ME 241D THEORY OF SHELLS VIBRATION AND STABILITY
Spring 1994C.R. SteeleME 241DTHEORY OF SHELLSVIBRATION AND STABILITYCourse OutlinePART 1 Waves and vibrationI. Introduction1. Plane wavesa. Vibrationb. Wave propagation2. Beam with and without an elastic foundation3. Ring, extensional and inextensional vibrationII. Circular Plate1. Waves and vibration in rectangular coordinates2. Bessel function solutions for polar coordinatesIII. Shallow shell1. Derivation from von Kármán equations2. Waves and vibration in rectangular coordinates3. Bessel function solutions for polar coordinatesIV. Fluid-Elastic interactionPART 2. StabilityTexts:C.R. Steele and C. Balch, ME241A,B Course NotesR. Szilard, Theory and Analysis of Plates, McGraw-Hill, 1975J. 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. INTRODUCTION1. PLANE WAVESBy way of introduction and/or review of some basic ideas of dynamics , we firstconsider the "wave equation":Δ u ux x uy y uz z 1 ut tc2where 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:ux x 1 ut tc2The 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:εx 1 σx – ν σy σzEεz εy 1 σy – ν σx σzE 0so that:σz σy ν σx1–νgiving the result:εx 1 ν 1 – 2ν σx1–ν ENote: For Lamé parameters:λ 2µ 1–ν E1 ν 1 – 2νThe equation of equilibrium is: σx 2 u ρ x t 2For constant ρ, the displacement u can be eliminated and the final equation is:
2 σx 2 σx 1 x 2c12 t 2
where:c1 E 1–νρ 1 ν 1–2ν1/2is the "dilatational" velocity.Shear waves – For the propagation of shear waves in an elastic medium , thevelocity is:c2 Eρ 2 1 ν1/2Bar – For the axial motion of an elastic bar: 2 uA σx ρ A x t 2 uA σx E A xSo, for the cross-sectional area A and the density ρ constant, we again have the onedimensional wave equation: 2 σx 2 σx 1 x 2cb2 t 2where the "bar" velocity is:cb Eρ1/2Plane stress – For a flat plate, the constitutive relation is:σx Eεx ν εy1 – ν2For εy 0, we have the wave equation: 2 u ρ 1 – ν 2 2 u E x 2 t 2so the "plate" velocity is:
cp ρ1/2E1 – ν2The various speeds can vary substantially. The ratio of the dilatational speed tothe bar speed is:for ν – 1 1c121–ν 1 ν 1–2νcb26/5for ν 0for ν 1/4for ν 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:2c121 – ν 1 – ν2 1 ν1 ν 1–2ν1–2νcp2String under tension – The one-dimensional wave equation arises in many othersituations. For a string under tension the equation is: w 2 w 12 x cString t 2where w is the normal displacement and the speed is:cString TρA1/2in 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 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:εθ w 1 N θ – ν N xR Etwhere 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:
Nx 0Nθ p RFor conservation of the fluid the wall displacement is related to the axial displacement ofthe fluid: u – 2πRw xThe wall displacement can be eliminated, yielding the relation:π R2 up – Et2 R xNote 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/(2R). Finally the equation of equilibrium for thefluid is needed: p 2 u –ρ x t 2with which the wave equation in the same form as in the preceding cases is obtained,except that the speed is:cTube Et2R ρ1/2Thus, for a thin-walled tube, the speed of the fluid-elastic wave propagation is muchslower 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 isux x 1 ut tc2For the homogeneous case, i.e. when c is a constant, the solutions are straightforward.The solutions periodic in time are obtained by assuming the formu x, t Re e iω t f ( x ) f ( x ) cos ω tThen fmust satisfy the equationf ″ – ω fc22i.e.ωf A sin ωc x B cos c x
For a slab with free faces at x 0, L, the solution is:where the natural frequencies are at:f A sin ωcxω πn cLn 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:ω x–Lσ x, t e iω t C2 sincω x–LcC2iωtu x, t ecoscωETherefore the ratio of the input stress to the output displacement is:u L, tσ 0, t1 cEω sinωLcFrom this it is clear that a substantial output is obtained at the resonant frequencies:ωL n π for n 1, 2, .cThe 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.
