Civil Engineering Reference
In-Depth Information
p(i),v
z
(i),σ
zz
(i),σ
xz
(i),σ
yz
(i),i
=
1
,
2
,
3
,
4 be the contributions at the surface of the
layer of the four waves with a normalization factor
N
i
=
1 to the pressure, the normal
velocity and the related total stress components The four normalization coefficients are
obtained from the set
p
e
N
i
σ
zz
(i)
−
+
τ
zz
=
(10.111)
i
=
1
,
2
,
3
,
4
where
jω
i
p
e
=−
N
i
(u
z
(i)
+
w
z
(i))Z
0
/
cos
θ
=
1
,
2
,
3
,
4
(10.112)
N
i
σ
xz
(i)
=
0
,
i
=
1
,
2
,
3
,
4
N
i
σ
yz
(i)
=
0
,
(10.113)
i
=
1
,
2
,
3
,
4
jω
i
−
N
i
(u
z
(i)
+
w
z
(i))Z
0
/
cos
θ
=
N
i
p(i)
(10.114)
=
1
,
2
,
3
,
4
i
=
1
,
2
,
3
,
4
The vertical displacement of the frame is
N
i
u
z
(i)
.
10.8
Rayleigh poles and Rayleigh waves
When the symmetry axis
Z
and the axis
Z
normal to the surface are parallel, the porous
layer is invariant under rotations around
Z
. The Hankel transform can be used for circular
excitations and the Sommerfeld representation for the point sources in air. When
Z
and
Z
are not parallel, the invariance disappears, and Equation (7.5) cannot be replaced by
the Sommerfeld integral. For similar reasons, the Hankel transform cannot be used for
circular excitations. An excitation by a line source can be used to create Rayleigh waves
which propagate in the direction perpendicular to the line source. The source can be a
thin bar (blade) bonded on the porous surface on an axis
which moves in the normal
direction. A rough model for the source is a normal stress distribution applied to the
frame, centered on
, and given by
τ
zz
(y,t)
=
g(r)h(t)
(10.115)
where
r
is the distance from the line source. The stress field can be rewritten
∞
∞
1
4
π
2
h(ω)g(q
r
)
exp
(jω(t
τ
zz
(r,t)
=
−
q
r
r))ω
d
q
r
d
ω
(10.116)
−∞
−∞
where
h
and
g
are Fourier tansforms given by
∞
h(ω)
=
h(t)
exp
(
−
jωt)
d
t
(10.117)
−∞
