DISPERSION CURVES - Unesp
Transcript of DISPERSION CURVES - Unesp
DISPERSION DISPERSION CURVESCURVES
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Elisabetta ManconiUniversity of Parma, Italy
Overview
Aim of this lecture is to give anintroduction to dispersion curves andwave properties.
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� Theoretical background � Introduction to waves in simple structures� Examples� Summary� References
Background to waves in structures
� Alternative viewpoint to modes.
� Vibration propagate through structure as waves.
� Transmission of vibration/energy flow throughstructures.
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� Useful approach, especially at higher frequencies.
� FEA is inappropriate for large sized structures andhigh frequency analysis because the computationalcost becomes prohibitive. In particular, in order toobtain accurate predictions at high frequencies thesize of the elements should be of the order of thewavelength, resulting in impossibly large computersmodels in many cases.
� Knowledge of wave properties is of importance in anumber of applications such structure borne sound,energy methods, waveguide structures, shockresponse, non-destructive-testing, structural-health-monitoring, etc.
Wave properties
� Characteristics of a typical harmonic wave
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( ) ( ) or y cos( )i t kxw x We A t kxω ω−= = −
A: amplitude of wave (m)ω: radial frequency (rad/s)f : cyclic frequency (Hz)λ: wavelength (m)k: wavenumber (1/m)c: phase velocity (m/s)T : period (s)ω=2πf; T=1/f; λ=2 π/k
� Dispersion relations/dispersion curves
� Phase velocity
� Group velocity
Dispersion curves
0Lw wµ+ =ɺɺ
( )L x
is the displacement response variable
is a self-adjoint differential stiffness operator
( ),w x t
µ the mass per unit length
Here we are considering structures which areuniform and unbounded in one direction.Equations of free vibration can be written as [1]:
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e.g. equation of motion of a string under tension
Under the passage of a time harmonic wavewhere k is the wavenumber and W is
the displacement amplitude of the wave, thewavenumber is a solution to the dispersionequation
( ) ( ), i t kxw x t Weω −=
( )2 2 0p k µω− =where is a real-valued function and is often apolynomial. The solutions thus come in pairs, ±k,representing the same wave mode traveling in the ±xdirections.
2( )p k
e.g. equation of motion of a string under tension2 2
2 20
w wT
x tµ∂ ∂+ =
∂ ∂
Each solution to the dispersion equation gives onebranch of the dispersion curves and represents awave which can be any of the following forms:
(1) propagating: k is pure real;(2) evanescent: k is pure imaginary;(3) attenuating oscillatory: k is complex.
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Different types of waves, see also [2,3,4]
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Wavenumber
The wave speed (phase velocity ) is
The group velocity is defined as
see also [2,3,4].
ck
ω=
gck
ω∂=∂
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Longitudinal wave propagation in a bar
Consider a uniform infinite elastic bar lying along the x-axis. The parameter of the beam are: cross-sectionalarea A, Young’s modulus E and density ρ. Assume thelongitudinal displacement is u(x).
Equation of motion is
Assume harmonic solution
2 2
2 20
u uE
x tρ∂ ∂− =
∂ ∂
22
2 0i t d u
u ue E udx
ω ρω= ⇒ + =
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This has solutions of the form:
Substituting it gives (dispersion relation)
There are two real roots for k. These are real: k = ±kL
with1/2
Lkρω =
2 2 0Ek ρω− =
( ) ikxnu x A e=
the longitudinal wavenumber.Hence the general solution is
The first and second terms represent respectivelywaves varying harmonically in space and time,propagating to the left/right of the bar.
LkE
ω=
( ) ( )1 2( , ) L Li t ik x i t ik xu x t Ae A eω ω− += +
Dispersion relation
Phase velocity
Group velocity
1/2
LkE
ρω =
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1/2
L
Ec
k
ωρ = =
1/2
Notice that the phase velocity is independent offrequency and wavelength so all harmonic wavestravel at the same velocity so longitudinal waves arenon-dispersive .
