Vibration and Shock Handbook 04

Vibration and Shock Handbook 04 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

4 Distributed-Parameter Systems Clarence W de Silva The University of British Columbia 4.1 4.2 Introduction Transverse Vibration of Cables 4.3 Longitudinal Vibrations of Rods 4-13 4.4 Torsional Vibration of Shafts 4-19 Wave Equation † General (Modal) Solution † Cable with Fixed Ends † Orthogonality of Natural Modes † Application of Initial Conditions Equation of Motion † 4-1 4-2 Boundary Conditions Shaft with Circular Cross Section Noncircular Shafts † Torsional Vibration of 4.5 Flexural Vibration of Beams 4-26 4.6 Damped Continuous Systems 4-50 4.7 Vibration of Membranes and Plates 4-52 Governing Equation for Thin Beams † Modal Analysis † Boundary Conditions † Free Vibration of a Simply Supported Beam † Orthogonality of Mode Shapes † Forced Bending Vibration † Bending Vibration of Beams with Axial Loads † Bending Vibration of Thick Beams † Use of the Energy Approach † Orthogonality with Inertial Boundary Conditions Modal Analysis of Damped Beams Transverse Vibration of Membranes † Rectangular Membrane with Fixed Edges † Transverse Vibration of Thin Plates † Rectangular Plate with Simply Supported Edges Summary This chapter presents the analysis of continuous (or distributed-parameter) mechanical vibrating systems In these systems, inertial, elastic, and dissipative effects are found continuously distributed in one, two, or three dimensions Examples such as strings, rods, shafts, beams, membranes, and plates are studied Modal analysis is carried out using the separation of time and space The orthogonality property of mode shapes is established Boundary conditions are derived Free vibration and forced vibration are analyzed 4.1 Introduction Often in vibration analysis, it is assumed that inertial (mass), flexibility (spring), and dissipative (damping) characteristics can be “lumped” as a finite number of “discrete” elements Such models are termed lumpedparameter or discrete-parameter systems Generally, in practical vibrating systems, inertial, elastic, and dissipative effects are found continuously distributed in one, two, or three dimensions Correspondingly, 4-1 © 2005 by Taylor & Francis Group, LLC 4-2 Vibration and Shock Handbook we have line structures, surface/planar structures, or spatial structures They will possess an infinite number of mass elements, continuously distributed in the structure, and integrated with some connecting flexibility (elasticity) and energy dissipation In view of the connecting flexibility, each small element of mass will be able to move out of phase (or somewhat independently) with the remaining mass elements It follows that a continuous system (or a distributed-parameter system) will have an infinite number of degrees of freedom (DoFs) and will require an infinite number of coordinates to represent its motion In other words, extending the concept of a finite-degree-of-freedom system as analyzed previously, an infinitedimensional vector is needed to represent the general motion of a continuous system Equivalently, a onedimensional continuous system (a line structure) will need one independent spatial variable, in addition to time, to represent its response In view of the need for two independent variables in this case, one for time and the other for space, the representation of system dynamics will require partial differential equations (PDEs) rather than ordinary differential equations (ODEs) Furthermore, the system will depend on the boundary conditions as well as the initial conditions Strings, cables, rods, shafts, beams, membranes, plates, and shells are example of continuous members In special cases, closed-form analytical solutions can be obtained for the vibration of these members A general structure may consist of more than one such member, and furthermore, boundary conditions (BCs) could be various, individual members may be nonuniform, and the material characteristics may be inhomogeneous and anisotropic Closed-form analytical solutions would not be generally possible in such cases Nevertheless, the insight gained by analyzing the vibration of standard members will be quite beneficial in studying the vibration behavior of more complex structures The concepts of modal analysis may be extended from lumped-parameter systems to continuous systems In particular, since the number of principal modes is equal to the number of DoFs of the system, a distributed-parameter system will have an infinite number of natural modes of vibration A particular mode may be excited by deflecting the member so that its elastic curve assumes the shape of the particular mode, and then releasing from this initial condition When damping is significant and nonproportional, however, there is no guarantee that such an initial condition could accurately excite the required mode A general excitation consisting of a force or an initial condition will excite more than one mode of motion However, as in the case of discrete-parameter systems, the general motion may be analyzed and expressed in terms of modal motions, through modal analysis In a modal motion, the mass elements will move at a specific frequency (the natural frequency), and bearing a constant proportion in displacement (i.e., maintaining the mode shape), and passing the static equilibrium of the system simultaneously In view of this behavior, it is possible to separate the time response and spatial response of a vibrating system in a modal motion This separability is fundamental to modal analysis of a continuous system Furthermore, in practice an infinite number of natural frequencies and mode shapes are not significant and typically the very high modes may be neglected Such a modal-truncation procedure, even though carried out by continuous-system analysis, is equivalent to approximating the original infinite-degree-of-freedom system by a finitedegree-of-freedom one Vibration analysis of continuous systems may be applied in the modeling, analysis, design, and evaluation of such practical systems as cables; musical instruments; transmission belts and chains; containers of fluid; animals; structures including buildings, bridges, guideways, and space stations; and transit vehicles, including automobiles, ships, aircraft, and spacecraft 4.2 Transverse Vibration of Cables The first continuous member which we will study is a string or cable in tension This is a line structure whose geometric configuration can be completely defined by the position of its axial line with reference to a fixed coordinate line We will study the transverse (lateral) vibration