Civil Engineering Reference
In-Depth Information
By the use of Equations (1.56) and (1.57) one obtains
ρ
∂
2
ϕ
∂t
2
p
=−
(1.58)
At an angular frequency
ω
(
ω
=
2
πf
,where
f
is frequency),
p
can be rewritten as
p
=
ρω
2
ϕ
(1.59)
As an example, a simple solution of Equation (1.56) is
A
ρω
2
ϕ
=
exp[
j(
−
kx
3
+
ωt)
+
α
]
(1.60)
In this equation,
A
and
α
are arbitrary constants, and
k
is the wave number
ω(ρ/K)
1
/
2
k
=
(1.61)
The phase velocity is given by
c
=
ω/
Re
k
(1.62)
and Im
(k)
appears in the amplitude dependence on
x
3
,exp
(
Im
(k)x
3
)
. In this example,
u
3
is the only nonzero component of
u
:
∂ϕ
∂x
3
=−
jkA
ρω
2
exp[
j(
u
3
=
−
kx
3
+
ωt
+
α)
]
(1.63)
The pressure
p
is
ρ
∂
2
ϕ
∂t
2
p
=−
=
A
exp[
j(
−
kx
3
+
ωt
+
α)
]
(1.64)
This field of deformation corresponds to the propagation parallel to the
x
3
axis of a
longitudinal strain, with a phase velocity
c
.
1.8
Wave equations in an elastic solid
A scalar potential
ϕ
and a vector potential
ψ
(ψ
1
,ψ
2
,ψ
3
)
can be used to represent dis-
placements in a solid
∂ϕ
∂x
1
+
∂ψ
3
∂x
2
−
∂ψ
2
∂x
3
u
1
=
∂ϕ
∂x
2
+
∂ψ
1
∂x
3
−
∂ψ
3
∂x
1
u
2
=
(1.65)
∂ϕ
∂x
3
+
∂ψ
2
∂x
1
−
∂ψ
1
∂x
2
u
3
=
In vector form, Equations (1.65) reduce to
u
=
grad
ϕ
+
curl
ψ
(1.66)