1/2
gL L
Ec c
k
ωρ ∂= = = ∂
Propagation of a longitudinal wave packet in an infinite rod
•
•The peaks move at speed c: phase velocity•Waveform repeat after one wavelength
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Snapshots of a wave at four different times
As you can see in the animation below, it is thedisturbance which travel, not the individual particles.In longitudinal waves the oscillations always move inthe same direction as the wave
Example: real valued longitudinal dispersion curves for a bar
(positive-going waves)
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Frequency
Wa
venu
mbe
r
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Torsional wave propagation in a circular bar
Consider again a uniform elastic bar. The parameter ofthe beam are: shear modulus G=E/2(1+ν) and densityρ (ν is the Poisson ratio). Assume the rotationaldisplacement is θ(x)
Equation of motion is
where, for a circular cross section of radius r
for a circular cross section of radius r.Assume harmonic solution
2 2
2 20pGJ I
x t
θ θρ∂ ∂− =∂ ∂
22
2 0i t
p
de GJ I
dxω θθ θ ρ ω θ= ⇒ + =
4
; =12p p
rI J I
π=
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This has solutions of the form:
Substituting it gives (dispersion relation)
There are two real roots for k. These are real: k = ±k T
with1/2
pIk
ρω
=
2 2 0pGJk Iρ ω− =
( ) ikxnx A eθ =
the torsional wavenumber.
Hence the general solution is
The first and second terms represent respectivelywaves varying harmonically in space and time,propagating to the left/right of the bar.
TkGJ
ω=
( ) ( )1 2( , ) T Ti t ik x i t ik xx t Ae A eω ωθ − += +
Dispersion relation
Phase velocity
Group velocity
1/2
pT
Ik
GJ
ρω
=
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1/2
pT
Ic
k GJ
ρω = =
Again the phase velocity is independent offrequency and wavelength so all harmonic wavestravel at the same velocity so longitudinal waves arenon-dispersive .
1/2
pgT T
Ic c
k GJ
ρω ∂= = = ∂
Bending wave propagation in a thin beam
Consider an infinite beam lying along the x axis. Asshown in the figure, we assume that plane sectionsremain plane and perpendicular to the longitudinal axis– this beam is called ‘Euler-Bernoulli beam’. At highfrequencies the assumptions are no longer valid andshear deformation and rotational inertia must beincluded.
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Parameters of the beam are:ρ density,E Young’s modulus,I second moment of area of cross-section aboutneutral axis (e.g. for rectangular section,I = bh3/12),A cross-sectional area.
For a uniform beam, E, I, ρ, A are constants, so
Assume a harmonic solution,
This has solutions of the form:
4 2
4 20
w wEI A
x tρ∂ ∂+ =
∂ ∂
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42
4 0i t d w
w we EI A wdx
ω ρ ω= ⇒ − =
( ) ikxnw x A e=
Substituting it gives (dispersion relation)
There are four roots for k. Two are real: k = ±kB andtwo imaginary: k = ±ikB with
the flexural wavenumber.
1/41/2 ρω =
B
Ak
EI
( ) nw x A e=
4 2 0EIk Aρ ω− =
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In flexural waves oscillations are alwaysperpendicular to the wave motion. An example of atransverse wave is shown below
General solution
Near field or evanescent waves are waves thatdecay exponentially with distance. They are verylocalised and don’t carry energy
( ) ( ) ( ) ( )1 2 3 4( , ) ω ω ω ω+ − + −= + + +B B B Bi t k x i t k x i t ik x i t ik xw x t Ae A e A e A e
+ - travelling wavesNearfield or evanescent
Dispersion relation
Phase velocity
Group velocity
1/41/2
B
Ak
EI
ρω =
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1/4
1/2B
EIc
Aω
ρ =
Notice that the phase velocity depends on frequencyso flexural waves are dispersive : the phase speeddepends on frequency. Higher frequencycomponents propagate faster. As a consequence apulse “spreads out” or disperses:
2gB Bc c=
Propagation of a wave packet in an infinite beam
Real valued bending dispersion curves for a thin beam (positive going waves)
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Wa
venu
mbe
r
Frequency
Wave propagation in a thin plate plate
Plate bending waves
Consider a flat isotropic plate of thickness h lyingalong the x−y plane
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The transverse displacement w(x,y,t) satisfies thefollowing differential equation
where is the flexural rigidity and ν is the
Poisson ratio.
4 4 4 2
4 2 2 4 22 0
w w w wD h
x y x y tρ ∂ ∂ ∂ ∂+ + + = ∂ ∂ ∂ ∂ ∂
3
212(1 )
EhD
ν=
−
Assume a harmonic solution,
where and are the wavenumbers in the x and ydirections.
Substituting it gives (dispersion relation)
( )22 2 2 0x yD k k hρ ω+ − =
( , , ) yxik xik x i tw x y t We e eω=
xk yk
( )2 2 2cos , sin , , k k k k k k kθ θ= = = +
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Since ,angle θ being the heading of the wave, the flexuralwavenumber is
The plate free wavenumber is similar to that for abeam, apart from the factor (1−ν2).
The above theory is for thin plates (corresponding toEuler-Bernoulli beams). For thick plates sheardeformation must be included.
