Civil Engineering Reference
In-Depth Information
At any stage of the test
=−
L
L
0
ε
a
=−
V
V
0
ε
v
v
=
v
0
(1
−
ε
v
)
and
F
a
A
σ
=
σ
=
300 kPa
σ
=
σ
+
r
c
a
r
where the current area is
A
=
A
0
(1
−
ε
v
)/(1
−
ε
a
). From Eqs. (3.5) to (3.8),
1
3
ε
ε
=
ε
−
s
a
v
and
q
=
σ
a
−
σ
r
)
qp
1
3
(
σ
a
+
σ
r
)
(
=
=
2
=
p
−
u
or
F
a
A
q
=
p
=
1
3
q
−
+
p
0
u
where
p
0
300 kPa. The test results are given in the right-hand side of Table 7.3 and
are plotted in Fig. 7.13 as O
=
→
A.
Example 7.4: Interpretation of an undrained triaxial test
The first three columns in
Table 7.4 give data from an undrained triaxial compression test in which the cell
pressure was held constant at
300 kPa. At the start of the test the sample was
38 mm diameter and 76 mm long, the pore pressure was
u
0
=
σ
=
c
100 kPa and the specific
volume was
v
2.19.
For an undrained test
=
0 (by definition), but otherwise the calculations are the
same as those given in Example 7.3. The test results are given in the right-hand side of
Table 7.4 and are plotted in Fig. 7.13 as O
ε
=
v
→
B.
Example 7.5: Stress paths
The left-hand side of Table 7.5 gives the initial states and
increments of axial and radial total stresses for a set of drained and undrained tri-
axial stress path tests. In the drained tests the pore pressure was
u
0. The soil
can be assumed to be isotropic and elastic so that shearing and volumetric effects are
decoupled.
The stress paths corresponding to tests lasting for 10 hours are shown in Fig. 7.14.
The right-hand side of Table 7.5 gives the states at the start and at the end of each path.
For the undrained test
=
p
=
0 and shear and volumetric effects are
decoupled). For the drained tests the changes of
q
and
p
are found from Eqs. (7.13)
and (7.14).
δ
0 (because
δε
=
v
