[Braja m. das]_principles_of_foundation_engineerin(book_fi.org)

Principles of
Foundation Engineering

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CONVERSION FACTORS FROM ENGLISH TO SI UNITS
Length: 1 ft = 0.3048 m Stress: 11b/ft' = 47.88 N/m'
1 ft = 30.48 em llb/ft' = 0.04788 kN/m'
1 ft = 304.8 mm 1 U.S. ton/ft' = 95.76 kN/m'
1 in. = 0.0254 m 1 kip/ft' = 47.88 kN/m'
1 in. = 2.54 cm lIb/in' = 6.895 kN/m'
1 in. = 25.4 mm Unit weight: 1 1b/ft3 = 0.1572 kN/m'
Area: 1 fl' = 929.03 X Hr'm' l Ib/in' = 271.43 kN/mJ
1 ft' = 929.03 cm' Moment: 1 lb-ft = 1.3558 N' m
1 ft' = 929.03 X 10' mm'
1 in:! = 6.452 X 10-' m'
l Ib-in. = 0.11298 N·m
1 in' = 6.452 cm' Energy: 1 ft-Ib = 1.3558J
1 in:! = 645.16 mm' Moment of 1 in' = 0.4162 X 106 mm'
Volume: 1 ft" = 28.317 X 10-3 m' inertia: 1 in' = 0.4162 X 10-6 m'
ift' = 28.317 X 10' cm"
Section 1 in3 = 0.16387 X 10' mm'
,1 in] = 16.387 X 10-6 m'
lin' = 16.387 cm'
modulus: I in3 = 0.16387 X lW'm'
Force: lIb = 4.448 N
Hydraulic 1 ft/min = 0.3048 m/min
lIb = 4.448 X 10-3 kN
conductivity: 1 ft/min = 30.48 cm/min
lib = 0.4536 kgf
1 ft/min = 304.8 mm/min
1 kip = 4.448 kN
1 ft/sec = 0.3048 m/sec
1 U.S. ton = 8.896 kN
1 fI/sec = 304.8 mm/sec
lib = 0.4536 X 10' metric tall
1 in/min = 0.0254 m/min
1 Ib/n = 14.593 N/m
1 in./sec = 2.54 cm/sec
1 in/sec = 25.4 mm/sec
Coefficient of 1 in'/sec = 6.452 cm'/sec
consolidation: 1 in'/sec = 20.346 X 103 m'/yr
1 ft'/sec = 929.03 cm'/sec

Principles of
Foundation Engineering
Sixth Edition
Braja M. Das
THOMSON
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To our granddaughter, Elizabeth Madison

2
Geotechnical Properties of Soil 1
1.1 Introduction 1
1.2 Grain-Size Distribution 2
1.3 Size Limits for Soils 5
1.4 Weight-Volume Relationships 5
1.5 Relative Density 9
1.6 Atterberg Limits 12
1.7 Soil Classification Systems 13
1.8 Hydraulic Conductivity of Soil 21
1.9 Steady-State Seepage 23
1.10 Effective Stress 25
1.11 Consolidation 27
1.12 Calculation of Primary Consolidation Settlement 32
1.13 Time Rate of Consolidation 33
1.14 Degree of Consolidation Under Ramp Loading 40
1.15 Shear Strength 43
1.16 Unconfined CompressionTest 48
1.17 Comments on Friction Angle, '1" 49
1.18 Correlations for Undrained Shear Strength, c" 52
1.19 Sensitivity 53
Problems 54
References 58
Natural Soil Deposits and Subsoil Exploration 60
2.1 Introduction 60
Natural Soil Deposits 60
2.2 Soil Origin 60
vii

viii Contents
2.3 Residual Soil 61
2.4 Gravity Transported Soil 62
2.5 Alluvial Deposits 62
2.6 Lacustrine Deposits 65
2.7 Glacial Deposits 65
2.8 Aeolian Soil Deposits 67
2.9 Organic Soil 68
Subsurface Exploration 68
2.10 Purpose of Subsurface Exploration 68
2.11 Subsurface Exploration Program 68
2.12 Exploratory Borings in the Field 71
2.13 Procedures for, Sampling Soil 74
2.14 Observation of Water Tables 85
2.15 Vane ShearTest 86
2.16 Cone Penetration Test 90
2.17 Pressuremeter Test (PMT) 97
2.18 Dilatometer Test 99
2.19 Coring of Rocks 101
2.20 Preparation of Boring Logs 105
2.21 Geophysical Exploration 105
2.22 Subsoil Exploration Report 113
Problems 114
References 119
Shallow Foundations: Ultimate Bearing Capacity 121
3.1 Introduction 121
3.2 General Concept 121
3.3 Terzaghi's Bearing Capacity Theory 124
3.4 Factor of Safety 128
3.5 Modification of Bearing Capacity Equations for Water Table 130
3.6 The General Bearing Capacity Equation 131
3.7 Meyerhofs Bearing Capacity, Shape, Depth, and Inclination Factors 136
3.8 Some Comments on Bearing Capacity Factor, Ny, and Shape Factors 138
3.9 A Case History for Bearing Capacity Failure 140
3.10 Effect of Soil Compressibility 142
3.11 Eccentrically Loaded Foundations 146
3.12 Ultimate Bearing Capacity under Eccentric Loading-Meyerhofs
Theory 148
3.13 Eccentrically Loaded Foundation-Prakash and Saran's Theory 154
3.14 Bearing Capacity of a Continuous Foundation Subjected to Eccentric
Inclined Loading 161
Problems 165
References 168
� I

Ultimate Bearing Capacity of Shallow Foundations:
Special Cases 170
Contents ix
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth 170
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain
by Weaker Soil 177
4.4 Closely Spaced Foundations-Effect on Ultimate
Bearing Capacity 185
4.S Bearing Capacity of Foundations on Top of a Slope 188
4.6 Bearing Capacity of Foundations on a Slope 191
4.7 Uplift Capacity of Foundations 193
Problems 199
References 202
Shallow Foundations: Allowable Bearing Capacity
and Settlement 203
Ve�iicalsi�e;sincreaseinaSoilMassCausedby FoundationLoad. . 204
5.2 Stress Due to a Concentrated Load 204
5.3 Stress Due to a Circularly Loaded Area 205
5.4 Stress below a Rectangular Area 206
5.5 Average Vertical Stress Increase Due to a Rectangularly
Loaded Area 213
5.6 Stress Increase under an Embankment 216
ElasticSettlement 220
5.7 Elastic Settlement Based on the Theory of Elasticity 220
5.8 Elastic Settlement of Foundations on Saturated Clay 230
5.9 Improved Equation for Elastic Settlement 230
5.10 Settlement of Sandy Soil: Use of Strain Influence Factor 236
5.11 Range of Material Parameters for Computing Elastic
Settlement 240
5.U Settlement of Foundation on Sand Based on Standard
Penetration Resistance 241
5.13 General Comments on Elastic Settlement Prediction 246
5.14 Seismic Bearing Capacity and Settlement in Granular Soil 247
cons�lidationSettlement 252
5.15 Primary Consolidation Settlement Relationships 252
5.16 Three·Dimensional Effect on Primary Consolidation Settlement 254
5.17 Settlement Due to Secondary Consolidation 258
5.18 Field Load Test 260

x Contents
5.19 Presumptive Bearing Capacity 263
5.20 Tolerable Settlement of Buildings 264
Problems 266
References 269
6 Mat Foundations 272 .
6.2 Combined Footings 272
6.3 Common Types of Mat Foundations 275
6.4 Bearing Capacity of Mat Foundations 277
6.5 Differential Settlement of Mats 280
6.6 Field Settlement Observations for Mat Foundations 281
6.7 Compensated Foundation 281
6.S Structural Design of Mat Foundations 285
Problems 304
References 307
7 Lateral Earth Pressure 308
7.2 Lateral Earth Pressure at Rest 309
Active Pressure 312
7.3 Rankine Active Earth Pressure 312
7.4 A Generalized Case for RankineActive Pressure 315
7.5 Coulomb's Active Earth Pressure 323
7.6 Active Earth Pressure for Earthquake Conditions 328
7.7 Active Pressure for Wall Rotation about the Top: Braced Cut 333
7.S Active Earth Pressure for Translation of Retaining
Wall-Granular Backfill 334
7.9 General Comments on Active Earth Pressure 338
Passive Pressure 338
7.10 Rankine Passive Earth Pressure 338
7.11 Rankine Passive Earth Pressure: Inclined Backfill 344
7.12 Coulomb's Passive Earth Pressure 345
7.13 Comments on the Failure Surface Assumption for Coulomb's
Pressure Calculations 347
7.14 Passive Pressure under Earthquake Conditions 348
Problems 349
References 352
: I
!

Retaining Walls 353
"" �--. "'� . �. ,� ".�'-'
GravityandCantileverWalls 355
8.2 Proportioning Retaining Walls 355
Contents xi
8.3 Application of Lateral Earth Pressure Theories to Design 356
8.4 Stability of Retaining Walls 358
8.5 Check for Overturning 359
8.6 Check for Sliding along the Base 361
8.7 Check for Bearing Capacity Failure 364
8.8 Construction Joints and Drainage from Backfill 374
8.9 Some Comments on Design of Retaining Walls 377
M;Ch�;'k;Jry-ft�bTiii.;;rR:;'i;;i�i;'iWai[;'··..379
8.10 Soil Reinforcement 379
8.11 Considerations in Soil Reinforcement 380
8.12 General Design Considerations 382
8.13 Retaining Walls with Metallic Strip Reinforcement 383
8.14 Step-by-Step-Design Procedure UsingMetallic Strip Reinforcement 390
8.15 Retaining Walls with Geotextile Reinforcement 395
8.16 Retaining Walls with Geogrid Reinforcement 399
8.17 General Comments 402
Problems 404
References 407
Sheet Pile Walls 409
9.2 Construction Methods 413
9.3 Cantilever Sheet Pile Walls 414
9.4 Cantilever Sheet Piling Penetrating Sandy Soils 415
9.5 Special Cases for Cantilever Walls Penetrating a Sandy Soil 422
9.6 Cantilever Sheet Piling Penetrating Clay 423
9.7 Special Cases for Cantilever Walls Penetrating Clay 428
9.8 Anchored Sheet-Pile Walls 429
9.9 Free Earth Support Method for Penetration of Sandy Soil 430
9.10 Design Charts for Free Earth Support Method
(Penetration into Sandy Soil) 435
9.11 Moment Reduction for Anchored Sheet-PileWalls 440
9.12 Computational Pressure Diagram Method for Penetration
into Sandy Soil 443
9.13 Field Observations of an Anchored Sheet Pile Wall 447
9.14 Free Earth Support Method for Penetration of Clay 448
9.15 Anchors 452
9.16 Holding Capacity of Anchor Plates in Sand 454

xii Contents
9.17 Holding Capacity of Anchor Plates in Clay (cp = 0 Condition) 460
9.18 Ultimate Resistance of Tiebacks 460
Problems 461
References 464
10 Braced Cuts 466
10.2 Pressure Envelope for Braced-Cut Design 467
10.3 Pressure Envelope for Cuts in Layered Soil 471
10.4 Design of Various Components of a Braced Cut 472
10.5 Bottom Heave of a Cut in Clay 482
10.6 Stability of the Bottom of a Cut in Sand 485
10.7 Lateral Yielding of Sheet Piles and Ground Settlement 487
Problems 489
References 490
11 Pile Foundations 491
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
11.14
11.15
11.16
11.17
11.18
11.19
11.20
11.21
Introduction 491
Types of Piles and Their Structural Characteristics 493
Estimating Pile Length 502
Installation of Piles 504
Load Transfer Mechanism 508
Equations for Estimating Pile Capacity 509
Meyerhofs Method for Estimating Qp 512
Vesic's Method for Estimating Qp 515
Janbu's Method for Estimating Qp 516
Coyle and Castello's Method for Estimating Qp in Sand 520
Other Correlations for Calculating Qp with SPT and CPT Results
Frictional Resistance (Q,) in Sand 524
Frictional (Skin) Resistance in Clay 528
Point-Bearing Capacity of Piles Resting on Rock 531
Pile Load Tests 538
Comparison of Theory with Field Load Test Results 542
Elastic Settlement of Piles 543
Laterally Loaded Piles 546
Pile-Driving Formulas 562
Pile Capacity For Vibration-Driven Piles 568
Negative Skin Friction 570
Group Pile; 573
11.22 Group Efficiency 573
11.23 Ultimate Capacity of Group Piles in Saturated Clay 576
11.24 Elastic Settlement of Group Piles 580
521
..

