Geotechnical Engineering Unsaturated and Saturated Soils

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Geotechnical Engineering: Unsaturated and Saturated Soils

Geotechnical Engineering: Unsaturated and Saturated Soils Jean-Louis Briaud

Cover image: C Art Koenig, Photographer/Artist Cover design: Wiley This book is printed on acid-free paper. Copyright C 2013 by John Wiley & Sons, Inc. All rights reserved Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada 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, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600, 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 www.wiley.com/go/permissions. 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 the author shall be liable for damages arising herefrom. For general information about our other products and services, 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 publishes in a variety of print and electronic formats and by print-on-demand. Some material included with standard print versions of this book may not be included in e-books or in print-on-demand. If this book refers to media such as a CD or DVD that is not included in the version you purchased, you may download this material at http://booksupport.wiley.com. For more information about Wiley products, visit www.wiley.com. Library of Congress Cataloging-in-Publication Data: Briaud, J.-L. Introduction to geotechnical engineering : unsaturated and saturated soils / Jean-Louis Briaud. pages cm ‘‘Published simultaneously in Canada’’—Title page verso. Includes bibliographical references and index. ISBN 978-0-470-94856-9 (cloth : acid-free paper); 978-1-118-41574-0 (ebk.); 978-1-118-41826-0 (ebk.) 1. Geotechnical engineering–Textbooks. 2. Soil mechanics–Textbooks. I. Title. TA705.B75 2013 624–dc23 2013004684 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

CONTENTS Acknowledgments

xxi

CHAPTER 1

Introduction 1.1 Why This Book? 1.2 Geotechnical Engineering 1.3 The Past and the Future 1.4 Some Recent and Notable Projects 1.5 Failures May Occur 1.6 Our Work Is Buried 1.7 Geotechnical Engineering Can Be Fun 1.8 Units Problems Problems and Solutions

1 1 1 2 2 5 5 5 5 10 11

CHAPTER 2

Engineering Geology 2.1 Definition 2.2 The Earth 2.3 Geologic Time 2.4 Rocks 2.5 Soils 2.6 Geologic Features 2.7 Geologic Maps 2.8 Groundwater Problems Problems and Solutions

15 15 15 15 17 17 19 20 20 22 22

CHAPTER 3

Soil Components and Weight-Volume Parameters 3.1 Particles, Liquid, and Gas 3.2 Particle Size, Shape, and Color 3.3 Composition of Gravel, Sand, and Silt Particles 3.4 Composition of Clay and Silt Particles 3.5 Particle Behavior 3.6 Soil Structure 3.7 Three-Phase Diagram 3.8 Weight-Volume Parameters 3.9 Measurement of the Weight-Volume Parameters 3.10 Solving a Weight-Volume Problem Problems Problems and Solutions

26 26 26 28 28 29 30 30 31 32 33 34 35 v

vi

CONTENTS

CHAPTER 4

Soil Classification 4.1 Sieve Analysis 4.2 Hydrometer Analysis 4.3 Atterberg Limits and Other Limits 4.4 Classification Parameters 4.5 Engineering Significance of Classification Parameters and Plasticity Chart 4.6 Unified Soil Classification System Problems Problems and Solutions

46 46 47 50 53

CHAPTER 5

Rocks 5.1 Rock Groups and Identification 5.2 Rock Mass vs. Rock Substance 5.3 Rock Discontinuities 5.4 Rock Index Properties 5.5 Rock Engineering Properties 5.6 Rock Mass Rating 5.7 Rock Engineering Problems 5.8 Permafrost Problems Problems and Solutions

63 63 63 66 66 67 68 69 71 72 74

CHAPTER 6

Site Investigation, Drilling, and Sampling 6.1 General 6.2 Preliminary Site Investigation 6.3 Number and Depth of Borings and In Situ Tests 6.4 Drilling 6.4.1 Wet Rotary Drilling Method 6.4.2 Hollow Stem Auger Drilling Method 6.5 Sampling 6.5.1 Sample Disturbance 6.5.2 Common Sampling Methods 6.6 Groundwater Level 6.7 Field Identification and Boring Logs 6.8 Soil Names 6.9 Offshore Site Investigations 6.9.1 Offshore Geophysical Investigations 6.9.2 Offshore Geotechnical Drilling 6.9.3 Offshore Geotechnical Sampling Problems Problems and Solutions

80 80 80 80 81 81 82 83 83 84 85 87 88 89 94 95 99 99 100

CHAPTER 7

In Situ Tests 7.1 Standard Penetration Test 7.2 Cone Penetration Test 7.3 Pressuremeter Test 7.4 Dilatometer Test 7.5 Vane Shear Test

104 104 107 111 114 115

55 55 56 57

CONTENTS

7.6 7.7 7.8 7.9 7.10 7.11

7.12

7.13

CHAPTER 8

CHAPTER 9

Borehole Shear Test Plate Load Test California Bearing Ratio Test Pocket Penetrometer and Torvane Tests Pocket Erodometer Test Compaction Control Tests 7.11.1 Sand Cone Test 7.11.2 Rubber Balloon Test 7.11.3 Nuclear Density/Water Content Test 7.11.4 Field Oven Test 7.11.5 Lightweight Deflectometer Test 7.11.6 BCD Test Hydraulic Conductivity Field Tests 7.12.1 Borehole Tests 7.12.2 Cone Penetrometer Dissipation Test 7.12.3 Sealed Double-Ring Infiltrometer Test 7.12.4 Two-Stage Borehole Permeameter Test Offshore In Situ Tests Problems Problems and Solutions

vii 117 119 122 122 123 124 124 124 125 125 126 126 127 127 129 130 131 132 135 136

Elements of Geophysics 8.1 General 8.2 Seismic Techniques 8.2.1 Seismic Waves 8.2.2 Seismic Reflection 8.2.3 Seismic Refraction 8.2.4 Cross Hole Test, Seismic Cone Test, and Seismic Dilatometer Test 8.2.5 Spectral Analysis of Surface Waves 8.3 Electrical Resistivity Techniques 8.3.1 Background on Electricity 8.3.2 Resistivity Tomography 8.4 Electromagnetic Methods 8.4.1 Electromagnetic Waves 8.4.2 Ground-Penetrating Radar 8.4.3 Time Domain Reflectometry 8.5 Remote Sensing Techniques 8.5.1 LIDAR 8.5.2 Satellite Imaging Problems Problems and Solutions

151 151 151 151 153 154

Laboratory Tests 9.1 General 9.2 Measurements 9.2.1 Normal Stress or Pressure 9.2.2 Shear Stress 9.2.3 Water Compression Stress 9.2.4 Water Tension Stress 9.2.5 Normal Strain

172 172 172 172 172 173 173 179

155 156 160 160 160 161 161 162 162 165 165 165 166 166

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CONTENTS

9.3

9.4

9.5

9.6

9.7

9.8

9.9

9.10

9.11

9.12

9.13

9.14

9.15

9.16

9.17 9.18 9.19 9.20

9.2.6 Shear Strain 9.2.7 Bender Elements Compaction Test: Dry Unit Weight 9.3.1 Saturated Soils 9.3.2 Unsaturated Soils Compaction Test: Soil Modulus 9.4.1 Saturated Soils 9.4.2 Unsaturated Soils Consolidation Test 9.5.1 Saturated Soils 9.5.2 Unsaturated Soils Swell Test 9.6.1 Saturated Soils 9.6.2 Unsaturated Soils Shrink Test 9.7.1 Saturated Soils 9.7.2 Unsaturated Soils Collapse Test 9.8.1 Saturated Soils 9.8.2 Unsaturated Soils Direct Shear Test 9.9.1 Saturated Soils 9.9.2 Unsaturated Soils Simple Shear Test 9.10.1 Saturated Soils 9.10.2 Unsaturated Soils Unconfined Compression Test 9.11.1 Saturated Soils 9.11.2 Unsaturated Soils Triaxial Test 9.12.1 Saturated Soils 9.12.2 Unsaturated Soils Resonant Column Test 9.13.1 Saturated Soils 9.13.2 Unsaturated Soils Lab Vane Test 9.14.1 Saturated Soils 9.14.2 Unsaturated Soils Soil Water Retention Curve (Soil Water Characteristic Curve) Test 9.15.1 Saturated Soils 9.15.2 Unsaturated Soils Constant Head Permeameter Test 9.16.1 Saturated Soils 9.16.2 Unsaturated Soils Falling Head Permeameter Test for Saturated Soils Wetting Front Test for Unsaturated Soils Air Permeability Test for Unsaturated Soils Erosion Test 9.20.1 Saturated Soils 9.20.2 Unsaturated Soils Problems Problems and Solutions

180 180 181 181 181 184 184 184 185 185 190 190 190 191 192 192 192 193 193 193 193 193 195 195 195 196 196 196 197 198 198 199 202 202 204 206 206 206 206 206 208 209 209 211 212 213 214 215 215 217 218 221

CONTENTS

CHAPTER 10

CHAPTER 11

Stresses, Effective Stress, Water Stress, Air Stress, and Strains 10.1 General 10.2 Stress Vector, Normal Stress, Shear Stress, and Stress Tensor 10.3 Sign Convention for Stresses and Strains 10.4 Calculating Stresses on Any Plane: Equilibrium Equations for Two-Dimensional Analysis 10.5 Calculating Stresses on Any Plane: Mohr Circle for Two-Dimensional Analysis 10.6 Mohr Circle in Three Dimensions 10.7 Stress Invariants 10.8 Displacements 10.9 Normal Strain, Shear Strain, and Strain Tensor 10.10 Cylindrical Coordinates and Spherical Coordinates 10.11 Stress-Strain Curves 10.12 Stresses in the Three Soil Phases 10.13 Effective Stress (Unsaturated Soils) 10.14 Effective Stress (Saturated Soils) 10.15 Area Ratio Factors α and β 10.16 Water Stress Profiles 10.17 Water Tension and Suction 10.17.1 Matric Suction 10.17.2 Contractile Skin 10.17.3 Osmotic Suction 10.17.4 Relationship between Total Suction and Relative Humidity 10.17.5 Trees 10.18 Precision on Water Content and Water Tension 10.19 Stress Profile at Rest in Unsaturated Soils 10.20 Soil Water Retention Curve 10.21 Independent Stress State Variables Problems Problems and Solutions Problem-Solving Methods 11.1 General 11.2 Drawing to Scale as a First Step 11.3 Primary Laws 11.4 Continuum Mechanics Methods 11.4.1 Solving a Failure Problem: Limit Equilibrium, Method of Characteristics, Lower and Upper Bound Theorems 11.4.2 Examples of Solving a Failure Problem 11.4.3 Solving a Deformation Problem 11.4.4 Example of Solving a Deformation Problem 11.4.5 Solving a Flow Problem 11.4.6 Example of Solving a Flow Problem 11.5 Numerical Simulation Methods 11.5.1 Finite Difference Method 11.5.2 Examples of Finite Difference Solutions

ix 245 245 245 246 246 247 248 248 249 249 250 251 251 252 253 253 254 255 256 257 258 258 260 260 260 262 264 264 267 280 280 280 280 281

281 281 283 283 286 286 289 289 291

x

CONTENTS

CHAPTER 12

CHAPTER 13

11.5.3 Finite Element Method 11.5.4 Example of Finite Element Solution 11.5.5 Boundary Element Method 11.5.6 Discrete Element Method 11.6 Probability and Risk Analysis 11.6.1 Background 11.6.2 Procedure for Probability Approach 11.6.3 Risk and Acceptable Risk 11.6.4 Example of Probability Approach 11.7 Regression Analysis 11.8 Artificial Neural Network Method 11.9 Dimensional Analysis 11.9.1 Buckingham  Theorem 11.9.2 Examples of Dimensional Analysis 11.10 Similitude Laws for Experimental Simulations 11.10.1 Similitude Laws 11.10.2 Example of Similitude Laws Application for a Scaled Model 11.10.3 Example of Similitude Laws Application for a Centrifuge Model 11.11 Types of Analyses (Drained–Undrained, Effective Stress–Total Stress, Short-Term–Long-Term) Problems Problems and Solutions

294 300 304 304 305 305 308 310 312 313 314 315 315 316 317 317

Soil Constitutive Models 12.1 Elasticity 12.1.1 Elastic Model 12.1.2 Example of Use of Elastic Model 12.2 Linear Viscoelasticity 12.2.1 Simple Models: Maxwell and Kelvin-Voigt Models 12.2.2 General Linear Viscoelasticity 12.3 Plasticity 12.3.1 Some Yield Functions and Yield Criteria 12.3.2 Example of Use of Yield Criteria 12.3.3 Plastic Potential Function and Flow Rule 12.3.4 Hardening or Softening Rule 12.3.5 Example of Application of Plasticity Method 12.4 Common Models 12.4.1 Duncan-Chang Hyperbolic Model 12.4.2 Modified Cam Clay Model 12.4.3 Barcelona Basic Model 12.4.4 Water Stress Predictions Problems Problems and Solutions

345 345 345 346 347

Flow of Fluid and Gas Through Soils 13.1 General 13.2 Flow of Water in a Saturated Soil 13.2.1 Discharge Velocity, Seepage Velocity, and Conservation of Mass 13.2.2 Heads

370 370 370

317 317 319 319 321

347 348 349 350 351 352 352 353 353 353 354 355 357 357 358

370 371

CONTENTS

13.2.3 13.2.4 13.2.5 13.2.6

13.3

CHAPTER 14

Hydraulic Gradient Darcy’s Law: The Constitutive Law Hydraulic Conductivity Field vs. Lab Values of Hydraulic Conductivity 13.2.7 Seepage Force 13.2.8 Quick Sand Condition and Critical Hydraulic Gradient 13.2.9 Quick Clay 13.2.10 Sand Liquefaction 13.2.11 Two-Dimensional Flow Problem 13.2.12 Drawing a Flow Net for Homogeneous Soil 13.2.13 Properties of a Flow Net for Homogeneous Soil 13.2.14 Calculations Associated with Flow Nets 13.2.15 Flow Net for Hydraulically Anisotropic Soil 13.2.16 Flow and Flow Net for Layered Soils Flow of Water and Air in Unsaturated Soil 13.3.1 Hydraulic Conductivity for Water and for Air 13.3.2 One-Dimensional Flow 13.3.3 Three-Dimensional Water Flow 13.3.4 Three-Dimensional Air Flow Problems Problems and Solutions

Deformation Properties 14.1 Modulus of Deformation: General 14.2 Modulus: Which One? 14.3 Modulus: Influence of State Factors 14.4 Modulus: Influence of Loading Factor 14.5 Modulus: Differences Between Fields of Application 14.6 Modulus, Modulus of Subgrade Reaction, and Stiffness 14.7 Common Values of Young’s Modulus and Poisson’s Ratio 14.8 Correlations with Other Tests 14.9 Modulus: A Comprehensive Model 14.10 Initial Tangent Modulus Go or Gmax 14.11 Reduction of Gmax with Strain: The G/Gmax Curve 14.12 Preconsolidation Pressure and Overconsolidation Ratio from Consolidation Test 14.13 Compression Index, Recompression Index, and Secondary Compression Index from Consolidation Test 14.14 Time Effect from Consolidation Test 14.15 Modulus, Time Effect, and Cyclic Effect from Pressuremeter Test 14.16 Resilient Modulus for Pavements 14.17 Unsaturated Soils: Effect of Drying and Wetting on the Modulus 14.18 Shrink-Swell Deformation Behavior, Shrink-Swell Modulus

xi 371 371 372 373 373 374 375 375 375 377 379 379 380 381 382 382 384 386 387 388 391 401 401 402 402 403 405 405 406 408 408 411 412 413 415 416 418 419 420 422

xii

CONTENTS

CHAPTER 15

CHAPTER 16

14.19 Collapse Deformation Behavior Problems Problems and Solutions

424 426 429

Shear Strength Properties 15.1 General 15.2 Basic Experiments 15.2.1 Experiment 1 15.2.2 Experiment 2 15.2.3 Experiment 3 15.2.4 Experiment 4 15.2.5 Experiment 5 15.2.6 Experiment 6 15.3 Stress-Strain Curve, Water Stress Response, and Stress Path 15.4 Shear Strength Envelope 15.4.1 General Case 15.4.2 The Case of Concrete 15.4.3 Overconsolidated Fine-Grained Soils 15.4.4 Coarse-Grained Soils 15.5 Unsaturated Soils 15.6 Experimental Determination of Shear Strength (Lab Tests, In Situ Tests) 15.7 Estimating Effective Stress Shear Strength Parameters 15.7.1 Coarse-Grained Soils 15.7.2 Fine-Grained Soils 15.8 Undrained Shear Strength of Saturated Fine-Grained Soils 15.8.1 Weak Soil Skeleton: Soft, Normally Consolidated Soils 15.8.2 Strong Soil Skeleton: Overconsolidated Soils 15.8.3 Rate of Loading Effect on the Undrained Strength ′ and the SHANSEP Method 15.9 The Ratio su /σov 15.10 Undrained Shear Strength for Unsaturated Soils 15.11 Pore-Pressure Parameters A and B 15.12 Estimating Undrained Shear Strength Values 15.13 Residual Strength Parameters and Sensitivity 15.14 Strength Profiles 15.15 Types of Analyses 15.16 Transformation from Effective Stress Solution to Undrained Strength Solution Problems Problems and Solutions

443 443 443 443 444 444 444 444 445

Thermodynamics for Soil Problems 16.1 General 16.2 Definitions 16.3 Constitutive and Fundamental Laws 16.4 Heat Conduction Theory

445 447 447 448 448 448 449 450 451 451 453 454 454 455 456 456 458 458 459 461 462 463 463 464 465 472 472 472 473 473

CONTENTS

16.5 16.6 16.7 16.8 16.9

CHAPTER 17

Axisymmetric Heat Propagation Thermal Properties of Soils Multilayer Systems Applications Frozen Soils Problems Problems and Solutions

Shallow Foundations 17.1 Definitions 17.2 Case History 17.3 Definitions and Design Strategy 17.4 Limit States, Load and Resistance Factors, and Factor of Safety 17.5 General Behavior 17.6 Ultimate Bearing Capacity 17.6.1 Direct Strength Equations 17.6.2 Terzaghi’s Ultimate Bearing Capacity Equation 17.6.3 Layered Soils 17.6.4 Special Loading 17.6.5 Ultimate Bearing Capacity of Unsaturated Soils 17.7 Load Settlement Curve Approach 17.8 Settlement 17.8.1 General Behavior 17.8.2 Elasticity Approach for Homogeneouss Soils 17.8.3 Elasticity Approach for Layered Soils 17.8.4 Chart Approach 17.8.5 General Approach 17.8.6 Zone of Influence 17.8.7 Stress Increase with Depth 17.8.8 Choosing a Stress-Strain Curve and Setting Up the Calculations 17.8.9 Consolidation Settlement: Magnitude 17.8.10 Consolidation Settlement: Time Rate 17.8.11 Creep Settlement 17.8.12 Bearing Pressure Values 17.9 Shrink-Swell Movement 17.9.1 Water Content or Water Tension vs. Strain Curve 17.9.2 Shrink-Swell Movement Calculation Methods 17.9.3 Step-by-Step Procedure 17.9.4 Case History 17.10 Foundations on Shrink-Swell Soils 17.10.1 Types of Foundations on Shrink-Swell Soils 17.10.2 Design Method for Stiffened Slabs on Grade 17.11 Tolerable Movements

xiii 474 475 476 477 478 479 480

485 485 485 485 488 491 491 491 494 496 498 499 500 502 502 504 504 506 507 507 508 510 510 511 511 513 513 513 514 514 516 517 517 518 522

xiv

CONTENTS

CHAPTER 18

17.12 Large Mat Foundations 17.12.1 General Principles 17.12.2 Example of Settlement Calculations 17.12.3 Two Case Histories Problems Problems and Solutions

523 523 524 527 529 531

Deep Foundations 18.1 Different Types of Deep Foundations 18.2 Design Strategy 18.3 Pile Installation 18.3.1 Installation of Bored Piles 18.3.2 Nondestructive Testing of Bored Piles 18.3.3 Installation of Driven Piles 18.3.4 Pile Driving Formulas 18.3.5 Wave Propagation in a Pile 18.3.6 Wave Equation Analysis 18.3.7 Information from Pile Driving Measurements (PDA, Case, CAPWAP) 18.3.8 Suction Caissons 18.3.9 Load Testing (Static, Statnamic, Osterberg) 18.4 Vertical Load: Single Pile 18.4.1 Ultimate Vertical Capacity for a Single Pile 18.4.2 Miscellaneous Questions about the Ultimate Capacity of a Single Pile 18.4.3 Settlement of a Single Pile 18.5 Vertical Load: Pile Group 18.5.1 Ultimate Vertical Capacity of a Pile Group 18.5.2 Settlement of Pile Groups 18.6 Downdrag 18.6.1 Definition and Behavior 18.6.2 Downdrag on a Single Pile 18.6.3 Sample Downdrag Calculations 18.6.4 LRFD Provisions 18.6.5 Downdrag on a Group of Piles 18.7 Piles in Shrink-Swell Soils 18.7.1 The Soil Shrinks 18.7.2 The Soil Swells 18.8 Horizontal Load and Moment: Single Pile 18.8.1 Definitions and Behavior 18.8.2 Ultimate Capacity 18.8.3 Displacement and Maximum Moment: Long Flexible Pile 18.8.4 Displacement and Maximum Moment: Short Rigid Pile 18.8.5 Modulus of Subgrade Reaction 18.8.6 Free-Head and Fixed-Head Conditions 18.8.7 Rate of Loading Effect 18.8.8 Cyclic Loading Effect 18.8.9 P-y Curve Approach 18.8.10 Horizontal Loading Next to a Trench

553 553 553 555 555 558 560 561 562 563 566 570 571 575 575 580 584 587 587 589 592 592 592 593 595 596 597 597 598 598 598 599 599 601 602 602 603 604 605 606

CONTENTS

CHAPTER 19

xv

18.9

Horizontal Load and Moment: Pile Group 18.9.1 Overturning Moment 18.9.2 Ultimate Capacity 18.9.3 Movement 18.10 Combined Piled Raft Foundation Problems Problems and Solutions

606 607 607 609 609 612 616

Slope Stability 19.1 General 19.2 Design Approach 19.3 Infinite Slopes 19.3.1 Dry Sand 19.3.2 Dry c′ − ϕ ′ Soil 19.3.3 c′ − ϕ ′ Soil with Seepage 19.3.4 c′ − ϕ ′ Soil with Unsaturated Conditions 19.4 Seepage Force in Stability Analysis 19.5 Plane Surfaces 19.6 Block Analysis 19.7 Slopes with Water in Tensile Cracks 19.8 Chart Methods 19.8.1 Taylor Chart 19.8.2 Spencer Chart 19.8.3 Janbu Chart 19.8.4 Morgenstern Chart 19.9 Method of Slices 19.9.1 Ordinary Method of Slices 19.9.2 Bishop Simplified Method 19.9.3 Generalized Equilibrium Method 19.9.4 Critical Failure Circle 19.10 Water Stress for Slope Stability 19.10.1 Piezometric and Phreatic Surface 19.10.2 Water Stress Ratio Value 19.10.3 Grid of Water Stress Values 19.10.4 Water Stress Due to Loading 19.10.5 Seepage Analysis 19.11 Types of Analyses 19.12 Progressive Failure in Strain-Softening Soils 19.13 Shallow Slide Failures in Compacted Unsaturated Embankments 19.14 Reinforced Slopes 19.14.1 Reinforcement Type 19.14.2 Factor of Safety 19.15 Probabilistic Approach 19.15.1 Example 1 19.15.2 Example 2 19.15.3 Example 3 19.16 Three-Dimensional Circular Failure Analysis 19.17 Finite Element Analysis

649 649 649 650 651 651 651 652 652 654 654 654 655 655 657 658 659 661 662 664 665 667 667 667 668 668 668 668 668 669 669 670 670 670 671 671 671 672 672 674

xvi

CONTENTS

CHAPTER 20

CHAPTER 21

19.18 Seismic Slope Analysis 19.18.1 Pseudostatic Method 19.18.2 Newmark’s Displacement Method 19.18.3 Postearthquake Stability Analysis 19.18.4 Dynamic Finite Element Analysis 19.19 Monitoring 19.20 Repair Methods 19.20.1 Increase the Resisting Moment 19.20.2 Decrease the Driving Moment Problems Problems and Solutions

674 674 675 676 676 676 679 679 680 680 682

Compaction 20.1 General 20.2 Compaction Laboratory Tests 20.3 Compaction Field Tests 20.4 Compaction and Soil Type 20.5 Intelligent Roller Compaction 20.5.1 Soil Modulus from Vibratory Rollers 20.5.2 Roller Measurements as Compaction Indices 20.6 Impact Roller Compaction 20.7 Dynamic or Drop-Weight Compaction Problems Problems and Solutions

698 698 698 700 701 701 704

Retaining Walls 21.1 Different Types (Top-Down, Bottom-Up) 21.2 Active, At Rest, Passive Earth Pressure, and Associated Displacement 21.3 Earth Pressure Theories 21.3.1 Coulomb Earth Pressure Theory 21.3.2 Rankine Earth Pressure Theory 21.3.3 Earth Pressure Theory by Mohr Circle 21.3.4 Water in the Case of Compression Stress (Saturated) 21.3.5 Water in the Case of Tension Stress (Unsaturated or Saturated) 21.3.6 Influence of Surface Loading (Line Load, Pressure) 21.3.7 General Case and Earth Pressure Profiles 21.4 Special Case: Undrained Behavior of Fine-Grained Soils 21.5 At-Rest Earth Pressure 21.6 Earth Pressure Due to Compaction 21.7 Earth Pressures in Shrink-Swell Soils 21.8 Displacements 21.9 Gravity Walls 21.10 Mechanically Stabilized Earth Walls 21.10.1 External Stability 21.10.2 Internal Stability

705 706 707 710 710 716 716 716 717 717 719 720 722 722 722 723 723 724 725 726 726 727 729 729 730

CONTENTS

21.11 Cantilever Top-Down Walls 21.11.1 Depth of Embedment and Pressure Diagram 21.11.2 Displacement of the Wall, Bending Moment, and P-y Curves 21.12 Anchored Walls and Strutted Walls 21.12.1 Pressure Distribution 21.12.2 Pressure vs. Movement 21.12.3 Base Instability 21.12.4 Movement of Wall and Ground Surface 21.12.5 Anchors 21.12.6 Embedment Depth and Downdrag 21.12.7 P-y Curve Approach and FEM Approach 21.13 Soil Nail Walls 21.13.1 External Stability 21.13.2 Internal Stability 21.13.3 Wall Movement 21.13.4 Other Issues 21.14 Special Case: Trench Problems Problems and Solutions CHAPTER 22

Earthquake Geoengineering 22.1 Background 22.2 Earthquake Magnitude 22.3 Wave Propagation 22.4 Dynamic Soil Properties 22.5 Ground Motion 22.6 Seismic Hazard Analysis 22.7 Ground Response Analysis 22.7.1 One-Dimensional Solution for Undamped Linear Soil on Rigid Rock 22.7.2 One-Dimensional Solution for Damped Linear Soil on Rigid Rock 22.7.3 Layered Soils 22.8 Design Parameters 22.8.1 Site Classes A–E for Different Soil Stiffness 22.8.2 Code-Based Spectrum 22.8.3 Hazard Levels 22.9 Liquefaction 22.9.1 Phenomenon 22.9.2 When to Do a Liquefaction Study? 22.9.3 When Can a Soil Liquefy? 22.10 Seismic Slope Stability 22.11 Seismic Design of Retaining Walls 22.11.1 Seismic Design of Gravity Walls 22.11.2 Water Pressures on Walls during Earthquake 22.11.3 Seismic Design of MSE Walls 22.11.4 Seismic Design of Cantilever Walls 22.11.5 Seismic Design of Anchored Walls

xvii 732 733 733 735 736 737 738 739 740 742 745 746 746 748 751 751 751 752 754 784 784 784 786 786 786 789 792 792 793 793 794 795 795 797 797 797 797 797 801 802 802 804 805 805 805

xviii

CONTENTS

CHAPTER 23

CHAPTER 24

22.12 Seismic Design of Foundations Problems Problems and Solutions

806 807 809

Erosion of Soils and Scour Problems 23.1 The Erosion Phenomenon 23.2 Erosion Models 23.3 Measuring the Erosion Function 23.4 Soil Erosion Categories 23.5 Rock Erosion 23.6 Water Velocity 23.7 Geometry of the Obstacle 23.8 Bridge Scour 23.8.1 Maximum Scour Depth (zmax ) Analysis 23.8.2 Maximum Shear Stress at Soil–Water Boundary when Scour Begins 23.8.3 Final Scour Depth (zfinal ) Analysis for Constant Velocity Flow and Uniform Soil 23.8.4 Final Scour Depth (zfinal ) Analysis for a Velocity Hydrograph and Layered Soil 23.8.5 The Woodrow Wilson Bridge Case History 23.9 River Meandering 23.9.1 Predicting River Meandering 23.9.2 The Brazos River Meander Case History (Park 2007) 23.10 Levee Overtopping 23.10.1 General Methodology 23.10.2 Hurricane Katrina Levee Case History: New Orleans 23.11 Countermeasures for Erosion Protection 23.12 Internal Erosion of Earth Dams 23.12.1 The Phenomenon 23.12.2 Most Susceptible Soils 23.12.3 Criterion to Evaluate Internal Erosion Potential 23.12.4 Remedial Measures Problems Problems and Solutions

823 823 824 824 825 826 829 831 831 832

Geoenvironmental Engineering 24.1 Introduction 24.2 Types of Wastes and Contaminants 24.3 Laws and Regulations 24.4 Geochemistry Background 24.4.1 Chemistry Background 24.4.2 Geochemistry Background 24.5 Contamination 24.5.1 Contamination Sources 24.5.2 Contamination Detection and Site Characterization 24.5.3 Contaminant Transport and Fate

837 839 840 841 844 844 845 847 847 848 850 851 851 852 852 854 855 857 872 872 872 873 874 874 876 877 877 877 880

CONTENTS

24.6

Remediation 24.6.1 Risk Assessment and Strategy 24.6.2 In Situ Waste Containment 24.6.3 Soil Remediation 24.6.4 Groundwater Remediation Landfills 24.7.1 Waste Properties 24.7.2 Regulations 24.7.3 Liners 24.7.4 Covers 24.7.5 Leachate Collection 24.7.6 Landfill Slopes 24.7.7 Gas Generation and Management Future Considerations Problems Problems and Solutions

883 883 885 887 888 890 890 891 892 893 893 894 895 895 896 897

Geosynthetics 25.1 General 25.2 Types of Geosynthetics 25.3 Properties of Geosynthetics 25.3.1 Properties of Geotextiles 25.3.2 Properties of Geomembranes 25.3.3 Properties of Geogrids 25.3.4 Properties of Geosynthetics Clay Liners 25.3.5 Properties of Geofoams 25.3.6 Properties of Geonets 25.4 Design for Separation 25.5 Design of Liners and Covers 25.6 Design for Reinforcement 25.6.1 Road Reinforcement 25.6.2 Mechanically Stabilized Earth Geosynthetic Walls 25.6.3 Reinforced Slopes 25.6.4 Reinforced Foundations and Embankments 25.7 Design for Filtration and Drainage 25.8 Design for Erosion Control 25.9 Other Design Applications 25.9.1 Lightweight Fills 25.9.2 Compressible Inclusions 25.9.3 Thermal Insulation 25.9.4 Geosynthetics and Landfill Slopes Problems Problems and Solutions

904 904 904 905 905 908 909 910 911 912 913 913 915 915

24.7

24.8

CHAPTER 25

CHAPTER 26

xix

Soil Improvement 26.1 Overview 26.2 Soil Improvement without Admixture in Coarse-Grained Soils 26.2.1 Compaction 26.2.2 Dynamic Compaction

915 918 918 919 920 922 922 922 922 922 923 924 938 938 938 938 938

xx

CONTENTS

26.3

26.4

26.5

26.6

26.7

CHAPTER 27

26.2.3 Vibrocompaction 26.2.4 Other Methods Soil Improvement without Admixture in Fine-Grained Soils 26.3.1 Displacement–Replacement 26.3.2 Preloading Using Fill 26.3.3 Prefabricated Vertical Drains and Preloading Using Fill 26.3.4 Preloading Using Vacuum 26.3.5 Electro-osmosis 26.3.6 Ground Freezing 26.3.7 Hydro-Blasting Compaction Soil Improvement with Replacement 26.4.1 Stone Columns without Geosynthetic Sock 26.4.2 Stone Columns with Geosynthetic Encasement 26.4.3 Dynamic Replacement Soil Improvement with Grouting and Admixtures 26.5.1 Particulate Grouting 26.5.2 Chemical Grouting 26.5.3 Jet Grouting 26.5.4 Compaction Grouting 26.5.5 Compensation Grouting 26.5.6 Mixing Method 26.5.7 Lime Treatment 26.5.8 Microbial Methods Soil Improvement with Inclusions 26.6.1 Mechanically or Geosynthetically Stabilized Earth 26.6.2 Ground Anchors and Soil Nails 26.6.3 Geosynthetic Mat and Column-Supported Embankment Selection of Soil Improvement Method Problems Problems and Solutions

938 940 941 941 941 943 944 945 945 945 946 946 947 948 948 949 950 950 950 950 951 952 952 953 953 953 953 955 955 956

Technical Communications 27.1 General 27.2 E-Mails 27.3 Letters 27.4 Geotechnical Reports 27.5 Theses and Dissertations 27.6 Visual Aids for Reports 27.7 Phone Calls 27.8 Meetings 27.9 Presentations and PowerPoint Slides 27.10 Media Interaction 27.11 Ethical Behavior 27.12 Professional Societies 27.13 Rules for a Successful Career

962 962 962 963 963 963 964 965 965 966 966 967 967 967

References

969

Index

983

ACKNOWLEDGMENTS 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)

My colleagues also provided advice on many topics: • • • • • • • •

Marcelo Sanchez (Texas A&M University) Don Murff (Exxon) Jose Roesset (Texas A&M University) Giovanna Biscontin (Texas A&M University) Chuck Aubeny (Texas A&M University) Zenon Medina Cetina (Texas A&M University) Vincent Drnevich (Purdue) Chris Mathewson (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.

xxi

CHAPTER 1

Introduction

1.1

WHY THIS BOOK?

