Module 4 Lecture 21 Pore water pressure and shear strength - 5 ...

NPTEL- Advanced Geotechnical Engineering

Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1

Module 4

Lecture 21

Pore water pressure and shear strength - 5

Topics

1.3 SHEAR STRENGTH OF COHESIVE SOILS

1.3.1 Triaxial Testing in Clays

Consolidated drained test

Consolidated undrained test

Unconsolidated undrained test

1.3 SHEAR STRENGTH OF COHESIVE SOILS

The shear strength of cohesive soils can, generally, be determined in the laboratory by either direct shear test

equipment or triaxial shear test equipment; however, the triaxial test is more commonly used. Only the shear

strength of saturated cohesive soils will be treated here. The shear strength based on the effective stress can

be given by [equation 3] . For normally consolidated clays, and, for

overconsolidated clays, .

1.3.1 Triaxial Testing in Clays

The basic features of the triaxial test equipment were shown in Figure 4. 5. Three conventional types of tests

are conducted with clay soils in the laboratory:

1. Consolidated drained test or drained test (CD test or d test).

2. Consolidated undrained test (CU test).

3. Unconsolidated undrained test (UU test).

Each of these tests will be separately considered in the following sections.

Consolidated drained test

For the consolidated drained test the saturated soil specimen is first subjected to a confining pressure

through the chamber fluid; as a result, the pore water pressure of the sample will increase by . The

connection to the drainage is kept open for complete drainage so that becomes equal to zero. Then the

deviator stress (piston stress) is increased at a very slow rate, keeping the drainage valve open to allow

complete dissipation of the resulting pore water pressure . Figure 4.17 shows the nature of the variation

of the deviator stress with axial strain. From Figure 4.17, it must also be pointed out that, during the



application of the deviator stress, the volume of the specimen gradually reduces for normally consolidated

clays. However, overconsolidated clays go through some reduction of volume initially but then expand. In a

consolidated drained test, the total stress is equal to the effective stress since the excess pore water pressure

is zero. At failure, the maximum effective principal stress is where is the deviator

stress at failure. The minimum effective principal stress is .

From the results of a number of tests conducted using several specimens, the Mohr’s circles at failure can be

plotted as shown in Figure 4.18. The values of c and are obtained by drawing a common tangent to these

Mohr’s circles, which is the Mohr-Coulomb envelope. For normally consolidated clays (Figure 4.18a), we

can see that . Thus the equation of the Mohr-Coulomb envelope can be given by . The

slope of the failure envelope will give us the angle of friction of the soil. As shown by equation (5) for these

soils

Figure 4.17 Consolidation drained triaxial tests in clay (a) application of confining pressure (b)

application of deviator stress



The plane of failure makes an angle of with the major principal plane.

For overconsolidated clays (Figure 4.18b), . So, the shear strength follows the equation the values of can be determined by measuring the intercept of the failure envelope on the

shear stress axis and the slope of the failure envelope, respectively. To obtain a general relation between

, refer to Figure 4.19, from which

Figure 4.18 Failure envelopes for (a) normally consolidated and (b) overconsolidated

clays from consolidated drained triaxial tests



(20)

or

(21)

Note that the plane of failure makes an angle of with the major principal plane.

If a clay s initially consolidated by an encompassing chamber pressure of and allowed to swell

under a reduced chamber pressure of , the specimen will be overconsolidated. The failure envelope

obtained from consolidated drained triaxial tests of these types of specimens has two distinct branches, as

shown in Figure 4.20. Portion of the failure envelope has a flatter slope with a cohesion intercept, and

the portion represents a normally consolidated stage following the equation

Figure 4.19 Derivation of equation (21)



The shear strength of clays at very large strains is referred to as residual shear strength (i.e., the ultimate

shear strength). It has been proved that the residual strength of a given soil is independent of past stress

history

(22)

(i.e., the c components is 0). For triaxial tests,

(23)

Where

The residual friction angle in clays is of importance in subjects such as the long-term stability of slopes.

Consolidated undrained test

In the consolidated undrained test, the soil specimen is first consolidated by a chamber confining pressure

; full drainage from the specimen is allowed. After complete dissipation of excess pore water pressure, ,

generated by the confining pressure, the deviator stress is increased to cause failure of the specimen.

During this phase of loading, the drainage line from the specimen is closed. Since drainage is not permitted,

the pore water pressure (pore water pressure due to deviator stress, ) in the specimen increases.

Simultaneous measurements of and are made during the test. Figure 4.21 shows the nature of the

Figure 4.20 Failure envelope of a clay with perconsoldation pressure



variation of and with axial strain; also shown is the nature of the variation of the pore water pressure

parameter see equation (5 from chapter 4)] with axial strain. The value of A at failure, is

positive for normally consolidated clays and becomes negative for overconsolidated clays. Thus, is

dependent on the overconsolidated ratio. The overconsolidation ratio, OCR, for triaxial test conditions may

be defined as

(24)

Where is the maximum chamber pressure at which the specimen is consolidated and then allowed

to rebound under a chamber pressure of .