,-
I
(
/
11.25 Consolidation Settlement of Group Piles 581
11.26 Piles in Rock 584
Problems 584
References 588
· 12 Drilled-Shaft Foundations 591
Introduction 591
Types of Drilled Shafts 592
Construction Procedures 593
Other Design Considerations 598
Load Transfer Mechanism 598
Estimation of Load-Bearing Capacity 599
Contents xiii
U.l
U.2
U.3
U.4
U.5
U.6
U.7
U.8
U.9
U.I0
Drilled Shafts in Granular Soil: Load-Bearing Capacity
Drilled Shafts in Clay: Load-Bearing Capacity 613
Settlement of Drilled Shafts at Working Load 620
Lateral Load-Carrying Capacity-Characteristic Load
and Moment Method 622
602
U.11 Drilled Shafts Extending into Rock 631
Problems 635
References 639
13 Foundations on Difficult Soils 640
Collapsible Soil 640
13.2 Definition and Types of Collapsible Soil 640
13.3 Physical Parameters for Identification 641
13.4 Procedure for Calculating Collapse Settlement 645
13.5 Foundation Design in Soils Not Susceptible to Wetting 646
13.6 Foundation Design in Soils Susceptible to Wetting 648
Expansive Soils 649
13.7 General Nature of Expansive Soils 649
13.8 Laboratory Measurement of Swell 650
13.9
'
Classification of Expansive Soil on the Basis of Index Tests 655
13.10 Foundation Considerations for Expansive Soils 656
13.11 Construction on Expansive Soils' 661
Sanitary Landfills 665
13.U General Nature of Sanitary Landfills 665
13.13 Settlement of Sanitary Landfills 666
Problems 668
References 670

xiv Contents
14.; Soi/lmprovement and Ground Modification 672
14.2 General Principles of Compaction 673
14.3 Correction for Compaction of Soils with Oversized Particles 676
14.4 Field Compaction 678
14.5 Compaction Control for Clay Hydraulic Barriers 681
14.6 Vibroflotation 682
14.7 Precompression 688
14.8 Sand Drains 696
14.9 Prefabricated Vertical Drains 706
14.10 Lime Stabilization 711
14.11 Cement Stabilization 714
14.12 Fly-Ash Stabilization 716
14.13 Stone Columns 717
14.14 Sand Compaction Piles 722
14.15 Dynamic Compaction 725
14.16 Jet Grouting 727
Problems 729
References 731
Appendix A 735
Answers to Selected Problems 740
Index 745
..

,
L
Soil mechanics and foundation engineering have rapidly developed during the last
fifty years. Intensive research and observation in the field and the laboratory have
refined and improved the science of foundation design. Originally published in
1984,this text on the principles of foundation engineering is now in the sixth edi
tion.The use of this text throughout the world has increased greatly over the years;
it also has been translated into several languages. New and improved materials that
have been published in varibus geotechnical engineering journals and conference
proceedings have been continuously incorporated into each edition of the text.
Principles ofFULI/.daciall Ellgilleering is intended primarily for undergraduate
civil engineering students. The first chapter, on geotechnical properties of soil, re
views the topics covered in the introductory soil mechanics course, which is a pre
requisite for the foundation engineering course. The text is comprised of fourteen
chapters with examples and problems, one appendix, and an answer section for se
lected problems. The chapters are mostly devoted to the geotechnical aspects of
foundation design. Systeme International (SI) units and English units are used in
the text.
Because the text introduces civil engineering students to the application of the
fundamental concepts of foundation analysis and design, the mathematical deriva
tions are not always presented. Instead, just the final form of the equation is given.
A list of references for further inforI]1ation and study is included at the end of each
chapter.
Example problems that will help students understand the application of vari
ous equations and graphs are given in each chapter. A number of practice problems
also are given at the end of each chapter. Answers to some of these problems are
given at the end of the text.
Following is a brief overview of the changes from the fifth edition.
• Seve�al tables were changed to graphical form to make interpolation conve
nient.
• Chapter 1 on geotechnical properties of soil has several additional empirical re
lationships for the compression index. Also included are the time factor and
xv

xvi Preface
average degree of consolidation relationships for initial pore water pressure in-
creasing and having trapezoidal. sinusoidal. and triangular shapes.
.
•
<;:hapter 2 on natural deposits and subsoil exploratton has an expanded descnp
lion on natural soil deposits. Also incorporated mto thiS chapter are several re
cently developed correlations between:
- the consistency index. standard penetration number, and unconfined com
pression strength
- relative density, standard penetration number, and overconsolidation ratio
for granular soils
- cone penetration resistance. standard penetration number, and average grain
size
- peak and ultimate friction angles hased on the results of dilatometer tests
•
Chapter 3 on shallow foundations (ultimate bearing capacity) has added sec
tions on Prakash and Saran's theory on the ultimate bearing capacity of eccen
trically loaded foundations and hea;ing capacity of foundations due to eccentric
and inclined loading.
•
Chapter 4 on ultimate bearing capacity of shallow foundations (special cases)
has new sections on the effect on ultimate bearing capacity for closely spaced
foundations, bearing capacity of foundations on a slope. and ultimate uplift ca
pacity.
•
Chapter 5 on shallow foundations (allowahle bearing capacity and settlement)
has some more elctborate tahk, for t.:lastit: �C'ttle111ent calculation. Comparisons
between predicted and measured settlements of shallow foundations on granu
lar soils also are presented.
• Chapter 7 on lateral earth pressure has a new section on a generalized relation
ship for Rankine active pressure (retaining wall with inclined back face and in
clined granular backfill). Relationships for active and passive earth pressures
under earthquake conditions are presented. Also added in this chapter is the ac
tive earth-pressure theory for translations of a retaining wall with a granular
backfill.
• Chapter 8 on re�aining walls has several comparisons of predicted and observed
lateral earth pressure on retaining walls and mechanically stabilized earth walls.
• New design charts for design of sheet-pile walls using the free earth support
method has been incorporated into Chapter 9. .
•
Chapter 12 on drilled shaft foundations has some recently published results on the
load bearing capacity based on settlement in gravelly sand.
• Chapter 13 on foundations on difficult soils has a new classification system of ex
pansive soils based on free swell ratio.
• The fundamental concepts of jet grouting have beenincorporated into Chapter 14
on soil improvement and ground modification.
Foundation analysis and design, as my colleagues in the geotechnical engi
neering area well know, is not just a matter of using theories, equations, and graphs
from a textbook. Soil profiles found in nature are seldom homogeneous, elastic,
and isotropic. The educated judgment needed to properly apply the theories,
equations, and graphs to the evaluation of soils and foundation design cannot be
.,

J .
Preface xvii
overemphasized or completely taught in the classroom. Field experience must
suppletn"Jlt classroom work.
I am grateful to my wife for her continuous help during the past twenty-five
years for the development of the original text and five subsequent revisions. Her ap
parent inexhaustible energy has been my primary source of inspiration.
Over 35 colleagues have reviewed the original text and revisions over the
years.Their comments and suggestions have been invaluable. I am truly grateful for
their critiques.
Thanks are also due to Chris Carson, General Manager, Engineering, and the
Thomson staff: Kamilah Reid-Burrell and Hilda Gowans, Developmental Editors,
Susan Calvert, Editorial Content Manager, Renate McCloy, Production Manager,
Vicki Gould, Permission Coordinator,Angela Cluer, Creative Director,and Andrew
Adams, Cover Designer, for their interest and patience during the revision and
production of the manuscript.
Braja M. Das

. ,' , ' ,
Introduction
The design of foundations of structures such as buildings, bridges, and dams gener
ally requires a knowledge of such factors as (a) the load that will be transmitted by
the superstructure to the foundation system, (b) the requirements of the local build
ing code, (c) the behavior and stress-related deformability of soils that will support
the foundation system, and (d) the geological conditions of the soil under consider
ation. To a foundation engineer, the last two factors are extremely important be
cause they concern soil mechanics.
The geotechnical properties of a soil-such as its grain-size distribution, plas
ticity, compressibility, and shear strength-can be assessed by proper laboratory
testing. In addition, recently emphasis has been placed on the in sitll determination
of strength and deformation properties of soil, because this process avoids disturb
ing samples during field exploration. However, under certain circumstances, not all
of the needed parameters can be or are determined, because of economic or other
reasons. In such cases, the engineer must make certain assumptions regarding the
properties of the soil.To assess the accuracy of soil parameters-whether they were
determined in the laboratory and the field or whether they were assumed-the en
gineer must have a good grasp of the basic principles of soil mechanics. At the same
time, he or she must realize that the natural soil deposits on which foundations are
constructed are not homogeneous in most cases. Thus, the engineer must have a
thorough understanding of the geology of the area-that is, the origin and nature of
soil stratification and also the groundwater conditions. Foundation engineering is a
clever combination of soil mechanics, engineering geology, and proper judgment de
rived from past experience. To a certain extent, it may be called an art.
When determining which foundation is the most economical, the engineer
must consider the superstructure load, the subsoil conditions, and the desired toler
able settlement. In general, foundations
'
of buildings and bridges may be divided
into two major categories: (1) shallow foundations and (2) deep foundations. Spread
footing'; wall footings, and mat foundations are all shallow foundations. In most
shallow foundations, the depth of embedmelll can be equal to or less than three to
fOllr times the width of the foundation. Pile and drilled shaft foundations are deep
foundations. They are used when top layers have poor load-bearing capacity and
1

2 Chapter 1 Geotechnical Properties of Soil
when the use of shallow foundations will cause considerable structural damage or
instability. "The problems relating to shallow foundations and mat foundations are
considered in Chapters 3, 4, 5, and 6. Chapter 11 discusses pile foundations, and
Chapter 12 examines drilled shafts.
This chapter serves primarily as a review of the basic geotechnical properties of
soils. It includes topicssuch as grain-size distribution.plasticity,soil classification,effec
tive stress,consolidation,and shear strength parameters. It is based on the assumption
that you have already been exposed to these concepts in a basic soilmechanics course.
1.2 Grain-Size Distribution
In any soil mass, the sizes of the grains vary greatly.To classify a soil properly, you
must know its grain-size distribution. The grain-size distribution of coarse-grained
soil is generally determined by means of sieve analysis. For a fine-grained soil, the
grain-size distribution can be obtained by means of hydrometer analysis. The funda
mental features of these analyses are presented in this section.Fordetailed descrip
tions, see any soil mechanics laboratory manual (e.g.. Das, 2002).
Sieve Analysis
A sieve analysis is conducted by taking a measured amount of dry. well-pulverized
soil and passing it through a stack of progressively finer sie'es with a pan at the bot
tom.The amount ofsoil retained on each sieve is measured, and the cumulative per
centage of soil passing through each is determined. This percentage is generally
referred to as percent finer. Table 1.1 contains a list of U.S. sieve numbers and the
corresponding size of their openings. These sieves are commonly used for the analy
sis of soil for classification purposes.
Table 1. 1 U.S.StandardSieve Sizes
Sieve No.
4
6
8
10
16
20
30
40
50
60
80
100
140
170
200
270
Opening (mm)
4.750
3.350
2.360
2.000
1.180
0.850
0.600
0.425
0.300
0.250
0.180
0.150
0.106
0.088
0.075
0.053
;

il .
�
·il
100
80
1.2 Grain-Size Distribution 3
� 60 +---.----
e
ii
"'40
E
u
�
u
.. 20
10 0.1
Grain size. D(mm)
Figure 1.1 Grain-size distribu-
0.01 tioncurveofa coarse-grained
soil obtained from sieve analysis
The percent finer for each sieve, determined by a sieve analysis, is plotted on
semilogarithmicgraphpaper, as shown in Figure 1.1.Note that thegrain diameter,D, is
plotted onthe logarithmic scaleandthepercent finer is plottedonthe arithmetic scale.
Two parameters can be determined from the grain-size distribution curves of
coarse-grained soils: (1)theuniformity coefficient (C,,) and(2)thecoefficientof gra
dation, or coefficient ofCIIrv(/ture (Cel. These coefficients are
(Ll)
and
(1.2)
where DlO, D30, and D60 are the diameters corresponding to percents finer than 10,
3D,and 60%, respectively.
For the grain-size distribution curve shown in Figure 1.1, DIO = 0.08 mm,
D30 = 0.17 mm,and D60 = 0.57 mm.Thus,the values ofC" andCare
and
C =
0.57
= 7 13
" 0.08
.
0.17'
C
,
=
-(0-.5":::7:'::':('-0.-08-
= 0.63

4 Chapter 1 Geotechnical Properties ofSoil
Parameters C" and C, are used in the UnifiedSoil Classification System, which is de
scribed later in the chapter.
Hydrometer Analysis
Hydrometer analysis is based on the principle of sedimentation of soil particles in
water.This test involves the use of 50 grams ofdry,pulverized soiLA deflocculating
agent is always added to the soiL The most common deflocculating agent used for
hydrometer analysis is 125 cc of 4% solution of sodium hexametaphosphate. The
soil is allowed to soak for at least 16 hours in the deflocculating agent. After the
soaking period,distilled water is added, and the soil-deflocculating agent mixtureis
thoroughly agitated. The sample is then transferred to a l000-ml glass cylinder.
More distilled water is added to the cylinder to fill it to the 1000-ml mark, and then
the mixture is again thoroughly agitated. A hydrometer is placed in the cylinder to
measure the specific gravity of the soil-water suspension in the vicinity of the in
strument's bulb (Figure 1.2), usually over a 24-hour period. Hydrometers are cali
brated to show the amount of soil that is still in suspension at any given time t. The
largest diameter of the soil particles still in suspension at time t can be determined
by Stokes' law,
where
D = diameter of the soil particle
G, = specific gravity of soil solids
'1 = viscosityofwater
TL
e 1
Figure 1.2 Hydrometeranalysis
(1.3)
.. i
i

