Harmonic Standing-Wave Excitations of Simply-Supported Thick-Walled Hollow Elastic Circular Cylinders: Exact 3D Linear Elastodynamic Response
aa r X i v : . [ phy s i c s . c l a ss - ph ] J un Harmonic Standing-Wave Excitations of Simply-SupportedThick-Walled Hollow Elastic Circular Cylinders:Exact 3D Linear Elastodynamic Response
Jamal Sakhr and Blaine A. Chronik
Department of Physics and Astronomy,The University of Western Ontario,London, Ontario, Canada N6A 3K7 (Dated: June 26, 2020)
Abstract
The forced-vibration response of a simply-supported isotropic thick-walled hollow elastic circularcylinder subjected to two-dimensional harmonic standing-wave excitations on its curved surfacesis studied within the framework of linear elastodynamics. Exact semi-analytical solutions for thesteady-state displacement field of the cylinder are constructed using recently-published parametricsolutions to the Navier-Lam´e equation. Formal application of the standing-wave boundary condi-tions generates three parameter-dependent 6 × Keywords: thick-walled hollow elastic cylinders; simply-supported thick cylindrical shells; harmonicstanding-wave boundary stresses; forced vibration; linear elastodynamic response
Typeset by REVTEX 1 . INTRODUCTION
The vibration of an isotropic thick-walled hollow elastic circular cylinder is one of theclassical applied problems of elastodynamics and has been of longstanding general interestto applied mathematicians, acousticians, engineers, and physicists [1–4]. There is a vastliterature on the free vibration of finite-length isotropic hollow elastic circular cylinders,and while much of the fundamental work on the subject was carried out prior to the 1980s,there has, over the last three decades, been a steady stream of publications devoted todeveloping and testing various methodological approaches to obtaining natural frequenciesand mode shapes under a variety of end conditions (see, for example, Refs. [5–14] andreferences therein). The literature on the forced vibration of hollow elastic cylinders (finite-length, isotropic, circular, or otherwise) is much smaller in comparison. Forced-vibrationanalyses of thick-walled elastic cylinders based on the exact three-dimensional (3D) theoryof linear elasticity, in particular, remain scarce. In the context of forced-vibration analysesbased on the Navier-Lam´e equation of motion, the most notable and general work is due toEbenezer and co-workers [15] (ERP), who devised an exact series method to determine thesteady-state vibration response of a finite-length isotropic hollow elastic circular cylindersubjected to arbitrary axisymmetric excitations on its surfaces.As pointed out in many reviews (see, for example, Ref. [3]), many fundamental forced-vibration problems involving hollow elastic cylinders have not yet been studied or solvedusing the exact 3D theory of linear elasticity. A useful and analytically tractable modelproblem that has been surprisingly overlooked is the steady-state vibration response problemfor a simply-supported isotropic thick-walled hollow elastic circular cylinder subjected toarbitrary excitations on its curved surfaces. For a simply-supported cylinder, arbitrary asymmetric excitations on the curved surfaces can be naturally expressed as superpositionsof two-dimensional (2D) harmonic standing waves in the circumferential and axial directions.Thus, the precursor is to consider individual 2D harmonic standing-wave excitations on thecurved surfaces. This latter problem is theoretically significant in its own right since it is oneof the few model problems for which the effects of individual harmonic excitations on thecurved surfaces of the cylinder can be isolated and studied without having to incorporatenon-trivial corrections in order to simultaneously satisfy the end conditions. In general, theexact nature of these effects will be obscured by other excitations needed to generate the2esired end conditions. The free-vibration analog of the proposed problem, that is, the free-vibration problem fora simply-supported isotropic (thick-walled) hollow elastic circular cylinder, is an importantbenchmark problem in many numerical free-vibration studies (see, for example, Ref. [13]and references therein). Explicit analytical formulations and mathematical analyses of thisproblem are however not easy to find in the literature. Weingarten and Reismann [16] appliedthe method of eigenfunction expansions to this problem and obtained an implicit solutionin 1974. The free-vibration analog can also be extracted as a special case of the largelyoverlooked work of Prasad and Jain [17], who, already in the mid 1960s, considered theproblem of free harmonic standing waves in a simply-supported transversely isotropic hollowelastic circular cylinder. In the two aforementioned works, the authors did not actuallysolve their obtained frequency equations, and thus, did not explicitly obtain any naturalfrequencies or mode shapes. These have however since been obtained both indirectly andthrough the use of specialized methods tailored to solving the general free-vibration problemfor an isotropic hollow elastic circular cylinder [6, 8, 13, 14].Although hitherto unstudied, the problem of interest in this paper is not without prece-dent. Hamidzadeh et al. (see Ref. [3] and references therein) considered the problem of de-termining the resonant frequencies of an infinitely-long isotropic thick-walled elastic circularcylinder subjected to harmonic boundary stresses. Using a “frequency sweeping” procedure(see Ref. [3] for details), they calculated the resonant frequencies and subsequently com-pared them to the natural frequencies given in Ref. [18]; dissimilarities were observed forshort cylinders and fundamental resonant modes. The boundary-value problem consideredby Hamidzadeh et al. is the infinite-cylinder analog of the boundary-value problem that weseek to study in this paper.Before entering into details, it is useful to give a brief overview of the paper. In Section II, Incidentally, a corollary of ERP’s work [15] is that arbitrary axisymmetric excitations on the curvedsurfaces of a finite-length hollow elastic cylinder can be mathematically expressed as (finite or infinite)superpositions of one-dimensional harmonic standing waves in the axial direction. Curiously, the effect ofthe individual harmonics on the steady-state vibration response was not considered in Ref. [15]. It is a well-known and often-cited fact that the characteristics of free harmonic standing waves in a finite-length simply-supported hollow elastic cylinder are formally equivalent to those of free harmonic traveling waves in a corresponding cylinder of infinite length, the latter of which are well-studied (see, for example,Ref. [18]). Numerical solutions to the free standing-wave problem are thus readily available as a corollary(see Ref. [1] for further discussions).
3e define the boundary-value problem of interest. In Sections IV-VI, we construct, exploitingcertain known solutions to the Navier-Lam´e equation [19] (Section III), an exact semi-analytical 3D elastodynamic solution to the problem. The method of solution is direct anddemonstrates a general approach that can be applied to solve other similar forced-vibrationproblems involving elastic cylinders. The solution itself, although exact and given in closedform, involves six constants whose values are not given in explicit analytical form. It isin this sense that the obtained solution is a “semi-analytical” solution. Some numericalexamples are subsequently given in Sections VIII-IX, wherein the obtained solution is usedto study the steady-state frequency response in some example excitation cases. In each case,consistency with published natural frequency data is observed. More detailed conclusionsbased on our numerical investigations are summarized in Section XI.
II. MATHEMATICAL DEFINITION OF THE PROBLEM
