Waves Vibration Intro

8/13/2019 Waves Vibration Intro

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I. INTRODUCTION

1. PLANE WAVES

By way of introduction and/or review of some basic ideas of dynamics , we first

consider the "wave equation":

! u = u x x + u y y + u z z = 1

c 2 u t t

where x, y, z are the spatial coordinates of the medium and c is a property of the mediumwhich turns out to be the (local) speed of disturbance propagation. For plane motion, theequation is:

u x x = 1

c 2 u t t

The general equations of elasticity are considerably more complex, but in several specialcases, reduce to this form.

Plane strain – For the plane strain motion of a slab:

! x = 1

E " x – # " y + " z

! z = ! y = 1

E " y – # " x + " z = 0

so that:

! z = ! y = " ! x1 – "

giving the result:

! x = 1 + " 1 – 2 " # x

1 – " E

Note: For Lamé parameters:

! + 2 µ = 1 – " E

1 + "

1 – 2"

The equation of equilibrium is:

! ! x

! x = "

! 2 u

! t 2

For constant ! , the displacement u can be eliminated and the final equation is:



! 2

! x

! x 2 =

1

c 12

! 2

! x

! t 2



where:

c 1 = E 1 – !

" 1 + ! 1 – 2 !

1 / 2

is the "dilatational" velocity.

Shear waves – For the propagation of shear waves in an elastic medium , thevelocity is:

c 2 = E

! 2 1 + "

1/ 2

Bar – For the axial motion of an elastic bar:

!! x

A ! x = " A !2

u! t 2

A ! x = E A! u

! x

So, for the cross-sectional area A and the density ! constant, we again have the one-dimensional wave equation:

! 2! x

! x 2 = 1

c b2

! 2

! x

! t 2

where the "bar" velocity is:

c b = E

!

1/ 2

Plane stress – For a flat plate, the constitutive relation is:

! x = E 1 – "

2 # x + " # y

For " y = 0, we have the wave equation:! 2 u

! x 2 =

! 1 – " 2

E ! 2 u

! t 2

so the "plate" velocity is:



c p = E

! 1 – " 2

1/ 2

The various speeds can vary substantially. The ratio of the dilatational speed tothe bar speed is:

c 12

c b2

= 1 – !

1 + ! 1 – 2 ! =

" for ! # – 1

1 for ! = 0

6/5 for ! = 1/4

" for ! # 1/2

Thus for the usual metals and concrete, c1 is not much different from cb , however fornearly incompressible materials c1 >> cb. The ratio is similar for the plate velocity:

c 12

c p2 =

1 – ! 1 – ! 2

1 + ! 1 – 2 ! = 1 +

! 2

1 – 2 !

String under tension – The one-dimensional wave equation arises in many othersituations. For a string under tension the equation is:

! w

! x = 1

c String2

! 2 w

! t 2

where w is the normal displacement and the speed is:

c String = T

! A

1/ 2

in which T is the tension, ! is the density and A is the area of the string cross section.

Fluid in an elastic tube – Another interesting case is that of the flow of a fluid inan elastic tube, such as an artery. The assumptions are (1) the dominant velocity of thefluid is plane and in the direction of the axis (the long wavelength assumption), (2) theinertia of the tube wall is negligible, and (3) the fluid is incompressible. The stress-strainrelation for the tube is:

! " = w R

= 1 E t

N " – # N x

where w is the outward displacement, t is the thickness and R is the radius of the tubewall. For the wave response the axial stress resultant is zero, and the circumferentialstress resultant is determined by the local fluid pressure p:



N x = 0 N ! = p R

For conservation of the fluid the wall displacement is related to the axial displacement ofthe fluid:

! R 2 ! u

! x = – 2 ! R w

The wall displacement can be eliminated, yielding the relation:

p = – E t 2 R

! u! x

Note that this is similar to the stress-strain relation for a bar, except that the pressure hasthe opposite sign as the stress. Also note the effective Young's modulus is equal to thewall modulus reduced by the factor t /(2 R). Finally the equation of equilibrium for thefluid is needed:

! p! x

= – ! ! 2

u

! t 2

with which the wave equation in the same form as in the preceding cases is obtained,except that the speed is:

c Tube = E t 2 R !

1/ 2

Thus, for a thin-walled tube, the speed of the fluid-elastic wave propagation is much

slower than that for the elastic waves in the wall without fluid.

Vibration – As has been indicated, the equation for plane wares in an acoustic orelastic medium, axial motion of an elastic bar, or the transverse motion of a stretchedstring is

u x x = 1

c 2 u t t

For the homogeneous case, i.e. when c is a constant, the solutions are straightforward.The solutions periodic in time are obtained by assuming the form

u x , t = Re e i ! t f ( x ) = f ( x ) cos ! t

Then must satisfy the equation

f !

