chapter 5 - soil strength

CHAPTER 5

SHEAR STRENGTH OF SOIL

5.1 INTRODUCTION

The shear strength of soil is the shear resistance offered by the soil to overcomeapplied shear stresses. Shear strength is to soil as tensile strength to steel. Whenyou design a steel truss for a bridge, for example, you have to make sure thatthe tensile stress in any truss element is less than the tensile strength of steel, withsome safety factor. Similarly, in soil mechanics one has to make sure that the shearstress in any soil element underlying a shallow foundation, for example, is less thanthe shear strength of that particular soil, with some safety factor.

In soils the shear strength, f , is a function of the applied normal effective stress,. The MohrCoulomb failure criterion (discussed in the next section) provides arelationship between the two:

f = c + tan (5.1)

where c is the cohesion intercept of the soil and is the internal friction angle ofthe soil. These two parameters are termed the strength parameters of a soil. Theycan be obtained from laboratory and eld tests.

Consider the stability of the soil slope shown in Figure 5.1. Soil slopes usuallyfail in the manner shown in the gureat failure, there exists a failure surfacealong which the applied shear stress is equal to the shear strength of the soil. Letsassume that this particular soil has c = 10 kPa and = 30. Also, lets assumethat the applied normal effective stress and shear stress at point A are = 110 kPa

162

Applied Soil Mechanics: with ABAQUS Applications. Sam Helwany 2007 John Wiley & Sons, Inc. ISBN: 978-0-471-79107-2

DIRECT SHEAR TEST 163

Failure Surface

A

FIGURE 5.1 Shear strength concept.

and = 40 kPa, respectively. The shear strength offered by the soil at point A canbe calculated using (5.1):

f = c + tan = 10 + 110 tan 30 = 73.5 kPa

Note that at point A the applied shear stress is 40 kPa, which is less than theshear strength f = 73.5 kPa at the same point. This means that point A is not onthe verge of failure. The soil at point A will fail only when the applied shear stressis equal to the shear strength.

The shear strength of soil has to be determined accurately because it is crucialfor the design of many geotechnical structures, such as natural and human-madeslopes, retaining walls, and foundations (both shallow and deep). The shear strengthparameters can be measured in the eld using the vane shear test, for example. Theycan be obtained from correlations with the standard number N obtained from thestandard penetration test, or from correlations with the cone resistance obtainedfrom the cone penetration test. The shear strength parameters can also be measuredin the laboratory using direct shear and/or triaxial compression testing methodson undisturbed or reconstituted soil samples. These testing methods are describednext. Other laboratory shear tests are available primarily for research purposes,including simple shear tests and hollow cylinder triaxial tests.

5.2 DIRECT SHEAR TEST

To measure the frictional resistance between wood and steel, a wooden block anda steel block can be stacked vertically and placed on a rough surface as shown inFigure 5.2. A constant vertical load F is applied on top of the steel block. Then agradually increasing lateral load is applied to the steel block until it starts slidingagainst the wooden block. Sliding indicates that shear failure has occurred between

164 SHEAR STRENGTH OF SOIL

F

TA

Rough Surface

= F/A

= F/A

f = Tf /A

= T/A

= tan1 = tan1 (f /)

= T/A

(a) (b)

Steel

Wood

FIGURE 5.2 Shear test concept.

the two blocks. Thus, the applied shear stress at failure (f = Tf /A) is equal tothe shear resistance (or shear strength) between the blocks, where Tf is the appliedlateral force at failure and A is the cross-sectional area of the block. In referenceto Figure 5.2b, the friction factor between the two materials can be calculated as = tan = f /, where, is the friction angle between the two blocks and is the normal stress applied (= F/A). In this testing arrangement, termed directshear, the shear strength at a predetermined shear plane is measured at a constantnormal stress.

The direct shear test for soils uses the same concept as that illustrated above.The soil is placed in a split shear box that consists of two halves, as shownin Figure 5.3. The box has a square cross section and can accommodate a soil

T Soil

Actuator

Actuator

VerticalDisplacement

ShearDisplacement

F = constant

= F/A

= T/A

Shear Box

DisplacementTransducer

Disp. Transducer

FIGURE 5.3 Direct shear test apparatus.


specimen 10 cm 10 cm in plane and 2.5 cm in height. The upper half of theshear box is not allowed to move laterally, whereas the bottom half can slidelaterally by the action of a horizontal actuator. After placing the soil inside thebox, a loading plate is seated on top of the soil and a vertical load is appliedusing a vertical actuator. A gradually increasing lateral displacement (at a constantrate) is then applied to the bottom half of the box, via the horizontal actuator, togenerate shear stresses within the soil. The shear force, T , is measured using a loadcell attached to the piston of the horizontal actuator. The horizontal displacement(shear displacement) of the bottom half of the shear box is measured during shearingusing a horizontal displacement transducer. Also, the vertical displacement of theloading plate is measured during shearing using a vertical displacement transducer.Figure 5.4 is a photo of a direct shear apparatus.

The results of a direct shear test are plotted in the shear displacement versus shearstress plane such as the one shown in Figure 5.5a. The vertical displacement ofthe loading plate is plotted against the shear displacement as shown in Figure 5.5b.The test results shown in Figure 5.5a and 5.5b are typical for loose sands. Notethat the shear stress is calculated by dividing the measured shear force by thecross-sectional area of the soil specimen. Figure 5.5a shows that the shear stressincreases in a nonlinear manner as the shear displacement increases. The shear

FIGURE 5.4 Direct shear test apparatus. (Courtesy of Geocomp)


Compression

Dilation

Ver

tical

Disp

lace

men

t

Shear Displacement

Compression

Dilation

Ver

tical

Disp

lace

men

t

Shear Displacement

(b) (d )

Shear Displacement

f

Shear Displacement

f

Ultimate Strength Ultimate Strength

Peak Strength

(c)(a)

FIGURE 5.5 Typical direct shear test results on (a, b) loose sand and (c, d) dense sand.

stress approaches failure at = f , where f is the shear strength of the soil.Figure 5.5b shows a downward displacement of the loading plate, indicating soilcompression. Loose sand consists of loosely packed grains with large voids inbetween. During shearing, some of the voids in the shear zone and its vicinity willcollapse, causing the soil specimen to compress (downward plate displacement).

The behavior of dense sand during shearing is quite different from that of loosesand even though the two sands are assumed to be identical in terms of theirgradation and specic gravitythey differ only in relative density. In the sheardisplacement versus shear stress plane (Figure 5.5c) the dense sand exhibits greaterstrength, or peak strength, at an early stage during shearing. After reaching peakstrength, the shear stress decreases as the shear displacement increases until reach-ing an ultimate strength that is approximately the same as the ultimate strengthof the loose sand (Figure 5.5a). Figure 5.5d shows the direct shear test results fordense sand in the shear displacement versus vertical displacement plane. In thisgure, downward displacement of the loading plate indicates soil compression,and upward displacement of the plate indicates soil expansion (dilation). A pecu-liar characteristic is noted in the gure. At an early stage of shearing, the sandcompresses slightly and then starts to dilate until a later stage of shearing, whenit levels out as shown in the gure. Dense sand consists of densely packed grainswith small voids in between. During shearing, some of the grains will slide and rollon top of other densely packed particles in the shear zone and its vicinity, causingthe soil specimen to dilate (upward plate displacement).


The results of a direct shear test provide the shear strength (f ) of the soilat a specic normal stress (). The direct shear test is repeated several times onidentical soil specimens using different normal stresses. Typical direct shear testresults for sand using three different normal stresses are shown in Figure 5.6a. Theshear strength of the soil at different normal stresses can be determined from thegure as indicated by points 1, 2, and 3. The at-failure tests results (points 1, 2,

Shear Displacement (mm)

(kP

a)

1

2

3

(a)

= 10 kPa

= 30 kPa

= 50 kPa

00

10

20

30

40

50

1 2 3 4 5 6

37

(kPa)

(kP

a)

=

tan

MohrCoulomb Failure Criterion for Sand

1

2

3

(b)

0

10

20

30

40

50

0 10 20 30 40 50 60

FIGURE 5.6 Determination of the MohrCoulomb failure criterion for sand (direct sheartest).


and 3) are presented in the normal stress versus shear stress plane in Figure 5.6b.The three data points in Figure 5.6b are best tted with a straight line. This straightline is the MohrCoulomb failure criterion. The slope of this line is the internalfriction angle of the soil, , and its intercept with the shear stress axis is thecohesion intercept, c. The parameters c and are the strength parameters of thesoil. They are unique parameters for a given soil. For a sandy soil c is zero, thatis, the MohrCoulomb failure criterion passes through the origin, so f = tan .In Figure 5.6b the internal friction angle can be calculated from the slope of theMohr-Coulomb failure criterion: = tan1(f /) 37.

Figure 5.7a presents typical direct shear test results for clay under three differ-ent normal stresses. The MohrCoulomb failure criterion for this clay is shown inFigure 5.7b. The gure shows that this soil has a cohesion intercept of approxi-mately 9 kPa and an internal friction angle of approximately 26.5. With the helpof Figure 5.7b, one can dene the cohesion intercept (or cohesion) as the shearstrength of the soil at zero normal stress (or zero conning pressure). This meansthat clays have some shear strength even when they are not subjected to conningpressure. It also means that sands do not have any shear strength without conningpressure. That is why we cannot make shapes out of dry sand, although we cancertainly make shapes out of clay. For this reason sands and gravels are calledcohesionless, whereas clays are called cohesive.

