Environmental Engineering Reference
In-Depth Information
The inverse of the coefficients in the first
term (i.e.,
The water phase constitutive relation can be written by
substituting the triaxial stress conditions into Eq. 13.29:
E/
[3
(
1
2
μ)
]) is commonly referred to as the “bulk
modulus” of the soil (Lambe and Whitman, 1979).
The constitutive equation for the water phase can be
derived from Eq. 13.29:
−
d
σ
3
−
u
a
+
d
1
3
σ
1
−
σ
3
dV
w
V
0
=
3
E
w
d
u
a
−
u
w
H
w
d
u
a
−
u
w
H
w
E
w
d
σ
3
−
u
a
+
dV
w
V
0
=
3
+
(13.38)
(13.33)
The
du
a
and
d
u
a
−
u
w
terms in Eqs. 13.37 and 13.38
refer to the pore-air pressure and matric suction increments,
respectively, during triaxial loading. For triaxial testing in
the laboratory, isotropic loading is generally first applied fol-
lowed by uniaxial compression. Therefore, it may be useful
to separate the pore-air pressure and matric suction incre-
ments during the two loading conditions. This can be done
by superimposing Eq. 13.32 (i.e., isotropic loading) and
Eq. 13.34 (i.e., uniaxial loading) to form the soil structure
constitutive relationship for triaxial loading:
The soil will undergo equal deformation in all directions
(or isotropic compression) when subjected to an equal all-
around pressure provided the soil properties are also isotropic
(Fig. 13.9).
13.3.5 Uniaxial Loading
A total stress increment is applied to the soil in one direction
for uniaxial loading (e.g., the vertical direction), as illustrated
in Fig. 13.9. It is assumed that no shear stress is developed
on the
x-, y-,
and
z
-planes. The stress increase is applied
in the
y-
direction,
dσ
y
, while the total stress change in the
other two directions is zero (i.e.,
dσ
x
=
dσ
z
=
3
1
d
σ
3
−
u
ai
−
2
μ
dε
v
=
0). The soil
compresses in the
y
-direction and expands in the
x-
and
z-
directions. Applying these stress conditions to Eq. 13.26 gives
the soil structure constitutive equation for uniaxial loading:
E
+
d
1
3
σ
1
−
σ
3
−
u
au
H
d
u
a
−
u
w
i
+
d
u
a
−
u
w
u
3
3
1
d
1
3
σ
y
−
u
a
+
(13.39)
H
d
u
a
−
u
w
(13.34)
The water phase constitutive equation for uniaxial loading
is obtained from Eq. 13.29:
−
2
μ
E
3
dε
v
=
+
where:
du
ai
=
pore-air pressure increment during isotropic load-
ing,
d
u
a
−
u
w
H
w
E
w
d
1
3
σ
y
−
u
a
du
au
=
pore-air pressure
increment during uniaxial
dV
w
V
0
=
3
loading,
+
(13.35)
du
wi
=
pore-water pressure increment during isotropic
loading, and
13.3.6 Triaxial Loading
The triaxial loading conditions shown in Fig. 13.9 can be con-
sidered as a superposition of isotropic and uniaxial loading.
Isotropic loading applies to an all-around pressure change
of
dσ
3
, whereas the uniaxial pressure change applies to a
du
wu
=
pore-water pressure increment during uniaxial
loading.
Equation 13.39 can be used to compute the volumetric strain
increment during triaxial loading. Only the first and third
terms are used since the other terms drop out during isotropic
compression. On the other hand, the second and fourth terms
are used to compute the volumetric strain increment during
uniaxial compression while the other terms become zero. In
the end, volumetric strain increments during isotropic and
uniaxial compression can be summed to give the total vol-
umetric strain associated with triaxial loading. It should be
noted that Eq. 13.39 reverts to Eq. 13.37 by substituting
du
a
=
du
ai
+
du
au
and
du
w
=
du
wi
+
du
wu
into Eq. 13.39.
Similarly, the water phase constitutive relation can be
obtained by superimposing Eq. 13.33 (i.e., isotropic loading)
and Eq. 13.35 (i.e., uniaxial loading):
deviator stress
d
σ
1
−
σ
3
in the vertical or
y-
direction. The
x-, y-,
and
z-
planes are assumed to be principal planes. The
above stress conditions [i.e.,
dσ
x
=
dσ
z
=
dσ
3
and
dσ
y
=
dσ
3
+
d
σ
1
−
σ
3
] can be substituted into Eq. 13.26 to give
the soil structure constitutive equation for triaxial loading:
σ
3
+
σ
3
+
σ
1
−
σ
3
+
σ
3
3
3
1
d
−
2
μ
dε
v
=
−
u
a
E
H
d
u
a
−
u
w
3
+
(13.36)
Rearranging the above equation gives
d(σ
3
−
u
ai
)
+
d
1
3
1
d
σ
3
−
u
a
+
d
1
3
σ
1
−
σ
3
3
σ
1
−
σ
3
dV
w
V
0
=
3
E
w
−
2
μ
dε
v
=
E
d
u
a
−
u
w
i
H
w
d
u
a
−
u
w
u
H
w
−
u
au
+
H
d
u
a
−
u
w
3
+
(13.40)
+
(13.37)
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