= – " 2

c 2 f

i.e. f = A sin !

c x + B cos !

c x



For a slab with free faces at x = 0, L, the solution is:

= A sin ! xc

where the natural frequencies are at:

! = " n c L

n = 1, 2, 3,...

Forced motion of bar – Consider a bar (or acoustic tube) with a prescribedforcing at one end ( x = 0). The output is the displacement at the other end ( x = L). Forthe acoustic tube, the output displacement of the air provides the forcing for the soundradiated from the end of the tube. For a first approximation the impedance at the end issmall in comparison with the impedance in the tube. Therefore the end boundarycondition is s( L, t ) = 0. The solution for the stress and displacement is:

! x , t = ei " t

C 2 sin" x – L

c

u x , t = e i ! t c C 2! E

cos! x – L

c

Therefore the ratio of the input stress to the output displacement is:

u L , t

! 0, t= c

E " 1

sin " Lc

From this it is clear that a substantial output is obtained at the resonant frequencies:

! Lc

= n " for n = 1, 2, ...

The response function is shown in the following figures. This is the behavior in mostwind instruments such as the oboe, flute and the brass instruments. However, thewoodwinds use only the first two or three of the resonance frequencies. On the brassinstruments it is difficult to excite the fundamental; the second through the sixteenth areused extensively.



10000100010010

.01

.1

1

10

100

1000

requency Hz

Respons

u L)/

0)

Figure (a) – Response of acoustic tube, given by the transfer function defined by the

output displacement divided by the input pressure. This simulates a horn tuned to the keyof A.

43210- 1- 2

.01

.1

1

10

100

1000

Frequency octaves above fundamental

Respons

u L)/

0)

Figure (b) – This is the same response as in Figure (a) , but with the frequency scale in

octaves above the fundamental frequency.



On the other hand such instruments such as the clarinet and saxophone have a differentsort of mouthpiece for which the input is approximated by considering a prescribeddisplacement at the end x = 0. The response function is then:

u L , t

u 0, t = 1cos ! L

c

which gives the behavior shown in Figure (c). The resonance frequencies are at thevalues:

! Lc

= n "

2 for n = 1, 3, 5, ...

Thus the "octave key" on the clarinet and saxophone gives the second resonant frequencywhich is three times the frequency of the first.

43210- 1- 2

1

10

100

1000

10000

requency in octaves

Respons

u L)/

0)

Figure (c) – Response consisting of the ratio of displacements at the two ends. The

parameters are the same as in Figure (b). Note that for displacement forcing theresonance frequencies are at the odd integer multiples of the fundamental.

Pulse propagation in semi-infinite medium – If the form for f is chosen to be:

= e ± i !

c x



then the total solution is obtained in the traveling wave form:

u x , t = A sin ! t + xc

+ B sin ! t – xc

The node points, which are the points of zero displacement, of the first term are at x = –ct . Thus the points of constant displacement move in the negative x direction with the

phase velocity c. The second term gives a sinusoidal "wave train" which moves in thepositive x direction with the phase velocity c:

The general solution is

u ( x , t ) = f t + xc + g t – xc

where f and g are arbitrary functions. Thus generally, the solution is the superpositionof two deformation distributions which move with the velocity c in opposite directions.The solution to various transient problems is readily obtained. For example, consider asemi-infinite slab initially at rest with a prescribed pressure on the face. The conditionsare:

u ( 0, t ) = h ( t ) u ( x , 0 ) = u t ( x , 0 ) = 0

For this the term for the waves incoming from + ! must be excluded, so f = 0 and thefunction g must be:

g (t ) = h ( t )

so the result is

u ( x , t ) =

h ( t – xc

) for x < c t

0 for x > c t

A step input:

h (t ) = 0 for t < 01 for t < 0

produces a step wave:

xt

u



Figure – A step wave moving to the right with the velocity c.

For a half-sine wave pulse acting on the face:

h ( t ) =

0 for t < 0

sin ! t for 0 < t < "

!

0 for "

! < t

the result is a half-sine wave moving into the medium:

xt

u

c

( )

Figure – A half-sine wave pulse traveling to the right.