'Y"=unit weight ofwater
L = effective length (i.e., length measured from the water surface in the cylinder to
the center ofgravity of the hydrometer;see Figure 1.2)
t = time
Soil particles having diameters larger than those calculated by Eq. (1.3) would have
settled beyond the zone of measurement. In this manner, with hydrometer readings
taken atvarioustimes,thesoilpercentfiner than agiven diameterD can be calculated
and agrain-sizedistributionplotprepared.The sieve and hydrometertechniques may
be combined for a soil having both coarse-grained and fine-grained soil constituents.
Size Limits for Soils
Several organizations have attempted to develop the size limits forgravel, sand, silt,
andclay on the basis ofthe grain sizes present in soils.Table 1.2 presents the size lim
its recommended by the American Association of State Highway and Transporta
tion Officials (AASHTO) and the Unified Soil Classification systems (Corps of
Engineers, Department of the Army, and Bureau of Reclamation).The table shows
that soilparticlessmallerthan 0.002 mm have been classified asclay. However,clays
by nature are cohesive and can be rolled into a thread when moist.This property is
caused by the presence ofclay minerals such as kaolinite, illite, and montmorillonite.
In contrast, some minerals, such as quartz andfeldspar, may be present in a soil in
particle sizes as small as clay minerals,but these particles will not havethe cohesive
property of clay minerals. Hence,they are called clay-size particles, not clayparticles.
1.4 Weight-Volume Relationships
In nature, soils are three-phase systems consisting of solid soil particles, water, and
air (or gas).To develop the weight-volume relationships for a soil, the three phases
can be separated as shown in Figure 1.3a. Based on this separation.the volume rela
tionships can then be defmed.
The voidratio, e, is the ratio of the volume ofvoids to the volume ofsoil solids
in a given soil mass,or
V,.
e=
V,
Table 1.2 Soil-Separate Size Limits
Classification system
Unified
AASHTO
Grain size (mm)
Gravel: 75 mm to4.75 mm
Sand:4.75 mm to0.075 mm
Siltand clay (fines): <0.075 mm
Gravel: 75 mm to 2 mm
Sand:2 mm to0.05 mm
Silt:0.05 mm to 0.002 mm
Clay: <0.002 mm
(1.4)

Volume
Note: V(I+ V".+ Vs= V
lV".+ Ws= W
T J
Volume �ht
T f" ,-----, t'�O
1 tV
1
Volume
Je f"1 V,,=wG.,
tV = 1
i
w ->-
1 (aj
(n) Unsaturated soil; Vs= I
(c) Saturated soil; Vs= I
Figure 1.3 Weight-volume relationships
where
Vu = volume ofvoids
V, = volume ofsoil solids
tv, w,
-1 -1
The porosity, 11, is the ratio of the volume of voids to the volume of the soil
specimen. or
where
V = total volume of soil
V,.
11:::: -
V
(1.5)

Moreover,
V,.
n = v,. = v,. = -::-:-_v."-'= =
V v,+v,. V, v,. l+e
-+-
V, v,
e
(1.6)
The degree ofsaturation, S, is the ratio of the volume of water in the void
spaces to the volume ofvoids,generally expressed as a percentage, or
where
v'v = volume ofwater
v
S(%) = � X 100
Vv
Note that, for saturated soils, the degree ofsaturation is 100%.
The weight relationships are moisture content, moist unit weight,
weight, andsaturated unit weight, often defined as follows:
where
.
(
Ww
MOIsture content = w %) =
W
X 100
,
W, = weight of the soil solids
"'", = weight of water
where
M · . .
h WOlst umt welg t = 'Y =
V
W = total weight of the soil specimen = W, + W".
The weight ofair, W.. in the soil mass is assumed to be negligible.
W
Dry unit weight = Yd = v'
(1.7)
dry unit
(1.8)
(1.9)
(1.10)
When a soil mass is completely saturated (i.e., all the void volume is occupied
by water), the moist unit weight of a soil [Eq. (1.9)] becomes equal to the saturated
unit weight (y,,, ). So y = y", if V,. = 1;,..
More useful relations can now be developed by consideringa representativesoil
specimen in which the volume ofsoil solids is equal to u/lity, as shown in Figure 1.3b.
Note that ifV, = 1, then,from Eq. (1.4),Y" = e, and the weight ofthe soil solids is
where
G, = specific gravity ofsoil solids
y". = unit weight of water (9.81 kNjm'. or 62.4 Ib/ft')

Also,from Eq. (1.8), the weight of water W,.. = wW,. Thus, for the soil specimen un
derconsideration. W.,. = wW, = wG;y.,.. Now,forthe general relation formoist unit
weight given in Eq. (1.9),
W W, + W;, G;yw(1 +w)
y=-= =
V V,+v" 1 +e
Similarly, the dry unit weight [Eq. (1.10)] is
W, W, G,yw
Yd = - = = --
V V,+Vv l+e
From Egs. (1.11) and (1.12), note that
Y
I'll =
1 + 'W
If a soil specimen is completely saturated. as shown in Figure 1.3c. then
/� :-:: t:
Also, for this case,
Thus,
e = wG, (for saturated soil only)
The saturated unit weight ofsoil then becomes
. .... . "",i:.W.. G,1., +lIYw'
:y.;,�. v.' +' V. =
l' +', . ,,' .' , -,
(1.11)
(1.12)
(1.13)
(1.14)
(1.15)
Relationships similar to Eqs. (1.11), (1.12), and (1.15) in terms of porosity can
also be obtained by considering a representative soil specimen with a unit volume.
These relationships are
and
Y = G,y".(l - n) (1+ 10)
Yd = (1 - n)G,y.,.
y,,, = [(1 - Il )G, + n]y".
(1.16)
(1.17)
(1.18)

1.5 Relative Density 9
Table 1.3 Specific Gravities ofSome Soils
Typo of Soil
Quartz sand
Silt
Clay
Chalk
Loess
Peat
G.
2.64-2.66
2.67-2.73
2.70-2.9
2.60-2.75
2.65-2.73
1.30-1.9
Exceptfor peat and highlyorganicsoils,the general range ofthe values ofspe
cific gravity of soil solids (G,) found in nature is rather small.Table 1.3 gives some
representative values. For practical purposes, a reasonable value can be assumed in
lieuofrunning a test.
Relative Density
In granular soils, the degree of compaction in the field can be measured according
to the relativedensity, defined as
where
erna, = void ratio of the soil in the loosest state
emin = void ratio in the densest state
e = in situvoid ratio
(1.19)
The relative densitycan also be expressed in terms ofdryunit weight, or
where
D (
) { Yd - Yd{m',) }Yd(m,,)
, 0/0 = -- X 100
1'd(max) - I'd(min) Yd
Yd = in situ dry unit weight
Yd(m",) = dry unit weight in the denseststate; that is,when the void ratio is em"
Yd(min) = dry unit weight in the loosest state; that is,when the void ratio is em"
(1.20)
The denseness of a granularsoil is sometimes related to the soil's relative den
sity.Table 1.4 gives a general correlation of the denseness and D,. For naturally oc
curringsands,the magnitudes ofem" and em" [Eq. (1.19)] may vary widely.The main
reasons for such wide variations are the uniformity coefficient, Cu, and the round
neSS ofthe particles.

.'
....
Table 1.4 Denseness ofa Granular Soil
Relative density, 0,(%) Description
0-20
20-40
40-60
60-80
80-100
1.1
Very loose
Loose
Medium
Dense
Very dense
A 0.25 ft3 moistsoilweighs'3().8lb.
Given G, = 2.7. Determine the
a. Moist unit weight, l'
b. Moisturecontent, W
c. Dry unitweight,I'd
d. Voidratio,e
e. Porosity,n
f. Degree ofsaturation, S
Solution . '.
Parta,.Moist UnitWeigtit
FromEq.(f.?);�· .: "� .
.
Part b: Moisture.Content .
FromEq.(1.8),
Part c: Dry UnitWeigbt
FromEq. (1.10),
, .

I
I
7.5 Relative Density 1 1
Part d: Void Ratio
From Eq. (1.4),
fl�rte; Porosity
From Eq. (1.6),
V,
e=-
V,
W, 28.2
V;
=
G,,,/w
=
(2.67)(62.4)
=
0.25 - 0.169 =
0.479e
0.169 .
0.169 ft3
e 0.479
n=
1 + e
=
1 + 0.479
= 0.324
Partf: Degree of Saturation
FromEq. (1.7),
Example 1.2
v"
S(%) = v.-x 100
v
v,. = v - v, = 0.25 - 0.169 = 0.081 ft'
V =
Woo
=
30.8 - 28.2
= 0 042 f 3w
62 4 ' t
111' .
s
=
0.042 x 100= 51 9%0.081
• •
A soil has a void ratio of0.72, moisture content = 12%, and G, = 2.72. Deter
mine the
a. Dryunit weight (kN/m3)
b. Moist unit weight (kN/m3)
Co Weight ofwater in kN/m' to be added to make the soil saturated
Solution
Part a: Dry Unit Weight
FromEq. (1.12),
_
G,,,/w _ (2.72)(9.81) _ 3
Yd - -- - 1 0 72 - 15.S1kNjm
1 + e + .

Part b: Moist Unit Weight
From Eq.(1.11),
G,'Yw(1 + IV) (2.72)(9.81)(1 + 0.12) 3
'Y= = =
17.38 kN/m
1 + e 1 + 0.72
Part c: Weight of Water to be Added
From Eq.(1.15),
(G, + elY", (2.72 + 0.72)(9.81) ,
'Y,,, = =
1 0
=
19.62 kN/m'
1 + e + .72
Waterto be added = 'Y,,, - 'Y=
19.62 - 17.38 =
2.24 kN/m3
Example 1.3
•
The maximum and minimum dry unit weights of a sand are 17.1 kN/m3 and
14.2 kN/m3, respectively.The sand in the field has a relative density of 70% with
a moisture content of 8%. Determine the moist unit weight of the sand in the
field.
Solution
From Eq.(1.20),
[ 'Yd-'Yd(mioJ ]['Yd(m�)]Dr
=
'Yd(max)-'Yd(min) ----:y:;-
0.7 =
['Yd-14.2 )[17.1]17.1 - 14.2 'Yd
'Yd=
16.11 kN/m3
'Y=
'Yd(1 + w) =
16.11(1 +
1
�) = 17.4 kN/m3
1.6 Atterberg Limits
•
When a clayeysoil is mixed with an excessive amount ofwater.it may flow like a semi
liqllid. Ifthesoil isgraduallydried.it will behave like a plastic. semisolid. orsolidmater
ial,depending onits moisture content.The moisturecontent,in percent.at whichthesoil
changes from a liquid to a plastic state is defined as the liqllid limit (LL). Similarly.the

I ,
1
1.7 Soil Classification Systems 13
Volume ofthe
soil-water
mixture
Solid
state
, , ,
I Semisolid I Plastic
,
'Semiliquid
, ,
I state I state I state ]ncrease of1-----+,---......------.--_)1 moisture content
,
,
,
,
,
,
,
,
,
,
tsL tLLhL, ,
'-__----',___--'-_____-"__
-+
Moisture
content
Figure 1.4 Definition ofAtterberg limits
moisturecontent,inpercent,atwhichthesoilchangesfrom a plasticto a semisolid state
and froma semisolidto a solid state are defined astheplastic limit (PL) andtheshrink
age limit (SL),respectively.These limits are referred to asAtterberglimits (Figure 1.4):
• The liquid limit of a soil is determined by Casagrande's liquid device (ASTM
Test Designation D-4318) and is defined as the moisture content at which a
groove closure of 12.7 mm (1/2 in.) occurs at 25 blows.
• The plastic limit is defined as the moisture content at which the soil crumbles
when rolled into a thread of3.l8 mm (1/8 in.) in diameter (ASTM Test Designa
tion D-4318).
• The shrinkage limitis defined as the moisturecontent at which the soil does not
undergo any further change in volume with loss ofmoisture (ASTM Test Desig
nation D-427).
The difference between the liquid limit and the plastic limit of a soil is defined
as theplasticityindex (PI),or
PI '" LL -:-PL
1.7 Soil Classification Systems
(1.21)
Soil classification systems divide soils into groups and subgroups based on common
engineering properties such as thegrain-size distribution, liquid limit, and plasticlimit.
The two major classification systems presently in use are (1) the AmericanAssociation
ofStateHighwayand Transportation Officials (AASHTO) System and (2) the Unified
Soil Classification System (alsoASTM).TheAASHTO system is used mainlyfor the
classificationofhighway subgrades. It is not used in foundation construction.

