Geoscience Reference
In-Depth Information
Equation (1.146) can be used to define the Skempton
coefficient
B
:
1
K
=
−
1
V
∂
V
1 139
∂
P
p
,
T
p
u
=
BP
1 148
1
K
u
=
1
V
∂V
∂
−
1 140
P
m
,
T
∂
p
B
≡
1 149
∂
P
m
,
T
Obviously, it is harder to deform a rock if some fluid
(for instance, water) is resisting the deformation. Conse-
quently, we have the following property:
K
S
≥
B
=
K
u
−
=
1
−
K
K K
u
K
. The
undrained regime is defined by replacing the porous
material by an equivalent elastic material with a bulk
modulus
K
u
(the shear modulus remains unchanged).
In the undrained regime, Hooke
K
u
≥
1 150
α
K
u
1
−
K K
S
As
K
S
≥
K
u
≥
K
,0
≤
B
≤
1, and 0
≤
p
u
≤
P
. The Skempton
coefficient can be used to determine the effect of a change
of the confining pressure on a change on the fluid pres-
sure in undrained conditions. Two extreme cases can be
considered. When
K
u
K
, we have
B
0, for instance,
when gas is present in the pore space of the porous mate-
rial. In this case, an increase of the confining pressure has
no effect on the fluid pressure because of the very high
compressibility of the gas. The second case is when
K
0,
K K
S
,
K
u
. In this case, we have
α
,
B
1. This
is the case of a very compressible soil with a very incom-
pressible fluid like water.
To understand the implication of the drained and
undrained regimes in terms of deformation, we can think
about a tank filled by a water-saturated sand. We apply at
t
= 0 a mechanical load on the top of the sand (think
about a building built very quickly on a porous soil).
The deformation that follows can be considered using
two steps: in the first step, the deformation is instantane-
ous and undrained with a bulk modulus
K
u
and a shear
modulus
G
. In this step, we can compute the undrained
fluid pressure everywhere, and the fluid has no time to
move. In the second step, the fluid starts to flow in
response to the generated (undrained) fluid pressure dis-
tribution. This flow is responsible for a delayed mechan-
ical response: the soil creeps over time because of the
flow of the pore water until an equilibrium situation is
reached. Our poroacoustic analysis was proceeding along
the same lines: we first determined the undrained fluid
pressure associated with the stress or deformation associ-
ated with the propagation of the seismic waves (using an
undrained bulk modulus; see Figure 1.19), and then, we
let the fluid flows in response to the fluid pressure distri-
bution. The flow of water is responsible for the streaming
current density defined as the advective drag of the
effective charge density
Q
V
contained in the pore water
(Figure 1.19).
'
s law is therefore
given by
2
3
G
T
ij
=
K
u
−
ε
δ
ij
+2
G
ε
1 141
kk
ij
We can look for the bulk deformation of the material in
the undrained regime (the pure shear components are
unchanged with respect to the drained case). This yields
T
kk
=3
K
u
ε
kk
1 142
and therefore,
1
3
K
u
T
kk
=
P
K
u
ε
kk
=
−
1 143
What about the fluid pressure in the undrained case?
We can still apply the complete form of Hooke
s law by
writing that the fluid pressure is equal to the undrained
fluid pressure. The effective stress law yields
'
1
K
P
ε
kk
P
,
p
u
=
−
−
α
p
u
1 144
Equating Equations (1.143) and (1.144) yields
K
u
=
1
P
K
P
−
α
p
u
1 145
and therefore,
p
u
=
K
u
−
K
P
1 146
α
K
u
p
u
=
K
−
K
u
α
ε
kk
1 147