Pulse propagation in finite slab – For the finite slab, initially at rest, with thepressure on one face

u (0, t ) = h ( t )

while the other face is stress-free

u ( L, t ) = 0

the solution is obtained by adding appropriate semi-infinite slab solutions. (1) First thereis the pulse moving to the right, exactly as in the semi-infinite medium:

u ( x , t ) = h ( t – xc

)

(2) The boundary condition at x = L can be satisfied by adding an image pulse travelingto the left:

– h t + x - 2 Lc

(3) Then for the boundary condition at x = 0, use a second image traveling to theright:



+ h t – x + 2 L

c

The procedure continues indefinitely. The complete solution consists of an infinite trainof pulses moving to the left and an infinite train of pulses moving to the right:

u ( x , t ) = h t – xc

– h t + x – 2 L

c +

+ h t – x + 2 Lc

– h t + x - 4 Lc

+ ...

where h { z } = h ( z ) for z > 0 and zero for z < 0 . Thus, for the half-sine-waveimpact, the solution looks something like this at successive times:

xct

u

c

( )

L c

mage pulse

(a) Before pulse arrives at left boundary ( x = L), the solution is the same as for a semi-infinite medium.

x

u

ot l

(b) After the pulse front arrives at the left boundary, it is partially canceled by the left–

moving image pulse.

c

x

u

c



(c) At a later time, all that is in the physical domain is the image pulse. However, it willbe canceled at x = 0 by an image pulse, not shown, coming from the left side.

x

u

(d) As the image pulse arrives at the left face, it will be canceled by a second image pulsetraveling to the right, which is of the same sign as the original pulse.

A short time later after figure (d), only the second image pulse is in the physical domainand we are back to the situation in Figure (a). Of particular practical importance, acompressive pulse on one face causes a reflected tensile pulse from the free face of theslab. Since metals are weaker in tension, typical failure is fracture near the free surface.The same situation occurs in a laminated composite which is subjected to a compressivepressure pulse on one face. The failure will be a delamination occurring near the backface due to the tensile image pulse.

Inhomogeneous Medium – We now consider the inhomogeneous slab for whichthe velocity c varies with the distance. A general solution cannot be obtained, however,the separation of variables still gives the solution periodic in time

u x , t = Re e i ! t f ( x )

!

+ " 2

c 2 f = 0

So an asymptotic expansion, often called WKB, valid for sufficiently high frequencies,can be readily obtained in the form:

f = e –i!" ( x ) f 0 + 1i !

f 1 – 1! 2

f 2 + . . .

Substituting into the equation and equating the coefficient of each power of # to zero

gives the sequence of equations:

(! "

)2

= 1c 2

2 ! "

f 0"

+ ! #

f 0 = 0

2 ! "

f 1"

+ ! #

f 1 = f 0#



B = – f 1

! f 0 at x = 0 = 0

Then at the other boundary x = L, the condition is:

f ( L ) = cos n ! A sin " f 0 + cos " f 1#

which gives the correction term:tan ! " ! = – f 1

# f 0

Thus, the natural frequencies are at

! = n "

dxc

0

L – 1

2n " c 1/ 2 ( c 1/2 )

#

dx + O ( " n ) – 3

0

L

The maximums of f are at:

! " #

cos !" f 0 + f 0#

sin !" – f 1 " #

sin !" = 0

cot

!" =

f 1 " #

– f 0#

!" #

f 0

which gives the envelope of the modes

( f )max = f 0 1 + (!

"

f 1 )2 – ( f 0

"

)2

2 (#! "

f 0 )2

+ O # – 4

Thus, the envelope is essentially the same for all modes with # sufficiently large. Forexample, take the distribution of sound speed:

c = c 0 e ! x

which gives the result for the natural frequencies:



! c o

= " n #

1 – e –" L 1 +

( 1– e –" L ) ( 1 – e " L )

8 n 2 # 2 + O 1

n 4 # 4

So if kL is not too large, the result is accurate for all natural frequencies n = 1,2,3, .......

The envelope of the modes is:

( f )max = e ! x /2 1 + ( 1 – e –! x )2

8n 2" 2

(e ! x – 1 )2

16 – e 2 ! x + O 1

n 4 " 4



2. Beam with and without and elastic foundation

In the preceding section, the "wave equation" was considered. There are twoimportant properties of this equation: (1) Discontinuities or disturbances propagate witha certain velocity (which is the property of hyperbolic partial differential equations in

general) and (2) a propagating pulse retains the same form when the velocity is constant.This latter behavior is referred to as nondispersive . Generally, and particularly for thebeam equations, a propagating disturbance pulse will change its form as it propagates inthe medium. Such a medium is called dispersive . The beam is an example of a mediumwhich is dispersive. The equation from the elementary beam theory (Euler-Bernoulli) is,for a beam with constant properties and no lateral load:

EI! 4 w

! x 4 + ! A

! 2 w

! t 2 + k w = 0

in which EI is the bending stiffness, ! A is the mass per unit length, and k is the elasticfoundation stiffness.

