Part 4

RAYLEIGH AND LAMB

WAVES

Rayleigh Surface Wave

x1x2

x3

surface wave

x3

x1

Partial Wave Decomposition Displacement potential: = ∇ϕ + ∇×u ψ Wave equations:

2 2

2 2 2 2sd2 2 2 2

sdandk k

c t c t1 ∂ ϕ 1 ∂

∇ ϕ = = − ϕ ∇ = = −∂ ∂

ψψ ψ

Wave velocities:

sd2 andc cλ+ μ μ

= =ρ ρ

Wave propagates in the 1x direction:

2 1 3 22

0 0, 0,ux∂

= ⇒ = ψ = ψ = ψ = ψ∂

Simplified wave equation:

2 2

2d2 2

1 30k

x x∂ ϕ ∂ ϕ

+ + ϕ =∂ ∂

2 2

2s2 2

1 30k

x x∂ ψ ∂ ψ

+ + ψ =∂ ∂

Axial displacement (tangential to the surface)

11 3

ux x∂ϕ ∂ψ

= −∂ ∂

Transverse displacement (normal to the surface):

33 1

ux x∂ϕ ∂ψ

= +∂ ∂

Normal stress in the propagation direction:

2 2 2 2

11 2 2 2 1 31 3 1( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ −∂ ∂∂ ∂ ∂

Normal stress parallel to the surface:

2 2 2 2

33 2 2 2 1 31 3 3( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ +∂ ∂∂ ∂ ∂

Shear stress:

2 2 2

31 13 2 21 3 1 3(2 )

x x x x∂ ϕ ∂ ψ ∂ ψ

τ = τ = μ + −∂ ∂ ∂ ∂

We seek solution in the following form

1 1( )3( ) i k x tF x e − ωϕ =

1 1( )3( ) i k x tG x e − ωψ =

1 RR

k kcω

→ =

From the wave equation,

2 2 2

2 2 2Rd d2 2 2

1 3 30 ( )Fk k k F

x x x∂ ϕ ∂ ϕ ∂

+ + ϕ = ⇒ = −∂ ∂ ∂

2 2 2

2 2 2s sR2 2 2

1 3 30 ( )Gk k k G

x x x∂ ψ ∂ ψ ∂

+ + ϕ = ⇒ = −∂ ∂ ∂

Solutions:

2 2

3 R d3( ) x k kF x Ae− −= and 2 2

3 sR3( ) x k kG x Be− −= The potential functions can be rewritten as follows

d 3 R 1( )x i k x tAe e−κ − ωϕ =

s 3 R 1( )x i k x tBe e−κ − ωψ =

2 2d R dk kκ = − and 2 2

s sRk kκ = −

The surface wave solution is a combination of coupled partial (longitudinal and shear) waves.

The partial wave amplitude ratio (B /A) is determined

by the condition that the surface is traction free.

Both partial waves are evanescent, i. e., R s dk k k> > or R s dc c c< <

Boundary conditions:

Both normal and tangential stresses vanish on the surface ( 3 0x = )

2 2 2 2

33 2 2 2 1 31 3 3( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ +∂ ∂∂ ∂ ∂

2 2s RRd( 2 ) 2k i k= λ+ μ κ ϕ − λ ϕ − μ κ ψ

2 2 2

31 13 2 21 3 1 3(2 )

x x x x∂ ϕ ∂ ψ ∂ ψ

τ = τ = μ + −∂ ∂ ∂ ∂

2 2d R s R2 ( )i k k= − μ κ ϕ − μ κ + ψ

At the surface, without the common term of R 1( )i k x te − ω :

2 2

s RRd2 2

d R s R

( 2 ) 200 2 ( )

k i k ABi k k

⎡ ⎤λ+ μ κ − λ − μ κ⎡ ⎤ ⎡ ⎤⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦− μ κ −μ κ +⎣ ⎦

2 2 2 2 2 2 2 2

s sR R R Rd d( 2 ) 2 ( 2 ) 2 ( )k k k k k kλ+ μ κ − λ = μ − λ+ μ = μ −μ = μ κ +

2 2s s RR

2 2d R s R

( ) 200 2 ( )

k i k ABi k k

⎡ ⎤κ + − κ⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥

− κ − κ +⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

Characteristic equation:

2 2 2 2s s dR R( ) 4 0k kκ + − κ κ =

Rayleigh Velocity

s

d

1 2( )2(1 )

cc

− νξ = =

−ν

Rs

cc

η =

Exact Rayleigh equation:

6 4 2 2 28 8(3 2 ) 16(1 ) 0η − η + − ξ η − − ξ = Approximate expression:

0.87 1.12

1+ ν

η ≈+ν

Shear Velocity / Longitudinal Velocity

Ray

leig

h V

eloc

ity /

Shea

r Vel

ocity

0.88

0.89

0.9

0.91

0.92

0.93

0.94

0.95

0.96

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

approximationexact

Partial Wave Composition The partial wave amplitude ratio:

2 2s d RR

2 2s R s R

22

k i kBA i k k

κ + κ= = −

κ κ +

By substituting the Rayleigh wave number Rk from the characteristic equation:

d

s

B i iA

κ= − = − ζ

κ

( ) 1.5 1.6vζ = ζ ≈ −

d 3xA e−κϕ = and s 3xi A e−κψ = − ζ

(without the common R 1( )i k x te − ω term)

Displacements Axial displacement (tangential to the surface):

d 3 s 31 R s1 3

( )x xu i A k e ex x

−κ −κ∂φ ∂ψ= − = − ζκ∂ ∂

Transverse displacement (normal to the surface):

d 3 s 33 d R3 1

( )x xu A e k ex x

−κ −κ∂φ ∂ψ= + = − κ − ζ∂ ∂

3

3 d R

1 R s0x

u ki iu k=

κ − ζ= = − ζ

− ζκ

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2Depth / Rayleigh Wavelength

Nor

mal

ized

Par

ticle

Dis

plac

emen

t.

normal (u3)

tangential (u1)

ν = 0.3

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 20 0.5 1 1.5 2Depth / Rayleigh Wavelength

Nor

mal

ized

Par

ticle

Dis

plac

emen

t.

normal (u3)

tangential (u1)

ν = 0.3

Stresses

d 3 s 313 R d2 ( )x xi A k e e−κ −κτ = − μ κ −

d 3 s 32 233 sR( )( )x xA k e e−κ −κτ = μ + κ −

13 R d2 233 sR

2i k ik

τ κ= − = − ζ

τ + κ

2

s d R2 211 d 3 s 3R d 2 2

s R

4( 2 ) exp[ ] exp[ ]

kA k x x

k

⎧ ⎫μκ κ⎪ ⎪⎡ ⎤τ = − λ+ μ −λκ −κ − −κ⎨ ⎬⎣ ⎦ κ +⎪ ⎪⎩ ⎭

Depth / Rayleigh Wavelength

Nor

mal

ized

Stre

ss

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

τ

13τ

τ33

11

ν = 0.3

LAMB PLATE WAVES

Lowest-order Lamb waves in a plate

a) symmetric

b) asymmetric

x1

x2

x3

d

wave direction

2

Potential Method 1 1 1 3 3 2 1 1 3 3exp[ ( )] exp[ ( )]d d d dA i k x k x A i k x k xϕ = + + −

2 23 1d dd dk k k i= − = κ or 2 2

1d d dk kκ = −

1 1 1 3 3 2 1 1 3 3exp[ ( )] exp[ ( )]s s s sB i k x k x B i k x k xψ = + + −

2 23 1s s ssk k k i= − = κ or 2 2

1s ssk kκ = −

ks1 = kd1 = k, 2 2d dk kκ = − , 2 2

s sk kκ = −

3 1 3 1cosh( )exp( ) sinh( )exp( )s d a dA x i k x A x i k xϕ = κ + κ 3 1 3 1sinh( )exp( ) cosh( )exp( )s s a sB x i k x B x i k xψ = κ + κ

2 1sA A A= + , 2 1aA A A= − , 2 1sB B B= − and 2 1aB B B= +

Boundary Conditions

33 3 33 3 31 3 31 3( ) ( ) ( ) ( ) 0x d x d x d x dτ = = τ = − = τ = = τ = − =

normal stress:

2 2 2 2

33 2 2 2 1 31 3 3( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ +∂ ∂∂ ∂ ∂

shear stress:

2 2 2

31 13 2 21 3 1 3(2 )

x x x x∂ ϕ ∂ ψ ∂ ψ

τ = τ = μ + −∂ ∂ ∂ ∂

Example of Boundary Conditions Normal stress on the upper surface of the plate: 33 3( ) 0x dτ = =

2 2 2 2 2

33 2 2 2 21 31 3 1( ) 2( )d

s

cx xc x x x

∂ ϕ ∂ ϕ ∂ ψ ∂ ϕτ μ = + + −

∂ ∂∂ ∂ ∂/

Partial derivatives (without the common 1exp( )i k x :

2

23 32

1[ cosh( ) sinh( )]s d a dk A x A x

x∂ ϕ

= − κ + κ∂

2

23 32

3[ cosh( ) sinh( )]s d a dd A x A x

x∂ ϕ

= κ κ + κ∂

2

3 31 3

[ cosh( ) sinh( )]s s s a si k B x B xx x∂ ψ

= κ κ + κ∂ ∂

Normal stress on the upper surface of the plate:

2 2 2 233 ( )cosh( ) ( )sinh( )

2 cosh( ) 2 sinh( ) 0

s d s s d a

s s s s s a

k d A k d A

i k d B i k d B

τ μ = +κ κ + +κ κ

+ κ κ + κ κ =

/

Full 4×4 Boundary Condition Matrix

0000

s

a

s

a

AABB

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

a

2 2 2 2

2 2 2 2

2 2 2 2

( )cosh( ) ( )sinh( ) 2 cosh( ) 2 sinh( )

( )cosh( ) ( )sinh( ) 2 cosh( ) 2 sinh( )

2 sinh( ) 2 cosh( ) ( )sinh( ) ( )cosh( )

2 si

s d s d s s s s

s d s d s s s s

d d d d s s s s

d

k d k d i k d i k d

k d k d i k d i k d

i k d i k d k d k d

i k

+ κ κ +κ κ κ κ κ κ

+ κ κ − +κ κ κ κ − κ κ=

κ κ κ κ − + κ κ − + κ κ

− κ

a

2 2 2 2nh( ) 2 cosh( ) ( )sinh( ) ( )cosh( )d d d s s s sd i k d k d k d

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥κ κ κ + κ κ − + κ κ⎢ ⎥⎣ ⎦

by summing and subtracting the 1st and 2nd rows as well as the 3rd and 4th rows

2 2

2 2

2 2

2 2

( )cosh( ) 0 2 cosh( ) 0

0 ( )sinh( ) 0 2 sinh( )

0 2 cosh( ) 0 ( )cosh( )

2 sinh( ) 0 ( )sinh( ) 0

s d s s

s d s s

d d s s

d d s s

k d i k d

k d i k d

i k d k d

i k d k d

⎡ ⎤+ κ κ κ κ⎢ ⎥

+κ κ κ κ⎢ ⎥= ⎢ ⎥

κ κ − + κ κ⎢ ⎥⎢ ⎥κ κ − + κ κ⎢ ⎥⎣ ⎦

a

Symmetric subsystem:

2 2

2 2( )cosh( ) 2 cosh( ) 0

02 sinh( ) ( )sinh( )s d s s s

sd d s s

k d i k d ABi k d k d

⎡ ⎤+κ κ κ κ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

κ κ − +κ κ ⎣ ⎦⎢ ⎥ ⎣ ⎦⎣ ⎦

Asymmetric subsystem:

2 2

2 2( )sinh( ) 2 sinh( ) 0

02 cosh( ) ( )cosh( )s d s s a

ad d s s

k d i k d ABi k d k d

⎡ ⎤+κ κ κ κ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

κ κ − +κ κ ⎣ ⎦⎢ ⎥ ⎣ ⎦⎣ ⎦

Characteristic Equations Symmetric subsystem:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s s d s dk d k d+ κ κ − κ κ κ = Asymmetric subsystem:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s d d s sk d k d+ κ κ − κ κ κ = Group velocity:

gck

∂ω=

∂

pg p

cc c k

k∂

= +∂

1

pg

p

p

cc c

c

=∂ω−∂ω

Potential Amplitude Distributions #1 Potential amplitude ratios:

ands s a aA B A B/ / from the boundary conditions Symmetric waves: 3 1cosh( )exp( )s dA x i k xϕ = κ

3 12 22 sinh( ) sinh( )exp( )

