is the impact strength required of the geosynthetic to safely handle the impact loading? Solution 26.1 We assume that the stone has a unit weight of 26 kN/m3 : Estone = Wh =

𝜋d3 𝜋 × 0.063 𝛾h = × 26000 × 1.5 = 4.41 J 6 6

Estone 4.41 = = 0.294 Fsoil 15

944

26 GEOSYNTHETICS

We must satisfy: Egeosyn Estone ≤ Fsoil Freduc Egeosyn ≥ 0.294 × 5 = 1.47 J The geosynthetic must have an impact strength at least equal to 1.47 J. Problem 26.2 A 0.5 m-thick layer of base course has been placed on top of a geotextile. Trucks with tire pressures equal to 600 kPa will travel on top of the base course during construction. The stones are 60 mm in diameter and fairly sharp, such that the product S1 S2 S3 in Eq. (26.24) is equal to 0.3. If the geotextile strength reduction factor is 4.5 (Eq. (26.39)), what is the required ultimate strength of the geotextile to safely avoid puncture? Solution 26.2 Fstone = pda 2 S1 S2 S3 = (600 + 0.5 × 20) × 0.062 × 0.3 = 0.659 kN P 0.659 kN St = = = 3.5 𝜋d 0.06𝜋 m kN Tultimate = Tallowable × RF = 3.5 × 4.5 = 15.8 m In this situation, a geotextile rated at 15.8 kN/m is needed. Problem 26.3 A landfill owner is considering replacing a 1 m-thick layer of compacted clay with a 15 mm-thick GCL as part of the design of a new landfill liner. The landfill has an area of 7.5 hectares and the fee collected per cubic meter of waste is $90. How much additional income does the owner stand to collect from the saving in the thickness of the liner? Solution 26.3 The change in height after replacing the compacted clay layer with the GCL: ΔH = 1 − 0.015 = 0.985 m For a landfill area of 7.5 hectares and a fee of $90 per m3 : Additional income per hectare = 90 × 10,000 × 0.985 = $886,500∕ha $886,500 Total income = × 7.5 ha = $6,648, 750 ha Problem 26.4 A geosynthetic clay liner and a compacted clay liner are being compared. The GCL is 15 mm thick and has a hydraulic conductivity of 10–11 m/s; the CCL is 500 mm thick and has a hydraulic conductivity of 10–9 m/s. The water level is 1 m above the top of the liner and the pressure head is assumed to be zero on the bottom side of the liner. Calculate the amount of water going through the GCL and the CCL. Solution 26.4 Using Darcy’s law, and assuming that the hydraulic gradient is the total head divided by the thickness of the GCL and a flow through a unit area, the amount of water through the GCL is: q = kiA = (1 × 10−11 )

(

) 1.015 (1 × 1) 0.015

26.9 OTHER DESIGN APPLICATIONS

945

q = 7 × 10−10 m3 ∕s Using the same procedure, the flow rate through the CCL is: q = kiA

) 1.500 (1 × 1) 0.500 q = 3 × 10−9 m3 ∕s = (1 × 10−9 )

(

Problem 26.5 A geosynthetic clay liner has a bentonite clay layer with the following shear strength characteristics: c′ = 0 and 𝜑′ = 10∘ . It is placed on the side slope of a landfill that has an 18∘ angle with the horizontal. The plan is to cover the GCL uniformly with 20 m of waste weighing 10 kN/m3 . What cohesion c′ must be developed by needle-punching in the GCL to have a factor of safety of 1.5 against failure in the bentonite? Use the infinite slope equation from Chapter 20, Section 20.3. Solution 26.5 Figure 26.1s shows the illustration of the infinite slope. Note that W is the weight of the wedge, T is the shear force, and N is the normal force. H is the height of the waste, L is the length of the wedge, and 𝛽 is the inclination of the slope. L H W T β

N

Figure 26.1s Illustration of infinite slope.

Based on the equilibrium condition and the definition of factor of safety, FS can be calculated: FS =

tan 𝜑′ c′ + tan 𝛽 𝛾H sin 𝛽 cos 𝛽

Here, 𝛾 is the unit weight of the waste and 𝜑′ is the friction angle. To achieve a factor of safety of 1.5, the cohesion developed by needle-punching has to satisfy the following equation: tan 10∘ c′ 1.5 + tan 18∘ 10 × 20 × sin 18∘ cos 18∘ Therefore, the cohesion developed by needle-punching must be c′ = 56 kPa. Problem 26.6 Design a 6 m-high geosynthetic-reinforced MSE wall. The vertical spacing between geosynthetic layers is 0.5 m, the backfill is sand with a unit weight of 20 kN/m3 and a friction angle of 34∘ , a surcharge of 20 kN/m2 is applied on the ground surface at the top of the wall, and the geosynthetic is a geogrid. The soil on which the wall is being built is a very stiff clay with an undrained shear strength su equal to 100 kN/m2 and a friction angle of 25∘ . The soil behind the wall is a sandy clay with a unit weight of 20 kN/m3 and a friction angle of 30∘ . Assume reasonable values for all other parameters needed for the design. Solution 26.6 (Figure 26.2s) H = 6m Sv = 0.5 m Δ𝜎v = 20 kN∕m2 Use a minimum reinforcement length L = 4.2 m as the length-to-height ratio of the reinforced wall (should be no less than 0.7).

946

26 GEOSYNTHETICS

Assume that the first layer of geosynthetics is placed at 0.25 m from the finished grade. Consider the ultimate strength resistance of the geosynthetic (Tult ) as 170 kN/m. q = 20 kN/m2

6m

Sandy Clay φ′s = 30° γs = 20 kN/m3

Sand Backfill φ′b = 34° γb = 20 kN/m3

1m

Very Stiff Clay Su = 100 kPa, φ′f = 25°

Figure 26.2s Retaining wall.

a. External stability, earth pressures. ka =

1 − sin 𝜑b 1 − sin 30 = = 0.33 1 + sin 𝜑b 1 + sin 30

Then the active load generated by the horizontal soil pressure Pa1 and the traffic surcharge Pa2 can be computed as: Pa1 =

Ka × 𝛾s × H 2 0.333 × 20 × (6)2 = = 120 kN∕m 2 2

• Located 2 m above the bottom of the wall (xa1 = 2 m, as shown in Figure 26.3s). Pa2 = Ka × q × H = 0.333 × 20 × 6 = 40 kN∕m • Located 3 m above the bottom of the wall (xa2 = 3 m as shown in Figure 26.3s). We can now calculate the sliding and overturning stability (ignoring the traffic surcharge). q = 20 kN/m2

Pa1 = 40 kN/m

6m Pa1 = 120 kN/m

xa2 = 3 m

xa1 = 2 m

o Ps

120 kPa

20 kPa

Figure 26.3s Pressure diagram on retaining walls.

b. External stability, sliding analysis. Using the LRFD approach and no traffic surcharge: 𝜑 W tan 𝜑′f ≥ 𝛾Pa1 or 𝜑𝛾b HL tan 𝜑′f ≥ 𝛾Pa1

26.9 OTHER DESIGN APPLICATIONS

Using 𝜑 as 0.85, 𝛾 as 1.25, and L as 4.2 m, we have: 0.85 × 20 × 6 × 4.2 × tan 25 ≥ 1.25 × 120 428.4 kN∕m ≥ 150kN∕m ∴OK Using the LRFD approach and the traffic surcharge: 𝜑 W tan 𝜑′f ≥ 𝛾1 Pa1 + 𝛾2 Pa2 or 𝜑𝛾b HL tan 𝜑′f ≥ 𝛾Pa1 Using 𝜑 as 0.85, 𝛾 as 1.25 for the dead load and 𝛾 as 1.75 for the live load, and L as 4.2 m, we have: 0.85 × 20 × 6 × 4.2 × tan 25 ≥ 1.25 × 120 + 1.75 × 40 428.4 kN∕m ≥ 220kN∕m ∴OK c. External stability, overturning analysis. Overturning around the toe (point O) of the wall with no traffic surcharge: 𝜑𝛾b H L2 𝛾Pa1 H 𝜑WL 𝛾Pa1 H 0.85 × 20 × 6 × 4.22 1.25 × 120 × 6 ≥ or ≥ or ≥ 2 3 2 3 2 3 899.6 kN ≥ 300 kN ∴OK Overturning around the toe (point O) of the wall with traffic surcharge: 𝜑(W + qL)L 𝛾1 Pa1 H 𝛾2 Pa2 H ≥ + or 2 3 2 0.85(20 × 6 × 4.2 + 20 × 4.2)4.2 1.2 × 120 × 6 1.75 × 40 × 6 ≥ + 2 3 2 1076 kN ≥ 510 kN ∴OK d. External stability, bearing capacity analysis. The eccentricity of the wall applied forces can be calculated as: W × e + q × L × e = Mov or (W + q × L) × e = Pa1 × xa1 + Pa2 × xa2 P × xa1 + Pa2 × xa2 120 × 2 + 40 × 3 e = a1 = = 0.61 m (W + q × L) (20 × 6 × 4.2 + 20 × 4.2) This eccentricity cannot be outside of the central one-third of the footing, which is: L 4.2 = = 0.7 m ∴OK 6 6 This means that there is no tension underneath the footing. The active length, according to Meyerhof’s distribution, is: e≤

Lactive = L − 2 × e = 4.2 − 2 × 0.61 = 2.98 m The bearing pressure is: p = (𝛾b × H + q) ×

L 4.2 = (20 × 6 + 20) × = (169.1)weight + (28.2)traffic = 197.3 kPa Lactive 2.98

The bearing capacity of the existing soil can be calculated according to the Skempton chart: qu = Nc Su + 𝛾b D = 7.5 × 100 = 750 kPa Checking for bearing capacity failure: 𝜑 × qbc ≥ 𝛾1 p1 + 𝛾2 p2 or 0.5 × 750 ≥ 1.25 × 169.1 + 1.75 × 28.2 375 kPa ≥ 260.7 kPa ∴OK e. Internal stability, pull-out failure. No traffic surcharge: Tmax = sv 𝜎ah = sv ka 𝜎v′ 1 − sin 𝜑r 1 − sin 34 ka = = = 0.283 1 + sin 𝜑r 1 + sin 34

947

26 GEOSYNTHETICS

Note that for an MSE wall built with geosynthetics, the kr and ka ratio are the same according to AASHTO LRFD (Figure 26.4s).

Me we tal b ar lde d w ma ts ire & gr ids

Meta l strip s

*Geosynthetics

0

Coefficient of lateral stress ratio = kr /ka 1 1.2 2.5 1.7

0

Depth below top of wall, Z

948

6000 mm

1 1.2 * Does not apply to polymer strip reinforcement

Figure 26.4s Coefficient of lateral stress ratio.

The results of T max at different heights are shown in Table 26.1s. T max1 is due to the soil weight (sv 𝜎 ah , active earth pressure) and T max2 is due to the traffic surcharge (sv ka × 20kN∕m2 ). Summary of calculation of T max

Table 26.1s

Layer no. 1 2 3 4 5 6 7 8 9 10 11 12

Depth (m)

ka

kr

σv (kPa)

𝜎 ah (kPa)

T max1 (kN/m)

T max2 (kN/m)

T max-total (kN/m)

0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75

0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283

0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283 0.283

5.0 15.0 25.0 35.0 45.0 55.0 65.0 75.0 85.0 95.0 105.0 115.0

1.414 4.241 7.068 9.895 12.722 15.549 18.376 21.204 24.031 26.858 29.685 32.512

0.71 2.12 3.53 4.95 6.36 7.77 9.19 10.60 12.02 13.43 14.84 16.26

2.83 2.83 2.83 2.83 2.83 2.83 2.83 2.83 2.83 2.83 2.83 2.83

3.53 4.95 6.36 7.77 9.19 10.60 12.02 13.43 14.84 16.26 17.67 19.08

Using the ultimate limit state procedure, we have: 𝛾1 Tmax 1 + 𝛾2 Tmax 2 = 𝜙 Tpullout

26.9 OTHER DESIGN APPLICATIONS

949

The active length of the reinforcement strip required to resist the pull-out load is: 𝛾T + 𝛾2 Tmax 2 Tpullout = 1 max 1 𝜑 ( ′ ) ( ) Tpullout 𝛾1 𝜎 ah × sv + 𝛾2 ka q × sv La = = 2 × fmax × b 2 × 𝜑 × 𝜎 ′ v × tan 𝛿 ( 𝜑) Lmax = (H − z) × tan 45 − 2 ( ′ ) ( ) ( 𝛾1 𝜎 ah × sv + 𝛾2 ka q × sv 𝜑) + Ltotal = Lmax + La = (H − z) × tan 45 − 2 2 × 𝜑 × 𝜎 ′ v × tan 𝛿 However, in construction practice, the length is often taken as constant throughout the height of the wall. The longest value of Ltotal is at the top of the wall (z = 0). Then: ( ′ ) ( 𝛾1 𝜎ah × sv + (𝛾2 ka q × sv ) 𝜑) + Ltotal = H × tan 45 − 2 2 × 𝜑 × 𝜎v′ × tan 𝛿 Default values for pullout friction factor, F* 1.2 + log Cu ≤ 2.0

20 (t/St) 0

Grid bearing member 6000 mm

t

Ribbed steel strips F* = tanΦf

Geogrids F* = 0.67 tanΦf

Geotextiles F* = 0.67 tanΦf

Smooth steel strips F* = 0.4

D

Steel grids F* = 10 (t/St)

Depth below top of wall z or z0 (mm)

Not to scale

LO St St

Figure 26.5s Default values for pull-out friction factor.

The resistance factor (𝜑) and load factor (𝛾 ) are taken as 0.9 and 1.5, respectively. The coefficient of friction (tan 𝛿) is computed according to AASHTO LRFD using Figure 26.5s. Based on Figure 26.5s, the friction factor for geogrids is equal to 0.67 times the tangent of the friction angle of the reinforced backfill (𝜑f ). Therefore: F ∗ = 0.67 tan 𝜑f = 0.67 tan 34 = 0.452 ( ′ ) ( ) 𝛾1 𝜎 ah × sv + 𝛾2 ka q × sv 𝜑) + = H × tan 45 − 2 2 × 𝜑 × 𝜎 ′ v × 0.67 tan 𝜑f (

Ltotal

950

26 GEOSYNTHETICS

) (1.5 × 0.283 × 20 × 0.25 × 0.5) + (1.75 × 0.283 × 20 × 0.5) ( 34 + Ltotal = 6 × tan 45 − 2 2 × 0.9 × 5 × 0.452 Ltotal = 3.19 m + 0.26 m + 1.22 m = 4.67 m This is slightly larger than the minimum requirement for the length of reinforced soil mass of 0.7 H or 4.2 m. f. Internal stability, yield of reinforcement. No traffic surcharge: Tult Tallow = RFID × RFCR × RFCBD Consider RF ID, RF CR and RF CBD as 1.55, 2.15, and 1.32, respectively: Tult T 170kN∕m = ult = = 38.6kN∕m 1.55 × 2.15 × 1.32 4.4 4.4 = 34.7 kN∕m

Tallow = 𝜑Tallow

Using the ultimate limit state analysis, we have: 𝛾1 Tmax 1 + 𝛾2 Tmax 2 = 𝜙Tallow 𝛾T + 𝛾2 Tmax 2 Tallow = 1 max 1 𝜑 1.5 × sv × 𝜎ah 1.75 × sv × ka q Tallow ≥ + 0.9 0.9 The maximum horizontal strength required is at the bottom of the wall, so we will check that layer of soil reinforcement: 1.5 × 0.5 × 32.5 1.75 × 0.5 × 0.283 × 20 Tallow ≥ + or Tallow ≥ 27 + 5.5 = 32.5kN∕m 0.9 0.9 Tallow = 38.6 kN∕m > 32.5 kN∕m 𝜑Tallow = 34.7kN∕m > 𝛾Tmax -total = 29.3kN∕m Here we compare either the factored or the unfactored resistance to the factored or unfactored loads. Detail calculations are shown in Table 26.2s. Table 26.2s

Summary of calculation for strength

Layer no. Depth (m) T max-total (kN/m) Ltotal (m) T ult (kN/m) T allow (kN/m) 𝜑 T allow (kN/m) 𝛾 T max -total Check 1 2 3 4 5 6 7 8 9 10 11 12

0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75

3.53 4.95 6.36 7.77 9.19 10.60 12.02 13.43 14.84 16.26 17.67 19.08

4.18 3.64 3.53 3.48 3.46 3.44 3.43 3.42 3.41 3.41 3.40 3.40

170.0 170.0 170.0 170.0 170.0 170.0 170.0 170.0 170.0 170.0 170.0 170.0

38.6 38.6 38.6 38.6 38.6 38.6 38.6 38.6 38.6 38.6 38.6 38.6

34.8 34.8 34.8 34.8 34.8 34.8 34.8 34.8 34.8 34.8 34.8 34.8

6.0 8.1 10.2 12.4 14.5 16.6 18.7 20.9 23.0 25.1 27.2 29.3

OK OK OK OK OK OK OK OK OK OK OK OK

Problem 26.7 A layer of geotextile is placed 1 m below a 2 m-wide strip footing. The footing rests on the surface of a loose sand with a friction angle equal to 30∘ and a pressuremeter limit pressure of 500 kPa within a depth equal to one footing width below the footing. The geogrid has an ultimate tensile strength of 100 kN/m.

26.9 OTHER DESIGN APPLICATIONS

951

a. Calculate the percent increase in ultimate bearing capacity between the case of no geogrid and the case with geogrid. b. If the geogrid has a global friction factor K (Eq. (26.7)) of 0.3, what length of geogrid is required to safely anchor the geogrid on each side of the footing? Solution 26.7 Figure 26.6s shows the foundation without geogrid. The ultimate bearing capacity of the foundation without geogrid is equal to pL : pu1 = pL pu

Strip footing

(PL)

O B

Figure 26.6s Illustration of foundation failure without geogrid.

Figure 26.7s shows the foundation with geogrid. The bearing capacity is calculated as: pu2 = pL +

2T B

Strip footing

pu

O B B/2 Geosynthetic layers

Figure 26.7s Illustration of foundation failure with geogrid.

In this problem, pL = 500 kPa, B = 2 m, and T = 100 kN∕m; hence: pu1 = PL = 500 kPa 2T 2 × 100 pu2 = PL + = 500 + = 600 kPa B 2 Using the geogrid improves the ultimate bearing capacity by 20%. The required length of geogrid can be calculated: Fus = 2Le K𝜎v′ tan 𝜙′ Assume that the soil unit weight is 20 kN/m3 . The geogrid is buried 1 m beneath the foundation; therefore: 𝜎v′ = 𝛾h = 20 × 1 = 20 kPa Fus 100 Le = = = 14.4 m 2 × 0.3 × 20 × tan 30∘ 2K𝜎v′ tan 𝜙′ So, the length of geogrid required to safely anchor the geogrid on each side of the footing is 14.4 m. Problem 26.8 A 30 m-wide, 7 m-high embankment is placed on soft clay with a geotextile between the surface of the soft clay and the embankment fill. The purpose of the geotextile is to increase the bearing capacity and reduce the settlement, but it is also used for separation, drainage, and filtering. The bottom of the embankment settles along an arc of circle with 1 m of settlement at the center and a negligible amount at the edges. What will be the tension load in the geotextile if its modulus is 500 kN/m?

952

26 GEOSYNTHETICS

Solution 26.8 Modulus, E = 500 kN∕m Settlement, s = 1m Width of embankment, L = 30 m Radius R of the circle is: R=

L2 302 = = 112.5 m 8s 8×1

Deformed length L′ : L′ = 2RArc sin Strain 𝜀 in the geosynthetic:

L = 30.09 m 2R

( ) 𝜀=

Arc sin

4s L

4s L

−1=

L′ − L 30.09 − 30 = = 0.003 L 30

Tension T equal to: T = E𝜀 = 500 × 0.003 = 1.5kN∕m Problem 26.9 A geotextile has an ultimate tensile strength of 100 kN/m and a maximum flow rate capacity of 8 × 10−7 m3 ∕s per meter of geotextile. What are reasonable values of the allowable tensile strength and allowable flow rate for this geotextile? Solution 26.9 Ultimate tensile strength, Tult = 100 kN∕m Maximum flow rate, qult = 8 × 10−7 m3 ∕sper meter of geotextile. The strength reduction factors take into account installation damage ID, creep CR, and chemical and biological degradation CBD. They are RF ID , RF CR , and RF CBD . They average respectively 1.55, 2.15, and 1.32: Tallow =

Tult 100 = = 23.05 kN∕m RFID × RFCR × RFCBD 1.55 × 2.15 × 1.32

The flow reduction factors take into account soil clogging and blinding, creep reduction of void space, adjacent materials intruding into the geotextile void space, chemical clogging, and biological clogging. They are RF SCB , RF CR , RF IN , RF CC , and RF BC . Their respective average values are: 4.41 (RF SCB ), 1.83 (RF CR ), 1.1 (RF IN ), 1.25 (RF CC ), and 2.2 (RF BC ). qult 8 × 10−7 8 × 10−7 = = (RFSCB × RFCR × RFIN × RFCC × RFBC ) 4.41 × 1.83 × 1.1 × 1.25 × 2.2 24.41 −7 3 = 0.33 × 10 m ∕s per meter of geotextile

qallow =

Problem 26.10 A construction site has a 30 m-long erodible slope with an angle of 6∘ . Silt fences are required. a. b. c. d.

How many silt fences are needed? If the 10-year rainstorm generates 100 mm/hr, what is the flow rate to be handled per meter of width of the fence? Calculate the height of the fence so that it can safely handle two 10-year storms each lasting 3 hours. Posts are placed every 3 m and the fence is allowed to deflect 0.2 m at its center. Estimate the tension in the fence fabric.

Solution 26.10 a. Number of silt fences: Lmax (m) = 36.2e−11.1 tan 𝛼

26.9 OTHER DESIGN APPLICATIONS

953

𝛼 = 6∘ ⇒ Lmax = 11.3 m ⇒ For 30 m long slope, 3 silt fences are needed. b. Flow rate per meter of fence: Q(m3 ∕hr) = C × I(m∕hr) × A(m2 ) C = 0.5 ⎫ ⎪ I = 0.1 ⎬ ⇒ Q = 0.565(m3 ∕hr∕m of fence) A = 1 × 11.3⎪ ⎭ c. Height of fence:

) H × 1(m) tan 𝛼 The time t (duration of 10 yr rain storm) is 3 hours. ( ) H 0.565 × 3 × 2 = H ⇒ H = 0.6 m ∘ tan 6 d. Tension in the fence geosynthetic fabric: 0.2 = 0.067, small enough to use the simplified equation: 3 𝛾 HL2 9.81 × 0.6 × 32 T= w ⇒T= = 16.55 kN∕m 16s 16 × 0.2 V(m3 ) = Qt = H

(

Problem 26.11 Derive Eq. (26.56) for silt fences. Solution 26.11 The average pressure on the fence is: 1 𝛾 H2 2 w The corresponding load on the fence is pL, where L is the length between posts. The resistance comes from the tension T in the fence geosynthetic. The component of T in the direction of the load is T sin𝛼 (Figure 26.8s). For equilibrium: p=

pL = 2T sin 𝛼 O P = Water pressure on fence

T

R

α α

α

α L/2

Post

A

L/2

B

S

C Post

Fence

Figure 26.8s Plan view of deformed silt fence.

In triangle OAC (Figure 26.8s), sin 𝛼 is given by: sin 𝛼 =

L∕2 R

T

954

26 GEOSYNTHETICS

So, T = pR The radius R is given by using triangle OAC (Figure 26.8s): R2 = (R − s)2 +

( )2 L 2

Because s is small compared to R and L, this gives: R= Then: T=p

L2 8s

𝛾 H L2 L2 = w 8s 16s

Problem 26.12 A 7 m-high, 60 m-wide embankment is to be built on a layer of soft clay with a water table at the ground surface. The soft clay is 5 m thick and the increase in stress in the clay layer can be taken as the pressure under the embankment because the clay layer is thin compared to the width of the embankment. The clay layer has the following consolidation characteristics: eo = 1.1, 𝛾 = 19kN∕m3 , Cc = 0.5. Two options are considered for the embankment fill: soil fill and geofoam fill. The soil fill has a unit weight of 20 kN/m3 and the geofoam fill 2 kN/m3 . What will be the settlement of the embankment in each case? Which fill type will have the shortest time to reach 90% consolidation? Solution 26.12 Using the consolidation theory: s=H

𝜎 ′ + Δ𝜎 Cc log ov ′ v 1 + e0 𝜎ov

a. Option 1: Soil fill. The increase in stress in the clay layer is: Δ𝜎v = 20 × 7 = 140 kPa The initial effective stress in the middle of the clay layer is: ′ 𝜎ov = 2.5 × 19 − 2.5 × 9.81 = 23 kPa

Therefore: s=5

0.5 23 + 140 log = 1.01 m 1 + 1.1 23

b. Option 2: Geofoam. The increase in stress in the clay layer is: Δ𝜎v = 2 × 7 = 14 kPa The initial effective stress in the middle of the clay layer is still: ′ 𝜎ov = 2.5 × 19 − 2.5 × 9.81 = 23 kPa

Therefore: s=5

0.5 23 + 14 log = 0.25 m 1 + 1.1 23

So, the settlement is reduced by a factor of 4 but the time to reach 90% consolidation is unchanged; the time required for the settlement to take place does not depend on the stress level, but rather on the drainage length and the properties of the compressing layer: H2 t = Tv dr cv Problem 26.13 A building refrigerated at −5∘ C is being designed on a soil with a high water table. The concern is the cost of the power (watts) to maintain the difference in temperature across the foundation. Two alternatives are considered. The first consists of

26.9 OTHER DESIGN APPLICATIONS

955

a relatively inexpensive 100 mm-thick concrete slab on grade on top of the soil. The second one consists of the same slab on grade on top of a 150 mm-thick geofoam. The concrete has a thermal resistance R value equal to 0.9m2 .C/W per meter of thickness and the geofoam 35m2 .C/W per meter of thickness. Calculate the amount of power required in each case to maintain the difference in temperature at −5∘ C above the slab and 0∘ C on top of the soil. Solution 26.13 The heat flow is defined as:

ΔQ ΔT (W) = kA Δt Δx

where: k = Material thermal conductivity(W∕m∘ C) A = Cross-sectional area ΔT∕Δx = Temperature gradient In terms of thermal resistance, the preceding equation can be written as: ΔQ AΔT (W) = Δt R 2 ∘ where R(m . C∕W) is the thermal resistance per meter thickness of the material. Note that the thermal resistance of a layered system is equal to the sum of each layer’s thermal resistance. Assuming a unit area of the slab (A = 1 m2 ), for the case of the concrete slab only, the heat flow is: ΔQ 1×5 (W) = = 55.5 Watt Δt 0.9 × 0.1 For the case of the concrete slab + 150 mm of geofoam: ΔQ 1×5 (W) = = 0.93 Watt Δt (0.9 × 0.1 + 35 × 0.15) The use of 150 mm of geofoam can reduce the power usage by 98.3%. Problem 26.14 A Styrofoam coffee cup holds coffee at 80∘ C. Your hand holding the coffee cup is at 30∘ C. Assuming a steady-state heat transfer in the cup wall, what is the R rating per meter of the Styrofoam if the amount of heat released from the coffee cup through the wall of the cup is 45 W? What is the thermal conductivity of the Styrofoam if the cup wall is 1.5 mm thick? Solution 26.14 Assume a coffee cup with an internal radius r1 = 40 mm, an external radius r2 = 41.5 mm and a height of 160 mm (Figure 26.9s).

r1

r2

Figure 26.9s Coffee cup dimensions.