Wave propagation – This is most definitely not a "wave equation", since it isparabolic and not hyperbolic. However, we can easily obtain a solution which gives awave propagating along the beam in the form:

w x , t = e i ! t – n x

where # is the frequency and n is the wave number , which is the reciprocal of the spatialwave length $ :

! = 2 "

n

Substitution into the equation, with the foundation stiffness k = 0, gives the frequency-wave number relation :

EI n 4 – ! A " 2 = 0

The amplitude of the wave is constant when the argument of the exponential is constant.This gives a point moving with constant velocity, which is called the phase velocity :

x

t = v phase = !

n

For the "wave equation" this is just the propagation velocity c. For the present beamproblem, the phase velocity is:

v phase = ! n

= EI

" A1/ 2n = EI

" A3/4

! 1/2



x

=

x

= 1

>

x

= 2 >

1

Figure – An indication of dispersion. At each different time the spatial distribution is

different.

From the general viewpoint, t he "wave" equation is a rather special case governing waves which are nondispersive. Generally, the energy propagates with the group velocity:

vg

= d !

d n

which for beam is twice the phase velocity:

EI n 4 – ! A " 2 = 0 v p = "

n = EI

! A n

v g = d " dn

= 2 EI

! A n = 2 v p

Thus if the transient problem is considered, in which the beam is initially at rest at time t= 0, and then the end of the beam is given a sinusoidal displacement or forcing:

w x , t = 0 for t ! 0

sin " t for t > 0



The resulting transient solution can be interpreted as consisting primarily of the steady-state wave propagating along the beam from the input end to the point moving with thegroup velocity. Ahead of the point moving with the group velocity, the amplitudedecreases, as indicated in the sketches:

p

x

x

p

Front moves with = 2 vg

FrontSteady-State

= > 01

For some other systems, the group velocity is less than the phase velocity v g < v p .There are even cases where the phase and group velocities have opposite sign.

Now the elastic foundation is added. The dispersion relation is:

! A " 2 = EI n 4 + k

which gives the phase velocity:

! n

= E I

" A n 2 + k

" A n 2

The new feature due to a non zero foundation modulus k is that the frequency has a nonzero value # 0 for zero wave number. This is referred to as the cutoff frequency , sincethere are no real values for the wave number, and thus no propagating waves, forfrequencies less than this value.



n

With

Foundation

Without

Foundation

utoffFrequency

The value of the cutoff frequency is:

! 0 = k

" A

which is simply the resonance of the beam as a rigid body oscillating on the foundation.This is a significant frequency as far as the general beam response is concerned. Asshown in the sketch, a point load oscillating with a frequency less than # 0 will cause a"local" response, while a frequency greater than # 0 causes the generation of propagatingwaves.

P 0 cos ! t

ssssssssssssssssssssssssssssssssss

quasi. - static

propagatingwaves

Another aspect of the effect of the foundation is seen in the following sketch of the phaseand group velocities:



minphasevelocity

n

p

vp

g

n

Figure – Phase and group velocities for the Euler-Bernoulli beam on an elasticfoundation. At the minimum of the phase velocity, the phase and group velocities are

equal.

A load moving with a constant velocity vL along the beam on an elastic foundation willhave a steady-state solution, if the load velocity is not equal to the minimum phasevelocity. For lower velocities the response is localized to the moving load. For highervelocities, waves will be generated which have the same phase velocity as the load. Thewave with the group velocity faster than the phase velocity will move ahead of the load,while the wave with the slower group velocity will trail behind the load.

Cylindrical shell analogy – The beam on an elastic foundation is analogous tothe cylindrical shell with axisymmetric deformation. In the static case the analogy is

exact; in the dynamic it is approximate. The radius of the cylinder is r , the thickness is t ,and the reduced thickness is c. In the context, there should be no confusion with the timeand velocity previously designated by these symbols.

Beam on foundation Cylindrical shellBending stiffness

EIBending stiffness

Etc 2 = Et 3/12(1- % 2)Foundation stiffness

kCircumferential stiffness

Et /r 2

Axial bar stiffness EA

Axial shell stiffness E & rt

The dispersion relation for the cylindrical shell is therefore:

! t " 2 = E t c 2 n 4 + E t

r 2

or solving for the frequency:



bar

eam on elasticfoundation

cylindrical shell

plate

Figure – Sketch of frequency-wave number relation for the cylindrical shell. The barand beam on the elastic foundation are good approximations, but the intersection of thetwo branches does not occur. The "bar" mode for low frequencies becomes the beammode for high frequencies, while the "breathing" mode for low wave number becomes

the "plate" mode for high frequencies.



xe x

e

y

e

e y

Figure – Unit vectors in the fixed x – and y – directions and in the r – and ( – directions.

