Introduction

The design of foundations of structures such as buildings, bridges, and dams gener­ally requires a knowledge of such factors as (a) the load that will be transmitted bythe superstructure to the foundation system, (b) the requirements of the local build­ing code, (c) the behavior and stress-related deformability of soils that will supportthe foundation system, and (d) the geological conditions of the soil under consider­ation. To a foundation engineer, the last two factors are extremely important be­cause they concern soil mechanics.

The geotechnical properties of a soil-such as its grain-size distribution, plas­ticity, compressibility, and shear strength-can be assessed by proper laboratorytesting. In addition, recently emphasis has been placed on the in situ determinationof strength and deformation properties of soil, because this process avoids disturb­ing samples during field exploration. However, under certain circumstances, not allof the needed parameters can be or are determined, because of economic or otherreasons. In such cases, the engineer must make certain assumptions regarding theproperties of the soil. To assess the accuracy of soil parameters-whether they weredetermined in the laboratory and the field or whether they were assumed-the en­gineer must have a good grasp of the basic principles of soil mechanics. At the sametime, he or she must realize that the natural soil deposits on which foundations areconstructed are not homogeneous in most cases. Thus, the engineer must have athorough understanding of the geology of the area-that is, the origin and nature ofsoil stratification and also the groundwater conditions. Foundation engineering is aclever combination of soil mechanics, engineering geology, and proper judgment de­rived from past experience. To a certain extent, it may be called an art.

When determining which foundation is the most economical, the engineermust consider the superstructure load, the subsoil conditions, and the desired toler­able settlement. In general, foundations of buildings and bridges may be dividedinto two major categories: (1) shallow foundations and (2) deep foundations. Spreadfootings, wall footings, and mat foundations are all shallow foundations. In mostshallow, foundations, the depth of embedment can be equal to or less than three tofour times the width of the foundation. Pile and drilled shaft foundations are deepfoundations. 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 orinstability. The problems relating to shallow foundations and mat foundatiollS areconsidered in Ch.apters 3, 4, 5, and 6. Chapter 11 discusses pile foundations,Chapter 12 examines drilled shafts.

This chapter serves primarily as a review of the basic geotechnical properties ofsoils. It includes topics such as grain-size distribution, plasticity, soil classification, effec~tive stress, consolidation, and shear strength parameters. It is based on the assumptionthat you have already been exposed to these concepts in a basic soil mechanics course~

Grain-Size Distribution

In any soil mass, the sizes of the grains vary greatly. To classify a soil properly,nlust know its grain-size distribution. The grain-size distribution of coarse-grableasoil is generally determined by means of sieve analysis. For a fine-grained soil., thegrain-size distribution can be obtained by means of hydrometer analysis. The funda­mental features of these analyses are presented in this section. For detailed descrip­tions, 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-pulverizedsoil and passing it through a stack of progressively finer sieves with a pan at the bt)t­

tom. The amount of soil retained on each sieve is measured, and the cumulativecentage of soil passing through each is determined. This percentage is generallyreferred to as percent finer. Table 1.1 contains a list of U.S. sieve numbers and thecorresponding size of their openings. These sieves are commonly used for the analy­sis of soil for classification purposes.

Table 1.1 U.S. Standard Sieve Sizes

Sieve no.

468

1016203040506080

100140170200270

Opening (mm)

4.7503.3502.3602.0001.1800.8500.6000.4250.3000.2500.1800.1500.1060.0880.0750.053

I

I

1.2 Grain-Size Distribution 3

Figure 1. 1 Grain-size dis­tribution curve of a coarse­

0.01 grained soil obtained fromsieve analysis

1 0_1

Grain size, D (mm)

-..~~~

1\

~

\\~

~

:~\11I ~

I ,: \.

; i\,I I \I I "I I "II I ,~,

To10

100

80,-...4-J..c:::~

-a3~ 60~

e~v~

t.H 404-J~VU~v

0-1 20

The percent finer for each sieve, determined by a sieve analysis, is plotted onsemilogarithmic graph paper, as shown in Figure 1.1. Note that the grain diameter, D, isplotted on the logarithmic scale and the percent finer is plotted on the arithmetic scale.

Two parameters can be determined from the grain-size distribution curves ofcoarse-grained soils: (1) the uniformity coefficient (Cu ) and (2) the coefficient ofgra­dation, or coefficient ofcurvature (Cc ). These coefficients are

(1.1)

and

(1.2)

where DIO' D30, and D60 are the diameters corresponding to percents finer tllan 10,30, and 600/0, respectively

For the grain-size distribution curve shown in Figure 1.1, D10 == 0.08 mill,D30 == 0.17 mID, and D60 == 0.57 mm. Thus, the values of Cu and Cc are

0.57Cu = 0.08 = 7.13

ane}

0.172

C == == 0.63c (0.57) (0.08)

4 Chapter 1 Geotechnical Properties of Soil

Parameters Cu and Cc are used in the Unified Soil Classification System, IS

scribed later in the chapter.

Hydrometer Analysis

Hydrometer analysis is based on the principle of sedimentation of soil particles inwater. This test involves the use of 50 grams of dry, pulverized soil. Aagent is always added to the soil. The most common deflocculating agent used forhydrometer analysis is 125 cc of 4% solution of sodium hexametaphosphate. Thesoil is allowed to soak for at least 16 hours in the deflocculating agent. After thesoaking period, distilled water is added, and the soil-deflocculating agent isthoroughly agitated. The sample is then transferred to a 1000-ml glassMore distilled water is added to the cylinder to fill it to the 1000-ml mark,the mixture is again thoroughly agitated. A hydrometer is placed in themeasure the specific gravity of the soil-water suspension in .the vicinity ofstrument's bulb (Figure 1.2), usually over a 24-hour period. Hydrometers arebrated to show the amount of soil that is still in suspension at any given time t.largest diameter of the soil particles still in suspension at time t can be detern1illedby Stokes' law,

(1.3)

where D == diameter of the soil particleGs == specific gravity of soil solids

TJ == viscosity of water

I

'··1.··.···'·

c'

II

\

I1

Ijj~~ 3~~~' I

I

I

I

IFigure 1.2 Hydrometer analysis ·:I·~·..

~L

1.4 Weight-Volume Relationships 5

'Yw == unit weight of waterL == effective length (i.e., length measured from the \vater su.rface in the

cylinder to the center of gravity of the hydrometer; see Figure 1.2)t == time

Soil particles having diameters larger than those calculated by Eq. (1.3) would havesettled beyond the zone of measurement. In this manner, with hydrometer readingstaken at various times, the soil percent finer than a given diameter D can be calculatedand a grain-size distribution plot prepared. The sieve and hydrometer techniques maybe combined for a soil having both coarse-grained and fine-graine(j soil constituents.

Size Limits for Soils

Several organizations have attempted to develop the size limits for gravel, sand, silt,and clay on the basis of the grain sizes present in soils. Table 1.2 presents the size lim­its recommended by the American Association of State Highwa)T and Transporta­tion Officials (AASHTO) and the Unified Soil Classification s~{stems (Corps ofEngineers, Departnlent of the Army, and Bureau of Reclamation). The table showsthat soil particles smaller than 0.002 mm have been classified as cl~y. However, claysby nature are cohesive and can be rolled into a thread when moist. This property iscaused by the presence of clay minerals SUCll as kaolinite, illite, and montmorillonite.In contrast, some minerals, such as quartz and feldspar, may be present in a soil inparticle sizes as small as clay minerals, but these particles will not llave the cohesiveproperty of clay minerals. Hence, they are called clay-size particles, not clay particles.

Weight-Volume Relationships

In nature, soils are three-phase systems consisting of solid soil particles, water, andair (or gas). To develop the weight-volume relationships for a soil, the three phasescan be separated as shown in Figure 1.3a. Based on this separation, the volume rela­tionships can then be defined.

The void ratio, e, is the ratio of the volume of voids to the volume of soil solidsin a given soil mass, or

(1.4)

Table 1.2 Soil-Separate Size Limits

Classification system

Unified

AASHTO

Grain size (mm)

Gravel: 75 mm to 4.75 mmSand: 4.75 mm to 0.075 mmSilt and clay (fines): <0.075 mmGravel: 75 mm to 2 mmSand: 2 mm to 0.05 mmSilt: 0.05 nun to 0.002 mmClay: <0.002 mm

6 Chapter 1 Geotechnical Properties of Soil

Note: Va + Vw + Vs = VW w + W s = W

Volume

v

Weight

w

Volume

(a)

WeightT

W =0

tWfWG;YW

W s = G/yw

~

Volume

T TVv = e t1 vw=wGs

tV = 1

~(b) Unsaturated soil; Vs = 1

Volume

(c) Saturated soil; V s = 1

Fig.'.Ire 1.3 Weight-volume relationships

Note:Vw = wGs = Se

where Vv == volume of voidsv:. == volume of soil solids

The porosity, n, is the ratio of the volull1e of voids to the volume of the soilspeCImen, or

(1.5)

1.4 Weight-Volume Relationships 7

where V == total volume of soil

Moreover,

e

1 + e(1.6)

The degree of saturation, 5, is the ratio of the volume of water in the voidspaces to the volume of voids, generally expressed as a percentage, or

(1.7)

(1.8)

where Vw == volume of water

Note that, for saturated soils, the degree of saturation is 1000/0.The weight relationships are moisture content, moist unit weight, dry unit

weight, and saturated unit weight, often defined as follows:

WwMoisture content = w(%) = W X 100

s

where Ws == weight of the soil solidsWw == weight of water

. . . h WMOist umt welg t = Y = V

where W == total weight of the soil specimen == Ws + Ww

The weight of air, Wa, in the soil mass is assumed to be negligible.

WDry unit weight = Yd = V

S

(1.9)

(1.10)

When a soil mass is completely saturated (i.e., all the void volume is occupiedby water), the moist unit weight of a soil [Eq. (1.9)] becomes equal to the saturatedunit weight (Ysat). So Y == Ysat if Vv == Vw·

More useful relations can now be developed by considering a representative soilspecimen in which the volume of soil solids is equal to unity, as shown in Figure 1.3b.Note that if v:, == 1, then, from Eq. (1.4), Vv == e, and the weight of the soil solids is

where Gs = specific gravity of soil solids'Yw == llnit weight of water (9.81 kN/m3

, or 62.4 Ib/ft3)

Also, froin Eq. (1.8), the weight of water Ww == wWs. Thus, for the soil specimen un­der consideration, Ww == wWs == wGsYw. Now, for the general relation for moist unitweigllt given in Eq. (1.9),

8 Chapter 1 Geotechnical Properties of Soil

(1.11)

Similarly, the dry unit weight [Eq. (1.10)] is

Relationships similar to Eqs. (1.11), (1.12), and (1.15) in terms of porosity canalso be obtained by considering a representative soil specimen with a unit volume.These relationships are

~I

I

iIf

t

(

~

II,

I',l'

f

'if

IIIj

Ij\

Ii

I~

II

i

I

1

(1.12)

(1.13)

(1.14)

(1.15)

(1.16)

(1.17)

(1.18)

(for saturated soil oniy)

'Ysat == [(1 - n )Gs + nJ'Yw

'Y == Gs'Yw(1 - n) (1 + w)

'Yd == (1 - n)Gs'Yw

Ww wGs'Ywl( == - == == wG

v 'Yw 'Yw s

'Yd == 1 + w

From Eqs. (1.11) and (1.12), note that

Also, for this case,

If a soil specimen is completely saturated, as shown in Figure 1.3c, then

Vv == e

Thus,

The saturated unit weight of soil then becomes

and

Ex~ept for peat and highly organic soils, the general range of the values of spe­cific gravity of soil solids (Gs ) found in nature is rather small. Table 1.3 gives somerepresentative values. For practical purposes, a reasonable value can be assumed inlieu of running a test.