Note that we used effective stresses in the discussion above when describing theMohrCoulomb failure criterion and the shear strength parameters of the soil. Thisis because the shear strength of soil is dependent on effective stresses rather thantotal stresses (recall the effective-stress principle). Also, this means that when weconduct a direct shear test on wet or saturated soils, we have to facilitate drainagewhile shearing the soil specimen to prevent the development of excess pore waterpressures in the soil. When the soil is saturated, the shear stress must be appliedvery slowly to prevent the development of excess pore water pressure. That waythe total stress is equal to the effective stress because the pore water pressure iskept equal to zero: = u = 0 = .

When measuring the shear strength of a soil using a direct shear test, one needsto duplicate the eld conditions of the soil being tested. Take, for example, thecase of the slope shown in Figure 5.1. To measure the shear strength at point A,we can estimate the normal stress at that point. This requires knowledge of thein situ unit weight of the soil and the location of the groundwater table. In the lab-oratory we can reconstitute a soil specimen in the direct shear box, aiming for thesame in situ soil density. The soil sample can be submerged in a water basin (notshown in Figure 5.3) to simulate eld conditions. A constant vertical stress equalto the normal stress calculated at point A is then applied to the soil specimen.The shearing stage should not start until equilibrium within the soil specimen isachieved. This means that if the soil is a clayey soil, we have to wait until theexcess pore water pressure generated as a result of stress application is dissipated.Shearing can then be applied until failure. The shear stress applied at failure shouldreect the true shear strength of the soil at point A.


1

2

3

(a)

= 10 kPa

= 30 kPa

= 50 kPa

(kP

a)

Shear Displacement (mm)0

0

10

20

40

30

50

60

1 2 3 4 5 6

c 9 kPa

(kPa)

(kP

a)

= c +

tan

MohrCoulomb Failure Criterion for Clay

(b)

1

2

3

00

10

20

40

30

50

60

10 20 30 40 50 60

26.5

FIGURE 5.7 Determination of the MohrCoulomb failure criterion for clay (direct sheartest).

Example 5.1 A dry sand sample is subjected to a normal stress = 20 kPa in adirect shear test (Figure 5.3). Calculate the shear force at failure if the soil sampleis 10 cm 10 cm in plane and 2.5 in height. The strength parameters of the sandare c = 0 and = 38.


=

1000 kPa

= 550

kPa

Rocky Wedge

Inclined Plane

FIGURE 5.8 Is the wedge going to slide?

SOLUTION: From the MohrCoulomb failure criterion we can calculate the shearstrength of the sand at = 20 kPa:

f = c + tan = 0 + 20 tan 38 = 15.6 kPaAt failure, the applied shear stress is equal to the shear strength of the soil (i.e.,

= f ); therefore, the shear force at failure is:T = A = 15.6 0.1 0.1 = 0.156 kN = 156 N.

Example 5.2 An intact rocky wedge is situated on an inclined plane as shown inFigure 5.8. Due to the self-weight of the wedge, a normal stress of 1000 kPa anda shear stress of 550 kPa are applied to the inclined plane. Determine the safety ofthe wedge against sliding given that the friction angle between the wedge and theinclined plane is 30.

SOLUTION: The shear strength offered at the wedgeplane interface can be cal-culated using the MohrCoulomb failure criterion with = 1000 kPa:

f = c + tan = 0 + 1000 tan 30 = 577.35 kPaTo avoid sliding, the applied shear stress must be less than the shear strengthoffered by the wedgeplane interface (i.e., < f ). Luckily, is 550 kPa, whichis smaller than f = 577.35 kPa. Therefore, the wedge is safe against sliding.

5.3 TRIAXIAL COMPRESSION TEST

The triaxial compression test is used to determine the shear strength of soil and todetermine the stressstrain behavior of the soil under different conning pressures.

TRIAXIAL COMPRESSION TEST 171

The test involves a cylindrical soil sample that is subjected to a uniform conningpressure from all sides and then subjected to an additional vertical load until failure.

Figure 5.9 shows the triaxial test apparatus schematically, and Figure 5.10 is aphoto of the apparatus. The cylindrical soil sample can have different dimensions.A typical triaxial soil specimen is 5 cm in diameter and 15 cm in height. Thesample is situated on top of the pedestal that is part of the base of the triaxialchamber, as shown in Figure 5.9. A loading plate is then placed on top of thespecimen. The pedestal, the soil specimen, and the top loading plate are carefullyenclosed in a thin rubber membrane. O-rings are used to prevent the conning uidfrom entering the soil specimen. Finally, the triaxial chamber is positioned on topof the base, the loading ram is lowered to the position shown in Figure 5.9, andthe triaxial chamber is lled with water.

As noted in Figure 5.9, two drainage tubes connect the top and bottom of thesoil specimen to the outside of the triaxial chamber. These tubes have valves thatare used to control drainage into the soil specimen. A third tube leads to the space

O-ring

Water

SoilSpecimen

RubberMembrane

TriaxialChamber

ConfiningPressure

DrainageLines

Pedestal

LoadingFrame

PorousStone

LoadingPlate

LoadingRam

LoadCell

Actuator

DisplacementTransducer(LVDT)

PorousStone

FIGURE 5.9 Triaxial test apparatus.


FIGURE 5.10 Triaxial test apparatus. (Courtesy of Geocomp)

inside the triaxial chamber, which is usually lled with water. This tube is used topressurize the conning uid (water) using pressurized air and a pressure regulator.

There are three types of triaxial compression tests: CD, CU, and UU. Eachtriaxial test consists of two stages: Stage I is the conditioning stage, during whichthe initial stress condition of the soil specimen is established; stage II is the shearingstage, during which a deviator stress is applied until failure. The designation of atriaxial test consists of two letters, the rst letter describes stage I and the seconddescribes stage II. Stage I can be either consolidated (C) or unconsolidated (U),and stage II can be either drained (D) or undrained (U). A triaxial CD test meansthat the soil specimen is allowed to consolidate in stage I of the triaxial test, andduring stage II the specimen is allowed to drain while being sheared. On the otherhand, a triaxial CU test means that the soil specimen is allowed to consolidatein stage I, and during stage II the specimen is not allowed to drain while beingsheared. Finally, the UU test means that the specimen is not allowed to consolidatein stage I and is not allowed to drain during shearing.

5.3.1 ConsolidatedDrained Triaxial Test

The consolidateddrained (CD) triaxial test is used to obtain the effective strengthparameters of soils. First, a soil specimen is saturated by circulating deaired water


through the specimen, from bottom to top, utilizing the two drainage tubes shownin Figure 5.9. After the specimen is fully saturated, the triaxial test is done intwo stages: a stress initialization stage and a shearing stage. In the rst stage aconning pressure is applied via the conning uid. Because the soil specimen isfully saturated, excess pore water pressure will be generated ( conning pressure).The soil specimen is allowed to consolidate by opening the two drainage valvesthroughout this stage. That will allow the excess pore water pressure to dissipategradually and the specimen to consolidate. The volume of the dissipated water canbe measured using a graduated ask. The volume of the dissipated water is equalto the volume change of the specimen because the specimen is fully saturated. Thevolumetric strain can be calculated by dividing the volume change by the initialvolume of the specimen. The consolidation curve for this stage can be plotted inthe time versus volumetric strain (v) plane, as shown in Figure 5.11.

In the shearing stage of the CD test a deviator stress d = 1 3 is appliedvery slowly while the drainage valves are opened, to ensure that no excess pore

3

33

3

3

3

Time

u

Time

Compression

v Time

Time

Consolidation

3

FIGURE 5.11 Stage I of a consolidated drained triaxial test: consolidation stage.


water pressure is generated. Consequently, the effective stresses are equal to thetotal stresses during this stage of the CD test. Because of its stringent loadingrequirements, the CD test may take days to carry out, making it an expensive test.Figure 5.12a shows stressstrain behavior typical of loose sand and normally con-solidated clay. Note that a is the axial strain and 1 3 is the deviator stress.The soil shows smooth nonlinear stressstress behavior reaching an ultimate shearstrength (1 3)f . Figure 5.12b shows the deformation behavior in the axial strain

Compression

Dilation

Compression

Dilation

a a

aavv

u = 0

d = 1 3

d = 1 3

3

(d )(b)

(c)(a)

3

3

3

(1 3)f

(1 3)f(1 3)ult

(1 3) (1 3)

FIGURE 5.12 Stage II of a consolidated drained triaxial test: shearing stage (a, b) loosesand; (c, d) dense sand.


(a) versus volumetric strain (v) plane typical of loose sand and normally consol-idated clay. In this gure the soil compresses as the shearing stress is increased.When the ultimate shear strength is approached, the curve levels out, indicatingthat the volumetric strain is constant (i.e., no further volume change). From themeasured deviator stress at failure (1 3)f , one can plot Mohrs circle, whichdescribes the stress state of the soil specimen at failure. Such a plot is shown inFigure 5.13. The MohrCoulomb failure criterion can be obtained by drawing aline that is tangent to Mohrs circle and passing through the origin. This is appli-cable to sands and normally consolidated clays since these soils have c = 0. Theslope of the MohrCoulomb failure criterion is the effective (or drained) frictionangle of the soil.