Also for a rigid body rotation:

u ! = C w = 0

which gives a constant ) ( . The curvature change measure is the derivative of the rotation,

which gives the bending moment:

! " =

# $ "

r #" M ! = EI " !

Note that this is not the actual change in the curvature of the ring. Specifically, the effect

of a uniform expansion is excluded. The external force per unit length of the ringreference line has the components qr and q ( .

y

x

Figure – Load components acting on ring.

The total potential energy consists of the strain energy of bending and stretching of thering and the potential of the external load components:



! u, w = Vd " 0

2 #

= 12

E I $ " 2 + E A % "

2 – q rw – q " u" r d " 0

2 #

in which V is the potential energy density, which in terms of u ( and w is:

V = 12

E I r 4

( ) – w !! + u " !

2 + 1

2 E A

r 2 ( )u "

! + w2 – q rw – q " u" r

in which derivatives with respect to ( are denoted by primes. The Euler-Lagrange equations, obtained by consideringvariations with respect to w and u ( , are the differential equations for the problem:

!2

! " 2

! V

! w ## –

!

! "

! V

! w # +

! V

! w= 0

– !! "

! V

! u " # + ! V

! u "

= 0

which are:

E Ir 4

( ) w !!!! – u " !!! + E A

r 2( ) w + u "

! = q r

EIr 4

( ) w !!! – u " !! – EA

r 2( ) w ! + u "

!! = q "

A very convenient method for the solution when the ring is complete is to use the Fourier series. The expansions when the radial load component is an

even function in ( are:

q r ( ) ! = q r n!n = 0, 1, ...

"

cos n ! q ! ( ) ! = q ! n!n = 1, ...

"

sin n !

with the corresponding expansions for the displacement components:

( )w ! = w n!n = 0, 1, ...

"

cos n ! u ! ( ) ! = u ! n!n = 1, ...

"

sin n !

When the series is substituted into the differential equations or into t he potential energy, all coupling terms disappear. Thus each harmonic can be treated

separately, and one obtains the algebraic relation between the Fourier coefficients:

E I r 4

n 4 + E Ar 2

E I r 4

n 3 + E Ar 2

n

E I r 4

n 3 + E Ar 2

n E I r 4

n 2 + E Ar 2

n 2

w nu ! n

= qr nq ! n

This matrix can be considered as a stiffness matrix , yielding the Fourier coefficients of the load from the Fourier coefficients of the displacement.



For the harmonic n = 0 this reduces to:

E Ar 2

0

0 0

w 0

u ! 0

= qr 0

q!

0

It follows that:

q ! 0 = 0

which is the condition that no resultant torque can act on the ring for static equilibrium,and that u ( 0 , which is the amplitude of rigid body rotation, is arbitrary. The axisymmetricradial expansion amplitude due to the uniform radial load is:

w 0 =

r 2

E A q r 0

For the harmonic n = 1, the equations are:

E I r 4

+ E Ar 2

1 1

1 1

w 1

u ! 1 =

qr 1

q ! 1

Thus for a solution:

q r 1 = q ! 1

which is the condition that the loading is self-equilibrating. The solution gives the sum of the displacement components:

E Ir 4

+ E Ar 2

w 1 + u ! 1 = q r 1 = q ! 1

When w1 + u ! 1 = 0, we have the rigid body translation previously discussed.



y

r

Resultantr

y

Resultant

(a) Load component q r1 > 0 producesresultant force in positive x-direction.

(b) Load component q ( 1 > 0 producesresultant force in negative x-direction.

Figure – Distribution of loading for components with n = 1. The distribution is self-equilibrating when q r1 = q ( 1.

Note that the stiffness factor for the harmonic n = 1 is about the same as for theaxisymmetric harmonic n = 0, since

EI r 4

+ EAr 2

= EAr 2

1 + rg

2

r 2

in which r g is the radius of gyration for the cross section. Thus for a thin ring, thebending stiffness factor EI is negligible for both the harmonics n = 0, 1.