1.5 Relative Density 9

Table 1.3 Specific Gravities of Some Soils

Tvpe of Soil

Quartz sandSiltClayChalkLoessPeat

2.64-2.662.67-2.732.70-2.92.60-2.752.65-2.731.30-1.9

Table 1.4 Typical Void Ratio, Moisture Content, and Dry Unit Weight for Some Soils

Type of soilVoid ratioe

Natural moisturecontent insaturated condition(%)

Dry unit weight, 1'd

Loose uniform sandDense uniform sandLoose angular-grained silty sandDense angular-grained silty sandStiff claySoft clayLoessSoft organic clayGlacial till

0.80.450.650.40.60.9-1.40.92.5-3.20.3

301625152130-502590-12010

14.51816191711.5-14.513.56-821

9211510212010873-928638-51

134

Table 1.4 presents some representative values for the void ratio, dry unit weight,and moisture content (in a saturated state) of some naturally occurring soils. Note thatin most cohesionless soils the void ratio varies from about 0.4 to 0.8.The dry unit weightsin these soils generally fall within a range of about 14-19 kN/m3 (90-120 Ib/ft3).

Relative Density

In granular soils, the degree of compaction in the field can be measured accordingto the relative density, defined as

(1.19)

where ema~ == void ratio of the soil in the loosest stateenlin == void ratio in the densest state

e == in situ void ratio

The values of emax are determined in the laboratory in accordance with the testprocedures outlined in the American Society for Testing and Materials, ASTM Stan­dards (2000, Test Designation D-4254).

10 Chapter 1 Geotechnical Properties of Soil

Table 1.5 Denseness of a Granular Soil

Relative density, D,(%) Description

The relative density can also be expressed in terms of dry unit weight, or

where Yd == in situ dry unit weight'Yd(max) == dry unit weight in the densest state; that is, when the void ratio is emin

'Yd(min) == dry unit weight in the loosest state, that is, when the void ratio is emax

The denseness of a granular soil is sometimes related to the soil's relative den­sity. Table 1.5 gives a general correlation of the denseness and Dr. For naturally oc­curring sands, the magnitudes of emax and emin [Eq. (1.19)] may vary widely. The mainreasons for such wide variations are the uniformity coefficient, Cu , and the round­ness of the particles.

I

I

I

I

I

(1.20)

Very looseLooseMediumDenseVery dense

()/ ) {'Yd - 'Yd(min) }'Yd(maX)Dr C}o == X 100

'Yd(max) - 'Yd(min) Yd

0-2020-4040-6060-8080-100

1.5 Relative Density 11

12 Chapter 1 Geotechnical Properties of Soil

Atterberg Limits

When a clayey soil is mixed with an excessive amount of water, it may flow like a semi­liquid. If the soil is gradually dried, it will behave like a plastic, semisolid, or solid mater­ial, depending on its moisture content. The moisture content, in percent, at which thechanges from a liquid to a plastic state is defined as the liquid limit (LL). Similarly, themoisture content, in percent, at which the soil changes from a plastic to a semisolid stateand from a semisolid to a solid state are defined as the plastic limit (PL) and the shrink­age limit (SL), respectively. These limits are referred to as Atterberg limits (Figure 1.4):

• The liquid limit of a soil is determined by Casagrande's liquid device (AsrrMTest Designation D-4318) and is defined as the moisture content at Wflicilgroove 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 crumbleswhen rolled into a thread of 3.18 mm (1/8 in.) in diameter (ASTMTest Designa­tion D-4318).

• The shrinkage limit is defined as the moisture content at which the soil does notundergo any further change in volume with loss of moisture (ASTM Test Desig~nation D-427).

The difference t~etween the liquid limit and the plastic limit of a soil is definedas the plasticity index (PI), or

(1.21)

Table 1.6 gives some representative values of the liquid limit and the plasticlimit for several clay lninerals and soils. However, Atterberg limits for different soilswill vary considerably, depending on the origin of the soil and the nature andamount of clay minerals in it.

Moisturecontent

Solidstate

, f I

, Semisolid 1 Plastic t SemiliquidI I IJ state 1 state I state Increase of

-----+-1----+-1-------+I---t..~ moisture content

I I 1I I I

, I 1I I I

I t II I I

I 1I I

I II

1 II

~ SL ~ PL

1 I

Volume of thesoil-water

mixture

Figure 1.4 Definition of Atterberg limits

1.7 Soil Classification Systems 13

Table 1.6 Typical Liquid and Plastic Limits for Some Clay Minerals and Soils

Description

KaoliniteIlliteMontmorilloniteBoston Blue clayChicago clayLouisiana clayLondon clayCambridge clayMontana clayMississippi gun1boLoessial soils in north and northwest China

Soil Classification Systems

Liquid limit

35-10050-100

100-8004060756639529525-35

Plastic limit

25-3530-6050-1002020252721183215-20

Soil classification systems divide soils into groups and subgroups based on commonengineering properties such as the grain-size distribution, liquid limit, and plastic limit.The two major classification systems presently in use are (1) the American AssociationofState Highway and Transportation Officials (AASHTO) System and (2) the UnifiedSoil Classification System (also ASTM). The AASHTO system is used mainly for theclassification of highway subgrades. It is not used in foundation construction.

AASHTO System

The AASHTOSoil Classification System was originally proposed by the HighwayResearch Board's Committee on Classification of Materials for Subgrades andGranular Type Roads (1945). According to the present form of this systenl, soils canbe classified according to eight major groups,A-l throughA-8, based on their grain­size distribution, liquid limit, and plasticity indices. Soils listed in groups A-I, A-2,andA-3 are coarse-grained materials, and those in groupsA-4,A-5,A-6, andA-7 arefine-grained materials. Peat, muck, and other highly organic soils are classified un­der A-8. They are identified by visual inspection.

TheAASHTO classification system (for soils A-I throughA-7) is presented inTable 1.7. Note that group A-7 includes two types of soil. For the A-7-5 type, theplasticity index of the soil is less than or equal to the liquid limit minus 30. For the A­7-6 type, the plasticity index is greater than the liquid limit nlinus 30.

For qualitative evaluation of the desirability of a soil as a highway subgradematerial, a number referred to as the group index has also been developed. Thehigher the value of the group index for a given soil, the weaker will be the soil's per­formance as' a subgrade. A group index of 20 or more indicates a very poor subgradematerial. The formula for the group index is

(1.22)

14 Chapter 1 Geotechnical Properties of Soil

Table 1.7 AASHTO Soil Classification System

Granular materials(35°k or less of total sample passing no. 200 sieve)

A-1 A-3 A-2

A-l-a A-l-b A-2-4 A-2-5 A-2-6 A,=2=7

50 max30 max 50 max 51 min15 max 25 max lOmax 35 max 35 max 35 max 35 rnax

General classification

Group classificationSieve analysis (% passing)

No. 10 sieveNo. 40 sieveNo. 200 sieve

For fraction passingNo. 40 sieve

Liquid limit (LL)Plasticity index (PI)

Usual type of material

Subgrade rating

6 maxStone fragments,gravel, and sand

NonplasticFine sand

40 max 41 min 40 max10 max 10 max 11 minSilty or clayey gravel and sand

Excellent to good

General classificationSilt-elay materials

(More than 350/0 of total sample passing no. 200 sieve)

40 max 41 min10 max 10 nlax

Mostly silty soilsFair to poor

40 max 41 min11 min 11 minMostly clayey soils

Group classification

Sieve analysis (O~ passing)No. 10 sieveNo. 40 sieveNo. 200 sieve

For fraction passingNo. 40 sieve

Liquid limit (LL)Plasticity index (PI)

Usual types of materialSubgrade rating

A-4

36 min

A-5

36 min

A-6

36 min

A-7A-7-Sa

A-7-6b

36 min

III1

I\

Ia If PI ~ LL - 30, the classification is A-7-5.b If PI > LL - 30, the classification is A-7-6.

where F200 == percent passing no. 200 sieve, expressed as a whole nllmberLL = liquid limitPI == plasticity index

When calculating the group index for a soil belonging to group A-2-6 oronly the partial group-index equation relating to the plasticity inldex:

, use

I

I

IGI == O.01(Fzoo - 15) (PI - 10)

The group index is rounded to the nearest whole number and written next to the soilgroup in parentheses; for example, we have

A-4'----.r--'

ISoil group

(5)~

Group index

I

I

1.7 Soil Classification Systems 15

v'/

//

~//V-line /

~

PI = 0.9 (LL - 8),

V/ CH

"/ /" / or

"y/ :<,/'

1:L/

/ "~/ or/ OL/ A-linev

PI = 0.73 (LL - 20)/ /CL-ML / //\. ~ ML MH

)f-

:7v or or

~-- OL OH/

40

30

50

20

oo 10 20 30 40 50 60 70 80 90 100

Liquid limit, LL

10

70

60

Figure 1.5 Plasticity chart

Unified System

The Unified Soil Classification System was originally proposed by A. Casagrande in1942 and was later revised and adopted by the United States Bureau of Reclama­tion and the U. S. Army Corps of Engineers. The system is currently used in practi­cally all geotechnical work.

In the Unified System, the following symbols are used for identification:

Symbol G s M C o Pt H L w p

Description Gravel Sand Silt Clay Organic silts Peat and highly High Low Well Poorlyand clay organic soils plasticity plasticity graded graded

TIle plasticit~{ chart (Figure 1.5) and Table 1.8 show the procedure for deter­mining the group symbols for various types of soil. When classifying a soil be sure toprovide the group nanle that generally describes the soil, along with the group sym­bol. Figllres 1.6, 1.7, and 1.8 give flowcharts for obtaining the group names for coarse­grained soil, inorganic fine-grained soil, and organic fine-grained soil, respectively.

Table 1.8 Unified Soil Classification Chart (after ASTM, 2000)

Soil Classification

Criteria for Assigning Group Symbols and Group Names Using Laboratory TestsaGroup

Symbol Group Nameb

Coarse-grained soilsMore than 50 0ib retained onNo. 200 sieve

Fine-grained soils50% or more passes theNo. 200 sieve

GravelsMore than 500;0 of coarsefraction retained on No.4sieve

Sands500/0 or more of coarsefraction passes No.4 sieve

Silts and ClaysLiquid limit less than 50

Silts and ClaysLiquid limit 50 or more

Clean GravelsLess than 5% finesc

Gravels with FinesMore than 12°ib finesc

Clean SandsLess than 5010 finesd

Sand with FinesMore than 120/0 finesd

inorganic

organic

inorganic

organic

Cu < 4 and/or 1 > Cc > 3e

Fines classify as ML or MH

Fines classify as CL or CH

Cu < 6 and/or 1 > Cc > 3C

Fines classify as ML or MH

Fines classify as CL or CH

PI > 7 and plots on or above "A" linej

PI < 4 or plots below "A" linej

Liquid limit - oven dried

L· 'dl" d . d < 0.75lqUl ImIt-not rle

PI plots on or above "A" line

PI plots below "N' line

Liquid limit - oven dried. 'd 1" d . d < 0.75LlqUI Iffilt- not rle

GW

GP

GM

GC

SW

SP

SM

SC

CL

ML

OL

CH

MH

OH

Well-graded gravelf

Poorly graded gravelf

Silty gravelf,g,h

Clayey gravelLg,h

Well-graded sandi

Poorly graded sandi

Silty sandg,h,i

Clayey sandg,h,i

Organic clayk, 1, m, n

Organic siltk,1, m, 0

Fat Clayk,l,m

Elastic siltk ,1, m

Organic clayk, l, m, p

Organic siltk, 1, m, q

Highly organic soils Primarily organic matter, dark in color, and organic odor PT Peat

aBased on the material passing the 75-mm. (3-in) sieve.bIf field sample contained cobbles or boulders, orboth, add "with cobbles or boulders, or both" togroup name.cGravels with 5 to 12 010 fines require dual symbols:GW-GM well-graded gravel with silt; GW-GC well­graded gravel with clay; GP-GM poorly gradedgravel with silt; GP-GC poorly graded gravel withclaydSands with 5 to 12% fines require dual symbols:SW-SM well-graded sand with silt; SW-SC well­graded sand with clay; SP-SM poorly graded sandwith silt; SP-SC poorly graded sand with clay

e _ _ (D30 )2

Cu - D60/DlO Cc - D D10 X 60

flf soil contains ~15°ib sand, add "with sand" to groupname.gIf fines classify as CL-ML, use dual symbol GC-GMorSC-SM.h If fines are organic, add "with organic fines" to groupname.i If soil contains ;?15 % gravel, add "with gravel" togroup name.j If Atterberg limits plot in hatched area, soil is a CL­ML, silty clay.

kIf soil contains 15 to 290/0 plus No. 200, add "withsand" or "with gravel," whichever is predominant.1If soil contains ~30% plus No. 200, predominantlysand, add "sandy" to group name.mlf soil contains ~30% plus No. 200, predominantlygravel, add "gravelly" to group name.npI ~ 4 and plots on or above "A" line.°PI < 4 or plots below "A" line.PPI plots on or above "A" line.qPI plots below "A" line.