“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 intended audience is anyone who is starting in the field of geotechnical engineering, including university students. 1.2

GEOTECHNICAL ENGINEERING

Geotechnical engineering is a young (∼100 years) professional field dealing with soils within a few hundred meters Too complex

Threshold of optimum simplicity

Too simple

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

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: • • • • • •

Foundations Slopes Retaining walls Dams Landfills Tunnels

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

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 0.001% of the soil that will provide the foundation support for the structure. Yet, on the basis of this extremely limited data, the geotechnical engineer must predict the behavior of the entire mass of soil. This is why geotechnical engineering is a very difficult discipline. 1

2

1 INTRODUCTION

(e.g., drill rigs for bored piles), and a number of ingenious ideas (e.g., reinforced earth walls). 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:

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 soilsrelated 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 Tower of Pisa was started in 1174 and completed in 1370. 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. 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

• 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? • 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

SOME RECENT AND NOTABLE PROJECTS

Among some notable geotechnical engineering projects and developments are the underpinning of the foundation of the Washington Monument in 1878 (Figure 1.2; Briaud et al.

16.80 m 11.23 m 24.38 m

4.11 m 12.57 m

12.57 m 13.41 m 38.54 m

Figure 1.2

The Washington Monument.

1.4 SOME RECENT AND NOTABLE PROJECTS

3

Figure 1.3 Culebra cut of the Panama Canal, 1913. (a: Courtesy of Fernando Alvarado; b: Courtesy of United States Geological Survey)

2009); the Panama Canal (1913) and its slope stability problems (Figure 1.3; Marcuson 2001); the Tower of Pisa (1310) and its foundation repair in 1990 (Figure 1.4; Jamiolkowski 2001); the locks and dams on the Mississippi River and their gigantic deep foundations (Figure 1.5); and airports built offshore, as in the case of the Tokyo Haneda airport runway extension (Figure 1.6). Among the most significant milestones

in the progress of geotechnical engineering are the discovery of the effective stress principle in saturated and then unsaturated soil mechanics; the development of laboratory testing and in situ testing to obtain fundamental soil properties; the combination of soil models with numerical methods to simulate three-dimensional behavior; the advent of geosynthetics and of reinforced soil, which is to geotechnical

Figure 1.4 The Tower of Pisa and its successful repair in 1995. (c: Courtesy of Dr. Gianluca De Felice (General Secretary), Opera Primaziale Pisana.)

4

1 INTRODUCTION

Figure 1.5 Lock and Dam 26 on the Mississippi River in 1990. (a: Courtesy of United States Army Corps of Engineers, b: Courtesy of Thomas F. Wolff, St. Louis District Corps of Engineers, 1981. c: Courtesy of Missouri Department of Transportation.)

Figure 1.6 Extension of the Tokyo Haneda airport in 2010. (Courtesy of Kanto Regional Development Bureau, Ministry of Land, Infrastructure, Transport and Tourism, Japan.)

1.8 UNITS

engineering what reinforced concrete is to structural engineering; and the development of instruments to monitor full-scale behavior of geotechnical engineering structures.

1.5

FAILURES MAY OCCUR

Failures do occur. The fact remains that it is not possible to design geotechnical engineering structures that will have zero probability of failure. This is because any calculation is associated with some uncertainty; because the geotechnical engineering profession’s knowledge, despite having made great strides, is still incomplete in many respects; because human beings are not error free; and because the engineer designs the geotechnical engineering structure for conditions that do not include extremely unlikely events such as an asteroid hitting the structure at the same time as an earthquake, a hurricane, and a 100-year flood during rush hour. Nevertheless, geotechnical engineers learn a lot from failures, because thorough analysis of what happened often points out weaknesses and needed improvement in our approaches. Some of the most notable geotechnical engineering failures have been the Transcona silo bearing capacity failure in 1913 (Figure 1.7), the Teton dam seepage failure in 1976 (Figure 1.8), and the failure of some of the New Orleans levees during Hurricane Katrina in 2005 (Figure 1.9).

1.6

OUR WORK IS BURIED

As Terzaghi is said to have noted, there is no glory in foundations. Indeed, most of our work is buried (Figure 1.10).

For example, everyone knows the Eiffel Tower in Paris, but very few know about its foundation (Figure 1.11; Lemoine 2006). The foundation was built by excavating down to the water level about 7 m deep—but the soil at that depth was not strong enough to support the 100 MN weight of the Tower, so digging continued. Because of the water coming from the River Seine, the deepening of the excavation had to be done using pressurized caissons (upside-down coffee cans, big ones!) so that the air pressure could balance the water pressure and keep it out of the excavation. Workers got into these 14 × 6 × 15 m caissons (Figure 1.12) and worked literally under pressure until they reached a depth where the soil was strong enough to support the Tower (about 13 m on the side closest to the river and about 8 m on the side away from the river). 1.7

GEOTECHNICAL ENGINEERING CAN BE FUN

Geotechnical engineering can be fun and entertaining, as the book by Elton (1999; Figure 1.13) on geo-magic demonstrates. Such phenomena as the magic sand (watch this movie: www.stevespanglerscience.com/product/1331?gclid=CNiW 1uu-aICFc9J2godZwuiwg), water going uphill, the surprisingly strong sand pile (Figure 1.13), the swelling clay pie (Figure 1.13), 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 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. 1.8

Figure 1.7 Transcona silo bearing capacity failure and repair (1913). (Courtesy of the Canadian Geotechnical Society.)

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 travelled by light in 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

6

1 INTRODUCTION

Figure 1.8 Teton Dam seepage failure (1976) (Photos by Mrs. Eunice Olson. Courtesy of Arthur G. Sylvester.)

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,

1.8 UNITS

Figure 1.9 New Orleans levee failures during the Katrina hurricane in 2005. (Courtesy of United States Army Corps of Engineers.)

Figure 1.10 A rendition of the geotechnical engineering world. (Courtesy of Hayward Baker Inc., Geotechnical Contractor.)

7

1 INTRODUCTION

Cross section A-A

4m 1m

54° 8m

13 m

54°

1m 4m 2m

7m 10 m 10 m

Pier 1

Ground surface

7m

13 m Sandy clay

5m 2m 2m

Fill

13 m

7m

8m

4m

Sand and gravel

6m 6m 6m

Plastic clay

Limestone

Pier 3

Pier 4

Plan view 6m

Pier 2 10 m

6m

10 m

10 m 6m

14 m 6m

N

Pier 3

Pier 1 141 m

A

A

in

0

se

m

m

0

er

iv

10

R e

10

8

0m

20

15

Pier 4

0 m ow

Fl

Figure 1.11

The Eiffel Tower foundation plan.

All piers have the same planar dimensions but different depths as shown in cross section AA.

1.8 UNITS

(a)

(b)

Figure 1.12

(c)

The Eiffel Tower foundation. (Photos b, c: Courtesy of the Mus´ee d’Orsay, Paris.)

Figure 1.13

Soil magic. (Courtesy of David J. Elton.)

9

10

1 INTRODUCTION

which is the force required to accelerate a mass of 1 kg to 1 m/s2 . 1 N = 1 kg × 1 m/s2 (1.1)

even where you are on Earth. Other units are shown in a table at the beginning of this book. Accepted multiples of units, also called SI prefixes, are:

This force is about the weight of a small apple. Humans typically weigh between 600 and 1000 N. Most often the kiloNewton (kN) is used rather than the Newton. The kilogram force is the weight of one kilogram mass. On Earth, the equation is: 1 kgf = 1 kg × 9.81 m/s2

terra giga mega kilo milli micro nano pico (An angstrom is 10−10 meter.)

(1.2)

The unit of stress is the kN/m2 , also called 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

1012 109 106 103 10−3 10−6 10−9 10−12

PROBLEMS

6m

36°

2.6 m

1.1 How would you decide if you have reached the threshold of optimum simplicity? 1.2 What was achieved by underpinning the 608 MN Washington Monument foundation from a 24.4 m square foundation to a 38.5 m square ring, as shown in Figure 1.2? 1.3 How would you go about deciding if the slopes of the Panama Canal are too steep? 1.4 What major geotechnical engineering problems come to mind for the extension of the Tokyo Airport? 1.5 Write a step-by-step procedure for the up-righting of the Transcona Silo. 1.6 For the 100 MN Eiffel Tower, calculate the average pressure under the foundation elements.

5m

3.6 m

7m

5.4 m

13 m

54°

14 m

Figure 1.1s

Foundation of the Eiffel Tower.

1.7 For the Tower of Pisa, calculate the pressure under the foundation, given that the foundation is a ring with a 19.6 m outside diameter and a 4.5 m inside diameter. Compare this pressure to the pressure obtained for the Eiffel Tower in problem 1.6. 19.6 m

4.5 m

Figure 1.2s

Tower of Pisa foundation.

1.8 UNITS

11

0.2 8

m

1.8 Calculate the pressure under your feet.

0.0

8m

0.0

9m

Figure 1.3s

Feet geometry.

1.9 What do you think caused the failure of the Teton Dam? What do you think might have avoided this problem? 1.10 Explain the magic behind Figures 1.13d and 1.13e. 1.11 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

1.12 What is the relationship between a kilopascal (kPa) and a pound per square foot (psf)? What is the net pressure in psf under the Eiffel Tower foundation?

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: • • • • •

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.

Problem 1.2 What was achieved by underpinning the 608 MN Washington Monument foundation from a 24.4 m square foundation to a 38.5 m square ring, as shown in Figure 1.2? Solution 1.2 By increasing the area of the foundation, the pressure under the Washington Monument was decreased. This allowed the construction of the column to be completed with greatly reduced settlement and avoided the overturning or collapse of the structure that would likely have occurred if no underpinning had been done. Problem 1.3 How would you go about deciding if the slopes of the Panama Canal are too steep?

12

1 INTRODUCTION

Solution 1.3 I would draw a free-body diagram of the mass that would be likely to fail, I would show all the external forces, and I would check the equilibrium of the system. I would also check the site and make observations of the slope as a function of time. If it had not already been built, I could observe neighboring slopes and make measurements. Problem 1.4 What major geotechnical engineering problems come to mind for the extension of the Tokyo Airport? Solution 1.4 Some of the problems associated with the extension of the Tokyo airport include: • • • •

Soil failure in the form of rotational sliding at the edges of the embankment. Excessive settlement of the embankment, and in particular differential movements. Erosion problems during storms. Earthquake-induced problems, as the airport is in a high-seismicity area.

Problem 1.5 Write a step-by-step procedure for the up-righting of the Transcona Silo. Solution 1.5 The following steps could be considered for the successful up-righting of the silo: • Build footings on top of which hydraulic jacks can be installed to raise the structure. Make sure the footings can resist the force necessary to lift the structure. • Lift the structure upward and start to backfill the failed soil. An alternative is to reinforce the existing failed soil. • Complete the reinforcement of the key locations beneath the silo. • Lower the jacks and allow the silo to rest on the reinforced earth. Problem 1.6 For the 100 MN Eiffel Tower, calculate the average pressure under the foundation elements. Solution 1.6 Pressure is force over area. The problem states that the Eiffel Tower exerts 100 MN of force on the foundation. From Figure 1.11, we know that the foundation of each leg of the Eiffel Tower is made of one rectangular foundation of 14 m by 6 m and three rectangular foundations of 10 m by 6 m. Therefore, the total area for the foundation of each leg is 14 m × 6 m + 3(10 m × 6 m) = 264 m2 . Assuming that the load is evenly distributed among the four legs, the load per leg is 100 MN divided by 4, or 25 MN. The average pressure per foundation element is 25000 = 94 264 Note that this pressure does not include the weight of the foundation. Weight of the largest foundation element:   2.6 × 3.6 5.4 × 8.6 + W = 25 × 14 × 7 + 3.6 × 6 + 2 2 = 25 × (98 + 21.6 + 4.68 + 23.22) × 6 = 221

Average pressure due to the weight of this foundation is: Pfoundation =

22125 = 263 14 × 6

which is much larger than the pressure due to the tower alone. Indeed, the weight of all the foundation elements is a lot more than the weight of the tower.

6m

36°

13

2.6 m

1.8 UNITS

5m

13 m

54°

5.4 m 7m

3.6 m

14 m

Figure 1.1s

Foundation of the Eiffel Tower.

If we assume a total unit weight of soil of 20 kN/m3 , this pressure Pfoundation is equivalent to the pressure created by a height of soil equal to 263 =1 hsoil = 20 Because 13 meters of soil were excavated, the weight of soil removed during the excavation was approximately equal to the weight of the foundation and the net pressure increase on the soil is Pnet = 94.6 kPa. However, the actual pressure under the biggest foundation element is Ptotal = 94.6 + 263 = 357 kPa. Problem 1.7 For the Tower of Pisa, calculate the pressure under the foundation, given that the foundation is a ring with a 19.6 m outside diameter and a 4.5 m inside diameter. Compare this pressure to the pressure obtained for the Eiffel Tower in problem 1.6. Solution 1.7 Pressure under the foundation =

142 × 102 = 469 285.21

If this pressure does not include the weight of the foundation, then Pnet = 496.8 kPa is the net pressure. Net pressure under the Eiffel Tower foundation = 94.6 kPa. The net pressure under the Tower of Pisa is about five times higher than the net pressure under the Eiffel Tower. Problem 1.8 Calculate the pressure under your feet. Solution 1.8 Effective area for one foot ≈ (0.28 − 0.08) × 0.09 = 0.018 m2 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 1.3s

Feet geometry.

14

1 INTRODUCTION

Problem 1.9 What do you think caused the failure of the Teton Dam? What do you think might have avoided this problem? Solution 1.9 The failure of the Teton Dam was likely due to seepage at the boundary between the dam and the abutment. This seepage led to piping in the dam and ultimately to its breach. One way to avoid such a problem is to build a wall penetrating into the abutment, called a key, to minimize the seepage at that interface. Problem 1.10 Explain the magic behind Figures 1.13d and 1.13e. Solution 1.10 The swelling clay pie 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. The sand pile at the top of the figure fails under the load applied (50 N) because the load exceeds the shear strength of the sand. The sand pile at the bottom of the figure 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 1.11 Are the following equations correct? Solution 1.11 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 1.12 What is the relationship between a kilopascal (kPa) and a pound per square foot (psf)? Solution 1.12

  0.22481 lb 1000 N × 1N 1 kPa = 1000 N/m2 =    = 20.9 psf  3.28 ft 2 2 1m × 1m

What is the net pressure in psf under the Eiffel Tower foundation? Total weight = 100 MN

Total area = (14 m × 6 m + 3(10 m × 6 m)) × 4 = 10 Pressure in kPa =

100 × 106 100 MN = 1056 m2 1056 N/m2

= 94697 N/m2 = 94

Pressure in psf = 94.7 kPa × 20.9(psf/kPa) = 1975

CHAPTER 2

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).

2.1

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.

2.2

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 2.1) 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 at the surface. The Earth has a radius of approximately 6400 km (Jefferis 2008). The first layer, known as the crust (Figure 2.2), 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 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 like they are on the Sun today. The cooling process started right away and has been progressing ever since. The present temperature gradient, shown in Figure 2.2, represents an average increase in temperature with depth of 15 degrees Celsius per kilometer in the crust, although the overall average is only 1 degree Celsius per kilometer. The gravity field is governed by the acceleration due to gravity (9.81 m/s2 on the 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 semiliquid and liquid layers below accumulate strains at various locations where they run into each other. 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 2.3).

2.3

GEOLOGIC TIME

Geologic time is a scale dividing the age of the earth (4600 million years) into 5 eras (Figure 2.4): 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 present) (Harland et al. 1989). Each era is subdivided into periods and then into epochs (Figure 2.5). The Quaternary era, for example, is divided into the Pleistocene period and the Holocene or Recent period. 15

16

2 ENGINEERING GEOLOGY

The Earth. (Courtesy of NOAA-NASA GOES Project.)

Figure 2.1

0

3000

Mass density (kg/m3)

Pressure (GPa)

Temperature (°C)

Free space

200

6000 0

400 0

4000

Atmosphere

8000

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 2.2

Earth temperature, pressure, and density.

North America

Europe Asia Africa South America

Figure 2.3

South America and Africa fit. (Courtesy of John Harvey.)

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.

100000

10000

1000

17

en oz oi c

65.5 C

251

M es oz oi c

Pa le oz oi c

ia n ec am br

542

Pr

o

4500

N

N

o

un iv

er s

15000

Ea rth

e

2.5 SOILS

10

100

Millions of years

Figure 2.4

Geologic time (eras).

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)

Devonian (359 – 416 My)

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

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

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

2.4

Ordovician (444 – 488 My)

Eocene (33.9 – 55.8 My) Triassic (201.6 – 251 My)

Paleocene (55.8 – 65.5 My)

Figure 2.5

Carboniferous (299 – 359 My)

Cambrian (488 – 542 My)

Geologic time (periods and epochs).

ROCKS

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 than concrete (sandstone). Older rocks are typically stronger than younger rocks. Figure 2.6 shows some of the main rock types.

2.5

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 2.7 shows some of those soils categories. Note that what the geotechnical engineer calls soil may be called rock

18

2 ENGINEERING GEOLOGY

Igneous rocks

Basalt

Gabbro

Granite

Obsidian

Pumlce

Sedimentary rocks

Breccia Conglomerate Limestone Sandstone

Shale

Metamorphic rocks

Gneiss

Marble Metaquartzite Schist

Slate

Minerals

Biotite Copper (black mica)

Halite

Figure 2.6

Figure 2.7

Magnetite

Diamond

Quartz

Feldspar

Silver

Gold

Talc

Main categories of rocks. (Courtesy of EDUCAT Publishers)

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

2.6 GEOLOGIC FEATURES

19

by the engineering geologist; this can create confusion during discussion and interpretation. 2.6

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 2.8) 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. Folds (Figure 2.9) 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).

Figure 2.8 Example of rock fault. (Courtesy of USGS U.S. Geological Survey.)

Figure 2.9 Example of anticline–syncline combination. (Photo by R. W. Schlische.)

Figure 2.10 Examples of sinkholes. (Left: Courtesy of R.E. Wallace, United States Geological Survey, USA,; Right: Courtesy of International Association of Certified Home Inspectors, Inc.)

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 2.10). 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 contours 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 2.11). 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 only coarse-grained soils heavy enough

20

2 ENGINEERING GEOLOGY

Figure 2.11

Example of meander migration.

not to be transported away are found there. In contrast, floodplains are a low-energy deposition environment where fine-grained soils are typically 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 2.12). 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.

2.7 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.

Figure 2.13 Example of geologic map. (Courtesy of National Park Service, NPS.)

Each rock area of a certain age is given a different color (Figure 2.13); 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.

2.8

Figure 2.12 Norton.)

Example of an alluvial fan. (Courtesy of Mike

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 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 2.14) is the surface of the water within the soil or rock

2.8 GROUNDWATER

21

Precipitation Infiltration Evaporation

Artesian pressure

GWT clay sand

Figure 2.14

where the water stress is equal to the atmospheric pressure (zero gauge 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 and the phreatic surface are the same. In some cases, though, they are different: artesian pressure refers to the case where the pressure in the water at some

Figure 2.15 Survey.)

Groundwater.

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 2.14). 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 2.15).

Example of flow due to artesian pressure. (Courtesy of USGS U.S. Geological

22

2 ENGINEERING GEOLOGY

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 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 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

Calculate the pressure at the center of the Earth. Calculate the temperature at the center of the Earth What is the depth of interest for most geotechnical engineering projects? List the Tertiary and Quaternary epochs. What happened about 10,000 years ago on the Earth? What are some of the consequences for soil and rock behavior today? What are the three main categories of rocks, and what is the origin of each category? What are the four main categories of soil sizes? How were each of these soils generated? What engineering geology features can you look for when you visit a site for a geotechnical engineering project? How can geologic maps be useful to the geotechnical engineer? Define the following terms: groundwater level, perched water, phreatic surface, aquifer.

Problems and Solutions Problem 2.1 Calculate the pressure at the center of the Earth. Solution 2.1 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: F = G.

m1 . m2 r2

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 2.1s). 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

2.8 GROUNDWATER

23

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. Therefore: Ma

Earth

dr

r

Mc

Mb

Figure 2.1s

The pressure P is P =

F A

Parameters definition.

where A is the area of the element, so: dm1 G.m2 ρ.dV G.m2 ρ.dr.A G.m2 G.m . 2 = . 2 = . 2 = ρ.dr. 2 2 A r A r A r r  m2 4 3 P = ρ.G. 2 .dr, where m2 = π r ρ r 3  4 P = π.G. ρ 2 .r.dr 3

dP =

Because the density of the Earth’s layers is not constant (see Figure 2.2), the pressure at the center of the Earth is:  3500×1000  6400×1000  6300×1000 4 P = π × 6.67 × 10−11 27002 rdr 50002 rdr + 120002 rdr + 3 6300×1000 3500×1000 0   3.5 × 106 6.3 × 106 6.4 × 106 N 2 −4 2 2 + 12.5r  72r  = 3.44 × 1011 2 = 344 GPa P = 2.79 × 10 + 3.645r  6 6 m 0 3.5 × 10 3.3 × 10

Note that in geotechnical engineering we calculate the pressure, also called vertical total stress, at a given depth z as:  P = γi Zi

Where γi is the unit weight of the Zi thick i th layer within the depth z. This is an approximation, as the unit weight Y = 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 2.2 Calculate the temperature at the center of the Earth. Solution 2.2 The temperature gradient is 15◦ Celsius per kilometer in the crust and 0.63◦ Celsius per kilometer in the mantle and the core. Therefore, the temperature at the center of the Earth is: Tcenter = 15 × 100 + 0.63 × 6300 = 5469 Problem 2.3 What is the depth of interest for most geotechnical engineering projects? Solution 2.3 The depth of interest for most geotechnical engineers is a few hundred meters. Problem 2.4 List the Tertiary and Quaternary epochs.

◦

24

2 ENGINEERING GEOLOGY

Solution 2.4 Holocene Pleistocene

0 to 10,000 years ago 10,000 to 1.8 million years ago

Pliocene Miocene

1.8 to 5.3 million years ago 5.3 to 23.8 million years ago

Oligocene Eocene

23.8 to 33.7 million years ago 33.7 to 54.8 million years ago

Paleocene

54.8 to 65 million years ago

Problem 2.5 What happened about 10,000 years ago on the Earth? What are some of the consequences for soil and rock behavior today? Solution 2.5 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 2.6 What are the three main categories of rocks, and what is the origin of each category? Solution 2.6 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 2.7 What are the four main categories of soil sizes? How were each of these soils generated? Solution 2.7 Soil class

Soil type Size (by USCS)

Coarse-grained soil Gravel Sand

75 mm to 4.75 mm 4.75 mm to 0.075 mm

Fine-grained soil

0.075 mm to 2 μm 1.0 24

10

23

22

21

0

10 10 10 Particle size (mm)

10

1

2

10

Figure 4.4

10

22

0

21

10 10 10 Particle size (mm) Silt

Sand

10

1

10

40 20

Coarse 1023 Clay

1022 1021 100 Particle size (mm) Silt

Sand

101

60 40

10

Cu>>1.0 24

23

10

22

21

0

10 10 10 Particle size (mm)

10

1

10

2

Hydrometer and hydrometer reading.

22

0

21

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 hatche and below d zone above hatche hatche d zone d zone

SP

SM

SM-SC

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

Below “A” in line or hatche below d zone hatche d zone

GP

GM

GM-GC

Above “A” line and above hatche d zone 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 4.17

Flowchart to classify a soil by the USCS.

The exact process for classifying a soil consists of a series of steps organized in a decision tree as shown in Figure 4.17. The first decision is based on the percent passing the no. 200 sieve (#200), which has an opening of 0.075 mm. If the soil has more than 50% particles by weight larger than 0.075 mm (#200), the soil is a coarse-grained soil. If the soil has more than 50% by weight smaller than 0.075 mm (#200), the soil is a fine-grained soil. For coarse-grained soils, if the percent by weight of the gravel-size particles is larger than the percent by weight of the sand-size particles, the soil is a gravel and the first letter is G. If not, the soil is a sand and the first letter is S. The second letter for a coarse-grained soil is W, P, M, or C. If the soil has less than 5% passing #200, it is clean and the second letter will be W or P, depending on the coefficient

of uniformity Cu and the coefficient of curvature Cc obtained from the particle size distribution curve. If the coarse-grained soil has more than 12% passing #200, the soil is dirty and the second letter will be M or C, depending on the Atterberg limits of the portion smaller than 0.425 mm; M will be selected if the soil plots below the A line on the plasticity chart and C if it plots above. If the percent passing #200 is between 5% and 12%, then a dual symbol will be required, as the soil is intermediate between clean and dirty. In this instance, the classification for the 12% case are obtained and the soil ends up with a dual symbol (e.g., GP-GC or SW-SM). For fine-grained soils, the plasticity index and the liquid limit are plotted on the plasticity chart and the dual symbol is read from the quadrant of the chart where the point is situated.

PROBLEMS 4.1 Calculate the thickness of the wire in the no. 200 sieve. 4.2 A dry sample of soil weighs 5 N. 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?

4.6 UNIFIED SOIL CLASSIFICATION SYSTEM

57

4.3 Why is the particle size of the particle size curve plotted on a log scale? Plot the particle size curve of problem 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. 4.4 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. 4.5 A cylindrical hydrometer has a radius of 20 mm and weighs 2 N. It is lowered into water mixed with fine soil particles. 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. 4.6 Explain the hydrometer analysis in your own words. Develop the equations necessary. 4.7 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

• Plasticity index • Liquidity index • Shrink-swell index 4.8 Classify the following soils: S1 S2 S3 S4 S5 (% finer) (% finer) (% finer) (% finer) (% finer) #4 #10 #40 #200 WL WP

52 38 18 8 17 11

52 38 18 2 NP NP

63 56 42 4 NP NP

98 90 47 20 32 26

100 97 82 70 48 34

Problems and Solutions Problem 4.1 Calculate the thickness of the wire in the no. 200 sieve. Solution 4.1 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 4.2 A dry sample of soil weighs 5 N. 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

58

4 SOIL CLASSIFICATION

• 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 4.2 The percent of coarse grain size particles is =

2 + 1.5 + 1 × 100 = 90% 5

2 × 100 = 40% 5 1.5 + 1 The percent of sand-size particles is = × 100 = 50% 5 0.5 The percent of fine grain size particles is = × 100 = 10% 5

The percent of gravel-size particles is =

Retained soil on sieve

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

Weight (N)

Accumulated weight (N)

2 1.5 1 0.5

2 3.5 4.5 5

Passing through sieve

Accumulated weight (%) 40 70 90 100

Weight (N) 3 1.5 0.5 0

Accumulated weight (%) 60 30 10 0

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

4.75 D60 = = 63 D10 0.075

Cc =

0.4252 D 2 30 = = 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 4.3 Why is the particle size of the particle size curve plotted on a log scale? Plot the particle size curve of problem 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 4.3 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 4.1s shows the particle size curve as percent finer vs. particle size on a log scale. Figure 4.2s 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 4.2s. The position of this point is the same on the scale of Figure 4.1s. The same approach applies to the 4.75 mm particle: log 4.75 = 0.677.

4.6 UNIFIED SOIL CLASSIFICATION SYSTEM

59

100 90 100

80 Percent finer (%)

90 Percent finer (%)

80 70 60 50 40 30

log (4.75) 5 0.677

60 50 40 30 20

20

log (0.075) 5 21.125

10

10 0 0.01

70

0.1

1

0 22.0

10

21.5

Particle size (mm) in log scale

Figure 4.1s

21.0

20.5

0.0

0.5

1.0

Log (Particle size (mm))

Percent finer vs. particle size on a log scale.

Figure 4.2s

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

Problem 4.4 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 4.4 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: v=



γs − γf 18μ



D2

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: v=



25.95 − 9.79 18 × 10−6



×



0.002 1000

2

= 3.59 × 10−6 (m/ sec) = 0.00359 (mm/ sec)

Problem 4.5 A cylindrical hydrometer has a radius of 20 mm and weighs 2 N. It is lowered into water mixed with fine soil particles. 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.

60

4 SOIL CLASSIFICATION

Solution 4.5 R 5 20 mm

0.62 m3

Water

1.0 m3

100 mm

15.92 kN 0.38 m3

Figure 4.3s

Soil

Hydrometer and three-phase diagram.

FBuoyancy = W V × γmixture = W

π × 0.042 × 0.1 × γmixture = 2 4 γmixture = 15.92 × 103 N/m3 = 15.92 kN/m3 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

9.81 × (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 4.6 Explain the hydrometer analysis in your own words. Develop the equations necessary. Solution 4.6 See Section 4.2 in this chapter. Problem 4.7 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 • Plasticity index • Liquidity index • Shrink-swell index Solution 4.7 • Plasticity index: PI = LL − PL = 55 − 25 = 30 • Liquidity index: LI = (w − PL)/PI = (22 − 25)/30 = −0.1 • Shrink-swell index: Iss = swell limit−shrinkage limit = 36 − 13 = 23

4.6 UNIFIED SOIL CLASSIFICATION SYSTEM

61

Problem 4.8 Classify the following soils: S1 (% finer)

S2 (% finer)

S3 (% finer)

S4 (% finer)

52 38 18 8 17 11

52 38 18 2 NP NP

63 56 42 4 NP NP

98 90 47 20 32 26

#4 #10 #40 #200 wL wP

S5 (% finer) 100 97 82 70 48 34

Solution 4.8 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 D 2 30 D10 × D60

The particle size distribution curves are drawn on Figures 4.4s to 4.8s and theclassification of the 5 soils is presented in the Table below. S1 (% finer) Sieve Opening (mm) 10 4.75 2 0.425 0.075 0.03 from hydrometer Other properties wL wP Ip Coarse fraction (%) Fine fraction (%) Gravel fraction (%) Sand fraction (%) D10 (mm) D30 (mm) D60 (mm) Cu Cc Classification

80 52 38 18 8 — 17 11 6 92 8 48 44 0.11 1.05 7 63.6 1.4 GW-(GC-GM)

S2 (% finer)

80 52 38 18 2 — NP NP NP 98 2 48 50 0.19 1.05 7 36.8 0.8 SP

S3 (% finer) Percent Finer 80 63 56 42 4 — NP NP NP 96 4 37 59 0.098 0.23 3.2 32.7 0.2 SP

S4 (% finer)

— 98 90 47 20 — 32 26 6 80 20 2 78 0.036 0.16 0.7 19.4 1.0 SM

S5 (% finer)

— 100 97 82 70 9 48 34 14 30 70 0 30

ML

62

4 SOIL CLASSIFICATION

M

100

S

G

20

Percent passing (%)

Percent passing (%)

40

D60

D30 D10

0 0.01

Figure 4.4s

0.1 1 Particle size (mm)

S

D30 D10

Figure 4.6s

D10

0 0.01

Figure 4.5s

Percent passing (%)

D60

0 0.01

20

D30

0.1 1 Particle size (mm)

10

Percent finer vs. log of particle size of sample S2.

M

S

G

Sample 4

0.1 1 Particle size (mm)

10

Percent finer vs. log of particle size of sample S3.

100

M

80 60 40 20

D60

D30 D10

0 0.01

Figure 4.7s

S

0.1 1 Particle size (mm)

G

80 60 40 20 0 0.01

Figure 4.8s

0.1 1 Particle size (mm)

10

Percent finer vs. log of particle size of sample S4.

Sample 5 Percent passing (%)

Percent passing (%)

20

40

D60

100

80

40

60

G

Sample 3

60

G

80

10

Percent finer vs. log of particle size of sample S1.

M

100

S

Sample 2

80 60

M

100

Sample 1

10

Percent finer vs. log of particle size of sample S5.

CHAPTER 5

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).

5.1

ROCK GROUPS AND IDENTIFICATION

A rock is a mixture of minerals (Sorrell and Sandstr¨om 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 5.1) 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 5.2) 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, or pressure, or a 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 5.3) 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 Figure 5.4 and 5.5a are very useful. Figure 5.4 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 rockforming mineral; use of a knife and one’s fingernail to test the strength; and observation of the cleavage, the color, and the luster. Figure 5.5a helps in identifying the rock itself. It distinguishes between rocks with a crystalline texture, rocks that have no grains visible and are uniformly smooth, and rocks with a clastic texture.

5.2

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. 63

64

5 ROCKS

Granite

Felspar

Basalt

Figure 5.1 Igneous rocks: (a) granite, (b) feldspar, (c) basalt. (Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.) Sandstone

Slate

Limestone

Siltstone

Figure 5.2 Sedimentary rocks: (a) sandstone, (b) slate, (c) limestone, (d) siltstone. (Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.) Gneiss

Marble

Figure 5.3 Metamorphic rocks: gneiss, marble. (Courtesy of Mineral Information Institute, an affiliate of the SME Foundation.)