AASHTO System
The AASHTO Soil Classification System was originally proposed by the Highway
Research Board's Committee on Classification of Materials for Subgrades and
GranularType Roads (1945). According to the present form ofthis system,soils can
be classified according to eight major groups,A-I through A-S,based on their grain
size distribution, liquilllimit, and plasticity indices. Soils listed in groups A-I,A-2,
and A-3 are coarse-grained materials,and those in groupsA-4,A-5.A-6,andA-7 are
fine-grained materials. Peat. muck, and other highly organic soils are classified un
der A-S.They are identified by visual inspection.
TheAASHTO classification system (for soils A-I through A-7) is presented in
Table 1.5. Note that group A-7 includes two types of soil. For the A-7-5 type, the
Table 1.5 AASHTO Soil Classification System
Granular materials
General <:lassification (35'% or less oftotal sample passing No. 200 sieve)
Group classification
Sieve analysis (% passing)
No. 10 sieve
No. 40sieve
No. 200 sieve
For fraction passing
No. 40sieve
Liquid limit (LL)
Plasticity index (PI)
Usualtype ofmaterial
Subgrade rating
A-1
A-1-a A-I-b
50 max
30 max 50max
15 max 25 max
6 max
Stone fragments,
gravel, and sand
A-3 A-2
A-2-4 A-2-S A-2-6
51 min
lO max 35 max 35 max 35 max
40 max 41 min 40max
Nonplastic lO max lO max 11 min
Finesand Silty or clayey gravel and sand
Excellent to good
Silt-clay materials
General classification (More than 35% of total sample passing No. 200 sieve)
Group classification
Sieve analysis (% passing)
No. 10 sieve
No.40 sieve
No.200 sieve
For fraction passing
No.40sieve
Liquid limit (LL)
Plasticity index (PI)
Usual types ofmaterial
Subgrade rating
'UPI '" LL - 30,the classification isA-7-5.
b [fPI > LL - 30.the classification isA-7-6.
A-4
36 min
40 max
lO max
A-S
36 min
41 min
lO max
Mostly silty soils
Fairto poor
A-6
36 min
40 max
A-7
A-7-5'
A_7_6 b
36 min
41 min
ll min 11 min
Mostly clayey soils
A-2-7
35 max
41 min
11 min

Symbol G
1.7 Soil Classification Systems 1 5
plasticity index ofthe soil is less than or equal to the liquid limit minus 30. For the
A-7-6 type,the plasticity index is greater thanthe liquid limit minus 30.
For qualitative evaluation of the desirability of a soi(as a highway subgrade
material, a number referred to as the group index has also been developed. The
higher thevalueofthe group index for a givensoil,the weaker will be the soil's per
formance as a subgrade.A group index of20or moreindicates a very poor subgrade
material.The formulafor thegroup index is
where
F,oo = percent passing no. 200 sieve,expressed as a whole number
LL = liquid limit
PI = plasticity index
(1.22)
When calculating the group index for a soil belonging to groupA-2-6 or A-2-7, use
only thepartial group-index equation relating to the plasticity index:
01 = 0.01(F200 - 15) (PI - 10) (1.23)
Thegroupindexis rounded to the nearest whole numberand written next to the soil
group in parentheses;for example, we have
Unified System
A-4 (5)------ -------
I Group index
Soil group
TheUnified Soil Classification System was originally proposed byA.Casagrande in
1942 and was later revised and adopted by the United States Bureau of Reclama
tion and the U.S. Army Corps ofEngineers.The system is currently used in practi
cally all geotechnical work.
In the Unified System,thefollowingsymbols are usedfor identification:
S M C o Pt H L W P
Description Gravel Sand Silt Clay Organic silts Peat and highly High Low Wen Poorly
and clay organic soils plasticity plasticity graded graded
The plasticity chart (Figure 1.5) and Table 1.6 show the procedure for deter
mining the group symbols for various types of soil. When classifying a soil be sure

16 Chapter 1 Geotechnical Properties o fSoil
70
60
"- 50
U-line /'
,
,
,
,
,
,.
"
PI = 0.9(LL - 8) //��0
"0
40.5
;:;. . ,' 0
,tL:§ 30
�
"
0: 20
10
0
0
,
CL - �1L "
''x��-,
- - - -
,
or
/ 0
ML
or
OL
A-line
PI = 0.73 (Ll - 20)
MH
or
OH
10 20 30 40 50 60 70 80 90 tOO
Liquid limit, LL
Figure 1.5 Plasticity cha�t
to provide the group name that generally describes the soil. along with the group
symbol. Figures 1.6, 1 .7, and 1.8 give flowcharts for obtaining the group names for
coarse-grained soil. inorganic fine�grajned soil. and organic fine-grained soil,
respecti't'Iy.
Example 1.4
Classify the following soil by theAASHTO classification system.
Percent passing No, 4 sieve = 92
Percent passing No, 10 sieve = 87
Percent passing No. 40 sieve = 65
Liquid limit = 22
Plasticityindex = 8
' "
Solution " , ... ,: .
Table 1.5 shows that it is a granular material because less than 35% is passing.a
No.200sieve.With LL = 22 (thatis,less than 40) and PI = 8 (thatis,less than.lO),
the soil falls in groupA-2-4. From Eq. (1.23),
..
GI = 0.01(�oo - 15) (PI - 10) = 0.01(30 - 15) (8 - 10)
= -0.3 = 0
The soil is A-2-4(O). •

...
-.j
Table 1.6 Unified Soil Classification Chart (after ASTM.2005)
Soil classification
Group
Criteria for assigning group symbols and group names using laboratory tests· symbol Group nameb
Gravels Clean Gravels C,, � 4 and l � C,. � 3" GW WeU-graded gravel!Coarse-grained soils
More than 5{)% rClilined on
No. 200sieve
More than 50% of connit!
[melion retained un No, 4
Less Ihnn 5". finesc
GruveJs with Fines
More than 12% finesc
ell < 4 andlor 1 > C,. > 3" GP Poorly graded gravelf
Fines classify as ML or MH OM Silty gravelJ.l�.h
sieve
Fines classify as CL or CH GC Clayey gravelf.�.h
Sands Clean Sand:- C,, � 6 and J .s:; C, .s:; 3e SW Well-grnded smid'
50% or more of coarse
fraction passes No.4 !-ieve
Less than 5',�;, finesd
Sand with Fines
C,, < 6 and/or l > Cr > 3c
Fines classify as ML or MH
SP Poorly graded sand'
SM Silty sand�·I1·1
More than [2% finesd Fines classify as CL or CH SC Clayey sandl!·h"
Sills and Clays Inorganic PI > 7 and plots on or above "A" line' CL Lean c1C1yk.1.1I1Fine-grained soils
50% or more passes the
No.200 sieve
Liquid limit less than 50 PI < 4 or plots below "A" line' ML Siltk.1,m
Organic Liquid Iimit-oven dried
. . 'd !
"
d
" < 0,75
IqUl Imlt-not fie
Silts and Clays Inorganic PI plots on or above "A" line
OL
CH
Organic c1ayk,l.nJ.1l
Organic silt);· I. III. I'
Fat clayLl.m
Liquid limit 50 or more PI plots below "A" line MH Elastic siltk.l·ru
HighlVorganicsoils
aBased on the material passing the 75-mm. (3-in) sieve.
blf field sample contained cobbles or boulders, or
both. add "with cobbles or boulders. or both" to
group name.
�Gravels with 5 to 12% fines require dual symbols:
OW-OM well-graded gravel with silt; OW-OC well
graded gravel with clay; OP-OM poorly graded
gravel with silt; OP-GC poorly graded gravel with
clay.
dSands with 5 to 12% fines require dual symbols:
SW-SM weJl-graded sand with silt; SW-SC wen
graded sand with clay; SP-SM poorly graded sand
with silt; SP-SC poorly graded sand with clay.
Organic Liquid limit-oven dried
L· 'd I" d . d
< 0.75 OH
Organic :lal·I.�)
IqUi lIDlt---not rIe Organic silt;·
I. m.lt
Primarily organi..· matter.dark in color, and organic odor PT Peat
'C - D 'D c .. (
D
,,,)',, - fW II! < �. I) x DIn b!l
f
ifsoilcontains � 15% sand,add "With sand" to group
name.
S If fines classify as CL·ML. use dual symbol GC-GM
or SC·SM.
b If fines are organic, add "with organic fines" to group
name.
I Ifsoil contains �15% gravel, add "with grave)" to
group name.
ilfAtterbcrg limits pint in hatched area, soil is a
CL-ML,silty clay.
It If soil contains 15 to 29% plus No. 200, add "with
sand" or "with gravel," whichever is predominant.
IIfsoil contains ."..30% plus No.200, predominantly
sand. add "sandy" to group name.
rolf soil contains ""'30% plus No. 200. predominantly
gravel, add "gravelly" to group name.
nPI � 4 and plots on or above "A" line.
°PI < 4 or plots below"/l:' linc.
PPI plots on or above "A" line.
qPI plots below "A" line.

-
'"
Gravel
% gravel >
% sand
Sand
% sand ?:.':
% gravel
Group Symhol Group Name
.-Y" <. 15�- .�and� Well'j,!raded gravel
< 5% lines � C,' � 4 and I � C, � 3 ) (;'� ;,. 15% sand ---+ Wcl1.gralkd gravel with santi.
�C - C _ l (,I'�< 15{iL�and-» !'m,rly groldeJ gnlYd
,, < 4 �ndIor l > .(' -" .
• ��;> 15% santl ---+ Pc.)(Jrly g.radcd gmvcl wilh �aml
C,, ;;'= 4 and l � C, ';; J
'-12% """(
<fines "" MLor MH � G.G�.-: 15';1 �ml1l-Jo- Wcll·gr:ldcd gravel wilhsilt
3 15% sunil -. Wdl-gri.lJcd !ra'eI with silt ,lLId sand
lints = CL. CH ---Jo- GW-GC�< 15'J. �aml-+ Well-graded gravel with clay (()r�ilty day)
C" < 4 and/or I :> C,. > J {"
(or CL-ML) ,7' 15':{· sunil -Jo- Well.grnded gravel with clay and sand (or silty clay i.lnd Mmd)
r . ' =
Ml MH �
(,p.('
M
"".: Iy;;. �lI11J ---+ Poorly graded grJvcI wilh ):ill
Incs • Of
. , <;..�, 15';;- sand ---+ POIlTI), gra(kd grilvcl wilh silt lind sand
tinl.!s ::: CL, CH �GI" (;C�'-: lY';· �and---+ PlMlrly graded gnwcl wilh day (orsill)' clay)
(or CL-ML) ;,. 15� sand---+- Punrlygraded grJvd wilh day and .�aml (or silty dllY :lIld sand,
.> 12% line�
�n"" � ML'" Mil -+ '''1 -____• .. I," · """ -+ SHt, " ,,"
. ... ,'. IS!;! .�anJ ---+ Silty gravel wilh sand
fines = CL or CH ---+ GC �� . 15':; �aml ---+ Clayey gravel
lines := CL-ML ---+G(,.(�-�
15�Yd lant.l
---+ Clnycy gravel -ilh �and
�< I::;'� �a11tl-+ ��lIy. daycy grilvel
,
.-: S'if. lines �C,, ;;"
C,, <
c" C;'
,-,,<> "",, (
�. 15'11 �;lIId -+ Silly. clayey gravel wIth �and
� .-: 15':; gravel-+ Wcll.grudcd sand
6 and I � C � J � S - '------.. '" 15';;, gravel -+ Wc1i·gmdcd sand with p:mvcl,
� 'I' �--,.. 15':t. !;ravcl -,.. Poorly graded sand
6 and/or I :> C" > J
S '---" "" 15% gmvd -. Puurly graded�i1ml with gmvcl
< lin.::; -= MLor MH -+-SW,S�.-: 15'�· �rale l -' Well.gradcdslllld with �ih
(, ilnd I "" C, "" 3
_
' , ' ,,� l.�(:'. �rllvel -' Wl'Il'gf'ouled sand willi sill ami gravel
lincs ,- CL. Cli ---Jo-S".,'il�< 15(�, grave! -+ Well.gradtd !and w�lh day (OTSI]I}, day)
.(or CL·ML) "" 15% gmvd -+ Well.graded sand With day and gravel (ur SIlly clay and grawl)
fines = ML or MH � SP.Si'lI« 15'.
�. gr:VcI -Jo- P"�Jrly grmlcd sand w�th s!lt .
C" < 6 nnd/or I > C,. '> :3 {' :"" 15'..l"' gravel -Jo- Poorly graded saud wtth Silt and gr:lvd
fines � CL.CIt --+SI'.S( ' �< 151H- gr:lvcl -. Poorly graded sand with day (Ill' silty day)
.
> 12% """ �
(orCL·ML) :,;, 15'ii gr,wc! � Poorly graded sand with c1"y ,nulgravel (or silty day ami gravel)
tines =
MLor MH �SM-----:" � 15% gri.lvd -. Silty s.md
::>: 15'';' g.ravd -Jo- Silty sand with �ravcl
fines "" CL orCH ---+- S(' ---!.<: 15% gravel -. Clayey sand
fines = CL.ML ---Jo- S('.SM
;<': 15% gravel -+ Clayey sand wilh gr.wel
�< 15% gl"dvcl -+ Silty. clayey sand
?-< 15% gravel -Jo- Silly. clayey sam! with gravel
Figure 1.6 Flowchart for classifying coarse-grained soils (more than 50% retained on No. 200 Sieve) (After ASTM.2(05)
.' •