10000HzFigure (a) – Response of acoustic tube, given by the transfer function defined by theoutput displacement divided by the input pressure. This simulates a horn tuned to the keyof ctaves12above34fundamentalFigure (b) – This is the same response as in Figure (a) , but with the frequency scale inoctaves 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, t1 u 0, tcos ωcLwhich gives the behavior shown in Figure (c). The resonance frequencies are at thevalues:ωL n πc2for 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 quency in octavesFigure (c) – Response consisting of the ratio of displacements at the two ends. Theparameters 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:f e iω xc
then the total solution is obtained in the traveling wave form:u x, t A sin ωt xc B sin ω t – xcThe 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 thephase velocity c. The second term gives a sinusoidal "wave train" which moves in thepositive x direction with the phase velocity c:The general solution isu ( x, t ) ft xc g t – xcwhere 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 ) ut ( x, 0 ) 0For 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 isu(x , t) h ( t – xc )for x c t0for x c tA step input:h (t ) 01for t 0for t 0produces a step wave:uc0ctx
Figure – A step wave moving to the right with the velocity c.For a half-sine wave pulse acting on the face:0for t 0for 0 t πωfor π tωh ( t ) sin ω t0the result is a half-sine wave moving into the medium:uc0c(t -ctπωx)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 faceu (0, t ) h ( t )while the other face is stress-freeu ( L, t ) 0the 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 - c2 L(3) Then for the boundary condition at x 0, use a second image traveling to theright:
h t – x c2 LThe 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:x–2Lu ( x , t ) h t – xc – h t c h t – x c2 L – h t x - c4 L .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:uc0(c t-ctπωLxc)Image pulse(a) Before pulse arrives at left boundary (x L), the solution is the same as for a semiinfinite medium.uTotalc0xc(b) After the pulse front arrives at the left boundary, it is partially canceled by the left–moving image pulse.uc0cLx
(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.ucLcx(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 timeu x, t Re e iω tf ( x )f″ ω f 0c22So an asymptotic expansion, often called WKB, valid for sufficiently high frequencies,can be readily obtained in the form:f e –iωζ ( x ) f 0 1 f 1 – 1 f 2 . . .iωω2Substituting into the equation and equating the coefficient of each power of ω to zerogives the sequence of equations:2′(ζ ) 1c2′″2 ζ f 0′ ζ f 0 0′″2 ζ f 1′ ζ f 1 f 0″
.The first equation gives the argument of the exponent, i.e. the phase function:xζ 0dxcwhile the second equation gives the amplitude function:′ –f0 A ( ζ )1/2 c ( x ) 1/2c (0 )The third equation gives the correction to the amplitude function:xf 1 1 f 0 c (0 )2f 0 f 0″ d x0Vibration, one-term approximation – For the vibration problem f 0 at x 0,L, use the expansion in the form:f A sinωζ f 0 – f 2 . cosωζω2 B cosωζ f 0 – f 2 . – sin ω ζω2f 1 – - f 3 .ω ω3f 1 – f 3 .ω ω3The one-term approximation isf A f 0 sinωζ B f 0 cosωζwhich has the simple solutionB 0ω ζ ( L ) n π for n 1, 2, 3, .Vibration, two-term approximation – To find the next approximation to thenatural frequencies takeωζ( L ) n π εf A f 0 sinω ζ f 1 cos ωζ B f 0 cosωζωTo satisfy the boundary conditions f 0 at x 0, the constants are:
f1ω f0B – 0at x 0Then at the other boundary x L, the condition is:f ( L ) cosnπ A sinε f 0 cosε f 1ωwhich gives the correction term:tan ε ε – f 1ω f0Thus, the natural frequencies are atω nπL0L– 12nπdxcc 1/2( c 1/2 )″dx O ( π n )– 30The maximums of f are at:′′ω ζ cosωζ f 0 f 0′ sinωζ – f 1 ζ sinωζ 0cotωζ ′f 1 ζ – f 0′′ωζ f 0which gives the envelope of the modes′( f )max f 0 1 22(ζ f 1 ) – ( f 0′ )′2 O ω–42 (ωζ f 0 )Thus, the envelope is essentially the same for all modes with ω sufficiently large. Forexample, take the distribution of sound speed:c c0 e κ xwhich gives the result for the natural frequencies:
( 1–e –κL ) ( 1 – e κL )ω κnπ1 O 1co 1 – e –κL228n πn 4π 4So 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 κ x/ 21 ( 1 – e –κ x )2 (e κ x – 1 )2– e 2κx O 1168n 2π 2n 4π 4