At failure,

=

=

Figure 4.21 Consolidation undrained triaxial test. (a) Application of confining pressure (b)

application of deviator stress



consolidated undrained tests on a number of specimens can be conducted to determine the shear strength

parameters of a soil, as shown for the case of a normally consolidated clay in Figure 4.22. The total-stress

Mohr’s circles (circles A and B) for two tests are shown by the broken lines. The effective-stress Mohr’s

circles C and D correspond to the total-stress circles Ai and B, respectively. Since C and D are effective-

stress circles at failure, a common tangent drawn to these circles will give the Mohr-Coulomb failure

envelope given by the equation . If we draw a common tangent to the total-stress circles, it will

be a straight line passing through the origin. This is the total-stress failure envelope, and it may be given by

(25)

Where is the consolidated undrained angle of friction.

The total-stress failure envelope for an over consolidated clay will be of the nature shown in Figure 4.23

and can be given by the relation

Figure 4.22 Consolidated undrained test results-normally consolidated clay



(26)

Where is the intercept of the total-stress failure envelope along the shear stress axis.

The shear strength parameters for overconsolidated clay based on effective stress, i. e., can be

obtained by plotting the effective-stress Mohr’s circle and then drawing a common tangent to

As in consolidated drained tests, shear failure in the specimen can be produced by axial compression or

extension by changing the loading conditions.

Unconsolidated undrained test

In unconsolidated undrained triaxial tests, drainage from the specimen is not allowed at any stage. First, the

chamber confining pressure is applied, after which the deviator stress is increased until failure occurs.

For these tests.

Total major principal stress

Total minor principal stress

Tests of this type can be performed quickly since drainage is not allowed. For a saturated soil, the deviator

stress at failure, is practically the same irrespective of the confining pressure (Figure 4.24). So, the

total-stress failure envelope can be assumed to be a horizontal line, and . The undrained shear strength

can be expressed as

Figure 4.23 Consolidated undrained test-total stress envelope for overconsolidated clay



Figure 4.24 Unconsolidated undrained triaxial test

Figure 4.25



(27)

This generally referred to as the shear strength based on concept.

The fact that the strength of saturated clay sin unconsolidated undrained loading conditions is the same

irrespective of the confining pressure can be explained with the help of Figure 4.25. If a saturated clay

specimen A is consolidated under a chamber confining pressure of and then sheared to failure under

undrained conditions, the Mohr’s circle at failure will be represented by circle no 1. The effective-stress

Mohr’s circle corresponding to circle no 1 is circle no. 2, which touches the effective-stress failure envelope.

If a similar soil specimen B, consolidated under a chamber confining pressure of is subjected to an

additional confining pressure of without allowing drainage, the pre water pressure will increase by .

We saw in chapter 4 that and, for saturated soils, .

Since the effective confining pressure of specimen B is the same as specimen A, it will fall with the same

deviator stress, . The total-stress Mohr’s circle for this specimen (i.e., B) at failure can be given by circle

no. 3. So, at failure, for specimen B

Total minor principal stress

Total major principal stress

The effective stresses for the specimen are as follows:

Effective major principal stress

Effective minor principal stress =

The above principal stresses are the same as those we had for specimen A. thus, the effective-stress Mohr’s

circle at failure for specimen B will be the same as that for specimen A, i.e., circle no 1.

The value of could be of any magnitude in specimen B; in all cases, would be the same

Example 1 Consolidated drained triaxial tests on two specimens of a soil gave the following results:

Test no. Confining pressure Deviator stress at failure

1 70 440.4

2 92 474.7

Determine the values of c and for the soil.

Solution From equation (21), ) . For test ;

. So,



(a)

Similarly, for test ; . Thus,

(b)

Subtracting equation (a) from (b),

Substituting in equation (a)

Example 2 A normally consolidated clay specimen was subjected to a consolidated undrained test. At

failure, , , and . Determine .

Solution Referring to Figure 4.26,

Hence

Again,

So,

Hence



Example 3 For a saturated clay soil, the following are the results of some consolidated drained triaxial tests

at failure:

Test no.

1 60 25.6

2 90 36.5

3 110 44.0

4 180 68.0

Draw a diagram, and from that determine for the soil.

Solution The diagram of is shown in Figure 4.27; this is a straight line, and the equation of it may

be written in the form

Figure 4.26



(a)

Now equation (20) can be written in the form

(b)

Comparing equations (a) and (b) we find and . From Figure 4.

27, and . So,

And

Figure 4.27

Module 4 Lecture 21 Pore water pressure and shear strength - 5 ...

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