Consider a simply-supported isotropic hollow elastic circular cylinder of length L andinner and outer radii R and R , respectively. The geometrical parameters { L, R , R } are allfinite, but otherwise arbitrary. Suppose the cylinder is subjected to time-harmonic stresseson its curved surfaces and furthermore that these stresses are spatially non-uniform and aresuch that the circumferential and longitudinal variations are also harmonic. In this paper,we shall work in the circular cylindrical coordinate system wherein all physical quantitiesdepend on the spatial coordinates ( r, θ, z ), which denote the radial, circumferential, andlongitudinal coordinates, respectively, and on the time t . Using this notation, the boundarystresses on the curved surfaces are: σ rr ( R , θ, z, t ) σ rr ( R , θ, z, t ) = AD cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (1a) σ rθ ( R , θ, z, t ) σ rθ ( R , θ, z, t ) = BE sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (1b) σ rz ( R , θ, z, t ) σ rz ( R , θ, z, t ) = CF cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (1c)4here σ rr ( r, θ, z, t ) is a normal component of stress, σ rθ ( r, θ, z, t ) and σ rz ( r, θ, z, t ) are shearcomponents of stress, {A , B , C , D , E , F } are prescribed constant stresses (each having unitsof pressure), k and m are prescribed non-negative integers, and ω > λ + 2 µ ) ∇ ( ∇ · u ) − µ ∇ × ( ∇ × u ) + b = ρ ∂ u ∂t , (2)where u ≡ u ( r, θ, z, t ) is the displacement field, λ > µ > , and ρ > b = ).The radial, circumferential, and longitudinal components of u shall here be denoted by u r ( r, θ, z, t ) , u θ ( r, θ, z, t ), and u z ( r, θ, z, t ), respectively.Although we have stated that the cylinder is simply supported, we have not yet specifiedthe boundary conditions at the flat ends of the cylinder, which are situated at z = 0and z = L . The classical simply-supported (SS) boundary conditions for the stress anddisplacement at the flat ends of the cylinder are: u r ( r, θ, , t ) = u r ( r, θ, L, t ) = 0 , (3a) u θ ( r, θ, , t ) = u θ ( r, θ, L, t ) = 0 , (3b) σ zz ( r, θ, , t ) = σ zz ( r, θ, L, t ) = 0 , (3c)where σ zz ( r, θ, z, t ) is the normal component of stress along the axis of the cylinder.Conditions (1a)-(1c) must be satisfied for all θ ∈ [0 , π ], z ∈ (0 , L ), and arbitrary t .Conditions (3a)-(3c) must be satisfied for all r ∈ [ R , R ], θ ∈ [0 , π ], and arbitrary t . Note Note that the first Lam´e constant λ need not be positive, but we have assumed it to be so for the purposesof this paper. t = t , and thus the problem as defined is not an initial-boundary-valueproblem. Note that, for forced motion, at least one of {A , B , C , D , E , F } must be non-zerowhen m = 0. If m = 0, then at least one of {A , C , D , F } is required to be non-zero inboundary conditions (1).For future reference, we cite here the cylindrical stress-displacement relations from thelinear theory of elasticity [3]: σ rr = ( λ + 2 µ ) ∂u r ∂r + λr (cid:18) ∂u θ ∂θ + u r (cid:19) + λ ∂u z ∂z , (4a) σ θθ = λ ∂u r ∂r + ( λ + 2 µ ) r (cid:18) ∂u θ ∂θ + u r (cid:19) + λ ∂u z ∂z , (4b) σ zz = λ ∂u r ∂r + λr (cid:18) ∂u θ ∂θ + u r (cid:19) + ( λ + 2 µ ) ∂u z ∂z , (4c) σ rθ = µ (cid:18) r ∂u r ∂θ + ∂u θ ∂r − u θ r (cid:19) = σ θr , (4d) σ rz = µ (cid:18) ∂u r ∂z + ∂u z ∂r (cid:19) = σ zr , (4e) σ θz = µ (cid:18) ∂u θ ∂z + 1 r ∂u z ∂θ (cid:19) = σ zθ . (4f)Relations (4), which provide the general mathematical connection between the componentsof the displacement and stress fields, will be used extensively in solving the above-definedboundary-value problem.Since the excitations are time-harmonic, relations (4) and solution uniqueness togetherimply that the response of the cylinder must necessarily be so as well. In other words, thedisplacement field u ( r, θ, z, t ) = e u ( r, θ, z ) sin( ωt ), where e u ( r, θ, z ) denotes the stationary ortime-independent part of the displacement field. It is the latter object that we ultimatelyseek to determine and to then study. 6 II. SOME PARAMETRIC SOLUTIONS TO THE NAVIER-LAM´E EQUATION
In the absence of body forces, the following parametric solutions to Eq. (2) can be ob-tained using a Buchwald decomposition of the displacement field (see Ref. [19] for details): u r = X s =1 a s J ′ n ( α s r ) I ′ n ( α s r ) + b s Y ′ n ( α s r ) K ′ n ( α s r ) h c s cos( nθ ) + d s sin( nθ ) i φ z ( z ) φ t ( t )+ nr a J n ( α r ) I n ( α r ) + b Y n ( α r ) K n ( α r ) h − c sin( nθ ) + d cos( nθ ) i χ z ( z ) χ t ( t ) , (5) u θ = nr X s =1 a s J n ( α s r ) I n ( α s r ) + b s Y n ( α s r ) K n ( α s r ) h − c s sin( nθ ) + d s cos( nθ ) i φ z ( z ) φ t ( t ) − a J ′ n ( α r ) I ′ n ( α r ) + b Y ′ n ( α r ) K ′ n ( α r ) h c cos( nθ ) + d sin( nθ ) i χ z ( z ) χ t ( t ) , (6)and u z = X s =1 γ s a s J n ( α s r ) I n ( α s r ) + b s Y n ( α s r ) K n ( α s r ) h c s cos( nθ ) + d s sin( nθ ) i d ψ z ( z )d z ψ t ( t ) , (7)where n is a non-negative integer and { a , a , a , b , b , b , c , c , c , d , d , d } are arbitraryconstants. The constituents of Eqs. (5)-(7) are as follows: (i) The constants α and α in the arguments of the Bessel functions are given by α = s (cid:12)(cid:12)(cid:12)(cid:12) κ − ρτ ( λ + 2 µ ) (cid:12)(cid:12)(cid:12)(cid:12) , α = s (cid:12)(cid:12)(cid:12)(cid:12) κ − ρτµ (cid:12)(cid:12)(cid:12)(cid:12) , (8)where κ ∈ R \{ } and τ ∈ R \{ } are free parameters. (ii) The correct linear combination of Bessel functions is determined by the relative valuesof the parameters { λ, µ, ρ, κ, τ } as given in Table I. (iii) In Eqs. (5)-(6), primes denote differentiation with respect to the radial coordinate r . (iv) The functions φ z ( z ), φ t ( t ), ψ z ( z ), ψ t ( t ), χ z ( z ), and χ t ( t ) are given by φ z ( z ) = ψ z ( z ) = E cos (cid:16)p | κ | z (cid:17) + F sin (cid:16)p | κ | z (cid:17) if κ < E exp ( −√ κz ) + F exp ( √ κz ) if κ > , (9)7 inear Combination s = 1 term s = 2 term { J n ( α s r ) , Y n ( α s r ) } κ > ρτ ( λ + 2 µ ) κ > ρτµ { I n ( α s r ) , K n ( α s r ) } κ < ρτ ( λ + 2 µ ) κ < ρτµ TABLE I: Conditions on the radial part of each term in Eqs. (5)-(7). φ t ( t ) = ψ t ( t ) = G cos (cid:16)p | τ | t (cid:17) + H sin (cid:16)p | τ | t (cid:17) if τ < G exp ( −√ τ t ) + H exp ( √ τ t ) if τ > , (10) χ z ( z ) = e E cos (cid:16)p | κ | z (cid:17) + e F sin (cid:16)p | κ | z (cid:17) if κ < e E exp ( −√ κz ) + e F exp ( √ κz ) if κ > , (11) χ t ( t ) = e G cos (cid:16)p | τ | t (cid:17) + e H sin (cid:16)p | τ | t (cid:17) if τ < e G exp ( −√ τ t ) + e H exp ( √ τ t ) if τ > , (12)where n E, F, G, H, e E, e F , e G, e H o are arbitrary constants. (v) The constant γ s in Eq. (7) is given by γ s = s = 1 κ (cid:16) κ − ρτµ (cid:17) if s = 2 . (13)Note that Eqs. (5)-(7) are valid so long as ( λ + 2 µ ) κ = ρτ (i.e., α = 0) and µκ = ρτ (i.e., α = 0); otherwise the radial parts must be modified as discussed in Ref. [19]. In thefollowing, these conditions will be satisfied, by construction. IV. GENERAL FORM OF THE DISPLACEMENT FIELDA. The General Case k = 0 When k = 0, general solutions suited to the boundary-value problem defined in Section IIcan be easily constructed from the parametric solutions given in Section III by identifying one8r a combination of the physical parameters { L, R , R , k, ω } with the (free) mathematicalparameters κ and τ . Let κ = − (cid:0) kπL (cid:1) and τ = − ω , and then choose particular solutionsdefined by taking E = 0 in Eq. (9), e E = 0 in Eq. (11), G = 0 in Eq. (10), e G = 0 in Eq. (12),and n = m in Eqs. (5)-(7). Noting the forms of Eqs. (5)-(7) and comparing (1a) with (4a),(1b) with (4d), and (1c) with (4e), we may immediately deduce that the axial and temporalparts of the displacement components are given by φ z ( z ) = ψ z ( z ) = F sin (cid:18) kπL z (cid:19) , χ z ( z ) = e F sin (cid:18) kπL z (cid:19) , (14) φ t ( t ) = ψ t ( t ) = H sin( ωt ) , χ t ( t ) = e H sin( ωt ) . (15)By defining a new set of arbitrary constants¯ A s ≡ a s c s F H, ¯ B s ≡ b s c s F H, e A s ≡ a s d s F H, e B s ≡ b s d s F H, ( s = 1 ,
2) (16a)¯ A ≡ a d e F e H, ¯ B ≡ b d e F e H, e A ≡ a c e F e H, e B ≡ b c e F e H, (16b)the following two independent particular solutions may be extracted from Eqs. (5)-(7): u r = X s =1 (cid:20) e A s n (cid:13) o + e B s n (cid:13) o(cid:21) − mr (cid:20) e A n (cid:13) o + e B n (cid:13) o(cid:21)! sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (17a) u θ = mr X s =1 (cid:20) e A s n (cid:13) o + e B s n (cid:13) o(cid:21)! − (cid:20) e A n (cid:13) o + e B n (cid:13) o(cid:21)! cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (17b) u z = (cid:18) kπL (cid:19) X s =1 γ s (cid:20) e A s n (cid:13) o + e B s n (cid:13) o(cid:21)! sin( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (17c)and u r = X s =1 (cid:20) ¯ A s n (cid:13) o + ¯ B s n (cid:13) o(cid:21) + mr (cid:20) ¯ A n (cid:13) o + ¯ B n (cid:13) o(cid:21)! cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (18a) u θ = − mr X s =1 (cid:20) ¯ A s n (cid:13) o + ¯ B s n (cid:13) o(cid:21)! + ¯ A n (cid:13) o + ¯ B n (cid:13) o! sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (18b) u z = (cid:18) kπL (cid:19) X s =1 γ s (cid:20) ¯ A s n (cid:13) o + ¯ B s n (cid:13) o(cid:21)! cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (18c)where n (cid:13) o = J ′ m ( α s r ) = mr J m ( α s r ) − α s J m +1 ( α s r ) I ′ m ( α s r ) = mr I m ( α s r ) + α s I m +1 ( α s r ) , (19a)9 (cid:13) o = Y ′ m ( α s r ) = mr Y m ( α s r ) − α s Y m +1 ( α s r ) K ′ m ( α s r ) = mr K m ( α s r ) − α s K m +1 ( α s r ) , (19b) n (cid:13) o = J m ( α r ) I m ( α r ) , n (cid:13) o = Y m ( α r ) K m ( α r ) , (19c) n (cid:13) o = n (cid:13) o = J m ( α s r ) I m ( α s r ) , n (cid:13) o = n (cid:13) o = Y m ( α s r ) K m ( α s r ) , (19d) n (cid:13) o = J ′ m ( α r ) = mr J m ( α r ) − α J m +1 ( α r ) I ′ m ( α r ) = mr I m ( α r ) + α I m +1 ( α r ) , (19e) n (cid:13) o = Y ′ m ( α r ) = mr Y m ( α r ) − α Y m +1 ( α r ) K ′ m ( α r ) = mr K m ( α r ) − α K m +1 ( α r ) . (19f)Note that particular solutions (17) and (18) automatically satisfy end conditions (3a)and (3b). Note also that (by virtue of (4c)) σ zz ( r, θ, z, t ) = F ( r, θ ) sin (cid:0) kπL z (cid:1) sin( ωt ), wherethe precise form of F ( r, θ ) is not relevant for our purposes, and thus displacements (17) and(18) as well automatically satisfy end conditions (3c). It can be deduced from inspection of(4), (17), and (18), that solution (17) is incompatible with boundary conditions (1a)-(1c),whereas solution (18) is compatible. Solution (18) therefore furnishes the general form of thedisplacement field appropriate to the boundary-value problem defined in Section II. Hence-forth, we switch to a less cumbersome notation by dropping the bars above the arbitraryconstants.The proper choices of Bessel functions in the radial parts of the displacement componentsdepend on the relative values of the material and excitation parameters; three cases can bedistinguished as listed in Table II. In order to provide a complete solution that involves onlyreal-valued Bessel functions, the problem will be solved separately for each of these threecases. Note that there are two degenerate cases not included in Table II: (i) ρω ( λ +2 µ ) = (cid:0) kπL (cid:1) ;and (ii) ρω µ = (cid:0) kπL (cid:1) . These two cases require special treatment and shall not be consideredhere. As a final remark, note that solution (18) is not valid when k = 0.10 ase Parametric Relationship ( k , ω ) Parametric Relationship ( κ , τ )1 ρω ( λ + 2 µ ) < ρω µ < (cid:18) kπL (cid:19) κ < ρτ ( λ + 2 µ ) and κ < ρτµ (cid:18) kπL (cid:19) < ρω ( λ + 2 µ ) < ρω µ κ > ρτ ( λ + 2 µ ) and κ > ρτµ ρω ( λ + 2 µ ) < (cid:18) kπL (cid:19) < ρω µ κ < ρτ ( λ + 2 µ ) and κ > ρτµ TABLE II: Parametric relationships defining three distinct sub-problems. In the second column,the relationship is expressed in terms of the physical excitation parameters k and ω , whereas inthe third column, the relationship is expressed in terms of the mathematical parameters κ and τ . B. The Special Case k = 0 It can be established (employing results from Ref. [19] or otherwise) that u r = 0 , u θ = 0 , u z = h AJ m ( αr ) + BY m ( αr ) i cos( mθ ) sin( ωt ) , (20)where α = p ρω /µ , is a solution to Eq. (2) that is furthermore compatible with boundaryconditions (1a)-(1c) when k = 0. Solution (20) therefore furnishes the general form of thedisplacement field appropriate to the boundary-value problem defined in Section II in thespecial case k = 0. V. ANALYTICS I: GENERAL CASE k = 0 A. Case 1: ρω ( λ + 2 µ ) < ρω µ < (cid:18) kπL (cid:19) In this case, κ < ρτ / ( λ + 2 µ ) and κ < ρτ /µ (c.f., Table II), and thus, according to TableI, linear combinations of { I m ( α s r ) , K m ( α s r ) } (and their derivatives) should be employed inthe radial parts of (18) (i.e., the modified Bessel functions should be chosen from (19)),where the constants α and α , as determined from Eq. (8), are: α = s(cid:18) kπL (cid:19) − ρω ( λ + 2 µ ) , α = s(cid:18) kπL (cid:19) − ρω µ . (21)11he constant γ s in Eq. (18c), as determined from Eq. (13), is given by γ s = s = 11 − h(cid:16) ρω µ (cid:17) / (cid:0) kπL (cid:1) i if s = 2 . (22)Inputting the above ingredients into (18), the displacement components take the form: u r = ( X s =1 (cid:20) A s (cid:18) mr I m ( α s r ) + α s I m +1 ( α s r ) (cid:19) + B s (cid:18) mr K m ( α s r ) − α s K m +1 ( α s r ) (cid:19)(cid:21) + mr h A I m ( α r ) + B K m ( α r ) i) cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (23a) u θ = − ( mr X s =1 h A s I m ( α s r ) + B s K m ( α s r ) i! + A (cid:20) mr I m ( α r ) + α I m +1 ( α r ) (cid:21) + B (cid:20) mr K m ( α r ) − α K m +1 ( α r ) (cid:21)) sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (23b) u z = (cid:18) kπL (cid:19) ( X s =1 γ s h A s I m ( α s r ) + B s K m ( α s r ) i) cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (23c)where the constants α s and γ s are given by Eqs. (21) and (22), respectively.We must now determine the values of the constants { A , A , A , B , B , B } in Eqs. (23a)-(23c) that satisfy boundary conditions (1a)-(1c). Substituting Eqs. (23a)-(23c) intoEqs. (4a), (4d), and (4e), and performing the lengthy algebra yields the stress components: σ rr ( r, θ, z, t ) = 2 µ ( X s =1 A s (cid:20)(cid:18) β s µ + m ( m − r (cid:19) I m ( α s r ) − α s r I m +1 ( α s r ) (cid:21) + A (cid:20) m ( m − r I m ( α r ) + α mr I m +1 ( α r ) (cid:21) + X s =1 B s (cid:20)(cid:18) β s µ + m ( m − r (cid:19) K m ( α s r ) + α s r K m +1 ( α s r ) (cid:21) + B (cid:20) m ( m − r K m ( α r ) − α mr K m +1 ( α r ) (cid:21) ) × cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (24a)where β s = λ " α s − γ s (cid:18) kπL (cid:19) + 2 µα s , s = 1 , rθ ( r, θ, z, t ) = − µ ( X s =1 A s (cid:20) m ( m − r I m ( α s r ) + α s mr I m +1 ( α s r ) (cid:21) + A (cid:20)(cid:18) α m ( m − r (cid:19) I m ( α r ) − α r I m +1 ( α r ) (cid:21) + X s =1 B s (cid:20) m ( m − r K m ( α s r ) − α s mr K m +1 ( α s r ) (cid:21) + B (cid:20)(cid:18) α m ( m − r (cid:19) K m ( α r ) + α r K m +1 ( α r ) (cid:21) ) × sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (25)and σ rz ( r, θ, z, t ) = µ (cid:18) kπL (cid:19) ( X s =1 A s (1 + γ s ) (cid:20) mr I m ( α s r ) + α s I m +1 ( α s r ) (cid:21) + A (cid:20) mr I m ( α r ) (cid:21) + X s =1 B s (1 + γ s ) (cid:20) mr K m ( α s r ) − α s K m +1 ( α s r ) (cid:21) + B (cid:20) mr K m ( α r ) (cid:21)) cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) . (26)Application of the boundary conditions then proceeds by substituting Eqs. (24), (25), and(26) into the LHSs of Eqs. (1a), (1b), and (1c), respectively, and then canceling identicalsinusoidal terms on both sides of the resulting equations. When m = 0, this procedure yieldssix conditions that can be compactly written as the following 6 × A B A B X A X B = S S , (27a)where, using a shorthand notation, the 3 × { A i , B i : i = 1 , } are A i = f m,i − v m +1 ,i g m,i − w m +1 ,i ( m − q m,i + mw m +1 ,i ( m − p m,i + mv m +1 ,i ( m − q m,i + mw m +1 ,i h m,i − w m +1 ,i p m,i + v m +1 ,i ) (1+ γ )( q m,i + w m +1 ,i ) q m,i , (27b) B i = e f m,i + e v m +1 ,i e g m,i + e w m +1 ,i ( m − e q m,i − m e w m +1 ,i ( m − e p m,i − m e v m +1 ,i ( m − e q m,i − m e w m +1 ,i e h m,i + e w m +1 ,i e p m,i − e v m +1 ,i ) (1+ γ )( e q m,i − e w m +1 ,i ) e q m,i , (27c)13nd the 3 × { X A , X B , S , S } are X A = A A A , X B = B B B , S = A − BC , S = D − EF . (27d)The shorthand notation employed in Eqs. (27b)-(27d) is as follows: f m,i e f m,i ≡ (cid:18) β R i µ + m ( m − R i (cid:19) I m ( α R i ) K m ( α R i ) , i = 1 , g m,i e g m,i ≡ (cid:18) β R i µ + m ( m − R i (cid:19) I m ( α R i ) K m ( α R i ) , i = 1 , h m,i e h m,i ≡ (cid:18) α R i m ( m − R i (cid:19) I m ( α R i ) K m ( α R i ) , i = 1 , p m,i e p m,i ≡ mR i I m ( α R i ) K m ( α R i ) , i = 1 , q m,i e q m,i ≡ mR i I m ( α R i ) K m ( α R i ) , i = 1 , v m +1 ,i e v m +1 ,i ≡ α I m +1 ( α R i ) K m +1 ( α R i ) , i = 1 , w m +1 ,i e w m +1 ,i ≡ α I m +1 ( α R i ) K m +1 ( α R i ) , i = 1 , A B C D E F ] ≡ (cid:20) A R µ B R µ C Lkπµ D R µ E R µ F Lkπµ (cid:21) . (27l)14 . Special Case: m = 0 When m = 0, u θ = 0, σ rθ = 0, and boundary conditions (1b) are identically satisfied.Application of boundary conditions (1a) and (1c) yields the 4 × f , − v , g , − w , e f , + e v , e g , + e w , v , (1+ γ ) w , − e v , − (1+ γ ) e w , f , − v , g , − w , e f , + e v , e g , + e w , v , (1+ γ ) w , − e v , − (1+ γ ) e w , A A B B = ACDF , (28)where the matrix elements are the evaluated zero- and first-order Bessel functions obtainedfrom substituting m = 0 in Eqs. (27e), (27f), (27j), and (27k).In this special case, the non-zero components of the displacement field reduce to: u r ( r, z, t ) = ( X s =1 α s h A s I ( α s r ) − B s K ( α s r ) i) sin (cid:18) kπL z (cid:19) sin( ωt ) , (29a) u z ( r, z, t ) = (cid:18) kπL (cid:19) ( X s =1 γ s h A s I ( α s r ) + B s K ( α s r ) i) cos (cid:18) kπL z (cid:19) sin( ωt ) , (29b)where constants α s and γ s are given by Eqs. (21) and (22), respectively, and the constants { A , A , B , B } are those obtained from solving Eq. (28). B. Case 2: (cid:18) kπL (cid:19) < ρω ( λ + 2 µ ) < ρω µ According to Tables I and II, the Bessel functions { J m ( α s r ) , Y m ( α s r ) } (and their deriva-tives) should in this case be employed in the radial parts of (18) (i.e., the unmodified Besselfunctions should be chosen from (19)). The displacement components thus take the form: u r = ( X s =1 (cid:20) A s (cid:18) mr J m ( α s r ) − α s J m +1 ( α s r ) (cid:19) + B s (cid:18) mr Y m ( α s r ) − α