1/41/2 h
kD
ρω =
( )2 2 2cos , sin , , x y x yk k k k k k kθ θ= = = +
In-plane waves
The in-plane displacement of a plate can bedescribed by two components u, v in the x, ydirectionsrespectively.
Equations of motion:
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Equations of motion:
a plane wave solution is
Put c=cosθ, s=sinθ, and substitute.
For non non-trivial solution, the determinant of thematrix must be zero. This gives:
'
(1 )L
Ec
ρ ν=
−
This can be simplified to
which has two roots.
These both correspond to non-dispersive waves, withphase speeds:
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The first of these waves (longitudinal wave) haspredominant motion in the direction of propagation.The second (transverse or shear wave) haspredominant motion perpendicular to the propagationdirection.
Dispersion curves for a thin isotropic plate
In thin isotropic plates, guided waves are classified into threetypes: extensional , flexural and shear waves. The classificationcan be given according to the direction of the displacement vectorfor wave propagating in the x direction. The waves involvingmotion of the medium in the x−z plane, are defined as extensionalor flexural. For these kinds of waves, the displacements of theparticles are directed primarily along the x and z axes respectively,that is y=0, x≠0, z≠0. On the other hand, waves which involvemotion of the medium primarily in the y direction, or in other termsx=z=0, y≠0, are classified as shear waves.
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Examples of symmetric and antisymmetric components of the displacements in the z direction [2].
Examples of symmetric and antisymmetric components of the displacements in the x and y directions [4].
Real valued dispersion curves for a thin isotropic plate(positive-going waves)
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Frequency
Wav
enum
ber
Bending wave propagation in a thin beam on elastic foundation
For a uniform beam, E, I, ρ, A are constants, so
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4 2
0w w
EI A swρ∂ ∂+ + =
This has solutions of the form:
Substituting it gives (dispersion relation)
4 20
w wEI A sw
x tρ∂ ∂+ + =
∂ ∂
( , ) ikx i tw x t Ae eω=
4 2 0EIk A sρ ω− + =
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This gives
hence there are four roots for k. In particular thereare only propagating wave solutions for frequenciesabove a ‘cut-off’ frequency
( ) ( )2 2
4 41,2 3,4;
A s A sk k i
EI EI
ρ ω ρ ω− −= ± = ±
s
The Euler Bernoulli beam on an elastic foundationhas two pairs of complex conjugate solutions belowthe cut-off frequency, which represent a set of 4attenuating oscillatory waves. Above there is a pairof propagating waves and a pair of evanescentwaves.
; c c
s
Aω ω ω
ρ>= =
Note that for
� propagating waves: k is pure real ( real andpositive);
� evanescent waves: k is pure imaginary ( realand negative);
� attenuating oscillatory waves: k is complex ( iscomplex).
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2k
2k
2k
In the latter case the solution form a complexconjugate pair and hence there are at least 4 roots tothe dispersion equation. A positive going wave eitherdecay in the positive x-direction, or, if , carriesenergy in the positive x-direction, i.e. it is one forwhich
{ } { }Im 0; 0 if Im 0gk c kk
ω∂< = > =∂
{ }Im 0k =
Dispersion curves for a thin beam on elastic foundation [1]
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( ) ( )2 2
4 41,2 3,4; ; c
A s A s sk k i
EI EI A
ρ ω ρ ωω
ρ− −
= ± = ± =
0.5
1
Re{k}
(k)
Rea
l(k)
Frequency
Dispersion curves for simple waveguides, positive-going waves only; - - - beam in bending and _____ on elastic
foundation.
0 0.2 0.4 0.6 0.8 1-1
-0.5
0
Im{k}
Frequency
Imag
(k)
Rea
l(k)
Bending wave propagation in a thin plate strip
Consider a rectangular plate Lx x Ly which has simplesupports on x = 0 and x = Lx as shown in the figure.
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Lx
Ly
Boundary conditions are:
Displacement at x = 0 and x = Lx equal to 0
bending moment at x = 0 and x = Lx equal to 00,
0 xx L
w
x =
∂ =∂
2
20,
0 xx L
w
x =
∂ =∂
thus the solutions have the form
substituting into the equation of motion
it gives (dispersion relation)
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( , , ) sin yik y i tmn
x
m xw x y t A e e
Lωπ
=
4 4 4 2
4 2 2 4 22 0
w w w wD h
x y x y tρ ∂ ∂ ∂ ∂+ + + = ∂ ∂ ∂ ∂ ∂
22
2 2 0m h
kπ ρ ω
+ − =
Hence
Now, substituting ky =0 into this equation gives the cut-off frequency for the m-th wave mode as
2 2 0yx
m hk
L D
π ρ ω + − =
2
2 for 1,2,3...yx
h mk m
D L
ρ πω
= − =
2
cmx
D m
h L
πωρ
=
Dispersion curves for a thin plate strip with simply supported edges
- flexural waves –(positive-going waves)
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Rea
l(k)
Frequency
Imag
(k)
Rea
l(k)
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Natural frequencies and modes in terms of waves.