5.2 ROCK MASS VS. ROCK SUBSTANCE

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, grey, or white

White, grey, 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 5.4 Identifying rock minerals. (From Goodman, 1989. Reprinted 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 sheared calcite or surfaces dolomite 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

Figure 5.5a Identification of rocks with crystalline texture. (From Goodman, 1989. Reprinted with permission of John Wiley & Sons.)

Green schist

65

66

5 ROCKS

No grains visible, uniformly smooth

Isotropic structure

Anisotropic structure

Harder than knife blade

No associated volcanic features

Hornfels or granulite

Softer than knife blade

Associated volcanic features

Spheroidal weathering

Soluble

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 5.5b Identification of rocks with no grains visible. (From Goodman, 1989. Reprinted with permission of John Wiley & Sons.) 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 5.5c Identification of rocks with clastic texture. (From Goodman, 1989. Reprinted with permission of John Wiley & Sons.)

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, 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.

5.3 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 (Figure 5.6 and Figure 5.7). Nonetheless, the two main types are joints and faults. Joints are created over

5.4

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

5.5 ROCK ENGINEERING PROPERTIES

67

Figure 5.6 Fissures and joints. (a: Courtesy of Lupin. c: Courtesy of Charles DeMets, University of Wisconsin-Madison. d: Courtesy of Alex Brollo.)

26 kN/m3 . The porosity of rock substance is at most 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.

5.5

Figure 5.7 A fault. (Courtesy of The United State Geological Survey USGS, USA)

ROCK ENGINEERING PROPERTIES

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

68

5 ROCKS

10 minutes and the rock pieces remaining after the test are weighed again. The ratio in percent of the weight after and before the test is the slaking durability index Isd . Rocks typically have Isd 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 to 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.

5.6

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 which 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 Iv 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 for an RQD higher than 90% and a velocity index higher than 0.8. Rock mass quality is very poor for an RQD less than 25% and a velocity index 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 5.1 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 Jn , the roughness of the joints Jr , the degree of alteration and filling of the joints Ja , the water inflow Jw , and the stress reduction factor SRF. Using these six parameters, the Q factor is derived with the following equation: Q=

RQD × Jr × Jw Jn × Ja × SRF

5.7 ROCK ENGINEERING PROBLEMS

69

Table 5.1 Rock Mass Rating (RMR) Geomechanics System (Waltham 1994) 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 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 7.24). 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.

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

117

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 7.17 Example of dilatometer test results. (Courtesy of Dr. Sylvano Marchetti, www.marchetti-dmt.it)

118

7 IN SITU TESTS

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 7.18 The vane shear test. (From Mayne et al. 2002. Courtesy of Professor Paul Mayne, Georgia Institute of Technology)

Figure 7.19

Field vane shear test. (Courtesy of Dr. Dimitrios P. Zekkos.)

7.7 PLATE LOAD TEST

119

Figure 7.20 Laboratory vane shear test. (a: Courtesy of ELE International, b: Courtesy of Impact Test Equipment Ltd)

Tmax

Failure surface

Torque (kN.m)

Vane

10 rapid rotations TRes

θmax

0

Figure 7.21

Rotation angle (°)

Vane shear test results.

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 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, Philionat and Zerhouni, 1993) is a similar tool.

7.7

PLATE LOAD TEST

The plate load test or PLT (Figure 7.25; 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 until the desired pressure p is reached. 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

7 IN SITU TESTS

Rod clamp Ring gear Bearing

Shear gauge

Borehole shear test device. (Courtesy of In-Situ Soil Testing, L.C.)

Torque arm Crank

Dynamometer

Fill port

Pull rod Base plate

Shear force divided by area of plates tf (kPa)

Figure 7.22

tf3

tf1 w8 sh1

Pull strap

sh2

sh3

Pressure applied on wall of borehole sN (kPa)

Figure 7.24 Shear head

Failure envelope given by stage BST

tf2

c (kPa)

120

Not to scale Gas lines

Shear plate

Figure 7.23 Borehole shear test device. (Courtesy of Professor Richard L. Handy, Handy Geotechnical Instruments, Inc.)

steps 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

Results of a borehole shear test.

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 7.26), 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 qc , the undrained shear strength su , the SPT blow

7.7 PLATE LOAD TEST

121

Figure 7.25 Plate load tests. (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

Displacement (mm)

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

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)

Figure 7.26

0

5

10

15

20

25

30

35

Time (min)

Results of load test for 0.3-m-diameter plate on medium dense silty sand.

count 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 (Briaud 2007). The soil modulus as measured during a plate test is obtained from the initial loading portion E0 (O to A on Figure 7.26) or from the slope of the reloading part of the unload-reload loop Er (B to

C on Figure 7.26). The equations to be used for E0 and Er , if the plate can be assumed to be rigid, are: (1 − ν 2 )πpD 4s (1 − ν 2 )π pD Er = 4s

E0 =

(7.13) (7.14)

where E0 is the initial modulus from a plate load test, v is Poisson’s ratio (to be taken as 0.5 if the plate test is

122

7 IN SITU TESTS

fast enough that no drainage can take place during the test and 0.35 if the 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 p is the pressure increment during the reload loop corresponding to the settlement increment s. In addition to obtaining the soil modulus, sometimes the modulus of subgrade reaction is calculated from the plate test, as follows: p (7.15) in kN/m3 K= s Note that K is not a soil parameter, since it depends on the size of the plate: K=

4E0 (1 − υ 2 )πD

(7.16)

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 7.26). 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 equation 7.17:  n t s1 = 1 (7.17) s2 t2

Figure 7.27 Company.)

CBR test in the field. (Courtesy of A F Howland

settlement s of 2.5 mm on a reference soil (crushed California limestone). The pressure necessary to create 2.5 mm of settlement 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: CBR =

100 × p(kPa) 6900

(7.19)

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 t E1 = 1 (7.18) E2 t2

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:

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.

Mr (kPa) = 10,000 × CBR su (kPa) = 11 × CBR

(7.20) (7.21)

where Mr is the resilient modulus and su is the undrained shear strength. 7.8 CALIFORNIA BEARING RATIO TEST The California bearing ratio test (CBR) is a form of plate test (Figure 7.27). 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

7.9 POCKET PENETROMETER AND TORVANE TESTS 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

7.10 POCKET ERODOMETER TEST

(a)

123

(b)

Figure 7.28 Pocket penetrometer and torvane: (a) Pocket penetrometer (see also this video: www.encyclopedia.com/video/PBo0UDVWhSo-hand-penetrometer-test.aspx). (b) Torvane (see also this video: www.encyclopedia.com/video/9Su3ehhLfwc-torvane-test.aspx)

test (PPT) (Figure 7.28) 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 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 7.28) 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 7.5, Figure 7.20) 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 is the most reliable. 7.10

POCKET ERODOMETER TEST

The pocket erodometer test (PET) (Figure 7.29, 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 7.29), measuring the distance x where the water reaches the floor, and using the following equation: 2H g

75 mm

H

d/2

d/2

Figure 7.29

(7.22)

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 7.30) to determine the erodibility category of the soil.

100 mm

X

x v0x = 

Pocket erodometer test.

124

7 IN SITU TESTS

7.11.1

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

Low erodibility IV < 1 mm

CH

SM 1

Rock

Very low erodibility V

No noticeable erosion

0.1 0.1

1

10

100

Velocity (m/s)

Figure 7.30

Erosion chart for various erosion depths from the PET.

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.

7.11

COMPACTION CONTROL TESTS

Soil compaction is one of many techniques of soil improvement and is discussed in Chapter 20. 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).

Sand Cone Test

The sand cone test (SCT; Figure 7.31) 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, 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. 7.11.2

Rubber Balloon Test

The rubber balloon test (RBT; Figure 7.32) follows exactly the same procedure as the sand cone method except that the 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

Figure 7.31 Field unit weight and water content by sand cone test. (b: Courtesy of Durham Geo Slope Indicator.)

7.11 COMPACTION CONTROL TESTS

Air valvle

Flat surface

Fill

Water Rubber balloon

Figure 7.32 Field unit for testing weight and water content by rubber balloon. (b: Courtesy of Humboldt Mfg. Co.)

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. 7.11.3

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 8.4.1). These photons go straight to the detector, or bump into the soil particles (Compton scattering) and deflect to arrive at the detector, or do 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 7.33); the detector is on the bottom side of the nuclear gage. This mode is preferred for density measurements. In the back-scatter mode, the nuclear gage sits on the soil surface and the source and detectors are on the bottom side of the gage (Figure 7.33). This is the mode used for 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. 7.11.4

Field Oven Test

The field oven (Figure 7.34) is a very simple instrument which 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

Nuclear test

Gauge

Gauge

Detectors Source Direct transmission

Figure 7.33

125

Detectors Source Backscatter

Nuclear density probe test for unit weight and water content.

126

7 IN SITU TESTS

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 Poisson’s ratio (range from 0.3–0.45, depending on soil type), F is the maximum force on the force versus time plot, D is the plate diameter, and  is the maximum displacement on the displacement versus time plot. For example, referring to the flexible plate LWD test in Figure 7.36, the modulus would be calculated as: E = 1(1 − 0.352 ) 7.11.6

Figure 7.34

Field oven test (FOT) for water content.

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. 7.11.5

Lightweight Deflectometer Test

The lightweight deflectometer (LWD) test (Figure 7.35) (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. 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 during the impact. The soil modulus is back-calculated from the knowledge of the peak force F and the peak deflection . The soil modulus E is calculated using the theory of elasticity: 4F (7.23) E = f (1 − ν 2 ) π D

4 × 7.5 = 76.3 MPa π × 0.2 × 0.55 × 10−3 (7.24)

BCD Test

A modulus E can also be obtained with a device called the BCD (Figure 7.37). 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 7.37). 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 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

Guide rod Drop weight Shock absorber Load cell Geophone Loading plate (a)

(b)

Figure 7.35 Falling weight deflectometer for soil modulus: (a) Principle. (b) Equipment. (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

127

2r Deflection, ∆ (mm)

Load, F (kN)

7.12 HYDRAULIC CONDUCTIVITY FIELD TESTS

GWL D

h1

h2

Casing

0 60

Time, t (ms)

Figure 7.36

Falling weight deflectometer data.

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 7.38), the hydraulic conductivity k is obtained from the equation:

600 mm

Acquisition processing & readout

Load-cell

2 mm

Strain-gage instrumented plate

150 mm (a)

Figure 7.37 ment.

(b)

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. 7.12.1

h1 h2 k= 11(t2 − t1 ) 2π r Ln

BCD test for soil modulus: (a) Principle. (b) Equip-

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 20 section 20.2). 7.12

Figure 7.38 Inflow well test in deep uniform soil. (After Hunt 1984.)

Borehole Tests

Borehole tests consist of drilling a borehole, changing the water level in the borehole, and recording the movement of the

(7.25)

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 7.38 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 (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

Phreatic surface during pumping

Q

Initial phreatic surface r R

Drawdown H h

Sand Well Impermeable

Figure 7.39 Pumping test in sand layer using one boring. (After Hunt 1984.)

128

7 IN SITU TESTS Phreatic surface during pumping

Q

In the case where the pervious layer to be tested is sandwiched between two impervious layers, where the uncased 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 7.41), the hydraulic conductivity k is obtained from the equation:

Initial phreatic surface R

Drawdown r1

H h2

h1

Sands Well

r2

R r k= 2π D(H − h) Q Ln

Impervious

Figure 7.40 Pumping test in sand layer using three borings. (After Hunt 1984.)

(as shown in Figure 7.39), the hydraulic conductivity k is obtained from the equation: R Q Ln r (7.26) k= π(H 2 − h2 ) 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 7.40). Then equation 7.26 becomes: r Q Ln 2 r1 (7.27) k= π(h22 − h21 )

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 ground-water 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 7.42). Then equation 7.28 becomes: r2 r1 k= 2π D(h2 − h1 ) Q Ln

R

Aquiclude (clayey soil) H h

Figure 7.41

(7.29)

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

r

Aquifer (sands)

(7.28)

Well

D

Pumping test in confined aquifer. (After Hunt 1984.)

7.12 HYDRAULIC CONDUCTIVITY FIELD TESTS

r

129

R

r1

Aquiclude (clayey soil)

r2

H

h Aquifer (sands)

h1 Well

Figure 7.42

Pumping test in confined aquifer using three borings. 600

Normally consolidated to lightly overconsoilidated

ui 400 ∆ us

∆ut

200

u ∆ud

0 1

u0 10

100

1000

10000

100000

–200

Pore pressure u(kPa)

Pore pressure u(kPa)

D

Well

600

400

ui ∆ut

200

u

u0

0 1 –200

Heavily overconsolidated

∆ us

∆ ud

10

100

1000

10000

100000

–400 Time t(s)

Time t(s)

Figure 7.43

Decay of excess pore pressure in piezocone dissipation test.

wall of the borehole, and quick conditions development in high-gradient situations. 7.12.2

h2

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 7.43a). In the case of heavily overconsolidated soils, the response shows first an increase in excess pore pressure followed by a decrease (Figure 7.43b). The reason for this dual behavior is that the total excess pore pressure ut has two components: one is due to the water stress response us to the mean all-around compression of the soil element (spherical stress tensor); the other is due to the water stress response ud to the shearing of the soil element (deviatoric stress tensor). When the soil element is subjected to an all-around mean pressure, us is always positive, but when the soil element is subjected to a shear stress, ud 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, ud is

positive, and both us and ud decrease as a function of time (Figure 7.43a). If, however, the soil element increases in volume during shearing, it is called dilatant, and ud is negative. The combination of us decreasing with time and ud increasing with time (becoming less negative) leads to a bump on the decay curve (Figure 7.43b). 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 7.44), which is well represented by the equation:

k(cm/s) =



1 251 t50 (s)

1.25

(7.30)

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 7.43a for a lightly overconsolidated clay. The time to 50% dissipation is found halfway between the initial value ui (t = 1s in Figure 7.43a) 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).

130

7 IN SITU TESTS

Hydraulic conductivity, k (cm/s)

1.E-01

Sand and gravel

1.E-02

Sand

1.E-03

u2 Silty sand to sandy silt

1.E-04 1.E-05

Silt

This is called Darcy’s law and is explained in Chapter 13 on flow through soils; v is the discharge velocity; and i is the hydraulic gradient, defined as the loss of total head ht of the flowing water per distance travelled z.

1.E-06 Clay

Parez & fauriel (1988)

1.E-07 1.E-08 0.1

1

10 100 t50 (sec)

1000

10000

Figure 7.44 Relationship between t50 and hydraulic conductivity for piezocone dissipation test. (From Mayne, Christopher, Berg, and DeJong 2002. Courtesy of Professor Paul Mayne, Georgia Institute of Technology, USA.)

7.12.3

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 (7.31)

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 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 in the soil surface and embedding and grouting the walls of the ring about 0.45 m below the surface (Figure 7.45). 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 10 on water stress for an explanation of how tension occurs in the soil water and Chapter 9 on laboratory tests for an explanation of how tensiometers work). As the water seeps through the unsaturated soil layer 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 13). Obtaining the hydraulic conductivity k from the SDRIT

i=

ht z

(7.32)

Conservation of mass leads to: Vf = vd AtC

(7.33)

where Vf 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 ht z

(7.34)

If the test is run long enough that the whole layer becomes saturated, then ht is the vertical distance from the bottom of the layer to the level of the water in the outer ring and z 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 water front has penetrated to a depth Dw below the top of the soil surface, the value of z is Dw and the value of ht is: ht = H + Dw + hp

(7.35)

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), ht is underestimated and k is overestimated. 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

7.12 HYDRAULIC CONDUCTIVITY FIELD TESTS Dial gages for heave measurement

131

Reference for swelling measurement

4.0 m 1.5 m

Outer ring

Tensiometer Inner ring

H

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 7.45 Sealed double-ring infiltrometer. (Courtesy of Professor Xiaodong Wang, University of Wisconsin, USA.)

swelling component would give an overestimated value of Vf and therefore an overestimated value of k. The volume of water Vs used to increase the volume of the soil through swelling is measured as follows: A reference beam is set up above the SDRI (Figure 7.44) and the vertical movement of the inner ring is recorded with respect to that beam (using dial gages, for example). The volume Vs corresponding to the vertical movement of the inner ring is subtracted from the volume of water Vt leaving the plastic bag to obtain the true volume Vf . flowing through the soil.

Figure 7.46a) 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):

7.12.4

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 5 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

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,

k1 =

π d2 h Ln 1 11D(t2 − t1 ) h2

(7.36)

132

7 IN SITU TESTS

Casing

d

H

D ≥ 5D Grout

Casing

d

H

D ≥ 5D Grout

L ≥ 5D Compacted clay

≥ 5D Compacted clay

(a)

(b)

Figure 7.46 Two-stage borehole permeameter: (a) Stage 1; (b) Stage 2. (Third picture: Courtesy of Craig Benson, University of Wisconsin.)

with

⎞   2 L L ⎠ A = d 2 Ln ⎝ + 1 + D D   L B = 8L(t2 − t1 ) 1 − 0.562e−1.57 D ⎛

(7.38)

9 L/D 5 1.5 Ratio of hydraulic conductivity kh/kv

k2 from stage 2 is calculated from the following equations (Hvorslev 1949): A h (7.37) k2 = Ln 1 B h2

(7.39)

where kh and kv are the hydraulic conductivity in the horizontal and vertical directions respectively. Then k2 /k1 is related to m through: ⎛ ⎞   2 L L ⎠ Ln ⎝ + 1 + D D k2 ⎞ =m ⎛ (7.41)   2 k1 mL mL ⎠ Ln ⎝ + 1+ D D

In equation 7.41, all quantities are known except m, which can therefore be obtained. Alternatively, m can be found by using Figure 7.47, 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): kh = mk1

kv = k1 /m

(7.42) (7.43)

L/D 5 1 5 L/D 5 2 3

1

2

Note that A is in m 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 (7.40) m= kv

7

1

1.2 1.4 1.6 1.8 Ratio of hydraulic conductivity k2/k1

2

Figure 7.47 Relationship between k1 /k2 and m for two-stage borehole permeameter. (After Daniel 1989.)

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). 7.13

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 8). 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 7.48) 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 7.49) 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.

7.13 OFFSHORE IN SITU TESTS

Lift lines Rod support guide

Cone rod Friction sleeve Weight Pore pressure sensor

Wheel drive Frame

CPT cone

Figure 7.48 Seabed units to deploy the CPT offshore. (a and b: Image courtesy Swan Consultants Ltd., Copyright EFS Danson 2005.)

Drilling fluid pressure Overshot knob Metering cylinder

Drill pipe

Borehole Landing ring

Remote memory unit

Drilling fluid

Cone rod Dril bit

Soil formation

Cone tip

Figure 7.49

“Dolphin” downhole system to deploy the CPT offshore. (Courtesy of FUGRO Inc.)

133

134

7 IN SITU TESTS Hoisting sensor

Signal cable

Extension pieve Locking strips

Electronics

Battery case

Skirt Vane assembly

Figure 7.50 Seabed system for vane shear test. (Image courtesy Swan Consultants Ltd., Copyright EFS Danson 2005.)

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 7.51

Influence of sample disturbance on vane shear results. (After Denk et al. 1981.)

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 7.49) or from a seabed platform (Figure 7.50). 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 7.51; 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.

7.13 OFFSHORE IN SITU TESTS

135

PROBLEMS

Pressure on cavity wall (kPa)

7.1 Assume that the blow count profile shown in Figure 7.4 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 N60 , the corrected profile for stress level N1 , and the corrected profile for silt content N ′ . Then create the combined corrected profile for energy, stress ′ level, and silt content, N1(60) . 7.2 A pressuremeter test gives the test curve shown in Figure 7.2s. Calculate the first load modulus E0 , the reload modulus of the first loop Er1 , 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

50 ΔR Relative increase inprobe radius (%) R0 0

10

Figure 7.2s

20

30

40

Pressuremeter test results.

7.3 Use the CPT profiles of Figure 7.8 to identify the main soil layers. Then classify the soil in each layer according to the CPT classification systems of Figure 7.10 and Figure 7.11. 7.4 Develop the equation for a rectangular vane that links the maximum torque Tmax to the undrained shear strength su of a fine-grained soil. 7.5 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. 7.6 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. 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? 7.7 A plate test gives the load settlement curve shown in Figure 7.26. 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. 7.8 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? 7.9 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. 7.10 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? 7.11 A sand cone apparatus is used to check the dry density of a compacted soil. The weight of dry sand used to fill the test hole and the funnel of the sand cone device is 8.7 N. The weight of dry sand used to fill the cone funnel is 3.2 N. The unit weight of the dry sand is calibrated to be 15.4 kN/m3 . The weight of the wet soil taken out of the test hole is 7.5 N and the water content of the soil from the test hole is 13.2%. Calculate the dry density of the compacted soil.

136

7 IN SITU TESTS

7.12 A lightweight deflectometer is used to obtain the modulus of the compacted soil. The plate is 200 mm in diameter and the results of the tests are shown in Figure 7.36. Calculate the modulus of deformation of the soil. What approximate stress level and strain level does it correspond to? 7.13 A borehole is drilled into a deep and uniform clay layer to a depth of 1.5 m. A 75 mm inside diameter casing is lowered to the bottom of the 100 mm diameter borehole and sealed to the borehole walls. The water is bailed out so that the water level starts 1 m below the groundwater level outside of the casing at time equal 0. Three days later the water level has risen 0.3 m in the casing. Calculate the hydraulic conductivity k of the clay layer. 7.14 A 10 m thick layer of silty sand is underlain by a deep layer of high-plasticity clay. The groundwater level is 2 m below the ground surface. A 100 mm diameter boring is drilled to a depth of 10 m and cased with a screen that allows the water to enter the borehole freely along the borehole walls. A pump is set up to pump the water out of the hole and reaches a steady state condition after 2 days; at that time it is able to maintain the water level in the hole at a depth of 6 m when the flow rate is 0.2 cubic meters per minute. Additional boreholes indicate that the radius of influence of the depressed water level is 9 m. Calculate the hydraulic conductivity of the silty sand layer. 7.15 A cone penetrometer dissipation test is performed at a depth of 15.2 m below the groundwater level in a silt deposit. The results of the tests are given in Figure 7.43a. Calculate the hydraulic conductivity of the silt layer. 7.16 A sealed double-ring infiltrometer is used to evaluate the field-scale hydraulic conductivity of a 1 m thick clay liner underlain by a free-draining layer of sandy gravel. The SDRI has a square outside ring that is 4 m by 4 m and an inside ring that is 1 m by 1 m. The wall of the outer ring is embedded and sealed 0.45 m below the ground surface and the wall of the inner ring is embedded and sealed 0.15 m below the ground surface. Water is poured into the infiltrometer to a height of 0.5 m above the ground surface and the inner ring is capped. After a period of one week, during which the liner below the infiltrometer becomes saturated and a steady-state flow develops, the daily volume of water flowing into the liner is 0.01 m3 as measured by a plastic bag connected to the sealed inside ring. The soil swells, and vertical movement measurements of the inside ring indicate that this swelling amounts to 0.004 m3 per day. Calculate the hydraulic conductivity of the liner. 7.17 A two-stage permeameter test is conducted to evaluate the vertical and horizontal hydraulic conductivity of a clay liner. In stage 1, a 0.1 m diameter borehole is drilled to a depth of 0.35 m. A 0.075 m inside diameter pipe is lowered to the bottom of the open borehole and sealed to the walls of the borehole. A 10 mm inside diameter graduated tube is placed on top of the 75 mm diameter pipe; then the pipe and the falling head permeameter fitted on top of it are saturated and the water seeps through the liner. After reaching a steady state, the following measurements are recorded. At time equal 0, the water is 0.6 m above the ground surface. After 30 minutes of infiltration, the water has dropped to a height of 0.5 m above the ground surface. In stage 2, a 75 mm borehole is advanced 0.2 m below the bottom of the stage 1 borehole (0.55 m below surface). The falling head permeameter test is repeated and the water level falls from 0.6 m above the ground surface at time equal 0 to 0.5 m above the ground surface in 5 minutes. Calculate the vertical and horizontal hydraulic conductivity of the clay liner. 7.18 Discuss the advantages and drawbacks of in situ tests versus laboratory tests.

Problems and Solutions Problem 7.1 Assume that the blow count profile shown in Figure 7.4 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 N60 , the corrected profile for stress level N1 , and the corrected profile for silt content N ′ . Then create the combined corrected profile for energy, stress level, and silt ′ content, N1(60) . Solution 7.1 The corrections of the SPT values are shown in Table 7.1s and are based on the following formulas: 

 Emeasured (J) Correction for energy level : N60 = Nmeasured × 285 (J)  0.5 100 Correction for stress level : N1 = Nmeasured × ′ σv0 (kPa)

7.13 OFFSHORE IN SITU TESTS

 Nmeasured − 15 Correction for silt content : N = 15 + 2 ⎛ ⎞  0.5 N60 × 100 − 15 ′ σ0v ⎜ ⎟ ′ Combined corrections : N1(60) = 15 + ⎝ ⎠ 2 

′

Table 7.1s Measured Depth

Corrected SPT Values

Energy level

Stress level

Silt

Combination

Nmeasured

Emeasured

N60

γ sat

′ σov

N1

N′

′ N1(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

The corrections of the SPT values are plotted on the graph shown in Figure 7.1s. SPT Blow Count (blow/0.3 m)

0

0

10

20

30

40

Depth (m)

5

10

Nmeasured 15

N60 N1 N′

20

N′1(60)

Figure 7.1s

Corrected SPT values.

50

137

138

7 IN SITU TESTS

Problem 7.2 A pressuremeter test gives the test curve shown in Figure 7.2s. 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?

Pressure on cavity wall (kPa)

Solution 7.2 1500 1250 1000 750 500 250

40 50 0 10 20 30 Relative increase inprobe radius ΔR (%) R0

Figure 7.2s

Pressuremeter test results.

According to the test results shown in Figure 7.2s, the following parameters are obtained: 1500 = 12423 kPa (0.18 − 0.017) 1500 = 28928 kPa The reload modulus of the first loopEr1 = (1 + 0.35) (0.12 − 0.05) The yield pressure py = 700 kPa The horizontal pressure p0h = 120 kPa The limit pressure pL = 1200 kPa

• First load modulus E0 = (1 + 0.35) • • • •

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 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 7.3 Use the CPT profiles of Figure 7.8 to identify the main soil layers. Then classify the soil in each layer according to the CPT classification systems of Figure 7.10 and Figure 7.11. Solution 7.3 A total of 10 layers are identifiable from the CPT profiles of Figure 7.8 and are shown in Figure 7.3s and Table 7.2s. Furthermore, the porewater pressure profile can be extended back to zero pressure and indicates that the water level is at a depth of 2.5 m below the ground surface. The classifications of the soil layers based on Figures 7.10 and 7.11 are presented in Table 7.2s, Figure 7.4s, and Figure 7.5s. At a coarser level, the stratigraphy can be simplified as shown in Figure 7.6s.

7.13 OFFSHORE IN SITU TESTS Tip Resistance qT (MPa) 0

10

20

30

Sleeve Friction fs (kPa) 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 7.3s

Soil layers. (Courtesy of Professor Paul Mayne, Georgia Institute of Technology)

Table 7.2s Depth Layer 1 2 3 4 5 6 7 8 9 10

qt

(m)

(Mpa)

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

4.0 0.8 10.0 26.0 13.0 26.0 37.0 28.0 0.9 10.0

(Bar) 40 8 100 260 130 260 370 280 9 100

Classification of Soil Layers fs

FR

(kPa)

(%)

Figure 7.10

Figure 7.11

3.40 4.30 0.70 0.70 1.10 0.60 0.60 0.60 2.70 2.00

Sandy silts & silt Clays Sands Sands Sands Sands Sands Sands Clayey silts Silty sands

Silty sand Clay Sand to silty sand Sand Sand to silty sand Gravelly sand Gravelly sand Gravelly sand Silty clay Silty sand

100 10 60 210 100 200 250 200 40 160

139

7 IN SITU TESTS

Robertson & Campanella (1983)

Cone Bearing, qt (bar)

1000

8 4, 6 and 7 Sands 5 3

100

10 Silty sands Sandy silts & silts

10

1

Clayey silts

9

2 Clays

0 0

Figure 7.4s

1

2 3 4 Friction Ratio, FR = fs/qt (%)

5

6

Soil classification based on CPT results. (Courtesy of Professor Paul Mayne, Georgia Institute of Technology)

Robertson et al. (1986)

1 = Sensitive clay

Zone 12 - sand to clayey sand

2 = Organic soil

1000 Zone 10 Cone bearing, qt (bar)

140

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 7.5s

Soil classification based on CPT results. (Courtesy of Professor Paul Mayne, Georgia Institute of Technology)

7.13 OFFSHORE IN SITU TESTS Sleeve friction fs (kPa)

Tip resistance qT (MPa) 0

10

20

30

40

0

200

400

Porewater pressure u2 (kPa) 600 –500

0

500

Soil classification by visual method

1 2 3 4 5 6 7 8 9 sand OC fissured clayey silt

5

Clean sand

Depth (m)

10

15 Soft silty clay 20

Sand

25

Figure 7.6s

SBT (1990)

1000

0

141

6 5 3 4 4 3 4 6 5 6 6 6 6 6 6 6 7 6 6 3 2 2 2 2 2 2 2 2 2 2 3 2 2 4

sand Silt mix

Organic clay or silt

sand

Simplified stratigraphy. (From Mayne 2007a, b, Courtesy of Professor Paul Mayne, Georgia Institute of Technology, USA.)

Problem 7.4 Develop the equation for a rectangular vane that links the maximum torque Tmax to the undrained shear strength su of a fine-grained soil. Solution 7.4

T

H

D

Figure 7.7s

Vane subjected to torque.

The failure surface around the vane is a cylinder with a diameter D and a height H. The torque generated from the sides of the cylinder is: D T1 = π DHsu 2

142

7 IN SITU TESTS

The torque generated by the top and bottom of the cylinder (ignoring the area occupied by the rod) is: T2 =



0

D 2

2.π.r.su .r.dr = 2.π.su

T = T1 + 2T2 = π DHsu T = π su .D

2

D H + 2 6

For vanes with H = 2D, the equation becomes:



T1 =



r3 3

 D2 0

= π.su

D3 12

D D3 + 2π.su . 2 12

7 π s D3 6 u

Problem 7.5 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 7.5 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 dFy

s

t

t

Pressure sensor

s

r

dFx y du

T u

x

pavg.r

Figure 7.8s

Applied stresses on vane.

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 7.8s. A free-body diagram of a quadrant of the failing soil mass gives the following equations:  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

(σ.r. sin θ + τ.r. cos θ )dθ = −σ.r. cos θ + τ.r. sin θ |02 = (τ + σ )r

7.13 OFFSHORE IN SITU TESTS

At failure:

τf = σ ′ tan ϕ σ ′ = p − τf → τf = (p − τf ) tan ϕ → τf =

143

tan ϕ p 1 + tan ϕ

From problem 7.4, we have: D3 D3 D + π τtop + π τbottom 2 12 12 3 D D3 tan ϕ D T = π DH p + π γ ′ z tan ϕ + π γ ′ (z + H ) tan ϕ 1 + tan ϕ 2 12 12 T = πDHτside

D3 D2 tan ϕ H p + (2z + H )γ ′ π tan ϕ 2 1 + tan ϕ 12   3 3 1 D D tan2 ϕ + (2z + H ) π γ ′ + π pD 2 H − T tan ϕ − T = 0 (2z + H )π γ ′ 12 12 2 √ 2 −B + B + 4AT tan φ = 2A D3 A = (2z + H )π γ ′ 12 3 D 1 B = (2z + H )π γ ′ + π pD 2 H − T 12 2

T =π

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 7.6 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. 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? Solution 7.6

T = 2(p + Su.l) Failure wedge p N

W tf

Su

L

B 0 < f < Su i

y P X

Figure 7.9s

Borehole shear test.

144

7 IN SITU TESTS

⎧ τf .B ⎪ ⎪ Fx = 0 → N sin i = fB + cos i ⎨ cos i ⎪ τf .B ⎪ ⎩ sin i Fy = 0 → p = N cos i + W + cos i τf B f.B N= + sin i sin i 1 2 W = γ B tan i 2   τf B τf B f.B 1 p= + sin i cos i + γ B 2 tan i + sin i sin i 2 cos i τf B f.B 1 p= + + γ B 2 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 B 2 , it is reasonable to neglect the influence of the weight of the wedge in calculating P: p∼

τf B f.B + + τf B tan i tan i tan i

By assuming i = 45◦ +ϕ /2 and using Mohr-Coulomb theory, we have: p∼ p∼

  f.B φ s B cos φ  + u  + su B cos φ tan 45 +  2 tan 45 + φ2 tan 45 + φ2 f.B  + 2su B  tan 45 + φ2

Shear stress on the failure surface

t

Su tf = Sucosw

f s

Figure 7.10s

Stress envelope.

If  = 30◦ for upper and lower limits of f, we will have: ⎧ f = 0 → p = 2su B ⎪ ⎪ ⎨ 

√  3 ⎪ ⎪ ⎩f = su → p = 2 + 3 su .B  √  3 2su B < p < 2 + su .B 3

P is the force needed to fail the wedge of soil above the borehole shear device. 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

7.13 OFFSHORE IN SITU TESTS

145

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 7.7 A plate test gives the load settlement curve shown in Figure 7.26. 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 7.7 The pressure versus displacement/width curve is shown in Figure 7.11s.