s Y m +1 ( α s r ) (cid:19)(cid:21) + mr h A J m ( α r ) + B Y m ( α r ) i) cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (30a) u θ = − ( mr X s =1 h A s J m ( α s r ) + B s Y m ( α s r ) i! + A (cid:20) mr J m ( α r ) − α J m +1 ( α r ) (cid:21) + B (cid:20) mr Y m ( α r ) − α Y m +1 ( α r ) (cid:21)) sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (30b)15 z = (cid:18) kπL (cid:19) ( X s =1 γ s h A s J m ( α s r ) + B s Y m ( α s r ) i) cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (30c)where α = s − (cid:18) kπL (cid:19) + ρω ( λ + 2 µ ) , α = s − (cid:18) kπL (cid:19) + ρω µ , (31)and γ s is again given by Eq. (22).The constants { A , A , A , B , B , B } in Eqs. (30a)-(30c) must as before be chosen soas to satisfy boundary conditions (1). Proceeding as in the previous case, we first obtaingeneral formulas for the radial components of the stress field. Substituting Eqs. (30a)-(30c)into Eqs. (4a), (4d), and (4e), and performing the lengthy algebra yields the required stresscomponents: σ rr ( r, θ, z, t ) = 2 µ ( X s =1 A s (cid:20)(cid:18) − η s µ + m ( m − r (cid:19) J m ( α s r ) + α s r J m +1 ( α s r ) (cid:21) + A (cid:20) m ( m − r J m ( α r ) − α mr J m +1 ( α r ) (cid:21) + X s =1 B s (cid:20)(cid:18) − η s µ + m ( m − r (cid:19) Y m ( α s r ) + α s r Y m +1 ( α s r ) (cid:21) + B (cid:20) m ( m − r Y m ( α r ) − α mr Y m +1 ( α r ) (cid:21) ) × cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (32a)where η s = λ " α s + γ s (cid:18) kπL (cid:19) + 2 µα s , s = 1 , σ rθ ( r, θ, z, t ) = − µ ( X s =1 A s (cid:20) m ( m − r J m ( α s r ) − α s mr J m +1 ( α s r ) (cid:21) + A (cid:20)(cid:18) m ( m − r − α (cid:19) J m ( α r ) + α r J m +1 ( α r ) (cid:21) + X s =1 B s (cid:20) m ( m − r Y m ( α s r ) − α s mr Y m +1 ( α s r ) (cid:21) + B (cid:20)(cid:18) m ( m − r − α (cid:19) Y m ( α r ) + α r Y m +1 ( α r ) (cid:21) ) × sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (33)16nd σ rz ( r, θ, z, t ) = µ (cid:18) kπL (cid:19) ( X s =1 A s (1 + γ s ) (cid:20) mr J m ( α s r ) − α s J m +1 ( α s r ) (cid:21) + A (cid:20) mr J m ( α r ) (cid:21) + X s =1 B s (1 + γ s ) (cid:20) mr Y m ( α s r ) − α s Y m +1 ( α s r ) (cid:21) + B (cid:20) mr Y m ( α r ) (cid:21)) cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) . (34)When m = 0, application of the boundary conditions (1) as described in Section V Ayields six conditions involving the constants { A , A , A , B , B , B } that can again be castin the form (27a), where, in the present case, the 3 × { A i , B i : i = 1 , } are A i = F m,i + V m +1 ,i G m,i + W m +1 ,i ( m − Q m,i − mW m +1 ,i ( m − P m,i − mV m +1 ,i ( m − Q m,i − mW m +1 ,i H m,i + W m +1 ,i P m,i − V m +1 ,i ) (1+ γ )( Q m,i − W m +1 ,i ) Q m,i , (35a) B i = e F m,i + e V m +1 ,i e G m,i + f W m +1 ,i ( m − e Q m,i − m f W m +1 ,i ( m − e P m,i − m e V m +1 ,i ( m − e Q m,i − m f W m +1 ,i e H m,i + f W m +1 ,i (cid:16) e P m,i − e V m +1 ,i (cid:17) (1+ γ ) (cid:16) e Q m,i − f W m +1 ,i (cid:17) e Q m,i , (35b)and the 3 × { X A , X B , S , S } are as given by (27d) and (27l). The shorthandnotation employed in Eqs. (35a) and (35b) is as follows: F m,i e F m,i ≡ (cid:18) − η R i µ + m ( m − R i (cid:19) J m ( α R i ) Y m ( α R i ) , i = 1 , G m,i e G m,i ≡ (cid:18) − η R i µ + m ( m − R i (cid:19) J m ( α R i ) Y m ( α R i ) , i = 1 , H m,i e H m,i ≡ (cid:18) − α R i m ( m − R i (cid:19) J m ( α R i ) Y m ( α R i ) , i = 1 , P m,i e P m,i ≡ mR i J m ( α R i ) Y m ( α R i ) , i = 1 , Q m,i e Q m,i ≡ mR i J m ( α R i ) Y m ( α R i ) , i = 1 , V m +1 ,i e V m +1 ,i ≡ α J m +1 ( α R i ) Y m +1 ( α R i ) , i = 1 , W m +1 ,i f W m +1 ,i ≡ α J m +1 ( α R i ) Y m +1 ( α R i ) , i = 1 , . (35i)
1. Special Case: m = 0 When m = 0, u θ = 0, σ rθ = 0, and boundary conditions (1b) are identically satisfied.Application of boundary conditions (1a) and (1c) yields the 4 × F , + V , G , + W , e F , + e V , e G , + f W , − V , − (1+ γ ) W , − e V , − (1+ γ ) f W , F , + V , G , + W , e F , + e V , e G , + f W , − V , − (1+ γ ) W , − e V , − (1+ γ ) f W , A A B B = ACDF , (36)where the matrix elements are the evaluated zero- and first-order Bessel functions obtainedfrom substituting m = 0 in Eqs. (35c), (35d), (35h), and (35i).In this special case, the non-zero components of the displacement field reduce to: u r ( r, z, t ) = − ( X s =1 α s h A s J ( α s r ) + B s Y ( α s r ) i) sin (cid:18) kπL z (cid:19) sin( ωt ) , (37a) u z ( r, z, t ) = (cid:18) kπL (cid:19) ( X s =1 γ s h A s J ( α s r ) + B s Y ( α s r ) i) cos (cid:18) kπL z (cid:19) sin( ωt ) , (37b)where constants α s and γ s are given by Eqs. (31) and (22), respectively, and the constants { A , A , B , B } are those obtained from solving Eq. (36). C. Case 3: ρω ( λ + 2 µ ) < (cid:18) kπL (cid:19) < ρω µ According to Tables I and II, the s = 1 term in each of the radial parts of Eqs. (18a)-(18c)should employ the modified Bessel functions { I m ( α r ) , K m ( α r ) } (and their derivatives)18hile the s = 2 terms should employ the Bessel functions { J m ( α r ) , Y m ( α r ) } (and theirderivatives). The remaining terms are unmodified from those of Case 2. The displacementcomponents thus take the form: u r = ( A (cid:20) mr I m ( α r ) + α I m +1 ( α r ) (cid:21) + B (cid:20) mr K m ( α r ) − α K m +1 ( α r ) (cid:21) + A (cid:20) mr J m ( α r ) − α J m +1 ( α r ) (cid:21) + B (cid:20) mr Y m ( α r ) − α Y m +1 ( α r ) (cid:21) + mr h A J m ( α r ) + B Y m ( α r ) i) cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (38a) u θ = − ( mr h A I m ( α r ) + B K m ( α r ) + A J m ( α r ) + B Y m ( α r ) i + A (cid:20) mr J m ( α r ) − α J m +1 ( α r ) (cid:21) + B (cid:20) mr Y m ( α r ) − α Y m +1 ( α r ) (cid:21)) × sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (38b) u z = (cid:18) kπL (cid:19) ( γ h A I m ( α r ) + B K m ( α r ) i + γ h A J m ( α r ) + B Y m ( α r ) i) × cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) , (38c)where α = s(cid:18) kπL (cid:19) − ρω ( λ + 2 µ ) , α = s − (cid:18) kπL (cid:19) + ρω µ , (39)and γ s is again given by Eq. (22).The constants { A , A , A , B , B , B } in Eqs. (38a)-(38c) must again be chosen so as tosatisfy boundary conditions (1). Proceeding as usual, we first obtain the pertinent compo-nents of the stress field. Substituting Eqs. (38a)-(38c) into Eqs. (4a), (4d), and (4e), and19erforming the necessary algebra yields the required stress components: σ rr ( r, θ, z, t ) = 2 µ ( A (cid:20)(cid:18) β µ + m ( m − r (cid:19) I m ( α r ) − α r I m +1 ( α r ) (cid:21) + A (cid:20)(cid:18) − η µ + m ( m − r (cid:19) J m ( α r ) + α r J m +1 ( α r ) (cid:21) + A (cid:20) m ( m − r J m ( α r ) − α mr J m +1 ( α r ) (cid:21) + B (cid:20)(cid:18) β µ + m ( m − r (cid:19) K m ( α r ) + α r K m +1 ( α r ) (cid:21) + B (cid:20)(cid:18) − η µ + m ( m − r (cid:19) Y m ( α r ) + α r Y m +1 ( α r ) (cid:21) + B (cid:20) m ( m − r Y m ( α r ) − α mr Y m +1 ( α r ) (cid:21) ) × cos( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (40)where the constants β and η are as given by Eqs. (24b) and (32b), respectively, σ rθ ( r, θ, z, t ) = − µ ( A (cid:20) m ( m − r I m ( α r ) + α mr I m +1 ( α r ) (cid:21) + A (cid:20) m ( m − r J m ( α r ) − α mr J m +1 ( α r ) (cid:21) + A (cid:20)(cid:18) m ( m − r − α (cid:19) J m ( α r ) + α r J m +1 ( α r ) (cid:21) + B (cid:20) m ( m − r K m ( α r ) − α mr K m +1 ( α r ) (cid:21) + B (cid:20) m ( m − r Y m ( α r ) − α mr Y m +1 ( α r ) (cid:21) + B (cid:20)(cid:18) m ( m − r − α (cid:19) Y m ( α r ) + α r Y m +1 ( α r ) (cid:21) ) × sin( mθ ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (41)20nd σ rz ( r, θ, z, t ) = µ (cid:18) kπL (cid:19) ( A (1 + γ ) (cid:20) mr I m ( α r ) + α I m +1 ( α r ) (cid:21) + A (1 + γ ) (cid:20) mr J m ( α r ) − α J m +1 ( α r ) (cid:21) + A (cid:20) mr J m ( α r ) (cid:21) + B (1 + γ ) (cid:20) mr K m ( α r ) − α K m +1 ( α r ) (cid:21) + B (1 + γ ) (cid:20) mr Y m ( α r ) − α Y m +1 ( α r ) (cid:21) + B (cid:20) mr Y m ( α r ) (cid:21)) cos( mθ ) cos (cid:18) kπL z (cid:19) sin( ωt ) . (42)When m = 0, application of the boundary conditions as described in Section V A yieldsa 6 × × { A i , B i : i = 1 , } are A i = f m,i − v m +1 ,i G m,i + W m +1 ,i ( m − Q m,i − mW m +1 ,i ( m − p m,i + mv m +1 ,i ( m − Q m,i − mW m +1 ,i H m,i + W m +1 ,i p m,i + v m +1 ,i ) (1+ γ )( Q m,i − W m +1 ,i ) Q m,i , (43a) B i = e f m,i + e v m +1 ,i e G m,i + f W m +1 ,i ( m − e Q m,i − m f W m +1 ,i ( m − e p m,i − m e v m +1 ,i ( m − e Q m,i − m f W m +1 ,i e H m,i + f W m +1 ,i e p m,i − e v m +1 ,i ) (1+ γ ) (cid:16) e Q m,i − f W m +1 ,i (cid:17) e Q m,i , (43b)and the 3 × { X A , X B , S , S } are as given by (27d) and (27l). The shorthandnotation employed for all matrix elements in Eqs. (43a) and (43b) is as previously definedby Eqs. (27e), (27h), (27j) and Eqs. (35d), (35e), (35g), (35i).
