Spring 1994 C.R. Steele

ME 241D THEORY OF SHELLS

VIBRATION AND STABILITY

Course Outline

PART 1 Waves and vibration I. Introduction 1. Plane waves a. Vibration b. Wave propagation 2. Beam with and without an elastic foundation 3. Ring, extensional and inextensional vibration II. Circular Plate 1. Waves and vibration in rectangular coordinates 2. Bessel function solutions for polar coordinates III. Shallow shell 1. Derivation from von Kármán equations 2. Waves and vibration in rectangular coordinates 3. Bessel function solutions for polar coordinates IV. Fluid-Elastic interaction PART 2. Stability Texts: C.R. Steele and C. Balch, ME241A,B Course Notes R. Szilard, Theory and Analysis of Plates, McGraw-Hill, 1975 J. Arbocz, M. Potier-Ferry, J. Singer, and V. Tvergaard, Buckling and Post Buckling, Lecture Notes in Physics 288, Springer, 1987. Werner Soedel, Vibrations of Shells and Plates, Marcel Dekker, Inc., 1993. Zdenek P. Bazant and Luigi Cedolin, Stability of Structures, Oxford, 1991.

I. INTRODUCTION 1. PLANE WAVES By way of introduction and/or review of some basic ideas of dynamics , we first consider the "wave equation":

Δ u = ux x + uy y + uzz = 1c2

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

ux x = 1c2

ut t The general equations of elasticity are considerably more complex, but in several special cases, reduce to this form. Plane strain – For the plane strain motion of a slab:

εx = 1E

σx – ν σy + σz

εz = εy = 1E

σy – ν σx + σz = 0

so that:

σz = σy = ν σx1 – ν

giving the result:

εx = 1 + ν 1 – 2ν σx1 – ν E

Note: For Lamé parameters:

λ + 2 µ = 1 – ν E 1 + ν 1 – 2ν

The equation of equilibrium is:

∂σx∂x = ρ ∂

2u∂t2

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

∂2σx

∂x 2 = 1

c12 ∂2σx

∂t2

where:

c1 = E 1 – ν ρ 1 + ν 1 – 2 ν

1/2

is the "dilatational" velocity. Shear waves – For the propagation of shear waves in an elastic medium , the velocity is:

c2 = E ρ 2 1 + ν

1/2

Bar – For the axial motion of an elastic bar:

∂∂x A σx = ρ A ∂2u∂t2

A σx = E A ∂u∂x

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

∂2σx

∂x 2 = 1

cb2 ∂2σx

∂t2

where the "bar" velocity is:

cb = Eρ

1/2

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

σx = E1 – ν2

εx + ν εy

For εy = 0, we have the wave equation:

∂2u∂x 2

= ρ 1 – ν2

E ∂2u∂t2

so the "plate" velocity is:

cp = E ρ 1 – ν2

1/2

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

c12

cb2

= 1 – ν 1 + ν 1 – 2 ν

=

∞ for ν →– 1

1 for ν = 0

6/5 for ν = 1/4

∞ for ν →1/2 Thus for the usual metals and concrete, c1 is not much different from cb , however for nearly incompressible materials c1 >> cb. The ratio is similar for the plate velocity:

c12

cp2 = 1 – ν 1 – ν2

1 + ν 1 – 2 ν = 1 + ν2

1 – 2 ν

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

∂w∂x = 1

cString2 ∂2w∂t2

where w is the normal displacement and the speed is:

cString = Tρ A

1/2

in which T is the tension, ρ 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 in an elastic tube, such as an artery. The assumptions are (1) the dominant velocity of the fluid is plane and in the direction of the axis (the long wavelength assumption), (2) the inertia of the tube wall is negligible, and (3) the fluid is incompressible. The stress-strain relation for the tube is:

εθ = wR

= 1E t

Nθ – ν Nx

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

Nx = 0 Nθ = p R For conservation of the fluid the wall displacement is related to the axial displacement of the fluid:

π R 2 ∂u∂x = – 2 π R w

The wall displacement can be eliminated, yielding the relation:

p = – E t2 R ∂u

∂x

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

∂p∂x = – ρ ∂

2u∂t2

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

cTube = E t2 R ρ

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 or elastic medium, axial motion of an elastic bar, or the transverse motion of a stretched string is

ux x = 1c2

ut 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 eiωt f ( x ) = f ( x ) cos ω t Then f must satisfy the equation

f ″ = – ω 2

c 2 f

i.e.

f = A sin ωc x + B cos ωc x

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

f = A sin ω xc

where the natural frequencies are at:

ω = π n c

L n = 1, 2, 3,...