11111" Willi

Group Symbol Group Name

GRAVEL%gravel >

%sand

<5% fines~ Cu ~ 4 and 1 ~ Cc ~ 3 • GW~ <15% sand ---.. Well-graded gravel~ ----. ~15% sand~Well-graded gravel with sand

Cu < 4 and/or 1 > Cc > 3-----------... GP~ <15% sand~ Poorly graded gravel----. ~15% sand~ Poorly graded gravel with sand

~ fines =ML or MH~ GW-GM~ <15% sand~Well-graded gravel with siltC ~ 4 and 1 ~ C ~ 3 ------.. ~15% sand~Well-graded gravel with silt and sand

<U C fines =CL, CH, • GW-GC~ <15% sand~Well-graded gravel with clay (or silty clay)

(or CL.ML) ~15% sand~Well-graded gravel with clay and sand (or silty clay and sand)5-12% fines

< fines =ML or MH~ GP-GM~ <15% sand~ Poorly graded gravel with siltCu < 4 and/or 1 > Cc > 3 ~15% sand~ Poorly graded gravel with silt and sand

fines =CL, CH, • GP-GC~ <15% sand~ Poorly graded gravel with clay (or silty clay)(or CL-ML) ~15% sand ---.. Poorly graded gravel with clay and sand (or silty clay and sand)

fines = ML or MH~ GM oc::::::::::=:: <15% sand~ Silty gravel~15% sand~ Silty gravel with sand

> 12% fines fines = CL or CH ----+ GC ac::::::::::::: <15% sand ---. Clayey gravel~15% sand~ Clayey gravel with sand

fines = CL-ML • GC-GM~ <15% sand ---.. Silty, clayey gravel~15% sand~ Silty, clayey gravel with sand

SAND%sand ~%gravel

<5% fines~ Cu ~ 6 and 1 ~ Cc ~ 3 • SW~ <15% gravel---+ Well-graded sand~ ----. ~15% gravel ---+ Well-graded sand with gravel

eu < 6 and/or 1 > Cc > 3-----------•• SP~ <15% gravel---+ Poorly graded sand------.. ~15% gravel --. Poorly graded sand with gravel

~fines = ML or MH~ SW-SM~ <15% gravel---+ Well-graded sand with siltC ~ 6 and 1 ~ Cc ~ 3 ------.. ~15%gravel---+ Well-graded sand with silt and gravel

<u fines =CL, CH, • SW-SC~ <15% gravel~Well-graded sand with clay (or silty clay)

(or CL-ML) ~15% gravel---+ Well-graded sand with clay and gravel (or silty clay and gravel)5-12% fines

< fines =ML or MH --.. SP-SM~ <15% gravel~ Poorly graded sand with siltCu < 6 and/or 1 > Cc > 3 ~ 15% gravel ---. Poorly graded sand with silt and gravel

fines =CL, CH, • SP-SC~ <15% gravel~ Poorly graded sand with clay (or silty clay)(or CL-ML) ~15% gravel---" Poorly graded sand with clay and gravel (or silty clay and gravel)

fines =ML or MH-+ SM oc::::::::::=:: <15% gravel~ Silty sand~ 15% gravel~ Silty sand with gravel

> 12% fines -~=::-----------.... fines =CL or CH ----+ SC -=:::::::::t <15% gravel---.. Clayey sand~15% gravel~ Clayey sand with gravel

fines =CL-ML • SC-SM~ <15% gravel---.. Silty, clayey sand~15% gravel~ Silty, clayey sand with gravel

Figure 1.6 Flowchart for Classifying Coarse-Grained Soils (More Than 500/0 Retained on No. 200 Sieve) (after ASTM, 2000)

Group Symbol Group Name

Inorganic

. <30% plus No. 200~ <15% plus No. 200 -------------.. lean clay

< ........ 15-29% plus No. 200~ %sand ~% gravel • lean clay with sandPI > 7 and plots --.. CL ~ %sand <% gravel • lean clay with gravelon or above <%sand ~% gravel~ <15% gravel • Sandy lean clay"N' -line ~30% plus No. 200 ~ ~15% gravel • Sandy lean clay with gravel

%sand <% gravel~ <15% sand • Gravelly lean clay~ ~15% sand • Gravelly lean clay with sand

<<30% plus No. 200~ <15% plus No. 200 • Silty clay

15-29% plus No. 200~ %sand ~% gravel • Silty clay with sand4 ~ PI ~ 7 and~ CL-ML ~ %sand <% gravel --------.. Silty clay with gravelplots on or above <%sand ~% gravel~ <15% gravel ... Sandy silty clay"A" -line ~30% plus No. 200 ~ ~15%gravel • Sandy silty clay with gravel

%sand <% gravel~ <15% sand ~ Gravelly silty clay~ ~ 15% sand • Gravelly silty clay with sand

LL<50

IL~50

<30% plus No. 200~ <15% plus No. 200 ... Silt

< ........ 15-29% plus No. 200~ %sand ~% gravel • Silt with sandPI < 4 or plots ~ ML ~ %sand <% gravel --------.. Silt with gravelbelow "A" -line <%sand ~% gravel~ <15% gravel • Sandy silt

~30% plus No. 200 ~ ~ 15% gravel • Sandy silt with gravel%sand <% gravel~ <15% sand ... Gravelly silt

. ( LL-ovendried ) ~~ 15% sand • Gravelly silt with sandOrganIc LL-not dried < 0.75 ----+ OL --+ See figure 1.5

<30% plus No. 200~ <15% plus No. 200 • Fat clay

< ~ 15-29% plus No. 200~ %sand ~% gravel • Fat clay with sandPI plots on or ---+. CH ~ %sand <% gravel -----+. Fat clay with gravelabove "A" -line <%sand ~% gravel~ < 15% gravel • Sandy fat clay

~30% plus No. 200 ~~ 15% gravel • Sandy fat clay with gravel%sand <% gravel~ < 15% sand ... Gravelly fat clay

~~ 15% sand • Gravelly fat clay with sandInorganic

<30% plus No. 200~ <15% plus No. 200 • Elastic silt

< ~ 15-29% plus No. 200~ %sand ~% gravel • Elastic silt with sandPI plots below ---.. MR ~ %sand <% gravel -----+. Elastic silt with gravel"A" -line <%sand ~% gravel~ < 15% gravel Sandy elastic silt

~30% plus No. 200 ~~15% gravel Sandy elastic silt with gravel%sand <% gravel~ <15% sand Gravelly elastic silt

~~ 15% sand • Gravelly elastic silt with sand

. ( LL- ovendried )OrganIc LL-not dried < 0.75 ----+ OR ----. See figure 1.5

Figure 1.7 Flowchart for Classifying Fine-Grained Soil (50% or More Passes No. 200 Sieve) (after ASTM, 2000)

-

Group Symbol

OL

Group Name

<30% plus No. 200~ <15% plus No. 200 ------.---------... Organic clay

L ~ 15-29% plus No. 200~ %sand ~% gravel .. Organic clay with sand-----. %sand <% gravel .. Organic clay with gravel

PI ;::: 4 and plots on < %sand ;?;% gravel~ <15% gravel • Sandy organic clayor above "A" -line ~30% plus No. 200 -----. ~15%gravel .. Sandy organic clay with gravel

%sand <% gravel~ <15% sand • Gravelly organic clay-----. ~ 15% sand • Gravelly organic clay with sand

<30% plus No. 200~ <15% plus No. 200 • Organic silt

L ~ 15-29% plus No. 200~ %sand ~% gravel .. Organic silt with sand~ %sand <% gravel -----.... Organic silt with gravel

PI < 4 and plots < %sand ;?;% gravel~ <15% gravel • Sandy organic siltbelow "A" -line ~30% plus No. 200 -----. ~15% gravel • Sandy organic silt with gravel

%sand <% gravel~ <15% sand .. Gravelly organic silt-----. ~ 15% sand .. Gravelly organic silt with sand

Plots on orabove "A" -line L

<30%PIUSNO.200~ <15% plus No. 200 .. Organic clay~ 15-29% plus No. 200~ %sand ~% gravel .. Organic clay with sand

-----.. %sand <% gravel -----... Organic clay with gravel

< %sand ;:::% gravel~ <15% gravel .. Sandy organic clay~30% plus No. 200 -----. ~15% gravel .. Sandy organic clay with gravel

%sand <% gravel~ <15% sand .. Gravelly organic clay-------.. ~ 15% sand .. Gravelly organic clay with sand

OH

Plots below"A"-line L

<30% plus No. 200~ <15% plus No. 200 .. Organic silt~ 15-29% plus No. 200~ %sand ~% gravel .. Organic silt with sand

~ %sand <% gravel -----:... Organic silt with gravel

< %sand ;3% gravel -~~--. <15% gravel .. Sandy organic silt~30% plus No. 200 -----. ~15% gravel • Sandy organic silt with gravel

%sand <% gravel~ <15% sand .. Gravelly organic silt-------.. ~ 15% sand • Gravelly organic silt with sand

Figure 1.8 Flowchart for Classifying Organic Fine-Grained Soil (50% or More Passes No. 200 Sieve) (after ASTM, 2000)

20 Chapter 1 Geotechnical Properties of Soil

Hydraulic Conductivity of Soil

The void spaces, or pores, between soil grains allow water to flow through them. Insoil mechanics and foundation engineering, you must know how much water is flow­ing through a soil per unit time. This knowledge is required to design earth dams, de­termine the quantity of seepage under hydraulic structures, and dewater foundationsbefore and during their construction. Darcy (1856) proposed the following equation(Figure 1.9) for calculating the velocity of flow of water through a soil:

(1.24)

In this equation, v == Darcy velocity (unit: em/sec)k == hydraulic conductivity of soil (unit: em/sec)i == hydraulic gradient

(

f1~

Ii

~~

I~

II

II!

I!

I

I

I

1.8 Hydraulic Conductivity of Soil 21

B

----....----....

----.... Direction of flow

~Figure 1.9 Definitionof Darcy's law

Table 1.9 Range of the Hydraulic Conductivity for Various Soils

Type of soil

Medium to coarse gravelCoarse to fine sandFine sand, silty sandSilt, clayey silt, silty clayClays

The hydraulic gradient is defined as

Hydraulicconductivity, k(em/sec)

Greater than 10- 1

10- 1 to 10- 3

10-3 to 10-5

10-4 to 10- 6

10-7 or less

· I1h1 ==-

L(1.25)

where I1h == piezometric head difference between the sections at AA and BBL == distance between the sections at AA and BB

(Note: Sections AA and BB are perpendicular to the direction of flow.)Darcy's law [Eq. (1.24)] is valid for a wide range of soils. However, with mate­

rials like clean gravel and open-graded rockfills, the law breaks down because of theturbulent nature of flow through them.

The value of the hydraulic conductivity of soils varies greatly. In the labora­tory, it can be determined by means of constant-head or falling-head permeabilitytests. The constant-head test is more suitable for granular soils. Table 1.9 providesthe general range for the values of k for various soils. In granular soils, the value de­pends primarily on the void ratio. In the past, several equations have been proposedto relate the value of k to the void ratio in granular soil:

(1.26)

22 Chapter 1 Geotechnical Properties of Soil

(1.27)

(1.28)

In these equations, k1 and k2 are the hydraulic conductivities of a given soil atratios el and ez, respectively.

According to their experimental observations, Samarasinghe, Huang,Drnevicll (1982) suggested that the hydraulic conductivity of normally consolidatedclays could be given by the equation

n

k == C_e

- (1.29)1 + e

where C and n are constants to be determined experimentally.

For clayey soils in the field, a practical relationship for estimating the hy­draulic conductivity (Tavenas et aI., 1983) is

(1.31)

(1.30)

kk

h< 1.5

v

eo - elog k = log ko - C

k

where k h == hydraulic conductivity for flow in the horizontal direction

For varved clays, the ratio of k h/ kv may exceed 10.