For the higher harmonics n " 2, however, if the bending stiffness terms are setequal to zero, the stiffness matrix is singular, which indicates that the amplitude ofdisplacement will be substantially larger than for the harmonics n = 0, 1. The exactsolution for the displacement coefficients is:

w n = qr n E I

r 4 n 2 + E A

r 2 n 2 – q ! n E I

r 4 n 3 + E A

r 2 n

Det

un =

– q r n E I

r4

n 3 + E A

r2

n + q ! n E I

r4

n 4 + E A

r2

Det

in which, after many terms cancel, the determinant reduces to:

Det = E A

r 2 E I

r 4 n 2 n 2 – 1 2

Thus for the thin ring, for which r g << r , the approximate solution is:



w n ! qr n n 2 – q " n n

n 2 n 2 – 1 E Ir 4

u ! n " –

w n

n

The tangential stiffness EA cancels out and only the bending stiffness EI remains. Thesedisplacements for n " 2 are therefore of the order of magnitude ( r /r g)2 larger than those forn = 0, 1. The strain computed from these displacement components is:

! " n # 0

Thus accord to this approximation, the deformation is inextensional . Since for highvalues of n, the EI terms in the numerator of the expressions for the displacementcomponents are significant, this approximation is valid only for sufficiently low values ofn:

n << r

r g2

If the assumption of inextensional deformation is assumed from the beginning, thepotential energy density is simply:

! = " V d # 0

2 $

= $ r 12

E Ir 4

( )wn 2 ( )n 2 – 1 2 – ( ) q r n – 1n q# n w n !n = 2, 3, ...

Setting the derivative with respect to the amplitude wn directly to zero yields theapproximate result previously obtained.

Ring, thermal stress – An important case for which the loading is such that the EA terms cancel in both the numerator and denominator is thermal stress. For heating ofthe ring, the potential energy is:

V = 12

E I ! 2 + 12

E A ( )" # – $ T 2 – q rw – q # u# r

where * is the coefficient of thermal expansion and T is the temperature. The equationsare the same as before, but with the load terms replaced by:

qr ! q r + 1r E A " T

q ! " q ! – 1r E A d

d ! # T

The preceding exact results for the Fourier harmonic coefficients of the displacementcomponents yield the following values for thermal heating:



w n = –

r ! T n

n 2 – 1 for n " 1

u ! n = n r " T n

n 2 – 1 for n # 1

w 1 + u ! 1 = r " T

1

1 + r g2

r 2

for n = 1

So, unlike the force loading, the deformation from thermal heating for all harmonics isthe same order of magnitude. If the stress is computed from these displacementcomponents the result is perhaps surprising. Only the harmonic n = 1 has a non zerostress. The direct stress is:

! direct n = F n A

= ( ) E " # n – $ T n = 0 for n % 1

= E ! T 1 ( )rgr

2

1 + ( )rgr

2 for n = 1

while the bending stress is:

! bending n = z M n

I = z E " # n = 0 for n $ 1

= E ! T 1 zr

1 + ( )rgr

2 for n = 1

in which z is the distance from the reference line of the ring. While all harmonics ofheating produce a distortion of the ring, only the harmonic n =1 produces stress, and alow magnitude of stress at that. The ring simply adapts to the heating without significantstress.

Ring, vibration – For the vibration of the circular ring, the solution can beobtained in the form of a product of sinusoidal variation in space and time:

w ! , t = w n cos n ! cos " t u ! ! , t = u ! n sin n ! cos " t

Consequently, the equations are the same as in the static case, but with inertia termsadded to the load components:

qr ! q r + " A # 2 w

q! n " q ! n + # A $ 2 u!

The "dynamic stiffness matrix" for a each Fourier harmonic is:



E Ir 4

n 4 + E Ar 2

– ! A " 2 E Ir 4

n 3 + E Ar 2

n

E Ir 4

n 3 + E Ar 2

n E Ir 4

+ E Ar 2

n 2 – ! A " 2

w nu# n

= qr nq# n

For the harmonic n = 0 this reduces to:

E A

r 2 – ! A " 0

2 = 0

which defines the ring breathing frequency :

! 0 = 1r

E

" =

c bar r

Thus the period T 0 of the breathing frequency is the time required for a wave traveling atthe bar velocity to transverse the circumference of the ring:

T 0 = 2 ! rc bar

For the harmonic n = 1, the result is:

! – " !

! ! – " = ! – " 2 – ! 2 = – 2 ! " + "

2 = 0

in which

! =

" A

# 2

! = E I

r4

+ E A

r2

Thus the resonant frequency is given by:

! = 2 " that is:

! 1 = 2r

E

" 1 +

r g2

r 2

1 2 = ! 0 2 1 +

r g2

r 2

1 2 # ! 0 2

We see that the resonance for n = 1 is dominated by the tangential stiffness EA, i.e ., it is extensional , and about 40 % higher than the axisymmetric

resonance n =0.