At steady state, the amount of heat Q (W) released from the coffee cup, assuming that it is a long hollow cylinder, can be calculated as follows: ΔT ΔT Q(W) = −kA ( ) = k2𝜋r1 L ( ) r2 r r1 × ln r r1 × ln r2 1

2

with A(m ) = 2𝜋r1 L as the internal surface area of the cup.

1

956

26 GEOSYNTHETICS

k(W∕m∘ C) is the thermal conductivity of the cup wall. ΔT is the temperature difference between the inside and outside faces of the cup wall. The preceding equation can be rewritten in terms of thermal resistance: ΔT R 2∘ where R(m C∕W∕m) is the thermal resistance per meter of the cup wall and is equal to: ( ) r ln r2 1 R= k Based on the data given in the problem statement, the thermal resistance of the Styrofoam is: Q(W) = −2𝜋 L

R = −2𝜋L

(30 − 80) ΔT = −2𝜋 × 0.16 × = 1.117m2 .∘ C∕W∕m Q 45

The thermal conductivity of the Styrofoam is then: k(W∕m.∘ C) =

( ) ln

r2 r1

R

( ln =

0.0415 0.04

1.117

) = 0.033

CHAPTER 27

Soil Improvement

S

oil improvement is an alternative considered when the natural soil does not meet the engineering requirements for a project. As an example, if the soil is too weak to carry the structure on a shallow foundation, two options exist: deep foundations or soil improvement plus a shallow foundation. A soil improvement technique is sought that would make a shallow foundation feasible. If the deep foundation will cost $1,000,000, while the soil improvement will cost $250,000 and the shallow foundation $500,000, then the soil improvement alternative becomes attractive. Typically, in this case, the soil improvement technique is verified by in situ testing to demonstrate that a sufficiently improved soil strength and soil modulus can be reached so that a shallow foundation is viable. A very large number of methods are aimed at soil improvement; this chapter summarizes the main methods. For additional information, the following excellent references can be consulted: the state-of-the-art report, published by the ISSMGE Technical Committee on ground improvement and presented at the 2009 International Conference on Soil Mechanics and Geotechnical Engineering (Chu et al., 2009); the book by Moseley and Kirsch (2004); the NHI manual (Elias et al., 2006); and the website www.geotechtools.org (Schaefer and Berg, 2013). 27.1

OVERVIEW

Over the past 50 years, many different soil improvement techniques have been developed, and they continue to be developed and revised as the space available for human activities decreases. These methods have been classified by the ISSMGE Technical Committee on Ground Improvement as shown in Table 27.1. The word ground is used in that classification because it can incorporate rock, but because this book is limited to soil, the term soil improvement is used here. There are five major categories of soil improvement methods: 1. Soil improvement without admixture in coarse-grained soils

2. Soil improvement without admixture in fine-grained soils 3. Soil improvement with replacement 4. Soil improvement with grouting and admixtures 5. Soil improvement with inclusions

27.2 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN COARSE-GRAINED SOILS 27.2.1

Compaction

Compaction in this instance refers to roller compaction for shallow densification of soil deposits. The rollers used are static rollers, such as sheep-foot rollers for fine-grained soils or vibratory rollers for coarse-grained soils. Most rollers are cylindrical, but some are uneven rollers. The depth of compaction is at most 1 m and the degree of compaction is highest near the surface. Compaction is used to prepare pavement layers, retaining wall backfills, and embankment fills. This topic is covered in Chapter 21. 27.2.2

Dynamic Compaction

Because of the limited depth of conventional compaction techniques and the need to compact natural soils at larger depths, the idea of dropping a heavy weight from a height onto the soil surface was pioneered by Louis Menard (Menard and Broise, 1975). A typical combination would be a 20-ton weight dropping from a height of 20 m. This technique is best suited to compaction of coarse-grained soils. This topic, including the depth that can be reached and the improvement ratio versus depth, is covered in Chapter 21. 27.2.3

Vibrocompaction

The vibrocompaction method consists of lowering a cylindrical vibrator from a crane into the soil to densify the soil (Figure 27.1). A grid of 3–4 meters center to center is common. The vibrator is 2–5 m long and 0.3–0.5 m in diameter, and weighs 15–40 kN. The vibrations are generated in the horizontal direction by rotating eccentric masses.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

957

958

27 SOIL IMPROVEMENT

Table 27.1

Classification of soil improvement methods

Category

Method

Principle

A. Ground improvement without admixtures in noncohesive soils or fill materials

A1. Dynamic compaction

Densification of granular soil by dropping a heavy weight from air onto ground Densification of granular soil using a vibratory probe inserted into ground Shock waves and vibrations generated by blasting cause granular soil ground to settle through liquefaction or compaction Densification of granular soil using the shock waves and energy generated by electric pulse under ultra-high voltage Compaction of fill or ground at the surface or shallow depth using a variety of compaction machines Remove bad soil by excavation or displacement and replace it by good soil or rocks. Some lightweight materials may be used as backfill to reduce the load or earth pressure Fill is applied and removed to preconsolidate compressible soil so that its compressibility will be much reduced when future loads are applied Vacuum pressure of up to 90 kPa is used to preconsolidate compressible soil so that its compressibility will be much reduced when future loads are applied Similar to dynamic compaction except that vertical or horizontal drains (or together with vacuum) are used to dissipate pore pressures generated in soil during compaction DC current causes water in soil or solutions to flow from anodes to cathodes installed in soil Change the physical or mechanical properties of soil permanently or temporarily by heating or freezing the soil Collapsible soil (loess) is compacted by a combined wetting and deep explosion action along a borehole Hole jetted into soft, fine-grained soil and backfilled with densely compacted gravel or sand to form columns

A2. Vibrocompaction A3. Explosive compaction

A4. Electric pulse compaction

A5. Surface compaction (including rapid impact compaction) B. Ground improvement without admixtures in cohesive soils

B1. Replacement, displacement (including load reduction using lightweight materials)

B2. Preloading using fill (including the use of vertical drains)

B3. Preloading using vacuum (including combined fill and vacuum)

B4. Dynamic consolidation with drainage (including the use of vacuum)

B5. Electro-osmosis or electrokinetic consolidation B6. Thermal stabilization using heating or freezing

B7. Hydro-blasting compaction

C. Ground improvement with admixtures or inclusions

C1. Vibro replacement or stone columns

27.2 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN COARSE-GRAINED SOILS

Table 27.1 Category

(Continued)

Method

Principle

C2. Dynamic replacement

Aggregates are driven into soil by high-energy dynamic impact to form columns. The backfill can be either sand, gravel, stones, or demolition debris Sand is fed into ground through a casing pipe and compacted by vibration, dynamic impact, or static excitation to form columns Sand is fed into a closed-bottom, geotextile-lined cylindrical hole to form a column Use of piles, rigid or semirigid bodies, or columns that are either premade or formed in situ to strengthen soft ground Use of piles, rigid or semirigid columns/inclusions, and geosynthetic girds to enhance the stability and reduce the settlement of embankments Use of microbial materials to modify soil to increase its strength or reduce its permeability Unconventional methods, such as formation of sand piles using blasting, and the use of bamboo, timber, and other natural products Grout granular soil or cavities or fissures in soil or rock by injecting cement or other particulate grouts to either increase the strength or reduce the permeability of soil or ground Solutions of two or more chemicals react in soil pores to form a gel or a solid precipitate to either increase the strength or reduce the permeability of soil or ground Treat the weak soil by mixing it with cement, lime, or other binders in situ using a mixing machine or before placement High-speed jets at depth erode the soil and inject grout to form columns or panels Very stiff, mortar-like grout is injected into discrete soil zones and remains in a homogeneous mass to densify loose soil or lift settled ground

C3. Sand compaction piles

C4. Geotextile confined columns

C5. Rigid inclusions

C6. Geosynthetic-reinforced column or pile-supported embankment

C7. Microbial methods

C8. Other methods

D. Ground improvement with grouting-type admixtures

959

D1. Particulate grouting

D2. Chemical grouting

D3. Mixing methods (including premixing or deep mixing)

D4. Jet grouting

D5. Compaction grouting

960

27 SOIL IMPROVEMENT

Table 27.1 Category

E. Earth reinforcement

(Continued)

Method

Principle

D6. Compensation grouting

Medium-to high-viscosity particulate suspensions are injected into the ground between a subsurface excavation and a structure to negate or reduce settlement of the structure due to ongoing excavation Use of the tensile strength of various steel or geosynthetic materials to enhance the shear strength of soil and stability of roads, foundations, embankments, slopes, or retaining walls Use of the tensile strength of embedded nails or anchors to enhance the stability of slopes or retaining walls Use of the roots of vegetation to create and improve stability of slopes

E1. Geosynthetics or mechanically stabilized earth (MSE)

E2. Ground anchors or soil nails

E3. Biological methods using vegetation (Source: From Chu et al., 2009/IOS Press.)

Cone penetration resistance, MPa

100

Cannot be campacted

Can be compacted 10

Figure 27.1 Example of vibrocompactor. (Source: Courtesy of Earth Tech, LLC.)

Can be marginally compacted

1 0

0.5

1

1.5

2

2.5

3

Friction ratio, %

The frequency of vibration is in the range of 25–35 Hz with amplitudes between 10 to 30 mm. The vibrator typically reaches depths of 20–30 m, with 60 m being rare. Pipes go through the body of the vibrator and can supply water or air to the bottom of the vibrator to help with penetration if necessary. The soils best suited to use this technique are clean sands. If the fine content becomes higher than 10–15%, the vibrocompaction process becomes much less efficient (Mitchell and Jardine, 2002). Massarsch (1991) proposed a CPT-based chart indicating which soils are most applicable to vibrocompaction (Figure 27.2).

Figure 27.2 Soil suitability for vibrocompaction based on CPT. (Source: Courtesy of Dr. Rainer Massarsch.)

27.2.4

Other Methods

Other compaction methods include rapid impact compaction (Figure 27.3), explosive compaction, and electric pulse compaction. In rapid impact compaction (Watts and Charles, 1993), a tamper is pounded repeatedly on the ground surface. The weight is lifted about 1 m up in the air and dropped at a rate of around 40 drops per minute. The hammer weighs about 100 kN and has a diameter between 1.5 and 1.8 m.

27.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

961

a probe into the soil and discharging high voltage sparks at a rate of about 10 per minute. This recent method is as yet unproven.

27.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS 27.3.1

Displacement-Replacement

The displacement-replacement technique consists of simply excavating the weak soil (say, su < 20 kPa) and replacing it with stronger soil. Excavation depths beyond 8 m are uncommon; the method can be costly and environmentally unfriendly because of the amount of spoil to be disposed of. In the case of peat bogs, the backfill, which may be twice as heavy as the natural soil, can create very large settlements. Sometimes the backfill is made of lightweight material such as geofoam blocks (see Chapter 26, Section 26.3.5) to avoid excessive settlement and bearing capacity issues. 27.3.2

Figure 27.3 Example of rapid impact compactor. (Source: Courtesy of Menard Bachy, Inc.)

This technique is best for sands and gravels and is not suited for saturated silts and clays. Explosive compaction consists of setting a series of detonation charges in the deposit. These detonations create waves that propagate in the soil and compact it. This technique is not commonly used, but has the advantage of being relatively inexpensive. Electric pulse compaction consists of lowering

Preloading Using Fill

The technique of preloading using fill consists of loading the soil surface with a fill, as in the case of an embankment and a surcharge fill. It has been used for many years to shorten the time required to reach a certain settlement under the design load. Once the settlement is reached, the surcharge is withdrawn and the road can be paved, for example. It is important to note, in this respect, that the time tU to reach U percent of consolidation depends not on the height of the fill but on the drainage length H and the soil coefficient of consolidation cv . In other words, if it takes 5 years to reach 90% of the final settlement under a 5 m-high embankment, it will also take 5 years to reach 90% of the final settlement under a 10 m-high embankment. However, if it takes 5 years to reach 90% of the final settlement under a 5 m-high embankment, it will take a lot less time to reach that same settlement under a 10 m-high embankment. To find out what height hs must be added as a surcharge on top of an he high embankment to reach, say, 90% of the settlement of the embankment within a target time tt , use the following steps (Figure 27.4):

Surcharge Embankment

tt

hs

ho

t90

Embankment

he Soft clay

Time

Smax (emb.)

Cc, cv, eo Smax (emb. + surc.) Settlement

Embankment + Surcharge

Figure 27.4 Surcharge to accelerate embankment settlement.

962

27 SOIL IMPROVEMENT

1. Calculate the maximum settlement of the embankment smax(emb) . For a normally consolidated clay, the following equation can be used (see Chapter 18, Section 18.8.9): ( ′ ) 𝜎ov + Δ𝜎 ′ Cc log (27.1) smax (emb) = ho ′ 1 + eo 𝜎ov where ho is the height of the soft clay layer, Cc is the compression index from consolidation tests, eo is the ′ initial void ratio of the soft clay layer, 𝜎ov is the initial effective stress in the middle of the soft clay layer, and Δ𝜎 ′ is the increase in stress in the middle of the soft clay layer. 2. Choose the target time tt to reach smax(emb) . 3. Knowing the target time tt and the coefficient of consolidation cv of the soft clay layer, calculate the time factor T U corresponding to tt using the equation (see Chapter 18, Section 18.8.10): tc TU = t 2v (27.2) hd where hd is the drainage length. This drainage length is equal to the soft clay layer thickness if the water can only drain through the top or the bottom of the layer; equal to one-half of the layer thickness if the water can drain through the top and the bottom of the layer; and equal to half the horizontal distance between vertical drains if such drains are installed. 4. Then find the average percent consolidation U corresponding to the time factor T U by using the curve that links both parameters (Figure 27.5). Note that U is equal to: s(t) (27.3) U= smax where s(t) is the settlement after a time t and smax is the final settlement.

5. Knowing U, use Eq. (27.3) to calculate the maximum settlement smax(emb+surch) under the embankment plus the surcharge. In Eq. (27.3), U is known and s(t) is equal to the settlement under the embankment plus the surcharge after a time equal to the target time s(tt ). By design, this settlement is equal to the maximum settlement under the embankment only, smax(emb) : s(tt ) = smax (emb) (27.4) smax (emd+surch) =

hs =

Average degree of consolidation Uav in percent

C2 0

Curve C1

0.2

Curve C2

0.4

Curve C3

Two way drainage

(27.7)

where γs is the unit weight of the surcharge soil.

60

100

Δ𝜎 ′ 𝛾s

8. Note that if the surcharge is too high, a slope stability or bearing capacity problem arises for the side of the embankment. In that regard, the height of the surcharge hs max that would generate a bearing capacity failure in

C1

80

(27.5)

7. Finally, the height of the surcharge hs is the height that generates an increase in effective stress in the soft clay layer equal to Δ𝜎 ′ . Often, if the soft clay layer is shallow and not very thick compared to the width of the embankment, the increase in stress is equal to the pressure generated by the surcharge at the ground surface and the height of the surcharge is:

C3

40

U

6. Once the maximum settlement under the embankment and the surcharge smax (emb+surch) are known, Eq. (27.1) can be used to back-calculate the value of Δ𝜎 ′ induced by the surcharge: ) ( ( (1+eo )s ) max (emb+surch) ′ ho Cc Δ𝜎 ′ = 𝜎ov −1 (27.6) 10

0 20

smax (emb)

0.6 0.8 Time factor T

1

Curve C1

Curve C2

1.2

Curve C3

One way drainage

Figure 27.5 Average percent consolidation U versus time factor T U .

1.4

27.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

963

a clay of undrained shear strength su can be estimated by: 5.14su (27.8) hs max = 𝛾s 27.3.3 Prefabricated Vertical Drains and Preloading Using Fill The technique of using vertical drains and preloading consists of loading the soil surface with a fill while accelerating the consolidation process by installing prefabricated vertical drains (PVDs) or sand drains. Prefabricated vertical drains are also called wick drains or band drains. They are installed to decrease the drainage length hd , thereby reducing the time necessary for the consolidation settlement to take place. The drainage is then shifted from a vertical drainage problem involving the vertical hydraulic conductivity kv to a horizontal drainage problem between PVDs involving the horizontal hydraulic conductivity kh . For example, if a soft clay layer is 10 m thick and has one-way drainage, the drainage length will be 10 m and the time required for 90% of the final settlement will be t1 . If PVDs are installed on a grid with a center-to-center spacing equal to 2 m, the drainage length is controlled by the horizontal spacing and becomes much shorter, so the time t2 for 90% of the final settlement is dramatically reduced compared to t1 . Note that the ratio of the two times requires comparing the solution of the one-dimensional consolidation problem for the embankment on top of the layer without PVDs (see Chapter 12, Section 12.4.6) to the solution for the drainage around a grid of drains; this is often approximated by the radial consolidation problem (Moseley and Kirsch, 2004). PVDs are typically made of a filter material covering both sides of a corrugated plastic shell (Figure 27.6). The width may be 100 mm, the thickness 3-–4 mm, and the installed length can be 30 m. The flow rate out of such drains is in the

(a)

Figure 27.6 Prefabricated vertical drain. (Source: Courtesy of Layfield Environmental Systems, Layfield Group Limited, 11120 Silversmith Place, Richmond, British Columbia, Canada V7A 5E4.)

range of 2–8 liters per minute, but may decrease with time because of siltation, for example. At the same time, the actual flow through the PVD decreases with time as consolidation takes place. Installation of PVDs is done by tying one end of the PVD with an anchor inside a small-diameter pipe called a mandrel and pushing the mandrel vertically into the soil and dragging the PVD with it (Figure 27.7). Once at the required depth, the mandrel is withdrawn and the PVD is left in place. PVDs can be placed to depths of several tens of meters on a grid with spacing in the range of 1–2.5 m. The tops of the drains are bound by a drainage layer or drainage blanket (0.5–1 m thick) made of clean sand, and the water is pumped away from the site. The drainage blanket is often placed before the PVDs are installed and serves as a work platform for the equipment. One issue associated with the placement of PVDs is the development of a “smear zone” in soft clays at the boundary between the soil and the PVD. This

(b)

Figure 27.7 Installation of prefabricated vertical drains. (Source: Courtesy of Hayward Baker Geotechnical Construction.)

27 SOIL IMPROVEMENT

smear zone is a few PVD diameters thick and can reduce the permeability of the interface. The purpose of PVDs is to minimize the consolidation time t needed to reach a given percent consolidation U h taken as a ratio, not a percent, in Eq. (27.9). This time t can be calculated by using the Barron-Hansbo formula (Barron, 1948; Hansbo, 1981): ( ( ) ) ( ) dw dw2 1 Ln − 0.75 + Fs Ln (27.9) t= 8ch de 1 − Uh where de is the equivalent diameter of the PVD defined in Eq. (27.10), ch is the horizontal coefficient of consolidation, dw is the well influence diameter (taken as 1.05 s for an equilateral triangle spacing pattern and 1.13 s for a square spacing pattern) where s is the spacing between PVDs, and F s is a soil disturbance factor (taken as 2 for highly plastic sensitive soils but taken as zero if ch has been conservatively estimated or accurately measured): 2(a + b) (27.10) de = 𝜋 where a is the PVD thickness and b is the PVD width. 27.3.4

Preloading Using Vacuum

Sometimes the soil is so soft that a surcharge fill cannot be placed to a sufficient height to be useful. In this case, preloading by vacuum is an alternative. The method consists of applying a vacuum, thereby decreasing the water stress, increasing the effective stress, and compressing the soil. A vacuum of 0.8 atmosphere is commonly applied and is equal to about 4 m of

Failure line Deviator stress q′

964

Ko line Surcharge Vacuum

Effective mean stress p′

Figure 27.8 Stress path comparison between surcharge preloading and vacuum preloading. (Source: From Chu et al., 2009/IOS Press.)

soil surcharge. One difference between this method and a fill method is that for vacuum preloading, the increase in effective stress is applied isotropically, as opposed to anisotropically for the fill. Figure 27.8 shows the difference in effective stress path between a surcharge fill and vacuum preloading. The construction sequence consists of constructing a 0.3 m-thick sand blanket on the site, installing prefabricated vertical drains on a square grid (say, 1 m center to center), laying down a grid of geotextile-covered perforated pipes in the sand blanket to connect the PVDs to the vacuum pump, and covering the ground surface with a geomembrane to seal the soil volume. The vacuum pump is turned on and vacuum consolidation takes place. A variant of this process is shown in Figure 27.9.

Atmospheric pressure Impervious membrane

Vacuum gas phase booster

Vacuum air water pump

Air flow

Fill

Water treatment station

Draining layer Horizontal drains Peripheral trenches Isotropic consolidation

Vertical vacuum transmission pipes

Figure 27.9 Menard vacuum consolidation. (Source: Courtesy of Menard, Bridgeville, PA; www.menardusa .com.)

27.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

Electro-osmosis Electromigration

965

Water transport from anode to cathode Ion transport to the opposite electrode Heat

Soil particle

Soil particle Soil particle

Anode

Water velocity profile

Cathode

Soil particle Soil particle

Soil particle

Figure 27.10 Electro-osmosis in clays. (Source: Courtesy of C.J. Athmer-Terran Corporation.)

The vacuum preloading method works well when the soil is soft, low permeability, and relatively homogeneous. If clean sand layers are interbedded in a deposit of soft clay, the efficiency of the process decreases unless cutoff walls can be installed first. Also, because vacuum preloading is isotropic, compression occurs in all direction equally and horizontal shortening takes place. This leads to vertical cracks in the soil mass. 27.3.5

Electro-Osmosis

The electro-osmosis process was discovered in the early 1800s and applied to soils by Leo Cassagrande in the early 1940s. It is based on the fact that when a DC electrical current is established between two electrodes (e.g., steel bars) driven into fine-grained soil, the water flows from the anode (positive charge) to the cathode (negative charge) (Figure 27.10). The reason for this water movement is as follows. Clay particles are negatively charged and as such attract cations (positively charged) such as sodium, calcium and magnesium to their surfaces. When a DC current is established between two metal rods, the cations that line the surface of the clay particles start sliding toward the cathode by electrical attraction. The movement of this boundary layer of cations drags the bulk soil water with it. The water that accumulates at the cathode is drained away and the water content of the clay decreases, with an associated increase in strength and stiffness. 27.3.6

Ground Freezing

The technique of ground freezing (Figure 27.11) consists of freezing the soil by installing a network of steel pipes and circulating either brine water or liquid nitrogen. The temperature of the circulating brine water is typically −20 ∘ C; liquid nitrogen is much colder, at around −200 ∘ C. Brine is much less expensive, but nitrogen takes a lot less time to freeze the

Figure 27.11 Ground freezing. (Source: Courtesy of British Drilling and Freezing Co. Ltd.)

ground. As a result, brine is used for large projects, whereas nitrogen may be economical when time is more important than cost savings. The advantage of ground freezing is that it is applicable to almost all soil conditions as long as the soil is saturated. Recall, however, that when water turns to ice, it expands by 10%. Applications include tunneling, retaining walls, cutoff walls, and contamination remediation. 27.3.7

Hydro-Blasting Compaction

The hydro-blasting compaction technique is particularly well suited to the treatment of collapsible soils. It consists

966

27 SOIL IMPROVEMENT

of wetting the soil to induce collapse and then detonating explosives in sequence to shake the soil into a more compact arrangement.

27.4 27.4.1

SOIL IMPROVEMENT WITH REPLACEMENT Stone Columns Without Geosynthetic Sock

Stone columns, also called aggregate columns (Figure 27.12), are constructed by opening holes in the soil to be improved (say, 1 m diameter) down to a chosen depth (say, 10 m) and backfilling them with aggregates or crushed stones. Opening of the hole in which to place the stones is done by vibration or by jetting. In the vibration technique, a vibrating cylinder is used (see Section 27.2.3) and the stones are placed upon withdrawal and are compacted using the same vibrator. In the jetting technique, the hole is created by a probe inserted to the chosen depth and rotated out of the hole while jetting horizontally to enlarge the hole before the stones are placed. A third technique, called the rammed aggregate pier method, consists of opening a hole with an auger and compacting the stones in the open hole in 0.3 m-thick lifts. In all cases, a stone column is placed in the soil to reinforce it vertically. This column can carry vertical compression load, but very little uplift load and horizontal load. It can also carry shear load, as required for the stabilization of unstable slopes. This latter case is handled as a slope stability problem. The rest of this section deals with the vertical compression capacity and settlement of stone columns. The column can be considered as a large sample of gravel loaded in a manner similar to a triaxial test. Therefore, at failure of the column, the ratio between the vertical effective stress 𝜎1′ and the horizontal effective stress 𝜎3′ is given by: 𝜎1′ = Kp 𝜎3′

Figure 27.12 Stone column construction. (Source: Courtesy of Menard Bachy, Inc.)

The settlement can also be estimated using pressuremeter data. The horizontal relative expansion of the column is considered to be equal to the relative expansion of the pressuremeter for the same horizontal pressure: ΔB ΔR = (27.14) B R where B and ΔB are the initial diameter and increase in diameter of the stone column respectively, and R and ΔR are the radius and increase in radius of the pressuremeter probe at a pressure corresponding to pL divided by a chosen factor of safety against horizontal expansion failure. Therefore, ΔB can be obtained from Eq. (27.14). The volume involved in the barrel-like deformation shown in Figure 27.13 extends to a depth equal to about two times the diameter of the stone column (Hughes and Withers, 1974). Thus, the initial volume involved in the deformation is: Vo = 2B

(27.11)

where K p is the coefficient of passive earth pressure. In this large-scale triaxial test, 𝜎3′ is limited by the maximum horizontal pressure that the soil can resist. This is given by the effective stress limit pressure p′L of the pressuremeter test. The value of p′L can be obtained by performing a drained pressuremeter test (pressure steps lasting until the probe volume stabilizes) and assuming that the water stress uw is equal to the hydrostatic pressure: p′L = pL − uw

(27.12)

Therefore, the drained ultimate load on the stone column is: (27.13) Qu = Kp (pL − uw )A where Qu is the ultimate load on the stone column, pL is the limit pressure from a drained pressuremeter test, uw is the hydrostatic pressure at the PMT testing depth, and A is the cross-sectional area of the stone column. Of course, there is a beneficial effect that increases when the spacing between stone columns decreases; this observation makes Eq. (27.13) conservative.