1. Special Case: m = 0 When m = 0, u θ = 0, σ rθ = 0, and boundary conditions (1b) are identically satisfied.Application of boundary conditions (1a) and (1c) yields the 4 × f , − v , G , + W , e f , + e v , e G , + f W , v , − (1+ γ ) W , − e v , − (1+ γ ) f W , f , − v , G , + W , e f , + e v , e G , + f W , v , − (1+ γ ) W , − e v , − (1+ γ ) f W , A A B B = ACDF . (44)21he matrix elements in Eq. (44) are the evaluated zero- and first-order Bessel functionsobtained from substituting m = 0 in Eqs. (27e), (27h), (27j) and Eqs. (35d), (35e), (35g),(35i).In this special case, the non-zero components of the displacement field reduce to: u r = ( A α I ( α r ) − B α K ( α r ) − A α J ( α r ) − B α Y ( α r ) ) sin (cid:18) kπL z (cid:19) sin( ωt ) , (45a) u z = (cid:18) kπL (cid:19) ( γ h A I ( α r )+ B K ( α r ) i + γ h A J ( α r )+ B Y ( α r ) i) cos (cid:18) kπL z (cid:19) sin( ωt ) , (45b)where constants α s and γ s ( s = 1 ,
2) are given by Eqs. (39) and (22), respectively, and theconstants { A , A , B , B } are those obtained from solving Eq. (44). VI. ANALYTICS II: SPECIAL CASE k = 0 As discussed in Section IV B, a general displacement field compatible with the boundary-value problem defined in Section II in the special case k = 0 is given by particular solution(20). We need now only to determine the values of the constants { A, B } in (20) that satisfyboundary conditions (1a)-(1c). Substituting the components of solution (20) into Eqs. (4a),(4d), and (4e) immediately yields the stress components: σ rr ( r, θ, z, t ) = σ rθ ( r, θ, z, t ) = 0 , (46a)and σ rz ( r, θ, t ) = µ ( A (cid:20) mr J m ( αr ) − αJ m +1 ( αr ) (cid:21) + B (cid:20) mr Y m ( αr ) − αY m +1 ( αr ) (cid:21)) cos( mθ ) sin( ωt ) . (46b)Boundary conditions (1a) and (1b) are therefore satisfied identically. Application of bound-ary conditions (1c) yields two conditions that can be compactly written as the following2 × Q m, − W m +1 , e Q m, − f W m +1 , Q m, − W m +1 , e Q m, − f W m +1 , AB = CF , (47a)where the following shorthand notation is employed for the matrix elements of Eq. (47a): Q m,i e Q m,i ≡ mR i J m ( αR i ) Y m ( αR i ) , i = 1 , W m +1 ,i f W m +1 ,i ≡ α J m +1 ( αR i ) Y m +1 ( αR i ) , i = 1 , C ≡ C µ , F ≡ F µ . (47d)The general solution to system (47) is: A = (cid:16) e Q m, − f W m +1 , (cid:17) C − (cid:16) e Q m, − f W m +1 , (cid:17) F (cid:0) Q m, − W m +1 , (cid:1) (cid:16) e Q m, − f W m +1 , (cid:17) − (cid:16) e Q m, − f W m +1 , (cid:17) (cid:0) Q m, − W m +1 , (cid:1) , (48a) B = (cid:0) Q m, − W m +1 , (cid:1) F − (cid:0) Q m, − W m +1 , (cid:1) C (cid:0) Q m, − W m +1 , (cid:1) (cid:16) e Q m, − f W m +1 , (cid:17) − (cid:16) e Q m, − f W m +1 , (cid:17) (cid:0) Q m, − W m +1 , (cid:1) . (48b)When m = 0, (48) reduces to: A = Y ( αR ) F − Y ( αR ) C α h J ( αR ) Y ( αR ) − J ( αR ) Y ( αR ) i , (49a) B = J ( αR ) C − J ( αR ) F α h J ( αR ) Y ( αR ) − J ( αR ) Y ( αR ) i . (49b)Thus, in the special case k = 0, the displacement field is given by (20) with the constants { A, B } given by (48), which reduces to (49) when m = 0. VII. CONSISTENCY WITH THE ERP FIELD EQUATIONS
As an analytical check, we have verified that, in the special m = 0 case, the generalstress and displacement fields obtained from applying our method of solution agree with thegeneral axisymmetric field equations that would be obtained from applying the mathematicalframework of Ebenezer et al. (ERP) [15]. Demonstration of this consistency is somewhatintricate; the details are therefore consigned to Appendix A. Equivalency of our m = 0solution with the axisymmetric solution that would be obtained from the ERP method thendirectly follows from application of the boundary conditions. VIII. EXAMPLE 1
As a numerical example, we examine the steady-state frequency response of a thick-walledsteel cylinder whose geometric and material properties are specified in Table III. As in other23 arameter Numerical ValueLength ( L ) 0 .
300 mOuter radius ( R ) 0 .
100 mInner radius ( R ) 0 .
050 mMass density ( ρ ) 8000 kg/m Young modulus ( E ) 190 GPaPoisson ratio ( ν ) 0 . λ ) 110 GPaSecond Lam´e constant ( µ ) 73 . steady-state frequency-response analyses (see, for example, Ref. [15]), we shall here restrictattention to studying the behavior of the stationary displacement field at a few suitably-chosen representative points in the cylinder as a function of the excitation frequency. To doso, we numerically evaluate the formulas for the displacement field obtained in Section V.The only non-trivial numerical detail is the determination of the (frequency-dependent)solution constants { A s , B s : s = 1 , , } , which we obtain by numerically solving the linearsystems (27), (35), and (43) (or their m = 0 analogs) pointwise for each excitation frequency.An alternative is to use a symbolic algebra package, solve these linear systems symbolically,and then evaluate the results at the excitation frequencies of interest. While it is possibleto obtain exact analytical expressions for each of the solution constants (using a symbolicalgebra package or otherwise), the resulting expressions are too algebraically complicatedfor general use. Note that no numerical solution is required when the longitudinal wavenumber k = 0 since the equivalent solution constants { A, B } in this special case are given,in closed form, by (48).Using the parameter values given in Table III, the components of the displacement fieldwere computed at excitation frequencies that are integer multiples of 1 Hz with lower andupper bounds of 1 Hz and 50 kHz, respectively. In all calculations, the excitation amplitudes Aside from nodal or semi-nodal points, we are free to choose any point in the cylinder as a representativepoint. ode number Circumferential wave number n m = 0 m = 1 m = 2 m = 31 7.698 2.997 5.391 11.5372 10.686 6.515 8.141 13.1173 11.827 7.284 12.154 15.9544 12.838 9.731 12.896 19.0355 16.229 11.486 14.429 19.6416 17.542 13.563 16.625 19.9647 20.398 15.467 17.460 22.2228 24.963 15.828 20.912 23.8059 25.425 17.334 21.290 25.163TABLE IV: Natural frequencies (cid:8) f ( m ) n : n = 1 , . . . , (cid:9) of a simply-supported isotropic (thick-walled)hollow elastic circular cylinder having the geometrical and material properties given in Table III.All values are in units of kHz. The above frequencies were computed using the free-vibrationfrequency data given in Table 6 of Ref. [14]. were set as follows: A = B = C = 10 Pa and D = E = F = − Pa. Given that the cylinderis being forced to vibrate, we expect to observe large displacements (i.e., resonances) whenthe excitation frequency is close to one of the natural frequencies of the simply-supported cylinder. For circumferential wave numbers m = { , , , } , the first nine of these frequen-cies (cid:8) f ( m ) n : n = 1 , . . . , (cid:9) , computed using free-vibration frequency data from Ref. [14], aregiven in Table IV. The absolute value of the (stationary) displacement at the interior point( r, θ, z ) = (( R + R ) / , π/ , L/
7) is shown in Figs. 1-4 for various values of the longitu-dinal wave number k and circumferential wave numbers m = 0, 1, 2, and 3, respectively.For visual reference, the natural frequencies listed in Table IV are marked by ‘ X ’s on thefrequency axis of each subplot.In each case, we observe unmistakable resonances around (a subset of) the natural fre-quencies of the simply-supported cylinder. While the results may at first appear to beparticularly simple, there are several interesting features that should be noted. First, notethat each individual excitation (obtained by specifying a single pair of ( m, k ) values) gen-25 × -6 × -6 A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 × -6 Excitation Frequency (kHz) × -6 k = 5k = 4k = 3k = 1k = 2 FIG. 1: Frequency response of the displacement at the point ( r, θ, z ) = (( R + R ) / , π/ , L/
7) forvarious values of the longitudinal wave number k and circumferential wave number m = 0. Forreference, the natural frequencies listed in Table IV are marked by ‘ X ’s on the frequency axis ofeach subplot. erates a unique series of resonances, as opposed to producing just one resonance. In otherwords, a single harmonic excitation excites a set of resonant modes instead of exciting onlyone resonant mode. Unfortunately, there does not appear to be any mathematical rule forpredicting which resonances will be excited by a particular excitation. More precisely, if(for a given circumferential wave number m ) (cid:8) f ( m ) n : n ∈ Z + (cid:9) denotes the complete naturalfrequency spectrum and n f ( m ) j : j ∈ S ( m, k ) ⊂ Z + o denotes the subset of the natural fre-26 × -6 × -6 A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 Excitation Frequency (kHz) × -6 k = 2k = 4k = 3 k = 1 FIG. 2: Same as Figure 1 except the circumferential wave number m = 1. quency spectrum at which an excitation with wave numbers ( m, k ) generates resonances,then there appears to be no deterministic rule for predicting the set of mode numbers { j : j ∈ S ( m, k ) ⊂ Z + } given the wave numbers ( m, k ) of the excitation. Second, notethat the resonances generated by any single excitation generally have different widths; inother words, the resonant modes excited by a particular excitation generally possess differ-ent decay properties. Practically speaking, this means that the displacement response toany harmonic excitation will have a varying degree of significance in the neighborhoods ofthe associated resonant frequencies n f ( m ) j o . For example, when the circumferential wavenumber m = 1, the response to a standing-wave excitation with longitudinal wave number k = 1 is significant at more frequencies neighboring f (1)4 = 9 .