2. Beam with and without and elastic foundationIn 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 ingeneral) 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: 4 w 2 w ρA kw 0 x 4 t 2in which EI is the bending stiffness, ρA is the mass per unit length, and k is the elasticfoundation stiffness.EIWave 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:ωt–nxw x, t e iwhere ω is the frequency and n is the wave number, which is the reciprocal of the spatialwave length λ:λ 2nπSubstitution into the equation, with the foundation stiffness k 0, gives the frequencywave number relation:EI n 4 – ρ A ω 2 0The 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 v phase ωntFor the "wave equation" this is just the propagation velocity c. For the present beamproblem, the phase velocity is:v phase ω EIn ρA1/2n EIρA3/4 1/2ω
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 particulardeformation shape which can be described at one instant of time by a superposition ofwaves will not look the same at a later time. That is, the shape disperses. There is not asingle velocity which characterizes the behavior.Now consider a sum of two waves of equal amplitude:w x, t sin ω1 t – n1 x sin ω2 t – n2 xUsing the trigonometric identities, this can be rewritten into the form:w x, t 2 sin ω1 ω2 t – n1 n2 x22cosω1 –ω2 n1 –n2t–x22which shows that the two waves con be considered as one wave with the averagedfrequency and wave number, with a modulating envelope which moves with thedifferences. When the frequencies approach each other the phase velocity of theenvelope becomes what is called the group velocity:v envelope Limn 1 n 2ω2 – ω1 d ω v groupn2 –n1dnThis turns out to be the velocity of energy propagation of the wave, and so is veryimportant in transient wave response, probably more important than the phase velocity.Thus for a general medium, phase velocity depends on wave number. Waves ofdifferent wavelength propagate at different velocities. Pulse shape will distort withpropagation distance as indicated in the sketch.
t 0xt t1 0xt t2 t 1xFigure – An indication of dispersion. At each different time the spatial distribution isdifferent.From the general viewpoint, the "wave" equation is a rather specialGenerally, the energy propagates with the group velocity:case governing waves which are nondispersive.v g dωdnwhich for beam is twice the phase velocity:EI n 4 – ρ A ω 2 0EIvp ωn ρA nEI n 2 v pvg d ω 2dnρAThus 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 0for t 0sin ωtfor t 0
The resulting transient solution can be interpreted as consisting primarily of the steadystate 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:Vt tp1 0x"Front" moves with v2gvpxSteady-State"Front"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 A ω 2 EI n 4 kwhich gives the phase velocity:ω nE I n2 kρAρ A n2The new feature due to a non zero foundation modulus k is that the frequency has a nonzero value ω0 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.
0nThe value of the cutoff frequency is:kρAω0 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 ω0 will cause a"local" response, while a frequency greater than ω0 causes the generation of propagatingwaves.P 0 cos ωtssssssssssssssssssssssssssssssssssω ω0quasi. - staticω ω 0propagatingwavesAnother aspect of the effect of the foundation is seen in the following sketch of the phaseand group velocities:
Vp ωndωdnVVg gvpminphasevelocitynFigure – Phase and group velocities for the Euler-Bernoulli beam on an elasticfoundation. At the minimum of the phase velocity, the phase and group velocities areequal.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 isexact; 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 foundationBending stiffnessEIFoundation stiffnesskAxial bar stiffnessEACylindrical shellBending stiffnessEtc2 Et3/12(1-ν2)Circumferential stiffnessEt/r2Axial shell stiffnessEπrtThe dispersion relation for the cylindrical shell is therefore:or solving for the frequency:ρ t ω 2 E t c 2n 4 E tr2
ω 2 E 1 r 2c 2n 4ρ r2Thus the cut-off frequency is the ring breathing resonance:ω02 Eρ r2The wavelength is:Λ 2nπwhile the decay distance for static self-equilibrating edge loads is:δ π 2rcThus the frequency-wave number relation can be rewritten in the form:ω 2 ω02 1 2δΛ4Therefore, wavelengths of vibration or wave propagation which are long in comparisonwith the static decay distance will all have a frequency close to the ring breathingfrequency. 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 willthe frequencies be different. This causes two difficulties for direct numerical methods.First the standard modal analysis is most effective when the eigenfrequencies are distinctand not closely bunched. Secondly, a fine mesh must be used on the shell for finitedifference and finite element methods to capture the significant bending waves with thewavelength O(δ).
