Civil Engineering Reference
In-Depth Information
Figure 9.17
Results from tests on a soil - Example 9.2.
Notice that the tests were carried out with
p
constant so that
p
f
=
p
0
, the pore pressures
at failure
u
f
in the undrained tests are given by
p
f
=
p
f
=
p
0
+
p
f
u
f
=
p
f
−
p
0
−
u
0
−
From Eq. (9.10),
Mp
f
q
f
=
The points corresponding to isotropic compression and to failure at the critical state
are shown in Fig. 9.17; these are linked by lines that represent approximately the state
paths for the tests.
Example 9.3: Normalized critical states
The initial states and the critical states given
in Table 9.1 can be normalizedwith respect to the critical pressure
p
c
given by Eq. (9.13)
or with respect to the equivalent specific volume
v
λ
given by Eq. (9.12). The values
for the normally consolidated samples A and C are given in Table 9.3 and the points
representing the critical state and normal compression lines are given in Fig. 9.18.
The initial and final state points may be joined together as shown by a line that
represents approximately the state paths followed by the drained and undrained tests.
Notice that in both tests the value of
p
/
p
c
decreases from 2.11 to 1.00. In the drained
test this is because
p
c
increases from 284 to 600 kPa as the specific volume decreases
Table 9.3
Normalized initial and critical states of a soil - Example 9.3
Sample
Initial state
Critical state
p
0
p
c
p
0
/
p
c
q
f
p
f
p
c
q
f
/
p
c
p
f
/
p
c
q
f
/
p
f
v
λ
v
λ
(kPa)
(kPa)
(kPa)
(kPa)
(kPa)
A
600
284
2.11
3.25
588
600
600
0.98
1.00
0.98
3.10
C
600
284
2.11
3.25
278
284
284
0.98
1.00
0.98
3.10