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

σ x , t = eiωt C2 sinω x – L

c u x , t = eiωt c C2

ω E cos ω x – L

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

u L, t σ 0, t

= cEω

1sinωL

c

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

ωLc = n π for n = 1, 2, ...

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

10000100010010.01

.1

1

10

100

1000

Frequency Hz

Resp

onse

u(

L)/p

(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 key of A.

43210- 1- 2.01

.1

1

10

100

1000

Frequency, octaves above fundamental

Resp

onse

u(

L)/p

(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 different sort of mouthpiece for which the input is approximated by considering a prescribed displacement at the end x = 0. The response function is then:

u L, t u 0, t = 1

cos ω Lc

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

ωLc = n π2 for n = 1, 3, 5, ...

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

43210- 1- 21

10

100

1000

10000

Frequency in octaves

Resp

onse

u(

L)/u

(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 the resonance 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:

f = e± i ωc

x

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

u x , t = A sin ω t + xc + B sin ω 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 the positive 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 superposition of two deformation distributions which move with the velocity c in opposite directions. The solution to various transient problems is readily obtained. For example, consider a semi-infinite slab initially at rest with a prescribed pressure on the face. The conditions are:

u ( 0, t ) = h ( t ) u ( x , 0 ) = ut ( x , 0 ) = 0 For this the term for the waves incoming from + ∞ must be excluded, so f = 0 and the function 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 < 0

1 for t < 0

produces a step wave:

xct0

u

c

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 ω t for 0 < t < πω

0 for πω

< t

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

xct0

u

c

πω( )t - c

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

Pulse propagation in finite slab – For the finite slab, initially at rest, with the pressure 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 there is 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 traveling to the left:

– h t + x - 2 Lc

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

+ h t – x + 2 Lc

The procedure continues indefinitely. The complete solution consists of an infinite train of 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 Lc +

+ h t – x + 2 Lc – h t + x - 4 L

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

xct0

u

c

πω( )t - c

L c

Image pulse

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

infinite medium.

x0

u

c

c

Total

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

moving image pulse.

c

x0

u

c L

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

c

x

u

c L

(d) As the image pulse arrives at the left face, it will be canceled by a second image pulse

traveling 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 domain and we are back to the situation in Figure (a). Of particular practical importance, a compressive pulse on one face causes a reflected tensile pulse from the free face of the slab. 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 compressive pressure pulse on one face. The failure will be a delamination occurring near the back face due to the tensile image pulse. Inhomogeneous Medium – We now consider the inhomogeneous slab for which the 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 eiωtf ( x )

f ″ + ω

2

c2 f = 0

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

f = e –iωζ ( x ) f0 + 1

i ω f1 – 1

ω2 f2 + . . .

Substituting into the equation and equating the coefficient of each power of ω to zero gives the sequence of equations:

(ζ ′ )2 = 1

c2 2 ζ ′ f0

′ + ζ ″ f0 = 0

2 ζ ′ f1′ + ζ ″ f1 = f0

″

......................

The first equation gives the argument of the exponent, i.e. the phase function:

ζ = dxc

0

x

while the second equation gives the amplitude function:

f0 = A ( ζ′ )

– 1/2 = c ( x )

c ( 0 )1/2

The third equation gives the correction to the amplitude function:

f1 = 12 f0 c (0 ) f0 f0

″

0

x

d x

Vibration, one-term approximation – For the vibration problem f = 0 at x = 0, L, use the expansion in the form:

f = A sinωζ f0 – f2ω2

+ ... + cosωζ f1ω

– - f3ω3

+ ...

+ B cosωζ f0 – f2ω2

+... – sin ω ζ f1ω

– f3ω3

+...

The one-term approximation is

f ≈ A f0 sinωζ + B f0 cosωζ which has the simple solution

B = 0 ω ζ ( L ) = n π for n = 1, 2, 3, ...