ωplatebarBeam on elasticfoundationcylindrical shellnFigure – 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 becomesthe "plate" mode for high frequencies.
Ring, statics - The ring is important in its own right and serves to introduce theeffect of curvature. The circumferential angle is θ and the tangential and normalcomponents of displacement are uθ and w, as shown in the figure, while the in-planerotation of the ring reference line is denoted by χθ .wχθθuθFigure – 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 ofthe reference line.The rotation and strain of the ring are:χθ εθ wr θ uθr θ uθr wrFor a rigid body translation both the rotation and strain are zero, as they should be. Arigid body translation in the x-direction is:so the components are:v w er u θeθ C exw C cosθuθ – C sinθ
yeθereyθexxFigure – Unit vectors in the fixed x – and y – directions and in the r – and θ – directions.Also for a rigid body rotation:uθ Cw 0which gives a constant χθ. The curvature change measure is the derivative of the rotation,which gives the bending moment:κθ χθr θMθ EI κθNote that this is not the actual change in the curvature of the ring. Specifically, the effectof a uniform expansion is excluded. The external force per unit length of the ringreference line has the components qr and qθ.yqθqrθxFigure – 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:
2πΠ u, w 2π1 E I κ 2 E A ε 2 – qrw – q u r dθθ θθθ2Vd θ 00in which V is the potential energy density, which in terms of uθ and w is:V 1 EI2 r4( – w′′ uθ′ )2 12 E 2A ( uθ′ w )2 – qrw – qθ uθrrThe Euler-Lagrange equations, obtained by consideringvariations with respect to w and uθ , are the differential equations for the problem:in which derivatives with respect to θ are denoted by primes. 2 θ V2 w′′–which are:EIr4EIr4– V θ w′ V θ uθ′ V uθ V w 0 0( w′′′′ – uθ′′′ ) E A2 ( w uθ′ ) qrr( w′′′ – uθ′′ ) – EA2 ( w′ uθ′′ ) qθrA 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 aneven function in θ are:qr ( θ) qθ ( θqr n cos nθn 0, 1, . ) qθ n sin nθn 1, .with the corresponding expansions for the displacement components:w(θ ) uθ ( θw n cos nθn 0, 1, . ) uθ n sin nθn 1, .When the series is substituted into the differential equations or into the potential energy, all coupling terms disappear. Thus each harmonic can be treatedseparately, and one obtains the algebraic relation between the Fourier coefficients:E I n4 E AE I n 3 EA nr4r2r4r2E I n 3 E A n E I n 2 E A n2r4r2r4r2wnuθ nq qr nθnThis 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:EAr200qr 0w0 quθ 0θ00It follows that:qθ 0 0which 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 axisymmetricradial expansion amplitude due to the uniform radial load is:2w 0 r qr 0EAFor the harmonic n 1, the equations are:E I EAr4 r21 11 1qw1 qr 1uθ1θ1Thus for a solution:qr 1 qθ 1which is the condition that the loading is self-equilibrating. The solution gives the sum of the displacement components:EI EAr4r2w 1 uθ 1 qr 1 qθ 1When w1 uθ1 0, we have the rigid body translation previously discussed.