Figure 5.12c shows typical stressstrain behavior of dense sand and overcon-solidated clay in the axial strain (a) versus deviator stress plane. The soil showsnonlinear stressstress behavior reaching a peak shear strength (1 3)f at anearly stage of shearing. After the peak strength is reached, a sharp decrease instrength is noted. Then the stressstrain curve levels out, approaching an ultimatestrength (1 3)ult . Note that this ultimate strength is the same as the ultimatestrength of the loose sand if the two sands were identical (not in terms of theirrelative density). Figure 5.12d shows typical deformation behavior of dense sandand overconsolidated clay in the axial strain (a) versus volumetric strain (v)plane. Initially, the soil compresses as the shearing stress is increased. But shortly

= t

an

(kP

a)

(kPa)0

50

0 50 100

A

P

A

90

1 = 1

3 = 3

A

3 1

(1 3)/2

(1 + 3)/2

FIGURE 5.13 Determination of the MohrCoulomb failure criterion for sand (CD triaxialtest).


after that the soil starts expanding (dilating) as the axial strain increases. Whenthe ultimate shear strength is approached, the curve levels out, indicating that thevolumetric strain is constant.

We need only one CD test to determine the strength parameters of dense sand(because c = 0). From the peak deviator stress measured (1 3)f , we can plotMohrs circle, which describes the stress state of the soil specimen at failure asshown in Figure 5.13. The MohrCoulomb failure criterion can be obtained bydrawing a line that is tangent to Mohrs circle and passing through the origin. Theslope of the MohrCoulomb failure criterion is the effective (or drained) frictionangle of the soil.

For overconsolidated clays the cohesion intercept, c, is not equal to zero. There-fore, we will need to have the results of at least two CD triaxial tests on twoidentical specimens subjected to two different conning pressures. From the mea-sured peak deviator stress (1 3)f of these tests, we can plot Mohrs circles thatdescribe the stress states of the soil specimens at failure, as shown in Figure 5.14.The MohrCoulomb failure criterion in this case can be obtained by drawing aline that is tangent to the two Mohrs circles. The Mohr-Coulomb failure crite-rion will intersect with the shear stress axis at = c, as shown in Figure 5.14.This is the effective cohesion intercept of the overconsolidated clay. The slopeof the MohrCoulomb failure criterion is the effective friction angle of theoverconsolidated clay.

You recall from the strength of materials laboratory that when a concrete cylin-drical specimen was crushed between the jaws of a compression machine, thereexisted a failure plane making an angle with the horizontal. Soils exhibit sim-ilar behaviorthey also have a failure plane that occurs at failure in a triaxial

=c +

tan

c

(kPa)

(kP

a)

0

50

0 50

Test 1

Test 2

FIGURE 5.14 Determination of the MohrCoulomb failure criterion for overconsolidatedclay (CD triaxial test).


compression test. Let us calculate, both graphically and analytically, the orienta-tion of the failure plane. In reference to Figure 5.13, a sandy soil will have a zerocohesion intercept, and the MohrCoulomb failure criterion is tangent to Mohrscircle. The tangency point (point A in Figure 5.13) represents the stress state onthe failure plane at failure. The x and y coordinates of point A give, respectively,the normal and shear stresses exerted on the failure plane. We can measure thex and y coordinates at point A directly from the gure provided that our graphis drawn to scale. We can determine the failure plane orientation using the polemethod. For a triaxial compression test, the pole (the origin of planes) is locatedat point P in Figure 5.13. The orientation of the failure plane can be obtained byconnecting point P with point A. The orientation of the failure plane is the anglebetween line PA and the horizontal. We can use a protractor to measure the angle. This is the graphical solution.

The analytical solution for the angle can be obtained easily from Figure 5.13.Using simple trigonometry, you can show that

= 45 +

2(5.2)

As mentioned above, the x and y coordinates of point A give, respectively, thenormal stress (A) and shear stress (A) exerted on the failure plane at failure.Lets calculate those analytically. In reference to Figure 5.13 we can write

A =1 + 3

2

1 3

2sin (5.3)

A =1 3

2cos (5.4)

The friction angle can be calculated from

sin = 1 3

1 + 3(5.5)

Example 5.3 A CD triaxial compression test was conducted on a sand specimenusing a conning pressure of 42 kPa. Failure occurred at a deviator stress of 53 kPa.Calculate the normal and shear stresses on the failure plane at failure. Also calculatethe angle made by the failure plane with the horizontal. (a) Solve the problemgraphically and (b) conrm your solution analytically.

SOLUTION: (a) Graphical solution Given: 3f = 42 kPa and (1 3)f = 53kPa. We need to calculate the major principal stress in order to draw Mohrs circle:

1f = 3f + 53 kPa = 42 kPa + 53 kPa = 95 kPa

The radius of Mohrs circle is

(1 3)f2

= 532

= 26.5 kPa


The x-coordinate of the center of Mohrs circle is

(1 + 3)f2

= 95+422

= 68.5 kPa

Now we can plot Mohrs circle with its center located at (68.5 kPa, 0) and radius= 26.5 kPa, as shown in Figure 5.15.

Since the soil is sand, we have c = 0. The MohrCoulomb failure criterion isdrawn as a straight line passing through (0,0) and tangent to the circle as shown inthe gure. Using the same scale as that used in the drawing, we can now measurethe coordinates of the tangency point: The normal stress at point A is A 57 kPa,and the shear stress at point A is A 26 kPa. The orientation of the failure planecan be obtained by connecting pole P with point A. The orientation of the failureplane is the angle between line PA and the horizontal. Using a protractor, we canmeasure the angle 56.

(b) Analytical solution First, we calculate the friction angle of the sand using(5.5):

= sin1 1 3

1 + 3= sin1 53

95 + 42 = 22.75

Therefore,

= 45 +

2= 45 + 22.75

2= 56.38

(kPa)

(kP

a)

=

tan

= 22.75

= 56.380

50

0 50 100

A

P 68.5 kPa

1 = 95 kPa = 42 kPa3

= 42 kPa3

= 95 kPa1

90

A

A

A

A

FIGURE 5.15 Mohrs circle for a CD triaxial test on sand.


In reference to Figure 5.15, we can write

A =1 + 3

2

1 3

2sin = 95 + 42

2 95 42

2sin 22.75 = 58.25 kPa

A =1 3

2cos = 95 42

2cos 22.75 = 24.44 kPa

Note that there is a relatively good agreement between the graphical and analyticalsolutions. Solving problems graphically rst can reveal better ways of solving themanalytically. Also, sometimes it is more efcient to solve problems graphically.

Example 5.4 Three CD triaxial compression tests were conducted on three over-consolidated clay specimens using three conning pressures: 4, 20, and 35 kPa.Failure occurred at the deviator stresses of 19, 36 and 54 kPa, respectively. Deter-mine the shear strength parameters of the soil.

SOLUTION: Lets calculate the major principal stress at failure for each test. Thenlets determine the x-coordinate of the center of each Mohrs circle and its radius:

3f (kPa) (1 3)f (kPa) 1f (kPa) (1 + 3)f /2 (kPa) (1 3)f /2 (kPa)4 19 23 13.5 9.5

20 36 56 38 1835 54 89 62 27

Using the results from the fourth and fth columns, we can draw three Mohrscircles as shown in Figure 5.16. The failure envelope (MohrCoulomb failure crite-rion) is then established. The friction angle and the cohesion intercept are measuredfrom the gure as 23 and c 5 kPa, respectively.

NormalStress (kPa)

0

50

0 50 100Test 1

Test 3

Test 2

23kPa

4kPa

20kPa

56kPa

35kPa

89kPa

5 kPa

(kP

a)

23

FIGURE 5.16 CD triaxial compression tests on three overconsolidated clays.


5.3.2 ConsolidatedUndrained Triaxial Test

The consolidatedundrained (CU) triaxial test includes two stages: stage I, theconsolidation stage, and stage II, the shearing stage. In stage I the specimen isconsolidated under a constant conning pressure. The drainage valves are openedto facilitate consolidation. The volume change is measured and plotted againsttime as shown in Figure 5.17a. In stage II the soil specimen is sheared in anundrained condition. The undrained condition is realized by closing the drainagevalves, thus preventing the water from owing out of the sample or into the sample.The undrained condition makes it possible to apply the deviator stress in a muchfaster manner than in the consolidateddrained triaxial test. This makes the CU testmore economical than the CD test. During stage II there will be no volume changesince water is not allowed to leave the specimen. Therefore, the volumetric strain(v) is always equal to zero during this stage. Because the soil has the tendency tocompress (or dilate) during shearing, there will be increase (or decrease) in excesspore water pressure within the saturated specimen. The excess pore water pressureis measured using a pressure transducer connected to one of the drainage tubes.The presence of pore water pressure indicates that the total stress is different fromthe effective stress in a CU test.

(+)

()

a a

a a

Compression

v Time

(a)

(b) (d)

(c) (e)

1 3 1 3

(1 3)f

(1 3)f (1 3)ult

ud

(ud)f(+)

()

ud

(ud)f

FIGURE 5.17 (a) Stage I (consolidation) and (b e) stage II (shearing) of a consoli-datedundrained triaxial test (b, c) loose sand; (d, e) dense sand.