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 7.11s

Pressure versus displacement/width curve.

The soil modulus is calculated based on point A in Figure 7.11s using the following equation: E=

π(1 − v2 )pB π(1 − 0.352 ) × 0.36 π(1 − v2 )p = = = 62 MPa s 4s 4× B 4 × 0.004

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 property beneath the 3 m by 3 m square footing unless they are the same. Problem 7.8 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 7.8 The elastic settlement equation for a plate load test is: s=

I (1 − ν 2 )pB E

146

7 IN SITU TESTS

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 ν is the Poisson’s ratio. The modulus of subgrade reaction K is calculated as the ratio between the pressure and the settlement: K=

p = s

p

=

I (1−ν 2 )pB E

E I (1 − ν 2 )B

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 7.9 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 7.9 Based on equation 7.14, 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 s2 =  1n =  t 1

t2

10

0.04 1 438000

= 16.8 mm

So, the calculated settlement of the footing after 50 years under a pressure of 100 kPa is 16.8 mm. Problem 7.10 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 7.10 Using Figure 7.30 and a PET hole depth of 6 mm, the soil 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 7.12s). With 2 hours of overtopping at this rate, 160 mm of erosion is estimated.

100000

Very high erodibility I > 75 mm

10000 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

Sensitivity starts below 2.5 pF

2 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

Figure 9.5 Filter paper method for water tension measurement. (a) Filter paper (matric suction only). (b) Filter paper (total suction). (c) High-precision scale with hood. (d) A calibration curve. (From Bulut et al., 2001. Courtesy of Dr. Bulut)

extremely carefully and quickly, as the weights involved are very small and the relative humidity of the air in the laboratory can influence the weight of the filter paper when it is transferred from the sample chamber to the scale chamber. 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 9.6). 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

Teflon filling

Ceramic bulb Thermocouple

Chamber

6 mm

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 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

Temperature sensor

Figure 9.6 (a) Cross section of a thermocouple psychometer. (b) Thermocouple psychometer. (b: Courtesy of Wescor-Elitechgroup.)

Chilled mirror psychrometers can be used to give the total suction of a soil (Figure 9.7). 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

9.2 MEASUREMENTS

177

Tensiometers Tensiometers can be used to measure the water tension or matric suction in a soil (Figure 9.8). 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 9.9). 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 it simply translates the origin of reference by applying an air pressure equal to the water tension (Figure 9.10). 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 the air pressure is increased in steps, each pressure step will drive

Figure 9.7 Chilled mirror psychrometer. (Photo courtesy of Decagon Devices, Inc.)

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 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). Mercury manometer

Bourdon vacuum gauge

Pressure transducer

Manual observation

To chart recorder for continuous observation

Manual observation Ground surface

Porous cup

Figure 9.8 Tensiometers. (a) Tensiometer with pressure-vacuum gage. (b) Types of tensiometers. (c) Tensiometer with pressure transducer. (a: Courtesy of Envco Global. c: Courtesy of STEP Systems GmbH, www.stepsystems.de)

178

9 LABORATORY TESTS

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).

Air-tight chamber

Air pressure supply Soil specimen High air-

entry disk

Distilled water reservoir

Salt Solution Equilibrium (SSE)

Figure 9.9 Pressure plate apparatus: (a) 500 kPa pressure plate. (b) 1500 kPa pressure plate.

Tension Uw

0 Ua

Uw

Figure 9.10

Compression Natural state Axis translation Ua

Pressure plate apparatus

Axis translation for water tension determination.

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 9.11) will generate a certain relative humidity in the air above it. The higher the salt concentration is, 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 9.3 gives the osmotic suction for different salts and

Cover

Porcelain

Soil specimen

plate

Salt solution

Figure 9.11

Salt solution equilibrium containers for water tension determination.

9.2 MEASUREMENTS

179

Table 9.3 Osmotic Suction in kPa of Some Salt Solutions at 25◦ C Osmotic Suction in kPa at 25◦ C Molality (mol/kg) 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

Na Cl

KCl

NH4 Cl

Na2 SO4

CaCl2

Na2 S2 O3

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

MgCl2 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

(After Bulut et al. 2001.) 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 (6 pF). 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. 9.2.5

Normal Strain

A normal strain ε is defined in one direction as the change in length z divided by the initial length z between two points. A normal strain is measured either by measuring a displacement and a length (z/z) or by using a strain gage (ε). Measurements of length are done with a ruler or a set of calipers (Figure 9.12). Displacements are measured with mechanical

devices such as dial gages (Figure 9.12) or electrical devices such as LVDTs, DCDTs, and potentiometers. A linear variable differential transformer (LVDT) (Figure 9.13) 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 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

180

9 LABORATORY TESTS

Figure 9.12

Mechanical devices to measure displacement: (a) Calipers. (b) Dial gage.

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. 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 9.14) 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 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

Primary coil

Core

small anchor blocks solidly connected to the material that is deforming. Between these two anchors is a high-tensilestrength 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. 9.2.6

Shear Strain

A shear strain γ is defined for two perpendicular directions (x and y as shown in Figure 9.15). 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 9.15). It can be shown (Chapter 10) that the shear strain in this case is given by: γxy = ε1 − ε2

(9.1)

Secondary coils

9.2.7

Figure 9.13 Linear variable differential transformers: (a) Principle. (b) Device.

Bender Elements

A bender element (Figure 9.16) 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 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,

Y

X

2 1

Figure 9.14 Foil and vibrating wire strain gages. (a) Foil strain gage. (b) Model 4000 Vibrating Wire strain gage. (b: Courtesy of Geokon, Inc.)

Figure 9.15

Getting shear strain from two normal strain gages.

9.3 COMPACTION TEST: DRY UNIT WEIGHT

181

Direction of elements trip and soil particle movement

–voltage

+voltage Direction of shear wave propagation

Bearing plate

Figure 9.16

Bender elements: (a) Principle. (b) Device.

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 creates an electrical signal that can be measured. When the bender element generates a shear wave, the wave travels through the soil and reaches the 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

9.3 9.3.1

Saturated Soils

Most of the time, the soil in a compaction test is unsaturated. 9.3.2

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 γd vs. water content w curve (Figure 9.17). 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:

(9.2)

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.

Ww = w Ws 25

20 Dry unit weight gd (kN/m3)

COMPACTION TEST: DRY UNIT WEIGHT

gdmax

20

gdmax C

15 19

B

A

18

0

D

10 5

Wopt 4

8 12 16 Water content w (%)

Figure 9.17

20

0

Wopt 0

4

8 12 16 Water content w (%)

Compaction curve.

20

(9.3)

182

9 LABORATORY TESTS

Figure 9.18 Compaction equipment and test: (a) Compaction mold. (b) Compaction test. (c) Compaction hammer. (a and c: Courtesy of Forney LP, Hermitage, PA.)

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 9.18) 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 9.4 gives the detailed requirements. At the end, weigh the mold plus soil and calculate the soil weight Wt . The dry unit weight is obtained by: Wt γd = (9.4) Vt (1 − w) where γd is the dry unit weight, Wt is the total weight of the soil sample in the mold, Vt is the total volume of the sample, and w is the water content of the sample. The combination of γd 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 (Figure 9.17). 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 9.17 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 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 γd and the water content w is a function of the degree of saturation: γd =

Table 9.4 Compaction Requirements for Standard Proctor Compaction Test 102 mm diameter 116 mm high mold

152 mm diameter 116 mm high mold

3 soil layers 25 blows per soil layer Hammer weight 24.5 N Hammer drop height 305 mm Volume 9.43 × 10−4 m3 Total energy 600 kN.m/m3

3 soil layers 56 blows per soil layer Hammer weight 24.5 N Hammer drop height 305 mm Volume 21.2 × 10−4 m3 Total energy 600 kN.m/m3

SGs γw S + Gs w

(9.5)

This relationship can be demonstrated as follows: Ws γs Vs Gs γw SGs γw = = = Vv V V Vt Vv + V s 1+ S+ w v Vs Vv Vs SGs γw SGs γw = = (9.6) V W W S + Gs w S+ w w s W w W s Vs

γd =

Equation 9.6 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 9.19). The

9.3 COMPACTION TEST: DRY UNIT WEIGHT

183

21 S 5 0.6 S 5 0.8

S51

Lines of equal degrees of saturation

Dry unit weight, gd (kN/m3)

20 g dmax gd 5

SGS gw S1GSW

19 Modified proctor, 2700 kN*m/m3 18 g dmax

17 Standard proctor, 600 kN*m/m3 16

0

Figure 9.19

4

8 12 Water content, w (%)

wopt 16

20

Compaction curve for Standard and Modified Proctor Compaction Tests.

18 Dry unit weight, gd (kN/m3)

wopt

Table 9.5 Compaction Requirements for Modified Proctor Compaction Test

Saturation line S 5 1.0 17

16 Line of equal water tension

15 Increasing water tension 14 10

14

18

22

26

30

Water content w (%)

Figure 9.20

Compaction test and water tension lines.

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 9.20. 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. The MPCT is very similar to the SPCT except for the different requirements listed in Table 9.5. The data reduction is the same and the result is also the γd vs. w curve. The difference is that, due to the higher compaction effort (2700 kN.m/m3

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

compared to 600 kN.m/m3 ), the curve for the MPCT is located higher than the curve for the SPCT (Figure 9.19). 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 7 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 9.21. Both soil A and soil B have a total unit weight of 20 kN/m3 ,

184

9 LABORATORY TESTS

Soil A Volume (m3)

1

Soil B Weight (kN)

0.1

Air

0

0.25

Water

2.5

0.65

Solids

17.5

20

Volume (m3) 0.19

Air

0.1

Water

0

1

20 0.71

gt 5 20 kN/m3 gd 5 17.5 kN/m3

Figure 9.21

Weight (kN)

20

Solids

gt 5 20 kN/m3 gd 5 19 kN/m3

Three-phase diagram showing the usefulness of dry unit weight.

yet soil A has a dry unit weight of 17.5 kN/m3 whereas soil B has a total and 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).

modulus, and displays the modulus E. 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 9.23). The modulus obtained with the BCD corresponds to a reload modulus, to a mean stress level averaging about 50 kPa

9.4 COMPACTION TEST: SOIL MODULUS Acquisition, processing & readout

9.4.1 Saturated Soils Most of the time, the soil in a compaction test is unsaturated.

Load-cell Strain-gage instrumented plate

2 mm

150 mm

60

20

50

16

40

12

30 8

20

4

BCD modulus (MPa) 3 Dry unit weight (kN/m )

10 0 0

2

4

6

8

10

12

0 14

Water content, w (%)

Figure 9.23

Compaction equipment and test.

Dry unit weight, �d (kN/m3)

Figure 9.22 BCD apparatus to get soil modulus during a Proctor compaction test: (a) BCD principle. (b) BCD on Proctor mold.

Modulus, E (MPa)

The compaction test described in section 9.3 yields the dry unit weight γd 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 γd vs. w curve by using a device called the BCD (Figure 9.22). 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 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 152 mm diameter compaction mold (Figure 9.22). 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

600 mm

9.4.2 Unsaturated Soils

9.5 CONSOLIDATION TEST

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.

9.5 9.5.1

185

s 5 Total stress Draining water

s9

CONSOLIDATION TEST

uw : Water stress

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 9.24). 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 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 9.25). 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

Loading

Dial gauge or LVDT

Porous stone Soil specimen Porous stone

Water Soil skeleton feels the effective stress s9 Real sample

Rheological model

Figure 9.25

Consolidation model.

increase in effective stress (σ ′ = σ − uw in this case, because the soil is saturated) and a settlement of the soil; this is the process of consolidation (Figure 9.26). 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 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, Fig. 9.26) gives one point (vertical effective stress σ ′ and vertical strain ε) on the consolidation test stress-strain curve (stress vs. strain, Figure 9.27a). The upward curvature of this stress-strain curve and the lack of maximum stress or failure stress or strength is 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 void ratio e versus the decimal logarithm of the effective stress log σ ′ (Figure 9.27b). 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 9.27). As such, Cc is: e , (9.7) Cc =  log σ During each 24-hour loading step, the decrease in sample height H is recorded as a function of time t to be able to develop the H vs. t curve. The vertical strain ε is obtained by dividing the change in height H by the original height Ho of the sample. Figure 9.28 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:

Figure 9.24 Consolidation test and equipment: (a) Principle. (b) Sample in ring. (c) Complete setup. (b: Courtesy of Lev Buchko, P.E. // Timely Engineering Soil Tests, LLC. c: Courtesy of Humboldt Mfg. Co.)

cv = T

H2 t

(9.8)

where T is the time factor, H the drainage length, and t the time elapsed. The drainage length H is equal to the height

186

9 LABORATORY TESTS

1

s

uw s9

2

3

4

s

s

s

0

s

0

uw

½ uw

0

0

s 2 uw

s 2 ½ uw

s

s

s

Stress, s 2

1

s

3

4

s9

uW

24 hrs Time, t

Settlement, s

end of load step 24 hrs

Immediate compression

Time, t

Figure 9.26

Consolidation process.

Ho 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, Ho /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 onedimensional consolidation theory (see Chapters 11 and 14 for more on consolidation theory). This time factor is linked to the average percent consolidation U, defined as: U=

s(t) ue =1− smax ue max

(9.9)

where s(t) is the settlement at time t, smax is the settlement at a time equal to infinity, ue is the 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 9.29. 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 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 9.30). 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 correspond to water being expelled from the pores (consolidation). Cassagrande proposed the following way to find ε0 (Figure 9.30): 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: ε0 = ε1 − (ε2 − ε1 ) = 2 ε1 − ε2

(9.10)

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

9.5 CONSOLIDATION TEST

187

Effective vertical stress, �′ (kPa)

7000 6000 5000 4000 3000 2000 1000 0 0

0.02

0.04

0.06 0.08 Strain, �

0.1

0.12

0.14

(a) 0.6 0.55

Void ratio e

0.5 1 Cc

0.45

0.4

0.35 0.3

1

10

100 1000 Effective vertical stress, �′ (kPa)

10000

(b)

Figure 9.27

Consolidation test results: (a) Stress-strain curve. (b) e log σ ′ curve

beginning of the parabola satisfies equation 9.10. 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 9.30. 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. 9.8 is used to calculate cv . All other quantities are known, including T50 = 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 9.31). Then a straight line is fitted to the early part of the curve (AB on Figure 9.31). 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 9.31). The intersection of line AC with √ the ε vs. t curve corresponds to t90 . Once t90 is known, Eq. 9.8 is used to calculate cv . All other quantities are known,

including T90 = 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 longterm effective stress that the soil has been subjected to. The following procedure is recommended to obtain σp′ from the consolidation test (Figure 9.32). Choose the point of highest curvature on the ε vs. log σ ′ curve (Point A on Figure 9.32); 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

188

9 LABORATORY TESTS

Time (min) 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 Stress, �′

0

0

0.01

Strain, �

0.02 0.03

Strain, �

0.04 0.05 0.06

Stress, �′

0.05

Strain, �

0.06

0.07

Strain, �

0.08

0.09 0.084 Stress, �′

0.086

Strain, �

0.088 0.09

Strain, �

0.092 0.094 0.096

Figure 9.28

Vertical strain vs. time for three consolidation test loading steps.

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 9.32). 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 uw at the bottom of the sample is kept at 5 to 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) uw 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 uw 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

9.5 CONSOLIDATION TEST

Percent consolidation U (%)

0 20 40 60 80 100

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Time factor, T

20 40 60 80 100 0.001

0.01

0.1

1

Time factor, T

Figure 9.29

Percent consolidation vs. time factor.

0 «0 0.01

x x

0.02 Strain, «

Percent consolidation U (%)

0

t1

t2 = 4t1

t50

«50 0.03

0.04 «100 0.05

0.06

End of primary consolidation

0.1

1

10

100

1000

Log time (min)

Figure 9.30 1938.)

Log time method to obtain the coefficient of consolidation. (After Cassagrande

189

190

9 LABORATORY TESTS

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 10):

0 A

0.01

Strain, «

0.02

σ ′ = σ − α uw 0.03

B C

0.04

0.05

x

0.15x

Ït 90 0.06 0

10 20 30 Square root of time, (min1/2)

40

Figure 9.31 Square root of time method to obtain the coefficient of consolidation. (After Taylor 1948.)

prevents lateral deformations and may not represent the true deformation of the soil in the field. 9.5.2 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

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 10. 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 9.8).

9.6 9.6.1

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 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

0.6

σ′p

0.55 A

Void ratio e

0.5

α α

0.45

0.4

0.35

0.3

σ′p 1

Figure 9.32

10

(9.11)

100 1000 Effective vertical stress, �′ (kPa)

10000

Method to determine the preconsolidation pressure from the consolidation test.

9.6 SWELL TEST

9

Vertical strain, « (%)

8 7 6 5 4 3 2 1 0

0

1

Figure 9.34

2

3 4 5 Time, t (days)

Swell pressure

Figure 9.33

Swell test equipment.

8

Increase in volume

E Swell H O

A

Prevent swell

D

F I

C J

G

Collapse

Unsaturated Soils

The swell test (Figure 9.33; 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 of water, and measuring the vertical swell movement (vertical strain) as a function of time (Figure 9.34). 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 9.35, path CD on Figure 9.36). If a vertical load is applied, the

7

Swell test results: Vertical strain vs. time.

K

9.6.2

6

B

Vertical strain, « (%)

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 9.6.2.

191

Decrease in volume

Vertical total stress, log s

Figure 9.35 total stress.

Shrink-swell test results: Vertical strain vs. vertical

test is simply called a swell test (path DE on Figure 9.35, path CE on Figure 9.36). Note that after swelling, a regular consolidation test can be performed on the sample (path BC and EF on Figure 9.35). 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 9.35). 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. 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 9.35). During this test, the volume of the sample is maintained constant and equal to its initial volume. When

192

9 LABORATORY TESTS

W

Swell limit W (free swell test) Wsw

WSH shrink limit (free shrink test)

D E Shrinkswell index C γw/γd 1 B Unsaturated

Iss 5 (∆W)max

Gravel, clean sand W Dirty sand, siltIss Iss 5 0 2 10 % Iss 5 0

ted

ra atu

∆V/V ∆V/V Clay, low plasticity W Clay, high plasticity Iss 5 10 2 30 % Iss 5 30 2 70 %

W

(∆V/V)max

S

0

∆V/V ∆V/V

W (%)

∆V/V

W (%) Bentonite

76

Kaolinite

56%

38 24%

20

14 –0.15 0 0.17

∆V/V

–0.3

0

0.5

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.

9.7 SHRINK TEST 9.7.1 Saturated Soils The shrink test (Figure 9.37) consists of trimming a sample of soil into a cylindrical shape, measuring its dimensions, and recording its weight. The initial volume Vo and the initial weight Wo 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 Ws . The average water content w of

Relative change in volume, DV/V (%)

Figure 9.36 Shrink-swell test results: Water content vs. relative change in volume. (a) Idealized behavior. (b) Typical ranges. (c) Low-plasticity clay example.

Free shrink test for shrinkage limit of undisturbed

2 4 6 8

10 0

Figure 9.38 time.

10

20

30 40 Time, t (hrs)

50

60

Free shrink test result: Relative change in volume vs.

the sample at any time during the test is (W(t) − Ws)/Ws. The results of the test consist of a plot of the relative change in volume (V(t) − V0 )/V0 = V/V0 as a function of time t (Figure 9.38), and the water content w as a function of the relative change in volume V/V0 (Figure 9.36). 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 9.36(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. 9.7.2

Figure 9.37 sample.

0

Unsaturated Soils

The shrink test applies equally to saturated soils and unsaturated soils. In both cases the water is in tension throughout

9.9 DIRECT SHEAR TEST

0 2 Collapse due to inundation

Vertical strain, « (%)

the test. The soil may start as a saturated soil, but, as it dries, it goes through the air entry value uwe , at which point it becomes unsaturated. The shape of the relative volume change vs. time curve for the free shrink test (Figure 9.38) is similar to the shape of the relative volume change vs. time curve for the free swell test (Figure 9.34). 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 9.36a). 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.

193

4 6 8 10 12

1

10

100

1000

10000

Vertical total stress, s(kPa)

Figure 9.39

9.8 9.8.1

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. 9.8.2

Collapse test: Vertical strain vs. vertical stress.

COLLAPSE TEST

Unsaturated Soils

The collapse test (ASTM D5333) is the same for saturated and unsaturated soils. It is performed with the sample confined 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 9.39.

9.9 9.9.1

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 9.40). The 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 9.41). 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 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

194

9 LABORATORY TESTS

0 Water grooves

Loading yoke

Water inlet

Pressure pad

Porous stone Sample Horizontal force

Loading cell

Porous stone

Large container Water grooves

Figure 9.40 Direct shear test and equipment. (a) Principle. (b) Sample. (c) Complete setup. (b: Courtesy of Lev Buchko, P.E. // Timely Engineering Soil Tests, LLC.) s9

t t

Peak

Critical state

dv

dh (mm) dv

Figure 9.41

dh (mm)

Contracts

Direct shear test results stress-displacement curve.

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 15 on shear strength, the envelope is represented by the following equation: s′ = c + σ ′ tan φ ′ (9.12) 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

150 Shear stress t (kPa)

s9

Dilates

dh

w9 100 Right 50 0 c9 0

100 50 150 200 Normal effective stress s9 (kPa)

250

150 Shear stress t (kPa)

t

Residual

envelope, it is necessary to run at least two direct shear tests to obtain c′ and φ ′ for a given soil (Figure 9.42). When soils are subjected to shearing, they can increase in volume (dilate), decrease in volume (contract), or not change volume. If a soil dilates during shear, the shear strength

uw3 100 uw2

Þw9 Wrong

uw1

50 Þc9

0 0

200 50 100 150 Normal total stress s (kPa)

250

Figure 9.42 Example of direct shear test strength results for saturated soils.

9.10 SIMPLE SHEAR TEST

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. 9.9.2

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 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 9.2.4, 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: ′

σ = σ − α uw

(9.14)

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 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 uwe as shown in Chapter 10. 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 9.43. 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 9.43). The apparent cohesion capp is equal to: capp = −α uw

w9

200 stanw9

(9.15)

Right

100 Capp 5 (–auw–bua) tanw9 c9

0

(9.13)

0

c9 100 200 300 400 Normal effective stress s9 (kPa)

300 Shear stress t (kPa)

tan ψ = z/x

300 Shear stress t (kPa)

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 15 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.

195

500

–a3uw3 –a2uw2

200 Þw9 –a1uw1

Wrong

100 Þc9

0 0

100 200 300 400 Normal total stress s (kPa)

500

Figure 9.43 Example of direct shear test strength results for unsaturated soil.

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.

9.10 9.10.1

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 9.44). A vertical load is applied to the top of the sample and maintained constant during the test. This vertical

Top cap

Vertical stress Brass rings

Base pedestal

Shear stress

Figure 9.44 Simple shear test equipment: (a) Principle. (b) Complete setup. (b: Courtesy of GDS Instruments.)

196

9 LABORATORY TESTS

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:

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 15 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 z to the change in horizontal movement x (Eq. 9.13). 9.10.2

γ = x/ho

(9.16)

where x 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 9.45). The shear strength τf 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 τf 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. When soils are subjected to shearing, they can increase in volume (dilate), decrease in volume (contract), or not change volume. If a soil dilates during shear, the shear strength

t

s9 t

t

Peak

g

9.11 9.11.1

UNCONFINED COMPRESSION TEST Saturated Soils

The unconfined compression test (ASTM D2166) (Figure 9.46) 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 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 σ3 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 σ1 . The sample compresses and the vertical displacement h 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 = h/h. The result of an unconfined compression test

Load transducer Soil sample

Critical state g

«v

Dilates Contracts

Figure 9.45

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.

Displacement transducer

dv

s9

Unsaturated Soils

Simple shear test results.

g

Figure 9.46 Unconfined compression test equipment: (a) Principle. (b) Complete setup. (a: After Ian Smith. b: Courtesy of ELE International.)

9.11 UNCONFINED COMPRESSION TEST

160

197

qu

Axial stress, s (kPa)

140 120 100 80 E

60 40 20

«f

0

2

Shear stress, t (kPa)

0

4 6 Axial strain, «(%)

8

10

100 80

su

60 40

su 5 qu /2

qu

20 40

80

120

160

Total stress, s (kPa)

Figure 9.47

Unconfined compression test results.

is a complete total stress σ vs. strain ε curve for the soil sample under zero lateral confinement (Figure 9.47). 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 15. (9.17) su = qu /2 An unconfined compression modulus of deformation E can also be obtained from this test as: E = σ1 /ε

(9.18)

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. 9.11.2

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 ′

(9.19)

In the unconfined compression test, the horizontal total stress is zero, therefore: σh = 0 = σh′ + α uw and therefore σh′ = −α uw

(9.20)

Meanwhile, the vertical total stress at failure is equal to qu ; therefore: σv = qu = σv′ + α uw and therefore σv′ = qu − α uw (9.21) The shear strength s is given by the point of tangency between the effective stress Mohr circle and the shear strength envelope (Figure 9.48). Triangle ACD on Figure 9.48 is such that: CD = sin ϕ ′ = AD

0.5((qu − αuw ) − (−αuw )) 0.5((qu − αuw ) + (−αuw )) +

c′ tan ϕ ′ (9.22)

Which leads to uw =

0.5qu (sin ϕ ′ − 1) + c′ cos ϕ ′ α sin ϕ ′

(9.23)

198

9 LABORATORY TESTS

Uw 5 – Shear stress t

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 top of the frame. 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. 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. 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 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 σ3 . 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 σ3 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

qu for c 5 0, w9 5 30° 2s w9

C

B c9 A

D

c9 O –auw tanw9

qu–auw Normal effective stress, s9

Figure 9.48 Water tension and unconfined compression strength relationship.

Eq. 9.23 gives the water tension at failure in the unconfined compression test. If it is further assumed that c′ = 0, ϕ ′ = 30◦ , and α = S, then Eq. 9.23 becomes: uw = −

9.12 9.12.1

qu 2S

(9.24)

TRIAXIAL TEST Saturated Soils

The triaxial test (ASTM D5311) (Figure 9.49) 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 Load

Piston Cap Sealing ring

Fluid inside chamber Perspex cylinder

Porous disc Pore-pressure measurement and drainage

Soil specimen Membrane Cell pressure measurement

Figure 9.49 Triaxial test equipment: (a) Principle. (b) Equipment. (b: Courtesy of Geotechnical Testing Equipment Ltd., UK.)

9.12 TRIAXIAL TEST

vertical direction increases. This stress is the major principal stress σ1 : (9.25) σ1 = σ3 + Q/A where σ3 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 σ3 . 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 = h/h) where h is the initial height of the sample and h is the change in height of the sample. Figure 9.50 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 (σ1f − σ3 ). This failure stress, along with information on the water stress, is used to obtain the effective stress shear strength parameters c′ and φ ′ . This process requires use of the Mohr circle (see Figure 9.51 and Chapter 15). 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 3 stresses to 2 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 9.51). Because the failure envelope is described by two parameters c′ and ϕ ′ (Eq. 9.12), a minimum of two

199

triaxial tests at two different confining pressures (σ3 ) must be performed to obtain the effective stress cohesion intercept c′ and the effective stress friction angle ϕ ′ . Figure 9.52 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 − 2νσ3 )/ε

(9.26)

where E is the total stress modulus of deformation of the soil, σ1 and σ3 are the major and minor principal total stresses respectively, ν is 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: E′ = (σ1′ − 2νσ ′3 )/ε

(9.27)

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: p=

σ1 + σ3 2

or

p=

σv + σh 2

(9.28)

q=

σ1 − σ3 2

or

q=

σv − σh 2

(9.29)

where σv and σv 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′ ): σ1′ + σ3′ or 2 σ ′ − σ3′ q′ = 1 =q 2 p′ =

p′ = or

σv′ + σh′ 2 σ ′ − σh′ q′ = v =q 2

(9.30) (9.31)

where σv′ and σv′ are the vertical and horizontal total stresses in a triaxial test, for example. Examples of effective stress paths are shown in Figure 9.53 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. 9.12.2

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.

9 LABORATORY TESTS

200

200

Normally consolidated or loose

Deviator stress: s1–s3 (kPa)

Over consolidated or dense

150

150 100

0

100

Maximum or peak strength

50

0

2

4

Residual strength 6

50 0

10

8

75

0

2

4

Water stress (pore pressure), uw (kPa)

50

25

25

0

2

4

6

0

10

8

225

0

2

4

Deviator stress: s1–s3 (kPa)

10

8

200 Normally consolidated or loose

Over consolidated or dense

150

150 100

100

Maximum or peak strength

50

0

2

4

Residual strength 6

50 0

10

8

0

2

4

6

10

8

4

4

Normally consolidated or loose

Over consolidated or dense

2

2 0

6

225

200

0

10

8

Normally consolidated or loose

50

0

6

75 Over consolidated or dense

Volumetric strain, «n (%)

0

2

4

6

0

10

8

22

22

24

24

0

2

4

6

10

8

Figure 9.50 Triaxial test results (example stress-strain curves): (a) Consolidation, undrained test. (b) Consolidation, drained test. 250

250

s93f 5 100 kpa

150

200 Shear stress s (kPa)

200

s12s3 (kPa)

Deviatoric stress

200

s93f 5 60 kpa

100

s93f 5 30 kpa

50 0

2

4

6

Axial strain « (%)

Figure 9.51 soils.

8

100 50

c 5 10

0

F 5 308

150

10

c 5 10 kpa

0

0

100

200

300

Effective normal stress s9 (kPa)

Triaxial test results: Examples of Mohr circles and strength envelope for saturated

400

9.12 TRIAXIAL TEST

201

Shear stress τ (kPa)

150 φ′ 100 Right 50

c′

0 0

50

150 100 200 Normal effective stress σ′ (kPa)

300

250

Shear stress τ (kPa)

150

≠φ′

100

Wrong 50

≠ c′

0 0

uw1 50

uw2

uw1

100

150

uw2 200

250

300

Normal total stress �′ (kPa)

Half deviator stress q 5 q′ 5 (�1 2 �3)/2 (kPa)

Figure 9.52

Triaxial test results: Mohr circles and strength envelope for saturated soils.

Effective stress path (lightly overconsolidated) Effective stress path (normally consolidated) Effective stress path (heavily overconsolidated)

Total stress path

Mean effective stress p′ 5 (�′1 1 �′3)/2 (kPa)

Figure 9.53

Triaxial test results: Stress paths.

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). 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

202

9 LABORATORY TESTS

Shear stress t (kPa)

300

w9 200 Right σtanw9 100 capp 5 (2auw 2bua) tanw9 c′

0

100

0

c9 300

200

400

600

500

Normal effective stress s9 (kPa)

Shear stress t (kPa)

300 Þw9 200 Wrong 100

Þc9 0

–a1uw1

–a1uw1

–a2uw2 100

200

–a2uw2

300

400

600

500

Normal total stress s (kPa)

Figure 9.54

Triaxial test results: Mohr circles and strength envelope for unsaturated soils.

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. 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

(9.32)

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 9.54). The apparent cohesion capp is equal to: capp = −α uw − β ua

(9.33)

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).

9.13 9.13.1

RESONANT COLUMN TEST Saturated Soils

The resonant column test (ASTM D4015) is used to determine the dynamic small strain properties of a soil. Such results are applied 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

9.13 RESONANT COLUMN TEST

Suspending spring

Support stand RVDT guide bracket

Drive coil

LVDT casing LVDT core

RVDT

RVDT guide pins

Fluid bath

Torsional accelerometer

Mounting post

Permanent magnet Top drive plate

Top cap inner Soil specimen

Containment cell Rubber membrane 0-ring

Porous disc Base pedestal

Base plate

203

Drainage lines

Figure 9.55 Resonant column test: (a) Principle. (b) Equipment. (b: Courtesy of Geotechnical Research Lab, Dept. of Civil Engineering, University of British Columbia.)

Amplitude

cyclic torsion (Figure 9.55). 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 9.56). This

1 0.707

f1 fn f2

Figure 9.56 vibration.

