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# Craig's Soil Mechanics 7th Edition

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```plane is

q0 ¼Mp0 ð4:17Þ

whereM is the slope of OS0. Using Equations 4.6(a), 4.14, 4.16 and 4.17 the parameter
M can be related to the angle of shearing resistance �0:

sin�0 ¼ �
0
1 � �03
�01 þ �03

¼ 3M
6þM

If the projection of the critical-state line on the v–p0 plane is replotted on a v–ln p0

plane it will approximate to a straight line parallel to the corresponding normal
consolidation line (of slope –�) as shown in Figure 4.23. The equation of the critical-
state line, with respect to v and p0, can therefore be written as

v ¼ �� � ln p0 ð4:18Þ

where � is the value of v on the critical-state line at p0 ¼ 1 kN/m2.
Similarly, the equation of the normal consolidation line (NN) is

v ¼ N � � ln p0 ð4:19Þ

The critical-state concept 123

where N is the value of v at p0 ¼ 1 kN/m2. The swelling and recompression relation-
ships can be approximated to a single straight line (RR), of slope ��, represented by
the equation

v ¼ v� � � ln p0 ð4:20Þ

where v� is the value of v at p
0 ¼ 1 kN/m2.

Example 4.4

The following parameters are known for a saturated normally consolidated clay:
N ¼ 2:48, � ¼ 0:12, � ¼ 2:41 and M ¼ 1:35. Estimate the values of principal stress
difference and void ratio at failure in undrained and drained triaxial tests on specimens
of the clay consolidated under an all-round pressure of 300 kN/m2. What would be the
expected value of �0cv?
After normal consolidation to 300 kN/m2 ( p0c), the specific volume (vc) is given by

vc ¼ N � � ln p0c ¼ 2:48� 0:12 ln 300 ¼ 1:80

In an undrained test on a saturated clay the volume change is zero, and therefore
the specific volume at failure (vf) will also be 1.80, i.e. the void ratio at failure (ef)
will be 0.80.
Assuming failure to take place on the critical-state line,

q0f ¼Mp0f

slope
–λ

slope
–κ

N

v

vκ

S″

S″ N

N

In p′

R

1

R

Γ

Figure 4.23 Critical-state concept: e� ln p0 relationships.

124 Shear strength

and the value of p0f can be obtained from Equation 4.18. Therefore

q0f ¼M exp
�� vf
�

� �

¼ 1:35 exp 2:41� 1:80
0:12

� �
¼ 218 kN=m2 ¼ ð�1 � �3Þf

For a drained test the slope of the stress path on a q0�p0 plot is 3, i.e.

q0f ¼ 3ðp0f � p0cÞ ¼ 3
q0f
M

� p0c
� �

Therefore,

q0f ¼
3Mp0c
3�M ¼

3� 1:35� 300
3� 1:35 ¼ 736 kN=m

2

¼ ð�1 � �3Þf

Then,

p0f ¼
q0f
M

¼ 736
1:35

¼ 545 kN=m2

vf ¼ �� � ln p0f ¼ 2:41� 0:12 ln 545 ¼ 1:65

Therefore,

ef ¼ 0:65

�0cv ¼ sin�1
3M

6þM
� �

¼ sin�1 3� 1:35
7:35

� �
¼ 33�

4.6 RESIDUAL STRENGTH

In the drained triaxial test, most clays would eventually show a decrease in shear
strength with increasing strain after the peak strength has been reached. However, in
the triaxial test there is a limit to the strain which can be applied to the specimen. The
most satisfactory method of investigating the shear strength of clays at large strains
is by means of the ring shear apparatus [3, 8], an annular direct shear apparatus.
The annular specimen (Figure 4.24(a)) is sheared, under a given normal stress, on
a horizontal plane by the rotation of one half of the apparatus relative to the other;
there is no restriction to the magnitude of shear displacement between the two halves