Figure 5.17b shows stressstrain behavior typical of loose sand and normallyconsolidated clay in the axial strain (a) versus deviator stress plane. The soilshows a smooth nonlinear stressstress behavior reaching an ultimate shear strength(1 3)f . Figure 5.17c shows the corresponding behavior in the axial strain (a)versus pore water pressure (ud) plane. In this gure the soil has a tendencyto compress as the shearing stress is increased. This causes a positive change inpore water pressure, as shown in the gure. When the ultimate shear strength isapproached, the pore water pressure levels out, approaching its maximum value atfailure (ud)f . From the deviator stress measured at failure (1 3)f , one canplot a Mohrs circle that describes the total stress state of the soil specimen atfailure. Such a plot is shown in Figure 5.18. Knowing the value of the pore waterpressure at failure (ud)f , one can calculate the effective principal stresses atfailure: 1f = 1f - (ud)f and 3f = 3f - (ud)f . The corresponding effectiveMohrs circle is shown in Figure 5.18. Note that the two circles have the samesize and that the effective-stress Mohrs circle can be obtained from the total stressMohrs circle if the latter is shifted to the left a distance equal to (ud)f . It followsthat there are two MohrCoulomb failure criteria: one for total stresses and one foreffective stresses. The slope of the effective-stress MohrCoulomb failure criterionis the effective (or drained) friction angle of the soil, whereas the slope of thetotal stress MohrCoulomb failure criterion is the consolidatedundrained frictionangle (without a prime) of the soil.

Figure 5.17d shows stressstrain behavior typical of dense sand and overcon-solidated clay in the axial strain (a) versus deviator stress plane. The soil showsnonlinear stressstrain behavior, reaching a peak shear strength (1 3)f at anearly stage of shearing. After the peak strength is reached, a sharp decrease instrength is noted. Then the stressstrain curve levels out, approaching an ultimatestrength (1 3)ult . Figure 5.17e shows the corresponding behavior in the axial

0

50

0 50 100

Effective Stress

Total Stress

NormalStress (kPa)

(kP

a)

(ud)f

= ta

n

=

tan

3 13 1

FIGURE 5.18 Total and effective-stress MohrCoulomb failure criteria from CU triaxialtest results: loose sand (or NC clay).


strain (a) versus pore water pressure (ud) plane. Initially, the pore water pressureincreases as the shearing stress is increased. But shortly after that the pore waterpressure starts to decrease, and then becomes negative (suction), as the shearingstress increases. When the ultimate shear strength is approached, the curve levelsout and the pore water pressure reach its maximum negative value.

We need only one CU test to determine the total and effective shear strengthparameters of a dense sand (because c = c = 0). From the peak deviator stressmeasured (1 3)f , we can plot the total stress Mohrs circle that describes thetotal stress state of the soil specimen at failure, as shown in Figure 5.19. Knowingthat the pore water pressure at failure is -(ud)f , we can calculate the effectivestresses at failure and plot the effective-stress Mohrs circle as shown in the samegure. Note that the effective-stress Mohrs circle has the same diameter as the totalstress Mohrs circle. Also note that the effective-stress circle results from the totalstress circle by a shift =(ud)f from left to right. The total stress MohrCoulombfailure criterion can be obtained by drawing a line that is tangent to the totalstress Mohrs circle and passing through the origin. The slope of the total stressMohrCoulomb failure criterion is the consolidatedundrained friction angle ofthe soil. The angle can be obtained as

sin = 1 31 + 3 (5.6)

Also, the effective-stress MohrCoulomb failure criterion for dense sand can beobtained by drawing a line tangent to the effective-stress Mohrs circle and passingthrough the origin. The slope of the effective-stress MohrCoulomb failure criterionis the effective (or drained) friction angle of the soil. Equation (5.5) can be usedto calculate the angle .

= tan

kP

a

0

50

0 50

(ud)f

100

= tan

Effective Stress

3 1

Total Stress

150 NormalStress (kPa)

3 1

FIGURE 5.19 Total and effective-stress MohrCoulomb failure criteria from CU triaxialtest results: dense sand.


For overconsolidated clays, the cohesion intercept, c, is not equal to zero. There-fore, we will need to have the results of at least two CU triaxial tests on two identicalspecimens subjected to two different conning pressures. From the measured peakdeviator stress (1 3)f of these tests we can plot Mohrs circles that describethe total stress state of the soil specimens at failure as shown in Figure 5.20. Thetotal stress MohrCoulomb failure criterion in this case can be obtained by drawinga line that is tangent to the two Mohrs circles. The total stress MohrCoulombfailure criterion will intersect with the shear stress axis at = c, as shown inFigure 5.20. This is the cohesion intercept of the overconsolidated clay. The slopeof the MohrCoulomb failure criterion, is the consolidated-undrained friction angle of the overconsolidated clay. Figure 5.20 also shows the effective-stress Mohrscircles for the two tests and the corresponding effective-stress Mohr-Coulomb fail-ure criterion, from which the effective shear strength parameters c and , can beobtained.

Example 5.5 A consolidatedundrained triaxial test was performed on a densesand specimen at a conning pressure 3 = 23 kPa. The consolidated undrainedfriction angle of the sand is = 37.5, and the effective friction angle is = 30.Calculate (a) the major principal stress at failure, 1f ; (b) the minor and the majoreffective principal stresses at failure, 3f and 1f ; and (c) the excess pore waterpressure at failure, (ud)f . Solve the problem graphically and analytically.

SOLUTION: Graphical solution (a) As shown in Figure 5.21a, we can draw theMohrCoulomb failure criteria for both total stresses and effective stresses. To doso we need the consolidated undrained friction angle = 37.5 and the effectivefriction angle = 30. Both criteria pass through the origin because c and c forsand are equal to zero. Next, we plot the conning pressure 3 = 23 kPa on the

= c+ ta

n

(kP

a)

0

50

0 50 100

= c+ ta

n Effective StressTotal Stress

Test 1

Test 2

c c

150

(ud)f

3 31 1

(ud)f

NormalStress (kPa)

3 3 < 1 c1

FIGURE 5.20 Total and effective-stress MohrCoulomb failure criteria from CU triaxialtest results: OC clay.


NormalStress (kPa)

(kP

a)

0

50

0 50 100

Effective Stress

Total Stress

150

(a)

0

50

0 50 100

Effective Stress

Total Stress

150

(ud)f = 13 kPa(b)

NormalStress (kPa)

(kP

a) = 37.5

= 37.5

' = 30

' = 30

3 = 36 kPa3 = 23 kPa

3 = 23 kPa

1= 95 kPa 1 = 108 kPa

FIGURE 5.21 Consolidated undrained triaxial test results of a dense sand.

normal stress axis as shown in Figure 5.21b. Now we can draw a circle (by trial anderror) that passes through 3 = 23 kPa and touches the total stress MohrCoulombfailure criterion. This is Mohrs circle, representing the total stresses at failure. Theintersection of the circle with the horizontal axis is the major principal stress atfailure. We can measure the location of this intersection to get 1f 95 kPa.

(b) To determine the minor and major effective principal stresses at failure (3fand 1f , respectively), we can draw an effective-stress Mohrs circle that touchesthe effectivestress MohrCoulomb failure criterion as shown in Figure 5.21b.This effective-stress Mohrs circle has the same diameter as the total stress Mohrscircle. The intersections of the effectivestress Mohrs circle with the horizontalaxis can be measured from the gure to give us 3f 36 kPa and 1f 108 kPa.


(c) From Figure 5.21b we can measure the distance between 1f and 1f (orbetween 3f and 3f ) to estimate (ud)f , which is approximately 13 kPa (neg-ative = suction).

Analytical solution (a) From (5.6) we have

sin = sin 37.5 = 1f 3f1f + 3f =

1f 231f + 23 1f = 95 kPa

(b) The effective-stress Mohrs circle has the same diameter as the total stressMohrs circle:

1f 3f = 1f 3f = 95 23 = 72 kPa

From (5.5),

sin = 1f 3f

1f + 3f 1f + 3f =

1f 3fsin

= 72 kPasin 30

= 144 kPa

or 1f + 3f = 144 kPa, but 1f 3f = 72 kPa. Solving these two equationssimultaneously, we get 1f = 108 kPa and 3f = 36 kPa.

(c) From the effectivestress equation and in reference to Figure 5.21b,

(ud)f = 1f 1f = 95 kPa 108 kPa = 13 kPa

Note that the graphical solution is done rst to gain some insight into the prob-lem. Doing so makes it easier to seek an analytical solution. Also note that thegraphical solution is very simple and generally yields accurate answers if the draw-ing is done correctly and to scale, as illustrated above.

5.3.3 UnconsolidatedUndrained Triaxial Test

The unconsolidatedundrained (UU) triaxial test is usually performed on undis-turbed saturated samples of ne-grained soils (clay and silt) to measure theirundrained shear strength, cu. The soil specimen is not allowed to consolidate instage I under the conning pressure applied. It is also not allowed to drain duringshearing in stage II. Identical soil specimens exhibit the same shear strength underdifferent conning pressures, as indicated in Figure 5.22. It seems that applyingmore conning pressure to the soil specimen does not cause any increase in its shearstrength! This can be explained as follows: When a fully saturated soil specimen issubjected to additional conning pressure (total stress), it generates an equal excesspore water pressure, which means that the additional connement does not causeadditional effective conning pressure. The effective-stress principle indicates thatthe shear strength of the soil specimen depends on the effective conning pressure.


cu

Test 1 Test 2

MohrCoulomb Failure Criterion: f = cu

TotalStress

TotalStress

3f 1f 3f 1f

u = 0

FIGURE 5.22 Typical results of an unconsolidated undrained triaxial test.