Frequency

Rotation amplitude vs. frequency of induced

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 Js is: Js = Ms d2s /8 (9.34) where Ms 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: (9.35) 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:     Js J ωn L 2π fn L ω L 2π fn L or s = = n tan tan Jm vs vs Jm vs vs (9.36) where Js and Jm 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 fn is the natural frequency of the soil. In Eq. 9.36, Js , Jm , and L are known, fn is measured in the test, and vs can be back-calculated. Then the shear modulus is obtained from: G = ρ v2s

(9.37)

204

9 LABORATORY TESTS

For the case where there is no mass at the top, Jm = 0, then 2π fn L/vs = π/2, and then G = ρ v2s = 16 ρ fn2 L2

Lnx1 − Lnxn 2π D =√ n−1 1 − D2

(9.39)

In Eq. 9.39, 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 9.57) obtained during steady-state torsional vibration of the sample: D = (f2 − f1 )/2fn

(9.40)

Amplitude

where f2 , f1 , and fn are defined in Figure 9.57. 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

D5

d (2p)2 + d2

r

(9.38)

Eq. 9.38 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 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 9.57) 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:

ln A1 d5 1 n An+1

tmax

1

0.707

fn f2 Frequency (f 2f ) D5 2 1 2fn f1

Figure 9.57 Method to obtain damping ratio from resonant column test: (a) Logarithmic decrement. (b) Half-power bandwidth.

u

L

r

r g

dr

Figure 9.58 Shear stress and shear strain in a resonant column torsion test: (a) Shear stress. (b) Shear strain.

an average of the shear stress generated on the sample cross section. This shear stress is zero at the center of the sample (τcenter = 0) and maximum at the edge (τedge ) (Figure 9.58). The mean shear stress τ is related to the maximum torque T as follows: (9.41) τ = 2T/π re3 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 (γedge ), while the shear strain is zero along the axis of the cylindrical sample (γcenter = 0) (Figure 9.58). The mean shear strain in the sample is usually taken as. γ = re θ/L

(9.42)

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 9.59. 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 9.59): D = Ac /4π Ae

(9.43)

where D is the damping ratio, Ac is the area inside the cycle, and Ae is the area inside the triangle shown in Figure 9.59. 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 9.60). 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 . 9.13.2

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

9.13 RESONANT COLUMN TEST

Shear stress, t 1

Shear stress, t

Gmax

g92

1

g91

Ae

Shear strain, g9 Shear strain, g9 Ac 1 G

t G5 g

D=

Shear stress, g9 Ac 4pAe

G 1

7

10

6

Shear Modulus, G (MPa)

12

5

8

4 6

Shear modulus Pore pressure increase

3

4

2

2

Pore pressure increase, Du (kPa)

Figure 9.59 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, g (%) 10 9 Damping ratio, D (%)

8 7

Steady state Amplitude decay

6 5 4 3 2 1 0 0.0001

0.001

0.01

Strain amplitude (%)

Figure 9.60 Shear modulus vs. shear strain and damping coefficient vs. shear strain: (a) Shearing modulus. (b) Damping ratio.

205

206

9 LABORATORY TESTS

reduction changes. Indeed, the water stress is rarely measured during the resonant column test.

9.14 9.14.1

LAB VANE TEST Saturated Soils

The lab vane test or VST (Figure 9.61) 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 9.61). The lab vane is made of two perpendicular blades, each having a 2-to-1 height-towidth ratio. The width of the blades varies from 12 to 25 mm; the larger vanes are used in softer soils. The vane is pushed perpendicularly into the end of a sample until the tops of the blades are 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 testers measure the torque developed and the rotation angle (Figure 9.62). The peak value of the torque

is recorded as Tmax . The blade is then rotated at least 10 times rapidly and a new maximum torque value, Tres , 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 a 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 Tmax :   D 2 H Tmax = π su D + (9.44) 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 7.4. The residual undrained shear strength sur is obtained from the same formula using Tres :   D 2 H Tres = π sur D + (9.45) 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 Tmax . 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.

Motor for torque application

Soil sample Vane

9.14.2

Figure 9.61 Lab vane test equipment: (a) Principle. (b) Equipment. (a: Adapted from BS 1377-7: 1990.)

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.

9.15 SOIL WATER RETENTION CURVE (SOIL WATER CHARACTERISTIC CURVE) TEST 9.15.1 Vane

Failure Surface

Torque (kN.m)

Tmax 10 rapid rotations

TRes

0

Figure 9.62

θmax

Rotation angle (°)

Lab vane test results.

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 9.63). 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,

9.15 SOIL WATER RETENTION CURVE (SOIL WATER CHARACTERISTIC CURVE) TEST

50

207

Air entry stress uwt

Water content w (%)

40

30 w 5 Cwlog uw Cw

20

1 10 Saturated 0

Unsaturated

1

10

100

10000

1000

Water tension stress uw (kPa)

Soil water retention curve.

Figure 9.63

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 9.64. 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

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: dw = Cw d(log10 uw )

(9.46)

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

(9.47)

Volumetric water content : θw = Vw /V

(9.48)

Volumetric water content u* (%)

60

Free draining water uwt

50

Water unavailable to plants

Water available to plants

40

uwt 5 air entry water tension

uwt

*u 5

uwt

30

Total volume

Clay

Silt

20

Volume of water

Sand 10 0 1

10

100

1000

10000

100000

Water tension stress uw (kPa)

Figure 9.64

Example of soil water retention curve.

1000000

208

9 LABORATORY TESTS

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 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 (uwe ). 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 9.15.2 related to unsaturated soils. 9.15.2

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 3.9; the water tension measurement was described in section 9.2.4 and summarized in Table 9.2. 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 9.9); it makes use of the axis translation technique (Figure 9.10) 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 9.11) 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 as when it is wetting; this is called the hysteresis in the SWRC. Figure 9.65 shows the difference between the drying curve

40.0 Drying

35.0

Wetting Water content, w (%)

30.0 25.0 20.0 15.0 10.0 5.0 0.0

1

10

Figure 9.65

100 1000 10000 Water tension, uw (kPa)

100000

Drying and wetting hysteresis loop in the SWRC.

1000000

9.16 CONSTANT HEAD PERMEAMETER TEST

Drying

Wetting

Figure 9.66 Geometrical explanation of drying and wetting hysteresis in the SWRC.

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 9.66). 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 9.66. 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 9.65). Several stages are identifiable in the drying or wetting process, as shown in Figure 9.67. 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. The effective degree of saturation Se is defined for a given degree of saturation S as: Se =

S − Sr 1 − Sr

(9.49)

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: ⎛ ⎞ 1 if uw ≤ uwae −λ  ⎠ Brooks and Corey (1964) Se = ⎝ uw if uw ≤ uwae uwae (9.50)  m 1 van Genuchten (1980) Se =  n 1 + αuw with m = 1 − 1/n 

Fredlund and Xing (1994) θ = C(uw )θs

Degree of saturation, S (%)

100 Residual air content

80 60

Desaturation Zone

Residual Stage

Wetting path Drying path Residual conditions

40 Air entry value uwac

20 0 1

10

Figure 9.67

100 1000 10000 100000 1000000 Soil suction, uw (kPa)

Various stages in the SWRC.

(9.51) m

1  n Ln(e + uw /a (9.52)

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); θs 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 9.68) 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. 9.16

Saturation Zone

209

9.16.1

CONSTANT HEAD PERMEAMETER TEST Saturated Soils

The constant head permeameter (CHP) (ASTM D2434; Figure 9.69) 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

210

9 LABORATORY TESTS

1.2 wPI 5 % Passing #200*PI

Degree of saturation

1.0 0.8

wPI 5 50 40 20 30 10 15 5

0.6 wPI 5 0.1 3

0.4 D60 5 1 mm

D60 5 0.1 mm

0.2 0.0 0.1

Figure 9.68 NCHRP.)

1

10

100 1000 Matric suction (kPa)

10000

100000

1000000

SWRC as a function of percent passing #200 and plasticity index. (Courtesy of

Filter stone

Level maintained constant 1

k=

Water uw =

hp2

4V

uw

×

l ht2 – ht3

hp2 + hp3

×γo> 0

 

tπD2

2

2 ht1 ht2 ht3

Saturated soil

hp3

l

3

Filter stone Volume of water V collected in time t

V

Datum

Figure 9.69 Constant head permeameter equipment: (a) Principle. (b) Equipment. (b: Courtesy of ELE International.)

Velocity, v (m/s)

Volume, V (mm3)

500000 400000 300000 q 5 10–5m3/s

200000 1

100000 0

0

5

10

15 20 25 Time, t (s)

30

Figure 9.70

35

40

0.008 0.006 k 5 2.6 3 10–4m/s

0.004 1

0.002 0

0

0.5

1 1.5 2 2.5 Hydraulic gradient, i = Dh/L

3

Constant head permeameter test results.

during a corresponding time. Typical results are shown in Figure 9.70. 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 13 on flow through soils for an explanation of the following equations

and parameters.) Darcy’s law gives: v=ki

(9.53)

where v is the discharge velocity through the sample, k is the hydraulic conductivity, and i is the hydraulic gradient. The

9.16 CONSTANT HEAD PERMEAMETER TEST

hydraulic gradient in this case is given by: i = h/l

(9.54)

Unsaturated soil

where h is the loss of total head through the flow distance l (Figure 9.69). Also, conservation of mass gives: Q=vA

k = Q l/h A

(9.56)

where Q is the discharge (m3 /s), l is the flow length between 2 points in the sample, h is the loss of total head between the same 2 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 π D2 /4 where D is the sample diameter. The loss of total head h is ht2 − ht3 , as shown in Figure 9.69. Therefore, the hydraulic conductivity k is: k=

4V l 2 tπD ht2 − ht3

(9.57)

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. 9.16.2

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. 9.55), 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. 9.55. 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

1

k=

Water uw =

2

(9.55)

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 equations 9.53 through 9.55 then gives the value of k:

Level maintained constant

High air entry disk

hp2

D

l ht1

4V  

tπD2

×

hp2 + hp3

2

211

uw

l ht2 – ht3

+

× γw < 0

Air 4 pressure ua –

3 hp3

5 ht2

ht3

Figure 9.71

High air entry disk

hp4

V

Volume of water V collected in time t

Datum

Constant head permeameter test for unsaturated soils.

locations, using tensiometers or other appropriate devices. Figure 9.71 shows the diagram for an unsaturated steadystate permeability test. The water level is maintained on the upstream side (point 1) and the water starts flowing. 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 larger at point 3 than at point 2 (Figure 9.71) 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 9.72). 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: 4V l (9.58) k= tπD 2 (ht2 − ht3 )

212

9 LABORATORY TESTS

Note that the chemistry of the water makes a difference when running a permeability test. If the water that seeps through the soil has a much different salt chemistry than the sample water, the osmotic suction could be activated and lead to a different water tension in the sample than if the water seeping through the sample had the same chemistry than the water in the sample.

1.E-02 Hydraulic conductivity k (m/s)

1.E-03 ksat

1.E-04 1.E-05 1.E-06

ksat

1.E-07

Fine grain soil

9.17 FALLING HEAD PERMEAMETER TEST FOR SATURATED SOILS

1.E-08 1.E-09 Coarse grain soil

1.E-10 1.E-11 1.E-12

1

10

100

1000

10000 100000

Water tension u (kPa)

Figure 9.72 rated soils.

Constant head permeameter test results for unsatu-

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 points where the total heads ht2 and ht3 are measured. Note that Eq. 9.58 is the same as Eq. 9.57. 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. 9.58 is: uw =

hp2 + hp3 2

10−6 m/s).

24.7

24.7.1

Waste Properties

As mentioned in section 24.2 and Figure 24.2, 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 sitespecific 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.

LANDFILLS

Most landfills (Figures 24.19 and 24.20) 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 in1960, 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

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 prob

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 Synthetic liner Compacted clay liner

Perforated leachate collection pipe

Leachate collection sump with riser

Ground water

Modern landfill design

Figure 24.19

Cross section of a landfill

24.7 LANDFILLS

891

Figure 24.20 Landfill under operation. (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/)

Numbers ranging from 3 to 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 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

(24.32)

For σ > 30 kPa,

s = σ tan 33 = 0.65 σ

(24.33)

and 15 to 20% for old landfills (Sharma and Reddy 2004). The creep settlement equation is written as: Cα log H = Ho 1 + eo



tend tstart



(24.35)

where H is the creep settlement, Ho 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. 24.7.2

Regulations

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 section 17.8.9):   ′ Cc σov + σv (24.34) log H = Ho ′ 1 + eo σov

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 postclosure. 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= (24.36) Dγ

where H is the settlement, Ho is the initial thickness of the waste layer, Cc is the compression index, eo is the initial ′ is the effective vertical stress before loading, void ratio, σov ′ and σov + σv 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 10 to 15 years, with the magnitude of H/Ho as much as 50% for new landfills

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.

892

24 GEOENVIRONMENTAL ENGINEERING

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 recordkeeping 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. 24.7.3

5. Geomembrane layer to prevent liquid penetration into the underlying layers; this geomembrane must be at least 0.75 mm thick for a flexible membrane liner and 1.5 mm thick for a high-density polyethylene (Figure 24.22) 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

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 24.21): 1. Waste 2. Protective soil cover to minimize damage to the underlying geotextile 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

Riser & cleanout pipes

Figure 24.22 Installing a geomembrane in a bottom liner. (Courtesy of Layfield Environmental Systems, Layfield Group Limited, 11120 Silversmith Place, Richmond, British Columbia, Canada V7A 5E4.)

24.7 LANDFILLS

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. 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. 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 25. The hydraulic conductivity k of the liner must be less than 10−9 m/s. The hydraulic conductivity k of soils is discussed in sections 13.2.5 and 13.2.6. The measurement of k in the laboratory is discussed in sections 9.16 to 9.19 and in the field in section 7.12. 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 gauging 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 Atterberg limits and k is not always clear. Mitchell and Madsen (1987) concluded that permeation with hydrocarbons

893

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 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. 24.7.4

Covers

Covers (Figure 24.21) 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 away rodents and insects, controlling gas emissions, and improving aesthetics. The typical cross section of a cover consists of a series of layers (1 through 5) as follows: 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 24.23). 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 to 5% on either side of the center to provide natural drainage. 24.7.5

Leachate Collection

The amount of leachate that would go through a single compacted clay liner is given by: q=k

h A L

(24.37)

where q is the flow in m3 /s, k is the soil hydraulic conductivity in m/s, h 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. 24.37. The

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24 GEOENVIRONMENTAL ENGINEERING

Figure 24.23

General cross section of a landfill.

change in total head h 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 through a composite liner was studied by Giroud and Bonaparte (1989), who recommended the following equation: q=

a 0.1 k 0.88 hw A 170000

(24.38)

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 ± S

(24.39)

where I is the infiltration, P is the precipitation, R is the runoff, E is the evapotranspiration, and S is the change in water volume per unit time of the soil cover. All terms in Eq. 24.39 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. 24.39, 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.

24.7.6

Landfill Slopes

The topic of slope stability is covered in Chapter 19. In the case of a landfill, slope stability comes into play in a number of instances (Figure 24.24), 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 19. 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 24.7.1). 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 section 19.3). The stability 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,

Final cover Not too steep Waste

Slope-liner

Excavation

Figure 24.24

Slope stability design issues in a landfill

24.8 FUTURE CONSIDERATIONS

h of the clay layer is:

Shear stress t (kPa)

200

Textured HDPE on non woven geotextile

100

0

895

0

100

h= 24.7.7

GCL internal residual strength

200 300 Normal stress s (kPa)

400

Figure 24.25 Strength envelopes of various interfaces (After Sharma et al. 1997)

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 24.25) to find out which of the interfaces is the weak link for a given normal stress. A seismic slope analysis is also necessary (see section 19.18). An additional problem may arise when during construction of the landfill, the excavation proceeds through a clay layer with an underlying sand layer under artesian pressure (Figure 24.26). 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 (γ h), 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>

hwgw Fg

Clay unit weight, g

h Sand under artesian pressure

Figure 24.26

Blowout problem at bottom of excavation.

(24.40)

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: vent to the atmosphere, vent and burn with no energy recovery, and 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 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 to 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.

24.8

hw

γw hw Fγ

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 24.27). 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

896

24 GEOENVIRONMENTAL ENGINEERING

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 Aluminume trimmings beer & soda cans

Tires

Glass containers

Products

Figure 24.27

Recycling rate for various waste products in 2011 (Source: EPA).

Figure 24.28

Postclosure use of landfills.

household waste, but extends as well to the industrial sector, which is by far the largest generator of waste. Efforts for 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 24.28).

PROBLEMS 24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8 24.9 24.10 24.11

HDPE (natural Pet white bottles translucent & jars bottles)

How do you define waste? What is the biggest generator of waste in the United States? Are solid wastes solids? What are the four main categories of waste? How long can a high-level radioactive waste continue to be deadly? What are RCRA and CERCLA and what do they regulate? What does an OSHA level C mean and what does it require? What is the name of the smallest piece of matter and what are its components? What is the difference between an atom and a molecule? What is the difference between an ion, an anion, and a cation? In the case of concentration in water, how many parts per million (ppm) are in 1 mg/m3 ?

24.8 FUTURE CONSIDERATIONS

897

24.12 What is the difference between organic and inorganic materials? 24.13 What is the difference between atomic absorption spectrophotometry (AAS) and gas chromatography-mass spectrometry (GC-MS)? 24.14 What are ESA I, II, and III, and when are they used? 24.15 What is the difference between the LIF CPT, the MIP CPT, and the BAT CPT? 24.16 In contaminant transport, what is the difference between concentration and flux? 24.17 Choose some reasonable values of the parameters in the solution to contamination propagation (Eq. 24.29) 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 tot www.lmnoeng.com/Groundwater/transportStep.htm 24.18 How can you form a bottom barrier at depth for a waste containment system? 24.19 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. 24.20 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. 24.21 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. 24.22 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. 24.23 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. 24.24 An excavation is dug for a landfill in a 20 m thick stiff, high-plasticity clay layer underlain by a sand layer under artesian pressure. The unit weight of the clay is 19 kN/m3 and that of the sand is 20 kN/m3 . The artesian pressure is such that a casing through the clay into the sand has water rising 10 m above 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? Draw the effective vertical stress profile before and after excavation to that depth.

Problems and Solutions Problem 24.1 How do you define waste? Solution 24.1 Waste is unwanted or useless material. Problem 24.2 What is the biggest generator of waste in the United States? Solution 24.2 The industrial sector is the largest generator of waste. Problem 24.3 Are solid wastes solids? Solution 24.3 No. The term solid waste is misleading, as a solid waste can be a solid, a liquid, or a gas.

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24 GEOENVIRONMENTAL ENGINEERING

Problem 24.4 What are the four main categories of waste? Solution 24.4 The four main categories of waste are: solid wastes, hazardous wastes, radioactive wastes, and medical wastes. Problem 24.5 How long can a high-level radioactive waste continue to be deadly? Solution 24.5 The radiation penetration from high-level wastes, which are generated by defense or nuclear power plant activities, remains lethal for 10,000 years. Problem 24.6 What are RCRA and CERCLA and what do they regulate? Solution 24.6 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 24.7 What does an OSHA level C mean and what does it require? Solution 24.7 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. Problem 24.8 What is the name of the smallest piece of matter and what are its components? Solution 24.8 An atom is the smallest piece of matter. Atoms are made of protons, electrons, and neutrons Problem 24.9 What is the difference between an atom and a molecule? Solution 24.9 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 24.10 What is the difference between an ion, an anion, and a cation? Solution 24.10 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 24.11 In the case of concentration in water, how many parts per million (ppm) are in 1 mg/m3 ?

24.8 FUTURE CONSIDERATIONS

899

Solution 24.11 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 24.12 What is the difference between organic and inorganic materials? Solution 24.12 The difference between organic and inorganic compounds is that most organic compounds contain carbon, whereas most inorganic compounds do not. Problem 24.13 What is the difference between atomic absorption spectrophotometry (AAS) and gas chromatography-mass spectrometry (GC-MS)? Solution 24.13 GC-MS is used to identify the components of a chemical mixture. AAS is used to measure the molar concentration of chemicals. Problem 24.14 What are ESA I, II, and III, and when are they used? Solution 24.14 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. 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 24.15 What is the difference between the LIF CPT, the MIP CPT, and the BAT CPT? Solution 24.15 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 24.16 In contaminant transport, what is the difference between concentration and flux?

900

24 GEOENVIRONMENTAL ENGINEERING

Solution 24.16 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 24.17 Choose some reasonable values of the parameters in the solution to contamination propagation (Eq. 24.29) 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 24.17

C Cw (x, t) = 0 erfc 2



R x − vs t $d 4 Rd DH t



C0 = Constant contaminant concentration at point X = 0 and t = 0

X = Distance

Vs = Seepage velocity Rd = Retardation factor Rd = 1 +

ρd Kd where ρd is dry density and n is total porosity n

Kd = Partition coefficient

DH = Hydrodynamic dispersion DH = D ∗ + αL vs where αL 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)

• Propagation plot (Figure 24.1s)

1 t 5 1000 Days 0.8

Cw /C0

0.6 t 5 100 Days

0.4

0.2 t 5 10 Days 0

0

10

20

30

Distance, x (m)

Figure 24.1s

Propagation plot.

40

50

24.8 FUTURE CONSIDERATIONS

• Varying parameter αL (Figure 24.2s) 1 a 5 100 a 5 10 a51

0.8

Cw /C0

0.6

0.4

0.2

0

0

10

20

30

40

50

Distance, x (m)

Figure 24.2s

Propagation plot for different values of α L .

• Varying parameter νs (Figure 24.3s) 1 Vs 5 1.92e-5 m/s Vs 5 4e-5 m/s Vs 5 1.5e-5 m/s

0.8

Cw /C0

0.6

0.4

0.2

0

0

10

20

30

40

50

Distance, x (m)

Figure 24.3s

Propagation plot for different values of vs .

• Varying parameter ρd (Figure 24.4s) 1 r 5 1.6 g/cm3 r 5 1.1 g/cm3 r 5 1.8 g/cm3

0.8

Cw /C0

0.6

0.4

0.2

0

0

Figure 24.4s

10

20 30 Distance, x (m)

40

50

Propagation plot for different values of ρ d .

901

902

24 GEOENVIRONMENTAL ENGINEERING

Problem 24.18 How can you form a bottom barrier at depth for a waste containment system? Solution 24.18 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 24.19 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 24.19 H = H0

Cα log 1 + e0



tend tstart



= 15 m ∗ 0.06 ∗ log



20∗ 365 day 1 day



= 3.48 m

Problem 24.20 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 24.20 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 ). 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 A= = = 6.08 × 105 m2 15 m × 8 kN/m3 1.2 × 102 kN/m2 Problem 24.21 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 24.21 The amount of leachate that would go through a single compacted clay liner is calculated by q = K h L A, where q is the flow in m3 /s, k is the soil hydraulic conductivity in m/s, h 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: h = 0.6 + 0.4 = 1 m

A = 200 × 200 = 40000 m2 q=K

h 1 A = 10−9 × × 40000 = 6.67 × 10−5 m3 /s L 0.6

Problem 24.22 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.

24.8 FUTURE CONSIDERATIONS

903

Solution 24.22 Fadv = vC = nvs C = 3.7 × 10−7 × 1500 × 10−3 = 5.55 × 10−7 mg/m2 s Problem 24.23 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 24.23 q=

a 0.1 k 0.88 hw A 170000

Case 1: Operating conditions 0.1  3.2 × 10−6 × (10−9 )0.88 × 0.2 × 1 4047 = 1.73 × 10−15 m3 /s = 0.55 cm3 /year q= 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 24.24 An excavation is dug for a landfill in a 20 m thick stiff, high-plasticity clay layer underlain by a sand layer under artesian pressure. The unit weight of the clay is 19 kN/m3 and that of the sand is 20 kN/m3 . The artesian pressure is such that a casing through the clay into the sand has water rising 10 m above 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? Solution 24.24 The safe thickness h of the clay layer after excavation is: h=

9.81 × 30 γ w hw = = 10.33 m Fγ 1.5 × 19

Therefore, the safe excavation depth is 20—10.33 = 9.67 m.

CHAPTER 25

Geosynthetics

25.1

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 $7 m2 . The book by Koerner (2012) is an excellent reference on geosynthetics. 25.2

TYPES OF GEOSYNTHETICS

Geotextiles (Figure 25.1) are textiles made of synthetic fibers. The fibers are either woven together or tied together (nonwoven). Weaving consists of standard interlacing with textile machinery. In nonwoven 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 25.1) 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 25.1) are plastic grids that have a very open configuration; they have large holes between ribs. They 904

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 25.1) 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 to 6 mm thick when the bentonite is hydrated at water contents of 10 to 35%. They are used mostly as nearly impervious barriers to contain liquids or vapors. Geofoams (Figure 25.1) 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 3 to 4 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 25.1), like geogrids, are open netting geosynthetics made of plastic. They are different from geogrids 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 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 are literally bags made of geosynthetic material; they are usually filled with sand and used for erosion protection in lieu of rip rap. Their size is in the range of rip rap, and can be as large as 5 m3 .

25.3 PROPERTIES OF GEOSYNTHETICS

(a) Geotextiles

(b) Geomembranes

(d) Geosynthetic clay liners

(g) Geocells

Figure 25.1

(c) Geogrids

(e) Geofoam blocks

(f) Geonets

(h) Geobags

(i) Geocomposites

Examples of geosynthetics. (Photographs compliments of the Geosynthetic Institute.)

Geocomposites 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 25.1

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. 25.3.1

25.3

905

PROPERTIES OF GEOSYNTHETICS

The parameters used to characterize geosynthetics are much more numerous than and often different from those used for 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

Properties of Geotextiles

Physical Properties The unit weight of typical plastics varies from 9 to 13 kN/m3 . The unit weight of dry, clean geotextiles is between 3 and 7 kN/m3 —but that is not the way it is typically given. Instead, it is quoted as mass per unit area (ASTM D5261) with values between 150 and 750 g/m2 (Koerner 2012). The

906

25 GEOSYNTHETICS

Table 25.1

Applications for Some Geosynthetics

Geotextiles

Geogrids

Geomembranes

• • • • •

• • • • •

• • • • • •

Separation Roadway reinforcement Soil reinforcement Filtration Drainage

Reinforcement Roads Slopes Walls Foundations

Liners Ponds Canals Landfills, dry Landfills, wet Landfill covers

thickness of commonly used geotextiles is between 0.5 to 4 mm (ASTM D5199). 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 25.2). The slope is 41.5o 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 G=M 2

(25.1)

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 ratio of the force per unit length of fabric over the normal strain Steel plate

Length L Geotextile G5M

L 2

3

41.58 Figure 25.2

Flexure stiffness test for geotextile.

Geosynthetic Clay Liners • Liners

Geofoam • • • •

Lightweight fill Compressible inclusions Thermal insulations Drainage

generated. Numbers in the range of 30 kN/m to 150 kN/m are common. The average tensile strength of geotextiles (ASTM D4632) is in the range of 50 to 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 to 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 to 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 middle 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): (25.2) P = St π d 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

25.3 PROPERTIES OF GEOSYNTHETICS

Normal force Block Geotextile Shear force Soil

Figure 25.3 Direct shear test for soil-geotextiles interface shear strength.

site-specific materials. The accepted test (ASTM D5321) is a variant of the soil direct shear test in which the top part of the soil is replaced by a geotextile-covered block (Figure 25.3). 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 t ε1 = 1 (25.3) ε2 t2 where ε1 and ε2 are the strains reached in a time t1 and t2 respectively and n is the viscous exponent. Table 25.2 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 to 12%. The apparent opening size (AOS) of a geotextile is obtained through a test (ASTM D4751) in which glass beads Table 25.2

of uniform diameter are placed on top of the geotextile and wet-sieved through the geotextile. The diameter for which 95% of the beads by weight are retained on the geotextile is the AOS, designated as O95 . Typical values range from 0.01 mm to 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  (m2 /s) is used for in-plane flow instead of the hydraulic conductivity. They are defined as follows: kcp

q q q = h = (25.4) iAt h A t At q q q t= (25.5) Transmissivity  = kip t = t = iA iWt iW Permittivity  =

t

=

where q is the flow, A is the flow area perpendicular to the flow, i is the hydraulic gradient, h 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

=

f w

=

f w

=

ρf μw ρw μf

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 to 0.01 0.02 0.07 0.08 to 0.14

0 to 0.01 — 0.19 to 0.2 0.12 to 0.19

(From data by den Hoedt 1986.)

907

(25.6)

908

25 GEOSYNTHETICS

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, f and w are the transmissivity for the fluid and for water respectively, ρ f and ρ w are 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. 25.3.2

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 to 3 mm (ASTM D5199). The height of asperities for textured geomembranes can be 0.25 to 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 to 10% range, whereas strains to failure for geomembranes are in the 20 to 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 to 50 MPa depending on the type of polymer. The high peak strengths tend to lead to lower residual strengths, which can be 50 to 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 designs, especially for slope stability. It is measured with the same test as for geotextiles (Figure 25.3). A major difference exists between smooth membranes and textured membranes. Koerner (2012) quotes friction angles of 17o and 18o for fine sand and smooth HDPE, and 22o to 30o for fine sand and textured HDPE. Sometimes geomembranes are placed against geotextiles. Again Koerner (2012) quotes friction angles of 6o to 11o for geotextiles and smooth HDPE, and 19o to 32o for geotextiles and textured HDPE. The worst combination seems to be a woven monofilament on top of a smooth HDPE (6o ); the best combination appears to be a nonwoven, needle-punched geotextile on top of a textured HDPE (32o ). 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 resists 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 bevelededge 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 to 500 N for thin, nonreinforced geomembranes and 200 to 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 to 107 rads), and degradation due to hot and cold temperatures, chemical action, and biological action. Designers use reduction factors to take into account these factors, all of

25.3 PROPERTIES OF GEOSYNTHETICS

which affect the long-term strength and function of the geomembrane. 25.3.3

Properties of Geogrids

Physical Properties Geogrids are open-grid geosynthetics (Figure 25.1). The distance between ribs is in the range of 10 to 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 to 95%. The mass per unit area varies quite a bit, from 200 to 1000 g/m2 . Mechanical Properties The flexural stiffness G was defined in Eq. 25.1. 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 Fug is the strength of the geogrid at the field scale where all element strengths are integrated (ASTM D6637). The wide width tensile strength Fug of many geogrids is in the range of 20 to 140 kN/m (force per unit length of fabric) reached at a wide range of strains from 5% to 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 to 255 kN/m have been measured at small strains (1% to 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 25.3. 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 (Fug given by the manufacturer), or you can break the soil (Fus ). Breaking the soil means failure in shear at the interface between the soil and the geogrid. That ultimate pull-out load

909

can be written as: Fus = 2Le Kσv′ tan ϕ ′

(25.7)

where Fus 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 25.4). Note that two phenomena contribute to the ultimate load Fus 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 one 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

(25.8)

1 Fus = 2 × 0.005 × 7 × 18 × + 2 × 0.003 0.1   7 1 × 0.005 × 18 × + 0.0025 × 1− 0.1 0.2   1 7 × 1− × 0.005 × 900 × (25.9) 0.1 0.2

Fus = 12.6 + 3.59 + 74.81 = 91 kN/m

(25.10)

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 Top cover

Air bag

Load cell Jack

Geosynthetic

~1 m Soil

Clamp ~2 m

Figure 25.4

Laboratory container for pull-out tests on geogrid.

910

25 GEOSYNTHETICS

capacity, so the friction will be mobilized first and the bearing capacity last. Now we can calculate the global friction factor K in Eq. 25.7 knowing that the friction angle is 30o and the vertical effective stress at the depth of the geogrid is 40 kPa: 91 Fus = = 0.281 2Le σv′ tan ϕ ′ 2 × 7 × 40 × tan 30 (25.11) The geogrid covers an area Ag per meter of geogrid width, which is a very small fraction of the total area At : K=

  1 1 + 0.003 × 1 − × 0.005 Ag = 0.005 × 7 × 0.1 0.1 7 (25.12) = 0.35 + 0.10 = 0.45 m2 × 0.2 The total area At is: At = 7 × 1 = 7 m2

(25.13)

Therefore, the area covered by the geogrid is 6.4% (0.45/7) of the total area, yet it develops 28.1% (Eq. 25.11) 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 Fug ), 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 25.5). Delayed creep failure typically occurs within the range of 25% to 50% of the ultimate tension Fug . 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. 25.3.4

Properties of Geosynthetics Clay Liners

Physical Properties 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 25.6). 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 nonreinforced 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. Nonreinforced 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 to 6 kg/m2 , with 4 to 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

Tensile strain (%)

10 8 6

Creep failure

Bentonite clay

4 2 0 22

Creep 0

2

4

Log time (hours)

Figure 25.5

Creep behavior of geogrids.

Geotextile or geomembrane

Needle punch or stitch bonds

(a) Non reinforced GCL

Geotextile (b) Reinforced GCL

6

Figure 25.6

Geosynthetic clay liners cross section.

25.3 PROPERTIES OF GEOSYNTHETICS

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 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:   h A (25.14) 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, h 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. Sometime a layer of bentonite without GCL is added at the junction. From the long-term endurance point of view, freezethaw 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 25.2.1; for geomembranes, see section 25.2.2. The shear strength of the bentonite clay layer is measured by a direct shear test (ASTM D6243). Koerner

911

(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 needlepunched 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. 25.3.5

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 to 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 to 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 25.7 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

912

25 GEOSYNTHETICS

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 25.7

Stress-strain curves for geofoam. (After Negussey 1997.)

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 ) (25.15) Modulus of XPS geofoam E (MPa) = 60γGF (kN/m3 ) (25.16) Yield strength of EPS geofoam σy (kPa) = 500γGF (kN/m3 ) (25.17) Yield strength of XPS geofoam σy (kPa) = 800γGF (kN/m3 ) (25.18) 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 )

(25.19)

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 14, Eq. 14.9; section 15.8) 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 section 16.3). The R value is defined as: T(◦ C) ◦ (25.20) R( C.m2 /W) = q(J/s.m2 )

where R is the R value or thermal resistance in T is the difference in temperature on either side of the geofoam in o 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 o Celsius.m/Watt. The R′ value of geofoams varies from 20 to 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. o Celsius.m2 /Watt,

25.3.6

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 to 1.6 kg/m2 and the thickness from 4 to 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  (Eq. 25.5), which is related to the flow rate per unit width (q/W) by: =

q 1 × i W

(25.21)

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.