-
CO
LL < SO
LL ", so
Inorganic
Group Symbol
« 30% plus No. 2(X)�< 15% plus No. 200
15-29% plus No, 200�% sand � % gravel
PI > 7 and plots -----+- CL % sand < % fravel
<
% sand ill % gravcl�< 15% grave 'on or above
;., 30% plus No. 20CI ......... >- 15% gravel
"A"-line % $800 < % gravcl�< 15%sand
........ j;Io 15% sand
< 30% plus No. 21)1) �< 15% plus No. 200
4,.; PI <: 7 and
< 15·29% plus No. 200�% sand � % gravel.... "-"
----+- CL·Ml. ....... % sand < % ravel
plols on or nbove
<
% sand � % grnvel�< 15%grave
f
"A"-line � 30f�, plus No. 20U -........ ;;. 15% gravel
% sand < % gravel � < 15% sand
........ "'" 15% sand
< 30% plus Nil. 200�< 15% plus No. 200
PI < 4 I ' ML < 15·29% plus No. 200�% sand ;:z, % gravel
Of p OIS ------.. .......% sand < % ravel
below "A"-Iine <% sand ;;.% gravel � < 15% grave
f
.
;;. 30% plus Nil. 2UIl ........ ;;. 15% gravel
!
% sand < % gravel�< 15%sand
. LL-ovendried .......,.,, ;:z, 15% sand
Orgamc
LL
. < 0.75)-'01. ------..Scc tiuurc I 8
-notdncd
'" .
< 15% plw; No. 200
% sand � % gravel
< ]()" ph" No
. ""
�
15.29% plo, No. 200
"'t
�
l�l�;;'�,f"'''< % sand ;;,. % gravel
-.:q;;,. 15% gravel
_ ell
I < 15% ..nd
PI plo" 0"' 0'.
" ]0% plo, No. 2IX'
<
% ,,,,d < % <" " ""t;. 15% ",nd
above "A"-Iine
Of llnic " 1 � < 15% plusNo.200
�% sand ;a, % gra��:
10 g
'"" plo, No. _IX
"" 29% plus No, 200
....... % sand < % fro
.
< .
15·
< 15" .''''
PI plo
" belo
w -
-
<"]()% plo, No. 'IX'<""md < " .",,,1 ""t.. t5% .and
"An-line
Mil
"
< 15" .and
sand "" % gnlvel
�� 1 S% gravel
(LL uvendried
)Organic . < 0,75 -'OH --.Scc Pigure 1.8
LL-nul dried
Group Name
•lean clay
•Lean claywith sand
•Leanclay with gravel
•Sandy lean clay
•Sandy leanclaywith gravel
•Gravelly Jeanclay
•Gravelly leanclay with sand
. Silty clay
•Siltyclaywith sand
..-Siltyclay with gravel
.. Sandy silty clay
..Sandy silty clay with gravel
..Gravelly siltyclay
.. Gravelly sillyclay wilh sand
..Silt
..-Sill with clay
.. Sill withgravel
"Sandy silt
..Sandy silt with gravel
..Gravelly silt
...Gravelly silt with sand
..-Fatclay
..Fat clay with sand
..Fat clay with gravel
.. Sandy fat clay
..Sandy fatclay with gravel
•Gravelly [al clay
..GraveUy fat claywith sand
_Elastic silt
.. Elastic silt with sand
.. Elutic silt wilh grovel
Sandy elastic sill
Sandy elastic silt with gravel
Gravelly clastic sill
Gravelly elastic silt with sand
Figure 1.7 Flowchart for classifying fine-grained soil (50% or more passes No.200 Sieve) (After ASTM, 2(05)

N
C>
GroupSymbol
m<
< 30% plus No. 200
�< 15% plus No, 2()O
PI ;0. 4 and plots ........ %sand .-..:: % gravel
<
15·2
9% I1lu.� No,
2
f)(I�%s.:md � % grJ.vcI
' 00 "A" r <% sand :-' 'h.gr.VcI �< J5%gravclon or a ve - me
;;. 30% plus No. 200
--..,.. ..", l:'i'l> gravel
% sand � ,;:' gra'd �< 15% s..1.rn.I
....... :?; 15% sand
PI < 4 and plots
below "A"-lioc
; I';% plu� No. 20Cl
% sand ;:' % grave:
< ]0% pi" No. 200 "t;S.i91< p" " No. 21M'�I< ;,od < % g"'''
<. 15% gravel
< d > '�' I!I<IVcl
� -" 15Iii- graYd
% san .- �
� 15% sand
" 30'" p'"' No. 200 <'" ""od < <, ,,,,,,I�'" " .. '"""
< 15% rhl� Ntl. 2(X)
% s.md � % gravel
< 30'> p'"' No. 200 "t" .29'" " h" No. """'!:% "od < .. g"'"
Group Name
Organic day
..,....Organic clay with sand
..Organic clay with gravel
..Sandy organic clay
.. Sandy organic clay with gravel
� Gravelly organic clay
Gravelly organic clay with sand
Organic silt
Organic sill with sand
Or,gunic silt with gravel
Sandy organic sill
Sandy organic sill with gravel
Gmvelly organic silt
Gravelly organic silt with smld
..Organic clay
..Organic clay with santi
Org.anic clay with gravel
<
.
< 15% gmvel% santi ,� to;. gravel �,." 15';. gravel
P'o" 00 0'
< ". >ood
,",," "A"-O",
" 30" p'" No. 2IMl <.. "od < '� .",,' �� ,,% ,,""
.
< . O,g,o;",It
. d
< " ", p" " No. 2IMl <, "od " '> g"''''
• 0" '0;' ,H, �"lt
lf
""
< 30'" p'" No. 200 "tl 5.i9% ph" No. 200�.:"od < .. f"'''''
• S,OOy Dlg
,O!' :h,w;,h g"",1
OH
<
.
< "'" g"''''
• S
",dy o'g""" ' . ;It
.....-Sandyorganic clay
..Sandyorganic clay with grovel
.. Gravelly organic clay
.. Gravelly organic clay with santi
t Organic sill w�th san
I
Plotsbelow
< < 15% sand
., Gravelly orgamc
"
A"-O",
" ]0% pi" No. 200 % "od . <,. """ �" ,,% ,,00
% "',,' ,. <, g"'" �" " ", gr,,,1
• O.."Uy o'g"'!' :H' w;,h ,,""
Figure 1.8 Flowchart for classifying organic fine-grained soil (50% or more passes No.200 Sieve) (AfterASTM,2005)
< �

1 I
J
l .
1.8 Hydraulic Conductivity ofSoil 21
Example 1.5
Classify the following soil by the Unified Soil Classification System:
Percentpassing No. 10 sieve = 71
Percent passing No.200 sieve = 41
Liquidlimit = 31
Plasticity index = 12
· $c:illition
· we'are'giveqthat F,oo = 41, LL = 31, and PI = 12. Since 59% of the 5ample is
· retai�e!l'on a No. 200 sieve, the soil is a coarse-grained material.The percentage
Pa5$inga No.4 sieve is 82,50 18%. is retained onNo..4 sieve(gnivelfraction).The
cOanefraction passing a No. 4 sieve (sand fraction) is 59 - 18 = 41% (which is
mqre than 50% 'ofthe total coarse fraction). Hence, the specimen is a sandy soil.
Now, using Table 1.6 and Figure 1.5, we identify the group symbol of the
soihs sc.
Again: from Figure 1.6, since the gravel fraction is greater than 15%, the .
groupnameis clayey sand with gravel.
.
•
1.B Hydraulic Conductivity ofSoil
The void spaces, or pores. between soil grains allow water to flow through them. In
soH mechanicsand foundation engineering.you must knov how much water is flow
ing through a soil per unit time.This knowledge is required to design earth dams. de
termine the quantity ofseepage under hydraulicstructures. and dewater foundations
before and during their construction. Darcy {I856) proposed the following equation
(Figure 1.9) for calculating the velocity offlow ofwa�er through a soil:
v = ki
In this equation,
v = Darcyvelocity (unit: em/sec)
k = hydraulic conductivity of soil (unit: cm/sec)
i = hydraulic gradient
The hydraulic gradient is defined as
. t1h
, = -
L
where
tlh = piezometric head difference between the sections at AA and BB
L = distance between the sections atAA and BB
(Note:Sections AA and BB are perpendicular to the direction of flow.)
(1.24)
(1.25)

I
.,
,
.,
I•
,
A
. 'Dir�Clion
. ., of tlow ," '
-----...
L
B
B
Direction
of flow Figure 1.9 Definition of
Darcy's law .
Table 1.7 Range of the Hydraulic Conductivity forVarious Soils
Type of soil
Medium to coarse gravel
Coarse to fine sand
Fine sand,silty sand
Silt, clayey silt. silty clay
Clays
Hydraulic
conductivity, k
(em/sec)
Greater than 10-'!
10-1 to 10-'
10-' to 10-'
LO ' to IO "
10-7 or less
Darcy's law [Eq. (1.24)] isvalid for a wide range ofsoils. However, with mate
rials like clean gravel and open-graded rockfills,the law breaks down because ofthe
turbulent nature of flowthrough them.
The value of the hydraulic conductivity of soils varies greatly. In the labora
tory, it can be determined by means of constant-head or falling-head permeability
tests. The constant-head test is more suitable for granular soils.Table 1.7 provides
the general range for the values ofk for various soils. In granular soils, the value de
pends primarily on the void ratio. In the past,several equations have been proposed
to relate the value of k to the void ratio in granular soil.
However the author recommends the following equation for use (also see
Carrier,2003):
where
k = hydraulic conductivity
e = void ratio
e'
k oc --
1 + e
(1.26)
More recently.Chapuis(2004) proposed an empirical relationship for k in con
junction with Eq. (1.26) as
[ e' ]0.7825k(cm/s) = 2.4622 Dlo(l + e)
(1.27)
where D = effective size (mm).
� I

I '
1
1.9 Steady-State Seepage 23
Thepreceding equation is valid for natural.uniform sand andgravel topredict k
that is in the range of 10-1 to 10-3 cm/s. This can be extended to natural, silty sands
without plasticity. It is notvalid for crushedmaterialsorsiltysoilswith some plasticity.
According to their experimental observations, Samarasinghe, Huang, and
Drnevich (1982) suggested that the hydraulic conductivity ofnormally consolidated
clays could be given by the equation
en
k = C--
1 + e
where C and n are constants to be determined experimentally.
Steady-State Seepage
(1.28)
For most cases of seepage under hydraulic structures, the flow path changes direc
tion and is not uniform over the entire area. In such cases, one of the ways of deter
mining the rate ofseepage is by a graphical construction referred to as theflownet,
a concept based on Laplace's theory of continuity. According to this theory, for a
steady flow condition, the flow at any point A (Figure 1.10) can be represented by
the equation
where
a'h a'h a'h
kx-, + ky-, + k,-, = 0
ax ay az
(1.29)
k" ky, k, = hydraulic conductivity of the soil in the x, y. and z directions,
respectively
y
h = hydraulichead at pointA (i.e.,the head ofwaterthat a piezometerplaced
at A wouldshowwith the downstream water level as datum, as shown in
Figure 1.10)
Water le....el rPiezometers
.)
z
Figure 1.10 Steady-state seepage

I
.,
.'

I
24 Chapter 1 Geotechnical Properties of50;1
For a two-dimensional flow condition, as shown in Figure 1.10,
alII
- = 0jJly
so Eq. (1.29) takes the form
If the soil is isotropic with respect to hydraulic conductivity, k, = k, = k, and
(1.30)
(1.31)
Equation (1.31), which is referred to as Laplace's equation and is valid for confined
flow.represents two orthogonal sets of curves known asflolV lines and eqllipotellliai
lilies. A flow net is a combination of numerous equipotential lines and flow lines. A
flow line is a path that a water particle would follow in traveling from the upstream
side to the downstream side. An equipotential line is a line along which water, in
piezometers. would rise to the same elevation, (See Figure 1.10.)
In drawing a flownet.you need to establish the boul1dary conditions. For exam
ple, in Figure LlO. the ground surfaces on the upstream (00') and downstream
( DD') sides are equipotential lines. The hase of the dam below the ground surface,
D' BeD, is a 110 line. lbe LOp 01 the mck surface, E1'. is abo a now line. Once the
boundaryconditionsare established, a number ofnow lines and equipotential lines are
drawn by trial and error so that all the flow elements in the net have the same length
to-width ratio (LjB). In most cases, LjB is held to unity, that is, the flow elements are
drawn as curvilinear "squares." This method is illustrated by the flow net shown in
Figure 1.11.Note that all flow lines must intersect all equipotential lines at rightangles.
Once the flow net is drawn, the seepage (in unit time per unit length of the
structure) can be calculated as
Water level
: -:
Figure 1. 11 Flow net
Nr
q = khma'X-
V
11
J ,/
- - - - - - - - - - - r
hl!lI�
. .' ". '
Water level
(1.32)
/.