731 kHz than neighboring27 × -6 × -6 A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 × -6 Excitation Frequency (kHz) × -6 k = 1k = 2k = 3k = 4k = 5 FIG. 3: Same as Figure 1 except the circumferential wave number m = 2. f (1)1 = 2 .
997 kHz or f (1)7 = 15 .
467 kHz.One other noteworthy feature in Figs. 1-4 is the conspicuous absence of resonances in theneighborhoods of certain natural frequencies, in particular, around f (0)8 = 24 .
963 kHz (when m = 0), around f (1)2 = 6 .
515 kHz (when m = 1), around f (2)4 = 12 .
896 kHz (when m = 2),and around f (3)4 = 19 .
035 kHz (when m = 3). As it turns out, when m = 0, resonancesat these frequencies are produced by boundary stresses of type (1) with longitudinal wavenumber k = 0, as shown in Fig. 5. The same figure also shows that, when m = 0, noresonance associated with f (0)8 = 24 .
963 kHz is produced by such an excitation. Thus, when28
Excitation Frequency (kHz) × -6 × -6 A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 × -6 × -6 k = 4k = 3k = 2k = 1k = 5 FIG. 4: Same as Figure 1 except the circumferential wave number m = 3. m = 0, there exist resonant modes at certain frequencies that cannot be excited by harmonicboundary stresses of type (1).In Figs. 1-5, we used common vertical scales in all subplots in order to make it easierto compare the different cases. We should however mention that the amplitudes of theresonances are not all equal, and this is evident when one views the displacement responseoutside the common vertical range shown in the figures. Differences in amplitude not onlyoccur between the different excitation cases; the amplitudes of the resonances generated byeach individual excitation also vary. While it may be obvious to some readers, it is worth29 × -6 m = 0 × -6 m = 1 k = 0 A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 m = 2 Excitation Frequency (kHz) × -6 m = 3 FIG. 5: Frequency response of the displacement at the point ( r, θ, z ) = (( R + R ) / , π/ , L/
7) forvarious values of the circumferential wave number m and longitudinal wave number k = 0. emphasizing that the amplitudes, which inherently depend on the frequency resolution and on where in the cylinder the displacement is evaluated, should not be interpreted asresonance intensities. In the present context of a lossless (i.e., undamped) cylinder, theamplitudes are insignificant since, in theory, the amplitude of any resonance asymptoticallyapproaches infinity as the excitation frequency approaches the associated natural frequency. As previously stated, the components of the displacement field were computed at excitation frequenciesthat are integer multiples of 1 Hz with lower and upper bounds of 1 Hz and 50 kHz, respectively. Whena different frequency discretization is used (e.g., 2 Hz instead of 1 Hz), the numerical amplitudes change.
IX. SUPPLEMENTARY EXAMPLES
As supplementary examples, we study the frequency response of three different cylinders,each possessing the same geometry and Poisson ratio ν = 0 .
300 but differing in their massdensities and Young moduli. The cylinder geometry is fixed as follows: L = 0 .
500 m, R = 0 .
050 m, and R = 0 .
150 m. Thus, the mean radius R ≡ ( R + R ) / .
100 m, thethickness-to-radius ratio h/R = 1 .
00, and the length-to-radius ratio
L/R = 5 .
00. The massdensities and Young moduli of the three cylinders are given in Table V.
Cylinder Material ρ (kg/m ) E (GPa)Cadmium 8650 50Ruthenium 12370 447Rhenium 21020 463TABLE V: Three different cylinder materials having the same Poisson ratio ν = 0 . For each of the three above-defined cylinders, the displacement responses at the point( r, θ, z ) = (( R + R ) / , π/ , L/
7) to six different standing-wave excitations are shown inFig. 6. Note that the displacement responses to the different standing-wave excitations areoverlaid on each subplot with the understanding that they correspond to separate excitationcases. So, it should be understood that, for instance, the orange curve is the responseto an excitation with wave numbers ( m, k ) = (2 , m, k ) = (2 , X ’s on the frequency axis of each subplot. It isinteresting to note that each excitation excites the same resonant modes independent of boththe mass density and stiffness of the cylinder. For example, the fourth mode (correspondingto natural frequency f (2)4 ) is always excited by a standing-wave excitation with longitudinalwave number k = 0, whereas the fifth mode (corresponding to natural frequency f (2)5 )is always excited by a standing-wave excitation with longitudinal wave number k = 1.Although not shown here, the same conclusion is reached when considering other non-zerocircumferential wave numbers (i.e., m = 2). × -6 m = 2 Cadmium A b s o l u t e V a l ue o f t he S t a t i ona r y D i s p l a c e m en t ( m ) × -6 Ruthenium
Excitation Frequency (kHz) × -6 Rhenium k = 0k = 1k = 2k = 3k = 4k = 5
FIG. 6: Responses at the point ( r, θ, z ) = (( R + R ) / , π/ , L/
7) to various standing-wave excita-tions with wave numbers m and k as indicated for three different hollow cylinders. The cylinderspossess identical geometries and Poisson ratios (see text) but differ in their mass densities andYoung moduli (c.f., Table V). Natural frequencies pertinent to each case are marked by ‘ X ’s on thefrequency axis of each subplot. . NUMERICAL EPILOGUE: A COMMENT ON MODE ORTHOGONALITY It is worthwhile to comment in more detail on the subtle and perhaps counterintuitivenature of the result that each harmonic standing-wave excitation excites many resonantmodes. The excitations in the present problem are pure 2D boundary stresses that varyharmonically in the circumferential and axial directions. As such, these boundary stressesare characterized by two parameters: the circumferential wave number m and the axial wavenumber k . The resonant modes of the cylinder, on the other hand, are three-dimensional,and hence cannot be uniquely specified using only the wave numbers m and k . The keyto understanding the preceding numerical results is to recognize that the wave numbers m and k are insufficient for indexing (i.e., uniquely classifying) all the different vibrationmodes of a thick-walled cylinder (a fundamental problem that has been previously discussedin the context of free vibrations in Refs. [21–23]). For any given fixed values of m and k (and assuming that all other parameters are also fixed), there exists a countably-infiniteset M ( m, k ) = {M i ( m, k ) : i ∈ Z + } of physically distinct resonant modes each possessing aunique shape composed of m full (cosine) waves around the circumference of the cylinder and k half (sine) waves along the axis of the cylinder. A resonant mode M i ( m, k ) ∈ M ( m, k )is excited when the excitation frequency f is at (or close to) the mode’s respective resonantfrequency, which, say, is equal to the natural frequency f ( m ) j . It is important to note thatthere is no formal correspondence between the integers i , j , and k , and for no reason shouldthey be expected to possess equal values. If the values of m and k are fixed such that( m, k ) = ( m ∗ , k ∗ ) in excitations (1a)-(1c), then these excitations will necessarily excite allof the modes in M ( m ∗ , k ∗ ) (each at its respective resonant frequency) because all of themodes in M ( m ∗ , k ∗ ) have wave numbers m = m ∗ and k = k ∗ (by definition). Despite thefact that all members of M ( m ∗ , k ∗ ) are characterized by the same circumferential and axialwave numbers, the constituent modes are: (i) excited at different frequencies; (ii) physicallydistinct (i.e., possess unique shapes); and most importantly (iii) linearly decoupled (i.e., noconstituent mode is a linear combination of other constituent modes). In short, all membersof any M ( m, k ) are orthogonal despite their common wave numbers m and k . Hence, theobserved numerical results do not violate mode orthogonality. This is true by virtue of the fact that, for a simply-supported cylinder, there exists a countably-infiniteset of unique free-vibration modes for any given fixed values of m and k [18].
33o give a numerical example, we revisit Example 1 and explicitly compute the (stationary)shapes of the first three resonant modes in M (1 , {A , B , C , D , E , F } ,which (for simplicity) are here set as follows: A = C = D = F = 0 and B = E = 0 .