25.5 DESIGN OF LINERS AND COVERS

25.4

DESIGN FOR SEPARATION

Impact

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 25.8). 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. 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: Estone = Wh =

π d3 γh 6

(25.22)

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 Fs varying between 5 and 25 (Koerner 2012). This energy is then compared to the impact strength Egeosyn (Joules) of the geosynthetic (section 25.3.1). 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. Fr varies from as low as 1.1 to as high as 9. The design ensures that: Egeosyn Estone ≤ Fsoil Freduc

(25.23)

Designing against puncture requires estimating the force Fstone 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 = p da 2 S1 S2 S3

(25.24)

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 Fstone is then compared to the strength of the geosynthetic (Eq. 25.2) 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

Stone

Punching

Tear P Tire

b

W

H

913

y Stone

Figure 25.8

Modes of failure of geosynthetics in separation.

geosynthetic trapped between the two soil layers is subjected to a tension force Ftension , which is given by (Giroud 1981):   b 2 2 2y Ftension = 0.1pda f (ε) = 0.025pda + (25.25) b 2y 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 25.8). Then the value of Ftension is compared to the strength of the geosynthetic (Eq. 25.2) divided by the cumulative reduction factor.

25.5

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 25.9). 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 25.10) 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

Waste Protective layer

Leachate collection system

Sand & gravel Compactive clay

Figure 25.9

1m

Early liner cross section.

914

25 GEOSYNTHETICS

Waste Filter soil Gravel pipes Geosynthetic clay liner or compacted clay

Primary liner Secondary liner

Compacted clay Native soil

Figure 25.10

Leachate collection Geomembrane (1.5 mm) Geonet Geotextile Geomembrane (1.5 mm)

Example of a modern liner cross section.

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. 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 an 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 25.11. 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 25.3. 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 25.12. 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

400

200 (30)

300

GM/CCL (31)

200

(32) (19)

(41)

100 GM/GCL 0

(8,15)

(19) (4)

0

1

2 Life cycle stage

(a) Sand leak detection system

3

Leakage rate (lphad)

Leakage rate (lphd)

(21) GM

150 GM/CCL 100

(11)

(7) GM (27)

(12)

50 (6) 1

GM/GCL

(4) 2 Life cycle stage

(b) Geonet leak detection system

Figure 25.11 Leakage rates in landfill liners. (After Koerner 2012. Robert M. Koerner— Copyright Owner.)

3

25.6 DESIGN FOR REINFORCEMENT

Table 25.3

915

Minimum Requirements for Geomembrane Survivability Required Value Considering the Degree of Severity

Property

Low

Thickness (mm) Tensile strength (kN/m) Tear resistance (N) Puncture (N) Impact resistance (J)

0.63 7 33 110 10

Medium 0.75 9 45 140 12

High 0.88 11 67 170 15

Very High 1.00 13 90 200 20

(After Koerner 2012)

Top soil Cover soil

Vegetation erosion protection geotextile Geotextile Geonet

Landfill cover

Compacted clay or geosynthetic clay liner

Geomembrane (1 mm)

Gas collection layer Waste

Figure 25.12

Example of a landfill cover. (After Koerner, 2012.)

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). 25.6 25.6.1

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 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 25.4. 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). 25.6.2 Mechanically Stabilized Earth Geosynthetic Walls 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 bottomup 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 25.13). These walls can be built in such

916

25 GEOSYNTHETICS

900

Layer numbers

800

2 Layers @650 mm

15

Mean values Cost (dollars/m2)

14 13 4 Layers @500 mm

12 11

6.0 m

9

10

5

6

9 Layers @300 mm

4 1

G 600 500

MSE (Metal)

400

/BIN

B CRI

100

3

1

)

hetics

osynt

(Ge MSE

7

9

11

13

Height of wall (m)

Cost of retaining walls. (From Koerner 2012.)

MSE wall with geosynthetics reinforcement.

a fashion to heights reaching tens of meters and are less expensive than conventional gravity or cantilever retaining walls (Figure 25.14). 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. 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 25.4. Average load and resistance factors for an LRFD approach are also shown in Table 25.4

s

wall

5

2

Figure 25.14 Figure 25.13

wa

200

3 O

ity rav

300

8 7

lls

700

Table 25.4. Bearing capacity and slope stability are dealt with in Chapters 17 and 19 respectively. Sliding of the wall mass can be addressed through a factor of safety expressed as: 

Resisting forces W tan ϕ ′ = Fsliding =  Pa Driving forces

(25.26)

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: γ Pa = ϕ W tan ϕ ′

(25.27)

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

Some Possible Load and Resistance Factors for External and Internal Stability of Geosynthetic MSE Wall Load Factor

Stability

Design Issue

External Stability

Bearing capacity Slope stability Sliding Overturning Pull-out Breakage

Internal Stability

Factor of Safety

Resistance Factor

Dead

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

Live 1.75 1.75 1.75 1.75 1.75 1.75

25.6 DESIGN FOR REINFORCEMENT

taken around the bottom of the front of the wall (point O on Figure 25.13):  Resisting moments Wd = (25.28) Foverturning =  P Driving moments a xa

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: γ Pa xa = ϕ W d

(25.29)

where γ is the load factor and ϕ is the resistance factor. Internal Stability 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 25.4. Average load and resistance factors for an LRFD approach are also shown in Table 25.4. Pull-out capacity requires an understanding of the load distribution in the reinforcement. Figure 25.15 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 Tmax . Beyond Tmax , 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 Tmax . The force Tmax is calculated as: Tmax = sv σah

(25.30)

where Tmax 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. 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 Tpullout (kN/m) of the geosynthetic layer is given by: Tpull out = 2fmax La

fmax = σv′ tan δ

H

Lmax

Tpull out = F × Tmax

In the simple case where σ ah = Ka σv′ , where Ka = 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 to 0.5 meters. A load and resistance factor approach would consist of replacing Eq. 25.34 by: γ 2La σv′ tan δ = ϕsv σah

(25.35)

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

Tmax Y

La Lmax

Sv

La L

45+f/2

Figure 25.15

(25.33)

and the required safe length La of the geosynthetic sheet is given by: F sv σah La = (25.34) 2σv′ tan δ

Active pressure

Sv

(25.32)

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 tanδ is the coefficient of friction between the soil and the geosynthetic. Then the ratio between the load Tmax and the resistance Tpullout must satisfy a factor of safety F (Table 25.4):

T

s9v

(25.31)

where fmax 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 fmax is evaluated as follows:

L Z

917

Load in the reinforcement. (After Theisen, 1992.)

sah

918

25 GEOSYNTHETICS

in the reinforcement layer is taken as the width of the active wedge (see Chapter 22 and Figure 25.15):   ϕ′ (25.36) Lmax = (H − z) tan 45 − 2 Therefore, the final length of the geosynthetic layer L is: γ   sσ ϕ′ ϕ v ah + ′ L = (H − z) tan 45 − 2 2σv tan δ

(25.37)

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 L = H tan 45 − + ′ 2 2σv tan δ

(25.38)

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 load Tmax without yielding or rupturing. For this, we need to find the allowable tensile resistance of the geosynthetic layer Tallow . This allowable tensile resistance Tallow is obtained from the measured ultimate tensile resistance Tult and given by: Tult Tallow = (25.39) RF ID × RF CR × RF CBD where Tallow and Tult are the allowable and ultimate resistance of the geosynthetic layer (kN/m); and RFID , RFCR , and RFCBD 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 Tult of the geosynthetic. The required ultimate strength Tult of the geosynthetic is such that:

25.6.3

Reinforced Slopes

A distinction must be made here between manmade bottomup 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 19.14. 25.6.4

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 the surface of a soft clay with an undrained shear strength su (Figure 25.16). 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 25.16, moment equilibrium around O gives: pu B

T B = su π BB + TB or pu = 2π su + 2 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: 2π su + 2 2π su

γ Tult or = Tmax RF ID × RF CR × RF CBD ϕ γ (25.40) = Tmax × RF ID × RF CR × RF CBD ϕ

Tallow = Tult

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. 25.35. Because the overlap is located near the wall where the tension load is less than Tmax , a length equal to 1/2 La is typically used for the required overlap distance.

(25.41)

T B = 2 or T = π s B u

(25.42)

pu

Strip footing O B A1

B/2 A2 Geosynthetic layers su

Figure 25.16

Foundation reinforcement.

25.7 DESIGN FOR FILTRATION AND DRAINAGE

For a 2 m 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 25.17 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: (R − s)2 +

L2 = R 2 or (neglecting higher-order terms) 4 L2 R= (25.43) 8s

where L is the width of the embankment. Then the geosynthetic stretches from an initial length L to a deformed length L′ : L′ = 2R Arc sin

L 2R

(25.44)

which leads to a strain ε in the geosynthetic of: ε=

Arc sin(4s/L) −1 4s/L

(25.45)

T = Eε

(25.46)

where E is the geosynthetic modulus (section 25.3.1).

R

R L

s

Geosynthetic layers

Exaggerated deflection profile

Figure 25.17

The geotextile or geogrid required needs to have a tensile strength much higher than T because of the reduction factors (Eq. 25.39). 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. 25.7. The anchor length can be shortened by wrapping the geosynthetic around large embedded stones or timber cribbing. 25.7

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: k q (25.47) req = = t h 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, h is the drop of 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. 25.47, and seeking the geotextile that satisfies this requirement. Note that the permittivity of the geotextile has to be corrected for reduction factors as follows: allow =

and a tension T equal to:

Embankment reinforcement.

919

ult (RF SCB × RF CR × RF IN × RF CC × RF BC )

(25.48) where ψ allow is the permittivity that can be used in design, ψ ult is the permittivity quoted by the manufacturer, RFSCB is the reduction factor for soil clogging and blinding, RFCR is the reduction factor for creep reduction of void space, RFIN is the reduction factor for adjacent materials intruding into the geotextile void space, RFCC is the reduction factor for chemical clogging, and RFBC 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 =

allow req

(25.49)

25 GEOSYNTHETICS

For critical applications, this factor of safety should be as high as 5 to 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 25.3.1 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

(25.50)

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. 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: req = kt =

q iw

(25.51)

where req 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 geosynthetic 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). 25.8

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 23.8 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 25.7. 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 (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 25.18 gives a range of velocities that can be resisted by various forms of

6 Long term allowable velocity (m/s)

920

Hard armor

5 4

Soft armor and vegetation

3 2 1 0 1

Vegetation Barren soil 10 Flow duration (hr)

Figure 25.18 Allowable velocities for erosion control measures. (After Theisen 1992.)

25.8 DESIGN FOR EROSION CONTROL

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 25.19). 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) = 36.2 e−11.1 tan α

(25.52)

where α is the slope angle. If the length of 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 )

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:

V (m3 ) = Qt = H

E (kN/km2 .yr) = 10 × R × K × LS × C × P

(25.54)

where R is the dimensionless rainfall coefficient, K is the dimensionless soil erodibility factor, LS is the dimensionless

H

H tan α



×1m

(25.55)

where Q is obtained from Eq. 25.53, 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 to 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 to 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:

T =

γ H L2 γw H L2 or T = w if 4s/L is small ⎞ ⎛ 4s 16s Arc sin ⎜ L⎟ 16s ⎝ ⎠ 4s L

(25.56) 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 25.19). Typical values of T range from 5 to 30 kN/m.

Accumulated fine soil

Geotextile Water seeps through

(b) Cross section

Figure 25.19



(25.53)

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):

(a) Silt fences

921

Silt fences. (a: Courtesy of Robert Koerner, 2012)

Fence

Post L/Z R

L/Z R

(c) Plan view

922

25 GEOSYNTHETICS

25.9

given by:

OTHER DESIGN APPLICATIONS

25.9.1

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. 25.9.2

Thermal Insulation

Geofoams are among the best temperature insulators. Recall from Chapter 16, Eq. 16.8 that the R factor of an insulator is a measure of the resistance to temperature propagation and is

(25.57)

where dx is the thickness of the insulating layer in meters and kt is the thermal conductivity of the insulating material 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 25.5 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: A dQ dT = kt A = dT (25.58) 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. 25.9.4

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 26. Table 25.5

Depth over wall height (z/h)

dx kt

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 to 600 mm thick and tends to decrease the pressure as shown in Figure 25.20. The thicker the geofoam layer is, the lower the pressure 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). 25.9.3

R=

Lightweight Fills

R Factor for Various Materials

0 0.2

Material

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 25.20 Decrease in earth pressure by compressible inclusions. (After Horvath 1997)

Steel Ice Concrete Glass Water (25◦ C) Glass wool Air (25◦ C) Geofoam blocks (After Koerner 2012)

R Factor (M2 .K/W or M2 .C/W) 0.022 0.45 0.95 1.25 1.64 23.8 38.5 25 to 40

25.9 OTHER DESIGN APPLICATIONS

923

PROBLEMS 25.1 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? 25.2 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. 25.24 is equal to 0.3. If the geotextile strength reduction factor is 4.5 (Eq. 25.39), what is the required ultimate strength of the geotextile to safely avoid puncture? 25.3 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? 25.4 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. 25.5 A geosynthetic clay liner has a bentonite clay layer with the following shear strength characteristics: c′ = 0 and ϕ ′ = 10o . It is placed on the side slope of a landfill that has an 18o 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 section 19.3. 25.6 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 34o , 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 25o . The soil behind the wall is a sandy clay with a unit weight of 20 kN/m3 and a friction angle of 30o . Assume reasonable values for all other parameters needed for the design. 25.7 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. 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. 25.7) of 0.3, what length of geogrid is required to safely anchor the geogrid on each side of the footing? 25.8 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 reduction, 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? 25.9 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? 25.10 A construction site has a 30 m long erodible slope with an angle of 6◦ . Silt fences are required. a. How many silt fences are needed? b. If the 10-year rainstorm generates 100 mm/hr, what is the flow rate to be handled per meter of width of the fence? c. Calculate the height of the fence so that it can safely handle two 10-year storms each lasting 3 hours. d. 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. 25.11 Derive Eq. 25.56 for silt fences. 25.12 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 = 19 kN/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?

924

25 GEOSYNTHETICS

25.13 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 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 0.9 m2 .C/W per meter of thickness and the geofoam 35 m2 .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. 25.14 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?

Problems and Solutions Problem 25.1 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 25.1 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

4.41 Estone = = 0.294 Fsoil 15 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 25.2 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. 25.24 is equal to 0.3. If the geotextile strength reduction factor is 4.5 (Eq. 25.39), what is the required ultimate strength of the geotextile to safely avoid puncture? Solution 25.2

Fstone = p da 2 S1 S2 S3 = (600 + 0.5 × 20) × 0.062 × 0.3 = 0.659 kN St =

P 0.659 kN = = 3.5 πd 0.06π m

Tultimate = Tallowable × RF = 3.5 × 4.5 = 15.8 In this situation, a geotextile rated at 15.8 kN/m is needed.

kN m

25.9 OTHER DESIGN APPLICATIONS

925

Problem 25.3 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 25.3 The change in height after replacing the compacted clay layer with the GCL: H = 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

Total income =

$886,500 × 7.5 ha = $6,648,750 ha

Problem 25.4 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 25.4 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 

q = 7 × 10−10 m3 /s Using the same procedure, the flow rate through the CCL is: q = kiA = (1 × 10−9 )



 1.500 (1 × 1) 0.500

q = 3 × 10−9 m3 /s Problem 25.5

A geosynthetic clay liner has a bentonite clay layer with the following shear strength characteristics: c′ = 0 and ϕ ′ = 10o . It is placed on the side slope of a landfill that has an 18o 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 section 19.3. Solution 25.5 Figure 25.1s 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.

926

25 GEOSYNTHETICS

L H W T N

b

Figure 25.1s

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 γ H 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: 1.5 =

c′ tan 10◦ ◦ + tan 18 10 × 20 × sin 18◦ cos 18◦

Therefore, the cohesion developed by needle-punching must be c′ = 56 kPa. Problem 25.6 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 34o , 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 25o . The soil behind the wall is a sandy clay with a unit weight of 20 kN/m3 and a friction angle of 30o . Assume reasonable values for all other parameters needed for the design. Solution 25.6 (Figure 25.2s) H=6m Sv = 0.5 m

σv = 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). 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 5 20 kN/m2

6m

Sand Backfill w9b 5 34° gb 5 20 kN/m3

Sandy Clay w9s 5 30° gs 5 20 kN/m3

1m

Very Stiff Clay Su 5 100 kPa, w9f 5 25°

Figure 25.2s

Retaining wall.

25.9 OTHER DESIGN APPLICATIONS

a. External stability, earth pressures. ka =

1 − sin 30 1 − sin ϕb = = 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 =

0.333 × 20 × (6)2 Ka × γs × H 2 = = 120 kN/m 2 2

• Located 2 m above the bottom of the wall (xa1 = 2 m, as shown in Figure 25.3s). 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 25.3s).

q 5 20 kN/m2

Pa1 5 40 kN/m

6m Pa1 5 120 kN/m

xa2 5 3 m

xa1 5 2 m

o

120 kPa

Ps

Figure 25.3s

20 kPa

Pressure diagram on retaining walls.

We can now calculate the sliding and overturning stability (ignoring the traffic surcharge). b. External stability, sliding analysis. Using the LRFD approach and no traffic surcharge: ϕW tan ϕf′ ≥ γ Pa1

or

ϕγb HL tan ϕf′ ≥ γ Pa1

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 ≥ 150 kN/m

∴ OK

Using the LRFD approach and the traffic surcharge: ϕW tan ϕf′ ≥ γ1 Pa1 + γ2 Pa2

or

ϕγb HL tan ϕf′ ≥ γ Pa1

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 ≥ 220 kN/m

∴ OK

927

928

25 GEOSYNTHETICS

c. External stability, overturning analysis. Overturning around the toe (point O) of the wall with no traffic surcharge: γ Pa1 H ϕγb H L2 γ Pa1 H ϕWL ≥ or ≥ 2 3 2 3 899.6 kN ≥ 300 kN ∴ OK

or

0.85 × 20 × 6 × 4.22 1.25 × 120 × 6 ≥ 2 3

Overturning around the toe (point O) of the wall with traffic surcharge: ϕ(W + qL)L γ P H γ P H ≥ 1 a1 + 2 a2 or 2 3 2 0.85(20 × 6 × 4.2 + 20 × 4.2)4.2 1.25 × 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 e=

or

(W + q × L) × e = Pa1 × xa1 + Pa2 × xa2

Pa1 × xa1 + Pa2 × xa2 120 × 2 + 40 × 3 = = 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: e≤

4.2 L = = 0.7 m 6 6

∴ OK

This means that there is no tension underneath the footing. The active length, according to Meyerhof’s distribution, is: Lactive = L − 2 × e = 4.2 − 2 × 0.61 = 2.98 m The bearing pressure is: p = (γb × H + q) ×

L Lactive

= (20 × 6 + 20) ×

4.2 = (169.1)weight + (28.2)traffic = 197.3 kPa 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 375 kPa ≥ 260.7 kPa

or

0.5 × 750 ≥ 1.25 × 169.1 + 1.75 × 28.2

∴ OK

e. Internal stability, pull-out failure. No traffic surcharge: Tmax = sv σah = sv ka σv′ ka =

1 − sin 34 1 − sin ϕr = 0.283 = 1 + sin ϕr 1 + sin 34

25.9 OTHER DESIGN APPLICATIONS

929

Note that for an MSE wall built with geosynthetics, the kr and ka ratio are the same according to AASHTO LRFD (Figure 25.4s). Coefficient of lateral stress ratio = kr /ka 1 1.2 2.5 1.7

0

ba r d w ma ts ire & gr ids lde

we

Me

tal

l strip Meta

*Geosynthetics

Depth below top of wall, Z

s

0

6000 mm

1 1.2 * Does not apply to polymer strip reinforcement

Figure 25.4s

Coefficient of lateral stress ratio.

The results of Tmax at different heights are shown in Table 25.1s. Tmax1 is due to the soil weight (sv σ ah , active earth pressure) and Tmax2 is due to the traffic surcharge (sv ka × 20kN/m2 ). Table 25.1s Layer No. 1 2 3 4 5 6 7 8 9 10 11 12

Summary of Calculation of Tmax

Depth (m)

ka

kr

σv (kPa)

σ ah (kPa)

Tmax1 (kN/m)

Tmax2 (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

Using the ultimate limit state procedure, we have: γ1 Tmax 1 + γ2 Tmax 2 = φTpullout

Tmax-total (kN/m) 3.53 4.95 6.36 7.77 9.19 10.60 12.02 13.43 14.84 16.26 17.67 19.08

25 GEOSYNTHETICS

The active length of the reinforcement strip required to resist the pull-out load is: γ1 Tmax 1 + γ2 Tmax 2 ϕ Tpullout (γ σ ′ × sv ) + (γ2 ka q × sv ) La = = 1 ah 2 × fmax × b 2 × ϕ × σv′ × tan δ  ϕ Lmax = (H − z) × tan 45 − 2  ′ × sv ) + (γ2 ka q × sv ) ϕ  (γ1 σah + Ltotal = Lmax + La = (H − z) × tan 45 − 2 2 × ϕ × σv′ × tan δ

Tpullout =

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:  ′ × sv ) + (γ2 ka q × sv ) ϕ  (γ1 σah Ltotal = H × tan 45 − + 2 2 × ϕ × σv′ × tan δ The resistance (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 25.5s. Based on this figure, the friction factor is equal to the tangent of the friction angle of the reinforced backfill (ϕ r ). Therefore: F ∗ = tan δ = tan ϕb = 0.6745

Default values for pullout friction factor, F* 1.2 1 log Cu ≤ 2.0

20 (t/St) 0

Not to scale

Grid bearing member 6000 mm

t

Figure 25.5s

Ribbed steel strips F* 5 tanFf

Geogrids F* 5 0.67 tanFf

Geotextiles F* 5 0.67 tanFf

Smooth steel strips F* 5 0.4

D

Steel grids F* 5 10 (t/St)

Depth below top of wall zorz0 (mm)

930

LO St St

Default values for pull-out friction factor.

25.9 OTHER DESIGN APPLICATIONS

931

 ′ × sv ) + (γ2 ka q × sv ) ϕ  (γ1 σah Ltotal = H × tan 45 − + 2 2 × ϕ × σv′ × tan δ   34 (1.5 × 0.283 × 20 × 0.25 × 0.5) + (1.75 × 0.283 × 20 × 0.5) + Ltotal = 6 × tan 45 − 2 2 × 0.9 × 5 × 0.6745

Ltotal = 3.19 m + 0.17 m + 0.81 m = 4.18 m

However, the required length of reinforced soil mass is 0.7 H or 4.2 m. f. Internal stability, yield of reinforcement. No traffic surcharge: Tult Tallow = RF ID × RF CR × RF CBD Consider RFID , RFCR , and RFCBD as 1.55, 2.15, and 1.32, respectively: Tult T 170 kN/m = ult = = 38.6 kN/m 1.55 × 2.15 × 1.32 4.4 4.4 = 34.7 kN/m

Tallow = ϕ Tallow

Using the ultimate limit state analysis, we have: γ1 Tmax 1 + γ2 Tmax 2 = φTallow

γ1 Tmax 1 + γ2 Tmax 2 ϕ 1.5 × sv × σah 1.75 × sv × ka q ≥ + 0.9 0.9

Tallow = Tallow

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 + 0.9 0.9 = 38.6 kN/m > 32.5 kN/m

Tallow ≥ Tallow

or

Tallow ≥ 27 + 5.5 = 32.5 kN/m

ϕTallow = 34.7 kN/m > γ Tmax -total = 29.3 kN/m Here we compare either the factored or the unfactored resistance to the factored or unfactored loads. Detail calculations are shown in Table 25.2s. Table 25.2s

Summary of Calculation for Strength

Layer No. Depth (m) Tmax-total (kN/m) Ltotal (m) Tult (kN/m) Tallow (kN/m) ϕTallow (kN/m) γ Tmax-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

932

25 GEOSYNTHETICS

Problem 25.7 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. 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. 25.7) of 0.3, what length of geogrid is required to safely anchor the geogrid on each side of the footing? Solution 25.7 Figure 25.6s 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 25.6s

Illustration of foundation failure without geogrid.

Figure 25.7s 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 25.7s

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 pu2 = PL +

2T 2 × 100 = 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′ = γ h = 20 × 1 = 20 kPa Le =

100 Fus = = 14.4 m ′ ′ 2Kσv tan φ 2 × 0.3 × 20 × tan 30◦

25.9 OTHER DESIGN APPLICATIONS

933

So, the length of geogrid required to safely anchor the geogrid on each side of the footing is 14.4 m. Problem 25.8 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 reduction, 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? Solution 25.8 Modulus, E = 500 kN/m Settlement, s = 1 m

Width of embankment, L = 30 m Radius R of the circle is: R=

302 L2 = = 112.5 m 8s 8×1

Deformed length L′ : L′ = 2RArc sin

L = 30.09 m 2R

Strain ε in the geosynthetic: Arc sin ε=



4s L

4s L



−1=

L′ − L 30.09 − 30 = = 0.003 L 30

Tension T equal to: T = Eε = 500 × 0.003 = 1.5 kN/m Problem 25.9 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 25.9 Ultimate tensile strength, Tult = 100 kN/m Maximum flow rate, q = 8 × 10 – 7 m3 /s per meter of geotextile. ult

The strength reduction factors take into account installation damage ID, creep CR, and chemical and biological degradation CBD. They are RFID, RFCR, and RFCBD. They average respectively 1.55, 2.15, and 1.32: Tallow =

Tult 100 = = 23.05 kN/m RF ID × RF CR × RF CBD 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 RFSCB , RFCR , RFIN , RFCC , and RFBC . Their respective average values are: 4.41 (RFSCB ), 1.83 (RFCR ), 1.1 (RFIN ), 1.25 (RFCC ), and 2.2 (RFBC ). qallow =

qult 8 × 10−7 8 × 10−7 = = (RF SCB × RF CR × RF IN × RF CC × RF BC ) 4.41 × 1.83 × 1.1 × 1.25 × 2.2 24.41

= 0.33 × 10−7 m3 /s per meter of geotextile

934

25 GEOSYNTHETICS

Problem 25.10 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 25.10 a. Number of silt fences: Lmax (m) = 36.2e−11.1 tan α ◦

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 ⎬ ⇒ Q = 0.565 (m3 /hr/m of fence) I = 0.1 ⎪ ⎭ A = 1 × 11.3

c. Height of fence:

V (m3 ) = Qt = H



H tanα



× 1(m)

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 H L2 9.81 × 0.6 × 32 T = w ⇒T = = 16.55 kN/m 16s 16 × 0.2 Problem 25.11 Derive Eq. 25.56 for silt fences. Solution 25.11 The average pressure on the fence is: p=

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 25.8s). For equilibrium: pL = 2T sin α In triangle OAC (Figure 25.8s), sinα is given by: sin α =

L/2 R

25.9 OTHER DESIGN APPLICATIONS

935

O P 5 Water pressure on fence

R

a a

T

T a

a L/2

Post

A

L/2

B

S

C Post

Fence

Figure 25.8s

Plan view of deformed silt fence.

So, T = pR The radius R is given by using triangle OAC (Figure 25.8s): R 2 = (R − s)2 +

 2 L 2

Because s is small compared to R and L, this gives: R= Then: T =p

L2 8s

L2 γ H L2 = w 8s 16s

Problem 25.12 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 = 19 kN/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 25.12 Using the consolidation theory: s=H

Cc σ ′ + σv log ov ′ 1 + e0 σov

a. Option 1: Soil fill. The increase in stress in the clay layer is: σv = 20 × 7 = 140 kPa The initial effective stress in the middle of the clay layer is: ′ = 2.5 × 19 − 2.5 × 9.81 = 23 kPa σov

936

25 GEOSYNTHETICS

Therefore: s=5

23 + 140 0.5 log = 1.01 m 1 + 1.1 23

b. Option 2: Geofoam. The increase in stress in the clay layer is: σv = 2 × 7 = 14 kPa The initial effective stress in the middle of the clay layer is still: ′ = 2.5 × 19 − 2.5 × 9.81 = 23 kPa σov

Therefore: s=5

23 + 14 0.5 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 25.13 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 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 0.9 m2 .C/W per meter of thickness and the geofoam 35 m2 .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 25.13 The heat flow is defined as:

Q T (W ) = kA t x

Where: ◦

k = Material thermal conductivity (W/m. C)

A = Cross-sectional area

T/x = Temperature gradient In terms of thermal resistance, the preceding equation can be written as: AT Q (W) = t R where R (m2 .◦ 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: 1×5 Q (W) = = 55.5 Watt t 0.9 × 0.1 For the case of the concrete slab + 150 mm of geofoam: 1×5 Q (W) = = 0.93 Watt t (0.9 × 0.1 + 35 × 0.15)

25.9 OTHER DESIGN APPLICATIONS

937

The use of 150 mm of geofoam can reduce the power usage by 98.3%. Problem 25.14 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 25.14 Assume a coffee cup with an internal radius r1 = 40 mm, an external radius r2 = 41.5 mm and a height of 160 mm.

r1 r2

Figure 25.9s

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: T T   = k2π r1 L   Q(W) = −kA r2 r r1 × ln r1 × ln 2 r1 r1 with A (m2 ) = 2π r1 L as the internal surface area of the cup. k (W/m◦ C) is the thermal conductivity of the cup wall. T 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: Q(W) = −2π L

T R

where R (m2 ◦ C/W/m) is the thermal resistance per meter of the cup wall and is equal to:   r ln 2 r1 R= k Based on the data given in the problem statement, the thermal resistance of the Styrofoam is: R = −2π L

T (30 − 80) ◦ = −2π × 0.16 × = 1.117 m2 . C/W/m Q 45

The thermal conductivity of the Styrofoam is then: ln ◦

k(W/m. C) =



r2 r1 R



ln =



0.0415 0.04 1.117



= 0.033

CHAPTER 26

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 alternatives 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 web site www.geotechtools.org (Schaefer 2013). 26.1

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 26.1. 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 938

3. Soil improvement with replacement 4. Soil improvement with grouting and admixtures 5. Soil improvement with inclusions 26.2 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN COARSE-GRAINED SOILS 26.2.1

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 is highest near the surface. Compaction is used to prepare pavement layers, retaining wall backfills, and embankment fills. This topic is covered in Chapter 20. 26.2.2

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 20. 26.2.3

Vibrocompaction

The vibrocompaction method consists of lowering a cylindrical vibrator from a crane into the soil to densify the soil (Figure 26.1). A grid of 3 to 4 meters center to center is common. The vibrator is 2 to 5 m long and 0.3 to 0.5 m in diameter, and weighs 15 to 40 kN. The vibrations are generated in the horizontal direction by rotating eccentric masses. The frequency of vibration is in the range of 25 to 35 Hz with amplitudes between 10 to 30 mm. The vibrator typically

26.2 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN COARSE-GRAINED SOILS

Table 26.1

939

Classification of Soil Improvement Methods (Chu et al. 2009)

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. 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.

B. Ground improvement without admixtures in cohesive soils

A2. Vibrocompaction A3. Explosive compaction A4. Electric pulse compaction A5. Surface compaction (including rapid impact compaction) 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 C2. Dynamic replacement

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

(Continued)

940

26 SOIL IMPROVEMENT

Table 26.1

(Continued)

Category

Method

Principle

D. Ground improvement with grouting-type admixtures

D1. Particulate grouting

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 homogenous mass to densify loose soil or lift settled ground. 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.

D2. Chemical grouting

D3. Mixing methods (including premixing or deep mixing) D4. Jet grouting D5. Compaction grouting

D6. Compensation grouting

E. Earth reinforcement

E1. Geosynthetics or mechanically stabilized earth (MSE) E2. Ground anchors or soil nails E3. Biological methods using vegetation

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 26.2). 26.2.4

Figure 26.1 LLC.)

Example of vibrocompactor. (Courtesy of Earth Tech,

reaches depths of 20 to 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 of this technique are clean sands. If the fine content becomes higher than 10 to 15%,

Other Methods

Other compaction methods include rapid impact compaction (Figure 26.3), 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. 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 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.

26.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

26.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

100 Cone penetration resistance, Mpa

941

26.3.1

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 section 25.3.5) to avoid excessive settlement and bearing capacity issues.

Cannot be campacted

Can be compacted 10

Can be marginally compacted

26.3.2

1 0

0.5

1

1.5

2

2.5

Displacement–Replacement

3

Friction ratio, %

Figure 26.2 Soil suitability for vibrocompaction based on CPT. (Courtesy of Dr. Rainer Massarsch)

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 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 26.4): 1. Calculate the maximum settlement of the embankment smax(emb) . For a normally consolidated clay, the following equation can be used (see section 17.8.9):   ′ Cc σov + σ ′ smax(emb) = ho (26.1) log ′ 1 + eo σov

Figure 26.3 Example of rapid impact compactor. (Courtesy of Menard Bachy, Inc.)

where ho is the height of the soft clay layer, Cc is the compression index from consolidation tests, eo is the ′ is the initial initial void ratio of the soft clay layer, σov effective stress in the middle of the soft clay layer, and σ ′ 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 TU corresponding to tt using the equation (see section 17.8.10): tc (26.2) TU = t 2v hd where hd is the drainage length.

942

26 SOIL IMPROVEMENT

Surcharge Embankment

Time

tt

t90

hs

ho

Soft clay

Smax (emb.)

Embankment

he Cc, cv, eo

Smax (emb. 1 surc.) Settlement

Embankment 1 Surcharge

Surcharge to accelerate embankment settlement.