( )sinh( )d d

s ss s

i k dA x i k xk d

κ κψ = κ

+κ κ

Asymmetric waves: 3 1sinh( )exp( )a dA x i k xϕ = κ

3 12 22 cosh( ) cosh( )exp( )

( )cosh( )d d

a ss s

i k dA x i k xk d

κ κψ = κ

+κ κ

Stress and Displacement Distributions #2 displacement components:

11 3

ux x∂ϕ ∂ψ

= −∂ ∂

33 1

ux x∂ϕ ∂ψ

= +∂ ∂

#3 stress components:

2 2 2 2

11 2 2 2 1 31 3 1( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ −∂ ∂∂ ∂ ∂

2 2 2 2

33 2 2 2 1 31 3 3( ) 2 ( )

x xx x x∂ ϕ ∂ ϕ ∂ ϕ ∂ ψ

τ = λ + + μ +∂ ∂∂ ∂ ∂

2 2 2

31 13 2 21 3 1 3(2 )

x x x x∂ ϕ ∂ ψ ∂ ψ

τ = τ = μ + −∂ ∂ ∂ ∂

Dispersion Curves

normalized velocity: s

cc

η =

normalized frequency: ss

2 f dk dcπ

Ω = =

0

1

2

3

4

5

0 1 2 3 4 5 6Normalized Frequency

Nor

mal

ized

Pha

se V

eloc

ity

symmetric modes asymmetric modes

A0

A1 A2

S0

S1 S2

0

1

2

0 1 2 3 4 5 6Normalized Frequency

Nor

mal

ized

Gro

up V

eloc

ity

symmetric modes asymmetric modes

A0 A1 A2

S0

S1 S2

(steel, cd = 5,295 m/s, cs = 3,141 m/s)

Asymptotic Behavior Low-Frequency Asymptotes Lowest-Order Symmetric Mode (s0):

20lim and

(1 )p g pd

Ec c cω →

= =ρ − ν

2 2 2 2( ) tanh( ) 4 tanh( ) 0s s d s dk d k d+ κ κ − κ κ κ =

2 2 2 2( ) 4 0s s d s dk d k d+ κ κ − κ κ κ =

2 2 2 2 2( ) 4 0s dk k+ κ − κ =

2 2 2d dk kκ = − and 2 2 2

s sk kκ = −

2 2 2 2 2 2(2 ) 4 ( ) 0s dk k k k k− − − =

4 2 2 4 4 2 24 4 4 4 0s s dk k k k k k k− + − + =

2 2 4 2 24 4 0s s dk k k k k− + + =

42

2 24( )s

s d

kkk k

=−

2

22 1 ss

d

cc cc

= −

1 22 2

sd

cc

− ν=

− ν

21 2 2 2 2(1 )2 12 2 1 (1 ) (1 )s sc c c− ν μ + ν μ

= − = = =− ν − ν − ν ρ − ν ρ

plate2(1 )Ec c= =

− ν ρ

Lowest-Order Asymmetric Mode (a0):

4 20lim and 2

3 (1 )p g pd

Ec d c cω →

= ω =ρ − ν

2 2 2 2( ) tanh( ) 4 tanh( ) 0s d d s sk d k d+ κ κ − κ κ κ = 1st-order approximation

0lim tanh( ) ...θ→

θ = θ +

2 2 2 2( ) 4 0s d d s sk d k d+κ κ − κ κ κ =

2 2 2 2 2( ) 4 0s sk k+κ − κ =

2 2 2( ) 0sk −κ = or 0sk = (only at 0ω = )

3rd-order approximation

3

0lim tanh( )

3θ→

θθ = θ −

3 3 3 32 2 2 2( ) ( ) 4 ( ) 0

3 3d s

s d d s sd dk d k d

κ κ+κ κ − − κ κ κ − =

2 2 2 22 2 2 2 2 2(2 ) (1 ) 4 ( )(1 ) 0

3 3d s

s sd dk k k k k

κ κ− − − − − =

4 2 2 2 2 2 2 2 2 2 2 23 (2 ) ( ) 4 ( ) 0s s sdk k k k k d k k k d− − − + − =

after further algebraic manipulation:

4 4 2 2 4 2 2 2 2 4 2 2 4 23 (4 4 )( ) 4 ( 2 ) 0s s s s sdk k k k k k k d k k k k k d− − + − + − + =