Steady-State Seepage

where k == hydraulic conductivity at void ratio eko == in situ hydraulic conductivity at void ratio eoCk == conductivity change index ~ OISeo

For clayey soils, the hydraulic conductivity for flow in the vertical and horizon­tal directions may vary substantially. The hydraulic conductivity for flow in the ver­tical direction (kv ) for in situ soils can be estimated from Figure 1.10. For marineand other massive clay deposits,

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- Jmining the rate of seepage is by a graphical construction referred to as the flow net, I

~

1.9 Steady-State Seepage 23

PI + CF = 1.25(both decimal)

10-10

klJ (m/sec)

PI =plasticity indexCF =clay fraction

1.2

0.4 L-_'-"'~~--'--L...&."""""---'-"'-..--I....--I..--"""""~-------

10-1J

2.0

Figure 1.10 Variation of in situ k v for clay soils (after Tavenaset aL, 1983)

y

Piezometers

..--------. '.

z

Figure 1.11 Steady-state seepage

a concept based on Laplace's theory of continuity. According to this theory, for asteady flow condition, the flow at any point A (Figure 1.11) can be represented bythe equation

(1.32)a2h a2h a2h

k X- 2 + kY

-2

+ k Z-2

== 0ax oy az

where kx , ky , k z == hydraulic conductivity of the soil in the x, y, and z directions,respectively

24 Chapter 1 Geotechnical Properties of Soil

h == hydraulic head at point A (i.e., the head of water that a piezometerplaced at A would show with the downstream water level as datum, asshown in Figure 1.11)

For a two-dimensional flow condition, as shown in Figure 1.11,

a2h-== 0a2y

so Eq. (1.32) takes the form

a2h a2hk

X-

2+ k

Z-

2== 0

ax azIf the soil is isotropic with respect to hydraulic conductivity, kx == k z == k, and

I

I

I

I

I

(1.34)

IEquation (1.34), which is referred to as Laplace's equation and is valid for confinedflow, represents two orthogonal sets of curves known as flow lines and equipotentiallines. A flow net is a combination of numerous equipotential lines and flow lines. Aflow line is a path that a water particle would follow in traveling from the upstreamside to the downstream side. An equipotential line is a line along which water, inpiezometers, would rise to the same elevation. (See Figure 1.11.)

In drawing a flow net, you need to establish the boundary conditions. For exam­ple, in Figure 1.11, the ground surfaces on the upstream (00') and downstream(DD') sides are equipotential lines. The base of the dam below the ground surface,0'BCD, is a flow line. The top of the rock surface, EF, is also a flow line. Once theboundary conditions are established, a number of flow lines and equipotential lines aredrawn by trial and error so that all the flow elem~nts in the net have the same length­to-width ratio (LjB). In most cases, LjB is held to unity, that is, the flow elements aredrawn as curvilinear "squares." This method is illustrated by the flow net shown in Fig­ure 1.12. Note that all flow lines must intersect all equipotential lines at right angles.

Water level'Y : p..... _

Water level

Figure 1. 12 Flow net

I

I

It.I·.••:.c c~ .

-·~::.I'·"j,

I

I

1.10 Effective Stress 25

Once the flow net is drawn, the seepage, in unit time per unit length of thestructure, can be calculated as

(1.35)

where Nt == number of flow channelsNd == number of drops

n == width-to-Iength ratio of the flow elements in the flow net (BIL)hmax == difference in water level between the upstream and downstream sides

The space between two consecutive flow lines is defined as a flow channel, and thespace between two consecutive equipotential lines is called a drop. In Figure 1.12,Nt == 2, Nd == 7, and n == 1. When square elements are drawn in a flow net,

(1.36)

Effective Stress

Consider the vertical stress at a point A located at a depth hI + hz below the gro'undsurface, as shown in Figure 1.13a. The total vertical stress at A is

(J == h1'Y + hz'Ysat (1,.37)

where Y and Ysat are unit weights of soil above and below the water table, respecti\rely.

The total stress is carried partially by the pore water in the void spaces and I)ar­tially by the soil solids at their points of contact. For example, consider a wavy planeAB drawn through point A (see Figure 1.13a) that passes through the points of con­tact of soil grains. The plan of this section is shown in Figure 1.13b. The small dots inthe figure represent the areas in which there is solid-to-solid contact. If the SUlll ofthese areas equals A', the area filled by water equals XY - A. The force carriedby the pore water over the area shown is then

Fw == (XY - A')u

where u == pore water pressure == ywhz

(1.38)

(1.39)

Now let Fi , Fz, ... be the forces at the contact points of the soil solids, as ShOWll inFigure l.i3a. The sum of the vertical components of these forces over a horizlon­tal area XY is

Fs == IFI(v) + F 2(v) + .. · (1.40)

where F1(v), F2(v)' • • . are the vertical components of forces FI , F2, ••• , respectiv'ely

According to the principles of statics,

26 Chapter 1 Geotechnical Properties of Soil

Saturatedunit weight = 1'sat

Areas ofsolid-to-solidcontact

• • ...•• • ..

• •, . 1# ,• y.'• •,

•'1

114--.. -X-~·I

(b)

I

I1

II

-

Saturated unitweight = T'satA _

Flow of water

(c)

I

I

I

I

~h

iii

-I(a)

,f4-.... -x

Figure 1.13 Calculation of effective stress

The term u' in Eq. (1.41) is generally referred to as tIle vertical effective stress.Note that the quantity a is very small; thus,

where a == A'/ XY == fraction of the unit cross-sectional area occupied by solid­to-solid contact

a' = Fs/(XY) = verticle component of forces at solid-to-solid contactpoints over a unit cross-sectional area

or

(u)XY == (XY - A')u + Fs

so

u == (1 - a)u + u' (1.41)

1.10 Effective Stress 27

(1.42)

Observe that effective stress is a derived quantity. Also, because the effectivestress u' is related to the contact between the soil solids, changes in effective stresswill induce changes in volume. Effective stress is responsible for producing frictionalresistance in soils and rocks. For dry soils, U == 0; hence, u == u'.

For the problem under consideration in Figure 1.13a, u == hz'Yw( 'Yw == unitweight of water). Thus, the effective stress at point A is

u' == u - U == (hi/' + h'Z/'sat) - hz'Yw

== h(y + hz( 'Ysat - 'Yw) == h1'Y + h2y' (1.43)

where 'Y' == effective, or submerged, unit weight of soil== 'Ysat - 'Yw

From Eq. (1.15),

Gs'Yw + e'Yw'Ysat == 1 + e

so

G/yw + eyw yw(Gs - 1)'Y' = 'Ysat - 'Yw = 1 + e - 'Yw = 1 + e (1.44)

For the problem in Figures 1.13a and 1.13b, there was no seepage of water inthe soil. Figure 1.13c shows a simple condition in a soil profile in which there is up­ward seepage. For this case, at point A,.

a == h i yw + hz'Ysat

and

u == (hi + h2 + h)yw

Thus, from Eq. (1.42),

0-' == (J - u == (h1yw + h2Ysat) - (hi + hz + h)'Yw

== hz( 'Ysat - 'Yw) - hyw == hzy' - h'Yw

or

if' = h2( Y' - :2Yw ) = h2 ( Y' - i'Yw) (1.45)

Note in Eq (1.45) that h/h2 is the hydraulic gradient i. If the hydraulic gradient isvery high, s that y' - iyw becomes zero, the effective stress will become zero. In otherwords, ther is no contact stress between the soil particles, and the soil will break up.This situati n is referred to as the quick condition, or failure by heave. So, for heave,

(1.46)

28 Chapter 1 Geotechnical Properties of Soil

where icr == critical hydraulic gradient

For most sandy soils, icr ranges from 0.9 to 1.1, with an average of about unity.

Depth, Z

o Effective... ." .. . ..' .. stress, (T'

. . : ..' ~ .. ' : ~:..:: .--:" (kN/m2)

: '.' :,':'::' :. Dry sand: Gs:~:2.65;. e ,:, (t6:'3m

~\... \ .. ~ .;.";:.: :.., .:\..;.: \ .. ~~.;..::...: :..,' .:'•• , , .•.• •,•.• , •• ,., ., .• ,te.o !.,.\.~ ,. ••••••••••:, • !.,.\.~e:-, ••••••": : , ~....... • JJ : ••••• • • CI ~ .•~••• ~ ~-:.; ..•, :.•--:.: .... :. ••~••• ~:~ ay:. _";.:.'- :. .~:l

3 ill .~I:..• : ~ ': ., ~ .;.:..:. ~ I:..• :Gs = 2.7 i ~ .;.:••~ .~•• ••• It ~ •••:.:.: •• • :. ~ " ••• ~ 30% •• •.:. ~ , ·1

J·\...~:: ~ .;..... .- .. ,.... ...~ I W = 0 .- •• , ••••.I .: ~ • ~ •••••• :..; ••••.:. .: ~ -;. - :..; •••••~Jt.. .. , .'.... :-:.. ........., .'.... :-: _., ..... "..~ I··· ...,..... , ' '. 76·38- .,..... C, .. ··'·.. _~.. , •• : I ••••': , I. -= . , •. · ..': , ... ... ..-... ••.. -. -'-•• -, .- ___.-,. -.-. ~ I.-.~_, _.ft ;

\ '/ / ...... , \ '" '/ / ...... , \ '""'/, '" - - / ....... '" --

'" ,./ - \./ Rock ./ '\ - - \ ./ ...... ,/ , -'" ....... -- " ....... _/ -- ...: ...... -- '" " ....... _/ '" ....... -

1. 11 Consolidation 29

Consolidation

In the field, when the stress on a saturated clay layer is increased-for example, bythe construction of a foundation-the pore water pressure in the clay will increase.Because the hydraulic conductivity of clays is very small, some time will be requiredfor the excess pore water pressure to dissipate and the increase in stress to be trans­ferred to the soil skeleton. According to Figure 1.15, if flo- is a surcharge at theground surface over a very large area, the increase in total stress at any depth of theclay layer will be equal to liu.

However, at time t = 0 (i.e., immediately after the stress is applied), the excesspore water pressure at any depth du will equal flu, or

flu == flh{yw == flo- (attimet == 0)

Immediately afterloading:time t = 0

-!-T~hi

----l

Figure 1.15 Principles of consolidation

30 Chapter 1 Geotechnical Properties of Soil

Hence, the increase in effective stress at tinle t == 0 will be

D.u' == D.u - flu == 0

Theoretically, at time t == 00, when all the excess pore water pressure in the claylayer has dissipated as a result of drainage into the sand layers,

Llu == 0 (at time t == (0)

Then the increase in effective stress in the clay layer is

!lu' == au - flu == au - 0 == auThis gradual increase in the effective stress in the clay layer will cause settlementover a period of time and is referred to as consolidation.

Laboratory tests on undisturbed saturated clay specimens can be conducted(ASTM Test 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 aring, with one porous stone at the top and one at the bottom of the specimen(Figure 1.16a). A load on the specimen is then applied so that the total verticalstress is equal to u. Settlement readings for the specimen are taken periodicallyfor 24 llours. After that, the load on the specimen is doubled and more settlementreadings are taken. At all times during the test, the specimen is kept under ,vater.The procedure is continued until the desired limit of stress on the clay specimenis reached.

Based on the laboratory tests, a graph can be plotted showing the variation ofthe void ratio e at the end of consolidation against the corresponding vertical effectivestress a'. (On a semilogarithmic graph, e is plotted on the arithmetic scale and a' onthe log scale.) The nature of the variation of e against log a' for a clay specimen isshown in Figure 1.16b. After the desired consolidation pressure has been reached, thespecimen can be gradually unloaded, which will result in the swelling of the specimen.The figure also shows the variation of the void ratio during the unloading period.

From the e-Iog u' curve shown in Figure 1.16b, three parameters necessary forcalculating settlement in the field can be determined:

1. The preconsolidation pressure, (T~, is the maximum past effective overburdenpressure to which the soil specimen has been subjected. It can be determined byusing a simple graphical procedure proposed by Casagrande (1936). The proce­dure involves five steps (see Figure 1.16b):3. Determine the point 0 on the e-Iog a' curve that has the sharpest curvature

(i.e., the snlallest radius of curvature).b. Draw a horizontal line GA.c. Draw a line OB that is tangent to the e-log a' curve at O.d. Draw.a line OC that bisects the angle AOB.e. Produce the straight-line portion of the e-Iog a' curve backwards to intersect

OC. This is point D. The pressure that corresponds to point D is the precon­solidation pressure u~.