Figure 26.4

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 the horizontal distance between vertical drains if such drains are installed. 4. Then find the average percent consolidation U corresponding to the time factor TU by using the curve that links both parameters (Figure 26.5). Note that U is equal to: s(t) (26.3) U= smax

smax(emb+surch) =

′ σ ′ = σov 10

max(emb+surch) ho Cc

Average degree of consolidation Uav in percent

C3 C1

60 80 100

C2 0

0.2

Curve C1

0.6 0.8 Time factor T

0.4

Curve C1

Curve C1

Two way drainage

Figure 26.5

(26.6)

where γ s is the unit weight of the surcharge soil. 8. Note that if the surcharge is too high, a slope stability or bearing capacity problem arises for the side of the

0

40

−1

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 σ ′ . Often, if the soft clay layer is 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: σ ′ (26.7) hs = γs

(26.4)

20

(26.5)

6. Once the maximum settlement under the embankment and the surcharge smax(emb+surch) is known, Eq. 26.1 can be used to back-calculate the value of σ ′ induced by the surcharge:     (1+eo )s

where s(t) is the settlement after a time t and smax is the final settlement. 5. Knowing U, use Eq. 26.3 to calculate the maximum settlement smax(emb+surch) under the embankment plus the surcharge. In Eq. 26.3, 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)

smax(emb) U

1

Curve C1

1.2

Curve C1 Curve C1

One way drainage

Average percent consolidation U versus time factor TU .

1.4

26.3 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

943

embankment. In that regard, the height of the surcharge hs max that would generate a bearing capacity failure in a clay of undrained shear strength su can be estimated by: 5.14su (26.8) hs max = γs 26.3.3 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-tocenter 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 section 11.4.6) 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 26.6). The width may be 100 mm, the thickness 3 to 4 mm, and the installed length can be 30 m. The flow rate out of such drains is in the

(a)

Figure 26.6 Prefabricated vertical drain. (Courtesy of Layfield Environmental Systems, Layfield Group Limited, 11120 Silversmith Place, Richmond, British Columbia, Canada V7A 5E4.)

range of 2 to 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 26.7). 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 to 2.5 m. The tops of the drains are bound by a drainage layer or drainage blanket (0.5 to 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 smear zone

(b)

Figure 26.7 Installation of prefabricated vertical drains. (Courtesy of Hayward Baker Geotechnical Construction.)

26 SOIL IMPROVEMENT

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 Uh taken as a ratio, not a percent, in Eq. 26.9. This time t can be calculated by using the Barron-Hansbo formula (Barron 1948; Hansbo 1981):       d 1 d2 (26.9) t = w Ln w − 0.75 + Fs Ln 8ch de 1 − Uh where de is the equivalent diameter of the PVD defined in Eq. 26.10, 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 Fs 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): de =

2(a + b) π

(26.10)

where a is the PVD thickness and b is the PVD width. 26.3.4

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

Failure line Deviator stress q′

944

Ko line Surcharge Vacuum

Effective mean stress p′

Figure 26.8 Stress path comparison between surcharge preloading and vacuum preloading. (After Chu et al. 2009.)

to about 4 m of 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 26.8 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 26.9. The vacuum preloading method works well when the soil is soft, low permeability, and relatively homogeneous. If clean

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 26.9 usa.com)

Menard vacuum consolidation. (Courtesy of Menard, Bridgeville, PA; www.menard

26.4 SOIL IMPROVEMENT WITHOUT ADMIXTURE IN FINE-GRAINED SOILS

Electroosmosis Electromigration

945

Water transport from anode to cathode Ion transport to the opposite electrode Heat

Soil particle

Soil particle Soil particle Water velocity profile

Anode

Cathode

Soil particle Soil particle

Figure 26.10

Soil particle

Electro-osmosis in clays. (Courtesy of C.J. Athmer—Terran Corporation.)

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. 26.3.5

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 26.10). 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. 26.3.6

Ground Freezing

The technique of ground freezing (Figure 26.11) consists of freezing the soil by installing a network of steel pipes and circulating either brine water or liquid nitrogen. The temperature of circulating brine water is typically −20C; liquid nitrogen is much colder, at around −200C. Brine is much less expensive, but nitrogen takes a lot less time to freeze the 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

Figure 26.11 Ground freezing. (Courtesy of British Drilling and Freezing Co. Ltd.)

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. 26.3.7

Hydro-Blasting Compaction

The hydro-blasting compaction technique is particularly well suited to the treatment of collapsible soils. It consists of wetting the soil to induce collapse and then detonating explosives in sequence to shake the soil into a more compact arrangement.

946

26 SOIL IMPROVEMENT

26.4 26.4.1

SOIL IMPROVEMENT WITH REPLACEMENT

Therefore, the drained ultimate load on the stone column is:

Stone Columns without Geosynthetic Sock

Stone columns, also called aggregate columns (Figure 26.12), 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 (section 26.2.3) 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′

(26.13)

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. 26.13 conservative. 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: B R = B R

(26.14)

where B and B are the initial diameter and increase in diameter of the stone column respectively, and R and R 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, B can be obtained from Eq. 26.14. The volume involved in the barrel-like deformation shown in Figure 26.13 extends to a depth equal to about 2 times the diameter of the stone column (Hughes and Withers 1974). Thus, the initial volume involved in the deformation is:

(26.11)

where Kp 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 pL′ of the pressuremeter test. The value of pL′ 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: pL′ = pL − uw

Qu = Kp (pL − uw )A

Vo = 2B

π B2 4

(26.15)

If, during the deformation of the column under load, the volume of stone experiences a volume change V, then the volume V of the deformed column under load will be: V = Vo + V

(26.16)

B s91

(26.12)

s

s93 5

2B ?

Stone column DB

Figure 26.12 Stone column construction. (Courtesy of Menard Bachy, Inc.)

Figure 26.13

p9L F

Soft clay

Expansion of a stone column under load.

26.4 SOIL IMPROVEMENT WITH REPLACEMENT

947

The deformed volume V is also equal to: π V = (2B − s) (B + B)2 4

(26.17)

where s is the settlement of the stone column. This settlement s is then given by: ⎛ ⎞  V 1 + ⎜ Vo ⎟ ⎟ ⎜ (26.18) s = 2B ⎜1 −   ⎟ ⎝ B 2 ⎠ 1+ B

The relative increase in stone column diameter B/B is obtained from Eq. 26.14 using a ratio R/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 V/V can be obtained from a triaxial test on the stone column material. The value of V/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: Q = σ1′

π B2 4

(26.19)

If V is 0 and if R/R is small, then Eq. 26.18 reduces to: R s = 4B R

(26.20)

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 (section 26.3.3). 26.4.2

Stone Columns with Geosynthetic Encasement

More recently, geosynthetic encasement, in the form of a large sock (Figure 26.14), 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 value of Tu ranges from 25 to 60 kN/m and that of E from 30 to 150 kN/m. The modulus E is defined as: E=

T ε

(26.21)

Figure 26.14 Stone column with geotextile encasement. (Courtesy of HUESKER Inc.)

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 to 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: Soil fails laterally pu1 = kp σ3′ = kp (pL′ + pgeo ) (26.22) Geotextile fails in hoop tension   r pu2 = kp σ3′ = kp 2G + pgeo f ro

(26.23)

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 pL′ , 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 is the pressure contributed by the geotextile when stretched at r/ro , pgeo f 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),

948

26 SOIL IMPROVEMENT

pL9

Natural soil

Stone column

Geotextile p

Dr

r0 T

Figure 26.15

26.4.3 T

Pressure and tension in the geotextile encasement.

and r/ro is the relative increase in radius of the stone column. The expression 2Gr/ro is the pressure contributed by the soil outside the geotextile for a radial strain r/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 r/ro is given by (Figure 26.15): pgeo 2r = 2T = 2Eε = 2E

r ro

or

strain at failure for the limit pressure. Note that the product E r/ro cannot be larger than the tensile capacity Tu 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 simplified method for hand calculations.

pgeo = E

r (26.24) ro2

Then the ultimate load on the encased stone column corresponding to this failure mechanism is:   r (26.25) Qu1 = kp pL′ + E 2 π ro2 ro The value of kp is obtained from the friction angle of the stone column material, pL′ from a pressuremeter test in the natural soil, E from the geotextile material, ro from the size of the stone column, and r/ro as 0.41 to correspond with the

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 26.16). The same range of weight, drop height, and pounder diameter are used for both DC and DR. If the energy used is high (200 to 400 kJ/m3 ) and the soil is softer (PMT limit pressure 100 to 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 to 250 kJ/m3) and the soil is stronger (PMT limit pressure 250 to 700 kPa), then the craters are limited in depth, DC take place, and the degree of improvement is lower.

26.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES You might have heard the words grout, concrete, cement, and mortar: what are they, and what is the difference?

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 26.16

Dynamic replacement method (After Chu et al. 2009).

26.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES

and then injecting the grout. The spacing between boreholes is in the range of 1 to 2 m and the hydraulic conductivity of the soil for which this technique is applicable is 10−2 to 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 to 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:

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 26.17 shows the range of applicability of various grouting techniques. 26.5.1

N1 =

N3 =

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,

Percent finer by weight (%)

Clay

80

D10(soil) D65(grout)

Silt

or

N2 =

D10(soil) D95(grout)

D10(soil) D90(grout)

+ k1

w/c P + k2 FC Dr

Sand

Gravel

Boulder

Compensation grouting Compaction grouting

60

Ultra fine cements Water-glass solution (low viscosity) Economical

20 0

Cement suspensions

Uneconomical

0.002 0.006 0.02

0.06

(26.27)

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 waterto-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 N3 is larger than 26.

Jet grouting

40

(26.26)

where D10(soil) is the grain size of the soil corresponding to 10% fines, and D65(grout) and 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:

Particulate Grouting

100

949

0.2

0.6

2

6

20

60

Grain size (mm)

Figure 26.17 Range of application of grouting techniques (After Keller 2012; www.keller grundbau.com/download/pdf/en/Keller_66–01E.pdf).

950

26 SOIL IMPROVEMENT

26.5.2

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.

26.5.3

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 (Figure 26.17). Jet grouting consists of drilling a borehole down to the desired depth. The drill bit has a diameter in the range of 100 to 150 mm. Once the required depth is reached, a horizontal high-pressure jet (∼20 MPa) is generated to erode the soil laterally. The rod is withdrawn while rotating (Figure 26.18). Insertion of injecting tool into a drilled hole

Injection of high velocity cement, slurry, and air

Figure 26.18

This erosion process generates a larger-diameter hole (1 to 1.5 m) that is then filled with grout.

26.5.4

Compaction Grouting

Compaction grouting (Figure 26.19; 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 to 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 to 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 to 3.5 m center to center.

26.5.5

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

Jet grouting (After Hayward Baker, Inc.).

26.5 SOIL IMPROVEMENT WITH GROUTING AND ADMIXTURES

Drill rod

Compaction grout casing

Figure 26.19

Compaction grouting. (Courtesy of Arizona Repair Masons Inc.)

Casing

951

Casing

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 26.20 Example of SCM excavation support construction sequence. (Courtesy of JAFEC USA, Inc. Geotechnical Constructors)

of soils, but mostly in fine-grained soils, although difficulties have been encountered in soft clays (Chu et al. 2009). 26.5.6

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 26.20) 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 unconfined compressive strength

952

26 SOIL IMPROVEMENT

in the 2000 kPa range. The modulus of deformation of SCM varies in the range of 100 to 1000 MPa and can be estimated by the following equation (Briaud and Rutherford 2010): ESoil Cement (kPa) = 12,900(fc′ (kPa))0.41 26.5.7

(26.28)

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 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 26.21). 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 26.21 Lime stabilization of pavement layers. (Photo by James Cowlin/Asphalt Busters, Phoenix, AZ.)

Figure 26.22 Ettringite crystals. (Courtesy of www.sciencedirect .com/science/article/pii/S0008884698001379)

26H2 O), which is a highly expansive mineral. Ettringite crystals are needle like (Figure 26.22) 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).

26.5.8

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 26.23 and 26.24) and can increase the shear strength of the sand by a factor of 2 (Ng et al. 2012).

26.6 SOIL IMPROVEMENT WITH INCLUSIONS

953

26.6.3 Geosynthetic Mat and Column-Supported Embankment Geosynthetic mat and column-supported embankments (GMCSs) (Figure 26.25) 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 2013):

Calcite

Sand particle

Sand particle

50.0 mm

Figure 26.23 Light microscopic image of calcite crystals, produced by ureolytic bacteria, cementing two sand particles. (Courtesy of Salwa Al-Thawadi.)

26.6 26.6.1 Earth

s ≤ 0.67H + a

SOIL IMPROVEMENT WITH INCLUSIONS

s ≤ 1.23H − 1.2a

Mechanically or Geosynthetically Stabilized

s ≤ a + 3 meters

Mechanically stabilized earth (MSE) walls are covered in section 21.10. Geosynthetically stabilized earth (GSE) walls are covered in section 25.6.2. 26.6.2

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:

Ground Anchors and Soil Nails

Ground anchor walls are covered in section 21.12 and soil nail walls are covered in section 21.13.

(a)

(26.29)

where H is he 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. 26.29) are set to ensure that proper arching will develop in the embankment through the bridging

(b)

Figure 26.24 Microbial-induced calcite precipitation: (a) Bacteria and calcium chloride. (b) Bricklike product. (Courtesy of Ginger Krieg Dosier, bioMASON Inc.)

954

26 SOIL IMPROVEMENT

Surcharge, q n H 5 Hb 1 HEmb

1

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 26.25 Geosynthetic reinforced column supported embankment. (Courtesy of Professor Vernon Schaefer, [email protected])

layer to transfer all the embankment weight to the columns. Arching is also ensured by selecting a material for the bridging layer that satisfies a number of criteria (Schaefer 2013). 3. Determine the column load Qcol knowing the height of the embankment: Qcol = (γ H + q)A

(26.30)

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 26.25). 4. Design the piles to safely carry Qcol . See sections 18.4 and 18.5. 5. Calculate the tension load in the geosynthetic layer. This tension load has two components: the tension load T1 due to the vertical load transferred from the embankment to the columns through the bridging layer, and the tension load T2 due to the tendency of the embankment to spread laterally. Filz et al. (2012) recommend the following expression for calculating the value of T1 : 6T1

3



σ A − (6T1 − EGS ) net soil p



=0

(26.31)

where T1 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 T2 is obtained from: T2 =

1 K γ H 2 + qKa H 2 a

(26.32)

where T2 is the tension load per unit length of embankment in the geosynthetic due to the embankment tendency to spread laterally, Ka 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 section 25.3.3). 7. Calculate the embedment length Le of the geosynthetic layer: (T1 + T2 )F (26.33) Le ≥ γ H(tan δ1 + tan δ2 )

26.7 SELECTION OF SOIL IMPROVEMENT METHOD

where F is the factor of safety, γ is the unit weight of the embankment soil, H is the height of the embankment, and tanδ1 and tanδ 2 are the coefficients of friction 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 section 18.5). 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 section 19.14). A factor of safety of 1.3 to 1.5 is common.

955

26.7 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 Khouri (2000) proposed a software product called Soil and Site Improvement Guide to optimize the choice. More recently, Schaefer (2013) optimized the decision process through freeware available at www.geotechtools.org.

PROBLEMS 26.1 Three soils have the following CPT characteristics. Can they be vibrocompacted?

Soil Soil 1 Soil 2 Soil 3

Point Resistance

Friction Ratio

10 MPa 8 MPa 15 MPa

1.2% 0.5% 2%

26.2 . An embankment was built as shown in Figure 26.1s. a. What is the maximum settlement of the embankment? b. How much time is required for 90% of that settlement to occur? c. 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.)

Surcharge

Hsurcharge

Fill 7m Sand γ = 20 kN/m3 Cc 1 + eo

= 0.2

Clay

γ = 18 kN/m3 Cv = 1.8 × 10–3cm2/sec.

5m

Dense gravel

Figure 26.1s

Highway embankment for preloading problem.

26.3 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 26.2s.

956

26 SOIL IMPROVEMENT

C L

Compacted fill γt = 19.6 kN/m3

15 m 6m

1

2

Normally consolidated clay γt = 16.5 kN/m3 Cv = Ch = 0.0093 m2/day

18.3 m

Sand

Figure 26.2s

Highway embankment for prefabricated vertical drain problem.

26.4 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 a conservative value 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. 26.5 Repeat problem 26.4 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. 26.6 How would you make cement? How do you make lime? What is the difference between cement and lime? 26.7 A soil has a D10 equal to 2 mm. The grout used to strengthen it has a D65 of 60 μ m and a D95 of 130 μ m. Can particulate grouting be successful? 26.8 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 (T1 ), and the tension due to lateral spreading (T2 ). 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.

Problems and Solutions Problem 26.1 Three soils have the following CPT characteristics. Can they be vibrocompacted?

Soil

Point Resistance

Friction Ratio

10 MPa 8 MPa 15 MPa

1.2% 0.5% 2%

Soil 1 Soil 2 Soil 3

Solution 26.1 According to Massarsch guidelines, and Figure 26.2: • Soil 1—Marginal • Soil 2—Yes • Soil 3—No

26.7 SELECTION OF SOIL IMPROVEMENT METHOD

957

Problem 26.2 An embankment was built as shown in Figure 26.1s. a. What is the maximum settlement of the embankment? b. How much time is required for 90% of that settlement to occur? c. 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.) Surcharge

Hsurcharge

Fill 7m Sand γ = 20 kN/m3 Cc 1 + eo

= 0.2

γ = 18 kN/m3

Clay

Cv = 1.8 × 10–3cm2/sec.

5m

Dense gravel

Figure 26.1s

Highway embankment for preloading problem.

Solution 26.2 a. Maximum Settlement of the Embankment Smax = H

Cc σ ′ + σ ′ log ov ′ 1 + eo σov

′ σov = 2.5 × 18 − 2.5 × 9.81 = 20.48 kPa

σ ′ = 20 × 7 = 140 kPa ∴ Smax = 5 × 0.2 × log

20.48 + 140 = 0.89 m 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 2.52

= 0.448

S6 months (fill+surcharge) Smax(fill+surcharge)

Because we want S6 months (fill + surcharge) = Smax (fill), then: U=

Smax (fill) Smax(fill+surcharge)

  ′ + σ ′ Cc 20.48 + σ 0.89 σov Smax (F + S) = log = 5 × 0.2 × log = Ho ′ 0.732 1 + eo σov 20.48

958

26 SOIL IMPROVEMENT



20.48 + σ 1.216 = log 20.48



σ = 316.3 kPa = σ (F ) + σ (S)

316.26 kPa = 20 × 7 + 20 × h h = 8.81 m

Problem 26.3 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 26.2s. 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 26.2s

Highway embankment for prefabricated vertical drain problem.

Solution 26.3 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 Fs 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: 1.277s 2 (Ln(17.07s) − 1.75) × 2.303 0.0744 9.108 = s 2 (Ln(17.07s) − 1.75) 360 =

This equation is solved by trial and error and gives a center-to-center spacing of s = 2.2 m. Problem 26.4 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

26.7 SELECTION OF SOIL IMPROVEMENT METHOD

959

pressuremeter tests are performed and a conservative value 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 26.4 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 333 Qu = = 166.6 kN F.S. 2 R s = 4B R E = 1700 kPa and v = 0.35

Qallowable =

R P 1 + v Pl = = R 2G E F.S. R 1.35 150 = = 0.06 R 1700 2 s = 4 × 0.06 = 0.24 m

Problem 26.5 Repeat problem 26.4 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. Solution 26.5 Figure 26.3s illustrates this problem. Fill

5m

Soft silt layer

1m

10 m

Stone columns

Figure 26.3s

Dense sand

Illustration of the soil profile.

Failure mechanism 1: Soil fails laterally The ultimate pressure the stone column can carry is calculated as: pu1 = kp (pL′ + pgeo ) where kp is the passive earth pressure coefficient of the soil, pL′ 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. kp =

1 + sin 38 1 + sin ϕ ′ = = 4.2 ′ 1 − sin ϕ 1 − sin 38

pL′ = pL − uw = 150 − 5 × 9.81 = 101 kPa

960

26 SOIL IMPROVEMENT

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: r = ro × ε = 0.5 × 0.41 = 0.205 m pgeo = E

0.205 r = 150 × = 123 kPa ro 2 0.52

Hence, the ultimate load the stone column can carry is: Qu1 = kp (pL′ + pgeo )π ro 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:   r + pgeo f pu2 = kp 2G ro where kp is the passive earth pressure coefficient of the soil, G is the shear modulus of the soil outside the geotextile, r/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=

T 60 kN/m = = 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. Problem 26.6 How would you make cement? How do you make lime? What is the difference between cement and lime? Solution 26.6 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 26.7 A soil has a D10 equal to 2 mm. The grout used to strengthen it has a D65 of 60 μ m and a D95 of 130 μ m. Can particulate grouting be successful? Solution 26.7 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.

26.7 SELECTION OF SOIL IMPROVEMENT METHOD

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In this problem, grouting is feasible because: D10(soil)

=

2 mm = 33.3 > 24 60 μ m

N1 =

D65(grout)

N2 =

D10(soil) 2 mm = 15.4 > 11 = D95(grout) 130 μ m

Problem 26.8 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 (T1 ), and the tension due to lateral spreading (T2 ). 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 26.8 The load per column is: Qcol = (γ H + q)A

Qcol = (20 × 7 + 0) × 2 × 2 = 560 kN The tension T1 due to the bridging effect between columns is given by:   σ × Asoil =0 6T1 3 − (6T1 − EGS ) net p    0.8 × 20 × 7 × 4 − π × 0.52 3 6T1 − (6T1 − 60) =0 π ×1 T1 3 − 114.6(T1 − 10) = 0 which gives a tension T1 equal to:

T1 = −14 kN/m

The tension T2 due to the lateral spread of the embankment is given by: 1 K γ H 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 T2 =

So, the lateral spreading effect is much more severe than the bridging effect in this case.

CHAPTER 27

Technical Communications

27.1

GENERAL

The most important concepts in technical communications are: 1. 2. 3. 4.

Be brief Be clear Be right technically 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 expos´e, 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 for the vocabulary as well as the sophistication of the technical content and thought process. “Be right technically” 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. “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. 962

27.2

E-MAILS

E-mail 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 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.

27.5 THESES AND DISSERTATIONS

27.3

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.

27.4

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

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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.

27.5

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 to 75 characters (5 to 10 words) are best. 2. Cover page. The cover page should include the title, the author’ (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. 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 you

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27 TECHNICAL COMMUNICATIONS

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 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. 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 name of the first author, followed by the 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; name of publisher and the publisher’s location; inclusive page numbers. If the reference is a web site address (URL), the reference is organized as follows: • Author if any is credited, copyright or posting date, title, the address/URL of the web site 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.

27.6

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

27.8 MEETINGS

the same symbols but different colors, the data points will be undistinguishable 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 gauge 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. 27.7

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. E-mail or phone call: that is the question! Most of the time, e-mails 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 e-mail 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 e-mail 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 e-mail. Most people appreciate being

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told unpleasant truths “in person” by a telephone call rather than reading them in e-mail. 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 a follow-up e-mail to restate and formally memorialize the points covered in the conversation. 27.8

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 e-mails during the meeting. All are distracting (and discourteous) and should not be allowed during meeting. Motions and votes are very valuable because the decision becomes extremely clear. It takes place by 1. A motion is proposed by someone. 2. The motion is seconded by a second person. If not seconded, the motion dies. 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. Although motions may seem cumbersome at times, they are very useful in case of arguments after the fact. Remember

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27 TECHNICAL COMMUNICATIONS

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. 27.9 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 something 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, movies 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 movies 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. 9. Keep an eye on your audience to see if you are getting blank stares or interested looks. Adjust 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. 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: 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 is a 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 27.6). 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 movies is entertaining and holds the audience’s attention. If you intend to use movies, make sure that they work and double-check them right before your presentation.

27.10

MEDIA INTERACTION

The media has essentially three forms: the written press, the audio press, and 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 printed. 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 differently. 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

27.13 RULES FOR A SUCCESSFUL CAREER

ceiling; then start with the obvious while formulating the rest of your answer in your mind.

27.11

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. It often makes it smooth interaction 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!

27.12

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/). For the world scene it is the International Society for Soil Mechanics and Geotechnical

967

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.

27.13

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 ten successes to erase one 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 Ten” 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: 10. Choose the relentless pursuit of excellence as a way of life. 9. Be curious. The discovery process is a fountain of youth. 8. Work hard but balance your interests (fun, family, sport, art, world news). 7. Make lots of friends. Nurture your public relations. 6. Look for solutions and not who is to blame. Leave that to the judge. 5. Be firm in your decisions, but always be fair and polite. 4. Treat others as you wish to be treated, and you will lead by example. 3. Communication is the best way to solve problems. Convince through logic and data. 2. Surround yourself with smart people and positive role models. 1. Pursue your dreams with vision and perseverance.

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INDEX A abutment scour, 836 accurate method, 584 acoustic impedance, 153 active retaining walls, 716–717 active zone, 478, 597 activity (soil parameter), 57 adobe, 88 adsorbed water layer, 29 aeolian soil, 88 aggregate columns, 946 air air entry value, 270 air permeability test (for unsaturated soils), 214–215 flow of, in unsaturated soil, 382–388 in situ air sparging (ISAS), 888 allowable stress design (ASD), 490 alluvial fans, 20 alluvium, 20, 88 American Society for Testing and Materials (ASTM), 49 anchor bond length, 741 anchored walls, 735–746, 805 anchoring length, 919 anchors, retaining walls and, 740–742 angle of repose, 651 antennas, 162 anticlines, 19 approach velocity, 842 aquifer, 22 area ratio factors, 253–254 artesian pressure, 21 artificial neural network (ANN) method, 314–315 associated displacement retaining walls, 716–717 associated flow rule, 352 ASTM Procedure, 884–885

atomic absorption spectrophotometry (AAS), 876 at-rest earth pressure, 724–725 attenuation relationship, 790 Atterberg, Albert, 53 Atterberg limits, 49, 53–56 augercast piles, 557 automatic hammers, 104 Avogrado, Amadeo, 874 Avogrado number, 874 axisummetric heat propagation, 474–475

B backward erosion, 852 band drains, 943 Barcelona Basic Model (BBM), 355–357 Barentsen, Pieter, 107 base grouting, 555 base instability, retaining walls and, 738 bathymetry, 94 battered piles, 553 BCD test, 126–127, 184 bearing capacity, 918 bells, 555 bender elements, 180–181 bentonite, 88 Bessel correction, 306 Biaud-Tucker SPT method, for driven piles in coarse-grained soils, 580 bioremediation (BR), 889–890 Bishop simplified method, 664–665 block analysis, 654 block failure, 588–589 body wave magnitude, 784 Boltzmann, Ludwig, 349 bored piles, 553, 555–558. See also pile installation borehole in situ tests, 127–129 983

984

INDEX

borehole shear test (BST), 117–119 borings field identification and boring logs, 87–88 site investigation, drilling, and sampling, 80–81 bottom barriers, 886–887 bottom-up retaining walls, defined, 716 bottom-up slopes, manmade, 918 boundary element method (BEM), 304 Boussinesq, Joseph, 2 Boutwell, Gordon, 131 bracketed duration, 789 Brazos River meander case history, 845–847 breaking the soil, 909 bridge scour case history, 841–844 defined, 831 explained, 831–841 8 Buckingham theorem, 315–316 bulbs of pressure, 509 bulk modulus, 346 buoyancy, underwater foundations and, 582 buoyancy force, 33 burping the tremie, 555

C calcareous sands, 88 calcium oxide, 952 caliche, 88 California bearing ratio test (CBR), 122 Cam Clay model, 354–355 cantilever, 727 cantilever edge distance, 518 cantilever gravity, 727 cantilever retaining walls, 805 cantilever top-down walls, 732–735 capillary zone, 423 CAPWAP method, 570 Case Method, 568–569 Cassagrande, Arthur, 58, 185, 186 cations, 28 Celsius, 6, 472 Celsius, Anders, 472 cement, 949 cementation, 403 centrifuge model, similitude laws application, 317–318 characteristics, 281 characteristic site period, 793 characteristic value, 491 chart approach, 506–507

chemical grouting, 950 chilled mirror psychrometers, 176–177 classification parameters, soil, 56–57, 58 clastic rocks, 68 clay composition, 27–28 clay liners, geosynthetic, 910–911 clear water scour, 831 cliff, 19 code approach, earthquake geoengineering, 795–797 Code of Federal Regulations, 873 cohesive soils, 453 collapse deformation behavior, 424–425 collapse test, 193 collapsible soils, 19, 88 colluvial fans, 20 colluvium, 88 combined piled raft foundation (CPRF), 609–612 compaction dynamic or drop-weight compaction, 707–710 earth pressure retaining walls due to, 725–726 field tests, 700 generally, 698 impact roller compaction, 706–707 intelligent roller compaction, 701–706 laboratory tests, 698–700 soil improvement and, 938 soil type and, 701 compaction control tests BCD test, 126–127 field oven test, 125–126 generally, 124 lightweight deflectometer (LWD) test, 126 nuclear density/water content test, 125 rubber balloon test (RBT), 124–125 sand cone test (SCT), 124 See also compaction test compaction grouting, 950 compaction test dry unit weight, 181–184 soil modulus, 184–185 compensation grouting, 950 Comprehensive Environmental Response, Compensation and Liability Act (CERCLA), 874 compressible inclusions, 922 compression index, 407–408 compressive strength, 443 concentrated leak, 852 concrete shear strength properties and, 448 soil improvement and, 949 conduction, 472

INDEX

cone penetration test (CPT), 107–111 cone penetrometer dissipation test (CPDT), 129 confinement effect, 403 conservation of mass, 370 consolidated undrained direct shear test (CUDS), 450–451 consolidated undrained simple shear test (UUSS), 450–451 consolidation settlement magnitude, 510–511 time rate, 511 consolidation test compression index, recompression index, and secondary compression index from, 407–408 defined, 185–190 preconsolidation pressure and overconsolidation ratio from, 413–416 time effect from, 416–418 constant gradient procedure, 188 constant head permeameter test, 209–212 constant rate of strain procedure, 188 constitutive laws, 280 constrained modulus, 346 contaminants, types of, 872–873. See also geoenvironmental engineering continuous bridge, 522 continuous control compaction, 701 contractile skin, 256, 257–258 contraction scour, 835 contractive soil, 30, 129 convection, 472 conventional compaction, 698 coring, of rock, 73 Coulomb, Charles, 2, 717 Coulomb earth pressure theory, 717–719 course-grained soils, shear strength properties and, 448–449, 451–452 covers, for landfills, 893 covers, geosynthetics, 913–915 crack openings, 676 creep, 348, 407–408 creep compliance function, 348–349 creep settlement, 511–513 critical circle, 667 critical damping, 787 critical hydraulic gradient, 374–375 critical plane, 652 cross hole sonic logging, 558 cross hole test, 155–156 cross-plane, 907 cryosuction process, 479 Culman, Carl, 2

985

cumulative distribution function, 306 cyclic loading effect, 583, 604–605 cyclic modulus, 401 cyclic stress ratio (CSR), 797 cylindrical coordinates, 250–251

D damper, 347–348 Daniel, David, 130 Darcy, Henry, 2, 371 Darcy’s Law, 130, 318, 371–372, 880 dashpot, 347–348 Da Vinci, Leonardo, 2 deconvolution, 794 deep cement mixing, 951 deep foundations combined piled raft foundation (CPRF), 609–612 design strategy, 553–555 downdrag, 592–597 horizontal load and moment, pile group, 606–609 horizontal load and moment, single pile, 598–606 pile installation, 555–575 piles in shrink-swell soils, 597–598 seismic design, 806–807 types of, 553 vertical load, pile group, 587–591 vertical load, single pile, 575–587 deep soil mixing, 951 deep water, 833 deformation properties collapse deformation behavior, 424–425 common values of Young’s modulus and Poisson’s ratio, 406–407 compression index, recompression index, and secondary compression index from consolidation test, 407–408 correlations with other tests, 408 deformation problems, solving, 283–286 generally, 401 initial tangent modulus (Gmax ), 411–412 modulus, defining, 402 modulus, modulus of subgrade reactions, and stiffness, 405–406 modulus, time effect, and cyclic effect from pressuremeter test, 418–419 modulus and differences between fields of application, 405 modulus and influence of loading factors, 403–405 modulus and influence of state factors, 402–403 modulus as comprehensive model, 408–411 modulus of deformation, generally, 401–402

986

INDEX

deformation properties (continued) preconsolidation pressure and overconsolidation ratio from consolidation test, 413–416 reduction of Gmax with strain (G/Gmax curve), 412–413 resilient modulus for pavements, 419–420 shrink-swell deformation behavior, shrink-swell modulus, 422–424 time effect from consolidation test, 416–418 unsaturated soils and effect of drying and wetting on the modulus, 420–422 dense nonaqueous phase liquids (DNAPLs), 877 depth of compaction, 709 design methods, prediction methods versus, 583–584 deterministic analysis, 312 diagenetic bonds, 453 dielectric constant, 162 diesel hammers, 560 diffusivity, 473 dikes, erosion and, 847–850 dilatancy test, 87 dilatant, 129 dilatant structure, of soil, 30 dilatometer test (DMT), 114–115 dimensional analysis, 315–316 dip, 19 direct current differential transformer (DCDT), 179 direct shear test, 193–195 direct strength equations, 491–494 discharge velocity, 211, 370 discounted anchor length, 741 discrete element method (DEM), 304–305 dispersed structure, of soil, 30 dispersion curve, 159 dispersive clays, 88 displacement-replacement technique, for soil improvement, 941 displacements, 249 downdrag, 592–597 drainage, geosynthetics and, 907, 919 drained analysis, 463, 669 drawing, to scale, 280 drilled piers, 553, 555 drilled shafts, 553, 555 drilling hollow stem auger drilling method, 82–83 wet rotary drilling method, 81–82 See also site investigation, drilling, and sampling Drucker-Prager criterion, 351 dry soil, 26 dry strength test, 87 dry unit weight, 31, 181–184