problem; that is, the vibration in a direction perpendicular to its axis and in a single plane Applications will include stringed musical instruments, overhead transmission lines (of electric power or telephone signals), drive systems (belt drives, chain drives, pulley ropes, etc.), suspension bridges, and structural cables carrying cars (e.g., ski lifts, elevators, overhead sightseeing systems, and cable cars) © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-3 As usual, we will make some simplifying assumptions for analytical convenience However, the results and insight obtained in this manner will be useful in understanding the behavior of more complex systems containing cable-like structures The main assumptions are: The system is a line structure The lateral dimensions are much smaller compared with the longitudinal dimension (normally in the x direction) The structure stays in a single plane and the motion of every element of the structure will be in a fixed transverse direction ð yÞ: The cable tension ðTÞ remains constant during motion In other words, the initial tension is sufficiently large that the variations during motion are negligible ›v Variations in slope ðuÞ along the structure are small Hence, for example, u ø sin u ø tan u ¼ : ›x A general configuration of a cable (or string) is shown in Figure 4.1(a) Consider a small element of length dx of the cable at location x; as shown in Figure 4.1(b) The equation (Newton’s Second Law) of motion (transverse) of this element is given by f ðx; tÞdx T sin u ỵ T sinu ỵ duị ẳ mxịdx ›2 vðx; tÞ ›t ð4:1Þ in which vðx; tÞ ¼ transverse displacement of the cable f ðx; tÞ ¼ lateral force per unit length of cable mxị ẳ mass per unit length of cable T ¼ cable tension u ¼ cable slope at location x: Note that the dynamic loading f ðx; tÞ may arise due to such causes as aerodynamic forces, fluid drag, and electromagnetic forces, depending on the specific application y Force per unit length = f(x,t) T x x+dx l x Mass per unit length = m(x) (a) y f.dx T θ+ ∂q dx ∂x m.dx θ T (b) FIGURE 4.1 x x+dx x (a) Transverse vibration of a cable in tension; (b) motion of a general element © 2005 by Taylor & Francis Group, LLC 4-4 Vibration and Shock Handbook Using the small slope assumption we have sin u ứ u and sinu ỵ duị ứ u ỵ du with u ẳ v=x and du ẳ v=x2 ịdx as dx ! 0: On substitution of these approximations into Equation 4.1 and canceling out dx, we obtain mðxÞ ›2 vðx; tÞ ›2 vðx; tÞ ẳT ỵ f x; tị t x2 4:2ị Now consider the case of free vibration where f ðx; tÞ ¼ 0: We have ›2 vðx; tÞ › vx; tị ẳ c t x with p c ẳ T=m 4:3ị 4:4ị Also, assume that the cable is uniform so that m is constant 4.2.1 Wave Equation The solution to any equation of the form (Equation 4.3) will appear as a wave, traveling either in the forward (positive x) or in the backward (negative x) direction at speed c: Hence, Equation 4.3 is called the wave equation and c is the wave speed To prove this fact, first, we show that a solution to Equation 4.3 can take the form vx; tị ẳ v1 x ctị 4:5ị First, let x ct ẳ z: Hence, v1 x ctị ẳ v1 zị: Then, v1 dv z › v1 dv ›z ¼ and ¼ dz ›x dz ›t ›x ›t with ›z ›z ¼ and ¼ 2c ›x ›t It follows that › v1 ›2 v1 ¼ v001 and ¼ c2 v001 ›x ›t where v001 ¼ d2 v1 dz Clearly, then, v1 satisfies Equation 4.3 Now, let us examine the nature of the solution v1 ðx ctÞ: It is clear that v1 will be constant when x ct ¼ constant: However, the equation x ct ¼ constant corresponds to a point moving along the x axis in the positive direction at speed c: What this means is that the shape of the cable at t ¼ will “appear” to travel along the cable at speed c: This is analogous to the waves we observe in a pond when excited by dropping a stone Note that the particles of the cable not travel along x: it is the deformation “shape” (the wave) that travels Similarly, it can be shown that vx; tị ẳ v2 x ỵ ctị â 2005 by Taylor & Francis Group, LLC ð4:6Þ Distributed-Parameter Systems 4-5 is also a solution to Equation 4.3 and this corresponds to a wave that travels backward (negative x direction) at speed c: The general solution, of course, will be of the form vx; tị ẳ v1 x ctị ỵ v2 x ỵ ctị 4:7ị which represents two waves, one traveling forward and the other backward 4.2.2 General (Modal) Solution As usual, we look for a separable solution of the form vx; tị ẳ Yxịqtị 4:8ị for the cable/string vibration problem given by the wave equation 4.3 If a solution in the form of Equation 4.8 is obtained, it will be essentially a modal solution This should be clear from the separability itself of the solution Specifically, at any given time t; the time function qðtÞ will be fixed and the structure will have a shape given by YðxÞ: Hence, at all times the structure will maintain a particular shape YðxÞ and this will be a mode shape Also, at a given point x of the structure, the space function YðxÞ will be fixed and the structure will vibrate according to the time response qðtÞ: It will be shown that qðtÞ will obey the simple harmonic motion of a specific frequency This is the natural frequency of vibration corresponding to that particular mode Note that, for a continuous system, there will be an infinite number of solutions of the form of Equation 4.8 with different natural frequencies The corresponding functions YðxÞ will be “orthogonal” in some sense Hence, they are called normal modes (normal meaning perpendicular) The systems will be able to move independently in each mode and this collection of solutions in the form of Equation 4.8 will be a complete set With this qualitative understanding, let us now seek a solution of the form of Equation 4.8 for the system Equation 4.3 Substitute Equation 4.8 in Equation 4.3 We obtain YðxÞ d2 qðtÞ d Yaị ẳ c qtị dt dx2 or d2 Yxị d2 qtị ẳ ẳ 2l2 YðxÞ dx c qðtÞ dt ð4:9Þ In Equation 4.9, since the left-hand terms are a function of x only and the right-hand terms are a function of t only, for the two sides to be equal in general, each function should be a constant (that is independent of both x and t) This constant is denoted by 2l2, which is called the separation constant and is designated to be negative There are two reasons for this If this common constant were positive, the function qðtÞ would be nonoscillatory and transient, which is contrary to the nature of undamped vibration Furthermore, it can be shown that a nontrivial solution for YðxÞ would not be possible if the common constant were positive The unknown constant l is determined by solving the space equation (mode shape equation) of Equation 4.9; specifically d2 Yxị ỵ l2 Yxị ẳ dx2 4:10ị and then applying the BCs of the problem There will