Waves Used in NDE
Learning Objectives
Bulk Compressional, Shear Waves
Rayleigh (surface) waves
Lamb (plate) waves
Compressional and Shear Wave Contact Transducers
P-wave transducer
S-wave transducer
fluid couplant
"solid" couplant
Angle Beam Shear Wave Transducer
P-wave crystal and backing
reflected
P- and S-waves
(suppressed)
θ p1
θs2
weld inspections
plastic wedge
SV-wave
Mode Conversion
fluid
ρ ,c
p1
1
incident
P-wave
v
p
θ p1
θ p1
v
inc
p
reflt
v
solid
ρ , cp2 , c s2
2
θ
x
p
trans
p2
θ
y
Generalized
Snell's law:
reflected
P-wave
v
s2
s
trans
transmitted
P-wave (φ)
transmitted
S-wave (ψ)
sin θ p1 sin θ p2 sin θ s2
=
=
c p1
c p2
cs2
Critical Angles
P
θp1
P
P
θp1
θp1
P
S
S
−1 ⎛ c p1 ⎞
θ p1 < sin ⎜ ⎟
⎝ c p2 ⎠
inhomogeneous
P-wave
−1 ⎛
inhomogeneous
P, S waves
c p1 ⎞
−1⎛ c p1 ⎞
< θ p1 < sin ⎜ ⎟
sin ⎜
⎟
⎝ cs 2 ⎠
⎝ c p2 ⎠
⎛ c p1 ⎞
(θ CR )1 = sin ⎜⎜ ⎟⎟
⎝ cp2 ⎠
−1
−1 ⎛ c p1 ⎞
θ p1 > sin ⎜
⎟
⎝ cs 2 ⎠
⎛ c p1 ⎞
⎟
c
⎝ s2 ⎠
(θ CR )2 = sin −1 ⎜
Rayleigh (Surface)Wave Transducer
θ p1
angle chosen such that
c p1
sin θ p1 =
cR 2
stress-free surface
Rayleigh wave
In the late 1800's Lord Rayleigh looked for a wave
confined near the stress-free surface of an elastic solid
of the form:
x
φ = A exp[−αy]exp[ik ( x − ct )]
ψ = B exp[−βy ]exp[ik ( x − ct )]
y
2
∂
φ
1
∇ 2φ − 2 2 = 0
c p ∂t
By satisfying the equations of motion
and the stress-free boundary
conditions τ yy = τ xy = 0 on y = 0
2
∂
ψ
1
∇2ψ − 2 2 = 0
cs ∂t
Rayleigh found that the wave speed, c, must satisfy
(
2−c /
2
)
2 2
cs
− 4 1 − c / cp 1 − c / cs = 0
2
2
2
2
Rayleigh's equation
(2 − c
2
/
) −4
2 2
cs
1− c /
2
2
cp
1− c /
2
2
cs
=0
There is always one real root of this equation, c = c R
where c R < c s
cR ≅
A good approximation of this root is:
0.862 +1.14ν
cs
1+ ν
ν ... Poisson's ratio
The Rayleigh wave travels about 90% of the shear wave speed
displacements
ux
uy
ux
uy
stresses
τyy
τxx
τxx
τyy ,τxy
τxy
The depth of penetration is a function of the frequency
Lamb (Plate) Waves
c = c(ω )
ω =2πf … frequency
(rad/sec)
2h
If one looks for solutions of the form
φ = f (y ) exp[ik ( x − ct )]
ψ = g( y) exp[ik ( x − ct )]
then solutions of the following two types are found:
extensional waves
f = A cosh(α y)
g = Bsinh(βy)
(a)
y
2h
flexural waves
10
x
10
(b)
f = A′ sinh (αy )
g = B′ cosh(βy )
satisfying the boundary conditions τ yy = τ xy = 0
on y = ±h gives the Rayleigh-Lamb equations:
4ω 2αβ
tanh (βh ) ⎡
=⎢ 2 2 2
tanh (αh ) ⎢ c ω / c + β 2
⎣
(
ω
c2
α=
1− 2
c
cp
⎤
⎥
2
⎥
⎦
±1
)
+ … extensional waves
- … flexural waves
ω
c2
,β=
1− 2
c
cs
There are multiple solutions of these equations. For each
solution the wave speed, c, is a different function of
frequency. Each of these different solutions is called a "mode"
of the plate.
consider the extensional waves
tanh ⎡ 2π fh 1/ c 2 − 1/ cs2
⎣
tanh ⎡ 2π fh 1/ c 2 − 1/ c 2p
⎣
2πfh
>> 1
If we let kh =
c
then both tanh functions are
and we find
(2 − c
2
/
⎤ 4 1 − c2 / c2 1 − c2 / c2
s
p
⎦=
2
2 2
⎤
2 − c / cs )
(
⎦
(high frequency)
≅1
) =4
2 2
cs
1 − c / cs 1 − c / c p
2
2
2
2
so we just have Rayleigh waves on both stress-free surfaces:
In contrast for kh <<1 (low frequency)
tanh (αh ) ≅ αh
we find
tanh (βh) ≅ βh
and the Rayleigh-Lamb equation reduces to
(
2−c /
2
)
2 2
cs
(
= 4 1 − c / cp
2
which can be solved for c to give
c = c plate =
E
ρ 1−ν 2
(
)
2
)
8000
fundamental
extensional
mode
7000
c
6000
phase 5000
velocity
4000 c = c
plate
(m/s)
c = cR
3000
2000
fundamental
flexural
mode
1000
0
0
1
2
symmetric modes
anti-symmetric modes
3
4
5
6
frequency-thickness (MHz-mm)
7
8