𝜋B2 4

(27.15)

B σ′1

s

σ′3 =

2B ?

Stone column ΔB

p′L F

Soft clay

Figure 27.13 Expansion of a stone column under load.

27.4 SOIL IMPROVEMENT WITH REPLACEMENT

967

If, during the deformation of the column under load, the volume of stone experiences a volume change ΔV, then the volume V of the deformed column under load will be: V = Vo + ΔV

(27.16)

The deformed volume V is also equal to: 𝜋 V = (2B − s) (B − ΔB)2 (27.17) 4 where s is the settlement of the stone column. This settlement s is then given by: ( ) ⎛ ⎞ 1 + ΔV ⎜ ⎟ Vo s = 2B ⎜1 − ( (27.18) )2 ⎟ ΔB ⎜ ⎟ 1+ B ⎠ ⎝ The relative increase in stone column diameter ΔB/B is obtained from Eq. (27.14) using a ratio ΔR/R from a pressuremeter test at a pressure corresponding to pL divided by a chosen factor of safety against horizontal expansion failure. The relative change in volume ΔV/V can be obtained from a triaxial test on the stone column material. The value of ΔV/V is the one that corresponds to a vertical stress 𝜎1′ applied at the top of the stone column. Therefore, the settlement s corresponds to a top load Q equal to: 𝜋B2 (27.19) 4 If ΔV is 0 and if ΔR/R is small, then Eq. (27.18) reduces to: Q = 𝜎1′

ΔR (27.20) R Another possible mode of failure is sliding along the sides of the column as a pile. The rules of design for piles can be used in this case, assuming that the failure will take place in the soft clay rather than the stone column at the vertical friction interface. Besides strengthening the soft soil, stone columns act as large drains. When the surface is loaded, the water squeezes out of the soil horizontally (because the drainage length is shorter in that direction), drains into the stone column, and is collected at the surface. The design of stone columns as drains follows the same process as for prefabricated vertical drains (see Section 27.3.3). s = 4B

27.4.2 Stone Columns with Geosynthetic Encasement More recently, geosynthetic encasement, in the form of a large sock (Figure 27.14), has been used to increase the horizontal resistance and therefore vertical capacity of the stone column. Because improved horizontal drainage is also an attribute of stone columns, the geosynthetic used is a geotextile that provides a filter between the native soil (often soft clay) and the stone column material. The critical factors for the encasement are the tensile capacity of the geotextile Tu (kN∕m) and its modulus E(kN∕m). The

Figure 27.14 Stone column with geotextile encasement. (Source: Courtesy of HUESKER Inc.)

value of T u ranges from 25 to 60 kN/m and that of E from 30 to 150 kN/m. The modulus E is defined as: T E= (27.21) 𝜀 where T is the force applied per meter of fabric and 𝜀 is the corresponding tensile strain. Note that for geotextiles, the tensile strain at failure 𝜀f is very large, in the range of 25–70%. Therefore, it is likely that the soil would fail before the geotextile sock did. The failure mechanism may involve failure of the column aggregate, failure of the geotextile encasement, or failure of the soil laterally. Because of the large strains required for the geotextile to fail, this failure mechanism is not likely. The ultimate pressure that can be placed at the top of the encased stone column is given by: ( ) (27.22) Soil fails laterally pu1 = kp 𝜎3′ = kp p′L + pgeo Geotextile fails in hoop tension ( ) Δr ′ + pgeo f pu2 = kp 𝜎3 = kp 2G ro

(27.23)

where pu1 is the ultimate pressure that can be placed at the top of the stone column if the soil fails first by reaching the soil effective stress horizontal limit pressure p′L , kp is the coefficient of passive earth pressure of the soil, 𝜎3′ is the horizontal stress generated by the combination of geotextile and soil, pgeo

968

27 SOIL IMPROVEMENT

pL′

The value of kp is obtained from the friction angle of the stone column material, p′L from a pressuremeter test in the natural soil, E from the geotextile material, ro from the size of the stone column, and Δr/ro as 0.41 to correspond with the strain at failure for the limit pressure. Note that the product E Δr/ro cannot be larger than the tensile capacity T u of the geotextile. The settlement calculations become rather cumbersome in close form and are best handled by numerical simulations starting with elasticity. Alexiew et al. (2003) and Raithel et al. (2005) proposed a simplified method for hand calculations.

Natural soil

Stone column

Geotextile p r0 T

Δr T

Figure 27.15 Pressure and tension in the geotextile encasement.

27.4.3 is the pressure contributed by the geotextile when stretched at Δr∕ro , pgeof is the pressure contributed by the geotextile at failure of the geotextile, G is the shear modulus of the soil outside the geotextile (soil being improved), and Δr/ro is the relative increase in radius of the stone column. The expression 2GΔr∕ro is the pressure contributed by the soil outside the geotextile for a radial strain Δr/ro . Because the geotextile strain at failure is often very large, pu1 is likely to control. The pressure pgeo contributed by the geotextile at a relative increase in radius of the stone column equal to Δr∕ro is given by (Figure 27.15): Δr Δr or pgeo = E 2 (27.24) pgeo 2r = 2T = 2E𝜀 = 2E ro ro Then the ultimate load on the encased stone column corresponding to this failure mechanism is: ( ) Δr ′ Qu1 = kp pL + E 2 𝜋ro2 (27.25) ro

Dynamic Replacement

Dynamic replacement (DR) starts by placing a blanket of aggregates on top of the soil to be improved. Then a dynamic compaction (DC) operation is performed, creating craters that are filled with aggregates to form a plug. More pounding takes place on top of the plug at the same locations; the craters deepen and more aggregates are placed in the open hole. The process is repeated until the crater decreases in depth. In this fashion a column of compacted aggregates is formed in place (Figure 27.16). The same range of weight, drop height, and pounder diameter are used for both DC and DR. If the energy used is high (200–400 kJ/m3 ) and the soil is softer (PMT limit pressure 100–400 kPa), then the craters are deep, DR takes place, and the degree of improvement is high. In contrast, if the energy is lower (50–250 kJ/m3) and the soil is stronger (PMT limit pressure 250–700 kPa), then the craters are limited in depth, DC take place, and the degree of improvement is lower.

1 - Create crater by light pounding 2 - Fill crater with granular material to form plug 3 - Continue pounding and penetration of column 4 - Fill crater with granular material, repeat 3 and 4 until DR column reaches design depth 3

1 2

4

Figure 27.16 Dynamic replacement method. (Source: From Chu et al., 2009/IOS Press.)

27.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES

27.5.1

27.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES

Percent finer by weight (%)

Clay

80

Silt

Particulate Grouting

Particulate grouting refers to grouting coarse-grained soils by injecting the grout under gravity or under pressure into the soil voids. It also refers to grouting fissures in rocks and cavities such as sinkholes. Particulate grouting consists of opening a borehole down to the desired depth, sealing it, and then injecting the grout. The spacing between boreholes is in the range of 1–2 m and the hydraulic conductivity of the soil for which this technique is applicable is 10−2 –10−5 m/s. The tube a manchettes (TAM) technique can be used to inject the grout into the soil under pressure. The TAM consists of a casing with holes at regular intervals (say, 0.5 m) covered by rubber sleeves. Two packers inside the TAM casing are inflated, one above the holes and one below; then the grout can be injected through that hole to force the rubber sleeve to lift off and allow the grout to flow into the adjacent soil under pressure. The pumping rate for particulate grouting can vary from 0.1 to 25 m3 /hr under a pressure of 0.5–10 MPa. The groutability of soils is often evaluated through a ratio N of the soil grain size to the grout grain size. For example:

You might have heard the words grout, concrete, cement, and mortar: what are they, and what is the difference? Cement is made of calcium and silicon. If you want to make cement in your kitchen, you mix powdered limestone (calcium carbonate, CaCO3 ) and powdered clay (mostly silica, SiO2 ) and heat it to 1450∘ C,; you will get a hard piece of rock out of the oven. (Note that the oven in your kitchen is very unlikely to be able to reach such high temperatures.) If you then grind that piece of rock into a very fine powder, you will have a crude cement. When you add water to that very dry cement powder, an exothermic reaction (generates heat) called hydration takes place and produces calcium silicate hydrate, which is the main source of cement strength. Cement is the binder in concrete, mortar, and grout. Concrete is the combination of cement, water, sand, gravel, and even larger aggregates. Mortar and grout are combinations of cement, water, and fine sand. The difference between mortar and grout is that typically grout will be more fluid than mortar. Sometimes grout is simply cement and water. Different grouting techniques are used depending on the type of soil to be improved (Warner, 2004). For gravels and coarse sands, the grout is injected by gravity or under pressure and fills the soil voids; the smaller the D50 of the soil, the finer, the more fluid, and the less viscous the grout has to be. These techniques include particulate grouting and chemical grouting. For fine sands and fine-grained soils, the grout is placed in a hole made in the soil to be improved. These techniques include jet grouting, compaction grouting, and compensation grouting. Also, for fine-grained soils, the soil can be mixed with grout that acts as a drilling fluid; this is soil mixing. Figure 27.17 shows the range of applicability of various grouting techniques.

100

N1 =

D10(soil)

or

D65(grout)

N2 =

D10(soil) D95(grout)

N3 =

Sand

D10(soil) D90(grout)

+ k1

w∕c P + k2 FC Dr

Gravel

Boulder

Compensation grouting Compaction grouting

60

20 0

Ultra fine cements Water-glass solution (low viscosity) Economical

Cement suspensions

Uneconomical

0.002 0.006 0.02

0.06

(27.26)

where D10( soil) is the grain size of the soil corresponding to 10% fines, and D65(grout) D90(grout) are the grain size of the grout corresponding to 65 and 90% fines, respectively. Mitchell and Katti (1981) state that grouting is feasible if N1 > 24 and not feasible if N1 < 11. Karol (2003) states that grouting is feasible if N2 > 11 and not feasible if N2 < 6. Groutability also depends on how fluid the grout is and what injection pressure is applied. Akbulut and Saglamer (2002) proposed a more complete expression that reflects the influence of these parameters:

Jet grouting

40

969

0.2

0.6

2

6

20

60

Grain size (mm)

Figure 27.17 Range of application of grouting techniques. (Source: From Keller, 2012; www .kellergrundbau.com/download/pdf/en/Keller_66–01E.pdf.).

(27.27)

970

27 SOIL IMPROVEMENT

where k1 and k2 are soil-specific factors (0.5 and 0.01 for the soil tested by Akbulut and Saglamer), w/c is the water-to-cement ratio of the grout, FC is the fine content, P is the grout pressure in kPa, and Dr is the soil relative density. The soil is considered groutable if N 3 is larger than 26.

100–150 mm. Once the required depth is reached, a horizontal high-pressure jet (∼20MPa) is generated to erode the soil laterally. The rod is withdrawn while rotating (Figure 27.18). This erosion process generates a larger-diameter hole (1–1.5 m) that is then filled with grout.

27.5.2

27.5.4

Chemical Grouting

Chemical grouting makes use of any grout that is a pure solution with no particles in suspension. Because it does not have any solids in suspension, it can penetrate finer soils. Whereas the groutability of soils by particulate grouts depends on the grain size of the solids in the grout, the groutability of chemical grouts depends on their viscosity. Chemical grout can be used in soils as fine as coarse silt. 27.5.3

Jet Grouting

Particulate and chemical grouts permeate the soil and fill the voids with grout. These techniques apply mostly to coarse-grained soils. For fine-grained soils, it is not possible for the grout to penetrate the voids, because they are too small. Instead, the approach consists of creating columns of grout in place. This is done by jet grouting, or compaction grouting, or compensation grouting, or soil mixing. Note that these techniques are also applicable to coarse-grained soils (see Figure 27.17). Jet grouting consists of drilling a borehole down to the desired depth. The drill bit has a diameter in the range of Insertion of injecting tool into a drilled hole

Injection of high velocity cement, slurry, and air

Compaction Grouting

Compaction grouting (Figure 27.19; Al-Alusi, 1997) consists of drilling a hole with a small casing to the depth where grouting is to start, and then injecting very stiff grout with 25 mm slump or less (decrease in height in a standard cone test) under 3–7 MPa pressure while withdrawing the grout casing. The grout injection is performed at discrete locations and forms bulbs of grout that are 0.3–0.6 m thick. The grout does not penetrate the soil voids, but instead displaces and densifies the soil around the bulb. A sudden drop in pressure often indicates soil fracture. The spacing between grouting holes is in the range of 2.5–3.5 m center to center. 27.5.5

Compensation Grouting

Compensation grouting is used to minimize the amount of soil deformation potentially created by excavation and tunneling. It consists of injecting a volume of grout that compensates for the volume of soil displaced so that the adjacent ground surface or buildings do not deflect excessively. The grout can be injected by intrusion grouting, fracture grouting, or compaction grouting. The method is used in many different types Completion of a subsurface superjet column

Figure 27.18 Jet grouting. (Source: After Hayward Baker, Inc.)

27.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES

Compaction grout casing

Drill rod

Casing

971

Casing

Figure 27.19 Compaction grouting. (Source: Courtesy of Arizona Repair Masons Inc.)

of soils, but mostly in fine-grained soils, although difficulties have been encountered in soft clays (Chu et al., 2009). 27.5.6

Mixing Method

The mixing method consists of mixing soil with grout in place. The grout serves as the slurry for the drilling process and the soil-grout mixture creates a strengthened column in situ. The technique is called deep soil mixing (DSM) or deep

cement mixing (DCM) or soil cement mixing (SCM). The drilling tool is usually a paddle auger (Figure 27.20) about 1 m in diameter; several side-by-side augers can be used at one time. Examples of construction with SCM include walls for deep excavation in soft clays, flow barriers, or simply forming a block of strengthened soil mass. The ratio of grout to soil varies from 0.15 to 0.4. Soils usually have compressive strengths less than 200 kPa and concrete more than 20,000 kPa; in SCM, the soil-cement mixtures have

5

1 2 4 3

1

Positioning of auger tool

2

Drilling and mixing soil with cement grout

3

Bottom mixing

4

Withdrawing while continuing soil mixing

5

Complete mixed product column

Figure 27.20 Example of SCM excavation support construction sequence. (Source: Courtesy of JAFEC USA, Inc. Geotechnical Constructors.)

972

27 SOIL IMPROVEMENT

unconfined compressive strength in the 2000 kPa range. The modulus of deformation of SCM varies in the range of 100–1000 MPa and can be estimated by the following equation (Briaud and Rutherford, 2010): ( )0.41 ESoil Cement (kPa) = 12,900 fc′ (kPa) (27.28) 27.5.7

Lime Treatment

If you want to make lime, you take a piece of natural limestone rock (CaCO3 ); heat it to about 1000 ∘ C, which drives the carbon (CO2 ) out of the limestone, and then grind the leftover piece of rock. You will have a white powder called lime or calcium oxide (CaO). If you then mix this white powder with a wet clay, it will hydrate, reabsorb carbon dioxide, and turn back into limestone. The difference between cement and lime is that lime does not strengthen as rapidly as cement; also, it is not as strong and is more brittle than cement. The strengthening of the lime-soil mixture is accompanied by a decrease in water content of the clay, an increase in pH (more alkaline), an increase in plastic limit, a decrease in plasticity index, and a decrease in shrink-swell potential. The lime affects the electrostatic field around the clay particles, which tend to flocculate and assume a more granular structure. The typical amount of lime added to a clay ranges between 2% and 8%. The design of the mix and the impact on the soil properties are given in Little (1999). The clay-lime mixture has unconfined compression strengths between 700 and 1400 kPa and moduli between 200 and 3000 MPa. Lime treatment is often used to stabilize pavement foundation layers, and works best when the soil has at least 25% passing the 75 micron sieve and a plasticity index (PI) of at least 10. In the field, the lime is mixed with the surface soil and hydrated (Figure 27.21). There is one case in which using lime can be very counterproductive: this is the case where the soil to be stabilized contains a certain amount of sulfate in the form of gypsum (CaSO4 2H2 O). The addition of lime (CaO) and water (H2 O) to this type of soil will form ettringite (Ca6 Al2 (SO4 )3 (OH)12

After applying lime slurry to prepared soil, the machine is run in reverse to ensure thorough mixing to the specified depth.

Figure 27.21 Lime stabilization of pavement layers. (Source: Photo by James Cowlin/Asphalt Busters, Phoenix, AZ.)

Figure 27.22 Ettringite crystals. (Source: Courtesy of www .sciencedirect.com/science/article/pii/S0008884698001379.)

26H2 O), which is a highly expansive mineral. Ettringite crystals are needle-like (Figure 27.22) and when mixed with water can swell to 250% of their initial height and destroy pavements. If the total soluble sulfate level is greater than about 0.3% in a 10-to-1 water-to-soil solution, additional precautions to guard against this sulfate reactions, such as swell tests, may be warranted (Little, 1999). 27.5.8

Microbial Methods

Certain naturally occurring bacteria are able to generate material that can either plug the soil voids (bio-plugging) or cement particles together (bio-cementation). Water-insoluble microbial slime is produced by facultative anaerobic and microaerophilic bacteria to plug the soil voids. Bio-plugging can decrease the hydraulic conductivity of the soil by a factor of 2 (Ng et al., 2012). Calcite is produced by ureolytic bacteria that precipitate calcium carbonate. Bio-cementing increases the shear strength of the soil by cementing the particles together. This process is called microbial-induced calcite precipitation (MICP) and works best with sand particles. It takes place when the urease enzyme produced by bacteria such as Bacillus megaterium decomposes urea by hydrolysis and produces ammonium. In turn, ammonium increases the pH and starts the precipitation of calcium carbonate. Calcium carbonate is the glue that cements the soil grains together (Figures 27.23 and 27.24) and can increase the shear strength of the sand by a factor of 2 (Ng et al., 2012).

27.6 SOIL IMPROVEMENT WITH INCLUSIONS

973

27.6.3 Geosynthetic Mat and Column-Supported Embankment Geosynthetic mat and column-supported embankments (GM-CSs) (Figure 27.25) are increasingly being used as a way to rapidly construct or widen embankments on soft soils. The construction proceeds by first constructing the columns to the required depth, and preferably to a strong layer; then covering them with a bridging layer made of interbedded select fill and geosynthetic layers (say, 1 m thick); and then completing the embankment to the design height. The design process proceeds as follows (Smith, 2005; Schaefer and Berg, 2013):

Calcite

Sand particle

Sand particle

50.0 μm

Figure 27.23 Light microscopic image of calcite crystals, produced by ureolytic bacteria, cementing two sand particles. (Source: Courtesy of Salwa Al-Thawadi.)

1. Investigate the site to collect the properties of the natural soil. 2. Choose the depth and spacing of the columns. Identify the repeatable shape in plan view of the group of columns called the unit cell. The depth should be chosen such that the columns reach a strong layer. The spacing s should be smaller than the following values: s ≤ 0.67H + a s ≤ 1.23H − 1.2a

27.6 27.6.1 Earth

SOIL IMPROVEMENT WITH INCLUSIONS Mechanically or Geosynthetically Stabilized

Mechanically stabilized earth (MSE) walls are covered in Chapter 22, Section 22.10. Geosynthetically stabilized earth (GSE) walls are covered in Chapter 26, Section 26.6.2. 27.6.2

Ground Anchors and Soil Nails

Ground anchor walls are covered in Chapter 22. Section 22.12 and soil nail walls are covered in Chapter 22, Section 22.13.

(a)

s ≤ a + 3 meters (27.29) where H is the height of the embankment, a is the side of the individual square cap on top of the column. If there is no cap, a is taken as 0.89 times the diameter of the column (equivalent areas). Center-to-center spacings of between 2 and 5 column diameters are common. The conditions placed on the spacing (Eq. (27.29)) are set to ensure that proper arching will develop in the embankment through the bridging layer to transfer all the embankment weight

(b)

Figure 27.24 Microbial-induced calcite precipitation: (a) Bacteria and calcium chloride. (b) Bricklike product. (Source: Courtesy of Ginger Krieg Dosier, bioMASON Inc.)

974

27 SOIL IMPROVEMENT

Surcharge, q n 1

H = Hb + HEmb

Select fill and geosynthetic reinforcement

General embankment fill

HEmb Hb

dw

Upper sand layer(s), if present Soft soil layer(s)

Columns

Bearing layer s1

s1 s2

s2 s3

s3

Figure 27.25 Geosynthetic reinforced column supported embankment. (Source: Courtesy of Professor Vernon Schaefer, [email protected].)

to the columns. Arching is also ensured by selecting a material for the bridging layer that satisfies a number of criteria (Schaefer and Berg, 2013). 3. Determine the column load Qcol knowing the height of the embankment: Qcol = (𝛾H + q)A

(27.30)

where 𝛾 is the unit weight of the embankment fill, H is the height of the embankment, q is the traffic surcharge, and A is the tributary area of the column or unit cell (Figure 27.25). 4. Design the piles to safely carry Qcol . See Chapter 19, Sections 19.4 and 19.5. 5. Calculate the tension load in the geosynthetic layer. This tension load has two components: the tension load T 1 due to the vertical load transferred from the embankment to the columns through the bridging layer, and the tension load T 2 due to the tendency of the embankment to spread laterally. Filz et al. (2012) recommend the following expression for calculating the value of T 1 : ( ) 𝜎net Asoil 3 6T1 − (6T1 − EGS ) =0 (27.31) p where T 1 is the tension load per unit length of embankment in the geosynthetic due to the embankment vertical load, EGS is the modulus of the geosynthetic

layers (kN/m), 𝜎 net is the difference between the soil pressure above the geosynthetic and below the geosynthetic, Asoil is the area within the unit cell underlain by soil, and p is the column or pile cap perimeter. The tension load T 2 is obtained from: T2 =

1 K 𝛾H 2 + qKa H 2 a

(27.32)

where T 2 is the tension load per unit length of embankment in the geosynthetic due to the embankment tendency to spread laterally, K a is the coefficient of active earth pressure of the embankment soil, H is the embankment height, and q is the traffic surcharge. 6. Select a suitable geosynthetic. Geogrids are most commonly used for this application. The geosynthetic has two strengths that must be checked: the creep-limited strength at 5% strain and the allowable tensile strength (see Chapter 26, Section 26.3.3). 7. Calculate the embedment length Le of the geosynthetic layer: (T1 + T2 )F Le ≥ (27.33) 𝛾H(tan 𝛿1 + tan 𝛿2 ) where F is the factor of safety, 𝛾 is the unit weight of the embankment soil, H is the height of the embankment, and tan 𝛿 1 and tan 𝛿 2 are the coefficients of friction

27.7 SELECTION OF SOIL IMPROVEMENT METHOD

between the geosynthetic and the soil above and below in the bridging layer. 8. Calculate the total settlement s of the embankment, which includes the compression of the embankment soil under its own weight, the compression of the columns under load, and the settlement of the group of columns (see Chapter 19, Section 19.5). If the settlement is excessive, the spacing between columns can be reduced and the columns can be lengthened. 9. The lateral extent of the group of columns should be decided by stability analysis of the embankment slope reinforced by the columns (see Chapter 20, Section 20.14). A factor of safety of 1.3–1.5 is common.

975

27.7 SELECTION OF SOIL IMPROVEMENT METHOD Considering how many different methods exist for soil improvement, it is important to have a tool that can optimize the choice of method for the given situation. The factors to be considered include the soil type; the fine content and size; the soil strength and compressibility; the area and depth of treatment; the proposed structure; the settlement criteria; the availability of skills, equipment, and materials; and the cost of the possible techniques. Sadek and Khoury (2000) proposed a software product called Soil and Site Improvement Guide to optimize the choice. More recently, Schaefer and Berg (2013) optimized the decision process through freeware available at www.geotechtools.org.

Problems and Solutions Problem 27.1 Three soils have the following CPT characteristics. Can they be vibrocompacted? Soil

Point resistance (MPa)

Friction ratio (%)

10 8 15

1.2 0.5 2

Soil 1 Soil 2 Soil 3

Solution 27.1 According to Massarsch guidelines, and Figure 27.2: • Soil 1—Marginal • Soil 2—Yes • Soil 3—No Problem 27.2 An embankment was built as shown in Figure 27.1s. Hsurcharge

Surcharge Fill Sand γ = 20 kN/m3 Cc 1 + eo

= 0.2

Clay

γ = 18 kN/m3 Cv = 1.8 × 10–3cm2/sec.

7m

5m

Dense gravel

Figure 27.1s Highway embankment for preloading problem.

1. What is the maximum settlement of the embankment? 2. How much time is required for 90% of that settlement to occur? 3. How much surcharge is required to get the maximum settlement in 6 months? (Assume that the stress increase in the clay layer is equal to the stress at the bottom of the embankment.)

976

27 SOIL IMPROVEMENT

Solution 27.2 a. Maximum Settlement of the Embankment 𝜎 ′ + Δ𝜎 ′ Cc log ov ′ 1 + eo 𝜎ov ′ 𝜎ov = 2.5 × 18 − 2.5 × 9.81 = 20.48 kPa

Smax = H

Δ𝜎 ′ = 20 × 7 = 140 kPa 20.48 + 140 = 0.89 m ∴ Smax = 5 × 0.2 × log 20.48 b. Time Required for 90% Settlement Tv = Cv

t 2 Hdr

∴ t = Tv

2 Hdr

Cv

when U = 90%, Tv = 0.848 ∴ t = 0.848

2.52 = 2.94 × 107 sec . = 341 days 1.8 × 10−3 × 10−4

c. Surcharge Needed to Reach Maximum Settlement in 6 Months The surcharge needed to reach the maximum settlement in 6 months is calculated as follows: Tv = 1.8 × 10−7 × when Tv = 0.448, U = 73.2% U=

6 × 30 × 24 × 3600 = 0.448 2.52

S6 months (fill+surcharge) Smax (fill+surcharge)

Because we want S6 months (fill + surcharge) = Smax (fill), then: U=

Smax

Smax (fill)

Smax (fill+surcharge) ) ( 𝜎 ′ + Δ𝜎 ′ Cc 0.89 20.48 + Δ𝜎 log ov ′ = 5 × 0.2 × log = Ho (F + S) = 0.732 1 + eo 𝜎 ov 20.48 ( ) 20.48 + Δ𝜎 1.216 = log 20.48 Δ𝜎 = 316.3 kPa = Δ𝜎 (F) + Δ𝜎 (S)

316.26 kPa = 20 × 7 × 20 × h h = 8.81 m Problem 27.3 A highway embankment is to be built on a soft clay layer. What is the spacing of prefabricated vertical drains necessary to obtain 90% consolidation in 12 months? The PVDs are 100 mm wide and 4 mm thick, and are constructed on a square grid. The soil data are shown in Figure 27.2s. Compacted fill γt = 19.6 kN/m3

C L

15 m 6m

1

2

Normally consolidated clay γt = 16.5 kN/m3 Cv = Ch = 0.0093 m2/day

18.3 m

Sand

Figure 27.2s Highway embankment for prefabricated vertical drain problem.

