Civil Engineering Reference
In-Depth Information
6.4
Wave equations
The equations of motion (1.44) in an elastic solid without external forces are
ρ
∂
2
u
s
i
∂t
2
=
(λ
+
µ)
∂θ
s
2
u
s
i
∂x
i
+
µ
∇
i
=
1
,
2
,
3
(6.46)
Using Equation (1.76), these equations can be rewritten
ρ
∂
2
u
s
i
∂t
2
=
(K
c
−
µ)
∂θ
s
2
u
s
i
∂x
i
+
µ
∇
(6.47)
The equations of motion of the frame can be obtained by modifying Equation (6.47)
in the following way: by comparing Equations (6.2) and (1.21), where
λ
2
µ
,it
appears that an extra term
Q(∂θ
f
/∂x
i
)
must be placed on the right-hand side of Equation
(6.47), and
K
c
−
2
µ
and
µ
must be replaced by
P
and
N
, respectively. For an nonviscous
fluid the inertial force at the left-hand side of Equation (6.47) is given by Equation (6.33),
where
ρ
11
is equal to
ρ
1
+
ρ
a
=
K
c
−
ρ
a
∂
2
u
i
∂t
2
ρ
∂
2
u
s
i
∂t
2
ρ
a
)
∂
2
u
s
i
∂t
2
→
(ρ
1
+
−
i
=
1
,
2
,
3
(6.48)
with
ρ
a
=
φρ
0
(α
∞
−
1
)
. For a viscous fluid, in the frequency domain,
α
∞
in
ρ
a
is
νφ
jωq
0
G(ω)
(see Eqautions (5.50) - (5.57)). The inertial coupling term is
replaced by
α
∞
+
replaced by
−
ω
2
ρ
a
(u
s
i
−
u
i
)
→−
ω
2
ρ
a
(u
s
i
−
u
i
)
+
σφ
2
G(ω)jω(u
s
i
−
u
i
)
(6.49)
Equation (6.47) becomes
ω
2
ρ
a
u
i
ω
2
u
s
i
(ρ
1
−
+
ρ
a
)
+
N)
∂θ
s
Q
∂θ
f
u
i
)
2
u
s
i
+
σφ
2
G(ω)jω(u
s
i
−
=
(P
−
∂x
i
+
N
∇
∂x
i
−
(6.50)
i
=
1
,
2
,
3
In the same way, the following equations can be obtained for the air in the porous
material:
R
∂θ
f
Q
∂θ
s
ω
2
u
i
(φρ
o
u
i
)
ω
2
ρ
a
u
s
i
=
σφ
2
G(ω)jω(u
s
i
−
−
+
ρ
a
)
+
∂x
i
+
∂x
i
+
(6.51)
i
=
1
,
2
,
3
In vector form, Equations (6.50) and (6.52) can be rewritten
ω
2
u
s
(ρ
1
+
ω
2
ρ
a
u
f
−
ρ
a
)
+
(6.52)
∇∇
·
u
s
∇∇
·
u
f
2
u
s
jωσφ
2
G(ω)(
u
s
−
u
f
)
=
(P
−
N)
+
Q
+
N
∇
−
ω
2
(φρ
o
ρ
a
)
u
f
ω
2
ρ
a
u
s
−
+
+
(6.53)
u
f
u
s
jωσφ
2
G(ω)(
u
s
u
f
)
=
R
∇∇
·
+
Q
∇∇
·
+
−