Duncan-Chang model (DC model), 353–354 dunes, 20 durability, 72–73 dynamic compaction, 698 dynamic finite element analysis, 676 dynamic replacement (DR), 948 dynamic soil properties, earthquake geoengineering, 786 E earth dams, internal erosion of, 851–854 earth pressure retaining walls at-rest earth pressure, 724–725 defined, 716–717 due to compaction, 725–726 earth pressures in shrink-swell soils, 726 theories, 717–723 earthquake geoengineering design parameters, 794–797 earthquake, defined, 784 earthquake magnitude, 784–786 generally, 784 ground motion, 786–789 ground response analysis, 792–794 liquefaction, 797–801 seismic design of foundations, 806–807 seismic design of retaining walls, 802–805 seismic hazard analysis, 789–792 seismic slope analysis, 674–676 seismic slope stability, 801 seismic waves, 151–153 edge drop, 518 edge lift, 518 effective stress analysis, 463 saturated soils, 253 unsaturated soils, 252–253 effective stress analysis, 669 effective stress cohesion intercept, 451 effective stress method, for driven piles in fine-grained soil, 580 effective stress principle, 3 effective unit weight, 31 Eiffel Tower, 5, 528 elasticity defined, 345–347 deformation properties and, 401 elasticity approach for homogenous soils, shallow foundations, 504 elasticity approach for layered soils, shallow foundations, 504–506 electrical double layer, 29

INDEX

electrical resistivity techniques, 160–161 electric pulse compaction, 940 electromagnetic methods electromagnetic waves, 161–162 ground-penetrating radar (GPR), 162 time domain reflectometry (TDR), 162–165 electro-osmosis, 945 end-bearing piles, 553 engineering geology, generally defined, 15 Earth and universe age, 15 geologic features, 19–20 geologic maps, 20 geologic time, 15–17 groundwater, defined, 20–22 rocks, defined, 17 soil creation, 17–19 Environmental Protection Agency, 873, 882, 883–884 environmental site assessments (ESAs), 877 epicenter, 784 epicentral distance, 784 equilibrium equations, for two-dimensional analysis (calculating stresses), 246–247 Erdbaumechanik (Terzaghi), 2 erosion control, geosynthetics, 920–921 erosion of soils and scour problems bridge scour, 831–841 countermeasures for erosion protection, 850–851 erosion function, measuring, 824–825 erosion models, 824 erosion phenomenon, 823–824 internal erosion of earth dams, 851–854 levee overtopping, 847–850 river meandering, 844–847 rock erosion, 826–829 soil erosion categories, 825–826 water velocity, 829–831 Woodrow Wilson Bridge case history, 841–844 erosion test, 215–218 error function, 307 escarpments, 19 excess pore pressure, 286 exit gradient, 380 expansive soils, 88 expected earthquake, 794 explosive compaction, 940

F factor of safety, 76 failure (geomembrane), 908

987

failure problems, solving, 281–283 falling head permeameter test (for saturated soils), 212–213 fate, contaminant transport and, 880 faults, 19, 71, 784 FEM approach, 745–746 FHWA method for bored piles in coarse-grained soils, 578–580 for bored piles in fine-grained soils, 578 field oven test, 125–126 field values of hydraulic conductivity, lab values versus, 373 fill, preloading using, 941–943 filter paper method, water tension stress, 174–175 filter soil, 854 filtration, 907, 919 fine-grained soils, shear strength properties and, 453–456 finite difference method (FDM), 289–294 finite element method (FEM), 294–304, 674 first load modulus, 401 fissures, 71 fixed-head condition, 602–603 floating foundation, 523 flocculated structure, of soil, 30 floodplain deposits, 19 flow channel, 377 flow field, 377 flow net calculations for, 379–381 defined, 377 drawing, for homogenous soil, 377–378 flow and, for layered soils, 381–382 for hydraulically anisotrophic soil, 380–381 properties of, for homogenous soil, 378 flow of fluid/gas generally, 370 water and air in unsaturated soil, 382–388 water in saturated soil, 370–382 flow path, 371 flow problems, solving, 286–289 flow rule, 352 folds, 19 foundations. See deep foundations; shallow foundations foundations, geosynthetic, 918–919 Fourier, Jean Baptiste Joseph, 787 Fourier, Joseph, 473 Fourier acceleration spectrum, 787–789 Fourier’s Law, 473 Fourier spectrum, 787 free-head condition, 602–603 free span distance, 518 free swell, 55

988

INDEX

free swell limit, 422 friction piles, 553 frozen soils, 478–479 fundamental laws, 280

G gamma-gamma logging, 558 gas, generated by landfills, 895 general bearing capacity equation, 494–496, 499 generalized equilibrium method, 665–667 geobags, 904 geocells, 904 geochemistry background, 874–877 defined, 876 See also geoenvironmental engineering geocomposites, 905 geoenvironmental engineering contamination, 877–883 future considerations, 895–896 generally, 872 geochemistry background, 874–877 landfills, 890–895 laws and regulations, 873–874 remediation, 872, 883–890 types of wastes and contaminants, 872–873 See also geosynthetics geofoams, 904 geogrids, 904 geologic maps, 20 geologic time, 15–17 geomembranes, 904, 908, 913. See also geosynthetics geometry of the obstacle, 831 geonets, 904, 912 geophysics, elements electrical resistivity techniques, 160–161 electromagnetic methods, 161–165 generally, 151 remote sensing techniques, 165–166 seismic techniques, 151–159 geosynthetics clay liners, 904 compressible inclusions, 922 defined, 904 erosion control, 920–921 filtration and drainage, 919–920 geosynthetic mat and column-supported embankment, 953–955 landfill slopes, 922 lightweight fills, 922

liners and covers, 913–915 properties of, 905–913 reinforcement, 915–919 thermal insulation, 922 types of, 904–905 geotechnical centrifuge, 317–318 geotechnical engineering, generally defined, 1 failures, 5 foundations, 5 as fun, 5 past and future of, 2 recent and notable projects, 2–5 units of measure, 5–10 geotextiles, 904 governing differential equation (GDE), 882 Gow, Charles, 104 grains, 26 gravel composition, 27 particle size, shape, color, 26–27 gravimetric water content, 31–32 gravity walls, 727–729, 802–804 ground, 938 ground freezing, 945 ground motion, earthquake and, 786–789 ground-penetrating radar (GPR), 162 ground response, 792 ground response analysis, 792 ground rolls, 153 groundwater defined, 20–22 deformation properties and, 423 groundwater table, 20 remediation, geoenvironmental engineering, 888–890 site investigation, drilling, and sampling, 85–87 water stress conditions and, 679 (See also slope stability) group velocity, 157 grout, 949 grouted barriers, 885 grouting techniques, for soil improvement, 948–953

H hand shaking test, 87 hand tampers, 698 Handy, Richard, 117 hardening rule, 352–353 hardness (rock), 73 harmonic functions, 377 hazard level, 797

INDEX

heads, of water, 371 head (water), 371 heat conduction theory, 473–474 heat flow, 472 heat transfer rate, 472 heave and critical block, 380 high air entry porous stone, 173 histogram, 304 hollow stem auger drilling method, 82–83 hurricanes defined, 850 Hurricane Katrina levee case history, 848–850 hydration, 949 hydraulic conductivity defined, 371–372 of saturated soils, 371–373 of unsaturated soils, for water and for air, 382–384 hydraulic conductivity field tests borehole tests, 127–129 cone penetrometer dissipation test (CPDT), 129 generally, 127 sealed double-ring infiltrometer test (SDRIT), 130–131 two-stage borehole permeameter test (TSBPT), 131–132 hydraulic gradient, 371 hydraulic hammers, 560 hydro-blasting compaction, 945 hydrograph, 830 hydrometer analysis, 49, 50–53 hypocenter, 784

I ice lenses, 479 igneous rocks, 17, 68 impact hammers, 560 impedance log, 560 impulse response method, 559–560 incremental loading procedure, 185 independent stress state variables, 264 inertial interaction, 806 initial tangent modulus (Gmax ) defined, 411–412 reduction of Gmax with strain (G/Gmax curve), 412–413 inliers, 19 in-plane, 907 in situ air sparging (ISAS), 888 in situ flushing, 888 in situ tests, 80–81 borehole shear test (BST), 117–119 California bearing ratio test (CBR), 122 compaction control tests, 124–127

989

cone penetration test (CPT), 107–111 dialatometer test (DMT), 114–115 generally, 104 hydraulic conductivity field tests, 127–132 offshore, 132–134 plate load test (PLT), 119–122 pocket erodometer test (PET), 123–124 pocket penetrometer test (PPT), 122–123 pressuremeter test (PMT), 111–114 shear strength properties and, 450–452 soil modulus and correlation with, 408 standard penetration test (SPT), 104–107 torvane test (TVT), 122–123 vane shear test (VST), 115–117 in situ waste containment, 885 intelligent compaction, 698, 703 interaction factor method, 590 interface shear stress, 829 International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE), 49, 174, 609 ions, 160 ironing, 708 isomorphous, defined, 28 isomorphous substitution, 28

J Janbu chart, 658–659 jet grouting, 950 joints, 71 Joule, James Prescott, 472 joules, defined, 472 jumping jacks, 698 junction strength, 909

K karst, 19 Kelvin, 6, 472 Kelvin-Voigt model, 347–348 Khalili rule, 717 kilogram, 5 kilo-Newton, 10 kilo-Pascal, 10 kinematic interaction, 806

L laboratory tests air permeability test for unsaturated soils, 214–215 collapse test, 193

990

INDEX

laboratory tests (continued) compaction test, dry unit weight, 181–184 compaction test, soil modulus, 184–185 consolidation test, 185–190 constant head permeameter test, 209–212 direct shear test, 193–195 erosion test, 215–218 falling head permeameter test for saturated soils, 212–213 generally, 172 lab vane test, 206 measurements, 172–181 resonant column test, 202–206 shrink test, 192–193 simple shear test, 195–196 soil water retention curve (soil water characteristic curve) test, 206–209 swell test, 190–192 triaxial test, 198–202 unconfined compression test, 196–198 wetting front test for unsaturated soils, 213–214 lab values of hydraulic conductivity, field values versus, 373 lab vane test, 206 lacustrine deposits, 88 landfills, 890–895 Laplace equation, 377 latent heat, 475 laterite, 88 leachate collection, 893–894 levee overtopping, erosion and, 847–850 LIDAR (laser radar), 165 light nonaqueous phase liquids (LNAPLs), 876–877 lightweight deflectometer (LWD) test, 126 lime, 952 limit pressure, 353 limit states defined, 488–489 limit state design (LSD), 490 limit state function, 488 linear elasticity, 401 linear viscoelasticity, 347–349 liners, geosynthetic, 904, 913–915 liquefaction earthquake geoengineering, 797–801 sand liquefaction, 375 liquidity index, 57 liquid limit, 53 live bed scour, 831 load cyclic loading effect, 604–605 horizontal load and moment, pile group, 606–609 horizontal load and moment, single pile, 598–606

load and resistance factor design (LRFD), 490, 595–596 loading-collapse curve (LC curve), 356 loading rate, undrained strength and, 456 load settlement curve approach, 500–502 normal compression loading (NCL) curve, 355 one-way cyclic loading, 583 plate load test (PLT), 119–122 rate of loading effect, 603–604 surface loading and retaining walls, 722–723 testing, pile installation, 571–575 vertical load, pile group, 587–591 vertical load, single pile, 575–587 See also deep foundations; shallow foundations loam, 89 loess, 89 long flexible pile, 599–601 longitudinal distortion, 522 long-term analysis, 463, 669 love waves, 153 LPC-CPT method, 578 LPC-PMT method, 576–578

M machine drive power, 705 major principal stress, 245 manometer, 173 Marchetti, Silvano, 114 marl, 89 mat foundation defined, 385 large mat foundations, 523–528 matric suction, 256–257 maximum dry density, 183 maximum shear stress, scour and, 837–839 Maxwell, James, 348 Maxwell model, 347–348 meandering, by rivers, 844–847 MEANDER method, 844 meander migration, 19 mechanically stabilized earth (MSE) walls, 729–732, 805, 915–918 mechanical waves, 161 Menard, Louis, 111 metamorphic rocks, 17, 68 meter, 5 methane, 895 method of slices, 661–667 Michelangelo, 2 microbial methods, for soil improvement, 952 micropiles, 557

INDEX

minor principal stress, 245 Mississippi River, locks and dams of, 3 mixing method, for grouting, 951–952 mobility, 559 Modified Cam Clay (MCC) model, 354–355 Modified Proctor Compaction Test (MPCT), 181, 183–184 modulus of deformation. See deformation properties modulus of elasticity, 345–347 modulus of subgrade reaction, 405–406, 602 Mohr, Otto, 2, 247, 350 Mohr circle earth pressure theory, 720–721 in three dimensions, 248 for two-dimensional analysis, 247–248 Mohr-Coulomb yield criterion, 350 moment magnitude, 785 monitored natural attenuation (MNA), 889 montmorillonite, 89 Morgenstern chart, 659–661 mortar, 949 movement at depth in the slope, 676 movements of the slope surface, 676

N National Geotechnical Experimentation Site, Texas A&M University, 485 natural unit weight, 31 negative pore pressure, 251 net increase in stress, 524 net settlement, 524 neutral point, 592 Newmark’s chart, 509–510 Newmark’s displacement method, 675–676 Newton, 6, 10 nodes, 289 nonaqueous phase liquids (NAPLs), 876–877 nonclastic rocks, 68 nondestructive testing (NDT), 558 nondispersive material, 157 normal compression loading (NCL) curve, 355 normality rule, 352 normal strain, 179–180, 249–250 normal stress, 245–246 nuclear density/water content test, 125 numerical simulation methods boundary element method (BEM), 304 discrete element method (DEM), 304–305 finite difference method (FDM), 289–294 finite element method (FEM), 294–304 numerical solutions, defined, 289

991

O offshore site investigations generally, 89–94 geophysical investigations, 94–95 geotechnical drilling, 95–98 geotechnical sampling, 99 in situ tests, 132–134 one-dimensional flow, 384–386 100-year flood, 830 one-way cyclic loading, 583 optimum water content, 183 organic clay/silt, 89 osmosis, 30 osmotic suction, 258 Osterberg load cell test, 572–573 outcrops, 19 outliers, 19 overconsolidated soil, 403, 448, 455 overconsolidation ratio (OCR), 408 overturning moment, 554, 607

P Panama Canal, 3 particles, of soil. See soil components particle velocity, 151–152 particulate grouting, 949 passive earth pressure retaining walls, 716–717 peak ground acceleration (PGA), 786 peak ground displacement (PGD), 786 peak ground velocity (PGV), 786 peat, 89 peel strength, 911 perched water, 22 permafrost, 20, 478 permanent set, 561 permeability, 373 permeance (geomembrane), 908 phase velocity, 157 phicometer, 119 phreatic surface, 21, 87, 667 pier scour, 832–834 piezometric surface, 21, 667–668 pile driving analyzer (PDA), 568–569 pile installation of bored piles, 555–558 information from pile driving measurements, 566–570 installation of driven piles, 560–561 load testing, 571–575 nondestructive testing of bored piles, 558–560 pile driving formulas, 561–562

992

INDEX

pile installation (continued) suction caissons, 570–571 wave equation analysis, 563–566 wave propagation in a pile, 562–563 piston samplers, 85 plane strain, 346 plane stress, 346 plasticity, 349–353, 353. See also soil constitutive models plasticity index, 53, 57, 58, 513 plastic limit, 53 plastic potential function, 352 plate load test (PLT), 119–122 plugging, 99 pocket erodometer test (PET), 123–124 pocket penetrometer test (PPT), 122–123 Poisson’s ratio, 73, 401–402, 406–407, 581–582. See also deformation properties Pole method, 247–248 polyaromatic hydrocarbons (PAHs), 882 polychlorinated biphenyls (PCBs), 876, 882 pore-pressure parameters (A and B), 458–459 pores, 26 pore water pressure, 173 positive pore pressure, 251 potential vertical rise (PVR) method, 514 precise method, 584 prediction methods, design methods versus, 583–584 prefabricated vertical drains (PVDs), 943–944 preloading using fill, 941–943 using vacuum, 943–944 pressuremeter test (PMT), 111–114, 418–419 pressure plate apparatus (PPA), 177–178 principal planes, 245 probability and risk analysis background, 305–308 probabilistic approach, 305 procedure for probability approach, 308–310, 312–313 risk and acceptable risk, 310–312 problem-solving methods artificial neural network (ANN) method, 314–315 continuum mechanics methods, 281–289 dimensional analysis, 315–316 drawing to scale and, 280 generally, 280 numerical simulation methods, 289–305 primary laws, 280 probability and risk analysis, 305–313 regression analysis, 313–314 similitude laws for experimental simulations, 317–319 types of analyses, 319

Proctor, Ralph, 181 progressive failure, 669 pseudostatic method, 674–675 psychrometers, 176–177 pull-out design, retaining walls and, 730–732, 748–749 pump and treat, 888 punching, 913 P waves, 152 P-y curve approach, 605–606, 733–735, 745–746

Q quantity of flow, 379 quick clay, 89, 375 quick sand, 89, 374–375

R radar satellite, 165 radiation, 472 raft foundations, 523 rammed aggregate pier method, 946 rams, 560 Rankine, William, 2, 719 Rankine earth pressure theory, 719–720 rapid impact compaction, 940 rare earthquake, 794 rate of loading effect, 603–604 Rayleigh waves, 153, 157 recompression index, 407–408 recovery ratio, 73 recurrence interval, 791 reflection, seismic, 153–154 refraction, seismic, 154–155 refractive index, 154 regression analysis, 313–314 relative humidity, total suction and, 258–260 relaxation, 348 relaxation modulus function, 348–349 reliability index, 306 remediation, geoenvironmental engineering, 872, 883–890 remolded shear strength, 446 remote sensing techniques, 165–166 replacement. See soil improvement residual shear strength, 446 residual soils, 89 residual strength, 461 residual stresses, 567 resilient modulus, 402, 419–420 resistivity tomography, 160–161 resonant column test, 202–206

INDEX

response spectrum, 787 retaining walls active, at rest, passive earth pressure, and associated displacement, 716–717 anchored walls and strutted walls, 735–746 at-rest earth pressure, 724–725 bottom-up, defined, 716 cantilever top-down walls, 732–735 displacements, 726–727 earth pressure due to compaction, 725–726 earth pressures in shrink-swell soils, 726 earth pressure theories, 717–723 gravity walls, 727–729 mechanically stabilized earth (MSE) walls, 729–732, 805, 915–918 seismic design of, 802–805 soil nail walls, 746–751 top-down, defined, 716 trenches, 751–752 undrained behavior of fine-grained soils, 723–724 return period, 791 revegetation, 920 Reynolds Number, 216 rib strength, 909 Richter scale, 784 risk analysis geoenvironmental engineering, 883–885 probability and risk analysis, 305–312 rivers Brazos River meander case history, 845–847 contraction scour, 835 river meandering, defined, 844 See also erosion of soils and scour problems river terraces, 20 road reinforcement, geosynthetics and, 815 rock erosion, 826–829 rock mass erosion, 828 rock quality designation, 73 rocks definitions, 17, 68 discontinuities in, 71 permafrost, 76 rock engineering problems, 74–76 rock engineering properties, 72–73 rock groups and identification, 68 rock index properties, 71–72 rock mass, defined, 68 rock mass rating, 73–74 rock mass vs. rock substance, 68–71 rock substance, defined, 68 rock substance erosion, 828

993

rubber balloon test (RBT), 124–125 rule of the middle third, 729

S salt solution equilibrium (SSE), 178–179 sampling disturbance, 83–84 methods, 84–85 offshore geotechnical sampling, 99 See also site investigation, drilling, and sampling sand composition, 27 particle size, shape, color, 26–27 sand cone test (SCT), 124 sand liquefaction, 375 San Jacinto Monument, 527–528 satellite imaging, 165–166 saturated flow, 382 saturated soil defined, 26 effective stress, 253 water flow in, 370–382 water stress predictions, 357–358 saturated unit weight, 31 saturation, 784 scaled model, similitude laws application (example), 318 Schmidt hammer, 73 scour problems. See erosion of soils and scour problems sealed double-ring infiltrometer test (SDRIT), 130–131 second, as unit of measure, 5 secondary compression index, 512–513 secondary consolidation, 407–408 secondary recompression index, 407–408 sedimentary rocks, 17, 68 seepage analysis, 668 seepage force, 371, 373–374, 652–653 seepage velocity, 211, 370 seismic cone test, 155–156 seismic dilatometer test, 155–156 seismic hazard analysis, 789–791 seismic reflection, 94, 153–154 seismic refraction, 94, 154–155 seismic slope analysis, 674–676 seismic slope stability, 801 seismic waves, 151–153 separation (geomembrane), 908, 913 service limit state, 489, 502 settlement consolidation settlement, magnitude, 510–511 consolidation settlement, time rate, 511

994

INDEX

settlement (continued) creep settlement, 511–513 example of settlement calculations, 524–527 general behavior, 502–504 geosynthetics and, 919 load settlement curve approach, 500–502 of piles, deep foundations, 584–587, 589–591 See also shallow foundations shale, 89 shallow foundations case history, 485 cost of, 553 definitions, 485 definitions and design strategy, 485–488 foundations on shrink-swell soils, 517–522 general behavior, 491 large mat foundations, 523–528 limit states, load and resistance factors, and factor of safety, 488–491 load settlement curve approach, 500–502 seismic design, 806–807 settlement, 502–513 shrink-swell movement, swelling pressures, and collapse movement, 513–517 tolerable movements, 522–523 ultimate bearing capacity, 491–500 SHANSEP method, 456–458 shape function matrix, 295 shear modulus, 346 shear strain, 180, 249–250 shear strength properties basic experiments, 443–445 estimating effective stress shear strength parameters, 451–454 estimating undrained shear strength values, 459–461 experimental determination of shear strength (lab tests, in situ tests), 450–451 generally, 443 pore-pressure parameters A and B, 458–459 residual strength parameters and sensitivity, 461–462 SHANSEP method, 456–458 shear strength, defined, 443 shear strength envelope, 447–449 strength profiles, 462–463 stress-strain curve, water stress response, and stress path, 445–447 transformation from effective stress solution to undrained strength solution, 463 types of analyses, 463 undrained shear strength for unsaturated soils, 458

undrained shear strength of saturated fine-grained soils, 454–456 unsaturated soils, 449–450, 458 shear stress, 245–246 Shelby tube sampler, 84 short rigid pile, 601–602 short-term analysis, 463, 669 short-term case, 454 shrinkage limit, 53, 57, 513 shrink-swell deformation behavior, 422–424 shrink-swell index, 57, 423–424, 513 shrink-swell modulus, 422–424 shrink-swell movement, shallow foundations and, 513–517, 514 shrink-swell soils deep foundations, piles in, 597–598 defined, 89 earth pressure, retaining walls, 726 foundations on, 517–522 shrink test, 192–193 sidescan sonars, 95 sieve analysis, 49–50 sign convention, for stresses and strains, 246 silt composition, 27–28 particle size, shape, color, 26–27 silt fences, 921 similitude laws, for experimental simulations, 317–319 simple shear test, 195–196 simply supported bridges, 522 sinkholes, 19 site classes, 795 site investigation, drilling, and sampling drilling methods, 81–83 field identification and boring logs, 87–88 generally, 80 groundwater level, 85–87 number and depth of borings and in situ tests, 80–81 offshore geophysical investigations, 94–95 offshore geotechnical drilling, 95–98 offshore geotechnical sampling, 99 offshore site investigations, 89–94 preliminary site investigation, 80 sampling disturbance, 83–84 sampling methods, 84–85 soil names, 88–89 slaking durability test, 72–73 slickensided clay, 89 slopes, geosynthetic, 918

INDEX

slope stability chart methods, 655–661 design approach, 649–650 finite element analysis, 674 generally, 649 infinite slopes, 650–652 method of slices, 661–667 monitoring, 676–679 plane surfaces, 654 probabilistic approach, 671–672 progressive failure in strain-softening soils, 669 reinforced slopes, 670 repair methods, 679–680 seepage force in stability analysis, 652–653 seismic slope analysis, 674–676 shallow slide failures in compacted unsaturated embankments, 669–670 slopes with water in tensile cracks, 654–655 three-dimensional circular failure analysis, 672–673 types of analyses, 668–669 water stress for slope stability, 667–668 slope stability, landfills, 894–895, 922 slurry trench barriers, 885 Snell’s law, 154 softening rule, 352–353 soil, generally creation of, 17–19 soil names, 88–89 (See also soil classification; soil components) stresses in three soil phases, 251–252 (See also stresses and strains) See also saturated soil; unsaturated soil soil cement mixing, 951 soil classification Atterberg limits, 53–56 classification parameters, 56–57 engineering significance of classification parameters and plasticity chart, 58 hydrometer analysis, 50–53 sieve analysis, 49–50 tests for, 49 Unified Soil Classification System, 49, 58–59 soil components composition of clay, silt, 27–28 composition of gravel, sand, silt, 27 particle behavior, 28–30 particles, liquids, and gas, 26 particle size, shape, color, 26–27 saturated, defined, 26 soil structure, 30 three-phase diagram of, 30–31

995

unsaturated, defined, 26 weight-volume parameters, 31–32 weight-volume parameters, measurement, 32–33 weight-volume parameters, solving problems of, 33–35 soil constitutive models common models, 353–358 elasticity, 345–347 linear viscoelasticity, 347–349 plasticity, 349–353 soil model, defined, 345 soil contact erosion, 852 soil erosion categories, 825–826. See also erosion of soils and scour problems soil improvement generally, 938 with grouting and admixtures, 948–953 with inclusions, 953–955 with replacement, 946–948 without admixture in coarse-grained soils, 938–940 without admixture in fine-grained soils, 941–945 soil modulus, compaction test, 184–185 soil nails, 679 soil nail walls, 746–751 soil remediation, geoenvironmental engineering, 887–888 soil water retention curve (soil water characteristic curve) test, 206–209 soil water retention curve (SWRC), 262–264 sonic echo method, 558–559 soundings, 80–81 sound waves, 152 specific gravity test, 33 specific heat, 473 specific surface, 373 spectral analysis of surface waves, 156 spectral analysis of surface waves (SASW), 156–157, 156–159 Spencer chart, 657–658 spherical coordinates, 250–251 split spoon sampler, 84 spread footing, 485 SPT blow count, 87, 104 staking durability test, 72 standard penetration test (SPT), 33, 104–107 Standard Proctor Compaction Test (SPCT), 181–184 standpipe, 173 standpipe piezometers, 86 static load tests, 571–572 Statnamic load test, 573–575 steam hammers, 560 steel sheet pile barriers, 885–886 stiffened slab on grade, 485, 517–519

996

INDEX

stiffness, 906 Stokes, George, 52 Stokoe, Ken, 156 stone columns, 946 strain gages, 180 strain hardening/softening, 353 strain rate, 404 strain tensor, 249–250 stresses and strains area ratio factors, 253–254 calculating stresses on any plane, equilibrium equations for two-dimensional analysis, 246–247 calculating stresses on any plane, Mohr circle for two-dimensional analysis, 247–248 cylindrical coordinates and spherical coordinates, 250–251 displacements, 249 effective stress (saturated soils), 253 effective stress (unsaturated soils), 252–253 generally, 245 independent stress state variables, 264 Mohr circle in three dimensions, 248 net increase in stress, 524 normal strain, shear strain, strain tensor, 179–180, 249–250 precision on water content and water tension, 260 sign convention for stresses and strains, 246 soil water retention curve (SWRC), 262–264 strain rate, 404 strains, defined, 249 stress, defined, 245 stresses in three soil phases, 251–252 stress history factor, 403 stress increase with depth, for shallow foundations, 508–510 stress invariants, 248–249 stress profile at rest in unsaturated soils, 260–262 stress-strain curves, 251 stress vector, normal stress, shear stress, stress tensor, 245–246 water stress profiles, 254–255 water tension and suction, 255–260 See also deformation properties; retaining walls; shear strength properties; soil constitutive models stress-strain curve, 445–447 strip footings, 385 structure, of soil, 30 strutted walls, 735–746 sub-bottom profilers, 95 submerged unit weight, 31 subsidence, 19, 22

Subsurface Contamination Reference Guide (US EPA), 882 suction, 26, 29, 251, 255 suction caissons, 570–571 suffusion, 852 S waves, 152 swelling pressure, 423 swell limit, 34, 191 swell test, 190–192 synclines, 19

T TAMU-Slab method, 518–519 Taylor chart, 655–657 tear, 913 temperature gradient, 472 tendon bond anchor, 741 tendon unbonded length, 740, 741 tensiometers, 177 tension strength, 443 Terzaghi, Karl, 2 Texas A&M University, 485 thermal conductivity, 472 thermocouple psychrometers, 176 thermodynamics for soil problems applications, 477–478 axisummetric heat propagation, 474–475 definitions, 472–473 frozen soils, 478–479 generally, 472 heat conduction theory, 473–474 multilayer systems, 476–477 thermal properties of soils, 475–476 thin-wall steel tube, 84 Thompson, William (First Baron Kelvin), 348 thread rolling test, 87 three-dimensional air flow, 387–388 three-dimensional circular failure analysis, 672–673 three-dimensional water flow, 386–387 three-phase diagram, of soil components, 30–31 till, 89 time domain reflectometry (TDR), 162–165 Tokyo Haneda airport, 3 tolerable movement, shallow foundations and, 522–523 top-down retaining walls, defined, 716 torvane test (TVT), 122–123 total (normal) stress analysis, 669 total stress analysis, 463, 669 total unit weight, 31

INDEX

toughness test, 87 Tower of Pisa, 2, 3, 487–488, 528 toxicity characteristics leaching procedure (TCLP), 876 transverse wave, 162 Trautwein, Steve, 130 trees, osmotic suction and, 260 trenches, retaining walls and, 751–752 Tresca yield criterion, 350 triaxial test, 198–202 true cohesion, 453 tuff, 89 two-dimensional flow problem, 375–377 two-stage borehole permeameter test (TSBPT), 131–132 2 to 1 method, 508 two-way cyclic loading, 583

U ultimate bearing capacity, 491–500 ultimate capacity, 599, 607–609 ultimate limit state, 489 unconfined compression test, 196–198 unconsolidated undrained triaxial test (UUT), 450–451 underreams, 555 undrained analysis, 463, 669 undrained behavior of fine-grained soils, retaining walls and, 723–724 undrained case, 454 undrained shear strength, 454 Unified Rock Classification System, 73 Unified Soil Classification System, 49, 58–59 unit cell, 953 United States Geological Service, 786 units of measure, 5–10 unit weight of solids, 30 unsaturated flow, 382 unsaturated soil defined, 26 effective stress, 252–253 formation and effect of drying and wetting on the modulus, 420–422 shear strength properties, 449–450, 458 stress profile at rest in, 260–262 three-phase soils, 1 ultimate bearing capacity of, 499–500 water and air flow in, 382–388 water stress predictions, 357–358 uplift force, on buried structures, 380 U.S. Resource Conservation and Recovery Act (RCRA), 891–892

997

V vacuum, preloading using, 943–944 Van der Waals forces, 29, 257 vane shear test (VST), 115–117 varved clay, 89 velocity hydrograph, 840 velocity index, 73 vibratory hammers, 560 vibratory rollers, 704–705 vibrocompaction, 938–940 viscous exponent, 122 Voigt, Woldemar, 348 Von Mises, Richard, 351 Von Mises criterion, 351

W waffle slab, 517, 518 wash hands test, 87 Washington Monument, 2, 525–526, 527–528 wastes, types of, 872–873. See also geoenvironmental engineering water, generally adsorbed water layer, 29 clear water scour, 831 compression stress, 173 deep water, 833 flow of, in saturated soil, 370–382 flow of, in unsaturated soil, 382–388 gravimetric water content, 31–32 perched water, 22 stresses and strains, precision on water content and water tension, 260 stress profiles, 254–255 stress response, 445–447 (See also shear strength properties) tension and suction, 255–260 tension stress, 173–176 water content, defined, 263 water content vs. strain curve, 513–514 water stress, in flow net, 380 water stress predictions, 357–358 water-vapor transmission, 908 wave amplitude, 152 wave equation analysis, 561, 562–566 wave frequency, 152 wave velocity, 151 weight-volume parameters generally, 31–32 measurement, 32–33

998

INDEX

weight-volume parameters (continued) solving problems of, 33–35 See also soil components wet rotary drilling method, 81–82, 112 wetting front test (for unsaturated soils), 213–214 wick drains, 943 wide width strength, 909 Woodrow Wilson Bridge, 841–844 work hardening/softening, 353 working stress design (WSD), 490 World Trade Center, 74–76, 528

Y yield horizontal seismic coefficient, 675 yielding, of soil, 350–352 yield stress, 448 Young, Thomas, 345, 401 Young’s modulus, 345–347, 401, 406–407. See also deformation properties

Z zone of influence, 507–508