Civil Engineering Reference
In-Depth Information
1.7
Wave equation in a fluid
In the case of an inviscid fluid,
µ
vanishes. The stress coefficients reduce to
σ
11
=
σ
22
=
σ
33
=
λθ
(1.49)
σ
12
=
σ
13
=
σ
23
=
0
p
,where
p
is the pressure. The bulk
modulus
K
, given by Equation (1.38), becomes simply
λ
:
The three nonzero stress elements are equal to
−
K
=
λ
(1.50)
The stress field (Equation 1.49) generates only irrotational deformations such as
=
0.
A representation of the displacement vector
u
in the following form can be used:
u
1
=
∂ϕ/∂x
1
,
2
=
∂ϕ/∂x
2
,
3
=
∂ϕ/∂x
3
(1.51)
where
ϕ
is a displacement potential.
In vector form, Equations (1.51) can be written as
u
=
∇
ϕ
(1.52)
Using this representation, the rotation vector
can be rewritten
1
2
curl
∇
ϕ
=
0
=
(1.53)
and the displacement field is irrotational.
Substitution of this displacement representation into Equation (1.46) with
µ =
0and
X
=
0 yields
ρ
∂
2
λ
∇∇
·
∇
ϕ
=
∂t
2
∇
ϕ
(1.54)
2
ϕ
, Equation (1.54) reduces, with Equation (1.50), to
Since
∇
·
∇
ϕ
=∇
K
∇
∂t
2
ϕ
2
ϕ
−
ρ
∂
2
∇
=
0
(1.55)
The displacement potential
ϕ
satisfies the equation of motion if
2
ϕ
=
ρ
∂
2
ϕ
K∂t
2
∇
(1.56)
If the fluid is a perfectly elastic fluid, with no damping,
K
is a real number.
This displacement potential is related to pressure in a simple way. From Equations
(1.49), (1.50) and (1.52),
p
can be written as
2
ϕ
p
=−
Kθ
=−
K
∇
(1.57)
