Civil Engineering Reference
In-Depth Information
given by Equation (11.74)
$
σ
ij
n
j
σ
ij
n
j
=
u
n
u
s
n
=
−
0
(13.12)
%
u
s
i
=
u
i
The first equation ensures the continuity of the total normal stresses at the interface.
The second equation expresses the fact that there is no relative mass flux across the
impervious interface. The third equation ensures the continuity of the solid displacement
vectors. Substituting Equation (13.12) into Equation (13.11), one obtains :
I
p
I
e
=
0;
this equation shows that the coupling between the poroelastic and the elastic media
is natural. Only the kinematic boundary condition
u
s
+
=
u
e
will have to be explicitly
imposed on
. In a finite element implementation, this may be done automatically through
assembling between the solid phase of the porous media and the elastic media.
13.4.2 Poroelastic - acoustic coupling condition
The acoustic medium is described in terms of its pressure field
p
a
.Let
a
and
a
denote
its volume and its boundary. Using pressure as the fluid variable and assuming harmonic
oscillations, the weak integral form of the fluid system is given by
p
a
δp
a
d
ρ
0
ω
2
∂p
a
∂x
i
∂(δp
a
)
∂x
i
∂p
a
∂n
δp
a
d
S
1
ρ
0
ω
2
1
ρ
0
c
0
1
δp
a
−
−
=
0
∀
a
a
(13.13)
where
δp
a
an arbitrary admissible variation of
p
a
,ρ
0
is the density in the acoustic domain,
c
0
is the speed of sound in the acoustic medium, and
∂/∂n
is the normal derivative. The
surface integral
ρ
0
ω
2
∂p
a
∂n
δp
a
d
S
1
I
a
=−
(13.14)
a
represents the virtual work done by the internal pressure at the boundary of the acoustic
domain due to an imposed motion on its surface.
Combining the weak formulations of the poroelastic and acoustic media, the boundary
integrals can be rewritten
∂p
a
1
ρ
0
ω
2
I
p
+
I
a
σ
ij
n
j
δu
s
i
d
S
−
φ(u
n
−
u
s
n
)δp
d
S
+
∂n
·
δp
a
d
S
(13.15)
=−
The positive sign of the last term of
I
p
I
a
is due to the direction of the normal vector
n
, inwards to the acoustic medium. The coupling conditions at the interface
are given
by Equation (11.72):
$
+
σ
ij
n
j
=−
p
a
n
i
∂p
a
∂n
=
1
ρ
0
ω
2
(13.16)
φ)u
s
n
+
φu
n
=
u
s
n
+
φ(u
n
−
u
s
n
)
(
1
−
%
p
a
p
=