Resultantyπqqr1XθXyqrResultantπqθ1(a) Load component qr1 0 produces(b) Load component qθ1 0 producesresultant force in positive x-direction.resultant force in negative x-direction.Figure – Distribution of loading for components with n 1. The distribution is selfequilibrating when qr1 qθ1.Note that the stiffness factor for the harmonic n 1 is about the same as for theaxisymmetric harmonic n 0, since2EI EA EA 1 rgr4 r2r2r2in which rg 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:qr n E I n 2 E A n 2 – qθ n E I n 3 E A nr4r2r4r2wn Det–qr n E I n 3 E A n qθ n E I n 4 E Ar4r2r4r2un Detin which, after many terms cancel, the determinant reduces to:Det E A E I n 2 n 2 – 1r2 r42Thus for the thin ring, for which rg r, the approximate solution is:
wn qr n n 2 – qθ n nn2 n2 – 1 E Ir4wnuθ n – nThe tangential stiffness EA cancels out and only the bending stiffness EI remains. Thesedisplacements for n 2 are therefore of the order of magnitude (r/rg)2 larger than those forn 0, 1. The strain computed from these displacement components is:εθ n 0Thus 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 rrg2If the assumption of inextensional deformation is assumed from the beginning, thepotential energy density is simply:Π 2π0V dθ π r n 2, 3, .1 E I ( w n )2 ( n 2 – 1 )2 –2 r4( qr n – 1n qθ n ) w nSetting 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 theEA terms cancel in both the numerator and denominator is thermal stress. For heating ofthe ring, the potential energy is:V 1 E I κ 2 1 E A ( εθ – α T )2 – qrw – qθ uθ r22where α 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 qr 1 E A α Tr1qθ qθ – r E A d α TdθThe preceding exact results for the Fourier harmonic coefficients of the displacementcomponents yield the following values for thermal heating:
w n – r α Tn for n 1n 2– 1uθ n n r α Tn for n 1n 2– 1w 1 uθ 1 r α T1 for n 1r21 gr2So, 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:σdirect n Fn E ( εθ n – α Tn ) 0 for n 1Arg 2r E α T1for n 1rg 21 r( )( )while the bending stress is:σbending n z Mn z E κθ n 0 for n 1Izr E α T1for n 1rg 21 r( )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 θ , t w n cos nθ cos ωtuθ θ , t uθ n sin nθ cos ωtConsequently, the equations are the same as in the static case, but with inertia termsadded to the load components:qr qr ρ A ω 2 wqθ n qθ n ρ A ω 2 uθThe "dynamic stiffness matrix" for a each Fourier harmonic is:
EI n4 EA –ρA ω2r4r2EI n3 EA nr4r2EI n3 EA nr4r2EI EA n2 –ρA ω2r4r2wnuθ n qr nqθ nFor the harmonic n 0 this reduces to:E A – ρA ω 2 00r2which defines the ring breathing frequency:E cbarrρω 0 r1Thus the period T0 of the breathing frequency is the time required for a wave traveling atthe bar velocity to transverse the circumference of the ring:T0 2c π rbarFor the harmonic n 1, the result is:α–λααα –λ2 α–λ2– α2 – 2 α λ λ 0in whichλ ρ A ωα E I E Ar4r22Thus the resonant frequency is given by:λ 2 αthat is:ω 1 r22E 1 rgρr212 ω0r22 1 gr212 ω0 2We 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 axisymmetricresonance n 0.Just as for the static loading
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.
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in stringer-stiffened shells because the postbuckling zone slope becomes less. As a result more instability is observed(see Figs. 6 and 7). Fig. 4. Buckling loads of the ring- stiffened and stringer-stiffened cylindrical shells with different width Fig. 5. Buckling and postbuckling of the ring- stiffened cylindrical shells with different width
Shells and Shell Theory A thin-walled cylindrical tank has high bending (flexural) stresses at the base. Use a finer mesh where there are discontinuities or abrupt changes in the structure. MAE456 Finite Element Analysis 20 Shells and Shell Theory For a cylindrical shell of radius R and thickness t, the localized bending dies out
Lecture 21 - Plates and shells Prof. K.J. Bathe MIT OpenCourseWare Timoshenko beam theory, and Reissner-Mindlin plate theory For plates, and shells, w, β x, and β y as independent variables. w displacement of mid-surface, w(x,y) A area of mid-surface p load per unit ar
results for fracture and fatigue of cracked plates under bending and out-of-plane shearing loads. 2 Crack Tip Fields Although most of the applications that motivate this work involve shells, the basic crack tip stress flelds will be described in terms of plate theory. Near the crack tip, the stress distribution in shells is the same as that .
The theory of Cosserat shells contains as a special case the linear theory of Cosserat plates. This theory is mostly formulated with the help of the introduction of one deformable director [124,125]. A variant with various directors is discussed, for example, in [251]. Applications of the Cosserat surface theory to sandwich plates are given in .
In the sequel we overview the most important details of the theory of plates and shells. 14.2 The basic equations of Kirchhoff plate theory The Kirchhoff plate theory is often called the theory of thin plates. We note that if the plate is relatively thick then the transverse shear deformation can be considered too. The rele-
A Course on Rough Paths With an introduction to regularity structures June 2014 Errata (last update: April 2015) Springer. To Waltraud and Rudolf Friz and To Xue-Mei. Preface Since its original development in the mid-nineties by Terry Lyons, culminating in the landmark paper [Lyo98], the theory of rough paths has grown into a mature and widely applicable mathematical theory, and there are by .