I
1. 10 Effective Stress 25
where
Nt = number of flow channels
Nd = number of drops
n = width-to-length ratio of the flow elements in the flow net (B/L)hm" = differencein water levelbetween theupstreamand downstreamsides
The space between two consecutive flow lines is defined as aflow channel. and the
space between two consecutive equipotential lines is called a drop. In Figure 1.11,
Nt = 2, Nd = 7, and n = 1. When square elements are drawn in a flow net,
Effective Stress
The total stress at a given point in a soil mass can be expressed as
u = u' + u
where
u = total stress
u' = effective stress
u = pore water pressure
(1.33)
(1.34)
The effective stress, u
'
, is the vertical component of forces at solid-to-solid contact
points over a unit cross-sectional area.Referringto Figure 1.12a, at pointA
u = yh, + y",h,
II = h,yw
where
Yw = unit weight of water
y,,, = saturated unit weight of soil
So
u' = (yh, + y".h,) - (h2yw)
= yh, + h,(y,,, - y",)
= yh, + y'h,
where y' = effective or submerged unit weight ofsoil.
(1.35)
For the problem in Figure 1.12a,there wasno seepage ofwater in the soil. Fig
ure 1.12b shows a simple condition in a soil profile in which there is upward seepage.
Forthis case, at point A,
and
u = (h, + h, + h)y",

T .:<>:··/., ·· : :' :. ' .:' >
th' > :':��?�;.
i
t:r��L;�L': : ,' ' ,' '. ' - t· ... . . . . :
. " :' : . ,
Saturated
Iz,
1
unit weight =
'Ysat
I-- x ---+/
(aJ
Figure 1.12 Calculation of effective stress
Thus.from Eq. (1.34).
Iz,
1
t
Saturated unit
weight =
'Ysat
A
t t tFlow ofwater
(b)
0" = cr - u = (h,y". + hzY,,,) - (h, + h, + h)y".
= h,(y", - y",) - hy", = h,y' - hy"
or
cr' = hz(y' -�',y",)= h,(y' - iyw) (1.36)
Note in Eq. (1.36) that hih, is the hydraulic gradient i. If the hydraulic gradient is
very high, so that y' - iyw becomes zero, the effective stress will become zero. In
other words, there is no contact stress between the 'soil particles, and the soil will
break up.Thissituation is referred to as the quick condition, orfailure by heave. So,
for heave.
(1.37)
wherei" = critical hydraulic gradient.
For most sandy soils,i" ranges from 0.9 to 1.1, with an average of about unity.
. . .. . . . . : ';": . ,
For the soil profile shown in Figure 1.13, determine the,total vertical stre�:pOte.
'
water pressure, and effective vertical stress at A, B, and C. ' '
Solution
The following table can now be prepared.
'.,

Point
1, 11 Consolidation 27
q' ;::;::: u - U
u(kN/m') u(kN/m') (kN/m')
o o
(4)(1'.) = (4)(14.5) = 58
58 + (1'..,)(5) = 58 + (17,2)(5) = 144
o
o 58
94.95(5) (y.) = (5) (9,81) = 49.05
A
<,:,::, " 'T':,:> ';''::', : : <, :,::, , ' , � ,> " '::
4 m Drysand"I" = 14.5 kN/m3
,
<, :,
, '..j.,:::
�at�r��ble , '
.
'
,
(> .:::,
Sm
L,. .
.. ..
.
' . ' . .,
'
'. .
...
... . .
Consolidation
.
.' . ;. .
Clay'Ysat = 17.2 kN/m3
,
.
.
.
.
..
'
. Figure 1.13 •
In the field, when the stress on a saturated clay layer is increased-for example, by
the construction of a'foundation-the pore water pressure in the clay will. increase,
Because the hydraulic conductivity ofclays is verysmall, some time will be required
fortheexcess pore water pressure to dissipate and the increase in stress to be trans
ferred to the soil skeleton, According to Figure 1.14, if do- is a surcharge at the
ground surface over a very large area, the increase in total stress at any depth of the
clay layer will be equal to du,
1 1 1 1
:::.::", �roun��atertable, .'
. '
. " ' , ' ,
� _ _ � _ _ _ _ � _ _ _ _ __ J� _ _ � _ _ _ _ � _ _ _ _ _ _ _
'
- , '
. Sand·
.Sand
Figure 1.14 Principles of consolidation
Immediatelyafterloading:timet=O

However, at timet = 0 (i.e.,immediately after the stress is applied), the excess
pore water pressure at any depth t::.u will equal CO', Or
t::.u = t::.h;y", = t::.0' (at tip1e t = 0)
Hence, the increase in effective stress at time t = 0 will be
t::.0" = t::.0' - t::.u = 0
Theoretically, at time t = x, when all the excess pore water pressure in the clay
layer has dissipated as a result of drainage into the sand layers,
t::.u = 0 (at time t = "' )
Then the increase in effective stress in the clay layer is
t::.0" = ,i0' - t::.u = t::.0' - 0 = D.O'
This gradual increase in the effective stress in the clay layer will cause settlement
over a period of time and is referred to as consolidation.
Laboratory tests on undisturbed saturated clay specimens can be conducted
(ASTMTest Designation D-2435) to determine the consolidation settlementcaused
by various incremental loadings. The test specimens are usually 63.5 mm (2.5 in.) in
diameter and 25.4 mm (1 in.) in height. Specimens are placed inside a ring, with one
porous stone at the top and one at the bottom of the specimen (Figure LISa). A
load on the specimen is then applied so that the total vertical stress is equal to 0'.
Settlement readings for the specimen are taken periodically for24 hours. After that,
the load on the specimen is doubled and more settlement readings are taken.At all
times during the test, the specimen is kept under water.The procedure is continued
until the desired limit of stress on the clay specimen is reached.
Based on the laboratory tests, a graph can be plotted showing the variation of
the void ratio e at theendofconsolidation against the corresponding vertical effective
stress (T'. (On a semilogarithmic graph, e is plotted on the arithmetic scale and (T' on
the log scale.) The nature of the variation of e against log (T' for a clay specimen is
shown in Figure 1.15b.After the desired consolidation pressure has been reached,the
specimen gradually can be unloaded,which will result in theswelling ofthe specimen.
The figure also shows the variation of the void ratio during the unloading period.
From the e-log (T' curve shown in Figure 1.15b, three parameters necessary for
calculating settlement in the field can be determined:
1. Thepreconsolidationpressure, a�. is the maximum past effective overburdenpres
sure to which the soil specimen has been subjected. It can be determined by
using a simple graphical procedure proposed by Casagrande (1936). The proce
dure involves five steps (see Figure 1.15b):
a. Determine the point 0 on the e-log (T' curve that has the sharpest curvature
(i.e.. the smallest radius of curvature).
b. Draw a horizontal line OA.
c. Draw a line OB that is tangent to the e-log 0" curve at 0.
d. Draw a line OC that bisects the angle AOB.
e. Producethe straight-line portion of the e-Iog 0" curvebackwards to intersect
Oc. This is point D. The pressure that corresponds to point D is the precon
solidation pressure (T;.

!I
I
i
il [
•
Dial
��:;]�;;::;llJ�!'-r--ttPOl'OUS stone
4----11+- Ring
;:<''-t-'-t---H-S'oilSpecimen
U.••J..t�i��iiit�iiiii �tporousstone
(a)
2.3
2.2
2.1
2.0
1.11 Consolidation 29
.g 1.9m
-:E! 1.8
�
1.7
1.6
1.5
1.4
10
Slope= C,
100
Effectivepressure.fT' (kN/m1)
(b)
400
Figure 1.15 (a) Schematic
diagram of consolidation
test arrangement:
(b) e-Iog <7
'
curve for a soft
clay from East S1. Louis.
lIIinois (Note: At the end
of consolidation. u = 0"')
Natural soil deposits can be normally consolidated or overconsolidated
(orpreconsolidaled). If the present effective overburden pressure u' = u� is
equal to the preconsolidated pressure <7; the soil is normally consolidated.
However, if u� < a�. the soil is overconsolidated,
2. The compression index, C" is the slope of the straight-line portion (the latter
part) of the loading curve, or
el - e2
c = .
, log u, - log ul
el - e2
10g(:D
(1.38)
where el and e, are the void ratios at the end of consolidation under effective
stresses "1 and if" respectively.

I
Voidratio.e
eo
0.42eo
,

Virgin------ 1 compression
Laboratory I rcurve.
consolidation I . SlopeCc
curve t �_ _ _ _ _ _ _ _ ...1 _ _
: I�I I
I I
I I
I I- - - - - - - - .,. -- , --
I I
I I
I I ,
L-----tl-..,II--_ Pressure,a
(logscale)
Figure 1.16 Construction ofvirgin
compression curve fornormally
consolidated clay
The compression index, as determined from the laboratory e-log u' curve,
will be somewhat different from that encountered in the field. The primary
reason is that the soil remolds itself to some degree during the field explo
ration.The nature of variation of the e-log u' curve in the field for a normally
consolidated clay is shown in Figure 1.16. The curve,generally referred to as
the virgin compression curve, approximately intersects the laboratory curve at
a void ratio of0.42eo (Terzaghi and Peck, 1967). Note that eo is the void ratio of
the clay in the field. Knowing the values ofeo and u;., you can easily construct
the virgin curve and calculate its compression index by using Eq. (1.38).
The value of C, can vary widely, depending on the soil. Skempton (1944)
gave an empirical correlation for the compression index in which
(1.39)
where LL= liquid limit.
Besides Skempton, several other investigators also have proposed correla
tionsforthe compression index. Some ofthose are given here:
Rendon-Herrero (1983):
C, = 0.141G;" C�
,
eo)"'8
Nagaraj and Murty (1985):
[LL(%)]C, = 0.2343
100
G
,
Park and Koumoto (2004):
C,. =
7 275371. 47 - 4. no
where no =: in situ porosity ofsoil.
(1.40)
(1.41)
(1.42)

. ,
1. 11 Consolidation 31
Worth and Wood (1978):
(PI(%»)C, = O.SG, 100
(1.43)
3. The swelling index, C" is the slope of the unloading portion of the e-Iog cr'
curve. InFigure USb, it is defined as
(1.44)
In most cases,the value of the swelling index is � to � of the compression index.
Following are some representative values ofC,/C, for natural soil deposits:
Description of soil
Boston Blue clay
Chicago clay
New Orleans clay
St. Lawrence clay
0.24-0.33
0.15-0.3
0.15-0.28
0.05-0.1
The swelling indexis also referred to as the recompression index.
The determination ofthe swelling index is important in the estimation of
consolidation settlement of overconsolidated clays. Inthe field, depending on
the pressure increase. an overconsolidated clay will follow an e-Iog cr' path
abc, as shown in Figure 1.17. Note that point a, with coordinates cr� and em
corresponds to the field conditions before any increase in pressure. Point b
corresponds to the preconsolidation pressure (cr�) of the clay. Line ab is ap
proximately parallel to the laboratory unloading curve cd (Schmertmann,
1953). Hence, if you know eo, cr�, cr;, C,. and C" you can easily construct the
field consolidation curve.
Void Sloperatio.e C1 u;
a
'0
0.42 eo
----
:'- b
Virgin] compressionJ rcurve,Laboratory SlopeC,"consolidation �
curve
,
, d
,
rSlope: Cs
�
- - --,--------- c
,
,
'---t-' _____� Pressure,cr'
(logscale)
Figure 1.17 Construction of field
consolidation curve for overconsolidated
clay

1. 12 Calculation of Primary Consolidation Settlement
The one-dimensional primary consolidation settlement (caused by an additional
load) of a clay layer (Figure 1.18) having a thickness H, may be calculated as
where
S, = primary consolidation settlement
!J.e = total change of void ratio caused by the additional load application
eo = void ratio of the clay before the application of load
For normally consolidated clay (that is, a� = a;)
O'� + o.u'
de = C, log ,
a"
where
a� = average effecti'e vertical stress on the clay layer
!J.a' = !J.a (thatis,added pressure)
Now,combining Eqs. (1.45) and (1.46) yields
For overconsolidated clay with a� + !J.a' .. a;,
a� + au'
!J.e = C,log ,
ao
(1.45)
(1.46)
(1.47)
(1.48)
. :.-, . Sand .'.,.
Averageeffectivepressurebeforeloadapplication= (J'; Figure 1.18 One-dimensional settlement
calculation
� I
'. -