Just as for the static loading, the behavior for n " 2 is quite different. Thedeterminant is:

Det = ( )! A " 2 2

– ! A " 2 ( ) E I

r 4 + E A

r 2 n 2 + E I

r 4 n 4 + E A

r 2

+ E A

r 2 E I

r 4 n 2 ( )n 2 – 1 2 = 0



100806040200

0e+0

1e+6

ircumferential harmonic n

e

ue

c

Hz

E x t e n s i o n a l

Inex tens iona l

Figure (b) – Resonant frequencies in a ring for high harmonics. Note that the twobranches come close but do not intersect. (steel, radius = 0.1m, cross section 0.01 x

0.01m, r /r g = 33) This is for the classical theory. Actually, transverse shear deformationis significant for values of n nearing the intersection of the two branches.

Ring, transient excitation – The transient excitation can be treated by the well-known and widely used method of modal expansion. The displacements are written inthe form:

( )w ! , t = w n ( ) t cos n ! !n = 2, 3, ...

u ! ( ) ! , t = u ! n ( ) t sin n ! !n = 2, 3, ...

in which the modal amplitudes, which in this case of the ring are just the Fouriercoefficients, depend on time. For the inextensional solution, the tangential component isrelated to the normal component of displacement as before:

u ! n = –

1n

w n

Hamilton's principle states that the first variation of the difference in the kinetic andpotential energies must be zero:



! " u , w = ! # t 1

t 2

# ( )T – V

0

2 $

d% d t = 0

After performing the integration over the space coordinate, the result is:

! = " # r 12

$ A

r 2 ( ) 1n 2

+ 1 w n2!

n = 2, 3, ...t 1

t 2

– 12

E Ir 4

( )w n 2 ( )n 2 – 1 2 – ( ) q r n – 1n q ! n w n

The modes are uncoupled and the Euler-Lagrange equation for mode n is:

w n +

!

n

2

w n =

Qn =

! n2

k n ( ) q r n –

1

n q " n

in which the model resonance and model stiffness are:

! n2 = E I

" A r 4 n 2 ( )n 2 – 1 2

n 2 + 1

k n = E I

r 4 ( )n 2 – 1 2

We take for an example a step pressure which acts on the one side of the ring attime t = 0:

qr ( ) ! , t = ""#$

%

P cos ! for – &

2 < ! < &

2 and t > 0

0 elsewhere

q ! ( ) ! , t = 0

The Fourier components of the load are:

q r n = – 2 P!

( ) – 1 n/2

–

1

n 2 – 1 for n = 2, 4, 6, ...

Thus the solution for the modal amplitude is:

w n ( ) t = – 2 P ( ) – 1 n /2 – 1 r 4

! ( )n 2 – 1 3 E I 1 – cos " n t



for the axisymmetric static solution, we have:

w 0 !"

#

$ %

&

1 – ' 2

( 0

2 =

r 3 ) ' 2

E

u0

= 0

Thus if the spin rate exceeds the breathing frequency, a divergence instability occurs.Generally, however, problems occur at far lower spin rates. For these equations it is not

possible to obtain vibration solutions in the same product form as for ! . = 0. However atraveling wave solution can be obtained for zero external loading:

w ! , t = Re w n e i ( ) " t – n ! = w n cos " t – n !

u ! ( ) ! , t = Re u ! n i e i ( ) " t – n ! = – u ! n ( )sin " t – n !

Substituting these into the differential equations gives the relation:

E I

r 4 n 4 + E A

r 2 – ! A " 2 + #

2 E I

r 4 n 3 + E A

r 2 n +2 ! A # "

E I

r 4 n 3 + E A

r 2 n + 2 ! A # " E I

r 4 n 2 + E A

r 2 n 2 – ! A " 2

w n

u $ n = 0

0

Setting the determinant equal to zero provides the relation between frequency w andwave number n. Note that the odd powers of # occur. For small spin rate, theapproximate inextensional solution is:

! 2 + 2 " ! – ! n 12 = 0

where # n1 is the inextensional resonance when = 0 and:

! = 2 " n

n 2 + 1

So for ! << # n1, the two roots are approximated by:

! a = ! n 1 – " ! b = – ! n 1 – "

Add the two wave solutions:

( )w ! , t = w n ( )cos " n 1 t – # t – n ! + ( )cos – " n 1 t – # t – n !

= 2 w n ( )cos ! t + n " ( )cos # n 1 t

u ! = 2 u ! n ( )sin " t + n ! ( )cos # n 1 t



For a stationary value of the total Lagrangian:

d d w n

( ) T – V = 0

which gives the result:

w n = q r n – 1n qq n

E I r 4

( )n 2 – 1 2 – ! A " 2

( ) n 2 + 6

Thus the amplitude is unbounded at the critical spin rates:

! n crit

2 = E I

" A r 4 ( )n 2 – 1 2

( )n 2 + 6

The harmonic n = 2 gives the minimum critical spin rate:

! 2 crit

2 = E I

" A r 4 910

For the conservative design of a device, the spin rate should be kept lower than this value.For more adventurous designs, the operating spin rate is between critical values.