be an infinite number of solutions for l; with corresponding natural frequencies v and mode shapes YðxÞ: The characteristic equation of 4.10 is p2 ỵ l2 ẳ 4:11ị which has the characteristic roots (or eigenvalues) p ẳ ^jl â 2005 by Taylor & Francis Group, LLC ð4:12Þ 4-6 Vibration and Shock Handbook The general solution is Yxị ẳ A1 e jlx ỵ A2 e2jlx ẳ C1 cos lx ỵ C2 sin lx ð4:13Þ Note that, since YðxÞ is a real function representing a geometric shape, the constants A1 and A2 have to be complex conjugates and C1 and C2 have to be real Specifically, in view of the fact that cos lx ẳ ejlx ỵ e2jlx = ị2 and sin lx ẳ ejlx e2jlx =2j ị, we can show that A1 ẳ C jC2 ị and A2 ẳ C ỵ jC2 ị For analytical convenience, we will use the real-parameter form of Equation 4.13 Note that we cannot determine both constants C1 and C2 using BCs Only their ratio is determined and the constant multiplier is absorbed into qðtÞ in Equation 4.8 and then determined using the appropriate initial conditions (at t ¼ 0) It follows that the ratio of C1 and C2 and the value of l are determined using the BCs Two BCs will be needed Some useful situations and appropriate relations are given in Table 4.1 4.2.3 Cable with Fixed Ends Let us obtain the complete solution for the free vibration of a taut cable that is fixed at both ends The applicable BCs are Y0ị ẳ Ylị ẳ ð4:14Þ where l is the length of the cable Substitution into Equation 4.13 gives C1 Ê ỵ C2 Ê ẳ C1 cos ll ỵ C2 sin ll ¼ Hence, we have C1 ¼ and C2 sin ll ẳ 4:15ị A possible solution for Equation 4.15 is C2 ¼ 0: However, this is the trivial solution, which corresponds to Yxị ẳ (i.e., a stationary cable with no vibration) It follows that the applicable, nontrivial solution is sin ll ¼ which produces an infinite number of solutions for l given by li ¼ ip l with i ẳ 1; 2; ; 4:16ị As mentioned earlier, the corresponding infinite number of mode shapes is given by Yi xị ẳ Ci sin ipx l ð4:17Þ Note: If we had used a positive constant l2 instead of 2l2 in Equation 4.9, only a trivial solution (with C1 ¼ and C2 ¼ 0) would be possible for YðxÞ: This further justifies our decision to use 2l2 : Substitute Equation 4.16 into Equation 4.9 to determine the corresponding time response (generalized coordinates) qi ðtÞ; thus d2 qi tị ỵ v2i qi tị ẳ dt 4:18ị in which vi ẳ li c ẳ â 2005 by Taylor & Francis Group, LLC ip l rffiffiffiffi T m for i ẳ 1; 2; ; 4:19ị Some Useful Boundary Conditions for the Cable Vibration Problem Type of End Condition Nature of End x ¼ x0 Fixed Boundary Condition vx0 ; tị ẳ x x Free T Modal Boundary Condition Yi x0 ị ẳ T vx0 ; tị ẳ0 x dYi x0 ị ẳ0 dx T vx0 ; tị kvx0 ; tị ẳ x T dYi x0 ị kYi x0 ị ẳ dx T ›vðx0 ; tÞ ›2 vðx0 ; tÞ kvx0 ; tị ẳ M x t T dYi x0 ị k v2i MịYi x0 ị ẳ dx ∂v ∂x Distributed-Parameter Systems TABLE 4.1 x Flexible T ∂v ∂x x k xo Flexible and inertial T M ∂v ∂x x k xo 4-7 © 2005 by Taylor & Francis Group, LLC 4-8 Vibration and Shock Handbook Equation 4.18 represents a simple harmonic motion with the modal natural frequencies vi given by Equation 4.19 It follows that there are an infinite number of natural frequencies, as mentioned earlier The general solution of Equation 4.19 is given by qi tị ẳ ci sinvi t ỵ fi ị 4:20ị where the amplitude parameter ci and the phase parameter fi are determined using two of the initial conditions of the system It should be clear that it is redundant to use a separate constant Ci for Yi ðxÞ in Equation 4.17, and that this may be absorbed into the amplitude constant in Equation 4.20 to express the general free response of the cable as vx; tị ẳ X ipx ci sin sinvi t ỵ fi ị l 4:21ị In this manner, the complete solution has been expressed as a summation of the modal solutions This is known as the modal series expansion Such a solution is quite justified because of the fact that the mode shapes are orthogonal in some sense, and what we obtained above were a complete set of normal modes (normal in the sense of perpendicular or orthogonal) The system is able to move in each mode independently, with a unique spatial shape, at the corresponding natural frequency, because each modal solution is separable into a space function, Yi ðxÞ, and a time function (generalized coordinate), qi ðtÞ: Of course, the system will be able to simultaneously move in a linear combination of two modes (say, C1 Y1 xịq1 tị ỵ C2 Y2 xịq2 tị), since this combination satisfies the original system Equation 4.3 because of its linearity and because each modal component satisfies the equation However, clearly, this solution, with two modes, is not separable into a product of a space function and a time function Hence, it is not a P modal solution In this manner, it can be argued that the infinite sum of modal solutions ci Yi ðxÞqi ðtÞ is the most general solution to the system (Equation 4.3) The orthogonality of mode shapes plays a key role in this argument and, furthermore, it is useful in the analysis of the system, as we shall see In particular, in Equation 4.21, the unknown constants ci and fi are determined using the system initial conditions, and the orthogonality property of modes is useful in that procedure 4.2.4 Orthogonality of Natural Modes A cable can vibrate at frequency vi while maintaining a unique natural shape Yi ðxÞ; called the mode shape of the cable We have shown that, for the fixed-ended cable, the natural mode shapes are given by sinðipx=lÞ with the corresponding natural frequencies, vi : It can be easily verified that > l l : ðl for i – j for i ẳ j 4:22ị In other words, the natural modes are orthogonal Equation 4.22 represents the principle of orthogonality of natural modes in this case Orthogonality makes the modal solutions independent and the corresponding mode shapes “normal.” It also makes the infinite set of modal solutions a complete set, or a basis, so that any arbitrary response can be formed as a linear combination of these normal mode solutions Orthogonality holds for other types of BCs as well To show this, we observe from Equation 4.9 that © 2005 by Taylor & Francis Group, LLC d2 Yi xị ỵ l2i Yi xị ẳ dx2 for mode i 4:23ị d2 Yj xị ỵ l2j Yj xị ẳ dx2 for mode j ð4:24Þ Distributed-Parameter Systems 4-9 Multiply Equation 4.23 by Yj ðxÞ; Equation 4.24 by Yi ðxÞ; subtract that second result from the first, and integrate with respect to x along the cable length from x ¼ to l: We obtain " # ðl ðl d2 Yj d2 Yi 4:25ị Yj 2 Yi dx ỵ l2i l2j ị Yi Yj dx ẳ dx dx 0 Integrating by parts, we obtain the results ðl ðl Yj ðl dY dYj d2 Yi dYi i dx dx ¼ Y j dx2 dx 0 dx dx Yi ðl dY dYj d2 Y j dYj i dx dx ¼ Yi 2 dx dx 0 dx dx l l Hence, the first term of Equation 4.25 becomes Yj dYj dYi Yi dx dx l