Civil Engineering Reference
In-Depth Information
2. Before the critical state there may be a peak state and after large strains clay soils
reach a residual state. The peak state is associated with dilation and the residual
state is associated with laminar flow of flat clay particles.
3. The critical states of soils in shear tests are given by
τ f
= σ f tan
φ c
(9.2)
σ f
e f =
e
C c log
(9.3)
4. The critical state strength of soil is uniquely related to its voids ratio or water
content so for undrained loading of saturated soil (i.e. at constant water content)
the undrained strength s u remains unchanged.
5.
If the soil is drained and effective stresses can be determined you can use effective
stress analyses and the critical state strength is given by
φ c . If the soil is saturated
and undrained you can use total stress analyses and the critical state strength is
given by the undrained strength s u .
6. To take account of different normal effective stresses and different voids ratios,
stresses, should be normalized with respect to the critical stress
σ c or the critical
voids ratio e λ given by
σ
e λ =
e
+
C c log
(9.7)
e
e
σ c =
log
(9.8)
C c
7. The critical states in shear tests are found also in triaxial tests and the critical state
line is given by
q f =
Mp f
(9.10)
ln p f
v f
= λ
(9.11)
φ c ,e and C c (or M ,
8. The critical state parameters
and
λ
) are material parameters:
they depend only on the nature of the soil grains.
9. The state of a soil is described by the distance of the voids ratio
effective stress
point from the critical state line and it is given by the either of the state parameters
S v or S s .
Worked examples
Example 9.1: Determination of critical state soil parameters A number of drained
and undrained triaxial tests were carried out on normally consolidated and overcon-
solidated samples of the same soil. Table 9.1 gives values for the stress parameters q f
and p f and the specific volume v f when the samples had reached failure at their critical
states.
The data are shown plotted in Fig. 9.16. Scaling from the diagram, M
=
0.98,
1.82 and p =
λ =
0.20. Substituting (say) v
=
600 kPa with
λ =
0.20 into Eq. (9.11)
we have
=
3.10.
 
 
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