Residual strength 125

of the specimen. The rate of rotation must be slow enough to ensure that the specimen
remains in a drained condition. Shear stress, which is calculated from the applied
torque, is plotted against shear displacement as shown in Figure 4.24(b).
The shear strength falls below the peak value and the clay in a narrow zone adjacent

to the failure plane will soften and reach the critical state. However, because of non-
uniform strain in the specimen, the exact point on the curve corresponding to the
critical state is uncertain. With continuing shear displacement the shear strength
continues to decrease, below the critical-state value, and eventually reaches a residual
value at a relatively large displacement. If the clay contains a relatively high propor-
tion of plate-like particles a reorientation of these particles parallel to the failure plane
will occur (in the narrow zone adjacent to the failure plane) as the strength decreases
towards the residual value. However, reorientation may not occur if the plate-like
particles exhibit high interparticle friction. In this case, and in the case of soils
containing a relatively high proportion of bulky particles, rolling and translation of
particles takes place as the residual strength is approached. It should be appreciated
that the critical-state concept envisages continuous deformation of the specimen as a
whole, whereas in the residual condition there is preferred orientation or translation of
particles in a narrow shear zone. The original soil structure in this narrow shear zone is
destroyed as a result of particle reorientation. A remoulded specimen can therefore be
used in the ring shear apparatus if only the residual strength (and not the peak
strength) is required.

Figure 4.24 (a) Ring shear test and (b) residual strength.

126 Shear strength

The results from a series of tests, under a range of values of normal stress, enable the
failure envelope for both peak and residual strength to be obtained, the residual
strength parameters in terms of effective stress being denoted c0r and �

0
r. Residual

strength data for a large range of soils have been published [13], which indicate that
the value of c0r can be taken to be zero. Thus, the residual strength can be expressed as

�r ¼ �0f tan�0r ð4:21Þ

4.7 PORE PRESSURE COEFFICIENTS

Pore pressure coefficients are used to express the response of pore water pressure to
changes in total stress under undrained conditions and enable the initial value of excess
pore water pressure to be determined. Values of the coefficients may be determined in
the laboratory and can be used to predict pore water pressures in the field under
similar stress conditions.

(1) Increment of isotropic stress

Consider an element of soil, of volume V and porosity n, in equilibrium under total
principal stresses �1, �2 and �3, as shown in Figure 4.25, the pore pressure being u0.
The element is subjected to equal increases in total stress ��3 in each direction,
resulting in an immediate increase �u3 in pore pressure.

The increase in effective stress in each direction ¼ ��3 ��u3
Reduction in volume of the soil skeleton ¼ CsVð��3 ��u3Þ

where Cs is the compressibility of the soil skeleton under an isotropic effective stress
increment.

Reduction in volume of the pore space ¼ CvnV�u3
where Cv is the compressibility of pore fluid under an isotropic pressure increment.
If the soil particles are assumed to be incompressible and if no drainage of pore fluid

takes place then the reduction in volume of the soil skeleton must equal the reduction
in volume of the pore space, i.e.

CsVð��3 ��u3Þ ¼ CvnV�u3
Therefore,

�u3 ¼ ��3 1
1þ nðCv=CsÞ
� �

Writing 1/[1þ n(Cv/Cs)] ¼ B, defined as a pore pressure coefficient,
�u3 ¼ B��3 ð4:22Þ

Pore pressure coefficients 127

In fully saturated soils the compressibility of the pore fluid (water only) is considered
negligible compared with that of the soil skeleton, and therefore Cv/Cs ! 0 and B! 1.
Equation 4.22 with B ¼ 1 has already been assumed in the discussion on undrained
strength earlier in the present chapter. In partially saturated soils the compressibility of
the pore fluid is high due to the presence of pore air, and therefore Cv/Cs > 0 and B < 1.
The variation of B with degree of saturation for a particular soil is shown in Figure 4.26.
The value of B can be measured in the triaxial apparatus. A specimen is set up under

any value of all-round pressure and the pore water pressure measured (after consoli-
dation if desired). Under undrained conditions the all-round pressure is then increased
(or reduced) by an amount```