which will vanish for common BCs Then, since li – lj for i – j, we have ðl Yi xịYj xịdx ẳ for i j We can pick the value of the multiplication constant in the general solution for YðxÞ; given by Equation 4.13, so as to normalize the mode shapes such that ðl Yi2 xịdx ẳ l which is consistent with the result 4.22 Hence, the general condition of orthogonality of natural modes may be expressed as > < for i j l 4:26ị Yi xịYj xịdx ẳ l > : for i ¼ j 4.2.4.1 Nodes When vibrating in a particular mode, one or more points of the system (cable) that are not physically fixed may remain stationary at all times These points are called the nodes of that mode For example, in the second mode of a cable with its ends fixed, there will be a node at the midspan This should be clear from the fact that the mode shape of the second mode is sin 2px=l which becomes zero at x ¼ l=2: Similarly, in the third mode, with mode shape sin 3px=l; there will be nodes at x ¼ l=3 and 2l=3: Example 4.1 If the cable tension varies along the length x; what is the corresponding equation of free lateral vibration? A hoist mechanism has a rope of freely hanging length l in a particular equilibrium configuration and carrying a load of mass M; as shown in Figure 4.2(a) Determine the equation of lateral vibration and the applicable BCs for the rope segment Solution With reference to Figure 4.1(b), Equation 4.1 may be modified for the case of variable T as 2T sin u ỵ T ỵ dTị sinu ỵ duị ¼ m dx © 2005 by Taylor & Francis Group, LLC ›2 v ›t ð4:27Þ 4-10 Vibration and Shock Handbook where f x; tị ẳ for free vibration Now, with the assumption of small u; and by neglecting the second-order product term dT du; we obtain T du þ u dT ¼ m x l ›v dx ›x Next, using u¼ ›v ; ›x du ¼ ›2 v ›T dx ; and dT ¼ ›x ›x and canceling dx; we obtain the equation of lateral vibration of a cable as m ›2 v ›2 v T v ẳT ỵ t x x x Longitudinal (axial) dynamics of the rope are negligible for the case of a stationary hoist Then, longitudinal equilibrium (in the x direction) of the small element of rope shown in Figure 4.2(b) gives y ð4:28Þ Mg (a) T+dT θ+dθ T ỵ dTị cosu ỵ duị T cos u mg dx ¼ For small u; we have cos u ứ and cosu ỵ duị ứ up to the first-order term in the Taylor series expansion Hence, dT ẳ mg dx x+dx x 4:29ị (b) Integration gives T ẳ T0 ỵ mgx 4:30ị with the end condition T mgdx FIGURE 4.2 (a) Free segment of a stationary hoist; (b) a small element of the rope T ¼ Mg at x ¼ Hence, T ¼ Mg ỵ mgx 4:31ị Note from Equation 4.29 that T=x ¼ dT=dx ¼ mg for this problem Substitute in (Equation 4.28) this fact and Equation 4.31 to obtain m ›2 v v v ẳ M ỵ mxịg ỵ mg ›t ›x ›x or ›2 v ¼ ›t M v v ỵx g ỵg x x m 4:32ị The BC at x ẳ is obtained by applying Newton’s Second Law to the end mass in the lateral ðyÞ direction This gives T0 ›vð0; tÞ v0; tị ẳM x t Now, using the fact that T0 ¼ Mg, we have the boundary condition g © 2005 by Taylor & Francis Group, LLC ›vð0; tị v0; tị ẳ x t Distributed-Parameter Systems 4-43 TABLE 4.4 Roots of the Frequency Equation for Bending Vibration of Beams First Three Roots li l End Conditions Pinned–pinned 4.5.7 p 2p 3p Fixed–fixed 4.730041 7.853205 10.995608 Free–free 4.730041 7.853205 10.995608 Fixed–pinned 3.926602 7.068583 10.210176 Fixed–free 1.875104 4.694091 7.854757 Fixed–sliding 2.365020 5.497804 8.639380 Pinned–free 3.926602 7.068583 10.210176 Bending Vibration of Beams with Axial Loads In practice, beam-type members that undergo flexural (transverse) vibrations may carry axial forces Examples are structural members such as columns, struts, and towers Generally, a tension will increase the natural frequencies of bending and compression will decrease them Hence, one way to avoid the excitation of a particular natural frequency (and mode) of bending vibration is to use a suitable tension or compression in the axial direction The equation of motion for the transverse motion of a thin beam subjected to an axial tension P may be easily derived by following the procedure that led to Equation 4.122 For simplicity, assume a thin beam subjected to a constant tensile force P: A small element dx of the beam is shown in Figure 4.16 The vertical component (in the positive y direction) of the axial force is 2P sin u ỵ P sinu ỵ duị ứ 2Pu ỵ Pu ỵ duị ẳ P du because the slope u ¼ ›v=›x is small Also, the change in slope is du ¼ ›2 v dx ›x It is clear that the previous equation of transverse dynamics of element dx has to be modified simply by adding the term Pð›2 v=›x2 Þdx to the f ðx; tÞdx side of the equation Then, the resulting equation of transverse vibration will be ! ›2 v ›2 ›2 v v rA ỵ EI 2 P ẳ f x; tị 4:184ị t x x ›x © 2005 by Taylor & Francis Group, LLC 4-44 Vibration and Shock Handbook dx x P θ q+dq P y, v(x,t) FIGURE 4.16 Beam element in transverse vibration and subjected to axial tension For modal analysis of a uniform beam, then, we use the equation of free motion rA v v v ỵ EI P ẳ t x x 4:185ị With a separable (modal) solution of the form vx; tị ẳ YðxÞqðtÞ ð4:186Þ q€ ðtÞ EI ðd4 Y=dx4 Þ P d2 Y=dx2 ị ẳ 2v2 ẳ2 rAY qtị 4:187ị we have which gives, as before, the time response equation of the generalized coordinates q tị ỵ v2 qtị ẳ ð4:188Þ and the mode shape equation EI d4 Y d2 Y P 2 rAv2 Y ¼ dx dx ð4:189Þ Note that the mode shape equation is still fourth order, but is different The analysis, however, may be done as before by using four BCs at the two ends of the beam to determine the natural frequencies (an infinite set) and the corresponding normalized mode shapes 4.5.8 Bending Vibration of Thick Beams In our derivation of the governing equation for the lateral vibration of thin beams (known as the Bernoulli –Euler beam equation), we neglected the following effects in particular: Deformation and associated lateral motion due to shear stresses Moment of inertia of beam elements in rotatory motion Note, however, that we did use the fact that shear forces ðQÞ are present in a beam cross section, even though the resulting deformations were not taken into account Also, in writing the equation for rotational motion of a beam element dx; we simply summed the moments to zero, without including the inertia moment These assumptions are valid for a beam whose cross-sectional dimensions are small compared with its length However, for a thick beam, the effect of shear deformation and rotatory inertia must be included in deriving the governing equation The resulting equation is known as the Timoshenko beam equation Important steps in the derivation of the equation of motion for the forced transverse vibration of beams, including the effects of shear deformation and rotatory inertia, are given now © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-45 x Bent Only x f(x,t) dx x+dx ∂2 θ M θ Q q +f = (a) y, v(x,t) x+dx x x ∂ t2 ∂ 2v ∂t2 M + Q+ ∂M ∂x δx ∂Q δx ∂x ∂v ∂x Bent and Sheared (b) y FIGURE 4.17 A Timoshenko beam element: (a) combined effect of bending and shear; (b) dynamic effects including rotatory inertia Consider a small element dx of a beam Figure 4.17(a) illustrates the contribution of the bending of an element and the deformation due to transverse shear stresses towards the total slope of the beam neutral axis Let u ¼ angle of rotation of the beam element due to bending f ¼ increase in slope of the element due to shear deformation in transverse shear (this is equal to shear strain) Then, the total slope of the beam element is v ẳuỵf x 4:190ị Here, v and x take the usual meanings, as for a thin beam Figure 4.17(b) shows an element dx of the beam with the forces, moments, and the linear and angular accelerations marked With the sign convention shown in Figure 4.17(b), the linear shear-stress–shearstrain relation can be stated as Q ¼ 2kGAf 4:191ị where the Timoshenko shear coefcient kẳ Average shear stress on beam cross section at x Shear stress at the neutral axis at x and G ¼ shear modulus The equation for translatory motion is f x; tịdx ỵ Q Q ỵ Q v dx ẳ rA dxị x t Hence, f x; tị â 2005 by Taylor & Francis Group, LLC ›Q ›2 v ¼ rA ›x ›t ð4:192Þ 4-46 Vibration and Shock Handbook The equation for the rotatory motion of the element, taking into account the rotatory inertia, is Mỵ M u dx M Q dx ẳ I r dxị ›x ›t which becomes ›M ›2 u Q ¼ Ir ›x ›t ð4:193Þ From the elementary theory of bending, as before M ẳ EI u x 4:194ị The relationship between the shear modulus and the modulus of elasticity is known to be E ẳ 21 ỵ nịG 4:195ị where n ẳ Poissons ratio This relation may be substituted if desired Manipulation of these equations yields EI ›4 v ›2 v E ›4 v r2 I ›4 v þ þ r A r I þ kG ›t ›x4 ›t kG ›x2 ›t ¼ f ðx; tÞ EI ›2 f ðx; tÞ rI f x; tị ỵ kGA x2 kGA t ð4:196Þ This is the Timoshenko beam equation for forced transverse motion Note that this equation is fourth order in time, whereas the thin beam equation is second order in time The modal analysis may proceed as before, by using the free f ẳ 0ị equation and a separable solution However, the resulting differential equation for the generalized coordinates will be fourth order in time and, as a result, additional natural frequency bands will be created The reason is the independent presence of shear and bending motions The differential equation of mode shapes will be fourth order in x and the solution procedure will be as before, through the use of four BCs at the two ends of the beam 4.5.9 Use of the Energy Approach So far, we have used only the direct, Newtonian approach in deriving the governing equations for continuous members in vibration Of course, the same results may be obtained by using the Lagrangian (energy) approach The general approach here is to first express the Lagrangian L of the system as L ẳ T p 2V 4:197ị where Tp ¼ total kinetic co-energy (equal to kinetic energy T for typical systems) V ¼ total potential energy Then, for a virtual increment (variation) of the system through incrementing the system variable, the following condition will hold: ðt ẵdL ỵ dW dt ẳ 4:198ị t1 where dL is the increment in the Lagrangian and dW is the work done by the external forces on the system due to the increment Finally, using the arbitrariness of the variation, the equation of motion, along with the BCs, can be obtained This approach is illustrated now © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-47 First, consider the free motion We can easily extend the result to the case of forced motion Kinetic energy is given by T¼ ðl ›v rA dx ›t ¼ ðl ›v rA ›t dx ð4:199Þ Potential energy due to bending results from work needed to bend the beam (angle denoted by u ¼ ›v=›x as usual) under bending moment M; thus !2 ðEnd ðl ›2 v M df ¼ EI V¼ dx ð4:200Þ ›x 2 End Before proceeding further, note the following steps of variation and integration by parts with respect to t: d ðt2 ›v t1 ›t dt ¼ ðt2 ›v ›v ðt2 ›v › dv d dt ¼ dt ¼ ›t t1 ›t t1 ›t ›t ›v dv ›t t2 t1 ðt2 ›2 v dv dt t1 ›t ð4:201Þ Note that we interchanged the operations d and ›=›t prior to integrating by parts Also, by convention, we assume that no variations are performed at the starting and ending times (t1 and t2 ) of integration Hence, dvt1 ị ẳ and dvt2 ị ẳ 4:202ị Similarly, variation and integration by parts with respect to x are done as follows: !2 ! ðl ðl ðl ›2 v ›2 v ›2 v ›2 v › ›v d dx d dx ¼ EI d dx ¼ EI EI 2 ›x ›x ›x ›x ›x ›x 0 " ¼ EI " ›2 v ›v d ›x ›x ›2 v ›v ¼ EI d ›x ›x #l #l " #l ðl › ðl › ›2 v ›v ›2 v ›v ›2 v › dv EI d dx ¼ EI d EI dx ›x ›x ›x ›x ›x ›x ›x ›x " › ›2 v EI dv x x #l ỵ l 2 v EI dv dx ›x2 ›x ð4:201aÞ Now, for the case of free vibration dW ẳ 0ị, substitute Equation 4.201 and Equation 4.201a in dT; and dV of Equation 4.199 and Equation 4.200, to obtain " # ðl ðt2 ðt2 ðl t2 ›2 v ›2 ›2 v ›v dv dL dt ¼0 ¼ dt dx rA þ EI dv þ dxrA ›t ›x ›x ›t 0 t1 t1 t1 ð4:203Þ " #l " #l 2 ð ðt2 t › v ›v v ỵ dt EI dV dt EI d ›x ›x ›x ›x2 t1 t1 Since Equation 4.203 holds for all arbitrary variations dvðtÞ; its coefficient should vanish Hence, rA ›2 v ›2 ›2 v ¼0 ỵ EI t x x 4:204ị which is the same Bernoulli– Euler beam equation for free motion as we had derived before The second integral term on the RHS of Equation 4.203 has no consequence We conventionally pick dvt1 ị ẳ and dvt2 ị ẳ at the time points t1 and t2 : The third integral term on the RHS of Equation 4.203 gives some BCs Specifically, if the slope BC ›v=›x is zero (i.e., fixed end), then the corresponding bending moment at the end is arbitrary, as expected However, if the slope at the boundary is arbitrary, then the bending moment EIð›2 v=›x2 Þ at the end should be zero (i.e., pinned or free end) The last integral term on the RHS of Equation 4.203 gives some other BCs Specifically, if the displacement BC v is zero (i.e., pinned or fixed end) then the corresponding shear force at the © 2005 by Taylor & Francis Group, LLC 4-48 Vibration and Shock Handbook end is arbitrary However, if the displacement at the boundary is arbitrary, then the shear force ð›=›xÞEIð›2 v=›x2 Þ at that end should be zero (i.e., free or sliding end) Next, consider a forced beam with force per unit length given by f ðx; tÞ: Then, the work done by f ðx; tÞdx in a small element dx of the beam, when moved through a displacement of dv; is f ðx; tÞdx dv ð4:205Þ Then, by combining Equation 4.205 with Equation 4.203, for arbitrary variation dv; we obtain the forced vibration equation rA v 2 v ẳ f