27.7 SELECTION OF SOIL IMPROVEMENT METHOD

977

Solution 27.3 The equation that gives the spacing is: t=

dw2 8ch

(

( Ln

dw de

)

) ( − 0.75 + Fs Ln

1 1 − Uh

)

where de is the equivalent diameter of the PVD defined as 2(a + b)∕𝜋, a and b are the width and thickness of the PVD, ch is the horizontal coefficient of consolidation, dw is the well influence diameter (taken as 1.05 s for an equilateral triangle spacing pattern and 1.13 s for a square spacing pattern) where s is the spacing between PVDs, and F s is a soil disturbance factor (taken as 2 for highly plastic sensitive soils but taken as zero if ch has been conservatively estimated or accurately measured). Using the parameters given in the problem, we get: ( ( ) ) ( ) (1.13s)2 1.13s 1 12 × 30 = Ln − 0.75 + 1 Ln 8 × 0.0093 2(0.1 + 0.004)∕𝜋 1 − 0.9 Note that an average value of Fs = 1 is used in this case: 360 =

1.277s2 (Ln(17.07s) − 1.75) × 2.303 0.0744

9.108 = s2 (Ln(17.07s) − 1.75) This equation is solved by trial and error and gives a center-to-center spacing of s = 2.2 m. Problem 27.4 A docking facility needs to have a 5 m-high fill built on top of a 10 m-thick soft silt layer underlain by dense sand. The groundwater level is at the ground surface. Stone columns 1 m in diameter and 10 m long are built. Long-term drained pressuremeter tests are performed and conservative values of the limit pressure and modulus are 150 kPa and 1700 kPa respectively. The friction angle of the gravel used for the columns is 38∘ . Calculate the load that can be safely carried by one column and the settlement of the top of the column under that load. Assume that no volume change takes place in the column. Solution 27.4 ( ) Qu = Kp pL − uw A Kp =

1 + sin 38 = 4.2 1 − sin 38

Qu = 4.2 × (150 − 5 × 9.8) × 0.52 𝜋 = 333 kN Q 333 Qallowable = u = = 166.6 kN F.S. 2 ΔR s=4B R E = 1700 kPa and v = 0.35 ΔR P 1 + v Pl = = R 2G E F.S. 1.35 150 ΔR = = 0.06 R 1700 2 s = 4 × 0.06 = 0.24 m Problem 27.5 Repeat problem 27.4 but this time the stone columns are encased in a geotextile with a stiffness E equal to 150 kN/m and a tensile strength of 60 kN/m.

978

27 SOIL IMPROVEMENT

Solution 27.5 Figure 27.3s illustrates this problem.

Fill

5m

Soft silt layer

1m

10 m

Stone columns

Dense sand

Figure 27.3s Illustration of the soil profile.

Failure Mechanism 1: Soil Fails Laterally The ultimate pressure the stone column can carry is calculated as: ( ) pu1 = kp p′L + pgeo where kp is the passive earth pressure coefficient of the soil, p′L is the drained pressuremeter limit pressure, and pgeo is the lateral confinement pressure generated by the geotextile at the strain corresponding to failure of the soil. 1 + sin 𝜑′ 1 + sin 38 = = 4.2 1 − sin 𝜑′ 1 − sin 38 p′L = pL − uw = 150 − 5 × 9.81 = 101 kPa kp =

The limit pressure is associated with a radial strain or hoop strain equal to 41%. We use this strain to calculate pgeo . Therefore, the deformation of the geotextile at failure is: Δr = ro × 𝜀 = 0.5 × 0.41 = 0.205 m Δr 0.205 = 123 kPa pgeo = E 2 = 150 × ro 0.52 Hence, the ultimate load the stone column can carry is: ( ) Qu1 = kp p′L + pgeo 𝜋ro 2 = 4.2 × (150 − 5 × 9.81 + 123) × 𝜋 × 0.52 = 739 kN As can be seen, the geotextile encasement more than doubles the ultimate load the stone column can carry. Failure Mechanism 2: Geotextile Fails in Hoop Tension The ultimate pressure the stone column can carry is calculated as: ( ) Δr pu2 = kp 2G + pgeo f ro where kp is the passive earth pressure coefficient of the soil, G is the shear modulus of the soil outside the geotextile, Δr/ro is the relative increase in radius of the stone column, and pgeo f is the confining pressure generated by the geotextile at failure. In this failure mechanism, the tensile strain of the geotextile at failure is calculated as: 𝜀=

60 kN∕m T = = 0.4 E 150 kN∕m

Note that when the hoop strain of the geotextile is 0.4, the soil is approximately at the limit pressure (radial or hoop strain of 0.41), so in this fortuitous case, failure mechanisms 1 and 2 are the same. Therefore, the ultimate load per stone column is 739 kN.

27.7 SELECTION OF SOIL IMPROVEMENT METHOD

979

Problem 27.6 How would you make cement? How do you make lime? What is the difference between cement and lime? Solution 27.6 Cement is made of calcium and silicon. To make cement in your kitchen, you mix powdered limestone (calcium carbonate, CaCO3 ) and powdered clay (mostly silica SiO2 ) and heat it to 1450∘ C; you will get a hard piece of rock. (Note that the oven in your kitchen is very unlikely to be able to generate this high a temperature.) If you then grind that piece of rock into a very fine powder, you will have a crude cement. When you add water to that very dry cement powder, an exothermic reaction (generates heat) called hydration takes places and produces calcium silicate hydrate, which is the main source of cement strength. Cement is the binder in concrete, mortar, and grout. To make lime, take a piece of natural limestone rock (CaCO3 ), heat it to about 1000∘ C to drive the carbon (CO2 ) out of the limestone, and then grind the leftover piece of rock; you will have a white powder called lime or calcium oxide (CaO). If you then mix this white powder with a wet clay, it will hydrate, reabsorb carbon dioxide, and turn back into limestone. The difference between cement and lime is that lime does not strengthen as rapidly as cement. It is weaker and more brittle than cement. Problem 27.7 A soil has a D10 equal to 2 mm. The grout used to strengthen it has a D65 of 60 μm and a D95 of 130 μm. Can particulate grouting be successful? Solution 27.7 N1 =

D10(soil) D65(grout)

or

N2 =

D10(soil) D95(grout)

According to one theory, grouting is feasible if N1 > 24 and not feasible if N1 < 11. According to another theory, grouting is feasible if N2 > 11 and not feasible if N2 < 6. In this problem, grouting is feasible because: D10(soil) 2 mm N1 = = = 33.3 > 24 D65(grout) 60 μm D10(soil) 2 mm = = 15.4 > 11 N2 = D95(grout) 130 μm Problem 27.8 A geosynthetic mat and column-supported embankment (GMCS) is used to build an embankment on soft clay. The embankment is 7 m high, built with a fill with a compacted unit weight of 20 kN/m3 and a friction angle of 32∘ . The columns are 1 m in diameter with no pile cap and are placed on a square 2 m center-to-center grid. The bridging layer is 1 m thick with two layers of geosynthetic. Calculate the load per column, the tension in the geosynthetic layers due to spanning across the columns (T 1 ), and the tension due to lateral spreading (T 2 ). Assume that the net difference in stress on either side of the geosynthetic layer is 80% of the pressure under the embankment and that the geosynthetic has a modulus equal to 60 kN/m. Solution 27.8 The load per column is: Qcol = (𝛾H + q)A Qcol = (20 × 7 × 0) × 2 × 2 = 560 kN The tension T 1 due to the bridging effect between columns is given by: ( ) 𝜎net × Asoil 6T13 − (6T1 − EGS ) =0 p ( ) 0.8 × 20 × 7 × (4 − 𝜋 × 0.52 ) 3 6T1 − (6T1 − 60) =0 𝜋×1 T13 − 114.6(T1 − 10) = 0 which gives a tension T 1 equal to: T1 = −14kN∕m

980

27 SOIL IMPROVEMENT

The tension T 2 due to the lateral spread of the embankment is given by: 1 K 𝛾H 2 + qKa H 2 a 1 − sin 𝜑 Ka = 1 + sin 𝜑 1 − sin 30 0.5 Ka = = = 0.333 1 + sin 30 1.5 1 T2 = × 0.333 × 20 × 72 + 0 = 163 kN∕m 2 So, the lateral spreading effect is much more severe than the bridging effect in this case. T2 =

CHAPTER 28

Technical Communications

28.1

GENERAL

The most important concepts in technical communications are: 1. 2. 3. 4.

Be brief. Be clear. Be technically correct. Be correct from the communication point of view.

“Be brief” is essential, as you may lose your reader or listener if your statement drags on. It is often better to be too short and entice your audience to come back to you for more than to be too long and boring. Of course, in the end, you wish to hit exactly the right length. In technical writing, 10-word sentences are about the best length. In oral communications, you have the choice between the 15-second sound bite, the 2-minute exposé, the 10-minute discourse, and the 1-hour lecture. Think about which one is right for the situation. “Be clear” requires that you put yourself in the shoes of your readers or listeners and aim at the sophistication level that most closely corresponds to their background. If you are unsure, assume a lower level and gradually increase the sophistication of the message. This sophistication level refers to the sophistication of the vocabulary as well as the sophistication of the technical content and thought process. “Be technically correct” is crucial in our field; it requires that any statement made be based on prior work by others or your own work. If not, it is necessary to acknowledge that your statement is based on your intuition or experience. If you use prior work in your statement, you must quote the source and respect intellectual property rights. “Be correct from the communication point of view” requires proper vocabulary, grammar, and diction, including being politically correct. Make sure to proofread your written work. Don’t forget that your listener may not speak your language, so be prepared to speak slowly and exercise patience when you get indications that you have lost contact.

28.2

EMAILS

Email has become a huge part of daily communications, because these messages are very convenient and time-efficient. They include the distribution lines (To, Cc, Bcc), the title, and the body of the message. In the distribution line, make sure that you copy those who truly need to see your message—and no more. The Bcc can be dangerous, as you are obviously hiding something from someone. Remember this golden rule of communication: It is always best to communicate in such a way that if your message were published on the front page of a major newspaper, you would not be embarrassed. Sometimes you will receive an unpleasant message. When you do, please follow this other golden rule: It is best not to answer unpleasant messages right away. In fact, it is often best not to answer them at all. Answering right away with another unpleasant message may give you a few seconds of pleasure, but days of agony later on. Unpleasant messages are best left to simmer for a few days (and it is often disturbing to the sender when such messages remain unanswered). An email signature with your complete title and contact information is important and convenient. It allows your reader to know who you are and gives your contact information in case a phone call is more appropriate as a response. However, if you do not wish to be contacted, or if your reader knows you well, these items are not useful and may convey a message of misplaced egocentric pride. When you write your name in your signature, write your first name in lower case and your family name in capital letters; you may be sending an email to someone in a country where it is not obvious which is your first name and which is your last name. Another problem may be that, in that other country, names are so different that your gender is not obvious. One trick is to answer by saying “Dear Dr. Something”: that way you do not have to decide. By the way, make sure that you include your country as part of your signature contact information.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

981

982 28.3

28 TECHNICAL COMMUNICATIONS

LETTERS

Letters are no longer very common. They are used for extraordinary and more formal occasions. Letter formats vary, but, generally speaking, you will have the letterhead at the top or bottom of the page (or both) with the text of your message in between. The letterhead has the name of your organization and general contact information. Start by indicating the date of the letter, then follow with the name, title, affiliation, and address of the person you are writing to. The greeting line should be Dear Mr. X, Dear Mrs. X, Dear Ms. X (for women not married or if you are not sure of their marital status), Dear Dr. X, Dear Professor X, and so on. If you have a choice of two titles, it is always best to choose the title indicating higher rank. Note that one uses Mr. President but Madame President, not Mrs. President. The text of your letter follows. This text should be two pages or less; letters that must be longer probably should be reports. Letters rarely have attachments unless they are cover letters.

28.4

GEOTECHNICAL REPORTS

A geotechnical design report communicates the site conditions, design, and construction information to the owner of the project, or to the owner’s representative. It is an essential part of the construction process and is used at the design stage, the construction stage, and after construction if there are claims. In any case, it should be clear, concise, and accurate. Although the content of a geotechnical report will vary depending on the type and complexity of the project, a typical report will contain at least the following information: 1. Summary of all site investigation data, including layering, groundwater conditions, and variability based on borings. 2. Geologic interpretation. 3. In situ test results. 4. Laboratory test results. 5. Interpretation and analysis of the subsurface data. 6. Predictive analyses, design. 7. Engineering judgment and recommendations, including solutions for possible problems. 8. Recommended special provisions and limiting conditions. The detailed data usually appears in appendices with all figures, including boring logs, soil profiles, and test results. Remember that when you have only a few borings at some distance from each other, it is unwise to infer the layering in between the borings unless confirmed through the geology of the site, geophysical methods, or other evidence. Though it is tempting to draw a continuous layering graph, it is best to use question marks between borings when suggesting a layering profile.

The report is written to help in the design of the project and so must be helpful to the person who will use it. At the same time, you have to be careful not to make statements that may hurt you or your company in the future. Although detailed calculations are not included in a geotechnical report, it is important to keep all your calculations, as you may have to go back to them later on. This is why it is important, when you make calculations, to clearly document the steps you took, why you assumed some values, how you came to a conclusion, and what published references you used. Once the geotechnical report is ready, have it proofread by a senior and experienced engineer. 28.5

THESES AND DISSERTATIONS

A thesis is usually required for a master’s degree, whereas a dissertation is produced by a candidate for a PhD degree. Both have the same typical organization. 1. Title. The title should reflect as precisely as possible the content of the work—no more, no less. It is better to have a longer, more descriptive title than a short and misleading or vague title. Overall, titles of about 50–75 characters (5–10 words) are best. 2. Cover page. The cover page should include the title, the author’s (or authors’) name(s), and the name of the institution. The date should also appear on the cover page. 3. Dedication. You may wish to dedicate your work to someone who is important to you. 4. Acknowledgments. This is where you thank those who have contributed to the work but are not the author(s). Don’t forget the name of the sponsoring organization and anyone in that organization who helped you in some fashion. 5. Table of contents (TOC). This is the first major step in writing a thesis or dissertation. The more time you spend on the table of contents, the less time you will have to spend writing and iterating. Start with major section and subsection titles. As you do so, think about the natural flow of the work. Then, for each subsection, write notes to yourself in bullet form identifying what you will talk about in each paragraph. The more detailed the TOC, the easier the body of the paper will be to write. A skimpy or poorly organized table of contents leads to many rewrites, frustration, and a feeling of not making progress. 6. Executive summary, abstract, or summary. Summaries or abstracts are a very important part of a thesis or dissertation, as people often do not take the time to read the details of your work. Describe the problem, summarize the important findings of each section (in order), and briefly state the most important conclusions. Usually there are no figures, tables, or photos in this part.

28.5 THESES AND DISSERTATIONS

7. Introduction. The purpose of the introduction is to answer the following questions: what, why, how, where, by whom, and for whom. Once these questions are answered, you can present a narrative outline of the thesis or dissertation. 8. Review of existing knowledge (literature review). It is important to collect and study existing information so that you do not repeat work that has already been well established. It is sometimes good to duplicate important experiments done by others, especially if there is some level of controversy regarding the techniques or findings, but overall this is not a great way to progress. Once you have summarized the existing knowledge, take the time to synthesize that knowledge, give your opinion and point out why your work was necessary or how it built on or extended previous work. 9. Experiments. A dimensional analysis is always a helpful initial step. If the experiment is a small-scale version of the full-scale prototype, scaling laws must be addressed and extrapolation of the results to the full scale explained. Experiments should be reported by first explaining what the purpose of the experiment was, then the design of the experiment, the description of the mechanical and electronic parts, the test procedure, the data acquisition, and the results. If the project included a large number of experiments, a table listing all the experiments should be presented. If there are too many parameters to report for each experiment, a number designation (e.g., T46) should be attributed to each one and the table should give all parameters. If there are too many results or figures to present in the main text, present a few strong examples in the main text and put all the results in an appendix. A summary table should be the first page in the appendix. The analysis of the results can appear here or in a separate section. 10. Numerical simulations. The motivation behind running numerical simulations should be outlined. The mesh size should be discussed first, demonstrating the reason for choosing the distance to the boundaries. The boundary conditions should be explained. The selection of the soil model and of the input parameters should be discussed next. A table summarizing the number of simulation cases helps readers understand the extent of the work and identify which parameters were varied. If the number of simulations is not too large, the results can be presented in the main text. If not, put the results in the appendix that starts with a summary table. The analysis of the results can be done here or in a separate section. 11. Analysis of data. This section makes use of all data accumulated to formulate a solution to the problem posed. Theory, measurements, engineering judgment, logic, and common sense all contribute to making the outcome and results as simple, sound, and useful as possible. It often takes a lot of effort to reach the optimum threshold of simplicity.

983

12. Conclusions. This is where you demonstrate your contribution to new knowledge in a succinct way. It is often convenient to go chapter by chapter and collect your conclusions from each part, arranging them in a consistent framework that shows progress in geotechnical engineering. I am reminded of two comments I received on my research in the early 1980s, one from my father and one from Geoff Meyerhof. My father, after patiently listening to my research work, looked straight at me and said: “So what?!” Meyerhof, after reading my early work on laterally loaded piles, said: “Too complicated!” So, while you have to deal with great complexity to solve the problem, in the end your goal should be to develop something “useful and simple.” 13. References. The purpose of a reference is to acknowledge the work of others and support your statement. Remember that in engineering, when you make a statement, you must have proof (experiment, theory, simulation, reference) or at least a factual basis for that statement; you may need to say that your statement is based on your experience or common sense or engineering judgment. The best way to quote a reference in the text is according to your institution’s mandated or preferred system; most use the author-date system, In this system, you use the last name of the first author, followed by the last name of the second author if there are only two authors, or by “et al.” if there are more than two authors, and followed by the year of publication. In the reference list, the full citation information for each source is given, organized in such a way that readers can easily track down and obtain the referenced publication. A typical presentation is: • Last name and initials of all authors, year of publication (in parentheses); title of paper, report, or book chapter; title of periodical, proceedings, or book (usually in italics); volume number and issue number; the publisher’s location and the name of the publisher; inclusive page numbers. If the reference is a website address (URL), the reference is organized as follows: • Author if any is credited, copyright or posting date, title, the address/URL of the website from which the piece was retrieved, and the date the material was accessed or downloaded. • If the reference is a CD, the reference citation is organized as follows: – Author(s), copyright date, title, medium, and producer/publisher and publisher’s location. 14. Appendices. The bulk of your data should appear in an appendix; you may need to use more than one. The front page of each appendix should explain what is in that appendix. This is where a summary table of tests or simulations becomes most useful.

984

28 TECHNICAL COMMUNICATIONS

28.6

VISUAL AIDS FOR REPORTS

Visual aids for reports may include figures, tables, and photographs. Figures showing graphs of data should have the two axes labeled with the spelled-out name of the variable, the letter symbol, and the unit in parentheses. The axes should have scales with about 20 tick marks and 4 or 5 numbers on the scale. The data points should be clearly identified. If you have more than one set of data points, use different symbols. Although color graphs are more appealing and easier to read, remember that a color graph created electronically may end up printed in black and white. In this case, if you have used the same symbols but different colors, the data points will be indistinguishable between sets. If a set of data points leads to a recommended design line, leave the data points with the recommended line; this will help the user gage the extent of the scatter and select a different value than your recommendation if necessary. If a regression line is drawn, indicate the equation of the line and the value of the coefficient of regression R2 . The size of the letters or numbers in a graph should be such that the graph can be easily read; the minimum size to ensure easy readability is about 1/20 of the size of the graph. Tables should have the name, symbol, and unit of the parameters at the top of each column or the beginning of each row. The caption of a table precedes the body of the table, whereas the caption of a figure or photo goes below the artwork (don’t ask me why!). For best results, photos must be sharp and high resolution. The rules about using visual aids that are not yours, or that are yours but that you have signed over to a publisher by signing a copyright agreement, vary from one source to another. For noncommercial purposes, the general rule is that the source of each visual aid must be acknowledged unless it is your own original work. The acknowledgment may be made simply by placing in the caption the name of the author and date of the publication where the visual aid was found (essentially, giving the reference citation). For commercial purposes, written permission must be obtained from the publisher of the visual aid, and that permission or credit line must be mentioned along with the acknowledgment of the source. Student work is noncommercial, but it is essential to get into the good habit of acknowledging any intellectual property you use that is not your own. People always appreciate when they are recognized and get upset when they are not. 28.7

PHONE CALLS

In all cases, one should prepare for a phone call—if nothing else, to anticipate questions and minimize cost. Know what you wish to achieve and have a plan on how to maneuver if the conversation goes in a different direction than you anticipated. Email or phone call?: that is the question! Most of the time, emails are very efficient, but there are some situations in which they are very dangerous, misleading, and inefficient. It is amazing to see how many different ways a given email may be interpreted by different people. There is a big difference between the written word and the spoken word. For example,

if an interaction might be contentious, it is best to pick up the phone. People tend to understand much better when spoken to than when written to. Reading an email can lead to a serious misunderstanding and an escalating response; it is often much easier to diffuse a misunderstanding on the telephone. There are also times when you simply have to be courageous enough to call the person rather than hiding behind an impersonal and cold email. Most people appreciate being told unpleasant truths “in person” in a telephone call rather than reading them in email. One might argue that some things must be in writing, and that is true. However, the best approach in those cases is to talk on the telephone and explain that the conversation will be followed by an email to restate and formally memorialize the points covered in the conversation.

28.8

MEETINGS

Three of the most important rules for efficient meetings are: 1. Do not interrupt anyone. 2. Be brief. 3. Be professional in your attitude toward your colleagues. Interrupting people when they speak is rude, but they have to respect your right to contribute as well by being brief. From time to time, someone may get under your skin, but it is important to remain calm under fire and concentrate on facts, data, logic, analysis, and reasoning to win your arguments rather than shouting or attacking someone personally. Accept that sometimes your point of view is not the view of the majority and that you are only a member of the team. In many situations, it is important to have the courage to change the things you can change, accept those that you cannot change, and have the wisdom to know the difference. If you are a participant in a meeting, speak up only when you really have something important to say—something that will advance the process. If you are presiding at the meeting, keep in mind the time allotted for each item on the agenda, have a plan if a discussion drags on for good reasons, cut off any unnecessary chat, and help the group stay focused on the topic by repeating during the discussion the original problem to be solved or question to be answered. Also, as the leader of the meeting, start by establishing some initial rules about cell phone use, side chats, and texting or answering emails during the meeting. All are distracting (and discourteous) and should not be allowed during a meeting. Motions and votes are very valuable because the decision becomes extremely clear. It takes place in the following sequence: 1. A motion is proposed by someone. 2. The motion is seconded by a second person. If not seconded, the motion dies.

28.10 MEDIA INTERACTION

985

3. Once the motion is seconded, a discussion period takes place during which you try to convince your colleagues that they should vote in a particular way. 4. The person presiding calls for the vote when the discussion is over and then the votes are recorded. This can be done by show of hands, voice call, or written ballots. The choices are yes, no, or abstain. For delicate matters, the vote may be secret, depending on the rules of the organization.

9. Keep an eye on your audience to see if you are getting blank stares or interested looks. Adjust your talk accordingly. 10. If there is no podium to lean on, you may find yourself on an open stage not knowing what to do with your hands and being self-conscious. A good trick in this situation is to grab a pen or a pointer. Both hands will naturally join to hold it, and you will not think about that any more.

Although motions may seem cumbersome at times, they are very useful in case of arguments after the fact. Remember that after any motion passes, there is usually a need for an action item: who will do what to implement the decision. Any action items or assignments should be included in the recorded minutes of the meeting. For further help in running meetings efficiently, consult Roberts’ Rules of Order.

PowerPoint presentations are subject to a fair amount of personal taste with regard to color, background, animations, and so on. However, there are some fundamental rules:

28.9 PRESENTATIONS AND POWERPOINT SLIDES If you are going to give an important lecture in front of many people, make sure that nothing is left to chance. The best presentations can be ruined if some technology fails to work. Here is a helpful checklist for presentation success: 1. Hook up your laptop to the local projector and check that it works properly. If you have to use someone else’s laptop, make sure you know how to use the basic functions on that laptop. 2. Check all your slides to make sure that they are exactly what you expect (equations are not changed, videos are working, and so on). 3. Bring a pointer or find out if one is available to borrow. 4. Know who will advance the slides. If you are not doing this yourself, what will be the signal to advance the slides? Constant use of a “next slide” request is not elegant; a sign of some sort between you and the projectionist is best, including when to start any videos or animations. 5. Keep an eye on time and pace yourself. It is best to practice the full presentation ahead of time and under “field” conditions to measure the time required. 6. The average time per slide is one minute; however, slides with only photos will go faster and slides with sample calculations will go slower. 7. Have a back-up plan if something fails to work. Can you project your voice without the microphone? Can you complete the presentation without slides, for example? Develop the talent of not requiring slides to guide your thoughts. 8. Have a special title slide and final slide that set your desired tone and reflect your personality.

1. Do not put too much information on the slide. Four bullets, or one graph (possibly two graphs) with explanation, or a couple of photos should be the maximum. 2. Graphs speak well to an engineering audience. These graphs follow the same rules as the figures of a report, thesis, or dissertation (see Section 28.6). With PowerPoint, the lettering should be even larger than for figures in a printed report. 3. An audience cannot absorb tables with more than 10 numbers in them at the normal rate of presentation. Generally speaking, tables are not a good way to convey an idea or a result in PowerPoint. 4. Equations may be necessary, but should be limited in length and complexity unless the audience is well versed in that aspect of the work. 5. The use of videos is entertaining and holds the audience’s attention. If you intend to use videos, make sure that they work and double-check them right before your presentation.

28.10

MEDIA INTERACTION

The media has essentially three forms: (1) the written press; (2) the audio press; and (3) the video press. In all cases, the most likely interaction will be an interview, although a written communication may be involved as well. This written part may be a press release or a letter to the editor sent to newspapers. In this case, you will have time to prepare and proofread your statements. Interviews for the written press are conducted in an informal setting, often by telephone, and are less stressful than audio and video interviews. For the written press, note that saying “off the record” is best avoided, as you have no insurance that your request will be honored. Always only say what you do not mind seeing in print. For the video press, you can have either a taped interview (that may be edited) or a live interview. While the possibility of having your statements edited may give you some level of confidence against mistakes, you should not behave dif-

986

28 TECHNICAL COMMUNICATIONS

ferently. Remember also that you most likely will not have an opportunity to edit your statements. Video editing is very time-consuming and not as easy as text editing. In preparing for the interview, take the time to review your notes and check your appearance in a mirror. Before the interview begins, ask the reporter about the line of questioning, including typical questions, and make sure the reporter knows how to pronounce your name and affiliation correctly. Find a way to be comfortable in front of the camera; the best way to do that is to ignore it completely. Just talk to the reporter as if you were chatting with him or her at the kitchen table. Overall, speak your mind, but do not say anything that you are not very sure of. Live TV interviews are an exercise in fast thinking and right thinking. Remember that the 15-second sound bite dominates the TV market. If you are uncomfortable with a question, find a way to answer by talking about what you really wish to talk about. To minimize errors in your answer and give yourself time to think, take a second before answering to look in the distance or at the ceiling; then start with the obvious while formulating the rest of your answer in your mind.