Natural soil deposits can be normally consolidated or overconsolidated

(or preconsolidated). If the present effective overburden pressure a' == u~ is

1.11 Consolidation 31

Dial gauge Q" Load

Water level

~--+---+-+- Porous stone

(J"', C.A.....-.. r---a.

'6ri A1-.... !2~~.

--po..1'-0,.....

~"'~ ......

............""'- .......

~......

C......

r\.......

' .......

(eb (J"{) rt[\ ' .......B

\ /Slope=Cc

(e3' us)1---r>.. '\I

I V;'f ~~~Slope = c-I I S (e4' lT4) _ (e2' lT2)

\

2.3

2.2

2.1

2.0~

.S 1.9+oJ~;...

'"'0 1.8~

1.7

1.6

1.5

1.410

(a)

100Effective pressure, 0-' (kN/m2)

(b)

400

Figure 1.16(a) Schematic diagram ofconsolidation test arrange­ment; (b) e-log (J"' curve fora soft clay from East St.Louis, Illinois (Note: At theend of consolidation,u == a')

equal to the preconsolidated pressure (T~ the soil is normally consolidated.However, if a~ < a~, the soil is overconsolidated.

The preconsolidation pressure ((T~) has been correlated with the index pa­rameters by several investigators. The U.S. Department of the Navy (1982) alsoprovided generalized relationships among (T~, LI, and the sensitivity of clayeysoils (St). The definition of sensitivity is given in Section 1.19. Figure 1.17 showsthe relationship between a~ and LI. Note that the latter-the liquidity index ofa soil-is defined as

LI = w - PLLL- PL

where w == in situ moisture contentLI == liquid limitPL == plastic limit

(1.47)

32 Chapter 1 Geotechnical Properties of Soil

2 3 4 56 8 10 20 30 40 160 80 10050

1\ \ \ ~1\

f\~2 \

, ,1\ 6 1\8Sf ,4

\ \ \ 1\r\ r\"\. i\.

" r'\~ ~,,'\~ ~ r\.

I"i'-

~~"\

~ '""r--..

'"'"~ ~

""I~~~

~ f".-"",~~'""'~~::~,

~-.........

I'-....I'-.~~~

f'."-0.2

-0.40.012345680.1 2345681.0

o

+0.2

+0.4

+1.2

+1.4

+0.8

+1.0

><Q)

"0 +0.6.5

Preconsolidation pressure, Pc (tonlft2)

0- 'C

Pa

Figure 1.17 Variation of (T~ witll LI (after U.S. Department of the Navy, 1982)Note: Pa == atmospheric pressure [~100 kN/m2 (1 U.S. ton/ft2

) ]

2. The compression index, Cc, is the slope of the straight-line portion (the latterpart) of the loading curve, or

(1.48)

where el and ez are the void ratios at the end of consolidation under effectiv·estresses a1 and a2, respectively

The compression index, as determined from the laboratory e-Iog a'curve, will be somewhat different from that encountered in the field. The pri­mary reason is that the sc~il remolds itself to some degree during the field ex­ploratio.n. The nature of variation of the e-Iog a' curve in the field for anormally consolidated clay is shown in Figure 1.18. Th.e curve, generally re~

ferred to as the virgin compression curve, approximately intersects the labo·­ratory curve at a void ratio of O.42eo (Terzaghi and Peck, 1967). Note that eo isthe void ratio of the clay in the field. Knowing the values of eo and u~, you caneasily construct the virgin. curve and calculate its compression index by usingEq. (1.48).

1.11 Consolidation 33

Void ratio, e\

Figure 1.18 Construction ofvirgin compression curve fornormally consolidated clay

Pressure, (J"~

(log scale)u}

curve

Laboratoryconsolidation

_____"____ (J"~ =: rr;\

\\

el

_\ VirginI \ compressionI~ curve,

\ Slope Cc

I \

\ \e2 ------------+---

i I~I 1 \

I I ~I I ~I II ,I I

------------+----1---I II II II I: I

0.42 eo

The value of Cc can vary widely, depending on the soil. Skempton (1944)gave an empirical correlation for the compression index in which

(1.49)

where LL == liquid limit

Besides SkemptC)ll, several other investigators have also proposed correlationsfor the compression indlex.

3. The swelling index, Cs' is the slope of the unloading portion of the e-Iog 0-'

curve. In Figure 1.16b, it is defined as

(1.50)

In most cases, the value of the swelling index is *to ! of the conlpression index.Following are some representative values of Cs/Ccfor natural soil deposits:

Description of soil

Boston Blue clayChicago clayNew Orleans claySt. Lawrence clay

0.24-0.330.15-0.30.15-0.280.05-0.1

34 Chapter 1 Geotechnical Properties of Soil

(Tfo

Figure 1.19 Construction offield consolidation curve foroverconsolidated clay

c

b

\\

\ Virgin

'\*-- compression\ curve,

\ Slope Cc

\

\\

~\

~

(TfC

IIII

Labor~toryconsolidationcurve

III

I dI

1 Slope

1 Cs------~-------------

IIIII

SlopeCs

-----~lI

Voidratio, e

The swelling index is also referred to as the recompression index.The determination of the swelling index is important in the estimation of

consolidation settlement of overconsolidated clays. In the field, depending onthe pressure increase, an overconsolidated clay will follow an e-log (I' pathabc, as shown in Figure 1.19. Note that point a, with coordinates (T~ and eo,corresponds to the field conditions before any increase in pressure. Point bcorresponds to the preconsolidation pressure «(T~) of the clay. Line ab is ap­proximately parallel to the laboratory unloading curve cd (Schmertmann,1953). Hence, if you know eo, a~, a~, Ce, and Cs' you can easily construct thefield consolidation curve.

,I!

IiI

Ili

1

Calculation of Primary Consolidation Settlement

The one-dimensional primary consolidation settlement (caused by an additionalload) of a clay layer (Figure 1.20a) having a thickness He may be calculated as

(1.51)

I- ~

1

III

!

where Sc = primary consolidation settlement .I[

Lle :::::: total change of void ratio caused by the additional load application -eo :::::: void ratio of the clay before the application of load

1.12 Calculation of Primary Consolidation Settlement 35

Added pressure == d,o-

~

(a)

Voidratio, e

Averageeffectivepressurebefore loadapplication

0-;

u'c

Voidratio,e -t- ':I Slope C,

~e I

_l_j _IIJ

I Pressure,cr'

0-; 0-; + ~o-' (log scale)

Normally consolidated clay

(b)

1Case 1 ~e

TI---.­I

Slope t

Cs IIIIIIIIIII

u ~ u ~ +au' ud+ au'Case 1 Case 2

Overconsolidated clay

(c)

Slope Cc

Pressure,[T'

(log scale)

Fig.:Jre 1.20 One-dimensional settlement calculation: (b) is for Eq. (1.53); (c) is forEqs.. (1.55) and (1.57)

Note that

~e . I .1

== 8 v == vertIca straIn+ eo

For normally consolidated clay, the field e-Iog a' curve will be like the onesho'wn in Figure 1.20b. If a~ == initial average effective overburden pressure on theclay' layer and aa' = average effective increase in pressure on the clay layercaused by the added load, the change in the void ratio caused by the increase inloa<l is

(T~ + ~(T'ile == Cc log--­

lT~(1.52)

36 Chapter 1 Geotechnical Properties of Soil

Now, combining Eqs. (1.51) and (1.52) yields

For overconsolidated clay, the field e-Iog a' curve will be like the one shown inFigure 1.20c. In this case, depending on the value of da', two conditions may arise.First, if u~ + au' < a~ then

a~ + da'ae == Cs log----a~

Combining Eqs. (1.51) and (1.54) gives

Second, if a~ < (T~ < a~ + da', then

Now, combining Eqs. (1.51) and (1.56) yields

(1.53)

(1.54)

(1.55)

(1.56)

In Section 1.11 (see Figure 1.15), we showed that consolidation is the result of thegradual dissipation of the excess pore water pressure from a clay layer. The dissipa­tion of pore water pressure, in turn, increases the effective stress, which induces set­tlement. Hence, to estimate the degree of consolidatic.n of a clay layer at some timet after the load is applied, you need to know the rate of dissipation of the excess porewater pressure.

Figure 1.21 shows a clay layer of thickness He that has highly permeable sandlayers at its top and bottom. Here, the excess pore water pressure at any point A atany time t afte-r the load is applied is llu == (llh)yw. For a vertical drainage condi­tion (that is, in the direction of z only) from the clay layer, Terzaghi derived the dif­ferential equation

77me Rate of Consolidation

a(Llu) a2(~u)--==c--at v dZ2

(1.57) I

I

(1.58)

1.13 Time Rate of Consolidation 37

(b)z=o

z=H

z == 2H ----------.. Liu

Tdh1.-

zGround

water table ...

r.!i~· Ii:~; ~:·~;~·~:).;~;~t:~~i\:.·:!~j~.:}\.::;:-;:-'~

Figure 1.21 (a) Derivation of Eq. (1.60); (b) nature of variation of du with time

where Cv == coefficient of consolidation, defined by

(1.59)

in which k == hydraulic conductivity of the clayLle == total change of void ratio caused by an effective stress increase of f1a-'eav == average void ratio during consolidationm v == volume coefficient of compressibility == f1e/[ f1a-' (1 + eav ) ]

Equation (1.58) can be solved to obtain f1u as a function of time t with the followingboundary conditions:

1. Because highly permeable sand layers are located at z == 0 and z == He' the ex­cess pore water pressure developed in the clay at those points will be immedi­ately dissipated. Hence,

f1u == 0 at z == 0

and

f1u = 0 at z = He = 2H

where H == length of maximum drainage path (due to two-way drainagecondition-that is, at the top and bottom of the clay)

2. At time t == 0,

38 Chapter 1 Geotechnical Properties of Soil

fiu == f1uo == initial excess pore water pressure after the load is appliedWith the preceding boundary conditions, Eq. (1.58) yields

A m~oo[2(~Uo), (MZ)J -M2yuU == L,,; SIn - e l'

m=O M H

where M :::: [(2m + 1)7TJ/2

m == an integer == 1, 2, ...Tv == nondimensional time factor == (Cvt) / H 2

(1.60)

(1.61)

The value of fiu for various depths (i.e., z == 0 to z == 2H) at any given time t

(and thus Tv) can be calculated from Eq. (1.60). The nature of this variation offlu is shown in Figures 1.22a and b. Figure 1.22c shows the variation of flul fluowith Tv and HIHe using Eqs.(1.60) and (1.61).

Determining the field value of Cv is difficult. Figure 1.23 provides a first-orderdetermination of Cv' using the liquid limit (U.S. Department of the Navy, 1971).

The average degree ofconsolidation of the clay layer can be defined as

(1.62) j

where SeCt) = settlement of a clay layer at time t after the load is appliedSc(max) = maximum consolidation settlement that the clay will undergo

under a given loading

If the initial pore water pressure (J).uo) distribution is constant with depth, asshown in Figure 1.22a, the average degree of consolidation can also be expressed as

or

SeCt)u==--Se(max)

12H 12H

o (lluo )dz - 0 (llu)dz

(2HJo

(lluo )dz

(1.63)

(lluo)2H - fH (llu)dz fH (llu)dz

U = (tiuo)2H = 1 - 2H(tiu

o)

Now, combining Eqs. (1.60) and (1.64), we obtain

U = Sc(t) = 1 _ m±oo(2-)e-M2Tv

Se(max) m=O M 2

(1.64)

(1.65)

1.13 Time Rate of Consolidation 39

~u ato/ t>

" /\

\

\

Lluo =\

constantwith depth

\\,,,II,

Impermeablelayer

(b)

Highly permeablelayer (sand)~u at

oit>",I

\\

\

Lluo =,

constantwith depth

IJ

II

II

~

I~

Highly permeablelayer (sand)

Highly permeablelayer (sand)

(a)

~

~~~

~~~ r-..

\\,,,'"""~~ -----~~

\\'\ '"" ~"~~r--..... -~ Tv=O

\ ~ \ "~",~ ~ .......... ~ \\\\ 1\" '" ~ ~ ~ Tv =0.1

.........

Tv= 1 \ \ \ \0.6 1\0.5 "\~0.4 " ,0.3 ~.2 ~~ ) \ \ ' 0.7\ \ \ " ", "",........

\ 0.8 \ \ \ \ \ '" '"\ , , , \ \ \0.9 I I J J ) ) )

J / 1/ / / / / / // / I / / / / // /// I / I /1

v /v / /'/

1// 1/ / / /' /' ~/ VI / //// ~

v~

/1V// /'/V~~ -~II;:V/~/~V- ~

-----//&~~~

I//~~~

2.0

1.5

HH

1.0c

0.5

oo 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Excess pore water pressure, Ji.uInitial excess pore water pressure, f1uo

(c)

Figure 1.22 Drainage condition for consolidation: (a) two-way drainage; (b) one-waydrainage; (c) plot of ilu/ iluo with Tv and H / He

40 Chapter 1 Geotechnical Properties of Soil

Figure 1.23 Range of Cv (after U.S. Department of the Navy, 1971)

\\

1\~

\r\.

'\ \.i\ \~

\~ Undisturbed samples: .

~ \ /V Cv in range of virgin compression

" V v

" ~ ""Cv in range of recompression

r\. lies above this lower limit

" "- ""'r\.. " '"~ " ~

"I\. " '"~ l~ ~

>-Completelyrem~ '"~ '"

"~

'"samples: ~

'"Cv lies below this upper limit~

-..~ ~-............. ~

~

""""'~

I

16014012080 100Liquid limit~ LL

6040

3

2

10-3

8

654

3

10-2

8

654

3

2

2

The variation of U with Tv can be calculated from Eq. (1.65) and is plotted inFigure 1.24. Note that Eq. (1.65) and thus Figure 1.24 are also valid when an imper­meable layer is located at the bottom of the clay layer (Figure 1.22). In that case, thedissipation of excess pore water pressure can take place in one direction only. Thelength of the maximum drainage path-is then equal to H == He-

The variation of Tv with U shown in Figure 1.24 can also be approximated by I7T(U% )2

Tv = 4" 100 (for U = 0-60%) (1.66) Iand

Tv == 1.781 - 0.933 log (100 - U%) (for U > 60%) (1.67)

1

ISivaram and Swamee (1977) also developed an empirical relationship be­

tween Tv and U that is valid for U varying from 0 to 100%. The equation is of theform

I

T2li = He

1.13 Time Rate of Consolidation 41

J4--------Eq. (1.66) ----------...4---- Eq. (1.67 )--..-t

1.0 I------.....---_--_--~--....__--+_--...,.._--__.__--__t

~o..,(.j

~

~ 0.6 t-----+­<l)

S~

0.4 t----t----t----t----t----+-----t----7fC----+-----f

0.2 t----t----+----+----+----:::lII....-=:----i-----+----+-----1

o ~ 10 20 30 40 50 60 70A.verage degree of consolidation, U(%)

80 90

Figure 1.24 Plot of time factor against average degree of consolitlation (fiuo == constant)

(1.68)

42 Chapter 1 Geotechnical Properties of Soil t

""',

ii

II

Il

I

Ii

II

I·:1:'..j~

I

1.14 Degree of Consolidation Under Ramp Loading 43

Degree of Consolidation Under Ramp Loading

The relationships derived for the average degree of consolidation in Section 1.13assume that the surcharge load per unit area (flu) is applied instantly at timet == O. However, in most practical situations, Llu increases gradually with time to amaximum value and remains constant thereafter. Figure 1.25 shows flu increasinglinearly with time (t) up to a maximum at time tc (a condition called ramp load­ing). For t ~ tc' the magnitude of Llu remains constant. Olson (1977) considered

z

Clay

(a)

Load per unitarea, flO"

~ llirne,t

Figure 1.25 One-dimensional consolidation due(b) to single ramp loading

44 Chapter 1 Geotechnical Properties of Soil

10

(1.69)

(1.71)

1.0

Time factor, Tv

0.1

Tv { 2 m=oo 1 2 }U == -,1- - L -4 [1 - exp(-M Tv)]Tc Tv m=O M

~r--::::~ - t--~

~~N - r--t--t--- r--... r-..... r-.-.r-- r-r----........., T = 10

~~~

r---..r----. ~ ~I-......... r-... K

~ '" '''~ 1\002 I\~~"" "'~~0~~ O.O~~0.1 0.,

""2

"~~ "'" 1\\ 0.5\ \ '\ \

"'"~ i'-. ~ 1\~ " \ \

f'..~

~~\ \ \ 1\ \ \\ 1\ \ 1\

~\\ \ \ \ \ \r\

~~ ~ \ \ \ \~ \

~~ ~\ \ \ \\ \

~~ 1\ 1\ \ \ \

\ \ ~~,1\ \ \ \ \ \~,

~""\ !\~

~~0 1\ \'-.....

o

20

40

60

80

1000.01

Figure 1.26 Olson's ran1p-loading solution: plot of U vs. Tv (Eqs. 1.69 and 1.70)

this phenomenon and presented the average degree of consolidation, U, in the fol­lowing form:

For Tv ~ Tc'

where m, M, and Tv have the same definition as ill Eq. (1.60) and where

Figure 1.26 shows the variation of U with Tv for various values of Tc' based onthe solution given by Eqs. (1.69) and (1.70).

••

1.15 Shear Strength 45

72kN/m2 ----71III

"-- .L...-- -----+Time,t

Shear Strength

The shear strength of a soil, defined in terms of effective stress, is

(1.72)

where u' == effective normal stress on plane of shearingc' == cohesion, or apparent cohesion

4>' == effective stress angle of friction

Equation (1.72) is referred to as the Mohr-Coulomb failure criterion. Thevalue of c' for sands and normally consolidated clays is equal to zero. For overcon­solidated clays, c' > o.

For most day-to-day work, the shear strength parameters of a soil (i.e., c' and<p') are determined by two standard la00ratory tests: the direct shear test and thetriaxial test.

46 Chapter 1 Geotechnical Properties of Soil

Direct Shear Test

Dry sand can be conveniently tested by direct shear tests. The sand is placed in ashear box that is split into two halves (Figure 1.28a). First a normal load is applied tothe specimen. Then a shear force is applied to the top half of the shear box to causefailure in the sand. The normal and shear stresses at failure are

J

II,II

I

Na-' == -

A

and

Rs ==-

A

where A == area of the failure plane in soil-that is, the cross-sectional area ofthe shear box

Several tests of this type can be conducted by varying the normal load. The angle offriction of the sand can be determined by plotting a graph of s against 0-' (== 0- fordry sand), as shown in Figure 1.28b, or

(1.73)

Shearstress

•II!

;1'LI

II

i..,·········I··.··.··.

"

(b)

Effective~_--I....~_~_--a.~_...-..~ nonnal

stress, if'

---

(a)

N

1

For sands, the angle of friction usually ranges from 26° to 45°, increasing withthe relative density of compaction. The approximate range of the relative density ofcompaction and the corresponding range of the angle of friction for various coarse­grained soils is shown in Figure 1.29.

Figure 1.28 Direct shear test in sand: (a) schematic diagram of test equipment;(b) plot of test results to obtaill tIle friction angle <P'

1.15 Shear Strength 47

45 r------....-..or-----~---____r----~---____,---~___.

40

~Q)

:g,--e-- 35ci.9-+-Ju:E~

0 30Q)

biJ~

----25

2.42.22.01.81.61.4

20 L-- --J...,. ""'--- ....J.- --.Io ......

1.2

Dry unit weight, Yd

Unit weight of water, Yw

Figure 1.29 Range of relative density and corresponding range of effective stress angle offriction for coarse-grained soil (modified from U.S. Department of the Navy, 1971)

Triaxial Tests

Triaxial compression tests call be conducted on sands and clays Figure 1.30a shows aschematic diagram of the triaxial test arrangement. Essentially, the test consists ofplacing a soil specimen confined by a rubber membrane into a lucite chamber andthen applying an all-around confining pressure ({}3) to the specimen by means ofthe chamber fluid (generally, water or glycerin). An added stress (f1u) can also beapplied to the specimen in the axial direction to cause failure (au == aUf at failure).Drainage from the specimen can be allowed or stopped, depending on the conditionbeing tested. For clays, three main types of tests can be conducted with triaxialequipment (see Figure 1.31):

1. Consolidated-drained test (CD test)2. Consolidated-undrained test (CD test)3. Unconsolidated-undrained test (UU test)

Consolidated-drained tests:

Step 1-. Apply chamber pressure U3. Allow complete drainage, so that the porewater pressure (u == uo ) developed is zero.

Step 2. Apply a deviator stress ~u slowly. Allow drainage, so that the pore waterpressure (u == ud) developed through the application of ~(J is zero. At

failure, 110- == 110-f; the total pore water pressure uf == U o + Ud == o.

48 Chapter 1 Geotechnical Properties of Soil

Piston

Porousstone --t-r--~.'£:.:....:t.:••.;.:~:~;,,~.••

I

u{ Effectivenormal stress

Consolidated-drained test(b)

Lucitechamber Shear

stress

Chamberfluid

Porousstone

~Valve c

i ,a;'0'3 3

To drainage and/orpore water pressuredevice

Schematic diagram of triaxialtest equipment

(a)

Rubbermembrane

Base---aM//.I'''''''

plate

Chamberfluid

Soilspecimen

Shearstress

Total stressfailureenvelope

~ ThW....-.----I. ----'__"'-- --A..-.. normal

u1 stress, (T

Shearstress

Shearstress

Effectivestressfailureenvelope

1/

TT s = Cuc 1-.l

0;' 0;' u{ Effective 0'33 3

normal stress, 0-'

Consolidated-undrained test

(c)

Figure 1.30 Triaxial test

Total stressfailure envelope(¢ = 0)

0'3 0'1 Normal stress(total),o-

Unconsolidated-undrained test

(d)

I1

II}

I

Ii1

I

I

1. 15 Shear Strength 49

. ... " ...

" .

. .. " ..

.. .. .. ... .

t

.' .. .

.. .

Figure 1.31 Sequence of stressapplication in triaxial test

So for consolidated-drained tests, at failure,

Major principal effective stress == CT3 + flCTf == CTl == CTl

Minor principal effective stress == CT3 == CT3

Changing CT3 allows several tests of this type to be conducted on various clay speci­mens. The shear strength parameters (c' arid cP') can now be determined by plottingMohr's circle at failure, as shown in Figure 1.3Gb, and drawing a common tangent tothe Mohr's circles. This is the Mohr-Coulomb failure envelope. (Note: For normallyconsolidated clay, c' ~ 0.) At failure,

(1.74)

Consolidated-undrained tests:

Step 1. Apply chamber pressure CT3. Allow complete drainage, so that the porewater pressure (u == uo ) developed is zero.

Step 2. Apply a deviator stress flCT. Do not allow drainage, so that the porewater pressure u == Ud -=I=- O. At failure, flCT == flCTf; the pore water pres­sure uf == Uo + Ud == 0 + ud(f)·

Hence, at failure,

Major principal total stress == CT3 + flCTf == CTl

Minor principal total stress == (T3

Major principal effective stress == ( CT3 + fi.CTf) - uf == 0"1

Minor principal effective stress = 0"3 - Uf = iT:'

Changing 0-3 permits multiple tests of this type to be conducted on several soilspecimens. The total stress Mohr's circles at failure can now be plotted, as shown in

50 Chapter 1 Geotechnical Properties of Soil

The pore pressure U a is the contribution of tile hydrostatic chamber pressure a3. Hence,

I

I

I

­, iI

1

I(i

(1.76)

(1.77)

(1.78)

(1.81)

(1.80)

Figure 1.30c, and then a common tangent can be drawn to define the failure enve­lope. This total stress failure envelope is defined by the equation

s == c + a tan 4J (1.75)

where c and 1J are the consolidated-undrained cohesion and angle of friction, re­spectively (Note: c ~ 0 for nornlally consolidated clays)

Sinlilarly, effective stress Mohr's circles at failure can be drawn to determinethe effective stress failure envelope (Figure 1.30c), which satisfy the relation ex­pressed in Eq. (1.72).

Unconsolidated-undrained tests:

Step 1. Apply chamber pressure a3. Do not allow drainage, so that the pore waterpressure (u == uo ) developed through the application of a3 is not zero.

Step 2. Apply a deviator stress da. Do not allow drainage (u == ud =1= 0). Atfailure~ lifT == liuf; the pore water pressure uf == Uo + ud(f)

For unconsolidated-undrained triaxial tests,

Major principal total stress == a3 + dat == al

Minor principal total stress == a3

The total stress Mohr's circle at failure can now be drawn, as shown in Fig­ure 1.30d. For saturated clays, the value of al - a3 == aaf is a constant, irrespectiveof the chamber confining pressure a3 (also shown in Figure 1.30d). The tangent tothese Mohr's circles will be a horizontal line, called the <p = 0 condition. The shearstress for this condition is

where Cu == undrained cohesion (or undrained shear strengtll)

The pore pressure developed in the soil specimen during the unconsolidated­undrained triaxial test is

where B == Skempton's pore pressllre parameter

Similarly, the pore parameter Ud is tlle result of the added axial stress /la, so

Ud == ~4~a (1.79)

where A ==- Skempton's pore pressure parameter

However,

Combining Eqs. (1.77), (1.78), (1.79), and (1.80) gives

U == Ua + Ud == BU3 + A(al - (3)

1.16 Unconfined Compression Test 51

The pore water pressure parameter B in soft saturated soils is unity, so

(1.82)

The value of the pore water pressure parameter A at failure will vary with the typeof soil. Following is a general range of the values of A at failure for various types ofclayey soil encountered in nature:

Type of soil A at failure

Sandy claysNormally consolidated claysOverconsolidated clays

0.5-0.70.5-1

-0.5-0

Unconfined Compression Test

The unconfined compression test (Figure 1.32a) is a special type of unconsolidated­undrained triaxial test in which the confining pressure U3 == 0, as shown in Fig­ure 1.32b. In this test, an axial stress flu is applied to the specimen to cause failure

Specimen

Shearstress

(a)Unconfinedcompressionstrength, qu

(c)

DegreeL...-. ---.. of

saturation

(b)

Tleu Total

~---------'----" normaltTl = t:.tTf stress

=qu

Figure 1.32 Unconfined compression test: (a) soil specimen; (b) Mohr's circle for thetest; (c) variation of qu with the degree of saturation

52 Chapter 1 Geotechnical Properties of Soil

(i.e., flu == !luf). The corresponding Mohr's circle is shown in Figure 1.32b. Notethat, for this case,

Major principal total stress == 110'f == qu

Minor principal total stress == 0

The axial stress at failure, AUf == qlO is generally referred to as thecompression strength. The shear strength of saturated clays under this condition(</J == 0), from Eq. (1.72), is

I

(1 I

The unconfined compression strength can be used as an indicator of the _"-"Jll...llU.LU" .....' ....... """

of clays.Unconfined compression tests are sometimes conducted on unsaturated soils.

With the void ratio of a soil specimen remaining constant, the unconfined compressionstrength rapidly decreases with the degree of saturation (Figure 1.32c). Figure 1.33shows an unconfined compression test in progress.

Figure 1.33 Unconfined conlpression test in progress (courtesy of Soiltest, Inc.,Lake Bluff, Illinois)

I

II

•ijII

·••~·I",',:,

1. 17 Comments on Friction Angle, 4J' 53

Comments on Friction Angle, <p'Effective Stress Friction Angle of Granular Soils

In general, the direct shear test yields a higher angle of friction compared with thatobtained by the triaxial test. Also, note that the failure envelope for a given soil isactually curved. The Mohr-Coulomb failure criterion defined by Eq. (1.72) is onlyan approximation. Because of the curved nature of the failure envelope, a soil testedat higher normal stress will yield a lower value of l/J'. An example of this relationshipis shown in Figure 1.34, which is a plot of ¢' versus the void ratio e for Chatta­choochee River sand near Atlanta, Georgia (Vesic, 1963). The friction angles shownwere obtained from triaxial tests. Note that, for a given value of e, the magnitude of¢' is about 4° to 5° smaller when the confining pressure 0"3 is greater than about70 kN/m2 (10 Ib/in2

), compared with that when (T3 < 70 kN/m2•

Effective S.tress Friction Angle of Cohesive Soils

Figure 1.35 shows the effective stress friction angle, <p', for several normally consoli­dated clays, obtained by conducting triaxial tests (Bjerrum and Simons, 1960). It can be

~(J)

...:s,--s- 40as

b1l§c:=o

"B:E(f)rf)<l)

.br.J)

(J)

~ 35 I------+-----+-~---.---~-+-------I------I~

~

45....---------.,.------r-------ro----..,..-----

e tan 1/ = 0.59 (70 kN/m2 (10 Ib/in2)

< lT3 < 550 kN/m2 (80 Ib/in2)

7 samples

1.21.11.00.9

Void ratio, e

0.80.7

30 "-- .l..- ........ ..L.- ..t..-~__.J...-_ __.,;:~

0.6

Figure 1.34 Variation of friction angle <p' with void ratio for ChattachoocheeRiver sand (after Vesic, 1963)

54 Chapter 1 Geotechnical Properties of Soil

• Scandinavian claysr + Clays from other countries

+ 1~r:4 +.~o Drained tests

---. . •. -. ~ --~~-+-+r----JL_ 0

0 l +

50

~ 40<1)

'"'t::J'-"

~30as

biJ~~ 20~0..0u·c 10~

oo 10 20 30 40 50 60 70

Plasticity index, PI

80 90 100

Figure 1.35 Varia­tion of friction angle¢' with plasticity in­dex for several clays(after Bjerrum andSimons, 1960)

(1.84)

seen from the figure that, in general, the friction angle cfJ' decreases with the increase inplasticity index. TIle value of cfJ' generally decreases from about 37°-38° with a tJ'..LII,.·<j."J'I'.-J1._

ity index of about 10 to about 25° or less with a plasticity index of about 100. Similar re­sults were also provided by Kenney (1959). The consolidated undrained friction angle(cfJ') of nornlally consolidated saturated clays generally ranges from 5°-20°.

The consolidated drained triaxial test was described in Section 1.15. Figure 1.36shows a schematic diagram of a plot of Jiu versus axial strain in a drained triaxial testflor a clay. At failure, for this test, f1CY = f1CYf. However, at large axial strain (i.e., theultimate strength condition), we have the following relationships:

Major principal stress: ul(ult) == U3 + Jiuult

Minor principal stress: CY3(ult) == CY3

At failure (i.e., peak strength), the relationship between CYI and CY3 is given byEq. (1.74). However, for ultimate strength, it can be shown that

, -, t 2 (45 + cfJ~)Ul(ult) - u3 an 2

where ¢; == residual effective stress friction angle

Figure 1.37 shows the general nature of the failure envelopes at peak strengthand ultimate strength (or residual strength). The residual shear strength of clays isilnportant in the evaluation of the long-term stability of new and existing slopesand the design of remedial measures. The effective stress residual friction angles 4»;of clays may be substantially smaller than the effective stress peak friction angle ¢'.Past research has shown that the clay fraction (i.e., the percent finer than 2 mi­crons) present in a given soil, CF, and the clay mineralogy are the two primary fac­t()rs that control </J;. The following is a summary of the effects of CF on </J;.

1. If CF is less than about 15%, then cfJ; is greater than about 25°.2. For CF > about 500/0, cP; is entirely governed by the sliding of clay minerals and

may' be in the range of about 10° to 15°.3. For kaolinite, illite, and montmorillonite, cfJ; is about 150, 100, and 50, respectively.

Illustrating these facts, Figure 1.38 shows the variation of 4>; with CF for several soils(Skempton, 1985).

I

I

.•.1··,

·,·······.·

.e

I

I

I"I:·:,'.,..•"

I

1.17 Comments on Friction Angle, cP' 55

Deviatorstress, !leT

lT3 == IT; == constant

L-- ~ Axial strain, cFigure 1.36 Plot of deviator stress vs.axial strain-drained triaxial test

Shear stress, 'T

Figure 1.37 Peak- and residual-strength envelopes for clay

40 _--.....,...-----r---...,-----,.---..,..-----r---..,..-----r---..,..---.....,

Sand Skempton (1985)

IT'-~1

Pa

Plasticity index, PI------'----- = 0.5 to 0.9Clay fraction, CF

----..,"'-

o " \--......... \

'" \\ \\ \

\ \\ a \\ \

\ "\ "-

\ ""0 0'" ' ..........." ........... '"00- __"'- o------~--'Q.. 0 0 00"'-......... a-----------0-----

•Kaolin

•Bentonite

1008060402001oooo-__....1- ....1-__---..1. ......... .......

oClay fraction, CF (%)

Figure 1.38 Variation of cP~ with CF (Note: Pa == atmospheric pressure)

56 Chapter 1 Geotechnical Properties of Soil

28

~24

'v"3---s" 20~

~~~ 16?

'.0'....I

:E~ 12,....~'I)1.1~ 8

4

Effectivenormal stress, 0-'

(kN/n12)

~ 100~ 400~ 700

o 100~ 400D 700

• 100... 400• 700

Clay sizefraction (CF)

(%)

~ 20

25 ~ CF ~ 45

~ 50

I

Figure 1.39 Variation of 4>; with liquid limit for some clays (after Stark, 1995) I

I300260220140 180

Liquid limit10060

O__~_~_......&....._--'--_.......&.. ----1__...1.0-_""",---""""'--""""'-"""""---_.---1

20

Figure 1.39 shows the variation of </J; with the liquid limit for some clays(Stark, 1995). It is important to note that

1. For a given clay, c/J; decreases with the increase in liquid limit.2. For a given liquid limit and given clay-size fractions present in the soil, the mag­

nitude of 0/; decreases with the increase in the normal effective stress, due to thecurvilinear nature of the failure envelope.

If.~

I

Correlations for Undrained Shear Strength, Cu :1;,

~Ille undrained shear strength, CU

' is an important parameter in the design of founda­tions. For normally consolidated clay deposits (Figure 1.40), the magnitude of Cu in­creases almost linearly with the increase in effective overburden pressure.

There are several empirical relations between Cu and the effective overburdenpressure (T~ in the field. Some of these relationships are summarized in Table 1.10.

Sensitivity

I'.','.itt" -

rOr luan)' naturally deposited clay soils, the unconfined compression strength is111 uch less when the soils are tested after remolding without any change in the mois­ture content. This property of clay soil is called sensitivity. The degree of sensitivity

'.·.···,'1·,,',·••'.",·.·,

.'C~

'1:)

-%

.~

1

I

1. 19 Sensitivity 57

y Groundwater table

Clay (T; == effective pressure

Cu = undrained shearstrength

Figure 1.40 Clay deposit

Table 1.10 Empirical Equations Related to Cu and Effective Overburden Pressure

Reference

Skempton (1957)

Chandler (1988)

Jamiolkowski et al.(1985)

Mesri (1989)

Bjerrum and Simons(1960)

Relationship

Cu(VST) == 0.11 + 0.0037PIu'o

PI == plasticity index (0/0 )cu(VST) == undrained shear strengthfrom vane shear test (See Chapter 3for details of vane shear test)

Cu(VST)-- ::::: 0.11 + O.0037PI

a'c(J"~ == preconsolidation pressure

Cu- == 0.23 ± 0.04a'c

C~ = 0.22(J"'o

c~ == f(LI)0-'o

LI = liquidity index[See Eq. (1.47) for definition]

Remarks

For normally consolidated clay

Canl be used for overconsolidated soilI

Accpracy ± 25 0/0Notlvalid for sensitive andfissired clays

For tightly overconsolidated claysI

See Figure 1.41 (p. 58) for normallyconsolidated clays

Ladd et al. (1977)

( ~ ) overconsolidated

(Cu

)

a~ normally consolidated

= (OCR)O.8

0-'OCR == overconsolidation ratio == _c

a'a

58 Chapter 1 Geotechnical Properties of Soil I0.4

0.3

Cu0.2

(T~

0.1

00 1 2 3

Liquidity index

4Figure 1.41 Variation of cu/(T~ with liq­uidity index (based on Bjerrum and Si­mons 1960)

I

I

is the ratio of the unconfined compression strength in an undisturbed state to that ina remolded state, or .1:·'·~

The sensitivity ratio of most clays ranges from about 1 to 8; however, highlyflocculent marine clay deposits may have sensitivity ratios ranging from about 10 to80. Some clays turn to viscous liquids upon remolding, and these clays are referredto as "quick" clays. The loss of strength of clay soils from remolding is caused pri­marily by the destruction of the clay particle structure that was developed duringthe original process of sedimentation.

d

I,j~'1

· .. ···.I····:··•.,·~·l.:.••.•'..