500 MPa.The results, obtained from use of our exact k = 0 solution (Section V), are shown in Fig. 7.As validation of these results, we note that the shapes of M (1 ,
1) and M (1 ,
1) are fullyconsistent with the corresponding free-vibration mode shapes given in Ref. [14]. To furtherexpose the differences in the shapes of M (1 ,
1) and M (1 , z = L/ i and j (as definedabove) generally differ from the axial wave number k (see Table VI). FIG. 7: (Top Row) Stationary shapes of the first three members of M (1 ,
1) for the cylinderconsidered in Example 1. The respective excitation frequencies are as indicated. (Bottom Row)Aerial views of the shapes of M (1 ,
1) and M (1 ,
1) at z = L/
2. For reference, the center point ofthe undeformed free cylinder is (in each case) marked by a black dot. Specifically, the first and fourth modes (respectively) shown in Fig. 4 of Ref. [14] in the “S-S n=1” case.Note that the circumferential wave number is denoted by the letter n in Ref. [14]. ode i j k M (1 ,
1) 1 1 1 M (1 ,
1) 2 4 1 M (1 ,
1) 3 7 1TABLE VI: Values of mode numbers i and j (as defined in the text) and axial wave number k forthe three modes shown in Fig. 7. Given the above comments, it is clear that no resonant mode can be individually excitedby excitations (1a)-(1c) for any given fixed values of the wave numbers m and k . XI. CONCLUSION
In summary, we have studied analytically the linear elastodynamic response of a simply-supported isotropic thick-walled hollow elastic circular cylinder subjected to 2D harmonicstanding-wave excitations on its curved surfaces. Our mathematical formulation employedthe exact 3D theory of linear elasticity. An exact semi-analytical solution for the steady-state displacement field of the cylinder was constructed using recently-obtained parametricsolutions to the Navier-Lam´e equation. In order to provide a solution that involves onlyreal-valued Bessel functions, the problem was solved separately in three distinct parameterregimes involving the excitation frequency. In each case, application of the standing-waveboundary conditions generates a parameter-dependent 6 × simply-supported cylinder. It is worthemphasizing that each standing-wave excitation generates a unique series of resonances(of varying widths), as opposed to generating just a single resonance. Put another way,each standing-wave excitation excites a set of (orthogonal) resonant modes as opposed toexciting only one resonant mode. Unfortunately, for any given value of the circumferentialwave number m (and assuming the values of all cylinder parameters are fixed), there isno way of predicting which modes will be excited given the value of the longitudinal wavenumber k .While the above numerical results provide important physical insight, they are cursoryin the sense that many fundamental aspects of the studied problem remain numerically un-explored. For instance, there is the fundamental problem of the “resonance dynamics”, thatis, how the resonances generated by any given standing-wave excitation evolve in frequencyspace under the variation of a material or geometric parameter. The resonant mode shapesand their evolution under parametric variation(s) is another fundamental aspect requiringsystematic investigation, and (as mentioned in the Introduction) the ultimate goal is totackle the more general forced-vibration problem involving arbitrary asymmetric surfaceexcitations – a problem for which the present work provides the necessary foundation. Wehope to explore these topics in future publications. Appendix A: General Axisymmetric Solution Using the ERP Field Equations
In this appendix, we show how our general axisymmetric solution may be reproducedfrom the general ERP field equations. For brevity, we shall here only provide the derivationfor the Case 1 sub-solution. The sub-solutions in the other two cases may be similarlyreproduced using the same logic. 36 . General ERP Field Equations
According to ERP [15], the following is an exact axisymmetric solution to Eq. (2), forarbitrary values of k rn ( n = 1 , , , . . . , N r ) and k zn ( n = 1 , , , . . . , N z ): u ( ERP ) z u ( ERP ) r = u (1) z u (1) r + u (2) z u (2) r + u (3) z u (3) r , (A1a)where u (1) z u (1) r = P cos(Ω z ) + N r X n =1 2 X s =1 P ns C ( k rn r ) cos( k zns z ) N r X n =1 2 X s =1 P ns ψ ns C ( k rn r ) sin( k zns z ) , (A1b) u (2) z u (2) r = QJ (Ω r ) + N z X n =1 2 X s =1 Q ns J ( k rns r ) cos( k zn z ) N z X n =1 2 X s =1 Q ns χ ns J ( k rns r ) sin( k zn z ) , (A1c) u (3) z u (3) r = RY (Ω r ) + N z X n =1 2 X s =1 R ns Y ( k rns r ) cos( k zn z ) N z X n =1 2 X s =1 R ns χ ns Y ( k rns r ) sin( k zn z ) , (A1d)Ω = s ρω ( λ + 2 µ ) , Ω = s ρω µ , (A1e) C ( k rn r ) = J ( k rn r ) + ζ n Y ( k rn r ) , C ( k rn r ) = J ( k rn r ) + ζ n Y ( k rn r ) , (A1f) P , Q , R , { P ns : n = 1 , , , . . . , N r ; s = 1 , } , { Q ns : n = 1 , , , . . . , N z ; s = 1 , } , { R ns : n = 1 , , , . . . , N z ; s = 1 , } are arbitrary constants (determined by the specificexcitation), and the remaining (frequency-dependent) constants are given by: k zn = s ρω ( λ + 2 µ ) − k rn , k zn = s ρω µ − k rn , n = 1 , , , . . . , N r (A1g) k rn = s ρω ( λ + 2 µ ) − k zn , k rn = s ρω µ − k zn , n = 1 , , , . . . , N z (A1h)37 n = k rn k zn , ψ n = − k zn k rn , n = 1 , , , . . . , N r (A1i) χ n = − k rn k zn , χ n = k zn k rn , n = 1 , , , . . . , N z . (A1j)The non-arbitrary constants { ζ n : n = 1 , , , . . . , N r } in (A1f) are chosen so as to satisfycertain conditions (see Ref. [15] for details). Since these constants will have no relevance inthe analyses that follow, it is sufficient for our purposes to leave them unspecified.The components of the stress field corresponding to the displacement field (A1) are thenas follows [15]: σ ( ERP ) rr = − P Ω λ sin(Ω z )+ N r X n =1 2 X s =1 P ns (cid:26)h ( λ + 2 µ ) ψ ns k rn − λk zns i C ( k rn r ) − µr ψ ns C ( k rn r ) (cid:27) sin( k zns z )+ N z X n =1 2 X s =1 Q ns (cid:26)h ( λ + 2 µ ) χ ns k rns − λk zn i J ( k rns r ) − µr χ ns J ( k rns r ) (cid:27) sin( k zn z )+ N z X n =1 2 X s =1 R ns (cid:26)h ( λ + 2 µ ) χ ns k rns − λk zn i Y ( k rns r ) − µr χ ns Y ( k rns r ) (cid:27) sin( k zn z ) , (A2) σ ( ERP ) zz = − P Ω ( λ + 2 µ ) sin(Ω z )+ N r X n =1 2 X s =1 P ns h − ( λ + 2 µ ) k zns + λψ ns k rn i C ( k rn r ) sin( k zns z )+ N z X n =1 2 X s =1 Q ns h − ( λ + 2 µ ) k zn + λχ ns k rns i J ( k rns r ) sin( k zn z )+ N z X n =1 2 X s =1 R ns h − ( λ + 2 µ ) k zn + λχ ns k rns i Y ( k rns r ) sin( k zn z ) , (A3)and σ ( ERP ) rz = − Q Ω µJ (Ω r ) − R Ω µY (Ω r )+ µ N r X n =1 2 X s =1 P ns h − k rn + ψ ns k zns i C ( k rn r ) cos( k zns z )+ µ N z X n =1 2 X s =1 Q ns h − k rns + χ ns k zn i J ( k rns r ) cos( k zn z )+ µ N z X n =1 2 X s =1 R ns h − k rns + χ ns k zn i Y ( k rns r ) cos( k zn z ) . (A4)38s noted in ERP [15], the harmonic time dependence has been dropped (for convenience)from all field equations.
2. Reduction and Equivalency of the ERP Field Equations
To enforce the vanishing of the longitudinal stress component (A3) at the ends of thecylinder (i.e., to satisfy condition (3c)), we set the arbitrary ERP constant k zn = ( nπ/L ) andprescribe P ns = 0 , ∀ n, s in Eq. (A3) and thereby in all of the remaining ERP field equations.The values of the constants ψ ns , k rn , and k zns appearing in the associated summands aresubsequently irrelevant. Unless the parameters are such that ρω / ( λ + 2 µ ) = ( nπ/L ) (asingular case that we have ignored, see comments at the end of Section IV A), we can alsoprescribe P = 0 in (A3) and thereby in (A2) and (A1b). Hence, under the condition that ρω / ( λ + 2 µ ) = ( nπ/L ) , the sub-solution (A1b) is zero, i.e., (cid:16) u (1) z , u (1) r (cid:17) = (0 , n in Eqs. (A1)-(A4) collapses to a single term that depends on the prescribed valueof n . Universally replacing the index n (in the ERP field equations) by k (the longitudinalwave number in our problem) then yields the following for the frequency-dependent ERPconstants: k rn −→ k rk = s ρω ( λ + 2 µ ) − (cid:18) kπL (cid:19) = iα Case 1 α Case 2 iα Case 3 , (A5) k rn −→ k rk = s ρω µ − (cid:18) kπL (cid:19) = iα Case 1 α Case 2 α Case 3 , (A6) χ n −→ χ k = − k rk k zk = α /i (cid:0) kπL (cid:1) Case 1 − α / (cid:0) kπL (cid:1) Case 2 α /i (cid:0) kπL (cid:1) Case 3 , (A7)39 n −→ χ k = k zk k rk = (cid:0) kπL (cid:1) /iα Case 1 (cid:0) kπL (cid:1) /α Case 2 (cid:0) kπL (cid:1) /α Case 3 , (A8)where the final equalities in (A5)-(A8) are with reference to constants { α , α } as definedby Eqs. (21), (31), and (39), for Cases 1, 2, and 3, respectively. a. Displacement Components For Case 1, the radial component of the ERP displacement field, which is the sum of thenon-vanishing radial components u (2) r and u (3) r in sub-solutions (A1c) and (A1d), reduces asfollows: u ( ERP ) r = ( X s =1 χ ks h Q ks J ( k rks r ) + R ks Y ( k rks r ) i) sin( k zk z )= ( α (cid:0) kπL (cid:1) h Q k i − J ( iα r ) + R k i − Y ( iα r ) i + (cid:0) kπL (cid:1) α h Q k i − J ( iα r ) + R k i − Y ( iα r ) i) sin (cid:18) kπL z (cid:19) = ( α (cid:0) kπL (cid:1) h ( Q k + iR k ) i − J ( iα r ) + 2 π R k K ( α r ) i + (cid:0) kπL (cid:1) α h ( Q k + iR k ) i − J ( iα r ) + 2 π R k K ( α r ) i) sin (cid:18) kπL z (cid:19) = ( α (cid:0) kπL (cid:1) h e Q k I ( α r ) + e R k K ( α r ) i + (cid:0) kπL (cid:1) α h e Q k I ( α r ) + e R k K ( α r ) i) sin (cid:18) kπL z (cid:19) = n α h A I ( α r ) − B K ( α r ) i + α h A I ( α r ) − B K ( α r ) io sin (cid:18) kπL z (cid:19) , (A9)where relations (A5)-(A8) have been employed in obtaining the second equality in (A9), thedefinition of the modified Bessel function of the second kind K n ( x ) = π i n +1 (cid:2) J n ( ix ) + iY n ( ix ) (cid:3) , n ∈ N (A10)40n obtaining the third equality, and the fundamental definition of the modified Bessel functionof the first kind I p ( x ) ≡ i − p J p ( ix ) , p ∈ R (A11)in obtaining the fourth equality. The last equality in (A9) ensues upon replacement of thearbitrary constants e Q ks ≡ Q ks + iR ks , e R ks ≡ (2 /π ) R ks ( s = 1 ,