28.12

28.11

28.13

ETHICAL BEHAVIOR

In the end, you have to answer to yourself and the dictates of your conscience. In any decision process, you are always free to choose what is right for you. Regardless of your decision, you will also have to face the consequences of that decision. You may get by with a few lucky ones, but you may also find yourself implicated in undeserved conflicts. There are close to 9 billion people on our planet and each one thinks differently—yet everyone thinks that they are right. This often makes smooth interactions very difficult. Nevertheless, there are reasonable guidelines governing ethical behavior. As an engineer making a decision, remember the following: • As engineers, we must uphold, as the highest priority, the safety of the general public within reasonable economic constraints. • If you are unsure about something, get advice from people whom you respect and who have a proven track record. • If at all possible, do not rush the decision. • In the process of deciding, reverse the roles; put yourself in the other person’s shoes and treat people the way you would like to be treated. Whatever you decide after reasonable thought, remember that you have done your best and you should not feel badly about it. If the outcome is unpleasant, do not quit: Keep fighting for what you think is right until it becomes right or you run out of energy. In any case, worrying and stressing are useless (and actually harmful to your health)—but that does not mean that you should take everything lightly. So, don’t stress and don’t worry; just prepare, plan, and concentrate. Easy to say but hard to do!

PROFESSIONAL SOCIETIES

In your life, you have two families: your blood family and your professional family. It is important to support your professional family by belonging to your professional organization. In the United States, it is the Geo-Institute (http:// www.asce.org/geo-institute). For the world scene it is the International Society for Soil Mechanics and Geotechnical Engineering (http://www.issmge.org/). By belonging to and being active in your professional society, you will participate in the work of technical committees, contribute to national decisions, and more generally strengthen and advance the practice of geotechnical engineering. In your work as a volunteer, you will be interacting and socializing with your peers; you will learn from them and you will teach them. You will also improve your technical communication skills, as you will naturally find yourself engaging in various types of communication. Being a member of your professional society ranks at the level of a family obligation; you should ask how you can help your professional society rather than ask what it can do for you. RULES FOR A SUCCESSFUL CAREER

A successful career is built on a series of demonstrated successes by an individual, either alone or as part of a team. In the performance of your job, remember when you make a decision of any sort that it will take 10 successes to erase 1 mistake from the minds of your peers. This is why it is always important to concentrate and plan. Also, before a challenging moment, remember that you may have been through similar tough moments before and done well; this recollection will give you added confidence and lower the stress. The following “Top 10 Rules” are some thoughts on what is important in a career. They have been inspired by discussion with many engineers over time, including Clyde Baker, and personal experiences as well: 1. Choose the relentless pursuit of excellence as a way of life. 2. Be curious. The discovery process is a fountain of youth. 3. Work hard but balance your interests (fun, family, sport, art, world news). 4. Make lots of friends. Nurture your public relations. 5. Look for solutions and not who is to blame. Leave that to the judge. 6. Be firm in your decisions, but always be fair and polite. 7. Treat others as you wish to be treated, and you will lead by example. 8. Communication is the best way to solve problems. Convince through logic and data. 9. Surround yourself with smart people and positive role models. 10. Pursue your dreams with vision and perseverance.

REFERENCES Aas, G., S. Lacasse, T. Lunne, and K. Hoeg. 1986. “Use of In Situ Tests for Foundation Design on Clay.” In: Proceedings of In Situ ’86, Use of In Situ Tests in Geotechnical Engineering, Blacksburg, VA: ASCE GSP 6, pp. 1–30. AASHTO (American Association of State Highway and Transportation Officials). 2007. “SI Units.” In: LRFD Bridge Design Specifications (5th ed.). Philadelphia, PA: ASTM. Washington, DC: Transportation Research Board, Academy of Sciences. Abdelmalak, R. 2007. “Soil Structure Interaction for Shrink-Swell Soils: A New Design Procedure for Foundation Slabs on Shrink-Swell Soils.” PhD dissertation, Department of Civil Engineering, Texas A&M University. Abdelmalak R. 2018. “Performance of Post-Tensioned Foundation Slab on Shrink-Swell Soils: A Case Study of the Briaud Tennis Court Slab, College Station, Texas.” In: Innovations in Geotechnical Engineering: A Symposium in Honor of Professor Jean-Louis Briaud (Ed. X. Zhang, P. Cosentino, and M. Hussein). ASCE Geo-Institute. Abramson, L.W., T.S. Lee, S. Sharma, and G.M. Boyce. 1996. Slope Stability and Stabilization Methods (1st ed.). New York: John Wiley & Sons. Abramson, L.W., T.S. Lee, S. Sharma, and G.M. Boyce. 2002. Slope Stability and Stabilization Methods. New York: John Wiley and Sons. ADSC-DFI. 2004. Drilled Shaft Inspector’s Manual (2nd ed.). Dallas, TX: The International Association of Foundation Drilling. Akbulut, S. and A. Saglamer. 2002. “Estimating the Groutability of Granular Soils: A New Approach.” Tunneling and Underground Space Technology 17(4): 371–380. Akrouch, G.A., J-L. Briaud, M. Sanchez, and R. Yilmaz. 2016. “Thermal Cone Test to Determine Soil Thermal Properties.” Journal of Geotechnical and Geoenvironmental Engineering, 142(3). Al-Alusi, H.R. 1997. “Compaction Grouting: From Practice to Theory.” Geo-Logan 1997, GSP-66: Grouting: Compaction, Remediation, and Testing July: 43–54. Alexiew, D., G.J. Horgan, and D. Brokemper. 2003. “Geotextile Encased Columns (GEC): Load Capacity and Geotextile Selection.” In: BGA International Conference on Foundations: Innovations, Observations, Design and Practice, pp. 81–90.

Allaire, A. 1916. “The Failure and Righting of a Million-Bushel Grain Elevator.” Paper no. 1363. Transactions of the American Society of Civil Engineers. Alonso, E.E., A. Gens, and A. Josa. 1990. “A Constitutive Model for Partially Saturated Soils.” Géotechnique 40(3): 405–430. Al-Thawadi, S.M. 2008. “High Strength In-Situ Biocementation of Soil by Calcite Precipitating Locally Isolated Ureolytic Bacteria.” PhD dissertation, Murdoch University, Western Australia. Anderegg, R. 1997. “Nichtlineare Schwingungen bei dynamischen Bodenverdichtern.” Dissertation 12419, Eidgenössische Technische Hochschule Zürich, Switzerland. Andersen, K.H., H.P. Jostad, and R. Dyvik. 2008. “Penetration Resistance of Offshore Skirted Foundations and Anchors in Dense Sand.” Journal of Geotechnical & Geoenvironmental Engineering 134(1): 106–116. Andrus, R.D., and K.H. Stokoe. 2000. “Liquefaction Resistance of Soils from Shear-Wave Velocity.” ASCE Journal of Geotechnical & Environmental Engineering 126(11): 1015–1022. Annandale, G.W. 1995. “Erodibility.” Journal of Hydraulic Research 33(4): 471–494. API (American Petroleum Institute). 2000. Recommended Practice for Planning, Designing and Constructing Fixed Offshore Platforms: Working Stresses Design (Recommended Practice 2A). Washington, DC: API. API (American Petroleum Institute). 2012. Design and Analysis of Station Keeping Systems for Floating Structures (Recommended Practice 2SK). Washington, DC: API. ARA-ERES. 2000. “Appendix DD-1: Resilient Modulus as Function of Soil Moisture—Summary of Predictive Models.” In: Guide for Mechanistic-Empirical Design of New and Rehabilitated Pavement Structures (NCHRP Report on Project 1-37a). Washington, DC: Transportation Research Board. ASCE (American Society of Civil Engineers). 2011. Minimum Design Loads for Buildings and Other Structures. Washington, DC: ASCE. ASTM (American Society for Testing and Materials). 1995. Standard Guide for Risk-Based Corrective Action Applied at Petroleum Release Sites (E1739). West Conshohoken, PA: ASTM.

Geotechnical Engineering: Unsaturated and Saturated Soils, Second Edition. Jean-Louis Briaud. © 2023 John Wiley & Sons, Inc. Published 2023 by John Wiley & Sons, Inc.

987

988

REFERENCES

ASTM (American Society for Testing and Materials). 2004a. “ASTM D4318 – Standard Test Methods for Liquid Limit, Plastic Limit, and Plasticity Index of Soils.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2004b. “ASTM D4943 – Standard Test Method for Shrinkage Factors of Soils by the Wax Method.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2005a. “ASTM D1586 – Standard Test Method for Standard Penetration Test (SPT) and Split-Barrel Sampling of Soils.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2005b. “ASTM D2937 – Standard Test Method for Density of Soil in Place by the Drive-Cylinder Method.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2005c. “ASTM D4959 – Standard Test Method for Determination of Water (Moisture) Content of Soil by Direct Heating.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2005d. “ASTM D854 – Standard Test Methods for Specific Gravity of Soil Solids by Water Pycnometer.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2006a. “ASTM D4253 – Standard Test Methods for Maximum Index Density and Unit Weight of Soils Using a Vibratory Table.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2006b. “ASTM D4254 – Standard Test Methods for Minimum Index Density and Unit Weight of Soils and Calculation of Relative Density.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2007. “ASTM D422 – Standard Test Method for Particle-Size Analysis of Soils.” Philadelphia, PA: ASTM. ASTM (American Society for Testing and Materials). 2013. “Section 04 Construction.” In: Annual Book of ASTM Standards. Philadelphia, PA: ASTM. Athanasopoulos, C.A. 2007. “Reducing the Seismic Earth Pressures on Retaining Walls by EPS Geofoam Buffers: Numerical Parametric Analyses.” In: Proceedings of Geosynthetics 2007. Washington, DC: IFAI Publications. Aubeny, C.P., and R.L. Lytton. 2004. “Shallow Slides in High Plasticity Clay Slopes.” ASCE Journal of Geotechnical & Environmental Engineering 130(7): 717–727. Austin, D.N., K.J. Wu, and D.F. White. 1993. “The Influence of Test Parameters and Procedures on the Tensile Modulus of Stiff Geogrids.” In: Geosynthetic Soil Reinforcement Testing Procedures (ASTM STP 1190). Philadelphia, PA: ASTM. Australian Standard. 1996. Residential Slabs and Footings (AS 2870). Sydney: Australia: Standard House. Averjanov, S.F. 1950. “About Permeability of Subsurface Soils in Case of Incomplete Saturation.” Engineering College 7. Baecher, G.B., and J.T. Christian. 2003. Reliability and Statistics in Geotechnical Engineering. New York: John Wiley & Sons, Inc. Baguelin, F., R. Frank, and J-F. Jezequel. 1982. “Parameters for Friction Piles in Marine Soils.” Paper presented at Second International Conference in Numerical Methods in Offshore Piling, Department of Civil Engineering, University of Texas, Austin, TX. Baguelin, F., J-F. Jezequel, and D.H. Shields. 1978. The Pressuremeter and Foundation Engineering. Clausthal-Zellerfeld, West Germany: Trans Tech Publications.

Bahmani, M.S., and J-L. Briaud. 2019. “Prediction of Settlement for Shallow Foundations on Sand.” CERGEP Report No.2 – US units, Civil Engineering, Texas A&M University. Bahmani, M.S., and J-L. Briaud. 2020. “Settlement of Shallow Foundations on Sand – A Database Study.” International Journal of Geoengineering Case Histories, 6(2): 1–17, doi: 10.4417/IJGCH-06-02-01. Bahmani, M.S., and J-L. Briaud. 2021.“Settlement of Shallow Foundations on Clay—A Database Study.” In: Proceedings of the Conference IFCEE 2021 and Geotechnical Special Publication 326 of the ASCE Geo-Institute. Baker, C.N., E.E. Drumright, J-L. Briaud, F.D. Mensah, and G. Parikh. 1993. Drilled Shafts for Bridge Foundations (Publication No. FHWA-RD-92-004). Washington, DC: Federal Highway Administration. Baker, W.F., C. Brown, J.J. Pawlikowski, and D.S. Rankin. 2013. “Tall Buildings and Their Foundations: Three Examples.” Paper presented at International Conference on Case Histories in Geotechnical Engineering. 5, Missouri University of Science and Technology. Baligh, M.M., V. Vivatrat, and C.C. Ladd. 1980. “Cone Penetration in Soil Profiling.” Journal of the Geotechnical Engineering Division 106(4): 447–461. Banks, D.C., W.E. Strohm, M. De Angulo, and R.J. Lutton. 1975. “Study of Clay Shale Slopes along the Panama Canal: Engineering Analyses of Slides and Strength Properties of Clay Shales along the Gaillard Cut.” Technical Report S-70-9. Vicksburg, MO: U.S. Army Engineer Waterways Experiment Station, Baracos, A. 1957. “The Foundation Failure of the Transcona Grain Elevator.” Engineering Journal 40(7): 973–977. Barker, R.M., J.M. Duncan, K.B. Rojiani, P.S.K. Ooi, C.K. Tan, and S.G. Kim. 1991. Manuals for the Design of Bridge Foundations (NCHRP Report 343). Washington, DC: Transportation Research Board, National Research Council. Barron, R.A. 1948. “Consolidation of Fine-Grained Soils by Drain Wells.” (Paper 2346), Transactions of the ASCE 113: 718–724. Bath, M. 1966. “Earthquake Energy and Magnitude.” In: Physics and Chemistry of the Earth, vol. 7 (Ed. L.H. Ahrens, F. Press, S.K. Runcorn, and H.C. Urey). New York, Pergamon Press, pp. 117–165. Bell, F.G. 2000. Engineering Properties of Soils and Rocks (4th ed.), Malden, MA: Blackwell Science Inc. Bell, F.G. 2007. Engineering Geology (2nd ed.). Burlington, UK: Elsevier. Bermingham, P., and M. Janes. 1989. “An Innovative Approach to Load Testing of High Capacity Piles.” In: Proceedings of the International Conference on Piling and Deep Foundations, vol. 1. Rotterdam: A.A. Balkema, pp. 409–413. Bernhardt, M., J-L. Briaud, D. Kim, M. Leclair, R. Storesund, S.G. Lim, R.G. Bea, and J.D. Rogers. 2011. “Mississippi River Levee Failures: June 2008 Flood.” International Journal of Geoengineering Case Histories 2(3): 127–162. doi: 10.4417/IJGCH-02-02-03. Bieniawski, Z.T. 1989. Engineering Rock Mass Classifications. New York: John Wiley & Sons. Bishop, A.W. 1955. “The Use of the Slip Circle in the Stability Analysis of Slopes.” Géotechnique 5(1): 7–17.

REFERENCES

Bishop, A.W., and D.J. Henkel. 1962. The Measurement of Soil Properties in the Triaxial Test (2nd ed.). London: Edward Arnold. Bjerrum, L. 1963. “Allowable Settlement of Structures.” In: Proceedings of the European Conference on Soil Mechanics Foundation Engineering. Wiesbaden, Germany, vol. 3, pp. 135–137. Bjerrum, L. 1972. “Embankments on Soft Ground.” In: Proceedings of the ASCE Conference on Performance of Earth-Supported Structures, Purdue University. vol. 2, pp. 1–54. Bjerrum, L., and A. Landva. 1966. “Direct Simple-Shear Test on a Norwegian Quick Clay.” Géotechnique 16(1): 1–20. Blatz, J., and K. Skaftfeld. 2003. “The Transcona Grain Elevator Failure: A Modern Perspective 90 Years Later.” In: Proceedings, 56th Canadian Geotechnical Conference, 4th Joint IAH-CNC/CGS Conference, Winnipeg, MB (29 Sept.–1 Oct. 2003), pp. 8-22–8-29. Blight, G.E. 1971. “Flow of Air through Soils.” ASCE Journal of the Soil Mechanics and Foundations Division 97(Sm4): 607–624. Bobet, A. 2010. “Numerical Methods in Geomechanics.” Arabian Journal for Science & Engineering 35(1B 2). Bolduc, L.C., P. Gardoni, and J-L. Briaud. 2008.“Probability of Exceedance Estimates for Scour Depth around Bridge Piers.” Journal of Geotechnical & Geoenvironmental Engineering 134(2). Bollaert, E. 2002. “Transient Water Pressures in Joints and Formation of Rock Scour Due to High-Velocity Jet Impact” (Communication No. 13). Lausanne, Switzerland: Laboratory of Hydraulic Constructions, École Polytechnique Fédérale de Lausanne. Bonaparte, R., D.E. Daniel, and R.M. Koerner. 2002. “Assessment and Recommendations for Improving the Performance of Waste Containment Systems” (EPA/600/R-02/099). Washington, DC: U.S. Environmental Protection Agency. Boore, D.M., W.B. Joyner, and T.E. Fumal. 1997. “Equations for Estimating Horizontal Response Spectra and Peak Acceleration from Western North American Earthquakes: A Summary of Recent Work.” Seismic Research Letters 68: 128–153. Bosshard, P. 2009. “Lessons from China’s Three Gorges Dam,” The Asia-Pacific Journal/Japan Focus, 7(48), Number 2, article ID 3262. Boutwell, G.P., and R.K. Derick. 1986. Groundwater Protection for Sanitary Landfills in the Saturated Zone. Paper presented at Waste Tech ’86, National Solid Waste Management Association, Chicago, IL. Bowders, J.J., and D.E. Daniel. 1987. “Hydraulic Conductivity of Compacted Clay to Dilute Organic Chemicals.” Journal of Geotechnical Engineering 113(12): 1432–1448. Bowles, J.E. 1996. Foundation Analysis and Design. New York: McGraw-Hill. Brandimarte, L., A. Montanari, J-L. Briaud, and P. D’Odorico. 2006. “Stochastic Flow Analysis for Predicting Scour of Cohesive Soils.” Journal of Hydraulic Engineering 132(5). Bray, J.D., R.B. Sancio, L.F. Youd, C. Christensen, O. Cetin, A. Onalp, T. Durgunoglu, J. P. C. Stewart, R.B. Seed, M.B. Baturay, T. Karadayilar, and C. Oge. 2001. “Documenting Incidents of Ground Failure Resulting from the August 17, 1999, Kocaeli, Turkey Earthquake.” Available at: Pacific Earthquake

989

Engineering Research Center, http://peer.berkeley.edu/turkey/ adapazari/index.html. Brebbia, C.A., J.C.F. Telles, and L.C. Wrobel. 1984. Boundary Element Techniques. Berlin: Springer-Verlag. Briaud, J.-L. 1992. The Pressuremeter. London: Taylor & Francis. Briaud, J.-L. 1997a. “SALLOP: Simple Approach for Lateral Loads on Piles.” Journal of Geotechnical & Geoenvironmental Engineering 123(10): 958–964. Briaud, J.-L. 1997b. “National Geotechnical Experimentation Sites at Texas A&M University, Clay and Sand: A Summary.” Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Briaud, J.-L. 2006a. “Bridge Scour.” Geotechnical News 24(3), September. Briaud, J.-L. 2006b. “Erosion Tests on New Orleans Levee Samples” (Internal Report). Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Briaud, J.-L. 2007. “Spread Footings in Sand: Load Settlement Curve Approach.” Journal of Geotechnical & Geoenvironmental Engineering 133(8), August. Briaud, J.-L. 2008. “Case Histories in Soil and Rock Erosion: Woodrow Wilson Bridge, Brazos River Meander, Normandy Cliffs, and New Orleans Levees” (The 9th Ralph B. Peck Lecture). Journal of Geotechnical & Geoenvironmental Engineering 134(10). Briaud, J.-L. 2012. “Summary Report.” Available at: http://ceprofs .tamu.edu/briaud/ accessed 31 March 2012. Briaud J.-L. 2013. “The Pressuremeter Test: Expanding Its Use.” 2013 ISSMGE Louis Menard Lecture. In: Proceedings of the 18th International Conference on Soil Mechanics and Geotechnical Engineering, Paris. Briaud J.-L. 2015a. “Scour Depth at Bridges: Method Including Soil Properties. I: Maximum Scour Depth Prediction.” Journal of Geotechnical and Geoenvironmental Engineering 141(2). Briaud J.-L. 2015b. “Scour Depth at Bridges: Method Including Soil Properties. II: Time Rate of Scour Prediction.” Journal of Geotechnical and Geoenvironmental Engineering 141(2). Briaud J.-L. 2018. “Normandy Cliff Stability: Analysis and Repair.” In: Proceedings of ISRM Conference, Geomechanics and Geodynamics of Rock Masses. St Petersburg, Russia (Ed. V. Litvinientko). Briaud, J.-L. 2021a. “40 Years of Full-Scale Infrastructure Testing at a National Site: Clay Site.” International Journal of Geoengineering Case Histories, 6(3): 1–24. doi: 10.4417/IJGCH-06-03-01. Briaud, J.-L. 2021b. “40 Years of Full-Scale Infrastructure Testing at a National Site: Sand Site.” International Journal of Geoengineering Case Histories, 6(3): 25–53. doi: 10.4417/IJGCH-06-03-02. Briaud, J.-L., R. Abdelmalak, and X. Zhang. 2010a. “Design of Stiffened Slabs on Grade on Shrink Swell Soils” (Keynote Lecture). In: Proceedings of the International Conference on Unsaturated Soils [UNSAT 2010], Barcelona, Spain. Rotterdam: Balkema. Briaud, J.-L., M. Ballouz, and G. Nasr. 2000. “Static Capacity Prediction by Dynamic Methods for Three Bored Piles.” Journal of Geotechnical & Geoenvironmental Engineering 126(7): 640–649. Briaud, J.-L., M. Ballouz, and G. Nasr. 2002. “Defect and Length Prediction by NDT Methods for Nine Bored Piles.”

990

REFERENCES

In: Proceedings of the International Deep Foundation Congress (GSP 116), Orlando, Florida (February 14–16, 2002). Briaud, J.-L., M. Bernhardt, and M. Leclair. 2011. “The Pocket Erodometer: A Simple Device to Estimate Soil Erodibility.” ASTM Geotechnical Testing Journal 34. Briaud, J.-L., M. Bernhardt, and M. Leclair. 2012. “The Pocket Erodometer Test: Development and Preliminary Results.” ASTM Geotechnical Testing Journal, 35(2). Briaud, J.-L., L. Brandimarte, J. Wang, and P. D’Odorico. 2007a. “Probability of Scour Depth Exceedance due to Hydrologic Uncertainty.” Georisk Journal for Assessment and Management of Risk for Engineered Systems and Geohazards 1(2): 77–88. Briaud, J.-L., and A. Chaouch. 1997. “Hydrate Melting around Hot Conductor.” Journal of Geotechnical & Geoenvironmental Engineering 123(7): 645–653. Briaud J.-L., M. Chedid, H.-C., Chen, and A. Shidlovskaya. 2017a. “Borehole Erosion Test.” Journal of Geotechnical and Geoenvironmental Engineering 143(8). Briaud J.-L., M. Chedid, and A. Shidlovskaya. 2016. “The Borehole Erosion Test.” I: Proceedings of the 8th International Conference on Scour and Erosion (Oxford, UK, 12–15 September 2016) (Ed. J. Harris, R. Whitehouse, and S. Mixon). Boca Raton, FL: CRC Press. Briaud, J.-L., H.-C. Chen, K.-A. Chang, Y.-A. Chung, N. Park, W. Wang, and P.-H. Yeh. 2007b. “Establish Guidance for Soil Properties-Based Prediction of Meander Migration Rate” (Report FHWA/TX-07/0-4378-1). College Station, TX: Zachry Department of Civil Engineering, Texas A&M University. Briaud J.-L., H.-C. Chen, A.V. Govindasamy, and R. Storesund. 2008. “Levee Erosion by Overtopping in New Orleans during the Katrina Hurricane.” Journal of Geotechnical and Geoenvironmental Engineering 134(5). Briaud, J.-L., H.C. Chen, K. Kwak, S-W. Han, and F. Ting. 2001a. “Multiflood and Multilayer Method for Scour Rate Prediction at Bridge Piers.” Journal of Geotechnical & Geoenvironmental Engineering 127(2): 114–123. Briaud, J.-L., H.-C. Chen, P. Nurtjahyo, and J. Wang. 2004. Pier and Contraction Scour in Cohesive Soils (NCHRP Report 516). Washington, DC: Transportation Research Board. Briaud, J.-L., H.M. Coyle, and H.M. Tucker. 1990. “Axial Response of 3 Vibratory and 3 Impact Driven H Piles in Sand.” Transportation Research Record (January). Briaud, J.-L., and G.Y. Felio. 1986. “Cyclic Axial Loads on Piles: Analysis of Existing Data.” Canadian Geotechnical Journal 23(3): 362–371. Briaud, J.-L., and E.E. Garland. 1985. “Loading Rate Method for Pile Response in Clay.” Geotechnical Engineering Journal 111(3): 319–335. Briaud, J.-L., and R.M. Gibbens. 1994. “Test and Prediction Results for Five Large Spread Footings on Sand.” FHWA Prediction Symposium (ASCE Geotechnology Special Publication No. 41). New York: ASCE, pp. 92–128. Briaud, J.-L., and R.M. Gibbens. 1999. “Behavior of Five Large Spread Footings in Sand.” Journal of Geotechnical & Geoenvironmental Engineering 125(9): 787–797. Briaud, J.-L., L.G. Huff, L.M. Tucker, and H.M. Coyle. 1984. “Evaluation of In Situ Test Design Methods for Vertically Loaded H Piles at Lock & Dam No. 26 Replacement Site” (Research Report 4690). USAE, Waterways Experiment Station.