Vibration, two-term approximation – To find the next approximation to the natural frequencies take

ωζ( L ) = n π + ε f ≈ A f0 sinω ζ + f1

ω cos ωζ + B f0 cosωζ

To satisfy the boundary conditions f= 0 at x = 0, the constants are:

B = – f1

ω f0 at x = 0 = 0

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

f ( L ) = cosnπ A sinε f0 + cosε f1ω which gives the correction term:

tan ε ≈ ε = – f1ω f0

Thus, the natural frequencies are at

ω = n π

dxc

0

L – 1

2nπ c1/2( c1/2 )″dx + O ( π n )–3

0

L

The maximums of f are at:

ω ζ ′cosωζ f0 + f0′ sinωζ – f1 ζ

′ sinωζ = 0

cotωζ = f1 ζ ′ – f0′

ωζ ′ f0

which gives the envelope of the modes

( f )max = f0 1 + (ζ′f1 )

2 – ( f0

′ )2

2 (ωζ ′ f0 )2

+ O ω–4

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

c = c0 eκx which gives the result for the natural frequencies:

ωco = κnπ1 – e–κL

1 + ( 1–e–κL ) ( 1 – eκL )8 n2π 2

+ O 1n4π 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 = eκx/2 1 + ( 1 – e–κx )2

8n2π 2 (e

κx – 1 )216 – e2κx + O 1

n4π 4

2. Beam with and without and elastic foundation In the preceding section, the "wave equation" was considered. There are two important properties of this equation: (1) Discontinuities or disturbances propagate with a 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 the beam equations, a propagating disturbance pulse will change its form as it propagates in the medium. Such a medium is called dispersive. The beam is an example of a medium which is dispersive. The equation from the elementary beam theory (Euler-Bernoulli) is, for a beam with constant properties and no lateral load:

EI ∂4w∂x 4

+ ρ A ∂2w∂t2

+ k w = 0

in which EI is the bending stiffness, ρA is the mass per unit length, and k is the elastic foundation stiffness. Wave propagation – This is most definitely not a "wave equation", since it is parabolic and not hyperbolic. However, we can easily obtain a solution which gives a wave propagating along the beam in the form:

w x , t = ei ω t – n x where ω is the frequency and n is the wave number, which is the reciprocal of the spatial wave length λ:

λ = 2πn Substitution into the equation, with the foundation stiffness k = 0, gives the frequency-wave number relation:

EI n4 – ρ A ω2 = 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:

xt = vphase = ωn

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

vphase = ωn = EIρA

1/2n = EIρA

3/4ω1/2

Thus the phase velocity is not constant, but depends on the wave number or frequency. Thus the wave for each frequency moves with a different phase velocity. A particular deformation shape which can be described at one instant of time by a superposition of waves will not look the same at a later time. That is, the shape disperses. There is not a single velocity which characterizes the behavior. Now consider a sum of two waves of equal amplitude:

w x , t = sin ω1t – n1 x + sin ω2t – n2 x Using the trigonometric identities, this can be rewritten into the form:

w x , t = 2 sin ω1+ω22 t – n1+n22 x cos ω1–ω22 t – n1–n2

2 x which shows that the two waves con be considered as one wave with the averaged frequency and wave number, with a modulating envelope which moves with the differences. When the frequencies approach each other the phase velocity of the envelope becomes what is called the group velocity:

venvelope = Limn1→n2 ω2 – ω1

n2–n1 = d ω

d n = vgroup

This turns out to be the velocity of energy propagation of the wave, and so is very important in transient wave response, probably more important than the phase velocity. Thus for a general medium, phase velocity depends on wave number. Waves of different wavelength propagate at different velocities. Pulse shape will distort with propagation distance as indicated in the sketch.

x

t = 0

x

t = t1 > 0

x

t = t 2 > t 1

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

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

v g = d ωd n

which for beam is twice the phase velocity:

EI n4 – ρ A ω2 = 0 vp = ωn = EI

ρ A n

vg = d ωdn

= 2 EIρ A

n = 2 vp

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 ωt 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 the group velocity. Ahead of the point moving with the group velocity, the amplitude decreases, as indicated in the sketches:

V p

x

x

p"Front" moves with = 2 vvg

"Front"Steady-State

t = t > 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:

ρ Aω2 = EI n4 + k which gives the phase velocity:

ωn = E Iρ A

n2 + kρ A n2

The new feature due to a non zero foundation modulus k is that the frequency has a non zero value ω0 for zero wave number. This is referred to as the cutoff frequency, since there are no real values for the wave number, and thus no propagating waves, for frequencies less than this value.