Standing waves
Standing waves may be created from two waves (withequal frequency, amplitude and wavelength) travellingin opposite directions. Unlike the travelling waves, thein opposite directions. Unlike the travelling waves, thestanding waves do not cause a net transport of energy(because the two waves which make them up arecarrying equal energy in opposite directions).
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Illustrative example: modes and waves in a simply supported finite
beam [5]
Consider a thin beam lying along the x-axis. The freemotion of a finite beam of consists of the superpositionof the 4 types of free wave
( ) ( ) ( ) ( )1 2 3 4( , ) ω ω ω ω+ − + −= + + +B B B Bi t k x i t k x i t ik x i t ik xw x t Ae A e A e A e
We refer to the wave amplitude of the positive goingwaves and to the negative going waves by
while we refer to the transfer matrices that describethe propagation of the waves from one point to anotherthrough appropriate wavenumber by
4 3
2 1
= =T T
A A
A A+ −
A A
0 0= =
0 0
ikx ikx
kx kx
e e
e e
−+ −
−
F F
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Now consider a beam of length L simply supported asshown in the figure.
The wave amplitudes at each end are denoted by Aand B. The reflection matrices at the boundaries are
A+ B+
B−A−
A B
and B. The reflection matrices at the boundaries arerespectively RA at the left end and RB at the right end.If the incident waves impinging upon the right end atx = 0 are the upstream wave A−, then reflected wavesof amplitude A+ will be generated and we will have therelation: A+ = RA A−. Similarly, if the incident wavesimpinging upon the left end are the downstream wavesof amplitude B+ we will have B− = RB B+. The waveamplitudes are related by
0 = ( )
0
0 = ( )
0
ikL
kL
ikL
kL
eL
e
eL
e
−+ + + +
−
−− − − −
−
=
= −
B A F A
A B F B
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Since the boundary conditions are the same at the leftand right ends, in this case we have RA= RB (note thatif the boundary conditions are different the reflectionmatrices are different). The reflection matrices areobtained applying the boundary conditions
0
2
20
01 1
01 1 1 10
1 0
x
x
wi i
w
x
=+ −
=
+ −
=− −
→ + =∂ − −= ∂−
= → =
A A
A R A R
Now, combining the wave reflection matrices andusing the transfer matrices, that is
A+ = RA A−, A − = F −(-L)B− , B− =RB B+ , B + = F + (L) A+
we obtain a characteristic equation
from which the natural frequencies of the system canbe found when the determinant of the left matrix iszero.
1 0
0 1A A+ − −
= → = − A R A R
( ) ( ) 0A BF L L− + + − − = R R F I A
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1/41/2
2
( ) ( ) 0 sin( ) 0
( 1,2,3,...)B
A B
Ak
EI
F L L kL
n n EIk n
L L Aρω
π πωρ
− +
=
− − = → = →
= → = =
R R F I
SUMMARY
In this lecture a short introduction to dispersioncurves and wave characteristics in simpleunbounded continuous structures was given. Thebackground and the basic steps to obtain dispersioncurves were discussed and few examples weregiven. Vibrations of a finite beam simply supported
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were also obtained using a wave approach.
What we have learned:
� Something about dispersion curves and how tointerpret them
� how to obtain dispersion relations anddispersion curves
� how to study vibrations in terms of waves andreflection at the boundaries
References
[1] B. R. Mace and E. Manconi, Wave motion anddispersion phenomena: Veering, locking and strongcoupling effects, Journal of the Acoustical Society ofAmerica, 131, 1015–1028, 2012.[2] K. Graff. Wave motion in elastic solids. DoverPublications, 1991.[3] F. Fahy, P. Gardonio, Sound and Structural
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[3] F. Fahy, P. Gardonio, Sound and StructuralVibration. - Radiation, Transmission and Response,Academic Press ,Oxford, England, 2007[4] B. A. Auld, Acoustic fields and waves in solids.Krieger Publishing Company, Malabar ,FL, 1990.[6] B. Mace, Wave Reflection and Transmission inBeam, Journal of Sound and Vibration, 97, 237-246,1984.