Briaud, J.L., I. Jung, A. Govindasamy, D. Kim, and J. Lee. 2018. “The Observation Method for Bridge Scour: Case Histories.” International Journal of Geoengineering Case Histories, 4(3): 185–202. doi:10.4417/IJGCH-04-03-03. Briaud, J.-L., and N.K. Kim. 1998. “Beam Column Method for Tieback Walls.” Journal of Geotechnical & Geoenvironmental Engineering 124(1): 67–69. Briaud J.-L., Y. Koohi, J. Nicks, and I. Jung. 2015. “San Jacinto Monument: New Soil Data and Analysis Including Subsidence.” Journal of Geotechnical and Geoenvironmental Engineering 141. ISSN 1090-0241/(0). Briaud, J.-L., Y. Li, and K. Rhee. 2006. “BCD: A Soil Modulus Device for Compaction Control.” Journal of Geotechnical & Geoenvironmental Engineering 132(1): 108–115. Briaud, J.-L., and Y. Lim. 1997. “Soil-Nailed Wall under Piled Bridge Abutment: Simulation and Guidelines.” Journal of Geotechnical & Geoenvironmental Engineering 123(11): 1043–1050. Briaud J.-L., and Y. Lim. 1999. “Tieback Walls in Sand: Numerical Simulation and Design Implications.” Journal of Geotechnical and Geoenvironmental Engineering 125(2): 101–110. Briaud, J.-L., M. Meriwether, and H. Porwol. 1983. “Pressuremeter Design of Retaining Walls.” Texas: Department of Transportation Report No. FHWA/TX-84/54+340-4f. Briaud, J.-L., and J. Miran. 1992a. The Cone Penetrometer Test (Publication No. FHWA-SA-91-043). Washington, DC: U.S. Department of Transportation. Briaud, J.-L., and J. Miran 1992b. The Flat Dilatometer Test (Publication No. FHWA-SA-91-044). Washington, DC: Federal Highway Administration, Office of Technology Applications. Briaud, J.-L., and A.M. Montalvo 2017. “Meander Migration: The Observation Method.” Canadian Geotechnical Journal, March 2017, https://doi.org/10.1139/cgj-2015-0639. Briaud, J.-L., J. Nicks, K. Rhee, and G. Stieben. 2007c. “The San Jacinto Monument Case History.” Journal of Geotechnical & Geoenvironmental Engineering 133(11): 1337–1351. Briaud J.-L., W.F. Powers, and D.E. Weatherby. 1998. “Should Grouted Anchors Have Short Tendon Bond Length?” Journal of Geotechnical and Geoenvironmental Engineering 124(2): 110–119. Briaud, J.-L., and C.J. Rutherford. 2010. “Excavation Support Using Deep Mixing Technology” (Keynote Lecture). In: Proceedings of the International Geotechnical Conference on Geotechnical Challenges in Megacities. Moscow, Russia (7–10 June 2010). St Petersburg, Russia: GRF. Briaud J.-L., I. Shafii, H-C. Chen, and Z. Medina-Cetina. 2019. “Relationship between Erodibility and Properties of Soils.” Research Report NCHRP 915, Transportation Research Board, National Academy of Sciences, Engineering and Medicine. Briaud, J.-L., B. Smith, K.-Y. Rhee, H. Lacy, and J. Nicks. 2009. “The Washington Monument Case History.” International Journal of Geoengineering Case Histories 1(3): 170–188. Available at: http://casehistories.geoengineer.org Briaud, J.-L., F. Ting, H-C. Chen, Y. Cao, S-W. Han, and K. Kwak. 2001b. “Erosion Function Apparatus for Scour Rate Predictions.” Journal of Geotechnical & Geoenvironmental Engineering 127(2): 105–113. Briaud, J.-L., F.C.K. Ting, H-C. Chen, R. Gudavalli, S. Perugu, and G. Wei. 1999. “SRICOS: Prediction of Scour Rate in Cohesive

REFERENCES

Soils at Bridge Piers.” Journal of Geotechnical & Geoenvironmental Engineering 123(4): 237–246. Briaud, J.L., and L.M. Tucker. 1984a. “Residual Stresses in Piles and the Wave Equation,” ASCE Symposium on Deep Foundations, San Francisco (October 1984). Briaud J.-L., and L.M. Tucker. 1984b. “Piles in Sand: A Method Including Residual Stresses.” Journal of Geotechnical Engineering 110(11): 1666–1680. Briaud J.-L., and L.M. Tucker. 1987. “Horizontally Loaded Piles Next to a Trench.” In: Proceedings of an ASCE Session on Foundations for Transmission Towers, Atlantic City, NJ (April 1987). Briaud J.-L., and L.M. Tucker. 1988. “Measured and Predicted Axial Response of 98 Piles.” Journal of Geotechnical Engineering 114(9): 984–1001. Briaud J.-L., and L.M. Tucker. 1997. Design and Construction Guidelines for Downdrag on Uncoated and Bitumen-Coated Piles (NCHRP Report 393). Washington, DC: Transportation Research Board, National Academy Press. Briaud, J.-L., L.M. Tucker, R.L. Lytton, and H.M. Coyle. 1985. Behavior of Piles and Pile Groups in Cohesionless Soils (FHWA-IRD-831038). Washington, DC: Federal Highway Administration. Briaud, J.-L., L.M. Tucker, and E. Ng. 1989. “Axially Loaded 5 Pile Group and Single Pile in Sand.” In: Proceedings of the 12th International Conference on Soil Mechanics and Foundation Engineering, Rio de Janeiro, vol. 2, pp. 1121–1124. Rotterdam: A. A. Balkema. Briaud J.-L., X. Zhang, and S. Moon. 2003. “Shrink Test: Water Content Method for Shrink and Swell Predictions.” Journal of Geotechnical & Geoenvironmental Engineering 129(7): 590–600. Brice, J.C. 1974. “Evolution of Meander Loops.” Geological Society of America Bulletin 85: 581–586. Brooks, R.H. and A.T. Corey. 1964. Hydraulic Properties of Porous Media (Hydrology Paper 3). Fort Collins, CO: Colorado State University. Brown, D.A., J.P. Turner, and R.J. Castelli. 2010. Drilled Shafts: Construction Procedures and LRFD Design Methods (NHI Course 132014, Geotechnical Engineering Circular no. 10, Federal Highway Administration Report no. FHWA NHI-10-016). Washington, DC: FHWA. Brunner, G.W. 2002. HEC-RAS River Analysis System Hydraulic Reference Manual (Version 3.1) (Report No. CPD-69). Davis, CA: U.S. Army Corps of Engineers, Institute for Water Resources, Hydrologic Engineering Research Center. Building Research Advisory Board (BRAB). 1968. National Research Council Criteria for Selection and Design of Residential Slabs-on-Ground (U.S. National Academy of Sciences Publication no. 1571). Washington, DC: National Academy Press. Bulut, R. 2003. “Measurement of Soil Suction.” Presentation at the Foundation Performance Association, Houston, TX, August 20. Bulut, R., S.M. Hineidi, and B. Bailey. 2002. “Suction Measurements: Filter Paper and Chilled Mirror Psychrometer.” In: Proceedings of the Texas ASCE 2002 Fall Meeting, Waco, TX. Bulut, R., R.L. Lytton, and W.K. Wray. 2001. “Soil Suction Measurements by Filter Paper.” In: Expansive Clay Soils and

991

Vegetative Influence on Shallow Foundations (ASCE Geotechnical Special Publication no. 115), pp. 243–261. Bulut, R. and W.K. Wray. 2005. “Free Energy of Water—Suction—in Filter Papers.” ASTM Geotechnical Testing Journal 28(4): 355–364. Burenkova, V.V. 1993. “Assessment of Suffusion in Noncohesive and Graded Soils.” In: Proceedings of the First International Conference on Geo-Filters, Karlsruhe, Germany. Rotterdam: Balkema. pp. 357–360. Burland J.B., M.B. Jamiolkowski, and C. Viggiani. 2009. “Leaning Tower of Pisa: Behavior after Stabilization Operations.” International Journal of Geoengineering Case Histories 1(3): 156–169. Burns, S.E. and P.W. Mayne. 1998. “Monotonic and Dilatory Pore Pressure Decay during Piezocone Tests.” Canadian Geotechnical Journal 35(6): 1063–1073. Byrne, R.J., D. Cotton, J. Porterfield, C. Wolschlag, and G. Ueblacker. 1998. Manual for Design and Construction Monitoring of Soil Nail Walls (Report FHWA-SA-96-69R). Washington, DC: Federal Highway Administration. CALTRANS. 1991. A User’s Manual for the SNAIL Program, Version 2.02—Updated PC Version. Sacramento, CA: California Department of Transportation, Division of New Technology, Material and Research, Office of Geotechnical Engineering. Canadian Geotechnical Society. 2007. Canadian Foundation Engineering Manual, (4th ed.). c/o BiTech Publishers. Cao, Y., J. Wang, J-L. Briaud, H-C. Chen, Y. Li, and P. Nurtjahyo. 2002. “EFA Tests and the Influence of Various Factors on the Erodibility of Cohesive Soils.” In: Proceedings of the First International Conference on Scour of Foundations. College Station, TX: Texas A&M University, Department of Civil Engineering. Cardimona, S. 1993. Electrical Resistivity Techniques for Subsurface Investigation (Internal Document). Rolla, MO: Department of Geology and Geophysics, University of Missouri-Rolla. Carman, P.C. 1938. “The Determination of the Specific Surface of Powders.” Journal of the Society for Chemical Industrial Transport 57: 225. Carrier, D.W., III. 2003. “Goodbye, Hazen; Hello, Kozeny-Carman.” Journal of Geotechnical & Geoenvironmental Engineering 129(11): 1054–1056. Carrier, W.D., and J.F. Beckman. 1984. “Correlations between Index Tests and the Properties of Remolded Clays.” Géotechnique 34(2): 211–228. Carroll, R.G., Jr., 1983. “Geotextile Filter Criteria.” In: TRR 916, Engineering Fabrics in Transportation Construction, Washington, DC, pp. 46–63. Carslaw, H.S., and J.C. Jaeger. 1947. Conduction of Heat in Solids (2nd ed.). Oxford: Oxford University Press. Cassagrande, A. 1938. “Notes on Soil Mechanics – First Semester.” Harvard University, unpublished. Cazzuffi, D., and S. Venesia. 1986. “The Mechanical Properties of Geotextiles: Italian Standard and Interlaboratory Test Comparison.” In: Proceeding of the Third Conference on Geotextiles, IFAI. Cedergren, H. 1967. Seepage, Drainage, and Flownets. New York: McGraw-Hill. Chadwick, W.L., A .Cassagrande, H.A. Coombs, M.W. Dowd, E.M. Fucik, R.K. Higginson, T.M. Leps, R.B. Peck, H.B. Seed,

992

REFERENCES

and R.B. Jansen. 1976. “Failure of Teton Dam.” Independent Panel to Review Cause of Teton Dam Failure, Report to U.S. Department of Interior and State of Idaho on Failure of Teton Dam. Washington, DC: U.S. Government Printing Office. Chapuis, R.P., and T. Gatien. 1986. “An Improved Rotating Cylinder Technique for Quantitative Measurements of the Scour Resistance of Clays.” Canadian Geotechnology Journal 23: 83–87. Chen, H-C. 2002. “Numerical Simulation of Scour around Complex Piers in Cohesive Soil.” In: Proceedings of First International Conference on Scour of Foundations College Station, TX (November 17–20, 2002), pp. 14–33. Chen, H-C., V.C. Patel, and S. Ju. 1990. “Solution of Reynolds-Averaged Navier Stokes Equations for Three-Dimensional Incompressible Flows.” Journal of Computational Physics 88(2): 305–336. Chen, T., and Y. Fang. 2008. “Earth Pressure due to Vibratory Compaction.” Journal of Geotechnical & Geoenvironmental Engineering 134(4): 437–444. Chow, V.T., D.R. Maidment, and L.W. Mays. 1988. Applied Hydrology. New York: McGraw-Hill. Christian, J.T., and W.D. Carrier III,. 1978. “Janbu, Bjerrum, and Kjaernsli’s Chart Reinterpreted.” Canadian Geotechnical Journal 15: 127. Chu, J., S. Varaksin, U. Klotz, and P. Menge. 2009. “Construction Processes.” In: Proceedings of the International Conference on Soil Mechanics and Geotechnical Engineering. Alexandria, Egypt: IOS Press. Chua, K.M., S. Gardner, and L.L. Lowery. 1987. “Wave Equation Analysis of a Vibratory Hammer-Driven Pile.” Paper presented at Offshore Technology Conference, Houston, TX (27–30 April 1987). Clayton, C.R.I., N.E. Simons, and M.C. Matthews. 1982. Site Investigation: A Handbook for Engineers. New York: Halsted Press. Clough, G.W., and T.D. O’Rourke. 1990. Construction Induced Movements of In Situ Walls: Design and Performance of Retaining Structures (Geotechnical Special Publications No. GSP 25), ASCE, pp. 439–470. Clough, R.W. 1960. “The Finite Element in Plane Stress Analysis.” In: Proceedings of the Second ASCE Conference on Electronic Computation, Pittsburgh, PA, pp. 345–378. Cornell, C.A., H. Banon, and A.F. Shakal. 1979. “Seismic Motion and Response Prediction Alternatives.” Earthquake Engineering & Structural Dynamics 7(4): 295–315. Couroyer, C. 2000. “Attrition of Alumina Catalyst Carrier Beads.” PhD thesis, University of Surrey, UK. Available at: www .engineering.leeds.ac.uk/ipse.old/research/ghadiri/ghadiri009 .shtml Cox, W.R., D.A. Dixon, and B.S. Murphy. 1983. Lateral Load Tests on 25.4 mm Diameter Piles in Very Soft Clay in Side by Side and In Line Groups (ASTM Special Technical Publication no. STP 835), pp. 122–140. Coyle, H.M., R.E. Bartoskewitz, and W.J. Berger. 1973. Bearing Capacity by Wave Equation Analysis—State of the Art. College Station, TX: Texas Transportation Institute, Texas A&M University. Coyle, H.M., and G.C. Gibbson. 1970. “Empirical Damping Constants for Sand and Clay.” Journal of the Soil Mechanics & Foundation Division, ASCE.

Crank, J. 1956. The Mathematics of Diffusion. New York: Oxford University Press. Croad, R.N. 1981. “Physics of Erosion of Cohesive Soils.” PhD thesis, Department of Civil Engineering, University of Auckland, Auckland, New Zealand. Crouch, S.L,. and A.M. Starfield. 1983. Boundary Element Methods in Solid Mechanics. London: Allen and Unwin. Cruz, E.C., and J-L. Briaud. 2021. “Settlement of Large Loaded Areas in Clays.” CERGEP Report No.14 – SI/US units, Civil Engineering, Texas A&M University. Cundall, P.A., and O.D.L. Strack. 1979. “A Discrete Numerical Model for Granular Assemblies.” Géotechnique 29(1): 47–65. Dalrymple, G.B. 1994. The Age of the Earth. Palo Alto, CA: Stanford University Press. Daniel, D.E. 1989. “In Situ Hydraulic Conductivity Tests for Compacted Clay.” Journal of Geotechnical Engineering 115(9): 1205–1226. Daniel, D.E., J.J. Bowders, and R.B. Gilbert. 1997. “Laboratory Hydraulic Conductivity Testing of GCLs in Flexible Wall Permeameters.” In Testing and Acceptance Criteria for Geosynthetic Clay Liners (ASTM STP 1308) (Ed. L.W. Well). Philadelphia, PA: ASTM, pp. 208–228. Davies, M.P., and R.G. Campanella. 1994. “Selecting Design Values of Undrained Strength for cohesionless soils.” In: Proceedings of the 47th Canadian Geotechnical Conference, Halifax, Nova Scotia, September 1994, BiTech Publishers. Vol. 1, pp. 176–186. Davis, R.O., and A.P.S. Selvadurai. 2002. Plasticity in Geomechanics. Cambridge: Cambridge University Press. De Alba, P., H.B. Seed, and C.K. Chan. 1976. “Sand Liquefaction in Large Scale Simple Shear Tests.” Journal of the Geotechnical Engineering Division, ASCE 102(GT9): 909–927. DeJong, J., and R.W. Boulanger. 2000. An Introduction to Drilling and Sampling in Geotechnical Practice (2nd ed.) [instructional video]. Department of Civil and Environmental Engineering, University of California-Davis. De Moor, J.J., R.T. Van Balen, and C. Kasse. 2007. “Simulating Meander Evolution of the Geul River (the Netherlands) Using a Topographic Steering Model.” Earth Surface Processes and Landforms 32(7): 1077–1093. Den Hoedt, G. 1986. “Creep and Relaxation of Geotextile Fabrics.” Journal of Geotextiles & Geomembranes 4(2): 83–92. Denk, E.W., W.A. Dunlap, W.R. Bryant, L.J. Milberger, and T.J. Whelan. 1981. “A Pressurized Core Barrel for Sampling Gas Charged Sediments.” In: Proceedings of the 13th Offshore Technology Conference, Richardson, TX, pp. 43–52. De Puy, M., and L.D. Alfaro. 2010. “Geotechnics of the Panama Canal Expansion.” GeoStrata Magazine—Geo Institute of ASCE, 14(6): 26–28, 30, 32–33. Desai, C.S. and J.F. Abel. 1972. Introduction to the Finite Element Method. New York: Van Nostrand Reinhold Company. De Wilde, P. 1996. Neural Network Models: An Analysis. London: Springer-Verlag. DNV (Det Norske Veritas) 2005. “Geotechnical Design and Installation of Suction Anchors in Clay.” Recommended Practice DNV-RP-E303. Donald, I.B. 1961. “The Mechanical Properties of Saturated and Partly Saturated Soils with Special Reference to Negative Pore Water Pressure.” PhD dissertation, University of London.

REFERENCES

Drucker, D.C., and W. Prager. 1952. “Soil Mechanics and Plastic Analysis or Limit Design.” Quarterly Journal of Applied Mathematics 10(2): 157–165. Duncan, J.M. 2010. “Managing Unstoppable Landslides in the Panama Canal.” GeoStrata Magazine, 14(6): 12–13. Duncan, J.M., and A.L. Buchignani. 1976. An Engineering Manual for Settlement Studies. Berkeley, CA: University of California at Berkeley. Duncan, J.M., and C.Y. Chang. 1970. “Non-Linear Analysis of Stress and Strain in Soils.” Journal of Soil Mechanics 96(Sm5): 1629–1653. Duncan, J.M., and R.B. Seed. 1986. “Compaction-Induced Earth Pressures under Ko-Conditions.” Journal of Geotechnical Engineering 112(1): 1–21. Duncan, J.M., and S.P. Wright. 2005. Soil Strength and Slope Stability. Hoboken, NJ: John Wiley & Sons. Elias, V., and I. Juran. 1991. Soil Nailing for Stabilization of Highway Slopes and Excavations (Publication FHWA-RD-89-198). Washington, DC: Federal Highway Administration. Elias, V., J. Welsh, J. Warren, R. Lukas, J.G. Collin, and R.R. Berg. 2006. Ground Improvement Methods (NHI-06-020). Washington, DC: Federal Highway Administration. Elton, D.J. 1999. Soils Magic (Geotechnical Special Publication no. 114). Reston, VA: American Society of Civil Engineers. Escario, V., J. Juca, and M.S. Coppe. 1989. “Strength and Deformation of Partly Saturated Soils.” In: Proceedings of the 12th International Conference on Soil Mechanics and Foundation Engineering, Rio de Janeiro, vol. 3, pp. 43–46. Eurocode 7. 2011. Geotechnical Design. Part 1: General Rules, Part 2: Ground Investigation and Testing (European Standard EN-1997). Brussels, Belgium: European Committee for Standardization. European Committee for Standardization. 2004. Geotechnical Investigation and Testing—Laboratory Testing of Soil Part 6: Fall Cone Test (ISO/TS 17892-6:2004). Paris, France: European Committee for Standardization. Fan, Q., H. Zhu, and X. Chen. 2011. “Key Issues in Rock Mechanics of the Three Gorges Project in China.” Journal of Rock Mechanics and Geotechnical Engineering 3(4):329–342, DOI: 10.3724/SP.J.1235.2011.00329. Fang, H.Y., ed. 1991. Foundation Engineering Handbook. New York: Van Nostrand Reinhold. Feda, J. 1966. “Structural Stability of Subsident Loess Soil from Prahadejuice.” Engineering Geology 1: 201–219. Fell, R., and J-J. Fry. 2005. “The State of the Art of Assessing the Likelihood of Internal Erosion of Embankment Dams, Water Retaining Structures and Their Foundations.” In: Internal Erosion of Dams and Their Foundations (Ed. R. Fell and J-J. Fry)/ London: Routledge. Fell, R., P. MacGregor, D. Stapledon, and G. Bell. 2005. Geotechnical Engineering of Dams. Leiden: Balkema. Fell, R., and C.F. Wan. 2005. “Investigation of Internal Erosion by the Process of Suffusion in Embankment Dams and Their Foundations.” In: Internal Erosion of Dams and their Foundations (Ed. R. Fell and J-J. Fry). London: Routledge, pp. 219–234. Fellenius, W. 1927. Erdstatische Berechnungen mit Reibung and Kohäsion. Berlin: Ernst.

993

FERUM. 2001. “Finite Element Reliability Using Matlab.” Available at http://www.ce.berkeley.edu/projects/ferum/index .html. Fetter, C.W. 1992. Contaminant Hydrogeology. New York: Macmillan. FHWA (Federal Highway Administration). 1984. Permanent Ground Anchors (Report FHWA-DP-68-1R). Washington, DC: FHWA. FHWA (Federal Highway Administration). 1998. Geotechnical Earthquake Engineering (FHWA HI-99-012). Washington, DC: FHWA . FHWA (Federal Highway Administration). 2006a. “Geotechnical Inputs for Pavement Design.” In: Geotechnical Aspects of Pavements Reference Manual (U.S. Department of Transportation Publication No. FHWA NHI-05-037, NHI Course No. 132040). Washington, DC: FHWA. FHWA (Federal Highway Administration) 2006b. “Soils and Foundations: Reference Manual - Volume I.” publication FHWA-NHI-06-088. Washington, DC: FHWA. FHWA (Federal Highway Administration). 2010. Drilled Shaft Manual (Publication No. FHWA-NHI-10-016). Washington, DC: FHWA. Filz, G., J. Sloan, M.P. McGuire, J. Collin, and M. Smith. 2012. “Column-Supported Embankments: Settlement and Load Transfer.” In: Geotechnical Engineering State of the Art and Practice: :Proceedings of the Geo-Institute Congress, Oakland, CA. Finno, R.J., and S.L. Gassman. 1998. “Impulse Response Evaluation of Drilled Shafts.” Journal of Geotechnical & Geoenvironmental Engineering 124(10): 965–975. Floss, R., and H-J. Kloubert. 2000. “Newest Innovations in Soil and Asphalt Compaction Technology.” Paper presented at International Workshop on Compaction of Soil, Granulates and Powders, Innsbruck (February 2000). Rotterdam: A.A. Balkema. Focht, J.A. Jr., F.R. Khan, and J.P. Gemeinhardt. 1978. “Performance of One Shell Plaza Deep Mat Foundation.” Journal of the Geotechnical Engineering Division, Vol. 104, No. GT5, pp. 593–608, Reston, VA: ASCE. Foster, M., R. Fell, and M. Spannagle. 2000a. “The Statistics of Embankment Dam Failures and Accidents.” Canadian Geotechnical Journal 37(5): 1000–1024. Foster, M., R. Fell, and M. Spannagle. 2000b. “A Method for Assessing the Relative Likelihood of Failure of Embankment Dams by Piping.” Canadian Geotechnical Journal 37(5), 1025–1061. Foster, R.J. 1988. Physical Geology (5th ed.). New York: Prentice Hall. Fourier, J.B. 1822. Théorie Analytique de la Chaleur. Trans. A. Freeman. New York: Dover. Frank, R. 1999. Calcul des Fondations Superficielles et Profondes. Paris: Presses de l’Ecole Nationale des Ponts et Chaussées, updated in 2013 with class notes to match the 2013 rules AFNOR. Frank, R., and S.R. Zhao. 1982. “Estimation par les paramètres pressiométriques de l’enfoncement sous charge axiale de pieux forés dans les sols fins.” Bulletin de Liaison du Laboratoire des Ponts et Chaussées 119: 17–24 (in French). Fredlund, D.G., and H. Rahardjo. 1993. Soil Mechanics for Unsaturated Soils. New York: John Wiley & Sons.

994

REFERENCES

Fredlund, D.G., and A. Xing, 1994. “Equations for the Soil-Water Characteristic Curve.” Canadian Geotechnical Journal 31: 521–532. Fredlund, D.G., A. Xing, and S. Huang. 1994. “Predicting the Permeability Function for Unsaturated Soils Using the Soil-Water Characteristic Curve.” Canadian Geotechnical Journal 31: 533–546. Freer, R. 2001. “The Three Gorges Project on the Yangtze River in China.” In: Proceedings of ICE, Civil Engineering Paper 12320. vol. 144, pp. 20–28. Funk, J.R. 2013. “Settlement of the Kansai International Airport Islands.” PhD dissertation, Civil Engineering, University of Illinois at Urbana-Champaign. Furudoi, T. 2010. “The Second Phase Construction of Kansai International Airport Considering the Large and Long-term Settlement of the Clay Deposits.” Soils and Foundations, 50(6). Gardner, W.R. 1958. “Some Steady State Solutions of the Unsaturated Moisture Flow Equation with Application to Evaporation from a Water Table.” Soil Science 85(4): 228–232. Garner, D. 2002. “Comparison of Suction and Water Content Determination.” Unpublished. Ghaboussi, J., and R. Barbosa. 1990. “Three-Dimensional Discrete Element Method for Granular Materials.” International Journal of Numerical & Analytical Methods in Geomechanics 14: 451–472. Gibbs, H.J., and J.P. Bara. 1962. Predicting Surface Subsidence from Basic Soil Tests (Field Testing of Soils, Special Technical Publication No. 322), pp. 231–247. Giroud, J-P. 1981. “Designing with Geotextiles.” Matériaux et Construction 14(82): 257–272. Giroud, J-P. 2010. “Development of Criteria for Geotextiles and Granular Filters.” (Prestigious Lecture). In: Proceedings of the 9th International Conference on Geosynthetics, Guarujá, Brazil. vol. 1, pp. 45–64. Giroud, J-P., and R. Bonaparte. 1989. “Leakage through Liners Constructed with Geomembranes: Geomembrane Liners.” Geotextiles & Geomembranes 8(1). Goble, G.G., F. Rausche, and G.E. Likins. 1993. Case Pile Wave Analysis Program—CAPWAP Manual. Cleveland, OH: GRL. Golder Associates. 1993. GOLDNAIL Soil Nailing Design Program. Seattle, WA: Author. Gong, Y., C. Aubeny, and J-L. Briaud. 2021. “Friction on Augercast Piles in Clays.” CERGEP Report No.13 – SI/US units, Department of Civil Engineering, Texas A&M University. Goodman, R.E. 1989. Rock Mechanics (2nd ed.). New York: John Wiley & Sons. Graham, W. 2008. “The Teton Dam Failure: An Effective Warning and Evacuation.” paper presented at ASDSO Annual Conference, Indian Wells, Association of State Dam Safety Officials. Gregersen, O. 1981. “The Quick Clay Landslide in Rissa, Norway. The Sliding Process and Discussion of Failure Modes.”, Publication No. 135. Oslo: Norwegian Geotechnical Institute. Griffiths, D.V., and P.A. Lane. 1999. “Slope Stability Analysis by Finite Elements.” Géotechnique 49(3): 387–403. GRLWEAP. 2012. Goble Rausche Likins Wave Equation Analysis Program—User’s Manual. Cleveland, OH: Pile Dynamics Inc. Gutenberg, B., and C.F. Richter. 1944. “Frequency of Earthquakes in California.” Bulletin of the Seismological Society of America 34(4): 1985–1988.