ω

n

ω0

WithFoundation

Without

Foundation

CutoffFrequency

The value of the cutoff frequency is:

ω 0 = kρ 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. As shown in the sketch, a point load oscillating with a frequency less than ω0 will cause a "local" response, while a frequency greater than ω0 causes the generation of propagating waves. P 0 cos ωt

ssssssssssssssssssssssssssssssssss

quasi. - static

propagatingwaves

ω ω< 0

ω ω> 0

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

min phasevelocity

n

V p = ωn

Vgvp

V g = ω

dnd

Figure – Phase and group velocities for the Euler-Bernoulli beam on an elastic

foundation. 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 will have a steady-state solution, if the load velocity is not equal to the minimum phase velocity. For lower velocities the response is localized to the moving load. For higher velocities, waves will be generated which have the same phase velocity as the load. The wave 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 to the 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 time and velocity previously designated by these symbols. Beam on foundation Cylindrical shell Bending stiffness

EI Bending stiffness Etc2 = Et3/12(1-ν2)

Foundation stiffness k

Circumferential stiffness Et/r2

Axial bar stiffness EA

Axial shell stiffness Eπrt

The dispersion relation for the cylindrical shell is therefore:

ρ t ω 2 = E t c 2n4 + E t r 2

or solving for the frequency:

ω 2= E ρ r 2

1 + r 2c2n 4

Thus the cut-off frequency is the ring breathing resonance:

ω02 = E

ρ r2

The wavelength is:

Λ = 2 π n

while the decay distance for static self-equilibrating edge loads is:

δ = π 2 r c Thus the frequency-wave number relation can be rewritten in the form:

ω 2 = ω02 1 + 2δ

Λ 4

Therefore, wavelengths of vibration or wave propagation which are long in comparison with the static decay distance will all have a frequency close to the ring breathing frequency. For the thin shell many modes of vibration have nearly the same frequency. Only when the wavelength is of the order of magnitude of the static decay distance will the frequencies be different. This causes two difficulties for direct numerical methods. First the standard modal analysis is most effective when the eigenfrequencies are distinct and not closely bunched. Secondly, a fine mesh must be used on the shell for finite difference and finite element methods to capture the significant bending waves with the wavelength O(δ).

ω

n

bar

Beam on elastic foundation

cylindrical shell

plate

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

the "plate" mode for high frequencies.

Ring, statics - The ring is important in its own right and serves to introduce the effect of curvature. The circumferential angle is θ and the tangential and normal components of displacement are uθ and w, as shown in the figure, while the in-plane rotation of the ring reference line is denoted by χθ .

w

uθ

θ

χθ

Figure – Displacement components tangential uθ and normal w to the ring reference line. The rotation χθ is positive corresponding to tension in the fibers on the positive w - side of

the reference line. The rotation and strain of the ring are:

χθ = ∂w

r ∂θ + uθ

r

εθ = ∂uθr ∂θ

+ wr

For a rigid body translation both the rotation and strain are zero, as they should be. A rigid body translation in the x-direction is:

v = w e r + u θ eθ = C ex so the components are:

w = C cosθ uθ = – C sinθ

x ex

e

θ

θ

y

erey

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

Also for a rigid body rotation:

uθ = C w = 0 which gives a constant χθ. The curvature change measure is the derivative of the rotation, which gives the bending moment:

κθ = ∂χθ

r ∂θ Mθ = EI κθ

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 ring reference line has the components qr and qθ.

y

x

q

qθ

r

θ

Figure – Load components acting on ring.

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

Π u, w = Vdθ 0

2π

= 12 E I κθ

2 + E A ε θ2 – qrw – qθ uθ r dθ

0

2π

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

V = 12 E Ir 4

( ) – w ′′ + uθ′ 2 + 12 E A

r2 ( ) uθ′ + w 2 – qrw – qθ uθ r

in which derivatives with respect to θ are denoted by primes. The Euler-Lagrange equations, obtained by considering variations with respect to w and uθ , are the differential equations for the problem:

∂2

∂θ 2 ∂V

∂w ′′ –

∂

∂θ ∂V

∂w ′ + ∂V

∂w = 0

– ∂

∂θ ∂V

∂uθ′ +

∂V

∂uθ

= 0

which are:

E I r 4( ) w ′′′′ – uθ′′′ + E A

r 2( ) w + uθ′ = qr

EI r 4( ) w ′′′ – uθ′′ – EA

r2( ) w ′ + uθ′′ = qθ

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 θ are:

qr ( ) θ = qr n∑n = 0, 1, ...

