Civil Engineering Reference
In-Depth Information
conditions are
σ
ij
n
j
p
i
n
i
=−
(13.25)
p
i
p
=
which express the continuity of the total normal stress and the continuity of the pressure
through the interface
. Since the pressure is imposed, the admissible variation
δp
will
fall to zero. Consequently, the surface integrals of Equation (13.8) simplify to
I
p
p
0
δu
s
n
d
S
=
(13.26)
p
0
on
,a
pressure excitation on the solid phase, given by Equation (13.8), needs to be applied.
This equation indicates that, in addition to the kinematic condition
p
=
13.4.6 Case of an imposed displacement field
In the case of an imposed displacement field
u
0
applied on
, used for instance to
simulate a piston motion, the boundary conditions are
u
i
n
=
u
n
φu
n
φ)u
s
n
+
u
s
n
=
0
u
s
i
=
u
i
(
1
−
−
⇒
(13.27)
u
i
=
u
s
i
The first condition expresses the continuity of the normal displacements between the
solid phase and the fluid phase. The second equation expresses the continuity between
the imposed displacement vector and the solid phase displacement vector. Since the
displacement is imposed, the admissible variation
δ
u
s
=
0
. Consequently, the surface
integrals of Equation (13.8) simplify to
I
p
=
0
(13.28)
This equation indicates that for an imposed displacement, the boundary conditions reduce
to the kinematic condition
u
s
=
u
0
on
.
13.4.7 Coupling with a semi-infinite waveguide
We consider here the coupling of a poroelastic medium with an infinite waveguide. An
example of application is the prediction of the absorption and transmission through a
nonhomogenous material made from patches of porous, solid and air media (Sgard 2002,
Atalla
et al
. 2003). The example of double porosity material is detailed in Section 13.9.5.
Let
p
a
,ρ
0
and c
0
denote the acoustic pressure, the density and speed of sound in the
waveguide, respectively. For a porous material placed in the wave guide, as represented
in Figure 13.2, the continuity conditions at the front
1
and rear
2
porous interfaces
are given by Equation (13.8)
I
i
=−
σ
ij
n
j
δu
s
i
d
S
−
φ(u
n
−
u
s
n
)δp
d
Si
=
1
,
2
(13.29)
i
i