Haberfield, C. 2010. “Engineering the Foundations for the Nakheel Tower.” In: Piling and Deep Foundations Middle East 2010. Dubai, UAE: Deep Foundations Institute. Habib, P. 1994. “Aspects géotechniques de l’accident du nouveau port de Nice.” Revue Française de Géotechnique, 65: 1–15. Hall, W.J., and N.M. Newmark. 1977. “Seismic Design Criteria for Pipelines and Facilities.” In: The Current State of Knowledge of Lifeline Earthquake Engineering, Proceedings of the ASCE. New York: ASCE, pp. 18–34. Handy, R.L. 1975. “Measurement of In-Situ Shear Strength.” In: Proceedings of the Conference on In Situ Measurement of Soil Properties, vol. 2, pp. 143–149. New York: ASCE. Handy, R.L. 1986. “Borehole Shear Test and Slope Stability.” In: Use of In Situ Tests in Geotechnical Engineering (Ed. S.P. Clemence). New York: ASCE, pp. 161–175. Hannigan, P.J. 1990. “Dynamic Monitoring and Analysis of Pile Foundation Installations.” In: Deep Foundations Institute Short-Course Text (1st ed.). Dubai, UAE: Deep Foundations Institute, p. 69. Hannigan, P.J., G.G. Goble, G. Thendean, G.E. Likins, and F. Rausche. 1998. Design and Construction of Driven Pile Foundations, Vols. 1 and 2 (FHWA Publication HI 97-014, NHI course nos. 13221 and 13222). Washington, DC: FHWA. Hansbo, S. 1981. “Consolidation of Fine-Grained Soils by Prefabricated Drains.” In: Proceedings of the 10th International Conference on Soil Mechanics and Foundation Engineering, vol. 3, Stockholm: Hanson, G.J. 1991. “Development of a Jet Index to Characterize Erosion Resistance of Soils in Earthen Spillways.” Transactions of ASAE 34(5): 2015–2020. Hardin, B.O. 1978 “The Nature of Stress-Strain Behavior for Soils.” In: Proceedings of the ASCE Geotechnical Engineering Division Specialty Conference on Earthquake Engineering and Soil Dynamics, vol. 1, pp. 3–89. Hardin, B.O., and V.P. Drnevich. 1972. “Shear Modulus and Damping in Soils: Design Equations and Curves.” Journal of Soil Mechanics, Foundations Division 98 (SM7): 667–692. Harland, W.B., A.V. Cox, P.G. Llewellyn, C.A. Pickton, A.G. Smith, and R. Walters. 1990. A Geologic Time Scale. New York: Cambridge University Press. Hart, R., P.A. Cundall, and J. Lemos. 1998. “Formulation of a Three-Dimensional Distinct Element Model—Part II. Mechanical Calculations for Motion and Interaction of a System Composed of Many Polyhedral Blocks.” International Journal of Rock Mechanics & Mining Sciences 25(3): 117–125. Hazen, A. 1892. Some Physical Properties of Sands and Gravels, with Special Reference to Their Use in Filtration (24th Annual Report, Massachusetts State Board of Health, Pub. Doc. No. 34), pp. 539–556. Heibaum, M.W. 2004. “Geotechnical Filters: The Important Link in Scour Protection.” In: Proceedings of the 2nd International Conference on Scour and Erosion (Ed. Y.M. Chiew et al.). Singapore: Nanyang Technology University, Maritime Research Center, pp. 13–28. Hemond, H.F., and E.J. Fechner-Levy. 2000. Chemical Fate and Transport in the Environment (2nd ed.). San Diego, CA: Academic Press. Hermana, R. 2001. “An Introduction to Electrical Resistivity in Geophysics.” American Journal of Physics 69(9).

REFERENCES

Hertlein, B.H. 2009. “Analysis Collateral: Information Needed for Foundation Integrity Test Data Analysis.” In: 2009 International Foundation Congress and Equipment Expo, Orlando, Florida, ASCE, pp. 528–535. Hickin, E.J., and G.C. Nanson. 1984. “Lateral Migration Rates of River Bends.” Journal of Hydraulic Engineering (ASCE) 110: 1557–1567. Hillen, M.M., S.N. Jonkman, W. Kanning, M. Kok, M.A. Geldenhuys, and M.J.F. Stive. 2010. “Coastal Defense Coast Estimate: Case Study of Netherlands, New Orleans, and Vietnam.” TU Delft Department of Hydraulic Engineering, ISBN 0169-6548. Hjulström, F. 1935. “The Morphological Activity of Rivers as Illustrated by River Fyris.” Bulletin of the Geological Institute Uppsala (23): 221. Hoek, E. 1994. “Strength of Rock and Rock Masses.” ISRM News Journal 2: 4–16. Hoffmans, G.J.C.M., and H.J. Verheij. 1997. Scour Manual. Rotterdam: Balkema. Hofland, B., J.A. Battjes, and R. Booij. 2005. “Measurement of Fluctuating Pressures on Coarse Bed Material.” Journal of Hydraulic Engineering 131(9): 770–781. Hogentogler, C.A,. and K. Terzaghi. 1929. “Interrelationship of Load, Road and Subgrade.” Public Roads 37–64. Hoit, M., C. Hays, and M. Mcvay. 1997. “The Florida Pier Analysis Program: Methods and Models for Pier Analysis and Design.” Journal of the Transportation Research Board 1569: 1–7. Holleman, D.A., and L.D. Olson. 2002. “Crosshole Sonic Logging ND Tomography Velocity Imaging of a New Bridge Foundation.” Paper presented at Structural Materials Technology Conference, American Society for Non-Destructive Testing, Cincinnati, OH (10–13 September 2012). Holtz, R.D., W.D. Kovacs, and T.C. Sheahan. 2011. An Introduction to Geotechnical Engineering (2nd ed.). Upper Saddle River, NJ: Pearson-Prentice Hall. Hong, G.T., C.P. Aubeny, and R.L. Lytton. 2010. “Lateral Earth Pressure on a Wall in Expansive Soils.” In: Proceedings of the International Conference on Unsaturated Soils (UNSAT 2010), Spain. (September 2010). Boca Raton, FL: CRC Press. Hooke, J.M. 1980. “Magnitude and Distribution of Rates of River Bank Erosion.” Earth Surface Processes and Landforms 5(2): 363–377. Hooke, J.M. 1984. “Changes in River Meanders: A Review of Techniques and Results of Analysis.” Progress in Physical Geography 8: 473–508. Horvath, J.S. 1991. “Using Geosynthetics to Reduce Earth Loads on Rigid Retaining Structures.” In: Proceedings Geosynthetics ’91 Washington, DC: IFAI Publications, pp. 409–423. Horvath, J.S., ed. 1994. “Polystyrene Foam in Below Grade Applications.” In: Proceedings of International Symposium. New York: Manhattan College. Hossain, K.M. 1996. “Load Settlement Curve Method for Footings in Sand at Various Depths, under Eccentric or Inclined Loads, and Near Slopes.” PhD dissertation, Texas A&M University, Department of Civil Engineering, College Station, TX. Houlsby, G. 1991. “How the Dilatancy of Soils Affects Their Behavior.” Invited Lecture, 10th European Conference on Soil Mechanics and Foundation Engineering, Florence, Italy.

995

Hughes, J.M.O., and N.J. Withers. 1974. “Reinforcing of Soft Cohesive Soils with Stone Columns.” Ground Engineering (May): 42–49. Hunt, B. 2001. “Measured Scour Depths and Velocities at the Old Woodrow Wilson Bridge in Washington DC.” Personal communication with author. Hunt, R. 1984. Geotechnical Engineering Investigation Manual. New York: McGraw Hill. Hunt, R. 1986. Geotechnical Engineering Techniques and Practices. New York: McGraw Hill. Hunt, R. 2005. Geotechnical Investigations Methods: A Field Guide for Geotechnical Engineers. Boca Raton, FL: CRC Press. Hvorslev, M.J. 1949. Subsurface Exploration and Sampling of Soils for Civil Engineering Purposes.” Vicksburg, MO: Waterways Experimental Station, Hvorslev, M.J. 1951. “Time Lag in the Observation of Ground-Water Levels and Pressures (U.S. Army Engineers Waterways Experiment Station, Vicksburg, MO).” In: International Congress of Soil Science vol. 2, pp. 37–48. Idriss, I.M. 1990. “Response of Soft Soil Sites During Earthquakes.” In: Proceedings of the H. Bolton Seed Memorial Symposium (Ed. J.M. Duncan). vol. 2, pp. 273–290. Idriss, I.M., and R.W. Boulanger. 2008. Soil Liquefaction during Earthquakes (Monograph MNO-12). Oakland, CA: Earthquake Engineering Research Institute. Idriss, I.M., and J.I. Sun. 1992. SHAKE91: A Computer Program for Conducting Equivalent Linear Seismic Response Analyses of Horizontally Layered Soil Deposits. Davis, CA: Center for Geotechnical Modeling, Department of Civil and Environmental Engineering, University of California at Davis. Isenhower, W.M., S.T. Wang, and L.G. Vasquez. 2018. Technical Manual for LPile. Available at: https://www.ensoftinc.com/ main/products/pdf/LPile%202016%20User%20Manual.pdf Iskander, M., S. Kelley, C. Ealy, and D. Roy. 2001. Class-A Prediction of Construction Defects in Drilled Shafts (TRB ID No. 01-0308). Washington, DC: Transportation Research Board. ISSMGE Technical Committee on Offshore Geotechnics. 2005. “Geotechnical and Geophysical Investigations for Offshore and Nearshore Developments.” Available at: http://www.ISSMGE .rg/ Jaky, J. 1944. “A nyugalmi nyomas tenyezoje [The Coefficient of Earth Pressure at Rest].” Magyar Mernok es Epitesz-Egylet Kozlonye, Hungary, pp. 355–358 [in Hungarian]. Jamiolkowski, M. 2001. “The Leaning Tower of Pisa, End of an Odyssey.” In: Proceedings of the 15th International Conference on Soil Mechanics and Geotechnical Engineering, Vol. 4, pp. 2979–2996. Rotterdam: Balkema. Jamiolkowski, M., S. Leroueil, and D.C.F. Lo Presti. 1991. “Design Parameters from Theory to Practice” (Theme Lecture). In: Proceedings of Geo-Coast ’91, pp. 1–41. Janbu, N. 1954. “Stability Analysis for Slopes with Dimensionless Parameters” (Harvard University Soil Mechanics Series No. 46). Doctor of Sciences thesis in the Field of Civil Engineering. Janbu, N. 1968. Slope Stability Computations (Soil Mechanics and Foundation Engineering Report). Trondheim, Norway: Technical University of Norway. Janbu, N. 1973. “Slope Stability Computations in Embankment-Dam Engineering.” In: Embankment-Dam

996

REFERENCES

Engineering (Ed. R.J. Hirschfeld and S.J. Poulos). New York: John Wiley & Sons, pp. 47–86. Jeanjean, P. 1995. “Load Settlement Curve Method for Spread Footings on Sand from the Pressuremeter Test.” PhD dissertation, Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Jeanjean, P. 2012. “State of Practice: Offshore Geotechnics Throughout the Life of an Oil and Gas Field.” In: Proceedings of the Geo-Institute Congress (Oakland, March 2012),. Reston, VA: ASCE. Jefferis, D. 2008. Exploring Our Solar System: The Earth, Our Home Planet. New York: Crabtree Publishing. Jennings, J.E., and K. Knight. 1975. “A Guide to Construction on or with Materials Exhibiting Additional Settlement Due to Collapse of Grain Structure.” In: Sixth Regional Conference for Africa and Soil Mechanics and Foundation Engineering, pp. 99–105. Jeong, S., and Briaud, J-L. 1994. “Nonlinear Three-Dimensional Analysis of Downdrag on Pile Groups.” Paper presented at ASCE Specialty Conference. “Settlement 94,” Texas A&M University (June 1994). Jeyapalan, J.K., and R. Boehm. 1986. “Procedures for Predicting Settlements in Sands.” In: Settlement of Shallow Foundations on Cohesionless Soils: Design and Performance. Reston, VA: ASCE, pp. 1–22. Jing, L., and J.A. Hudson. 2002. “Numerical Methods in Rock Mechanics.” International Journal of Rock Mechanics & Mining Sciences 39: 409–427. Jumikis, A.R. 1977. Thermal Geotechnics. New Brunswick, NJ: Rutgers University Press. Kagawa, T., B. Zhao, K. Miyakoshi, and K. Irikura. (2004). Modeling of 3D Basin Structures for Seismic Wave Simulations Based on Available Information on the Target Area: Case Study of the Osaka Basin, Japan.” Bulletin of the Seismological Society of America 94(4): 1353–1368. Kanamori. H. 1977. “The Energy Release in Great Earthquakes.” Journal of Geophysical Research 82: 2981–2987. Karlsrud, K., R. Nadim, and T. Haugen. 1986. “Piles in Clay under Cyclic Axial Load: Field Tests and Computational Modeling.” In: Proceedings of the Third International Conference on Numerical Methods in Offshore Piling, Nantes, France, pp. 165–190. Karol, R.H. 2003. Chemical Grouting and Soil Stabilization (3rd ed.). Boca Raton, FL: CRC Press. Katzenbach, R. 2012. International CPRF Draft Guidelines (ISS-MGE Technical Committee on Deep Foundations). London: ISSMGE. Kavazanjian, E. 1999. “Seismic Design of Solid Waste Containment Facilities.” In: Proceedings of the 8th Canadian Conference on Earthquake Engineering, Vancouver, BC, Canada (June), pp. 51–89. Kavazanjian, E., N. Matasovic, T. Hadj-Hamou, and P. J. Sabatini. 1997. Geotechnical Earthquake Engineering for Highways: Vol. 1: Design Principles and Vol. 2: Design Example (Geotechnical Engineering Circular No. 3, FHWA SA-97-076). Washington, DC: FHWA. Kavazanjian, E., J-N.J. Wang, G.R. Martin., A. Shamsabadi, I.P. Lam, S.E. Dickenson, and C.J. Hung. 2011. LRFD Seismic Analysis and Design of Transportation Geotechnical Features and Structural Foundations (NHI Course no. 130094 reference

manual, Geotechnical Engineering Circular no. 3, FHWA-NHI-11-032). Washington, DC: FHWA. Kearey, P., M. Brooks, and I. Hill. 2002. An Introduction to Geophysical Exploration (3rd ed.). Oxford: Blackwell. Keller. 2012. “Range of Application of Grouting Techniques.” Available at: http://www.kellergrundbau.com/download/pdf/en/ Keller_66-01E.pdf Kenney, T.C., and D. Lau. 1986. “Internal Stability of Granular Filters.” Canadian Geotechnical Journal 22(2): 215–223. Kerisel, J. 1985. “The History of Geotechnical Engineering up until 1700.” In: Proceedings of the Eleventh International Conference on Soil Mechanics and Foundation Engineering, San Francisco (Golden Jubilee Volume), Rotterdam: A. A. Balkema, pp. 3–93. Khalili, N., and M.H. Khabbaz. 1998. “Unique Relationship for the Determination of the Shear Strength of Unsaturated Soils.” Géotechnique 48(5): 681–687. Kim, D-S., W-S. Seo, and S-H. Lee. 2006. “Development of Modulus-Soil Moisture Model for Subgrade Soils Using Soil Suction Control Testing System.” In: Pavement Mechanics and Performance (GSP 154), Washington, DC: ASCE, pp. 256–263. Kim, K. 2010. “Numerical Simulation of Impact Roller for Estimating the Influence Depth on Soil Compaction.” Master’s thesis, Department of Civil Engineering, Texas A&M University, College Station, TX. Kirsten, H.A.D., J.S. Moore, L.H. Kirsten, and D.M. Temple. 1996. “Erodibility Criterion for Auxiliary Spillways of Dams.” Paper No. 962099. ASAE International Meeting, Phoenix, AZ. Koerner, R.M. 2012. Designing with Geosynthetics (6th ed.), vols. 1 and 2. Bloomington, IN: Xlibris Corp. Koerner, R.M., and T-Y. Soong. 2001. “Geosynthetic Reinforced Segmental Retaining Walls.” Geotextiles & Geomembranes 19(6): 359–386. Komine, H., and N. Ogata. 2004. “Predicting Swelling Characteristics of Bentonites.” Journal of Geotechnical & Geoenvironmental Engineering 130(8): 818–829. Kozeny, J. 1927. “Über kapillare Leitung des Wassers im Boden.” Wien Akademie Wiss. 136(2a): 271. Kramer, S.L. 1996. Geotechnical Earthquake Engineering. New York: Prentice Hall. Kulhawy, F.H., and J-R. Chen. 2007. “Discussion of ‘Drilled Shaft Side Resistance in Gravelly Soils’ by Rollins K. M., Clayton R. J., Mikesell R. C., Blaise B. C.” Journal of Geotechnical & Geoenvironmental Engineering 133(10): 1325–1328. Kulhawy, F.H., and C.S. Jackson. 1989. “Some Observations on Undrained Side Resistance of CIDH Piles.” In: Proceedings, Foundation Engineering: Current Principles and Practices, Vol. 2, pp. 1011–1025. Washington, DC: ASCE. Kulhawy, F.H., and P.W. Mayne. 1990. Manual on Estimating Soil Properties for Foundation Design (Electric Power Research Institute Report EL-6800). Palo Alto, CA: Electric Power Research Institute. Kwak, K., J-L. Briaud, Y. Cao, M-K. Chung, B. Hunt, and S. Davis. 2002. “Pier Scour at Woodrow Wilson Bridge and SRICOS Method.” In: Proceedings of the First International Conference on Scour of Foundations, Department of Civil Engineering, Texas A&M University, College Station, TX. Kwak, K., J-L. Briaud, and H-C. Chen. 2001. “SRICOS: Computer Program for Bridge Pier Scour.” In: Proceedings of the 15th International Conference on Soil Mechanics and Geotechnical Engineering, vol. 3, pp. 2235–2238. Rotterdam: A.A. Balkema.

REFERENCES

Ladd, C.C. 1991. “Stability Evaluation during Staged Construction: 22nd Terzaghi Lecture.” Journal of Geotechnical Engineering 117(4): 537–615. Ladd, C. C., and R. Foott. 1974. “New Design Procedure for Stability of Soft Clays.” Journal of Geotechnical Engineering 100(GT7): 763–786. Ladd, C.C., R. Foott, K. Ishihara, F. Schlosser, and H. Poulos. 1977. “Stress-Deformation and Strength Characteristics” (State of the Art Report). In: Proceedings of the 9th International Conference on Soil Mechanics and Foundation Engineering, Tokyo, vol. 2, pp. 421–494. Lagasse, P.F., P.E. Clopper, J.E. Pagán-Ortiz, L.W. Zevenbergen, L.A. Arneson, J.D. Schall, and L.G. Girard. 2009. Bridge Scour and Stream Instability Countermeasures: Experience, Selection and Design Guidance (Hydraulic Engineering Circular no. 23, 3rd ed., vols. 1 and 2). Washington, DC: Federal Highway Administration. Lagasse, P.F., J.D. Schall, and E.V. Richardson. 2001. Stream Stability at Highway Bridges (Hydraulic Engineering Circular no. 20, 3rd ed.). Washington, DC: Federal Highway Administration. Laliberte, G.E., and A.T. Corey. 1966. Hydraulic Properties of Disturbed and Undisturbed Clays (ASTM STP No. 417). West Conshohocken, PA: ASTM. Lambe, T.W., and R.V. Whitman. 1979. Soil Mechanics—SI Version. New York: John Wiley & Sons. Lancellotta, R., and C. Pepe. 1990. “Mechanical Behavior of Pisa Clay.” Research Report 2.1 and 2.2, Technical University of Torino. Landva, A.O., and J.L. Clark. 1990. Geotechnics of Wastefills: Theory and Practice (ASTM STP 1070), West Conshohocken, PA: ASTM., pp. 86–103. Lawson, W.D., P.W. Jayawickrama, T.A. Wood, and J.G. Surles. 2011. MSE Pullout Testing for RECO HA Ladder and HA Strip Reinforcements (Research Report for The Reinforced Earth Company, Vienna, VA). Texas Tech University, Department of Civil Engineering. Lay, T., H. Kanamori, C.J. Ammon, M. Nettles, S.N. Ward, R.C. Aster,… and S. Sipkin. 2005. “The Great Sumatra-Andaman Earthquake of 26 December 2004.” Science 308: 1127–1133. Lazarte, C.A., V. Elias, R.D. Espinoza, and P.J. Sabatini. 2003. Geotechnical Engineering Circular No. 7: Soil Nail Walls (Report No. FHWA0-IF-03-017). Washington, DC: Federal Highway Administration. Lemoine, B. 2006. La Tour de 300 Mètres. Paris: Société des Imprimeurs Lemercier. Li, X., L.M. Zhang, and D.G. Fredlund. 2009. “Wetting Front Advancing Column Test for Measuring Unsaturated Hydraulic Conductivity.” Canadian Geotechnical Journal 46(December): 1431–1445. Likins, G.E., and M. Hussein. 1988. “A Summary of the Pile Driving Analyzer Capacity Methods: Past and Present.” Paper presented at the 11th Pile Driving Analyzers Seminar, Cleveland, OH. Lisyuk, M., and V. Ulitsky. 2012. “Soil Structure Interaction.” In: Proceedings of the Russia-USA Geotechnical Summit (March) Oakland, CA. Little, D.N. 1999. “Evaluation of Structural Properties of Lime Stabilized Soils and Aggregates.” Available at: www.lime.org/

997

documents/publications/free_downloads/soils-aggregates-vol1 .pdf. Liu, J., X-T. Feng, X-L. Ding, J. Zhang, and D-M. Yue. 2003a. “Stability Assessment of The Three-Gorges Dam Foundation, China, Using Physical and Numerical Modeling—Part I: Physical Model Tests.” International Journal of Rock Mechanics & Mining Sciences 40(2003) 609–631. Liu, J., X-T. Feng, and X-L. Ding. 2003b. “Stability Assessment of the Three-Gorges Dam Foundation, China, Using Physical and Numerical Modeling—Part II: Numerical Modelling.” International Journal of Rock Mechanics & Mining Sciences 40(2003) 633–652. Liu, Z., J-S. L’Heureux, S. Glimsdal, and S. Lacasse. 2021. “Modelling of Mobility of Rissa Landslide and Following Tsunami.” Computers and Geotechnics 140(2021): 104388. Lo Presti D., M. Jamiolkowski, and M. Pepe. 2003. “Geotechnical Characterization of the Subsoil of Pisa Tower.” In: Characterization and Engineering Properties of Natural Soils (Ed. T.S. Tan). Lisse: Swets & Zeitlinger, ISBN 90 5809 537 1. Lowe, J., and L. Karafiath. 1960. “Stability of Earth Dams upon Drawdown.” In: Proceedings of the First Pan American Conference on Soil Mechanics and Foundation Engineering, Mexico City, pp. 537–552. Lowery, L.L. 1993. Pile Driving Analysis by the Wave Equation: Microwave Manual. Bryan, TX: Wild West Software. Lowery, L., T.J. Hirsch, and C.H. Samson, Jr. 1967. Pile Driving Analysis: Simulation of Hammers, Pile and Soils (Research Report 33-9, Project 2-5-62-33, Piling Behavior). College Station, TX: Texas Transportation Institute, Texas A&M University. Lu, N., and W.J. Likos. 2004. Unsaturated Soil Mechanics. Hoboken, NJ: John Wiley & Sons, Inc. Luettich, S.M., J.P. Giroud, and R.C. Bachus. 1992. “Geotextile Filter Design Guide.” Journal of Geotextiles & Geomembranes 11(4–6): 19–34. Lukas, R.G. 1995. Dynamic Compaction (Geotechnical Engineering Circular No.1, FHWA-SA-95-037). Washington, DC: Federal Highway Administration. Lundberg, G. 1939. “Elastische Berührung zweier Halbräume.” Forsch. Geb. Ingenieurwes. 10–5: 201–211. Lunne, T, and A. Kleven. 1981. “Role of CPT in North Sea Foundation Engineering.” In: Symposium on Cone Penetration Engineering Division, ASCE, pp. 49–75. Lupini, J.F., A.E. Skinner. and P.R. Vaughan. 1981. “The Drained Residual Strength of Cohesive Soils.” Géotechnique 31(2): 181–213. Lytton, R.L. 1994. “Prediction of Movement in Expansive Clays.” In: Proceedings of the Settlement Conference at Texas A&M University (ASCE Geotechnical Special Publication no. 40), vol. 2, pp. 1827–1845. MacDonald, M., and Soil Mechanics. 1994. “Study of Efficiency of Site Investigation Practices.” Project Report 60. Transport Research Laboratory. Wokingham: TRRL. Machan, G., and V. Bennett. 2008. Use of Inclinometers for Geotechnical Instrumentation on Transportation Projects: State of the Practice (Transportation Research Circular No. E-C129, October 2008). Makdisi, F.I., and H.B. Seed. 1978. “Simplified Procedure for Evaluating Dam and Embankment Earthquake Induced

998

REFERENCES

Deformation.” Journal of the Geotechnical Engineering Division 104(GT7): 849–867. Manso, P.F. 2006. “The Influence of Pool Geometry and Induced Flow Patterns in Rock Scour by High-Velocity Plunging Jets.” PhD thesis, Laboratory of Hydraulic Constructions, École Polytechnique Fédérale de Lausanne, Lausanne, Switzerland. Mansur, C.I., and R.I. Kaufman. 1962. “Dewatering.” In: Foundation Engineering (Ed. G.A. Leonards). New York: McGraw-Hill. Marchetti, S. 1975. “A New In Situ Test for the Measurement of Horizontal Soil Deformability.” In: Proceedings of the Conference on In Situ Measurement of Soil Properties (ASCE Special Conference, Raleigh, NC), vol. 2, pp. 255–259. Marcuson, W.F., III,. 2001. “Construction of the Panama Canal.” In: Proceedings of the 15th International Conference on Soil Mechanics and Geotechnical Engineering, vol. 3, pp. 2375–2390. Rotterdam: Balkema. Marcuson, W.F., and A.G. Franklin. 1983. “Seismic Design, Analysis, and Remedial Measures to Improve the Stability of Existing Earth Dams—Corps of Engineers Approach.” In: Seismic Design of Embankments and Caverns (Ed. T.R. Howard). New York: ASCE. Marinos, P., and E. Hoek. 2000. “GSI—A Geologically Friendly Tool for Rock Mass Strength Estimation.” In: Proceedings of the GeoEng2000 Conference, Melbourne, Australia, pp. 1422–1442. Massarsch, K.R. 1991. “Deep Soil Compaction Using Vibratory Probes.” In: Proceedings of Symposium on Design, Construction, and Testing of Deep Foundation Improvement: Stone Columns and Related Techniques (ASTM Special Technical Publication, STP 1089) (Ed. R C. Bachus), Philadelphia, PA: ASTM, pp. 297–319. Matasovic, N. 1993. “Seismic Response of Composite Horizontally-Layered Soil Deposits.” PhD dissertation, Civil Engineering Department, University of California-Los Angeles. Mayne, P.W. 1985. “Ground Vibrations During Dynamic Compaction.” In: Vibration Problems in Geotechnical Engineering. New York: ASCE, pp. 247–265. Mayne, P.W. 2007. “Cone Penetration Testing State-of-Practice.” NCHRP Project 20-05; Task 37-14: Synthesis 368 on Cone Penetration Test, National Cooperative Highway Research Program, National Academy of Sciences, Washington, DC. Mayne, P.W., B. Christopher, R. Berg, and J. DeJong. 2002. Subsurface Investigations—Geotechnical Site Characterization (Publication No. FHWA-NH-01-031). Washington, DC: National Highway Institute, Federal Highway Administration. Mayne, P.W., B.R. Christopher, and J. DeJong. 2001, Manual on Subsurface Investigations, National Highway Institute (Publication No. FHWA NHI-01-031). Washington, DC: Federal Highway Administration. Mayne, P.W., M.R. Coop, S.M. Springman, A. Huang, and J.G. Zornberg. 2009. “Geomaterial Behavior and Testing.” In: Proceedings of the 17th International Conference on Soil Mechanics and Geotechnical Engineering, London: ISSMGE, pp. 2777–2872. Mayne, P.W., and G.J. Rix. 1993. “Gmax—qc Relationships for Clays.” Geotechnical Testing Journal 16(1): 54–60. McDowell, C. 1956. “Interrelationships of Loads, Volume Change, and Layer Thickness of Soils to the Behavior of Engineering Structures.” In: Highway Research Board, Proceedings of the

