Download Module 4 Lecture 21 Pore water pressure and shear strength

NPTEL- Advanced Geotechnical Engineering
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
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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
.
Figure 4.17 Consolidation drained triaxial tests in clay (a) application of confining pressure (b)
application of deviator stress
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
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Figure 4.18 Failure envelopes for (a) normally consolidated and (b) overconsolidated
clays from consolidated drained triaxial tests
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
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Figure 4.19 Derivation of equation (21)
(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
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Figure 4.20 Failure envelope of a clay with perconsoldation pressure
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
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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
Figure 4.21 Consolidation undrained triaxial test. (a) Application of confining pressure (b)
application of deviator stress
(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,
=
=
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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 effectivestress 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
Figure 4.22 Consolidated undrained test results-normally consolidated clay
(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
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Figure 4.23 Consolidated undrained test-total stress envelope for overconsolidated 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.,
obtained by plotting the effective-stress Mohr’s circle and then drawing a common tangent to
can be
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
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Figure 4.24 Unconsolidated undrained triaxial test
Figure 4.25
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(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.
Deviator stress at failure
Confining pressure
1
2
Determine the values of c and
70
92
440.4
474.7
for the soil.
Solution From equation (21),
)
. For test
;
. So,
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(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
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Figure 4.26
Example 3 For a saturated clay soil, the following are the results of some consolidated drained triaxial tests
at failure:
Test no.
1
2
3
4
Draw a
60
90
110
180
diagram, and from that determine
Solution The diagram of
be written in the form
25.6
36.5
44.0
68.0
for the soil.
is shown in Figure 4.27; this is a straight line, and the equation of it may
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Figure 4.27
(a)
Now equation (20) can be written in the form
(b)
Comparing equations (a) and (b) we find
27,
and
. So,
and
. From Figure 4.
And
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