Therefore, since there was no increase in the effective conning pressure, therewould be no increase in shear strength.

As shown in Figure 5.22, the MohrCoulomb failure criterion is horizontal(u = 0) and it intersects the vertical axis at = cu. Note that cu is the undrainedshear strength of a soil and is equal to the radius of the total stress Mohrs circle,[i.e., cu = 1f 3f2 ]. The undrained shear strength is used appropriately to describethe strength of ne-grained soils subjected to rapid loading, during which drainageis not allowed; therefore, no dissipation of excess pore water pressures is possible.Examples of that include rapid construction of embankments on clay deposits orrapid loading of shallow foundations constructed on clay.

5.3.4 Unconned Compression Test

The unconned compression test is performed on unconned cylindrical speci-men of a cohesive soil to measure its unconned compression strength, qu. Theundrained shear strength, cu, is equal to one-half of the unconned compressionstrength, qu, as indicated in Figure 5.23. This test is a special case of the uncon-solidatedundrained triaxial test when performed without applying any conningpressure. The undrained shear strength obtained by the two tests for the samecohesive soil is theoretically identical. The unconned compression test is gener-ally performed on undisturbed specimens of cohesive soils at their natural watercontents. It is not possible to perform this test on cohesionless soils since they donot have shear strength at zero connement.

5.4 FIELD TESTS

The following is a brief description of eld tests that can be used to measure thein situ shear strength of soil.

FIELD TESTS 187

cuMohrCoulomb Failure Criterion: f = cu = qu/2

3 = 0

3 = 0

1 = qu

1 = qu

u = 0

FIGURE 5.23 Typical results of an unconned compression test.

5.4.1 Field Vane Shear Test

The vane shear test consists of a four-bladed vane that is pushed into the undisturbedcohesive soil at the bottom of a borehole and rotated to determine the torsionalforce required to cause a cylindrical soil specimen to be sheared by the vane.The torsional force is then correlated to the undrained shear strength, cu. This testprovides a direct and reliable measurement of the in situ undrained shear strength.A smaller handheld version of the vane shear test can be used to measure theundrained shear strength of samples recovered from a test boring. This is done byinserting the shear vane into the soil sample and twisting until failure. The undrainedshear strength of the soil is captured by an indicator mounted on the apparatus.

5.4.2 Cone Penetration Test

The cone penetration test (CPT) consists of a cylindrical probe with a cone tipthat is pushed continuously into the ground at a slow rate. The probe is instru-mented with strain gauges to measure the tip and side resistance while the probe isadvancing into the ground. The data are gathered continuously using a computer.The measured tip resistance is correlated with the undrained shear strength of thesoil at various depths.

5.4.3 Standard Penetration Test

The standard penetration test (SPT) consists of driving a sampler (a thin hollowcylinder) into the bottom of a borehole using a standard hammer dropped from a


standard distance. The number of blows required to drive the sampler a distanceof 30 cm into the ground is dened as the SPT blow count (N ). The blow count iscorrelated to the undrained shear strength for cohesive soils and the friction anglefor granular soils.

5.5 DRAINED AND UNDRAINED LOADING CONDITIONS VIA FEM

When saturated coarse-grained soils (sand and gravel) are loaded slowly, volumechanges occur, resulting in excess pore pressures that dissipate rapidly, due to highpermeability. This is called drained loading. On the other hand, when ne-grainedsoils (silts and clays) are loaded, they generate excess pore pressures that remainentrapped inside the pores because these soils have very low permeabilities. Thisis called undrained loading.

Both drained and undrained conditions can be investigated in a laboratory triax-ial test setup. Consider a soil specimen in a consolidateddrained (CD) triaxial test.During the rst stage of the triaxial test (Figure 5.11) the specimen is subjected toa constant conning stress, 3, and allowed to consolidate by opening the drainagevalves. In the second stage of the CD triaxial test (Figure 5.12) the specimen is sub-jected, by means of the loading ram, to a monotonically increasing deviator stress,1 3, while the valves are kept open. The deviator stress is applied very slowly toensure that no excess pore water pressure is generated during this stagehence theterm drained. Typical CD test results at failure can be presented using Mohrs circleas shown in Figures 5.13 and 5.14. The drained (or long-term) strength parametersof a soil, c and , can be obtained from the MohrCoulomb failure criterion asindicated in the gures. Note that c = 0 for sands and normally consolidated clays.These parameters must be used in drained (long-term) analysis of soils.

Now lets consider a soil specimen in a consolidatedundrained (CU) triaxial test(Figure 5.17). The rst stage of this test is the same as the CD testthe specimen issubjected to a constant conning stress, 3, and allowed to consolidate by openingthe drainage valves. In the second stage of the CU test, however, the specimen issubjected to a monotonically increasing deviator stress, 1 3, while the drainagevalves are closedhence the term undrained. The undrained condition means thatthere will be no volumetric change in the soil specimen (i.e., volume remainsconstant). It also means that excess pore water pressure will be developed inside thesoil specimen throughout the test. Measurement of the pore water pressure allowsfor effective stress calculations. Typical CU test results (at failure) for a normallyconsolidated clay can be presented using Mohrs circles as shown in Figure 5.24.The undrained (or short-term) strength parameter of a soil, cu, is the radius ofthe total stress Mohrs circle (note that u = 0). The parameter cu is termed theundrained shear strength of the soil and must be used in undrained (short-term)analysis of ne-grained soils. The effective-stress MohrCoulomb failure criterioncan be used to estimate the drained (or long-term) strength parameters c and ,as shown in Figure 5.24 (c = 0 for NC clay).

In a nite element scheme, the drained (or long-term) behavior of a soil can besimulated using coupled analysis, where the pore water pressure is calculated for a

DRAINED AND UNDRAINED LOADING CONDITIONS VIA FEM 189

(kP

a)

0

50

0 50 100

Effective Stress

Total Stress

NormalStress (kPa)

cu

13

u = 0

(ud)f

= cu

3 1

FIGURE 5.24 Drained and undrained strength parameters.

given load increment in each soil element and then subtracted from the total stressesto estimate the effective stresses in the element. These effective stresses control thedeformation and shear strength of the soil element according to the effective-stressprinciple. Constitutive models such as the cap and Cam clay models can be usedwithin the nite element framework to determine the deformation caused by theseeffective stresses.

Four main measures must be considered for a successful nite element analysisof soils considering their long-term (drained) behavior: (1) the initial conditions ofthe soil strata (initial geostatic stresses, initial pore water pressures, and initial voidratios), which will determine the initial stiffness and strength of the soil strata,must be estimated carefully and implemented in the analysis; (2) the boundaryconditions must be dened carefully: pervious or impervious; (3) the long-termstrength parameters of the soil must be used in an appropriate soil model; and(4) loads must be applied very slowly to avoid the generation of excess pore waterpressure throughout the analysis.

For undrained (or short-term) analyses, the aforementioned measures apply withthe exception of the last measurethe load can be applied very fast instead. Thisis one of the most attractive aspects of coupled analysis. Drained and undrainedanalyses (Examples 5.6 and 5.7, respectively) differ only in the way we apply theload: Very slow loading allows the excess pore water pressure generated to dissipateand the long-term-strength parameters to be mobilized, whereas fast loading doesnot allow enough time for the pore water pressure to dissipate, thus invoking theshort-term strength of the soil. This means that there is no need to input the short-term-strength parameters because the constitutive model will react to fast loadingin an undrained manner.

As an alternative method, undrained analysis can utilize the undrained shearstrength parameter, cu, directly in conjunction with a suitable constitutive model.If the cap model is used, the cap parameters and d can be calculated using cu


and u(= 0). This procedure can be used only for undrained (short-term) analy-sis. Unlike the preceding procedure, this alternative procedure cannot be used fordrained analysis and for cases falling between the drained and the undrained cases(partially drained).

Example 5.6 Hypothetical consolidated drained and consolidated undrained tri-axial test results for a clayey soil are given in Figures 5.25 and 5.26, respectively.The isotropic consolidation curve for the same soil is given in Figure 5.27. Usingthe nite element method, simulate the consolidateddrained triaxial behavior ofthis soil when subjected to a 210-kPa conning pressure. Consider a cylindricalsoil specimen with D = 5 cm and H = 5 cm as shown in Figure 5.28. The top andbottom surfaces of the specimen are open and permeable. The loading plate andthe pedestal are smooth; therefore, their interfaces with the clay specimen can beassumed frictionless. The clay specimen is normally consolidated and has a con-stant permeability k = 0.025 m/s and initial void ratio e0 = 0.889, correspondingto its 210-kPa conning pressure. Assume that the clay is elastoplastic, obeyingthe extended Cam clay model (Chapter 2).

SOLUTION(lename: Chapter5 Example6.cae): Lets determine the Cam claymodel parameters from the tests results provided in Figures 5.25 to 5.27. Fromthe elog p curve shown in Figure 5.27, we can determine the values of thecompression index Cc = 0.4 and the swelling index Cs = 0.06. Note that the mean

0.00 0.10 0.20 0.30 0.40 0.500

100

200

300

Dev

iato

r Stre

ss (k

Pa) 3 = 70 kPa

3 = 140 kPa

3 = 210 kPa

0.00 0.10 0.20 0.30 0.40 0.50Axial Strain

0.000.020.040.060.080.10V

olu

met

ric S

train

FIGURE 5.25 Hypothetical consolidated drained triaxial test results for a clayey soil.


0.00 0.10 0.20 0.30Axial Strain ( 101)

0

25

50

75

100

125

Dev

iato

r Stre

ss (k

Pa)

0.00 0.10 0.20 0.30Axial Strain ( 101)

0

25

50

75

100

125

Exce

ss P

ore

Wat

er P

ress

ure

(kPa)

3 = 70 kPa

3 = 140 kPa

3 = 240 kPa

FIGURE 5.26 Hypothetical consolidated undrained triaxial test results for a clayey soil.

10 100 1000Mean Effective Stress (kPa)

0.70

0.80

0.90

1.00

1.10

1.20

Voi

d Ra

tio

Cc = 0.4

Cs = 0.06

FIGURE 5.27 Hypothetical Isotropic consolidation test results for a clayey soil.


Axis of Symmetry

AxisymmetricFinite Element

Mesh

5 cm

5 cm2.5 cm

2.5 cm

x

z

FIGURE 5.28 Axisymmetric nite element mesh (one element).

effective stress in this gure is dened as p = (1 + 23)/3 (p = 3 for isotropicconsolidation). The parameter denes the elastic behavior of the soil in theCam clay model, and it is related to the swelling index through the equation =Cs/2.3 = 0.06/2.3 = 0.026. The parameter is related to the compression indexthrough = Cc/2.3 = 0.4/2.3 = 0.174.

The effective-stress Mohrs circles for the three consolidated drained triaxialtests and the three consolidated undrained tests are given in Figure 5.29. Theeffective-stress MohrCoulomb failure criterion is a straight line that is tangentto the circles as shown in the gure. The slope of this line is the effective frictionangle of the soil = 25.4. The Cam clay strength parameter M is related to the

= 25.4

(kPa)

(kP

a)

0

250

0 250 500

FIGURE 5.29 Effective-stress MohrCoulomb failure criterion for a clayey soil.


internal friction angle of the soil, , as follows:

M = 6 sin

3 sin (5.7)

Therefore,

M = 6 sin 25.4

3 sin 25.4 = 1

An alternative procedure for determining M is to plot the at-failure stresses of allsix triaxial tests in the p = (1f + 23f )/3 versus q = 1f 3f plane as shownin Figure 5.30. The data points are best tted with a straight line called the critical-state line. The slope of the critical-state line shown in the gure is the Cam claystrength parameter M .

In the Cam clay model, the yield surface size is fully described by the parameterp = (1 + 23)/3. The evolution of the yield surface is governed by the volumetricplastic strain pvol, which is a function of p. The relationship between

p

vol and pcan be deduced easily from an elog p line. The consolidation curve (the elogp line) is dened completely by its slope Cc (= 2.3), and the initial conditionsp0 (= 0) and e0. Note that , 0, and e0 are part of the input parameters requiredin the nite element program used herein.

The preconsolidation pressure, c, is also a required parameter. It is specied bythe size of the initial yield surface as shown in Table 5.1. Consider the initial yieldsurface that corresponds to p0 = 210 kPa (or p0/2 = 105 kPa), where p0 = 210 kPais equal to the conning pressure 3. In this example, c is assumed to be equalto the conning pressure, indicating that the clay is normally consolidated. If theclay were overconsolidated, its c can be set equal to its preconsolidation pressure,

0 100 200 300 4000

100

200

300

400

Undrained Test

Drained Test

M = 1

1

p = ( + 2 )/3 kPa1 3

q=

(

1

3)

kPa

FIGURE 5.30 Critical-state line for a clayey soil.


TABLE 5.1 Cam Clay Model Parameters

General Plasticity

(kg/m3) 1923 0.174k (m/sec) 0.025 Stress ratio, M 1w (kN/m3) 9.81 Initial yield surfacee0 0.889 size = p0/2 (kPa) 105

Elasticity Wet yield surface 0.026 size 1 0.28 Flow stress rate 1

which is greater than its conning pressure. A summary of the Cam clay modelparameters is given in Table 5.1.

A two-dimensional axisymmetric mesh is used with one element only, as shownin Figure 5.28. The element chosen is a pore uid/stress eight-node axisymmet-ric quadrilateral element with biquadratic displacement, bilinear pore pressure, andreduced integration. The boundary conditions of the nite element mesh shown inFigure 5.28 are as follows. On the bottom side, the vertical component of displace-ment is xed (uz = 0). The left-hand side of the mesh is a symmetry line (ux = 0).On the top surface a uniform downward displacement of 0.5 cm is applied veryslowly. During load application the top surface of the nite element mesh is madepervious. This means that the top and bottom of the soil specimen are allowed todrain due to symmetry about a plane passing through the midheight of the soilcylinder.

Similar to an actual consolidateddrained triaxial test, this nite element anal-ysis is carried out in two steps: a consolidation step and a shearing step. The rststep is a single increment of analysis with drainage allowed across the top surface.In this step the conning pressure of 210 kPa is applied at the top surface andthe right side of the mesh. During step 1, the geostatic command is invokedto make sure that equilibrium is satised within the soil specimen. The geostaticoption makes sure that the initial stress condition in the clay specimen falls withinthe initial yield surface of the Cam clay model. Step 2 is the shearing step withduration of 109 seconds. In this step the loading plate is forced to displace down-ward at a very small rate (5 1010 cm/s). This low rate of displacement is used toensure that the excess pore water pressure within the clay specimen is always zero.Automatic time stepping with a maximum pore water pressure change of 0.007 kPaper time increment is used. This procedure is useful for loading steps with verylong duration. When the loading starts in the beginning of the step, the rate of porewater pressure change is high, therefore, small time increments are used. Later,when the rate of pore water pressure change decreases, larger time increments areused. As an alternative, one can use a shorter duration for this loading step withxed-time stepping instead. However, we need to make sure that the excess porewater pressure during this step is kept equal to zero. This can be done easily byplotting the pore water pressure at the center of the soil specimen as a function oftime. If not zero, the duration of the shearing step can be increased.


0.00 0.10 0.20 0.30 0.400

50

100

150

200

250

300

Dev

iato

r Stre

ss (k

Pa) FEM-Cam Clay

FEM-Cap

Hypothetical CD Tests

0.00 0.10 0.20 0.30 0.40Axial Strain

0.000.020.040.060.080.10

Vol

umet

ric S

train

3 = 210 kPa

3 = 210 kPa

FIGURE 5.31 Comparison between hypothetical CD triaxial test results and nite elementprediction using Cam clay and cap models.

The calculated consolidateddrained stressstrain behavior of the clay specimenis compared with the hypothetical test data as shown in Figure 5.31. The axialstrain versus volumetric strain is also compared to the experimental data in thesame gure. Excellent agreement is noted in the gure. This means that the Camclay model embodied in the nite element method is capable of describing theconsolidateddrained triaxial behavior of normally consolidated clays.

Example 5.7 Using the same hypothetical test data and assumptions given inExample 5.6, predict the consolidatedundrained triaxial behavior of a triaxial soilspecimen subjected to a 210-kPa conning pressure.

SOLUTION(lename: Chapter5 Example7.cae): As described in Example 5.6, wehave estimated the parameters of the Cam clay model from the hypothetical testdata provided. Those parameters are used in this example as well. The solutionof this problem is identical to the solution of Example 5.6 with two exceptions:(1) the top and bottom of the soil specimen are made impervious (undrained), and(2) the shearing load is applied faster.

We will use the same nite element mesh (Figure 5.28) that was used in Example5.6. Similar to an actual consolidatedundrained triaxial test, this nite elementanalysis is carried out in two steps: a consolidation step and a shearing step. Inthe rst step, the consolidation step, an all-around conning pressure of 210 kPa


is applied while drainage is permitted across the top surface. During this step,the geostatic command is invoked to make sure that equilibrium is satisedwithin the clay specimen. The geostatic option makes sure that the initial stresscondition in the clay specimen falls within the initial yield surface of the Cam claymodel.

The second step, the undrained shearing step, has a duration of 100 seconds.In the beginning of this step the pervious boundary at the top surface is removed,making the top surface impervious by default. In this step the loading plate isforced to displace downward a distance of 0.127 cm at a displacement rate of1.27 103 cm/s. This rate of displacement, along with the impervious boundaries,will cause the excess pore water pressure to generate during shearing. Automatictime stepping with a maximum pore water pressure change of 0.7 kPa per timeincrement is used.

The consolidatedundrained stressstrain behavior calculated for the clay speci-men is compared with the hypothetical test data as shown in Figure 5.32. The axialstrain versus pore water pressure is also compared to the hypothetical test data inthe same gure. Excellent agreement is noted in the gure.

0.00 0.10 0.20 0.30 0.40 0.50Axial Strain ( 101)

Axial Strain ( 101)

0

30

60

90

120

150

Dev

iato

r Stre

ss (k

Pa) FEM-Cam Clay

FEM-Cap

Hypothetical CU Tests

0.00 0.10 0.20 0.30 0.40 0.500

30

60

90

120

150

Exce

ss P

ore

Wat

er P

ress

ure

(kPa)

3 = 210 kPa

3 = 210 kPa

FIGURE 5.32 Comparison between hypothetical CU triaxial test results and nite elementprediction using Cam clay and cap models.