Thirty Fifth Annual Meetings (Publication No. 426), Washington, DC: Transportation Research Board, pp. 754–772. Menard, L. 1963a. “Calcul de la force portante des fondations sur la base des résultats des essais pressiométriques.” Sols-Soils 2(5 & 6). Techniques Louis Menard. Menard, L. 1963b. “Calcul de la force portante des fondations sur la base des résultats des essais pressiométriques: seconde partie, résultats expérimentaux et conclusions.” Sols-Soils 2(6), Techniques Louis Menard. Menard, L., and Y. Broise. 1975. “Theoretical and Practical Aspects of Dynamic Consolidation.” Géotechnique 25(1): 3–18. Mercalli, G. 1883. “Vulcani e fenomeni vulcanici in Italia.” In: Geologia d’Italia (Ed. G. Negri, A. Stoppani, and G. Mercalli). Milan: Vallardi, pp. 217–218. Mesri, G., and M. Abdelghafar. 1993. “Cohesion Intercept in Effective Stress Stability Analysis.” Journal of Geotechnical Engineering 99(119): 1229–1249. Mesri, G., and J.R. Funk. 2015. “Settlement of the Kansai International Airport Islands.” Journal of Geotechnical and Geoenvironmental Engineering 141(2), ASCE. Meyerhof, G.G. 1951. “The Ultimate Bearing Capacity of Foundations.” Géotechnique 2(4): 301–332. Meyerhof, G.G. 1953. “The Bearing Capacity of Foundations under Eccentric and Inclined Loads.” In: Proceedings of the 3rd International Conference on Soil Mechanics and Foundation Engineering, Zurich, vol. 1, pp. 440–445. Meyerhof, G.G. 1955. “Influence of Roughness of Base and Ground Water Conditions on the Ultimate Bearing Capacity of Foundations.” Géotechnique 5: 227–242. Mindlin, R.D. 1936. “Force at a Point in the Interior of a Semi-Infinite Solid.” Physics 7: 195. Ministère de l’équipement, du logement, et des transports. 1993. Fascicule 62. Règles techniques de conception et de calcul des foundations des ouvrages de génie civil. Paris, France: Publications Eyrolles. Mitchell, J.K. 1976. Fundamentals of Soil Behavior. New York: John Wiley & Sons. Mitchell, J.K., and R.K. Katti. 1981. “Soil Improvement—State of the Art Report.” 10th ICSMFE, Stockholm, vol. 4, pp. 509–565. Mitchell, J.K., and F.T. Madsen. 1987. “Chemical Effects on Clay Hydraulic Conductivity.” In: Geotechnical Practice for Waste Disposal ’87 (Geotechnical Special Publication No. 13). Reston, VA: ASCE, pp. 87–116. Mitchell, J.K., and K. Soga. 2005. Fundamentals of Soil Behavior (3rd ed.). Hoboken, NJ: John Wiley & Sons. Mitchell, J.M., and F.M. Jardine. 2002. A Guide to Ground Treatment. London: CIRIA. Monaco, P., S. Marchetti, G. Totani, and M. Calabrese. 2005. “Sand Liquefiability Assessment by Flat Dilatometer Test (DMT).” In: Proceedings of XVI ICSMGE, Osaka, vol. 4, pp. 2693–2697. Mononobe, N., and H. Matsuo. 1929. “On the Deformation of Earth Pressure During Earthquakes.” In: Proceedings of the World Engineering Conference, vol. 9, p. 177. Moody, L.F. 1944. “Friction Factors for Pipe Flow.” Transaction of the American Society of Civil Engineers 66: 671–684. Mooney, M.A., and P.K. Miller. 2009. “Analysis of Lightweight Deflectometer Test Based on In Situ Stress and Strain Response.” Journal of Geotechnical & Geoenvironmental Engineering 135(2): 199–208.

REFERENCES

Morgenstern, N.R. 1963. “Stability Charts for Earth Slopes During Rapid Drawdown.” Géotechnique 13(1): 121–131. Morgenstern, N.R., and V.E. Price. 1965. “The Analysis of the Stability of General Slip Surfaces.” Géotechnique 15(1): 77–93. Moseley, M.P., and K. Kirsch. 2004. Ground Improvement. Abingdon: Spon Press. Moulton L. K., V. S. G. Hota, and G. T. Halvorsen. 1985. Tolerable Movement Criteria for Highway Bridges (FHWA/RD-85/107). Washington, DC: Federal Highway Administration. Munson, B.R., D.F. Young, and T.H. Okiishi. 2021. Fundamentals of Fluid Mechanics (9th ed.). Hoboken, NJ: John Wiley and Sons, Inc. National Cooperative Highway Research Program (NCHRP). 2008. NCHRP Project 12-70: Seismic Analysis and Design of Retaining Walls, Buried Structures, Slopes and Embankments, Recommended Specifications, Commentaries and Example Problems (NCHRP Report 611). Washington, DC: Transportation Research Board. NAVFAC, 1983. “Foundations and Earth Structures.” Naval Facilities Engineering Command, Publication DM-7.02. Negussey, D. 1997. Properties and Applications of Geofoam. Washington, DC: Society of the Plastics Industry. Nenad Bicanic. 2007. “Discrete Element Methods.” In: Encyclopedia of Computational Mechanics, Vol. 1: Fundamentals (Ed. E. Stein, R. de Borst, and T.J.R. Hughes). London: John Wiley & Sons, Ltd. Newmark, N.M. 1965. “Effects of Earthquakes on Dams and Embankments.” Géotechnique 15(2): 129–160. Ng, E., J-L. Briaud, and L.M. Tucker. 1988. Field Study of Pile Group Action in Sand Research Report FHWA-RD-88-QB2). Washington, DC: FHWA. Ng, W-S., M-L. Lee, and S-L. Hii. 2012. “An Overview of the Factors Affecting Microbial-Induced Calcite Precipitation and Its Potential Application in Soil Improvement.” World Academy of Science, Engineering and Technology 62: 723–729. Norme Française AFNOR. 2013. Justification des Ouvrages Géotechniques, Normes d’Application Nationale de L’Eurocode 7, NF P 94-261: Fondations Superficielles and NF P 94-262: Fondations Profondes. Paris: Association Française de Normalisation. Nurtjahyo, P.Y. 2003. “Chimera RANS Simulations of Pier Scour and Contraction Scour in Cohesive Soils.” PhD dissertation, Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Ohta, Y., and N. Goto. 1976. “Estimation of S-Wave Velocity in Terms of Characteristic Indices of Soil.” Butsuri-Tanko 29(4): 34–41. Okabe, S. 1926. “General Theory of Earth Pressure.” Journal of the Japanese Society of Civil Engineers 12(1): 123–134. Okabe, T. 1977. “Large Negative Friction and Friction-Free Pile Methods.” In: Proceedings of the 9th International Conference on Soil Mechanics and Foundation Engineering, Tokyo, pp. 679–682. O’Kelly, B.C., P.J. Vardanega, and S.K. Haigh. 2018. “Use of Fall Cones to Determine Atterberg Limits: A Review.” Géotechnique 68(10): 843–856. Olson, S.M. 1998. “Post-Liquefaction Shear Strength from Laboratory and Field Tests: Field Tests.” In: Proceedings, Shear Strength of Liquefied Soils. (Eds. T.D. Stark, S.M. Olson, S.L. Kramer, and T.L. Youd). Urbana, IL: University of Illinois at

999

Urbana-Champaign, pp. 130–152. (Available only on the World Wide Web at http://mae.ce.uiuc.edu/info/resources.html.) Olson, S.M., and T.D. Stark. 2002. “Liquefied Strength Ratio from Liquefaction Flow Failure Case Histories.” Canadian Geotechnical Journal 39: 629–647. O’Neill, M.W. 1983. “Group Action in Offshore Piles.” Paper presented at ASCE Specialty Conference on Geotechnical Engineering in Offshore Engineering, Austin, TX. O’Neill, M.W., and L.C. Reese. 1999. Drilled Shafts: Construction Procedures and Design Methods (Publication FHWA-If-99-025). Washington, DC: Federal Highway Administration. Osterberg, J.O. 1984. “A New Simplified Method for Load Testing of Drilled Shafts.” Foundation Drilling Magazine 23(6): 9–11. Othman, M.A., R. Bonaparte, and B.A. Gross. 1997. “Preliminary Results of Study of Composite Liner Field Performance.” In: Proceedings of the GRI-10 Conference, Folsom, PA: GII Publishing, pp. 115–142. Paikowsky, S.G. 2004. Load and Resistance Factor Design (LRFD) for Deep Foundations (NCHRP Report 507). Washington, DC: Transportation Research Board, National Academy of Sciences. Paikowsky, S.G., M.C. Canniff, N. Chelmsford, K. Lesny, A. Kisse, S. Amatya, and R. Muganga. 2010. LRFD Design and Construction of Shallow Foundations for Highway Bridge Structures (NCHRP Report 651). Washington, DC: Transportation Research Board, National Academy of Sciences. Parez, L., and R. Fauriel. 1988. “Le piezocone: Améliorations apportées à la reconnaissance de sols.” Revue Française de Géotechnique 44: 13–27. Park, N. 2007. “A Prediction of Meander Migration Based on Large-Scale Flume Tests in Clay.” PhD dissertation, Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Paul, D. 2013. “Teton Dam Failure – Lessons Learned.” Paper presented at conference, Brasilia, Brazil (21–24 May 2013). PDI. 2012. CAPWAP (Case Pile Wave Analysis Program). Cleveland, OH: Pile Dynamics Inc. Peck, R.B. 1969. “Deep Excavations and Tunneling in Soft Ground” (State of the Art Report). In: Proceedings of the 7th International Conference on Soil Mechanics and Foundation Engineering, Mexico City, pp. 215–290. Peck, R.B. 1985. “The Last Sixty Years.” In: Proceedings of the XI International Conference on Soil Mechanics and Foundation Engineering, San Francisco (Golden Jubilee Volume), 123–133. Rotterdam: A. A. Balkema. Peck, R.B., and F.G. Bryant. 1953. “The Bearing-Capacity Failure of the Transcona Elevator.” Géotechnique, 3(5): 201–208. Pereira, J.H.F., and D.G. Fredlund. 2000. “Volume Change Behavior of Collapsible Compacted Gneiss Soil.” Journal of Geotechnical & Geoenvironmental Engineering. 126(10): 907–916. Perzlmaier, S. 2005. “Hydraulic Criteria for Internal Erosion in Cohesionless Soils.” In: Internal Erosion of Dams and Their Foundations (Ed. R. Fell and J.-J. Fry). London: Routledge, pp. 179–190. Philiponnat, G. 1986. “Le Phicomètre: Essai de cisaillement direct in situ.” Revue Française de Géotechnique 35: 43–60. Philipponnat, G., and M. Zerhouni. 1993. “Interprétation de l’essai au Phicomètre.” Revue Française de Géotechnique 65: 3–28. Polshin, D.E., and R.A. Tokar. 1957. “Maximum Allowable Nonuniform Settlement of Structures.” In: Proceedings of the

1000

REFERENCES

4th International Conference, Soil Mechanics and Foundation Engineering. London: Butterworth-Heinemann, pp. 402–406. Post-Tensioning Institute (PTI). 2004. Design and Construction of Post-Tensioned Slabs-on-Ground (3rd ed.). Phoenix, AZ: Post-Tensioning Institute. Potts, D.M., and J.B. Burland. 2000. “Development and Application of a Numerical Model for Simulating the Stabilisation of the Leaning Tower of Pisa.” In: Developments in Theoretical Geomechanics (Ed. D.W. Smith and J.P. Carter). Rotterdam: Balkema. Potts, D.M., and L. Zdravkovic. 1999. Finite Element Analysis in Geotechnical Engineering. London: Thomas Telford Ltd. Poulos, H.G. 1988. Marine Geotechnics. London: Unwin Hyman Ltd. Poulos, H.G. 2009. “Tall Buildings and Deep Foundations – Middle East Challenges – Hauts bâtiments et fondations profondes – challenges du Moyen Orient.” In: Proceedings of the 17th International Conference on Soil Mechanics and Geotechnical Engineering, (Ed. M. Hamza et al.). IOS Press. doi:10.3233/978-1-60750-031-5-3173. Poulos, H.G. 2012. User’s Guide to DEFPIG: Deformation Analysis of Pile Groups. Sydney, Australia: Civil Engineering Department, University of Sydney. Poulos, H.G., and G. Bunce. 2008. “Foundation Design for the Burj Dubai – The World’s Tallest Building.” In: Proceedings of the 6th International Conference on Case Histories Iin Geotechnical Engineering, Arlington, VA (August 11–16, 2008). Reston, VA: American Society of Civil Engineers. Poulos, H.G., and E.H. Davis. 1980. Pile Foundation Analysis and Design. New York: John Wiley & Sons. Priest, S.D. 1993. Discontinuity Analysis for Rock Engineering. New York: Chapman & Hall. Puzrin A.M., E.E. Alonso, and N. Pinyol. 2010. Geomechanics of Failures. Berlin: Springer. Raithel, M., A. Kirchner, C. Schade, and E. Leusink. 2005. “Foundation of Constructions on Very Soft Soils with Geotextile Encased Columns—State of the art.” Geotechnical Special Publication: Innovations in Grouting and Soil Improvement 136: 1867–1877. Randolph, M.F. 1980. PIGLET: A Computer Program for the Analysis and Design of Pile Groups under General Loading Conditions (Soil Report TR91, CUED/D). Cambridge: University of Cambridge. Randolph, M.F., and B.S. Murphy. 1985. “Shaft Capacity of Driven Piles in Clay.” In: Proceedings of the Offshore Technology Conference, Houston, TX. Paper 4883, vol. 1, pp. 371–378. Randolph, M.F., and C.P. Wroth. 1978. “Analysis of Deformation on Vertically Loaded Piles.” Journal of Soil Mechanics & Foundation Engineering 2 (GT12): 1465–1488. Raudkivi, A.J. 1998. Loose Boundary Hydraulics. Rotterdam: Balkema. Rausche, F. 2002. “Modeling of Vibratory Pile Driving.” In: Proceedings of the International Conference on Vibratory Pile Driving and Deep Soil Compaction, Transvib 2002, pp. 21–32. Reese, L.C., and M.W. O’Neill. 1988. Drilled Shafts: Construction and Design (FHWA Publication No. HI-88-042). Washington, DC: Federal Highway Administration. Reese, L.C., F. Touma, and M.W. O’Neill. 1976. “Behavior of Drilled Piers under Axial Loading.” Journal of the Geotechnical Engineering Division, ASCE 102 (GT5): 493–510.

Reissner, H. 1924. “Zum Erdruck Problem” [Concerning the Earth Pressure Problem]. In: Proceedings of the First International Congress of Applied Mechanics. Delft, The Netherlands, pp. 295–311. Richards, A. 1988. “Vane Shear Strength Testing of Soils: Field and Laboratory Studies.” ASTM STP 1014, Philadelphia, PA: American Society for Testing and Materials. Richards, A.F., and H.M. Zuidberg. 1985. “In Situ Determination of the Strength of Marine Soils.” In: Strength Testing of Marine Sediments: Laboratory and In-Situ Measurements (ASTM STP 883). West Conshohocken, PA: American Society for Testing & Materials, pp. 11–40. Richardson, E.V., and S.M. Davis. 2001. Evaluating Scour at Bridges (Publication No. FWHA-IP-90-017, Hydraulic Engineering Circular No. 18). Washington, DC: Federal Highway Administration. Richter, C.F. 1935. “An Instrumental Earthquake Magnitude Scale.” Bulletin of the Seismological Society of America 25(1): 1–32. Rix, G.J., and K.H. Stokoe. 1991. “Correlation of Initial Tangent Modulus and Cone Resistance.” In: International Symposium on Calibration Chamber Testing, New York: Elsevier, pp. 351–362. Robert, H. 1986. Robert’s Rules of Order. New York: Bantam Books. p. 16. Robertson, P.K., and R.G. Campanella. 1983. “Interpretation of Cone Penetration Tests: Sands.” Canadian Geotechnical Journal 20(4): 719–733. Robertson, P.K., R.G. Campanella, D. Gillespie, and J. Greig. 1986. “Use of Piezometer Cone Data.” In: Use of In-Situ Tests in Geotechnical Engineering (GSP 6). Reston, VA: American Society of Civil Engineers, pp. 1263–1280. Robertson, P.K., and C.E. Wride. 1998. “Evaluating Cyclic Liquefaction Potential Using the Cone Penetration Test.” Canadian Geotechnical Journal 35: 442–459. Rogers. J.D. 2011. “Landslides of the Panama Canal.” Paper presented at the Pacific Section Meeting of the American Association for the Advancement of Science, San Diego, California. Available at: https://web.mst.edu/∼rogersda/ Panama-Canal-2012/Rogers-Landslides-Panama-Canalminimum.pdf Roscoe, K.H., and J.B. Burland. 1968. “On the Generalized Behaviour of ‘Wet’ Clay.” Engineering Plasticity 48: 535–609. Roscoe, K.H., A.N. Schofield, and C.P. Wroth. 1958. “On the Yielding of Soils.” Géotechnique 8(1): 22–52. Rutherford, C., G. Biscontin, and J-L. Briaud. 2005. “Design Manual for Excavation Support Using Deep Mixing Technology” (Internal Report). Zachry Department of Civil Engineering, Texas A&M University, College Station, TX. Sadek, S., and G. Khoury. 2000. “Soil and Site Improvement Guide: An Educational Tool for Engineered Ground Modification.” International Journal of Engineering Education 16(6): 499–508. Safir, L., and W. Safire. 1982. Good Advice. New York: Times Books. Samson, C.H., T.J. Hirsch, and L.L. Lowery. 1963. “Computer Study of Dynamic Behavior of Pilings.” Journal of the Structural Division, ASCE 89 (ST4), Proceedings Paper 3608. Sandström, Å. 1994. “Numerical Simulation of a Vibratory Roller on Cohesionless Soil.” (Internal Report). Stockholm, Sweden: Geodynamik.

REFERENCES

Sankey, J.E., M.J. Bailey, and B. Chen. 2008. “SeaTac Third Runway: Design and Performance of MSE Tall Wall.” In: New Horizons in Earth Reinforcement (Ed. J. Otani, Y. Miyata, and T. Mukunoki). London: Routledge. Sarma, S.K. 1973. “Stability Analysis of Embankments and Slopes.” Géotechnique 23(3): 423–433. Sarsby, R.W. 1985. “The Influence of Aperture Size and Particle Size on the Efficiency of Grid Reinforcement.” In: Proceedings of the 2nd Canadian Symposium on Geotextiles and Geomembranes, Edmonton, Canada. The Geotechnical Society of Edmonton, pp. 7–12. Saye, S.R., J.C. Volk, and P.C. Gerhart. 2000. “Design-Built I-95 Highway Reconstruction.” GeoStrata, Geo-Institute, ASCE. Schaefer, V.R., and R.R. Berg. 2013. “SHRP 2 R02: Geotechnical Solutions for Soil Improvement, Rapid Embankment Construction, and Stabilization of the Pavement Working Platform—Final Phase 2 Summary Report” (The Strategic Highway Research Program 2). Transportation Research Board of The National Academies. Web-based system. Available at: www.GeoTechTools.org Schalkoff, R.J. 1997. Artificial Neural Networks. New York: McGraw-Hill. Schmertmann, J.H. 1970. “Static Cone to Compute Static Settlement over Sand.” Journal of Soil Mechanics & Foundations Division 96(3): 1011–1043. Schmertmann, J.H. 1975. “Measurements of In Situ Strength.” In: Proceedings of the ASCE Specialty Conference on In Situ Measurements of Soil Properties, vol. 2, pp. 57–138. Schmertmann, J.H., J.P. Hartman, and P.R. Brown. 1978. “Improved Strain Influence Factor Diagram.” Journal of the Geotechnical Engineering Division 104: 1134. Schnabel, P.B., J. Lysmer, and H.B. Seed. 1972. SHAKE: A Computer Program for Earthquake Response Analysis of Horizontally Layered Sites (Report No. EERC 72-12). Berkeley, CA: Earthquake Engineering Research Center, University of California. Schofield, A.N., and C.P. Wroth. 1968. Critical State Soil Mechanics. New York: McGraw-Hill. Schofield, R.K. 1935. “The pF of the Water in Soil.” In: Transactions of the 3rd International Congress of Soil Science, vol. 2, pp. 37–48. Schwartz, D.P., and K.J. Coppersmith. 1984. “Fault Behavior and Characteristic Earthquakes: Examples from the Wasatch and San Andreas Fault Zones.” Journal of Geophysical Research 89(B7): 5681–5698. Seed, H.B., and I.M. Idriss. 1970. Soil Moduli and Damping Factors for Dynamic Response Analyses (Earthquake Engineering Research Center, Report No. EERC 70-10). Berkeley, CA: University of California. Seed, H.B., and I.M. Idriss. 1971. “Simplified Procedure for Evaluating Soil Liquefaction Potential.” Journal of the Soil Mechanics & Foundation Division, ASCE 107 (SM9): 1229–1274. Seed, H.B., R.B. Seed, F. Schlosser, F. Blondeau, and I. Juran. 1988. “The Landslide at the Port of Nice on October 16, 1979.” Earthquake Engineering Research Center, Report No. UCB/EERC-88/10. Seed, H.B., R.T. Wong, I.M. Idriss, and K. Tokimatsu. 1986. “Moduli and Damping Factors for Dynamic Analyses of

1001

Cohesionless Soils.” Journal of Geotechnical Engineering 112 (GT11): 1016–1032. Seed, R.B., and L.F. Harder. 1990. “SPT-Based Analysis of Cyclic Pore Pressure Generation and Undrained Residual Strength.” In: H. B. Seed Memorial Symposium, vol. 2, pp. 351–376. Berkeley, CA: BiTech Publishing, Ltd. Shackelford, C.D. 1994. “Waste-Soil Interactions That Alter Hydraulic Conductivity.” In: Hydraulic Conductivity and Waste Contaminant Transport in Soil (ASTM STP 1142) (Ed. S.J. Trautwein and E.D. Daniel). Philadelphia, PA: ASTM. Shackelford, C.D., C.H. Benson, T. Katsumi, T.B. Edil, and L. Lin. 2000. “Evaluating the Hydraulic Conductivity of GCLs Permeated with Non-Standard Liquids.” Geotextiles & Geomembranes 18: 133–161. Sharma, H.D. 2000. “Solid Waste Landfills: Settlements and Post-Closure Perspectives.” In: Proceedings of the ASCE National Conference on Environmental and Pipeline Engineering (Ed. R.Y. Surampali), pp. 447–455. Sharma, H.D., D.E. Hullings, and F.R. Greguras. 1997. “Interface Strength Tests and Application to Landfill Design.” In: Proceedings of the Conference on Geosynthetics ’97, Longbeach, CA: IFAI, pp. 913–924. Sharma, H.D., and K.R. Reddy. 2004. Geoenvironmental Engineering. Hoboken, NJ: John Wiley & Sons. Sherard, J.L. 1979. “Sinkholes in Dams of Coarse, Broadly Graded Soils.” In: Proceedings of the 13th International Congress on Large Dams, New Delhi, vol. 2, pp. 23–35. Paris: International Commission on Large Dams. Shidlovskaya, A. 2021. “Soil Erosion and its Mitigation: Improvements in Testing and Modelling.” PhD Dissertation, Zachry Department of Civil and Environmental Engineering, Texas A&M University, College Station, TX. Shidlovskaya, A., J-L. Briaud, and M. Mohammadrajadi. 2017. “St. Isaac’s Cathedral (St. Petersburg, Russia): Case History.” International Journal of Geoengineering Case Histories, 4(2): 113–133. Shields, A. 1936. “Anwendung der Ähnlichkeitsmechanik und der Turbulenzforschung auf die Geschiebebewegung.” Doktor-Ingenieurs dissertation, Technischen Hochschule, Berlin [in German]. Sjoberg, J. 1997. Estimating Rock Mass Strength Using the Hoek-Brown Failure Criterion and Rock Mass Classification: A Review and Application to the Aznalcollar Open Pit (Internal Report BM 1997-02). Luleå, Sweden: Luleå University of Technology. Skempton, A.W. 1951. “The Bearing Capacity of Clays.” In: Proceedings of the Building Research Congress, vol. 1, pp. 180–189. Skempton, A.W. 1954. “The Pore Pressure Coefficients A and B.” Géotechnique 4(4): 143–147. Skempton, A.W. 1985. “A History of Soil Properties, 1717–1927.” In: Proceedings, XI International Conference on Soil Mechanics and Foundation Engineering, San Francisco (Golden Jubilee Volume). Rotterdam: A. A. Balkema, pp. 95–121. Skempton, A.W., and D.H. MacDonald. 1956. “The Allowable Settlement of Buildings.” Proceeding of the Institution of Civil Engineers 3(5): 727–768. Smith, E.A.L. 1960. “Pile Driving Analysis by the Wave Equation.” Journal of Soil Mechanics and Foundation Engineering, SM4, 86: 35–61 (Discussions in 87: 63–75).

1002

REFERENCES

Smith, M.E. 2005. “Design of Bridging Layers in Geosynthetic Reinforced Column Supported Embankments.” PhD dissertation, Virginia Tech. Sorrell, C.A., and G.F. Sandström. 2001. Rocks and Minerals: A Guide to Field Identification. New York: Macmillan. Sotnikov, S.N. 1986. “Settlement of St. Isaac’s Cathedral: Design of Foundation on Soft Soils. Leningrad.” (Sotnikov S.N. . mevuzovski tematiqeski sbornik trudov: