Download 8. Surface waves and guided waves

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Surface waves and guided waves
4.1
Rayleigh surface wave
We consider a time harmonic plane wave propagating in the x1 -x2 plane and along the free surface x2 = 0
of an isotropic half space defined in the region x2 < 0. As seen in Sec.3.1, the displacements of the plane
wave in an isotropic media are expressed as a sum of three types of wave motions:
3
∑
ui =
A(m) d(m) exp(ik (m) p(m) · x)
(231)
m=1
where
k (1)
(m)
p3
(1)
d
= kp , k (2) = k (3) = ks
(232)
= 0 (m = 1, 2, 3)
= p
(1)
, d
(3)
(233)
(2)
= (0, 0, 1), d
=d
(3)
×p
(2)
.
(234)
The boundary condition on the free surface is given by τ2j = 0 (j = 1, 2, 3), namely,
τ2j =
3
∑
(m) (m)
ik (m) C2jkl dk pl
A(m) exp(ik (m) p(m) · x) = 0.
(235)
m=1
The above equations must be valid for all values of x1 , which implies
(1)
(3)
(2)
k (1) p1 = k (2) p1 = k (3) p1 = k.
(236)
From eqs.(232), (233) and (236), we have
√
√
p(1) = (k/kp , 1 − (k/kp )2 , 0), p(2) = p(3) = (k/ks , 1 − (k/ks )2 , 0)
and hence d(m) can be determined using eq.(234). Substitution of k (m) , p(m) and d(m)
condition (235) and some manipulation yields
√


 
2k 2 − ks2
0
2k kp2 − k 2
 A(1) /kp  


√
iµeikx1  k 2 − 2k 2
=
A(2) /k

2k ks2 − k 2
s
 (3) s  
√0
A
/k
2
2
s
0
0
ks k − k
s
(237)
into the boundary

0 
0
.

0
(238)
For a nontrivial solution, the determinant of the coefficients of A(1) /kp and A(2) /ks must vanish for
inplane motions and k = ks for antiplane motions.
The determinant for inplane motions is written as
√
√
(2k 2 − ks2 )2 + 4k 2 kp2 − k 2 ks2 − k 2 = 0.
(239)
The LHS of eq.(239) approaches from a negative value to zero as k → ∞ and takes a positive value at
k = ks . (If ks /k = ²(¿ 1) is substituted into eq.(239), then the LHS eq.(239) divided by 1/k 4 becomes
approximately 2((kp /ks )2 − 1)²2 < 0.) Thus eq.(239) has a real solution
for kp < ks < k√and it can
√
2
be proved that the solution is only one real root. For kp < ks < k, kα − k 2 becomes −i k 2 − kα2 to
satisfy the radiation condition for x2 → −∞. Then eq.(239) is transformed into the so-called Rayleigh
equation:
√
√
(2k 2 − ks2 )2 − 4k 2 k 2 − kp2 k 2 − ks2 = 0
(240)
which gives the velocity cR (cR < cs < cp ). After determining the velocity cR , or the wavenumber kR
from eq.(240), the ratio of amplitude A(2) /A(1) can be obtained as the eigenmode solution of eq.(238):
√
2 − k2
kR
(2)
2
2
2ik
k
R
s
p
A
−iks (ks − 2kR )
√
.
(241)
=
=
2
2
(1)
2 − k2
kp (2kR − ks )
A
2kR kp kR
s
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A general solution of the inplane displacement may thus be written in the form
{
}
2 √
kR √kR2 −kp2 x2 ks2 − 2kR
2
2
u1 = A(1)
e
e kR −ks x2 eikR x1
+
kp
2kR kp
√