,

:).;_:

~,

(1.85)

A moist soil has a void ratio of 0.7. The moisture content of the soil is 12%.If Gs == 2.7, determine the soil's (a) porosity, (b) degree of saturation, and(c) dry unit weight in kN/m3

.

For the soil described in Problem 1.1,a. What would be the saturated unit weight in kN/m3?b. How much water, in kN/m3

, needs to be added to the soil for completesaturation?

c. What would be the moist unit weight, in kN/m3, when the degree of satu-

ration is 70% ?A soil specimen has a volume of 1.75 ft3 and a mass of 193 lb. Let w == 15%and Gs == 2.67. Determine the specimen's (a) void ratio, (b) porosity, (c) dryunit weight, (d) moist unit weight, and (e) degree of saturation.A saturated soil specimen has w == 360/0 and 'Yd == 13.5 kN/m3

• Determinethe specimen's (a) void ratio, (b) porosity, (c) specific gravity of soil solids, and(d),saturated unit weight (in kN/n13

).

The laboratory test results on a sand are as follows: emax == 0.91, emin == 0.48, andGs = 2.67. What would be the dry and moist unit weights of this sand (in Ib/ft3

)

when compacted at a moisture content of lOOk> to a relative density of 65 % ?

1.1

1.2

1.3

1.4

1.6

1.7

1.8

1.9

~J

1.12

Problems 59

For a granular soil, y == 17 kN/m3, Dr == 600/0, W == 80/0, and Gs == 2.66. If emin

is 0.4, what is emax? What is the dry unit weight of the soil in the loosest state?Laboratory test results on six soils are given in the following table:

Sieve analysis-percent passing

Soil

Sieve No. A B C D E F

4 92 100 100 95 100 10010 48 60 98 90 91 8240 28 41 82 79 80 74

200 13 33 72 64 30 55Liquid limit 31 38 56 35 43 35Plastic limit 26 25 31 26 29 21

Classify the soils by the AASHTO Soil Classification System and give thegroup indices.Classify the soils given in Problem 1.7 by the Unified Soil Classification Sys­tem. Give group symbols and group names.The hydraulic conductivity of a sand was tested in the laboratory at a void ratioof 0.56 and was determined to be 0.14 em/sec. Estimate the hydraulic conduc­tivity of this sand at a void ratio of 0.79 by using Eqs. (1.26), (1.27), and (1.28).The in situ hydraulic conductivity of a clay is 5.4 X 10-6 em/sec at a void ratioof 0.92. What would be the hydraulic conductivity of the clay at a void ratio of0.72? Use Eq. (1.30).From the soil profile shown in Figure Pl.ll, determine the total stress, porewater pressure, and effective stress at A, B, C, and D.A sandy soil (Gs = 2.66) has void ratios of 0.42 and 0.97 in its densest andloosest states, respectively. Estimate the range of the critical hydraulic gradi­ent in this soil at which it might turn into quicksand.

A

Figure P1.11

60 Chapter 1 Geotechnical Properties of Soil

1.13 A normally consolidated clay layer 2.6 m thick has a void ratio of 1.3. LetLL == 41 and the average effective stress on the clay layer == 82 kN/m2

• Howmuch consolidation settlement would the clay layer undergo if the average ef­fective stress on it is increased to 120 kN/m2 as the result of the constructionof a foundation?

1.14 Assume that the clay layer in Problem 1.13 is preconsolidated. Letu~ == 95 kNjm2 and Cs == lj4Cc Estimate the consolidation settlement.

1.15 Suppose the clay in Figure Pl.ll is normally consolidated. A laboratory con­solidation test on the clay gave the following results:

.

'.'.'.·'::.1""

:}<

Pressure (kNJm2)

100200

Void ratio

0.9050.815

a. Calculate the average effective stress on the clay layer.b. Determine the compression index Cc•

c. If the average effective stress on the clay layer is increased ((T~ + ~(T') to115 kNjm2, what would be the total consolidation settlenlent?

1.16 For the clay soil in Problem 1.15c, Cv == 5.6 mm2jrnin. How long will it takethe clay to reach half the consolidation settlement? (Note: The clay layer inthe field is drained on one side only.)

1.17 A clay specimen 1 in. thick (drained on top only) was tested in the laboratory.For a given load increment, the time for 600/0 consolidation was 6 min. Howlong will it take for 50 0k consolidation for a similar clay layer in the field thatis 8 ft thick and drained on both sides?

1.18 A total of 60 mm consolidation settlement is expected in the two clay layersshown in Figure PI.18, due to a surcharge of du. Find the duration of sur­charge application at which 30 mm of total settlement would take place.

_* t t * ! ArT ttL1mt ~ Groundwater table

::/t:~~,1f:g})~/:·:j~;}:\\~:::{}/N:W;·:·~\f~r:·,~r~.i~iV·::::;{f~;}:\\!:'::~\r

2m Clay

t Cv = 2mm2

jmin

Figure P1. 18

··I,.~.·.·.··.·t

··,·1:···;···.

,t

".1··,.·'~t~

:Vt/'1

1m ::A)·~?{f!.;}:?~·:~0!·~:t\:·;\t="\::~;

Clay2m

).?\;.,:\li;:~r\::-~?( \:~:\'.~~-~. ::~{{?~:; ~:.~::~::.~ ?t.:.:(:. ". . :., :., : : ".. , '.' ... '.. ' .' .

(a)

Figure P1.19

60

(b)

Problems 61

70 kN/m2

Time, days

The coefficient of consolidation of a clay for a given pressure range wasobtained as 8 X 10-3 mm2/sec on the basis of one-dimensional consolida­tion test results. In the field, there is a 2-m-thick layer of the same clay(Figure Pl.19a). Based on the assumption that a uniform surcharge of70 kN/m2 was to be applied instantaneously, the total consolidation settle­ment was estimated to be 150 mm. However, during construction, the loadingwas gradual; the resulting surcharge can be approximated as shown in Fig­ure P1.19b. Estimate the settlement at t == 30 and t == 120 days after the be­ginning of construction.A direct shear test was conducted on a specimen of dry sand of area2 in. X 2 in~ The results are as follows:

Normal force (lb)

3355.266.2

Shear force at failure (lb)

20.735.840.2

Draw a graph of shear stress at failure vs. normal stress, and determine thesoil friction angle t:/J'.

1.21 A consolidated-drain~d triaxial test on a normally consolidated clay yieldedthe following results:

All-around confining pressure == 0"3 == 115 kN/m2

Added axial stress at failure == flu == 230 kN/m2

Determine the shear stress parameter <p'.1.22 A consolidated-drained triaxial test on a normally consolidated clay yielded a

friction angle <p' of 28°. If the all-around confining pressure during the testwas 15 Ib/in.2

, what was the major principal stress at failure?

62 Chapter 1 Geotechnical Properties of Soil

1.23 Following are the results of two consolidated-drained triaxial tests on a

Test I: 0"3 == 82.8 kN/m2; O"l(failure) == 329.2 kN/m2

Test II: 0"3 == 165.6 kN/m2; O"l(failure) == 558.6 kN/m2

Determine the shear strength parameters c' and ¢'.1.24 :: A consolidated-undrained triaxial test was conducted on a normally consoli­

dated saturated clay. The following are the test results:

U3 == 13 lb/in.2;

0"1 (failure) == 32 lb/in.2

The pore water pressure at failure, Ut, == 5.5 lb/in.2

Determine c, ¢, c', and <p'.

American Society for Testing and Materials (2000). Annual Book ofASTM Standards,Vol. 04.08, Conshohocken, PA.

Bjerrunl, L., and Simons, N. E. (1960). "Conlparison of Shear Strength Characteristics of Nor­mally Consolidated Clay," Proceedings, Research Conference on Shear Strength of Co­hesive Soils, ASCE, 711-726.

Casagrande,A. (1936). "Determination of the Preconsolidation Load and Its Practical Signif­icance," Proceedings, First International Conference on Soil Mechanics and FoundationEngineering, Cambridge, MA, Vol. 3, pp. 60-64.

Chandler, R. 1. (1988). "The In Situ Measurement of the Undrained Shear Strength of ClaysUsing the Field Vane," STP 1014, Vane Shear Strength Testing in Soils: Field and Labo­ratory Studies, ASTM, pp. 13-44.

Darcy, H. (1856). Les Fontaines Publiques de la Ville de Dijon, Paris.Das, B. M. (2002). Soil Mechanics Laboratory Manual, 6th ed., Oxford University Press, New

York.Highway Research Board (1945). Report of the Committee on Classification of Materials for

Subgrades and Granular Type Roads, Vol. 25, pp. 375-388.Jamiolkowski, M., Ladd, C. C., Germaine, J. T., and Lancellotta, R. (1985). "New Develop­

ments in Field and Laboratory Testing of Soils," Proceedings, XI International Confer­ence on Soil Mechanics and Foundation Engineering, San Francisco, Vol. 1, pp. 57-153.

Kenney, T. C. (1959). "Discussion," Journal of the Soil Mechanics and Foundations Division,American Society of Civil Engineers, Vol. 85, No. SM3, pp. 67-69.

Ladd, C. C., Foote, R., Ishihara, K., Schlosser, F., and Poulos, H. G. (1977). "Stress Deforma­tion and Strength Characteristics," Proceedings, Ninth International Conference on SoilMechanics and Foundation Engineering, Tokyo, Vol. 2,421-494.

Mesri, G. (1989). "A Re-evaluation of Su(mob) ~ 0.220"p Using Laboratory Shear Tests," Cana­dian Geotechnical Journal, Vol. 26, No.1, pp. 162-164.

Olson, R. E. (1977). "Consolidation Under Time-Dependent Loading," Journal of Geotechni­cal Engineering, ASCE, VoL 103, No. GT1, pp. 55-60.

Samarasinghe, A. M., Huang, Y. H., and Drnevich, \Z ~ (1982). "Permeability and Consolida­tion of Normally Consolidated Soils," Journal of the Geotechnical Engineering Divi­sion, ASCE, VoL 108, No. GT6, 835-850.

Schmertmann, J. H. (1953). "Undisturbed Consolidation Behavior of Clay," Transactions,American Society of Civil Engineers, Vol. 120, p. 1201.

I

I~·:I_.,···i...;

I

:.'.'1·'.·

(~

'~I·····..

2

I

I

References 63

Sivaram, B., and Swamee, P. (1977). "A Computational Method for Consolidation Coeffi­cient," Soils and Foundations, Tokyo, Vol. 17, No.2, pp. 48-52.

Skempton, A. W. (1944). "Notes on the Compressibility of Clays," Quarterly Journal of Geo­logical Society, London, Vol. C, pp. 119-135.

Skempton, A. WI (1957). "The Planning and Design of New Hong Kong Airport," Proceed­ings, The Institute of Civil Engineers, London, Vol. 7, pp. 305-307.

Skempton, A. W. (1985). "Residual Strength of Clays in Landslides, Folded Strata, and theLaboratory," Geotechnique, Vol. 35, No.1, pp. 3-18.

Stark, T. D. (1995). "Measurement of Drained Residual Strength of Overconsolidated Clays,"Transportation Research Record No. 1479, National Research Council, Washington,DC, pp. 26-34.

Tavenas, F., Jean, ~, Leblond, P., and Leroueil, S. (1983). "The Permeability of Natural SoftClays. Part II: Permeability Characteristics," Canadian Geotechnical Journal, Vol. 20,No.4, pp. 645-660.

Terzaghi, K., and Peck, R. B. (1967). Soil Mechanics in Engineering Practice, Wiley, New York.U.S. Department of the Navy (1971). Design Manual-Soil Mechanics, Foundations and Earth

Structures, NAVFAC DM-7, U.S. Government Printing Office, Washington, DC.U.S. Department of the Navy (1982). Soil Mechanics, NAVFAC DM7.1, U.S. Government

Printing Office, Washington, DC.Vesic, A. S. (1963). "Bearing Capacity of Deep Foundations in Sand," Highway Research

Record No. 39, National Academy of Sciences, Washington, DC., pp.112-154.