Eduardo Kausel-Fundamental solutions in elastodynamics_ a compendium-Cambridge University Press (2006)

162 Stiffness matrix method for layered media
Therefore, from Table 10.1, we have
{ } {
ũr K
SVP
11
K12
SVP
=
ũ z
(1)
K21 SVP K22
SVP
} −1 {
˜pr
˜p z
}
(1)
{
= 1 s
1
2 (1 − s2 ) ps − 1 2 (1 + } { }
s2 ) 1 1
2kµ ps − 1 2 (1 + s2 ) p 1 2 (1 − s2 ) 2π 0
{
1 s
1
2
=
(1 − }
s2 )
4π kµ ps − 1 2 (1 + (10.80)
s2 )
and
ũ θ(1) = 1
K ˜p SH θ(1)
( )( )
(10.81)
1 1 1
=
=
ks µ 2π 2π ks µ
Observe that, unlike the plane-strain case, the vertical components do not carry an imaginary
factor. Finally, the displacements in the spatial domain are
⎧ ⎫ ⎧
⎪⎨
u r ⎪⎬ ∫ ∞
⎪⎨
J ′ 1
1 kr J ⎫ ⎧ ⎫
1 0 ⎪⎬ ⎪⎨ ũ r ⎪⎬
1
u = u θ = T 1
⎪⎩ ⎪
kr
⎭ 0 ⎪⎩
J 1 J
1 ′ 0 ũ
⎪ ⎭ ⎪ θ kdk (10.82)
⎩ ⎪ ⎭
u z 0 0 J 1 ũ z (1)
that is
[
1
u r = (cos θ)
4πµ
∫ ∞
[
1
u θ = (− sin θ)
4πµ
[
1
u z = (cos θ)
4πµ
0
∫ ∞
0
∫ ∞
0
s 1 2 (1 − s2 )
s 1 2 (1 − s2 )
(
J 0 (kr) − J )
1(kr)
dk + 1
kr 2π µ
J 1 (kr)
dk + 1
kr 2π µ
ps − 1 2 (1 + ]
s2 )
J 1 (kr) dk
∫ ∞
0
1
s
∫ ∞
0
1
s
]
J 1 (kr)
dk
kr
(10.83)
(
J 0 (kr) − J ) ]
1(kr)
dk
kr
(10.84)
(10.85)
which can be evaluated by numerical integration or, as Chao does, by analytical means.
Example 10.7: Same as Example 10.6, but for a layer over an elastic half-space
The procedure is identical to that in Example 10.6. The only difference now is that in the
frequency–wavenumber domain, we must first solve the two systems
⎧
⎧ ⎫
ũ r1
1
⎪⎨
⎫⎪ ⎬
ũ z1
= ( K SVP) −1 1
⎪⎨ ⎪⎬ { }
0
ũθ1
and
= ( K SH) { }
−1 1 1
(10.86)
ũ ⎪⎩ r2 ⎪ 2π 0
ũ ⎭ ⎪⎩ ⎪⎭
θ2 2π 0
(1)
ũ z2 0
(1)