1. 13 Time Rate ofConsolidation 33
Combining Eqs. (1.45) and (1.48) gives
(1.49)
For overconsolidated clay, ifu� < u� < (T� + !lu'. then
(1.50)
Now, combining Eqs.(1.45) and (1.50) yields
s·c (1.51)
TIme Rate of Consolidation
In Section 1.11 (see Figure 1.14), we showed that consolidation is the result of the
gradual dissipation ofthe excess pore water pressure from a claylayer.The dissipa
tion ofpore water pressure,in turn, increases the effective stress, whichinduces set
tlement.Hence.to estimate the degree of consolidation of a clay layer at some time
taftertheloadisapplied,youneed to knowthe rate ofdissipation ofthe excesspore
water pressure.
Figure 1.19 shows a clay layer of thickness H, that has highlY permeable sand
layers at its top and bottom.Here,the excess pore water pressure at any pointA at
any time t after the load is applied is Ilu = (Ilh)y". For a vertical drainage condi
tion (that is, in the direction of z only) from the clay layer.Terzaghi derived the dif
ferential equation
Ground
a(Il11) a2(llll)-- = C --'-...,-'-
at v OZ2
"""------�.;lf
<: = 0
(h)
(1.52)
Figure 1.19 (a) Deriva
tion ofEq. (1.54);
(b) nature ofvariation of
UIi with time

where C, =coefficient of consolidation, defined by
in which
k = hydraulicconductivity of the clay
!le=totalchange ofvoid ratio caused by an effective stress increase of t.a'eO'= averagevoid ratio during consolidation
m, =volumecoefficient ofcompressibility = !lel[fla'(1 + eO')1
(1.53)
Equation (1.52) can be solved to obtain fluas a function oftime twiththe following
boundary conditions:
1. Because highly permeable sand layers are located at z=0 and z=H" the ex
cess pore water pressure developed in the clay at those points will be immedi
ately dissipated.Hence.
and
flu =0 at Z =0
flu =0 at z=H, =2H
where H =length of maximum drainage path (due to two-way drainage
condition-that is,at the top and bottom ofthe clay).
2. At time t =0, flu= fluo= initial excess pore water pressure after the load is
applied.With the preceding boundary conditions, Eq. (1.53) yields
where
n�X[2(flUO) . (MZ)] - M'T
flu= .... sm - e •
m=O M H
M=[(2m + 1).".V2
m =an integer = 1, 2, . . .
T, =nondimensional time factor = (C,t)IH'
(1.54)
(1.55)
The value of flufor various depths (i.e., z=0 to Z =2H) at any given time t (and
thus T,) can be calculated from Eq. (1.52). The nature of this variation of !luis
shown in Figures 1.20a and b. Figure 1.20c shows the variation of !lui!luowith T,
and HIH, using Eqs.(1.54) and (1.55).
The averagedegreeofconsolidationofthe clay layer can be defined as
U = Se(l) .
Sc(max)
(1.56)

Highlypermeable(sand).
..�, . .
.....: .
0.11 at
r'>O
'(,
. !iuo =
I
constantwithdepth,
,
1.13 Time Rate ofConsolidation 35
layer(b)
UU at
r' > O
'(,
!ilia =
I
constantwithdepthI
I
o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Excessporewaterpressure. .6.1/
Initalexcessporewaterpressure,.ll1l0
(e)
Figure 1.20 Drainage condition for consolidation: (a) two-way drainage; (b) one-way
drainage; (c) plot of !1ul!1uo with T, and HIH,
where
SoU) = settlement of a clay layer at time t after the load is applied
Seem,,) = maximum consolidation settlement that the clay will undergo under a
given loading
If the initial pore water pressure (!1uo) distribution is constant with depth, as
shown in Figure 1.20a, the average degree of consolidation also can be expressed as
SeCt)U = -- =
Sc(max)
(2H (2H
1 (!1uo)dz - 1 (!1u)dz
i2H
o
(!1uo)dz
(1.57)

or
(6.uo)2H - I'll(6.u)dz
U =----:---:-c--:-"-::::-- -- =1 -(6.uo)2H
Now,combining Eqs. (1.54)and (1.58),we obtain
r'H� (!>.u)dz
2H(6.uo)
Sc(t)
U;=--=lSc(max)
m � x
( 2 ) .
- 2: -,
e-M'T,m=O M
(1.58)
(1.59)
The variation of Uwith T, can be calculated from Eq. (1.59)and is plotted in
Figure 1.21.Note that Eq. (1.59)and thus Figure 1.21are also valid when an imper
meable layer is located at the bottom of the clay layer (Figure 1.20).In that case,the
dissipation of excess pore water pressure can take place in one direction only.The
length of the maximumdrainagepathis then equal to H= HeThe variation ofT, with Ushown in Figure 1.21can also be approximated by
and
"(U%)'T
,. = "4 100 (for U=0to 60%)
T,. =1.781 -0.933log (100 - U%) (for U > 60%)
(1.60)
(1.61)
Trapezoidal Variation Figure 1.22shows a trapezoidal variation of initial excess
pore water pressure with two-waydrainage.For this case the variation of Tv with U
will be the same as shown in Figure 1.21.
1.0
0.8
...:
� 0.6
9
u
<E
"
E 0.4;:::
1+1.----- Eq (1.60) -----...·�I·--Eq (1.61) ----+1
Sand Sand
o IO 20 30 40 50 60 70 80 90
Averagedegreeofconsolidation.U (%)
Figure 1.21 Plot of time factor against average degree of consolidation (auo = constant)

z
THc=2H
�
1.0 l
0.8
0.6
.....'
0.4
0.2
0
0 10
Sand
1.13 limeRate of Consolidation 37
Figure 1.22 Trapezoidal initial excess pore water-pressure
Sand distribution
z
Sand
THc=2H
�
Sand
Sand
j.--.jllo-+l
(.) (b)
For Figure 1.23a
Figure 1.23 Sinusoidal initial
excess pore water-pressure
distribution
ForFigure 1.23b
Figure 1.24 Variation
of U with T,
sinusoidal variation of
initial excess pore
20 30 40 50
U (%)
60 70 80 90 water-pressure
distribution
Sinusoidal Variation This variation is shown in Figures 1.23a and 1.23b. For the
initial excesspore water-pressure variationshown in Figure 1.23a,
,
7TZ
CJ.u = CJ.uo sin
2H
(1.62)
Similarly, for the case shown in Figure 1.23b,
7TZ
tJ.u = CJ.uo cos
4H
The variations of T" with U for these two cases are shown in Figure 1.24.
(1.63)

Sand
Sand
Sand
' .,' ,. ...
... .".;>"'-_ 1�.7,:;'" ��Ir;-�'II-
Rock I+- Juo -+-l Sand
1 .0 +--------------------------1
0.8
0.6
;...,-
0.4
0.2
o ��_,--,_-_.--,_-_,--,_-_,--,_-�
o 10 20 30 40 50 60 70 80
u (o/c)
Figure 1.25 Variation of U with T1.-triangular initial excess pore water-pressure
distribution
90
Triangular Variation Figures 1.25 and 1.26 show several types of initial pore water
pressurevariation and the variations ofTO' with the average degree ofconsolidation.
Example 1.7
A laboratory consolidation test on a normally consolidated clay showed the fol
lowing results:
Load. Au' (kN/m2)
140
212
Void ratio at the
end of consolidation. e
0.92
0.86
The specimen tested was 25.4 mm in thickness and drained on both sides. The
time required for the specimen to reach 50% consolidation was 4.5 min.
" ,
"
,

,
l
1.13 Time Rate ofConsolidation 39
Sand I-- .:1u" -.j
. :. ,'" ," " . . '. ...
.. , "
. : '. ',
'
"
Rock
" '" '. .. . : . '
, ,
'
;
1.0
-1--------------------------1
0.8
0.6
0.4
o 10 20 30 40 50 60 70 80
U {%)
Figure 1.26 Triangular initial excess pore water-pressure distribution-variation of U
with T;,
90
A similar clay layer in the field 2.8 m thick and drained on both sides, is sub
.jected. to a simiiar increase in average effective pressure (i.e" Uo = 140 IcN/m2
and Uo + Au' = 212 kN/m2). Determine
. .
a. the expected maximum primaryconsolidation settlement in the field.
b. the length of time required for the total settlement in the fielcl to reach
40 mm. (Assume a uniform initial increase in excess pore water p�essure
with depth.)
Solution
Part a .
For normally consolidated clay [Eq. (1.38)],
C =
el - e2
=
0.92 - 0.86 = 0.333,
log (:D log
G!�)

From Eq. (1.47),
S =
C,H, log
ero + f!,er'
=c 1 + eo Uo
(0.333) (2.8) 212
1 + 0.92
log
140
= 0.0875 m = 87.5 mm
Part b
From Eq. (1.56),the average degree of consolidation is
_
� _
� _
0
u - S -
87 5
(100) - 45.7 Yo
c(max) .
The coefficient of consolidation, C," can be calculated from the laboratory
test.FromEq. (1.55),
Cvt
T,, =
H'
For 50% consolidation (Figure 1.21), T,. = 0.197. t = 4.5 min. and H = HJ2 =
12.7 mm,so
H' (0.197) (12.7)' , .
C,. = TSO -t- =
4.5
= 7.061 mm-/mm
Again,for field consolidation. U = 45.7%. From Eg. (1.60)
But
or
'Tf (U%)' 'Tf (45.7)'
T
,
= '4 100
= '4 100
= 0.164
o
64
(2.8 X 1000)'
T H, ·
1
2
I = �, =
7.061
= 45,523 min = 3L6 days
Degree of Consolidation Under Ramp Loading
•
The relationships derived forthe average degree of consolidation in Section 1.13 as
sume that the surcharge load per unit area (f!,er) is applied instantly at time I = O.
However, in most practical situations. f!,er increases gradually with time to a maxi
mum value and remains constant thereafter. Figure 1.27 shows f!,er increasing lin
early with time (t) up to a maximum at time t,. (a condition called ramp loading).

z Au
1.14 Degree of Consolidation Under Ramp Loading 41
.,..,,-:'."7-:'",-, '7 ;-l!:-: - -- .,.., --;--,- -;- - --
Load per unit
area, .do-
" ', Sand
Clay
(a)
t" Time, t Figure 1.27 OneMdimensional consolidation due tu
(b) single ramp loading
For t "" t" the magnitude of ACT remains constant. Olson (1977) considered this
phenomenon and presented the average degree ofconsolidation, U,in the following
form:
and for Tv "" T"
T. { 2 m== 1
}U = � 1 - - 2: . [l - exp{-M'Tv)]
Tc 7;" m=O M
2 m='" 1
U = 1 - - 2: -, [exp{M'Tcl - l]exp{-M'T,)
Tc m=O M
wherem, M, and Tv have the same definiti,on as in Eq. (1.54) and where
T. =
C.t,
,
H'
(1.64)
(1.65)
(1.66)
Figure 1.28 shows the variation of U with T, for various values of T"., based on
the solution given by Eqs. (1.64) and (1.65).

0.01 0.1 1.0 10
Time factor.Til
Figure1.28 Olson's ramp-loading solution:plot of Vvs. T,. (Eqs. 1.64 and 1.65)
72 kN/m'
- - - - - -,,----
tc = IS days Time, t FfgUftl 1.29 . Ramp lo.di!'j�:!:'.?�
.,

Also,
Shear Strength
1. 15 Shear Strength 43
The shear strength of a soil, defined in terms ofeffective stress,is
where
u' = effective normal stress on plane of shearing
c' ::;; cohesion, or apparent cohesion
,,>' = effective stress angle of friction
(1.67)
Equation (1.67) is referred to as the Mohr-Coulomb fai/u,e criterion. The
value of c'
for sands and normally consolidated clays is equal to zero. For overcon
solidated clays,c' > O.
For most day-to-daywork,the shearstrengthparameters ofa soil (Le.,c' and </>')
are determined by twostandard laboratory tests:the directsheartest and the triax
ial test.
Direct Shear Test
Dry sand can be conveniently tested by direct shear tests. The sand is placed in a
shearbox that is split into twohalves (Figure 1.30a}.First a normal load is applied to
the specimen.Then a shear force is applied to the top halfof the shear box to cause
failure in the sand.The normal and shear stresses atfailure are
N
u' = -
A
and
R
s ::;; -
A
where A = area of the failure plane insoil-that is, the cross-sectional area of the
shear box.