x; tị ỵ EI t x x 4:206ị Note that external forces and moments applied at the ends of the beam can be incorporated into the BCs in the same manner 4.5.10 Orthogonality with Inertial Boundary Conditions It can be verified that the conventional orthogonality condition (Equation 4.163) holds for beams in transverse vibration, under common noninertial BCs When an inertia element (rectilinear or rotatory) is present at an end of the beam, this condition is violated A modified and more general orthogonality condition can be derived for application to beams with inertial boundary conditions To illustrate the procedure, consider a beam with a mass m attached at the end x ¼ l; as shown in Figure 4.18(a) A free-body diagram giving the sign convention for shear force Q acting on m is shown in Figure 4.18(b) The BCs at x ¼ l are: Bending moment vanishes, because there is no rotatory inertia at the end that is free Hence, EI vl; tị ẳ0 x l EI(x), ρ A(x) (a) ð4:207Þ Point Mass m x y, v(x,t) Q m (b) Q= v(l,t) ∂ ∂2v (l,t) EI ∂x ∂x FIGURE 4.18 (a) A beam with an end mass in transverse vibration; (b) free-body diagram showing the shear force acting on the end mass © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-49 Equation of rectilinear motion of the end mass m ›2 vl; tị vl; tị EI ẳ t ›x ›x ð4:208Þ In the usual manner, by substituting vx; tị ẳ Yi xịqi tị for mode i; along with q i tị ẳ 2v2i qi tị; into Equation 4.207 and Equation 4.208, with the understanding that the result should hold for any qi ðtÞ; we obtain the corresponding modal boundary conditions: d2 Yi lị ẳ0 dx2 4:209ị d d2 Yi lị EI ỵ v2i mYi lị ẳ dx dx2 ð4:210Þ for mode i: Now, we return to Equation 4.162: " #l " #l ðl d2 Yj dYj d2 Yi d d2 Yi d dYi d2 Yj 2 EI 2 ðvi vj Þ rAYi Yj dx Yj EI 2 Yi EI EI ¼ dx dx dx dx dx dx dx dx ð4:162Þ The second and the third terms of Equation 4.162 will vanish at x ¼ for noninertial BCs, as usual At x ¼ l; the third term will vanish in view of Equation 4.209 So, we are left with the second term at x ¼ l: Substitute Equation 4.210: d2 Y j d d2 Y d Yj EI 2i Yi EI dx dx dx dx ! l ẳ 2v2i v2j ịmYi lịYj ðlÞ Substitute Equation 4.211 into Equation 4.162 and cancel v2i v2j – for i – j: We obtain ( ðl for i – j rAYi Yj dx ỵ mYi lịYj lị ẳ aj for i ẳ j ð4:211Þ ð4:212Þ This is the modified and more general orthogonality property If the mass is at x ¼ 0; the direction of Q that acts on m will reverse and, hence, the second term in Equation 4.212 will become 2mYi ð0ÞYj ð0Þ: 4.5.10.1 Rotatory Inertia If there is a free rotatory inertia at x ¼ l; without an associated rectilinear inertia, then the shear force will vanish, giving d d2 Yi lị EI ẳ0 dx dx2 4:213ị The equation of rotational motion of J will give EI d2 Yi lị dY lị v2i J i ẳ dx2 dx ð4:214Þ Here, the second term in Equation 4.162 will vanish in view of Equation 4.213 Then, by substituting Equation 4.214 into the third term of Equation 4.162, we obtain the modified orthogonality relation ðl dY ðlÞ dYj ðlÞ rAYi Yj dx ỵ J i ẳ dx dx â 2005 by Taylor & Francis Group, LLC ( for i j aj for i ẳ j 4:215ị 4-50 4.6 Vibration and Shock Handbook Damped Continuous Systems All practical mechanical systems have some form of energy dissipation (damping) When the level of dissipation is small, damping is neglected, as we have done thus far in this chapter Yet, some effects of damping (e.g., the fact that at steady state the natural [modal] vibration components decay to zero leaving only the steady forcing component) are tacitly assumed even in undamped analysis The natural behavior of a system is expected to change due to the presence of damping In particular, the system’s natural frequencies will decrease (and be called damped natural frequencies) as a result of damping Furthermore, it is quite possible that a damped system would not possess “real” modes in which it could independently vibrate Mathematically, in that case, the modes will become complex (as opposed to real) and, physically, all the points of the system will not move, maintaining a constant phase at a given damped natural frequency In other words, a real solution that is separable in space ðxÞ and time ðtÞ may not be possible for the free vibration problem of a damped system Also, node points of an undamped system may vary with time as a result of damping With light damping, of course, such effects of damping will be negligible Since there are damped systems that not possess real natural modes of vibration, care should be exercised when extending the results of modal analysis from an undamped system to a damped one However, in some cases, the mode shapes will remain the same after including damping (even though the natural frequencies will change) This is analogous to the case of proportional damping, which was discussed in the section on lumped-parameter (multi-degree-of-freedom) vibrating systems The modal analysis of a damped system will become significantly easier if we assume that the mode shapes will remain the same as those for the undamped system Even when the actual type of damping in the system results in complex modes, for analytical convenience, an equivalent damping model that gives real modes is used in simplified analysis This is analogous to the use of proportional damping in lumped-parameter systems 4.6.1 Modal Analysis of Damped Beams Consider the problem of free damped transverse vibration of a thin beam, given by 2 v v v ỵ rA ẳ EI ỵL x x t t 4:216ị where L is a spatial differential operator (in x) Consider the following two possible models of damping: 1: L ¼ ›2 p ›2 E I dx2 ›x 2: L ¼ c ð4:217Þ ð4:218Þ Model corresponds to the Kelvin–Voigt model of material (internal) damping given by the stress– strain relation s ẳ E1 ỵ E p t 4:219ị where Ep is the damping parameter of the beam material Hence, we obtain the damped beam equation simply by replacing E in the undamped beam equation by E ỵ Ep ð›=›tÞ: Also, Ep is independent of the frequency of vibration for the viscoelastic damping model, but will be frequency dependent for the hysteretic damping model Modal analysis is done regardless of any frequency dependence of Ep and, in the final modal result for a particular modal frequency vi ; the appropriate frequency function for Epvị is used with v ẳ vi if the damping is of the hysteretic type It can be easily verified that the mode shapes of the damped system with model 4.217 are identical to those of the undamped system, regardless of whether the beam cross section is uniform or not In Model (Equation 4.218), the operator is a constant c: This corresponds to external damping of the linear viscous type, distributed along the beam length For