2) (A12)by a new set of arbitrary constants { A , A , B , B } as follows: e Q ks −→ (cid:18) kπL (cid:19) A s γ s = (cid:0) kπL (cid:1) A if s = 1 α A / (cid:0) kπL (cid:1) if s = 2 , (A13a) e R ks −→ − (cid:18) kπL (cid:19) B s γ s = − (cid:0) kπL (cid:1) B if s = 1 − α B / (cid:0) kπL (cid:1) if s = 2 . (A13b)Using the same logic (and again considering Case 1), the axial component of the ERPdisplacement field, which is the sum of the non-vanishing axial components u (2) z and u (3) z insub-solutions (A1c) and (A1d), similarly reduces as follows: u ( ERP ) z = QJ (Ω r ) + RY (Ω r ) + ( X s =1 h Q ks J ( k rks r ) + R ks Y ( k rks r ) i) cos( k zk z )= QJ (Ω r ) + h Q k J ( iα r ) + Q k J ( iα r ) i cos (cid:18) kπL z (cid:19) + RY (Ω r ) + h R k Y ( iα r ) + R k Y ( iα r ) i cos (cid:18) kπL z (cid:19) = QJ (Ω r ) + h ( Q k + iR k ) J ( iα r ) + ( Q k + iR k ) J ( iα r ) i cos (cid:18) kπL z (cid:19) + RY (Ω r ) − π h R k K ( α r ) + R k K ( α r ) i cos (cid:18) kπL z (cid:19) = QJ (Ω r ) + h e Q k I ( α r ) + e Q k I ( α r ) i cos (cid:18) kπL z (cid:19) + RY (Ω r ) − h e R k K ( α r ) + e R k K ( α r ) i cos (cid:18) kπL z (cid:19) = h QJ (Ω r ) + RY (Ω r ) i + (cid:18) kπL (cid:19) (cid:26) γ h A I ( α r ) + B K ( α r ) i + γ h A I ( α r ) + B K ( α r ) i(cid:27) cos (cid:18) kπL z (cid:19) . (A14)41he first bracketed term in the final line of (A14) is equivalent to the axial componentof (20) in the special m = 0 case. Comparing (A9) with (29a) and the cosine term of(A14) with (29b), we see that, in Case 1, the displacement components obtained from theERP method are identical to the displacement components obtained from our method ofsolution in the special m = 0 case. Using relations (A5)-(A8), definitions (A10)-(A11), andanalogs of substitutions (A13), and applying the same logic, it is straightforward to showthat the equivalence holds for Cases 2 and 3 as well. Thus, in the special m = 0 case, thegeneral displacement field obtained from our method of solution is identical to the generalaxisymmetric displacement field obtained from the ERP method. b. Stress Components For Case 1, the axisymmetric radial stress obtained from our method of solution is (omit-ting the sin( ωt ) factor): σ ( SC ) rr = ( X s =1 A s (cid:20) β s I ( α s r ) − µα s r I ( α s r ) (cid:21) + X s =1 B s (cid:20) β s K ( α s r ) + 2 µα s r K ( α s r ) (cid:21)) sin (cid:18) kπL z (cid:19) , (A15)which is immediately obtained upon substituting m = 0 into Eq. (24).Given the statements at the beginning of Section A 2, the ERP radial stress component(A2) reduces to the following: σ ( ERP ) rr = X s =1 Q ks (cid:26)h ( λ + 2 µ ) χ ks k rks − λk zk i J ( k rks r ) − µr χ ks J ( k rks r ) (cid:27) sin( k zk z )+ X s =1 R ks (cid:26)h ( λ + 2 µ ) χ ks k rks − λk zk i Y ( k rks r ) − µr χ ks Y ( k rks r ) (cid:27) sin( k zk z ) . (A16)Employing relations (A5)-(A8) and considering only Case 1, the frequency-dependent ERPconstants in (A16) simplify as follows: C k ( ω ) ≡ ( λ + 2 µ ) χ k k rk − λk zk = ( λ + 2 µ ) α (cid:0) kπL (cid:1) − λ (cid:18) kπL (cid:19) = β (cid:0) kπL (cid:1) , (A17a) C k ( ω ) ≡ ( λ + 2 µ ) χ k k rk − λk zk = 2 µ (cid:18) kπL (cid:19) = β (cid:0) kπL (cid:1) α , (A17b)42here the final equalities in (A17a)-(A17b) are with reference to the (frequency-dependent)constants { β , β } as defined by Eq. (24b). For future algebraic convenience, we also definethe frequency-dependent constants { D k ( ω ) , D k ( ω ) } as follows: χ k = α (cid:0) kπL (cid:1) i − ≡ D k ( ω ) i − , χ k = (cid:0) kπL (cid:1) α i − ≡ D k ( ω ) i − , (A17c)where the values of { χ ks : s = 1 , } in (A17c) are again pertinent to Case 1. Using thecompact notation of (A17) for the frequency-dependent ERP constants and keeping in mindthat we are here only considering Case 1, the ERP radial stress component (A16) thenreduces as follows: σ ( ERP ) rr = ( X s =1 Q ks (cid:20) C ks ( ω ) J ( iα s r ) − µr D ks ( ω ) i − J ( iα s r ) (cid:21) + X s =1 R ks (cid:20) C ks ( ω ) Y ( iα s r ) − µr D ks ( ω ) i − Y ( iα s r ) (cid:21)) sin (cid:18) kπL z (cid:19) = ( X s =1 ( Q ks + iR ks ) (cid:20) C ks ( ω ) J ( iα s r ) − µr D ks ( ω ) i − J ( iα s r ) (cid:21) − X s =1 π R ks (cid:20) C ks ( ω ) K ( α s r ) + 2 µr D ks ( ω ) K ( α s r ) (cid:21)) sin (cid:18) kπL z (cid:19) = ( X s =1 e Q ks (cid:20) C ks ( ω ) I ( α s r ) − µr D ks ( ω ) I ( α s r ) (cid:21) − X s =1 e R ks (cid:20) C ks ( ω ) K ( α s r ) + 2 µr D ks ( ω ) K ( α s r ) (cid:21)) sin (cid:18) kπL z (cid:19) = (cid:18) kπL (cid:19) ( X s =1 A s γ s (cid:20) C ks ( ω ) I ( α s r ) − µr D ks ( ω ) I ( α s r ) (cid:21) + X s =1 B s γ s (cid:20) C ks ( ω ) K ( α s r ) + 2 µr D ks ( ω ) K ( α s r ) (cid:21)) sin (cid:18) kπL z (cid:19) = ( X s =1 A s (cid:20) β s I ( α s r ) − µα s r I ( α s r ) (cid:21) + X s =1 B s (cid:20) β s K ( α s r ) + 2 µα s r K ( α s r ) (cid:21)) sin (cid:18) kπL z (cid:19) , (A18)where relations (A5)-(A6) have been employed in obtaining the first equality in (A18) anddefinitions (A10) and (A11) in obtaining the second and third equalities, respectively. Thefourth equality in (A18) derives from use of (A13), and the final equality in (A18) then43ollows from definitions (A17) and the fact that γ = 1 , γ = 1 (cid:0) kπL (cid:1) "(cid:18) kπL (cid:19) − ρω µ = 1 (cid:0) kπL (cid:1) α Case 1 − α Case 2 − α Case 3 . (A19)For Case 1 and longitudinal wave number k = 0, the axisymmetric shear stress obtainedfrom our method of solution is (again omitting the sin( ωt ) factor): σ ( SC ) rz = µ (cid:18) kπL (cid:19) ( X s =1 α s (1 + γ s ) h A s I ( α s r ) − B s K ( α s r ) i) cos (cid:18) kπL z (cid:19) , (A20)which is immediately obtained upon substituting m = 0 into Eq. (26).Given the remarks at the beginning of Section A 2, the ERP shear stress component (A4)reduces to the following: σ ( ERP ) rz = − µ Ω h QJ (Ω r ) + RY (Ω r ) i + µ ( X s =1 ( − k rks + χ ks k zk ) h Q ks J ( k rks r ) + R ks Y ( k rks r ) i) cos( k zk z ) . (A21)Employing relations (A5)-(A8) and considering only Case 1, the frequency-dependent ERPconstants in (A21) are as follows: ξ ks ( ω ) ≡ − k rks + χ ks k zk = α i − , if s = 1 α + (cid:0) kπL (cid:1) α ! i − , if s = 2= (cid:20) α s (1 + γ s ) γ s (cid:21) i − , (A22)where the final equality in (A22) derives from use of identity (A19). Then, using the compactnotation ¯ σ ( ERP ) rz ≡ − µ Ω h QJ (Ω r ) + RY (Ω r ) i , (A23a)¯ ξ ks ( ω ) ≡ iξ ks ( ω ) = α s (1 + γ s ) γ s , (A23b)and keeping in mind that we are considering Case 1, the ERP shear stress component (A21)44educes as follows: σ ( ERP ) rz = ¯ σ ( ERP ) rz + µ ( X s =1 ¯ ξ ks ( ω ) i − h Q ks J ( iα s r ) + R ks Y ( iα s r ) i) cos (cid:18) kπL z (cid:19) = ¯ σ ( ERP ) rz + µ ( X s =1 ¯ ξ ks ( ω ) h ( Q ks + iR ks ) i − J ( iα s r ) + 2 π R ks K ( α s r ) i) cos (cid:18) kπL z (cid:19) = ¯ σ ( ERP ) rz + µ ( X s =1 ¯ ξ ks ( ω ) h e Q ks I ( α s r ) + e R ks K ( α s r ) i) cos (cid:18) kπL z (cid:19) = ¯ σ ( ERP ) rz + µ (cid:18) kπL (cid:19) ( X s =1 ¯ ξ ks ( ω ) γ s h A s I ( α s r ) − B s K ( α s r ) i) cos (cid:18) kπL z (cid:19) = − µ Ω h QJ (Ω r ) + RY (Ω r ) i + µ (cid:18) kπL (cid:19) ( X s =1 α s (1 + γ s ) h A s I ( α s r ) − B s K ( α s r ) i) cos (cid:18) kπL z (cid:19) , (A24)where relations (A5)-(A8) have been employed in obtaining the first equality in (A24),and then (A10), (A11), (A13), and (A23), in obtaining the second, third, fourth, and fifthequalities, respectively.The first bracketed term in the final line of (A24) is consistent with the shear stresscomponent (46b) when m = 0. 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