44 Chapter 1 Geotechnical Properties ofSoif
Shearstress
N
.j.
'. ' . .
: ; ' ':'", .:- '
" . ' . , .... .. . " ,
·.::r �·
' . : :
(a)
--R
s = c' + cr
'
tancpo
�------------------
s,
sl_ _ _ _ _ _ _
Effective"'---'-''-'-:---'-,---"7--'"''''''normalu� CT
J
' O"
�
.
stress.cr '
0",
(b)
Figure 1.30 Direct shear test in sand: (a) schematic diagram of test equipment;
(b) plot of test results to obtain the friction angle <p'
Table 1.B Relationship between Relative Density anti Angle of Friction of
Cohesionless Soils
State of packing
Very loose
Loosl.!
Compact
Dense
Very dense
Relative
density
('10)
<20
�n-�()
40-60
60-80
>80
Angle of
friction, c/J' (deg.)
<30
30-35
35-40
40-45
>45
Several tests of this type can be conducted by varying the normal load.The an
gle offriction ofthe sand can be determined by plotting a graph ofs against (J" (= (J'
fordry sand), as shown in Figure 1.30b, or
4>' = tan-l(.!... ). u'
(1.68)
For sands, the angle of friction usually ranges from 26' to 45', increasing with
the relative density of compaction. A general range of the friction angle, 4>', for
sands is given inTable 1.8.
Triaxial Tests
Triaxial compression tests can be conducted on sands and clays Figure 1.31a shows a
schematic diagram of the triaxial test arrangement. Essentially. the test consists of
placing a soil specimen confined by a rubber membrane into a lucite chamber and
then applying an all-around confining pressure «(J'3) to the specimen by means of
the chamber fluid (generally. water or glycerin). An added stress (f),u) can also be

Porous
stone
Rubber
nembrane
Soil
specimen
Base
plare
Lucite
chamber
Chamber
fluid Shear
stress
Porous
4>'
stone
Chamber
fluid pore waler pressure
device
c
1L_-1�J..,__---_'-;---__-'-_
Effective
nonnal
U3 0"] uj
stress
Shear
stress
Shematic diagram of triaxial
test equipment
(aJ
Tolal stress
failure
envelope
Shear
stress
Consolidated-drained test
(b)
Effective
stress
failure
envelope
{ q,
Total
L__J___J_....J'-____..L_ nonnal
c
1 Effective
LJc----L------L------__-L;o nonnal
UJ
lure 1.31 Triaxial test
Shear
stress
T
stress, a
Consolidated-undrained [est
(c)
Total stress
failure envelope
(4) =0)
U3 uj u; u; stress, a'
s = ell
l L-L-_-'----'--_�
Unconsolidated-undrained test
(d)
Nonnal stress
(lOlal), u
45

I
.•.
.
. .....�
.. .. ... .'... :.
.
: ': � .-
....
.: :" ',
' :, .
: "
, > •
. ' "
', -.: .;,.,
.
: '.- ,
' " .
Figure 1.32 Sequence ofstress
application in triaxial test
applied to the specimen in the axial direction to cause failure (t>u = t>uf atfailure).
Drainage from the specimen can be allowed or stopped,depending on the condition
being tested. For clays, three main types of tests can be conducted with triaxial
equipment (seeFigure 1.32):
1. Consolidated-drained test (CD test)
2. Consolidated-undrained test (CU test)
3. Unconsolidated-undrained test (UU test)
Consolidated-Drained Tests:
Step 1. Apply chamber pressure U3' Allow complete drainage,sothat thepore
waterpressure (u = uo) developed is zero.
Step 2. Apply a deviatorstress t>u slowly.Allow drainage,so that the porewater
pressure (u = Ud) developed through the application of t>u is zero. At
failure, t>u = t>uf; the total pore water pressure uf = Uo + Ud = O.
So forconsolidated-drained tests, at failure,
Major principal effective stress = u3 + t>uf = u, = ui
Minor principal effective stress = U3 = uo
Changing U3 allows several tests of this type to be conducted on various clay speci
mens.The shear strength parameters (c' and "" ) can now be determined by plotting
Mohr'scircle at failure,as shown in Figure 1.31b, and drawing a common tangent to
the Mohr's circles. This is the Mohr-Coulomb failure envelope. (Note: For normally
consolidated clay,c' = 0.) At failure,
(1.69)
Consolidated-Undrained Tests:
Step 1. Apply chamber pressure U3' Allow complete drainage,so that the pore
water pressure (u = uo) developed is zero.
-1.
I

1.15 ShearStrength 47
Step 2. Apply a deviator stress t:..u. Do not allow drainage, so that the pore
water pressure U = Ud * O. At failure, t:..u = t:..uf; the pore water pres
sure uf = Uo + Ud = 0 + ud(fl'
Hence, at failure,
Major principal total stress = UJ + t:..uf = u,
Minor principal total stress = UJ
Major principal effective stress = ('<3 + t:..Uf) - Uf = ul
Minor principal effective stress = UJ - uf = uj
Changing U3 permits multiple tests of this type to be conducted on several soil
specimens.The total stress Mohr's circles at failure can now be plotted,as shown in
Figure 1.31c, and then a common tangent can be drawn to define thefailure enve
Jape. This totaJstressfailure envelope is defined by the equation
s = c + u tan </.> (1.70)
where c and </.> are the consolidated-undrained cohesion and angleoffriction, respec
tively. (Note: c = 0 for normally consolidated clays.)
Similarly, effective stress Mohr's circles at failure can be drawn to determine
the effective.stress failure envelope (Figure 1.31c), which satisfy the relation ex
pressed in Eq. (1.67).
Unconsolidated-Undrained Tests:
Step I. Apply chamber pressure u,. Do not allow drainage.so thatthe pore water
pressure (u = uo) developed through the application ofUJis not zero.
Step 2. Apply a deviator stress t:..u. Do not allow drainage (u = Ud * 0). At
failure, t:..u = t:..uf; the pore water pressure uf = Uo + u,/(I)
For unconsolidated-undrained triaxial tests,
Major principal total stress = UJ + t:..uf = u,
Minor principal total stress = U3
The total stress Mohr's circle at failure can now be drawn, as shown in
Figure 1.31d. For saturated clays, the value ofu, - UJ = t:..uf is a constant, irrespec
tive ofthe chamberconfiningpressure U3 (also shown in Figure 1.31d).The tangent to
these Mohr's circles will be a horizontal line, called the </.> = 0 condition. The shear
strength forthisconditionis
(1.71)
where c, = undrained cohesion (or undrained shear strength).
The pore pressure developed in the soil specimen during the unconsolidated
undrained triaxial test is
(1.72)

I
.,
I
1. 16
Theporepressure110 is the contribution of the hydrostatic chamberpressure eT,. Hence,
(1.73)
where B = Skempton's pore pressure parameter.
Similarly, the pore parameter lid is the result of the added axial stress !:leT, so
Ud = AdO"
where A = Skempton's pore pressure parameter.
However,
Combining Eqs. (1.72). (1.73). (1.74). and (1.75) gives
II = 110 + lid = BeT, + A(eTJ - eTJ)
(1.74)
(1.75)
(1.76)
The pore water pressure parameter B in soft saturated soils is approximately 1, so
(1.77)
The value of the pore water pressure parameter A at failure will vary with the type
of soil. Following is a general range of the values ofA at failure for various types of
clayey soil encountered in nature:
Type of soil
Sandy clays
Norman), consolidated clays
Overconsolidated clays
Unconfined Compression Test
A at failure
0.5-0.7
0.5-1
-0.5-0
The unconfined compression test (Figure 1.33a) is a special type of unconsolidated
undrained triaxial test in which the confining pressure eT3 = 0, as shown in Fig
ure 1.33b. In this test, an axial stress !:leT is applied to the specimen to cause failure
(i.e., !:leT = !:leTf). The corresponding Mohr's circle is shown in Figure 1.33b. Note
that, for this case,
Major principal total stress = !:leTf = q"
Minor principal total stress = °
The axial stress at failure, /leTr = qu, is generally referred to as the unconfined
compression strength. The shear strength of saturated clays under this condition
(q, = 0). from Eq. (1.67), is
s = c =
qu
" 2
(1.78)

1. 17
/;.u
<a)
Shear
stress
Tc"
1
003 = 0 u! = UU't
= qu
(b)
1. 17 Comments on Friction Angle, </>' 49
Specimen
Total
nennal
stress
Unconfined
compression
strength. qu
Degree
�---------------+ of
saturation
(e)
Figure 1.33 Unconfined compression test: (a) soil specimen; (b) Mohr's circle for the
test:(c) variation of qu with the deg.ree of saturation
The unconfined compression strength can be used as an indicator of the consistency
ofclays.
Unconfined compression tests are sometimes conducted on unsaturated soils.
With the void ratio ofa soil specimen remaining constant,the unconfined compression
strength rapidly decreases with the degree ofsaturation (Figure 1.33c).
Comments on Friction Angle, 4>'
Effective Stress Friction Angle of GranularSoils
In general, the direct shear test yields a higher angle of friction compared with that
obtained by the triaxial test. Also, note that the failure envelope for a given soil is
actually curved. The Mohr-Coulomb failure criterion defined by Eg. (1.67) is only
an approximation. Because ofthecurved,nature ofthe failure envelope, a soil tested
at highernormal stress willyield a lowervalue of4>'. Anexample ofthis relationship
is shown in Figure 1.34, which is a plot of 4>' versus the void ratio e for Chatta
choochee Riversand nearAtlanta, Georgia (Vesic, 1963).The friction angles shown
were obtained from triaxial tests, Note that, for a given value ofe, the magnitude of
4>' is about 4° to 5° smaller when the confining pressure u; is greater than about
70 kN/m' (10 lb/in'), compared with that when u; < 70 kN/m'(� 10 lb/in').

45
6 samples
� �
- elan4>'= 0.68 [0',< 70kN/m-(10 Ib/in')]3
'" 40 8
J
0;,
=
=
etan<1>' = 0.59 [70kN/m'(10 Ib/in')< u;< 550kN/m:!(80 Ib/in2)J30 +---�'-----�-----r--�-r�---r--�-'
0.6 0.7 0.8 0.9
Voidrario.I!
1.0 l.l 1.2
Figure 1.34 Variation of friction angle rb' "ilh void ratio for ChaH8choochee Riversand
(AfterVesie, 1963)
1.0
0.8
'"
0.6
=
in 0.4
0.2
0
5 10
o Kenney(1959)
• BjerrumandSimons(1960)
20 30 50 80 100 150
Plasticityindex(%)
Figure 1.35 Variation of sin 4>' with plasticity index (PI) for several normally
consolidated clays
Effective Stress Friction Angle ofCohesive Soils
FIgure 1.35shows the variation ofeffective stress friction angle,.p',forseveralnormally
consolidatedclays (Bejerrum and Simons, 1960;Kenney, 1959).It can be seen from the
figure that,in general. the friction angle .p' decreases with the increase in plasticityin
dex.Thevalue ofq,' generallydecreasesfrom about 37to 38'with aplasticity index of
about 10to about 25'or less with a plasticity index of about 100.The consolidated
';

1. 17 Comments on Friction Angle, </>' 51
Deviator
stress.Llu
(13 ::: (1) ::: constant
Figure 1.36 Plot ofdeviator stress versus
'----------+ Axial strain. E axial strain-drained triaxial test
Shear stress, T
T
c'
1 </>',
""'='--__:L_..L_______� Effective normalstress, (7"
Figure 1.37 Peak- and residual-strength envelopes for clay
undrainedfriction angle (q,) ofnormallyconsolidatedsaturatedclaysgenerallyranges
from 5 to 20°.
The consolidated drained triaxial test was described in Section 1.15.Figure 136
showsa schematicdiagramofa plot of t;(T versusaxial strain in a drained triaxial test
for a clay. At failure,for this test, t;(T = t;(Tf' However, at large axial strain (Le., the
ultimate strength condition),we have the following relationships:
Major principal stress:0-'1(011) = (T3 + t;(Toil
Minor principal stress: (T3(oit) = (T3
At failure (i.e., peak strength), the relationship between (Tl and (T3 is given by
Eq. (1.69). However,for ultimate strength, it can be shown that
(1.79)
where q,� = residual effective stress friction angle.
Figure 1.37 shows the general nature of the failure envelopes at peak strength
and ultimate strength (or residual strength).The residual shear strength of clays is
important in the evaluation of the long-term stability of new and existing slopes
and the design ofremedial measures.The effective stress residual friction angles q,;

40
Sand
-- -- .......
o ,
,
-- ......
,
,

0 " ,
, ,
, ,
, '
'.9 0.......
Skempton (1985)
,,'- = 1
PII
Plasticity index. PI ... 0.5 to 0.9
Clay fraction. CF
, -
'.......--Q. ��-OOO----O--...
...
-
- - -�-
- - _Q _ - - - -
•
K:lolin
•
Bentonite
O +---�---r--------'--------'-------'--------;
o 20 40 60 80 100
Clay fraction, CF (o/c I
Figure 1.38 Variation of c/J;, with CF (Note: Ptl = atmospheric pressure)
ofclaysmay be substantially smaller than the effectivestress peak friction angle ",.
Past research has shown that the clay fraction (i.e.,the percent finer than 2 microns)
presentin a given soil, CF, and the clay mineralogy are the two primaryfactors that
control 4>;.The following is a summary of the effects of CF on 4>;.
1. If CF is less than about 15%, then 4>; is greater than about 25'.
2. ForCF > about50%,4>; is entirely governed by the slidingofclayminerals and
may be in the range ofabout 10 to 15'.
3. For kaolinite,illite, and montmorillonite,</>; is about 15°, 10°, and 5', respectively.
Illustrating these facts,Figure 1.38 shows the variation of4>; with CFfor several soils
(Skempton, 1985).
Correlations for Undrained Shear Strength, Cu
The undrained shearstrength,c," is an important parameter in the design offounda
tions. For normally consolidated clay deposits (Figure 1.39), the magnitude ofCu in
creases almost linearly with the increase in effective overburden pressure.
For normally consolidated clays, Skempton (1957) has given the following cor
relation for the undrained shear strength:
C,,(VST)
= 0.11 + 0.OO37PI
(To
(1.80)

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