Example 5.8 Assuming that the clayey soil described in Example 5.6 is elasto-plastic, obeying the cap model (Chapter 2), predict the (a) consolidateddrained and(b) consolidatedundrained triaxial behavior of the soil when subjected to a 210-kPaconning pressure. Consider a cylindrical specimen with D = 5 cm and H = 5 cmas shown in Figure 5.28. The loading plate and the pedestal are smooth; therefore,their interfaces with the clay specimen can be assumed frictionless. The clay spec-imen is normally consolidated and has a constant permeability k = 0.025 m/s andinitial void ratio e0 = 0.889, corresponding to its 210-kPa conning pressure.SOLUTION: (a) Consolidateddrained triaxial condition (lename: Chapter5Example8a.cae) We need to determine the cap model parameters from the hypo-thetical test results provided in Figures 5.25 to 5.27. We use the same nite elementmesh (Figure 5.28) that was used in Example 5.6.

Six effective-stress Mohrs circles corresponding to failure stresses obtainedfrom the hypothetical triaxial test results (Figures 5.25 and 5.26) are plotted inFigure 5.29. Subsequently, the effective-stress MohrCoulomb failure criterion isplotted as a straight line that is tangential to the six circles. The soil strengthparameters = 25.4 and c = 0 kPa are obtained from the slope and intercept ofthe Mohr-Coulomb failure criterion.

For triaxial stress conditions, the MohrCoulomb parameters ( = 25.4 andc = 0 kPa) can be converted to DruckerPrager parameters (Chapter 2) as follows:

tan = 6 sin

3 sin for = 25.4 = 45

d = 18c cos

3 sin for c = 0 d = 0

An alternative procedure for determining and d is to plot the at-failure stressesof all six triaxial tests in the p = (1f + 23f )/3 versus q = 1f 3f plane asshown in Figure 5.33. The data points are best tted with a straight line whoseslope is equal to tan = 1; thus, = 45. The line intersects with the verticalaxis at d = 0. The cap eccentricity parameter is chosen as R = 1.2. The initialcap position, which measures the initial consolidation of the specimen, is taken as

plvol(0) = 0.0, which corresponds to p = 210 kPa. The transition surface parameter

= 0.05 is used. These parameters are summarized in Table 5.2.The cap hardening curve shown in Figure 5.34 was obtained from the isotropic

consolidation test results shown in Figure 5.27. From Figure 5.27 we can calculatethe plastic volumetric strain as

pv = 1 + e0 ln

p

p0= Cc Cs

2.3(1 + e0) lnp

p0

For this clayey soil we have = 0.174, = 0.026, p0 = 210 kPa, and e0 = 0.889;therefore,

pv =0.174 0.026

1 + 0.889 lnp

210= 0.07834 ln p

210


0 100 200 300 4000

100

200

300

400

Undrained Test

Drained Test

d = 0

p = ( + 2 )/3 (kPa)1 3

= 45

q=

(

1

3)

kPa

FIGURE 5.33 Determination of the cap model parameters d and for a clayey soil.

TABLE 5.2 Cap Model Parameters

General Plasticity

(kg/m3) 1923 d 0e0 0.889 (deg) 45

Elasticity R 1.2E (MPa) 182 Initial yield 0.0 0.28 0.05

K 1

0.00 0.03 0.06 0.09 0.12 0.15Plastic Volumetric Strain

0

300

600

900

1200

1500

Mea

n Ef

fect

ive

Stre

ss (k

Pa)

FIGURE 5.34 Determination of the cap model hardening curve.


This equation describes the evolution of the plastic volumetric strain (the hard-ening parameter) with the mean effective stress (Figure 5.34).

We will use the cap model parameters above to reproduce the stressstraincurves of the clayey soil subjected to a conning pressure of 210 kPa. This isdone by using one axisymmetric nite element in the same manner described inExample 5.6. The calculated consolidateddrained stressstrain behavior of theclay specimen is compared with the hypothetical test data as shown in Figure 5.31.The axial strain versus volumetric strain is also compared to the hypothetical testdata in the same gure. It can be noted from Figure 5.31 that the cap model canadequately simulate the behavior of the clayey soil on the elemental level.

(b) Consolidatedundrained triaxial condition (lename:Chapter5 Example8b.cae) As described above, we have estimated the parameters of the cap modelfrom the hypothetical test data. Those parameters will also be used here to predictthe consolidatedundrained triaxial behavior of the clayey soil. The solution of thisproblem is identical to the solution of the consolidateddrained triaxial test withtwo exceptions: (1) the top and bottom of the soil specimen are made impervious(undrained), and (2) the shearing load is applied faster.

The nite element mesh shown in Figure 5.28 is used here, too. The analysis iscarried out in two steps: a consolidation step and a shearing step. In the rst step,the consolidation step, a conning pressure of 210 kPa is applied while drainageis permitted across the top surface of the nite element mesh. During this step,the geostatic command is invoked to ensure equilibrium. The second step, theundrained shearing step, has a duration of 100 seconds, during which the perviousboundary at the top surface is canceled. The loading plate is forced to displacedownward a distance of 0.127 at a displacement rate of 1.27 103 cm/s. Thisloading will cause the excess pore water pressure to generate during shearing.Automatic time stepping with a maximum pore water pressure change of 0.7 kPaper time increment is used.

The calculated consolidatedundrained stressstrain behavior of the clay spec-imen is compared with the triaxial test data as shown in Figure 5.32. The axialstrain versus pore water pressure is also compared to the test data in the samegure. Excellent agreement is noted in the gure.

Example 5.9 : Back-Calculation of CD Triaxial Test Results Using Lades ModelUse Lades model parameters for the dense silty sand, obtained in Example 2.3(Table 2.5), along with the nite element method to back-calculate the stressstrainbehavior of the soil under CD triaxial conditions.

SOLUTION(lename: Chapter5 example9.cae): Using the nite element method,we simulate the consolidateddrained triaxial behavior of this soil under three con-ning pressures. A two-dimensional axisymmetric mesh is used with one elementonly (similar to Example 5.8). The element chosen is a four-node axisymmetricquadrilateral element. As in Example 5.8, the left-hand side of the mesh is a sym-metry line, and the bottom of the nite element mesh is on rollers. A uniform


downward displacement is applied slowly on the top surface of the mesh (strain-controlled triaxial test).

The analysis is carried out in two steps: a consolidation step and a shearing step.The rst step is a single increment of analysis in which the conning pressure isapplied at the top surface and the right side of the mesh. During step 1, the geo-static command is invoked to make sure that equilibrium is satised within the soilspecimen. The geostatic option makes sure that the initial stress condition in the soilspecimen falls within the initial yield surface of Lades model. Step 2 is a shearingstep in which the loading plate is forced to displace downward at a small rate.

In this simulation we use Lades model parameters for the dense silty sand,obtained in Example 2.3 (Table 2.5). Figure 5.35 presents the soils triaxial behav-ior under three conning pressures. The calculated stressstrain behavior is com-pared with the experimental data. Reasonable agreement is noted in the gure.Of particular interest is the dilative behavior in the volumetric strain versus axialstrain plane where the calculated results (Lades model) captured the importantdilation phenomenon. In Example 2.3 we presented a similar comparison betweenthe experimental data and spreadsheet calculations based on Lades model. It isnoteworthy that the nite element results obtained in the present example are inperfect agreement with the results of the spreadsheet analysis. This implies that theimplementation of Lades model in the nite element program used herein is donecorrectly.

Example 5.10 Plane Strain Test Simulation Using Lades Model Use Ladesmodel parameters for the dense silty sand, obtained in Example 2.3 (Table 2.5)along with the nite element method to predict the behavior of the soil specimenshown in Figure 5.36a. Assume that the soil specimen is innitely long in the z-direction and that a plane strain condition applies. Also assume that the specimenis subjected to a constant conning pressure of 172.37 kPa.

SOLUTION(lename: Chapter5 example10.cae): The two-dimensional planestrain nite element mesh used in this analysis is shown in Figure 5.36b. Themesh has 154 elements. The element chosen is a four-node bilinear plane strainquadrilateral element. As shown in Figure 5.36b, the bottom of the nite elementmesh is on rollers, simulating a frictionless interface between the soil and the bot-tom surface of the plane strain test apparatus. A uniform downward displacementis applied slowly on the top surface of the mesh (strain-controlled shear test).

The analysis is carried out in two steps: a consolidation step and a shearing step.The rst step is a single increment of analysis in which the conning pressure isapplied at the top surface and the sides of the mesh. During step 1, the geostaticcommand is invoked to make sure that equilibrium is satised within the soilspecimen. The geostatic option makes sure that the initial stress condition in the soilspecimen falls within the initial yield surface of Lades model. Step 2 is a shearingstep in which the loading plate is forced to displace downward at a small rate.