II. PLATES

The equation for the transverse bending of a plate which is constant thickness,attached to an elastic foundation, and made of a linearly elastic material is:

D !! w + k w + " h#

2 w

# t 2 = ( ) p x , y , t

in which k is the foundation modulus, p is the transverse pressure and D is the bendingstiffness:

D = E h c2

= E h 3

12 ( ) 1 – ! 2

For rectangular, cartesian coordinates x, y and for zero transverse pressure p = 0, atravelling wave solution can be found just as for the beam:

( )w x , y , t = e ( )i ! t – n x x – n y y



which yields the dispersion relation:

D ( )n x2 + n y

2 2 + k – ! h " 2 = 0

One of the easiest problems is to consider the free vibration of a plate with with simply-

supported edges.

z

y

x

s

Figure – Rectangular plate with simply-supported edges.

The values of the wave numbers are chosen to satisy the boundary conditions of simplesupport:

n x = m ! a n y = n !

b

where m and n are integers. The wave solutions are then added together to form theFourier series, which is complete in the space of the plate. A direct method for thevibration problem is to write the solution as product of a function of time and a functionof the space variables:

( )w x , y , t = cos ! ( )t W x , y

For the rectangular plate with simply-supprorted edges the spatial distribution can bewriteen as the double Fourier series:

( )W x , y = !m = 1

! w m , n ( )sin m " x

a # $ % & sin n " yb

!n = 1

!

Substituting this form of the solution into the differential equation or using the dispersionrelation easily yields the natural frequencies:

! h " m n2 = D ( ) m #

a2

+ ( )n # b

2 2+ k



Related with each natural frequency is a mode shape. The figure indicates the nature ofthe first four modes. When the foundation modulus is not equal to zero, there is a cut-offfrequency.

2 2 2 2

Figure – Vibration mode shapes. Shaded and white regions have the opposite directionof displacement. The integers are the values of m and n for each mode.

The equation for the mode shape in vibration can be factored:

( ) ! ! – " 4

w = 0

( ) ! + " 2 ( ) ! – "

2 w = 0

where:

! 4 = " A # 2 – k

D

Polar Coordinates – In polar coordinates one equation is

( ) ! + " 2 w = 0

or in expanded form:

!2

w

! r 2 +

1r

! w

! r

+ !

2w

r 2 ! " 2

+ # 2 w = 0

Thus it is convenient to use a Fourier series in the circumferential direction:

( )w r, ! = f n ( ) r! n = 0, 1,...

"

cos n !

The equation for the radial dependence for a particular harmonic is

f n" + 1

r f n' + ! " # $ %

2 – n2

r 2 f n = 0



which has the solutions in terms of Bessel functions:

f n ( ) r = An J n ( ) ! r + Bn Y n ( ) ! r

These functions are oscillatory for ! r > n , for ! > 0 .

The second equation

w = g n ( ) r! cos n !

where

g n!

+ 1r

g n"

– # $ % & ' 2

+ n2

r2

g n= 0

which has solutions in terms of the modified Bessel functions:

gn ( ) r = C n I n ( )! r + Dn K n ( )! r

These modified behave exponentially. Thus the total solution is

( )w x , y , t = e i ! t A n J n ( )" r + B n Y n ( ) " r!

+ C n I n ( ) ! r + D n K n ( ) ! r cos n "

The functions Y n and K n are singulkar at r =0.

So for a cruplete disc withont a hole the constants C n and D n must be yers.

For an inner and an onter boundary a # r # n, all four constants must be used to satisfythe boundary conditions. The displacement and stress quantities are computed from thefollowing formulas:

Rotation;

! r= –

" w

" r

! # =

" w

r " #

Curvature change <<< #### """ :



k r = ! " r

! rK # =

! " #

r ! # + 1

r " r

= – !

2 w

! r 2 = – 1

r 2 !

2 w

! # 2

– 1r

! w

! r

! r " = ! " r = –

##r

$% &

' ( )

1r

#w

#"

Moment–curvature relations:

u r = D ! r + " ! # u # = D ! # + " ! r

u ! r = u r ! = D 1 – " k r !

transverse shear:

Q r = – D!

! r" w

Q # = – D

!

r ! # " w

Effective transverse shears on boundary:

V r = Q r + 1r

! u r " ! "

V " = Q " = ! u r "

! r

Waves Vibration Intro

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