46 4 2 2 4 4 2 2 2 2 4 2 6 4 2 2 4

23 4 4 4 4 4 8 4 0s

s s s s s sd d dk k k k k k k k k k k k k k k k k k

d− + − + − + + − + =

42 4 4 2 2 2 2 4 2 4 2 2 4

23 4 4 4 4 0s

s s s s sd d dk k k k k k k k k k k k k k

d− + − + − + =

after neglecting all but the highest power terms in k (since k is large)

44 2 4 2

23 4 4 0s

s dk k k k k

d− + =

24 2 2 2

24 (1 )3

ss

d

cc d cc

= ω −

4 2 2 2 2 2 24 1 2 2 1(1 )3 2(1 ) 3 1s sc d c d c− ν

= ω − = ω− ν − ν

2 24

23 (1 )E dc ω

=ρ − ν

platef3

c dc c

ω= =

Low-Frequency Asymptotes of the Fundamental Lamb Modes

steel plate (cd = 5,900 m/s, cs = 3,200 m/s)

0

1

2

3

4

5

6

0 1 2 3 4 5Frequency × Thickness [MHz mm]

Phas

e V

eloc

ity [k

m/s]

.

fundamental symmetric mode (S0) thin-plate dilatational wave approximation fundamental asymmetric mode (A0) thin-plate flexural wave approximation

High-Frequency Asymptotes of the Fundamental Modes

Symmetric Mode:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s s d s dk d k d+ κ κ − κ κ κ =

Asymmetric Mode:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s d d s sk d k d+ κ κ − κ κ κ =

2 2d dk kκ = − is pure real since dc c<

2 2

s sk kκ = − is pure real since sc c<

lim tanh( ) 1θ→∞

θ =

2 2 2 2( ) 4 0s d sk k+κ − κ κ =

both fundamental modes approach the Rayleigh velocity

High-Frequency Asymptotes of the Higher-Order Modes

Symmetric Modes:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s s d s dk d k d+ κ κ − κ κ κ =

Asymmetric Modes:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s d d s sk d k d+ κ κ − κ κ κ =

2 2d dk kκ = − is pure real since dc c<

2 2

s sk kκ = − is pure imaginary since sc c>

tanh( ) tan( )i iθ = θ

0sκ =

sc c→

all higher-order modes approach the shear velocity

Cut-Off Frequencies

kcω

= , ss

kcω

= , dd

kcω

= , 2 2s sk kκ = − , 2 2

d dk kκ = −

c→∞ , 0k→

s si kκ = is pure imaginary, d di kκ = is pure imaginary

tanh( ) tan( )i iθ = θ

Symmetric Modes:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s s d s dk d k d+ κ κ − κ κ κ =

2 2 2 2( ) tan( ) 4 tan( ) 0s s d s dk k i k d k k k i k d− + =

4 sin( )cos( ) 0s s dk k d k d =

cos( ) 0dk d = or sin( ) 0sk d =

(2 1)2dk d n π

= − or sk d n= π

2 (2 1)2dd n λ

= − or 2 sd n= λ

(2 1)4

dc

cf nd

= − or 2

sc

cf nd

=

where n = 1, 2, ...

1 3 1 3( ) ( )u x u x= − and 3 3 3 3( ) ( )u x u x= − −

Asymmetric Modes:

2 2 2 2( ) tanh( ) 4 tanh( ) 0s d d s sk d k d+κ κ − κ κ κ =

4 cos( )sin( ) 0s s dk k d k d =

cos( ) 0sk d = or sin( ) 0dk d =

(2 1)2sk d n π

= − or dk d n= π

2 (2 1)2sd n λ

= − or 2 dd n= λ

(2 1)4

sc

cf nd

= − or 2

dc

cf nd

=

1 3 1 3( ) ( )u x u x= − − and 3 3 3 3( ) ( )u x u x= −

Cut-Off Frequencies

symmetric asymmetric

shear 2s

ccf nd

= (2 1)4

sc

cf nd

= −

dilatational (2 1)4

dc

cf nd

= − 2

dc

cf nd

=

general symmetric Lamb mode

pure transverse shear resonancepure transverse longitudinal resonance

general asymmetric Lamb mode

pure transverse shear resonancepure transverse longitudinal resonance