In this simulation we use Lades model parameters for the dense silty sand,obtained in Example 2.3 (Table 2.5). Figure 5.37 presents the deformed shape of


0 2 4 6 8 10

200

400

600

800

1000

1200

3 = 102.42 kPa

3 = 172.37 kPa

3 = 310.26 kPa

0 2 4 6 8 101

0

1

2

3

V (%

)

ExperimentalData

FEM

3 = 310.26 kPa

3 = 172.37 kPa

3 = 102.42 kPaExperimentalData

FEM

1 (%)

1 (%)

1

3 (K

Pa)

FIGURE 5.35 Stressstrain behavior of a dense silty sand: Lades model versus experi-mental data.

the plane strain soil model at an advanced stage of loading (postpeak behav-ior). Figure 5.38 shows the distribution of the vertical strains at an advancedstage of loading. Note the presence of a zone of concentrated strains thatresembles a failure plane. The severe distortion within this concentrated strainzone (also known as shear band ) is evident in Figure 5.37. Note that thezones above and below the shear band have a minimal amount of distortion.


(a) (b)

x

y

z

0.1 m

0.22

m

172.37 kPa

3

= 17

2.37

kPa

172.

37 k

Pa

FIGURE 5.36 Plane strain test conguration.

PROBLEMS

5.1 A direct shear test is performed on a sand specimen at a constant normalstress of 40 kPa. The measured shear stress at failure is 28.5 kPa. Calculatethe shear strength parameters of the sand. What is the shear stress at failureif the normal stress were 60 kPa?

5.2 The results of direct shear tests performed on a clay specimen indicated thatc = 30 kPa and = 22. What is the shear strength of this soil at = 0and = 100 kPa?

5.3 Refer to Figure 5.8. Due to the self-weight of the wedge a normal stress of1000 kPa and a shear stress of 620 kPa are applied to the inclined plane.What is the minimum friction angle between the wedge and the inclinedplane required to maintain equilibrium?

5.4 Express the MohrCoulomb failure criterion in terms of the minor and majorprincipal stresses. [Hint: In reference to Figure 5.13, on the failure plane atfailure we have

= 1 + 3

2

1 3

2sin and =

1 3

2cos ]

5.5 A CD triaxial compression test conducted on a sand specimen revealed that = 22. What is the deviator stress at failure if the conning pressure is50 kPa? Solve the problem graphically and analytically.

PROBLEMS 203

FIGURE 5.37 Deformed shape at an advanced stage (post failure).

FIGURE 5.38 Strain concentration along the failure surface.


5.6 A CD triaxial compression test conducted on a clay specimen revealed thatc = 5 kPa and = 22. What is the major principal stress at failure if theconning pressure is 35 kPa? Solve the problem graphically and analytically.

5.7 A consolidatedundrained triaxial test was performed on a sand specimenat a conning pressure 3 = 40 kPa. The consolidatedundrained frictionangle of the sand is = 28, and the effective friction angle is = 33.Calculate (a) the major principal stress at failure, 1f ; (b) the minor and themajor effective principal stresses at failure, 3f and 1f ; and (c) the excesspore water pressure at failure, (ud)f . Solve the problem graphically andanalytically.

5.8 A CD triaxial compression test was conducted on a sand specimen using aconning pressure of 50 kPa. Failure occurred at a deviator stress of 120kPa. Calculate the normal and shear stresses on the failure plane at failure.Solve the problem graphically and conrm your solution analytically.

5.9 A CU triaxial test was performed on a dense sand specimen at a conningpressure 3 = 40 kPa. The consolidatedundrained friction angle of the sandis = 39, and the effective friction angle is = 34. Calculate (a) themajor principal stress at failure, 1f ; (b) the minor and the major effec-tive principal stresses at failure, 3f and 1f ; and (c) the excess pore waterpressure at failure, (ud)f . Solve the problem graphically and analytically.

5.10 Refer to Example 5.6. Hypothetical consolidateddrained and consol-idatedundrained triaxial test results for a clayey soil are given inFigures 5.25 and 5.26, respectively. The isotropic consolidation curve forthe same soil is given in Figure 5.27. Using the nite element method, sim-ulate the consolidateddrained triaxial behavior of this soil when subjectedto a 140-kPa conning pressure. Consider a cylindrical soil specimen withD = 5 cm and H = 5 cm as shown in Figure 5.28. The top and bottomsurfaces of the specimen are open and permeable. The loading plate and thepedestal are smooth, therefore, their interfaces with the clay specimen canbe assumed frictionless. The clay specimen is normally consolidated andhas a constant permeability k = 0.025 m/s and initial void ratio e0 = 0.959,corresponding to its 140-kPa conning pressure. Assume that the clay iselastoplastic obeying the extended Cam clay model (Chapter 2).

5.11 Refer to Example 5.7. Using the same hypothetical test data and assump-tions given in Problem 5.10, predict the consolidatedundrained triaxialbehavior of a triaxial soil specimen subjected to 140-kPa conning pres-sure. Assume that the clay is elastoplastic, obeying the extended Cam claymodel (Chapter 2).

5.12 Refer to Example 5.8. Assuming that the clayey soil described in Problem5.10 is elastoplastic, obeying the Cap model (Chapter 2), predict the consol-idateddrained and the consolidatedundrained triaxial behavior of the soilwhen subjected to a 140-kPa conning pressure.

PROBLEMS 205

5.13 Obtain Lades model parameters (Chapter 2) for the soil described in Prob-lem 5.10. Use Lades model parameters, along with the nite element methodto predict the consolidateddrained triaxial behavior of the soil when sub-jected to a 140-kPa conning pressure.

5.14 The at-failure results of several CD and CU triaxial tests of a silty sandare presented in the p q plane as shown in Figure 5.39. The results ofan isotropic compression test performed on the same soil are shown inFigure 5.40. Estimate the Cam clay model parameters M,, and . Notethat you can obtain the compression index and the swelling index fromFigure 5.40 (not and ).

0 100 200 300 4000

125

250

375

500

(Cap Model)

d (Cap Model)

1M (Cam Clay Model)

p = (1 + 23 )/3 (kPa)

t = q

= (

1

3)

(kPa)

FIGURE 5.39 p versus q curve for silty sand.

Mean Effective Stress, (kPa) (log scale)

Voi

d Ra

tio

10 100 10000.29

0.30

0.31

0.32

0.33

0.34

0.35

Cc

Cs

FIGURE 5.40 elog p curve for silty sand.


5.15 Using the nite element method and the Cam clay model parameters for thesilty sand (Problem 5.14), predict the consolidateddrained triaxial behaviorof this soil when subjected to a conning pressure of 70 kPa. Note thatthe initial void ratio that corresponds to this conning pressure is 0.34(Figure 5.40).

5.16 Repeat Problem 5.15 for consolidatedundrained triaxial test conditions.

5.17 The at-failure results of several CD and CU triaxial tests of a silty sandare presented in the p t plane as shown in Figure 5.39. The results ofan isotropic compression test performed on the same soil are shown in

0 4 8 12 16 201

2

3

4

1/

3

1 (%)

4.5 kg/cm2

12.65 kg/cm2

3 = 0.94 kg/cm2

2 kg/cm2

0 4 8 12 16 20

2

1

0

1

2

3

4

5

6

1 (%)

v (%

)

3 = 0.94 kg/cm2

2 kg/cm2

4.5 kg/cm2

12.65 kg/cm2

FIGURE 5.41 Stressstrain behavior of loose Sacramento River sand. (Adapted fromLade, 1977.)

PROBLEMS 207

Figure 5.40. (a) Calculate the soils angle of friction, , and its cohesion, d,in the p t plane. (b) Using the results of the isotropic compression testperformed on the same soil (Figure 5.40), calculate the hardening curve forthe cap model assuming the initial conditions p0 = 70 kPa and e0 = 0.34.

5.18 Using the nite element method and the cap model parameters for thesilty sand (Problem 5.17), predict the consolidateddrained triaxial behav-ior of this soil when subjected to a conning pressure of 70 kPa. Notethat the initial void ratio that corresponds to this conning pressure is 0.34(Figure 5.40).

5.19 Repeat Problem 5.18 for consolidatedundrained triaxial test conditions.

5.20 The results of four consolidateddrained triaxial tests and one isotropic com-pression test on loose Sacramento River sand are shown in Figures 5.41and 5.42, respectively. Estimate Lades model parameters following the pro-cedure discussed in Chapter 2.

5.21 Using the nite element method and Lades model parameters obtainedin Problem 5.20, predict the consolidated drained triaxial behavior of aloose Sacramento River sand specimen subjected to a conning pressureof 2 kg/cm2.

5.22 From the results of the consolidated drained triaxial tests and the isotropiccompression test on loose Sacramento River sand shown in Figures 5.41and 5.42, respectively, estimate the Cam clay model parameters.

0 10 200

100

200

3

(kg/cm

2 )

v (%)

FIGURE 5.42 Isotropic compression behavior of loose Sacramento River sand. (Adaptedfrom Lade, 1977.)


5.23 Using the nite element method and the Cam clay model parameters obtainedin Problem 5.22, predict the consolidateddrained triaxial behavior of aloose Sacramento River sand specimen subjected to a conning pressure of2 kg/cm2.

5.24 Obtain the cap model parameters using the results of the consoli-dateddrained triaxial tests and the isotropic compression test on looseSacramento River sand shown in Figures 5.41 and 5.42, respectively.

5.25 Using the nite element method and the cap model parameters obtainedin Problem 5.24, predict the consolidateddrained triaxial behavior of aloose Sacramento River sand specimen subjected to a conning pressure of2 kg/cm2.

chapter 5 - soil strength

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