example, imagine a beam resting on a © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-51 foundation of viscous damping material For Model 2, it can be shown that the damped mode shapes are identical to the undamped ones, assuming that the beam cross section is uniform If the beam is nonuniform, the damped and the undamped mode shapes are identical if we assume that the damping constant c varies along the beam in proportion to the area of cross section AðxÞ of the beam We shall show this in the example given below Example 4.8 Perform the modal analysis for transverse vibration of a thin nonuniform beam with linear viscous damping distributed along its length and satisfying the beam equation ›2 ›2 v ›v ›2 v r Axịb r Axị ỵ EIxị ỵ ẳ0 x2 x2 ›t ›t ð4:220Þ Determine damped natural frequencies, modal damping ratios, and the response vðx; tÞ as a modal series expansion, given vx; 0ị ẳ dxị and v_ x; 0ị ¼ sðxÞ: Solution Substitute the separable solution vðx; tÞ ¼ YðxÞqðtÞ ð4:221Þ into Equation 4.220 We obtain d2 d2 YðxÞ qtị ỵ rAxịbYxị_qtị ỵ rAxịYxịqtị ẳ EI dx2 dx Group the functions of x and t separately and equate to the same constant v2 ; as usual: d2 d2 Yxị EI dx2 ẳ q tị ỵ b_qtị ẳ v2 dx rAxịYxị qtị 4:222ị We have d2 d2 Yxị v2 rAYxị ẳ EI dx2 dx 4:223ị q tị ỵ b_qtị ỵ v2 qtị ẳ ð4:224Þ and Note that Equation 4.223 is identical to that for the undamped beam Hence, with known BCs, we will obtain the same mode shapes Yi ðxÞ and the same undamped natural frequencies vi in the usual manner However, the equation of modal generalized coordinates qðtÞ given by Equation 4.224 is different from that for the undamped case ðb ¼ 0Þ: We write, for mode i q€ i ðtÞ þ 2zi vi q_ i ðtÞ þ v2i qi ðtÞ ¼ ð4:225Þ where zi ¼ b 2vi is the modal damping ratio for mode i: Damped natural frequencies are q vdi ẳ z2i vi 4:226aị 4:226bị Equation 4.225 can be solved in the usual manner, with initial conditions qi ð0Þ and q_ i ð0Þ determined a priori, using known vx; 0ị and v_ x; 0ị: â 2005 by Taylor & Francis Group, LLC 4-52 Vibration and Shock Handbook The modal series solution is vx; tị ẳ The initial conditions are X Yi ðxÞqi ðtÞ X Yi ðxÞqi 0ị ẳ dxị X Yi xị_qi 0ị ẳ sxị 4:227ị ð4:228Þ ð4:229Þ Multiply Equation 4.228 and Equation 4.229 by rAðxÞYj xị and integrate from x ẳ to l using the orthogonality condition ( ðl for i – j 4:230ị rAxịYi xịYj xịdx ẳ aj for i ẳ j We obtain qj 0ị ẳ l dxịrAxịdx aj 4:231ị q_ j 0ị ẳ l sxịrAxịdx aj ð4:232Þ This completes the solution for the free damped beam The forced damped case can be analyzed in the same manner as for the forced undamped case because the mode shapes are the same 4.7 Vibration of Membranes and Plates The cables, rods, shafts, and beams whose vibration we have studied thus far in this chapter are one-dimensional members or line structures These continuous members need one spatial variable ðxÞ; in addition to the time variable ðtÞ; as an independent variable to represent their governing equation of motion Membranes and plates are two-dimensional members or planar structures They need two independent spatial variables (x and y) in addition to time ðtÞ; for representing their dynamics A membrane may be interpreted as a two-dimensional extension of a string or cable In particular, it has to be in tension and cannot support any bending moment A plate is a two-dimensional extension of a beam It can support a bending moment Their governing equations will, therefore, resemble twodimensional versions of their respective one-dimensional counterparts Modal analysis will also follow the familiar steps, after accounting for the extra dimension In this section, we will give an introduction to the modal analysis of membranes and plates For simplicity, only special cases of rectangular members with relatively simple BCs will be considered Analysis of more complicated boundary geometries and conditions will follow analogous procedures, but requires a greater effort and produces more complicated results 4.7.1 Transverse Vibration of Membranes Consider a stretched membrane (in tension) that lies on the x–y plane, as shown in Figure 4.19 Transverse vibration vðx; y; tÞ in the z-direction is of interest By following a procedure that is somewhat analogous to the derivation of the cable equation, we can obtain the governing equation as " # ›2 vðx; y; tÞ 2 ẳc ỵ vx; y; tị 4:233ị ›t ›x2 ›y © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-53 a x b z, v(x,t) y FIGURE 4.19 A membrane or a plate in Cartesian coordinates with s T0 cẳ m0 4:234ị where T ¼ tension per unit length of membrane section (assumed constant) m0 ¼ mass per unit surface area of membrane For modal analysis, we seek a separable solution of the form vx; y; tị ẳ YxịZyịqtị Substitute Equation 4.235 into Equation 4.233 We obtain " # d2 YðxÞ d2 Zyị YxịZyịqtị ẳ c ỵ Yxị dx2 Zyị dy2 ð4:235Þ ð4:236Þ Since Equation 4.236 is true for all possible values of t; x; and y that are independent, the three function groups should separately equal the constants; thus d2 Y ẳ 2a2 Yxị dx2 or d2 Yxị ỵ a2 Yxị ẳ dx2 4:237ị d2 Zyị ẳ 2b2 Zyị dy or d2 Zyị ỵ b2 Zyị ẳ dy2 4:238ị q tị ẳ 2v2 qtị or q tị ỵ v2 qtị ẳ 4:239ị with v2 ẳ c2 a2 ỵ b2 ị 4:240ị The argument for using positive constants a2 ; b2 ; and v2 is similar to that we gave for the onedimensional case Next, Equation 4.237 and Equation 4.238 have to be solved using two end conditions for each direction, as usual This will provide an infinite number of solutions and bj ; and the corresponding natural frequencies v2ij ẳ ca2i ỵ b2j ị1=2 for i ẳ 1; 2; 3; and j ¼ 1; 2; 3; … along with the mode shape components Yi ðxÞ and Zj ðyÞ for the two dimensions â 2005 by Taylor & Francis Group, LLC 4:241ị 4-54 4.7.2 Vibration and Shock Handbook Rectangular Membrane with Fixed Edges Consider a rectangular membrane of length a and width b as shown in Figure 4.19 and with the four edges xed The BCs are v0; y; tị ẳ 0; va; y; tị ẳ 0; vx; 0; tị ẳ 0; vx; b; tị ẳ Using these in solving Equation 4.237 and Equation 4.238, as usual, we obtain Yi ðxÞ ¼ sin x Zj ðyÞ ¼ sin bj y " vij ẳ cẵa2i ỵ b2j 1=2 i2 j2 ẳ pc ỵ a b ip a jp with bj ¼ b with ¼ #1=2 for i ¼ 1; 2; 3; … and j ¼ 1; 2; ð4:242Þ ð4:243Þ ð4:244Þ Note that the spatial mode shapes are given by Yj xịZj yị ẳ sin 4.7.3 ipx jpy sin a b ð4:245Þ Transverse Vibration of Thin Plates Consider a thin plate of thickness h in a Cartesian coordinate system as shown in Figure 4.19 The usual assumptions as for the derivation of the Bernoulli–Euler beam equation are used In particular, h is assumed small compared with the surface dimensions (a and b for a rectangular plate) Then, shear deformation and rotatory inertia can be neglected, and