∞ cos nθ

qθ ( ) θ = qθ n∑n = 1, ...

∞ sin nθ

with the corresponding expansions for the displacement components:

( ) w θ = wn∑n = 0, 1, ...

∞ cos nθ

uθ ( ) θ = uθ n∑n = 1, ...

∞ sin nθ

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

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

E Ir 4

n4 + E Ar2

E Ir 4

n 3 + E Ar 2

n

E Ir 4

n 3 + E Ar 2

n E Ir4

n 2 + E Ar 2

n2 wn

uθ n =

qr nqθ 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 w0

uθ 0 =

qr 0qθ 0

It follows that:

qθ 0 = 0 which is the condition that no resultant torque can act on the ring for static equilibrium, and that uθ0 , which is the amplitude of rigid body rotation, is arbitrary. The axisymmetric radial expansion amplitude due to the uniform radial load is:

w0 = r 2E A

qr 0 For the harmonic n = 1, the equations are:

E Ir 4

+ E Ar 2

1 11 1

w 1u θ 1

= qr 1qθ 1

Thus for a solution:

qr 1 = qθ 1

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

E I r 4

+ E Ar 2

w1 + uθ 1 = qr 1 = qθ 1

When w1 + uθ1 = 0, we have the rigid body translation previously discussed.

y

X

qr

Resultant πqr1

yX

qθ

Resultant πq θ1

(a) Load component qr1 > 0 produces resultant force in positive x-direction.

(b) Load component qθ1 > 0 produces resultant force in negative x-direction.

Figure – Distribution of loading for components with n = 1. The distribution is self-equilibrating when qr1 = qθ1.

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

EIr4

+ EAr2

= EAr2

1 + rg2

r2

in which rg is the radius of gyration for the cross section. Thus for a thin ring, the bending 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 set equal to zero, the stiffness matrix is singular, which indicates that the amplitude of displacement will be substantially larger than for the harmonics n = 0, 1. The exact solution for the displacement coefficients is:

wn = qr n E I

r 4 n 2 + E A

r 2 n 2 – qθ n E I

r 4 n 3 + E A

r 2 n

Det

un = –qr n E I

r 4 n 3 + E A

r 2 n + qθ n E I

r 4 n 4 + E A

r 2Det

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 rg << r, the approximate solution is:

wn ≈ qr n n2 – qθ n n n2 n2 – 1 E I

r 4

uθ n ≈ – wn

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

εθ n ≈ 0 Thus accord to this approximation, the deformation is inextensional. Since for high values of n, the EI terms in the numerator of the expressions for the displacement components are significant, this approximation is valid only for sufficiently low values of n:

n << rrg

2

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

Π = ∫ V dθ 0

2 π = π r 12 E I

r4 ( ) wn 2 ( ) n2 – 1 2 – ( ) qr n – 1n qθ n wn ∑

n = 2, 3, ... Setting the derivative with respect to the amplitude wn directly to zero yields the approximate 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 of the ring, the potential energy is:

V = 12 E I κ2 + 12 E A ( ) εθ – α T 2 – qrw – qθ uθ r

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

qr ⇐ qr + 1r E A α T qθ ⇐ qθ – 1r E A d

dθα T

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

wn = – r α Tn

n2– 1 for n ≠ 1

uθ n = n r α Tn

n2– 1 for n ≠ 1

w1 + uθ 1 = r α T1

1 + rg2

r2

for n = 1

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

σdirect n = FnA

= ( ) E εθ n – α Tn = 0 for n ≠ 1

= E α T1 ( )

rgr

2

1 + ( ) rgr

2 for n = 1

while the bending stress is:

σbending n = z MnI

= z E κθ n = 0 for n ≠ 1

= E α T1 zr

1 + ( ) rgr

2 for n = 1

in which z is the distance from the reference line of the ring. While all harmonics of heating produce a distortion of the ring, only the harmonic n =1 produces stress, and a low magnitude of stress at that. The ring simply adapts to the heating without significant stress. Ring, vibration – For the vibration of the circular ring, the solution can be obtained in the form of a product of sinusoidal variation in space and time:

w θ, t = wn cos nθ cos ωt uθ θ, t = uθ n sin nθ cos ωt Consequently, the equations are the same as in the static case, but with inertia terms added to the load components:

qr ⇐ qr + ρ A ω 2 w qθ n ⇐ qθ n + ρ A ω 2 uθ

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

E I r 4

n 4 + E A r 2

– ρ A ω 2 E I r 4

n 3 + E A r 2

n

E I r 4

n 3 + E A r 2

n E I r 4

+ E A r 2

n 2 – ρ A ω 2 wn

uθ n =

qr nqθ n

For the harmonic n = 0 this reduces to:

E A r 2

– ρ A ω 02 = 0

which defines the ring breathing frequency:

ω 0 = 1r E ρ

= cbarr

Thus the period T0 of the breathing frequency is the time required for a wave traveling at the bar velocity to transverse the circumference of the ring:

T0 = 2 π rcbar

For the harmonic n = 1, the result is:

α – λ α

α α –λ = α – λ 2 – α2 = – 2 α λ + λ 2 = 0

in which

λ = ρ A ω 2

α = E I r 4

+ E A r 2

Thus the resonant frequency is given by:

λ = 2 α that is:

ω 1 = 2r Eρ

1 + rg2

r 2

1 2 = ω 0 2 1 + rg2

r 2

1 2 ≈ ω 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. The determinant is:

Det = ( )ρ A ω 2 2 – ρ A ω 2 ( ) E I r 4

+ E Ar 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

Because this is quadratic in ω2, for each harmonic n there are two distinct resonances. For the thin ring, there is a small root ω2

n1, given by:

ρ A ω n 12 n 2 + 1 ≈ E I

r 4 n 2 n 2 – 1 2

or:

ω n 1 ≈ ω 0 r gr n n 2 – 1

n 2 + 1

This is the inextensional frequency, dominated by the bending stiffness EI and much smaller than the breathing frequency w0. The large root of the equation is approximated by:

ωn 2 ≈ ω0 n 2 + 1 which is the extensional frequency, larger than the breathing frequency. The extensional and inextensional frequencies are shown in the figure for a ring with r/rg = 33. Note that the inextensional and extensional branches approach each other at n = r/rg. The exact and approximate values are very close for a ring of this geometry and thinner.

10864200

20000

40000

Circumferential harmonic n

Freq

uenc

y H

z

Extens ional

Inextensional

Figure (a) – First resonant frequencies in a ring. (Steel, radius = 0.1m, cross section 0.01 x 0.01m, r/rg = 33)

1008060402000e+0

1e+6

Circumferential harmonic n

Freq

uenc

y

Hz

Extens ional

Inextensional

Figure (b) – Resonant frequencies in a ring for high harmonics. Note that the two branches come close but do not intersect. (steel, radius = 0.1m, cross section 0.01 x 0.01m, r/rg = 33) This is for the classical theory. Actually, transverse shear deformation is 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 in the form:

( ) w θ, t = wn ( ) t cos nθ ∑n = 2, 3, ...

uθ ( ) θ, t = uθ n ( ) t sin nθ ∑n = 2, 3, ...

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

uθ n = – 1n wn Hamilton's principle states that the first variation of the difference in the kinetic and potential energies must be zero:

δ Π u, w =δ ∫ t1

t2

∫ ( ) T – V 0

2 π dθ dt = 0

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

Π = π ∫ r 12 ρ Ar2 ( ) 1

n2 + 1 wn

2∑n = 2, 3, ...

t1

t2

– 12 E I r4

( ) wn 2 ( ) n2 – 1 2 – ( ) qr n – 1n qθ n wn

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

wn + ωn2 wn = Qn = ωn

2

kn( ) qr n – 1n qθ n

in which the model resonance and model stiffness are:

ωn2 = E I

ρ A r4 n2 ( ) n2 – 1 2

n2 + 1

kn = E I

r4 ( ) n2 – 1 2

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

qr ( ) θ, t = ⎪⎪⎨⎧

⎩

P cos θ for – π2 < θ < π2 and t > 0

0 elsewhere

qθ ( ) θ, t = 0 The Fourier components of the load are:

qr n = – 2 Pπ

( ) – 1 n/2 – 1

n2 – 1 for n = 2, 4, 6, ... Thus the solution for the modal amplitude is:

wn ( ) t = – 2 P ( ) – 1 n/2 – 1 r4

π ( ) n2 – 1 3 E I 1 – cos ωnt

The amplitude decreases with n6, so the total series solution converges rapidly. Indeed the n = 2 term is dominant. Even for the bending moment, the series converges with n4. This is the situation for smooth static or dynamic loading of the ring. Modes other than n = 2 generally have little effect. Ring, rotating with constant angular velocity – For a variety of significant technical problems, including turbines, computer disks, tires, and saw blades, the spin velocity has a significant effect on the response.

y

x

θ

Ω

ψ

t

Figure – Ring rotating with angular velocity Ω . The coordinate θ is measured with

respect to a line fixed in the body. If the ring is rotating with the constant angular velocity Ω , and uθ and w are the displacements with respect to coordinates fixed in the body, then the kinetic energy density is:

T = 12 ρ A r + w 2 Ω + uθr2 + w 2

The significant terms retained for the linear theory are:

T = 12 ρ A uθ2 + 4 Ω w uθ + w 2 + Ω

2 w 2 + 2 r Ω

2 w

The coupling term gives the Coriolis acceleration effect, while the last term gives the centrifugal force radial loading. The equations are now:

E I r 4( ) w ′′′′ – uθ′′′ + E A

r 2( ) w + uθ′ + ρ ( ) A w – 2 Ω uθ – Ω

2 w = qr + ρ A r Ω

2

EI r 4( ) w ′′′ – uθ′′ – EA

r2( ) w ′ + uθ′′ + ρ ( ) A uθ + 2 Ω w = qθ

for the axisymmetric static solution, we have:

w0 ⎜⎛

⎝⎟⎞

⎠ 1 – Ω

2

ω02

= r3ρ Ω

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 Ω . = 0. However a traveling wave solution can be obtained for zero external loading:

( ) w θ, t = Re wn ei ( ) ωt – nθ = wn ( )cos ωt – nθ uθ ( ) θ, t = Re uθ n i ei ( ) ωt – nθ = – uθ n ( )sin ωt – nθ

Substituting these into the differential equations gives the relation:

E Ir 4

n4 + E Ar2

– ρ A ω2 + Ω2 E I

r 4 n 3 + E A

r 2 n +2 ρ A Ω ω

E Ir 4

n 3 + E Ar 2

n + 2 ρ A Ω ω E Ir4

n 2 + E Ar 2

n2– ρ A ω2 wn

uθ n = 0

0

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

ω2 + 2 β ω – ωn 12 = 0

where ωn1 is the inextensional resonance when Ω = 0 and:

β = 2 Ω nn2 + 1

So for Ω << ωn1, the two roots are approximated by:

ωa = ωn 1 – β ωb = – ωn 1 – β Add the two wave solutions:

( ) w θ, t = wn ( )cos ωn 1 t – β t – n θ + ( )cos – ωn 1 t – β t – n θ

= 2 wn ( )cos β t + n θ ( )cos ωn 1 t

uθ = 2 uθ n ( )sin β t + n θ ( )cos ωn 1 t

Thus in the body-fixed coordinates the same vibration mode is observed as in the ring without the spin. The difference is that the node points, i.e., the points at which w = 0, move with the velocity:

θnode = – βn = – 2 Ω

n2 + 1

So, if the spin is small in comparison with the resonance of the ring, the location of the node provides an excellent measure of the orientation of the ring with respect to inertial space:

θnode = – 2 Ωn2 + 1

This principle also holds for a rotating shell of revolution and provides the basis for the new hemispherical resonant gyroscope, being developed by Delco, Santa Barbara. Ring, rotating with space fixed load – If a load is acting on the rotating ring which is a static load fixed in space, then a steady-state deformation of the ring also stationary in space can generally be found. The Fourier harmonic of load is written as:

qr = qr n cos nψ = qr n ( )cos n Ω t + n θ

qθ = qθ n sin nψ = qθ n ( )sin n Ω t + n θ and the solution is in the form:

w = wn cos nψ = wn ( )cos n Ω t + n θ

uθ = uθ n sin nψ = uθ n ( )sin n Ω t + n θ The inextensional solution is:

uθ n = – wnn

and the integrated kinetic and potential energies are:

T = 12 ρ ( ) A n2 + 6 Ω2 wn2

V = 12 E I

r4 ( ) n2 – 1 2 wn2 – ( ) qr n – 1n qθ n wn

For a stationary value of the total Lagrangian:

d dwn

( ) T – V = 0

which gives the result:

wn = qr n – 1n qq n

E Ir4

( ) n2 – 1 2 – ρ A Ω2( ) n2 + 6

Thus the amplitude is unbounded at the critical spin rates:

Ωn crit2

= E Iρ A r4

( ) n2 – 1 2

( ) n2 + 6

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

Ω2 crit2

= E Iρ A r4

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 ΔΔ w + k w + ρ 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 bending stiffness:

D = E h c2 = E h312 ( ) 1 – ν2

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

( ) w x , y , t = e ( )i ω t – nx x – ny y

which yields the dispersion relation:

D ( ) nx 2 + ny 2 2 + k – ρ h ω 2 = 0 One of the easiest problems is to consider the free vibration of a plate with with simply-supported edges.

z

y

x

b

ss

sss

ss

ssa

Figure – Rectangular plate with simply-supported edges.

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

nx = m π a ny = n π

b

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

( ) w x , y , t = cos ω ( ) t W x , y

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

( ) W x , y = ∑m = 1

∞

wm, n ( )sin m π x a ⎛

⎝⎞⎠sin n π y

b ∑

n = 1

∞

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

ρ h ω m n2 = D ( ) m π

a 2 + ( ) n π b

2 2+k

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

1,1 1,2 2,1 2,2

y y y y

x x x x

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

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

( ) Δ Δ – α 4 w = 0

( ) Δ + α 2 ( ) Δ – α 2 w = 0 where:

α 4 = ρ A ω 2 – k

D

Polar Coordinates – In polar coordinates one equation is

( ) Δ + α2 w = 0 or in expanded form:

∂2 w

∂r 2 + 1r

∂w

∂r + ∂

2w

r2 ∂θ 2 + α2 w = 0

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

( ) w r, θ = fn ( ) r ∑ n = 0, 1,...

∞

cos nθ

The equation for the radial dependence for a particular harmonic is

fn" + 1r fn

' + ⎛⎝⎞⎠ α2 – n2

r2 fn = 0

which has the solutions in terms of Bessel functions:

fn ( ) r = An Jn ( ) α r + Bn Yn ( ) α r These functions are oscillatory for α r > n , for α > 0. The second equation

w = gn ( ) r ∑ cos nθ where

gn ″ + 1r gn

′ – ⎛⎝⎞⎠ α2 + n 2

r 2 gn = 0

which has solutions in terms of the modified Bessel functions:

gn ( ) r = Cn In ( ) α r + Dn Kn ( ) α r These modified behave exponentially. Thus the total solution is

( )w x , y , t = ei ω t An Jn ( ) α r + Bn Yn ( ) α r ∑

+ Cn In ( ) α r + Dn Kn ( ) α r cos nθ The functions Yn and Kn are singulkar at r =0. So for a cruplete disc withont a hole the constantsCn and Dn must be yers.

For an inner and an onter boundary a ≤ r ≤ n, all four constants must be used to satisfy the boundary conditions. The displacement and stress quantities are computed from the following formulas: Rotation;

χr = – ∂w

∂r χθ =

∂w

r ∂θ

Curvature change <<< ≤≤≤≤ ≥≥≥:

k r = ∂χ r ∂ r

K θ = ∂ χ θ

r ∂ θ + 1r χ r

= – ∂ 2 w

∂ r 2 = – 1

r 2 ∂ 2 w

∂θ 2 – 1r

∂ w

∂ r

κr θ = κθ r = –

∂

∂r ⎜⎛

⎝⎟⎞

⎠ 1r

∂w

∂θ

Moment–curvature relations: ur = D κr + ν κθ uθ = D κθ + ν κr u θ r = u r θ = D 1 – ν k r θ transverse shear:

Q r = – D ∂

∂ r Δ w

Q θ = – D ∂

r ∂θ Δ w

Effective transverse shears on boundary:

V r = Q r + 1r ∂u r θ

∂θ

V θ = Q θ = ∂u r θ

∂r