(
)
2

√ 2 2 
√ 2 2
2
2
k
−
2k
k
R
s
R
√
u2 = −iA(1)
− 1e kR −kp x2 +
e kR −ks x2 eikR x1 .
2 − k2


kp
2kp kR
s
(242)
(243)
It can be seen that the displacement decays exponentially with distance from the free surface and propagates in the x1 -direction, which is a surface wave called the Rayleigh wave. As seen in eq.(240), the
velocity cR is independent of the frequency and the Rayleigh wave is thus non-dispersive.
As seen in eq.(238), a nontrivial solution for antiplane motions can be obtained as k = ks . For k = ks ,
we have p(3) = (1, 0, 0), which is the propagation vector of the body wave with the constant amplitude
and no surface wave is generated in a half space.
4.2
Love surface wave
Let us consider a surface wave with the vibration polarization in the x3 direction in a layered media as
shown in Fig. 9. The displacements are assumed to be as follows:
u3 (x) = A(1) exp(iks p(1) · x) x ∈ D
ū3 (x) = A(2) exp(ik̄s p(2) · x) + A(3) exp(ik̄s p(3) · x) x ∈ D̄
(244)
(245)
where no incident wave is assumed. Since the equations of motion are satisfied by the displacement fields
in eqs.(244) and (245). The remaining requirement for the wave fields is the boundary conditions at
x2 = 0 and x2 = h:
u3
τ̄23
= ū3 , τ23 = τ̄23 for x2 = 0
= 0 for x2 = h
(246)
(247)
where τ23 = µ∂u3 /∂x2 . To satisfy the boundary conditions, the exponentials involved in the displacements
must be a common term. Then we have
(1)
=
(1)
p2
=
ks p1
(2)
(3)
k̄s p1 = k̄s p1 = k,
√
√
(3)
(2)
2
1 − (k/ks ) , p2 = −p2 = 1 − (k/k̄s )2 .
(248)
(249)
Using the above relations, the boundary conditions (246) and (247) are written in a matrix form:


  
−1
1
1
 A(1)   0 
√
√
√
 − µks 1 − (k/k )2

2
2
1 − (k/k̄s )
− 1 − (k/k̄s )
s
0
=
A(2)

 µ̄k̄s
√
√
√
√
 (3)   
2h
2h
i
1−(k/
k̄
)
−i
1−(k/
k̄
)
0
s
s
A
− 1 − (k/k̄s )2 e
0
1 − (k/k̄s )2 e
(250)
To have non-trivial solutions in eq.(250), the determinant of the matrix must be zero. The result is
√
( √
)
µ ks2 − k 2
2
2
.
(251)
tan h k̄s − k = i √
µ̄ k̄s2 − k 2
√
√
The above equation has a real solution in the interval ks < k < k̄s , if ks2 − k 2 is taken as −i k 2 − ks2
to satisfy the radiation condition of u3 as x2 → −∞. No real root exists if k̄s < ks . Consequently, the
characteristic equation (251) for Love waves can be written as
( √
)
√
√
( √
)
2 − k2
2
k
ω
µ
c
µ 1 − c2 /c2s
s
2
2
or tan
.
(252)
tan h k̄s − k = √
h
−1 = √
µ̄ k̄s2 − k 2
c
c̄2s
µ̄ c2 /c̄2s − 1
The above equation shows that the wave velocity c depends on the frequency and thus Love waves are
dispersive.
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Problem 4-1 Consider an antiplane motion in an isotropic homogeneous plate with the thickness 2h
as shown in Fig. 11. Assume that the displacement of an antiplane wave in a plate consists of upgoing
and downgoing waves defined by
(1)
(2)
u3 = u 3 + u3 =
2
∑
A(m) exp(iks p(m) · x)
(253)
m=1
√
where ks is the transverse wave number and p(m) = (k/ks , (−1)(m+1) 1 − (k/ks )2 ). The traction
free condition of τ23 = µ∂u3 /∂x2 = 0 is satisfied at the boundaries x2 = ±h. Obtain the dispersion
equation for antiplane waves in a plate in the following form:
√
ks2 − k 2 h = nπ/2 (n = 0, 1, 2, . . .)
(254)
x2
x 2 =h
D
(2)
(1)
u3
u3
x1
x 2 =-h
Figure 11: SH guided wave in a plate.
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