also normal stresses in the transverse direction ðzÞ can be neglected Furthermore, any end forces in the planar directions (x and y) are neglected The governing equation is " #2 ›2 vðx; y; tị 2 4:246ị ỵc ỵ vx; y; tị ẳ t x y with c2 ¼ E0 I Eh2 ¼ rA 12ð1 n2 Þr ð4:247Þ E ð1 n2 Þ ð4:248Þ where E0 ¼ I0 ¼ h3 ¼ second moment of area per unit length of section 12 4:249ị A0 ẳ h ¼ area per unit length of section r ¼ mass density of material E ¼ Young’s modulus of elasticity of the plate material n ¼ Poisson’s ratio of the plate material If we attempt modal analysis by assuming a completely separable solution of the form vx; y; tị ẳ YðxÞZðyÞqðtÞ in Equation 4.246, a separable grouping of functions of x and y will not be achieved in general However, the space and the time will be separable in modal motions Hence, we seek a solution of the form vðx; y; tị ẳ Yx; yịqtị â 2005 by Taylor & Francis Group, LLC ð4:250Þ Distributed-Parameter Systems 4-55 x y FIGURE 4.20 Mode i=1 , j=1 Mode i=2 , j=1 Mode i=1 , j=2 Mode i=2 , j=2 Mode i=3 , j=1 Mode i=3 , j=2 Mode shapes of transverse vibration of a simply supported rectangular plate We will obtain q tị ỵ v2 qtị ¼ ð4:251Þ 72 72 Yðx; yÞ l4 Yðx; yị ẳ 4:252ị and with the natural frequencies v given by v ẳ l2 c 4:253ị and 72 is the Laplace operator given by 72 ẳ â 2005 by Taylor & Francis Group, LLC 2 ỵ x ›y2 ð4:254Þ 4-56 Vibration and Shock Handbook and, hence, 72 72 is the biharmonic operation 74 given by 74 ¼ 4 4 ỵ2 2 ỵ x x ›y ›y ð4:255Þ The solution of Equation 4.252 will require two sets of BCs for each edge of the plate (as for a beam), but will be mathematically involved Instead of a direct solution, a logical trial solution that satisfies Equation 4.252 and the BCs is employed next for a simply supported rectangular plate The solution tried is in fact the correct solution for the particular problem 4.7.4 Rectangular Plate with Simply Supported Edges As a special case, we now consider a thin rectangular plate of length a; width b; and thickness h; as shown in Figure 4.19, whose edges are simply supported For each edge, the BCs are the displacement in zero and the bending moment about the edge zero Specifically, we have ! ›2 v 0 v ỵ n ẳ for x ¼ and a; # y # b vðx; y; tị ẳ and Mx ẳ E I x ›y ! ›2 v 0 › v þ n ¼ for y ¼ and b; # x # a ð4:256Þ vðx; y; tÞ ¼ and My ¼ E I ›y ›x where E0 and I are as given in Equation 4.248 and Equation 4.249 In this case, the mode shapes are found to be Yij x; yị ẳ sin ipx jpy sin a b for i ¼ 1; 2; … and j ẳ 1; 2; 4:257ị which clearly satisfy the BCs (Equation 4.256) and the governing model equation 4.252, with an infinite set of solutions for l given by ! j2 2 i lij ẳ p ỵ ð4:258Þ a2 b Hence, from Equation 4.253, the natural frequencies are i2 j2 vij ẳ p c ỵ a b ! ð4:259Þ where c is given by Equation 4.247 The overall response, then, is given by vx; y; tị ẳ 1 X X iẳ1 jẳ1 Aij sin vij t ỵ Bij cos vij t sin ipx ipy sin a b ð4:260Þ The unknown constants Aij and Bij are determined by the system initial conditions vðx; y; 0Þ and v_ ðx; y; 0Þ: The first six mode shapes of transverse vibration of a rectangular plate are sketched in Figure 4.20 Bibliography Benaroya, H 1998 Mechanical Vibration, Prentice Hall, Upper Saddle River, NJ Cao, Y., Modi, V.J., de Silva, C.W., and Misra, A.K., On the control of space platform-based variable geometry manipulator, Adv Astronaut Sci., 103, Part 1, 305– 326, 1999 Caron, M., Modi, V.J., Pradhan, S., de Silva, C.W., and Misra, A.K., Planar dynamics of flexible manipulators with slewing deployable links, J Guidance Control Dyn., 21, 4, 572–580, 1998 Chen, Y., Modi, V.J., de Silva, C.W., and Misra, A.K., On the performance of variable geometry manipulators: part I — three dimensional order N formulation, Acta Astronaut., 49, 1, –12, 2001 © 2005 by Taylor & Francis Group, LLC Distributed-Parameter Systems 4-57 Chen, Y., Modi, V.J., de Silva, C.W., and Misra, A.K., On the performance of variable geometry manipulators: part II — computer implementation and dynamical study, Acta Astronaut., 49, 1, 13– 30, 2001 Crandall, S.H., Karnopp, D.C., Kurtz, E.F., and Prodmore-Brown, D.C 1968 Dynamics of Mechanical and Electromechanical Systems, McGraw-Hill, New York den Hartog, J.P 1956 Mechanical Vibrations, McGraw-Hill, New York de Silva, C.W., A technique to model the simply supported Timoshenko beam in the design of mechanical vibrating systems, Int J Mech Sci., 17, 389 –393, 1975 de Silva, C.W., Dynamic beam model with internal damping, rotatory inertia and shear deformation, AIAA J., 14, 5, 676 –680, 1976 de Silva, C.W 2000 VIBRATION—Fundamentals and Practice, CRC Press, Boca Raton, FL de Silva, C W., Buyukozturk, O., and Wormley, D N., Postcracking compliance of RC beams, J Struct Div., Trans ASCE, 105, ST1, 35 –51, 1979 de Silva, C.W., and Wormley, D N., Material optimization in a torsional guideway transit system, J Adv Transp., 13, 3, 41 –60, 1979 de Silva, C.W., and Wormley, D.N., Torsional analysis of cutout beams, J Struct Div., Trans ASCE, 106, ST9, 1933–1946, 1980 de Silva, C.W., Wormley, D.N 1983 Automated Transit Guideways: Analysis and Design, D.C Heath and Co., Lexington, KY Dimarogonas, A 1996 Vibration for Engineers, 2nd ed., Prentice Hall, Upper Saddle River, NJ Inman, D.J 1996 Engineering Vibration, Prentice Hall, Englewood Cliffs, NJ Kim, S.W., Misra, A.K., Modi, V.J., Cyril, X., and De Silva, C.W., Modelling of contact dynamics of two flexible multi-body systems, Acta Astronaut., 45, 11, 669–677, 1999 Meirovitch, L 1986 Elements of Vibration Analysis, 2nd ed., McGraw-Hill, New York Rao, S S 1995 Mechanical Vibrations, 3rd ed., Addison-Wesley, Reading, MA Thomson, W.T., and Dahleh, M.D 1998 Theory of Vibration with Applications, 5th ed., Prentice Hall, Upper Saddle River, NJ Volterra E and Zachmanoglou, E.C 1965 Dynamics of Vibrations, Charles E Merrill Books, Columbus, OH Zhang, J., Modi, V.J., and De Silva, C.W., Parametric dynamical study of a novel manipulator, Adv Astronaut Sci., 110, 115–136, 2001 © 2005 by Taylor & Francis Group, LLC ... similarity with the cable vibration problem, and will conclude with an illustrative example © 2005 by Taylor & Francis Group, LLC 4-14 Vibration and Shock Handbook Box 4.1 TRANSVERSE VIBRATION CABLES... Taylor & Francis Group, LLC ›2 v ›t ð4:27Þ 4-10 Vibration and Shock Handbook where f x; tị ẳ for free vibration Now, with the assumption of small u; and by neglecting the second-order product term... Transverse vibration of a cable in tension; (b) motion of a general element © 2005 by Taylor & Francis Group, LLC 4-4 Vibration and Shock Handbook Using the small slope assumption we have sin u ứ u and