Vibrations Fundamentals and Practice ch06 Maintaining the outstanding features and practical approach that led the bestselling first edition to become a standard textbook in engineering classrooms worldwide, Clarence de Silva''s Vibration: Fundamentals and Practice, Second Edition remains a solid instructional tool for modeling, analyzing, simulating, measuring, monitoring, testing, controlling, and designing for vibration in engineering systems. It condenses the author''s distinguished and extensive experience into an easy-to-use, highly practical text that prepares students for real problems in a variety of engineering fields.
de Silva, Clarence W “Distributed Parameter Systems” Vibration: Fundamentals and Practice Clarence W de Silva Boca Raton: CRC Press LLC, 2000 Distributed-Parameter Systems Most of the vibration analysis examples encountered in the previous chapters assumed that the inertial (mass), flexibility (spring), and dissipative (damping) characteristics could be “lumped” as a finite number of “discrete” elements Such models are termed lumped-parameter or discreteparameter systems If all the mass elements move in the same direction, one has a one-dimensional system (i.e., “rectilinear” motion), with each mass having a single degree of freedom If the masses can move independently of each other, the number of degrees of freedom of such a one-dimensional system will be equal to the number of lumped masses, and will be finite In a “planar” system, each lumped mass will be able to move in two orthogonal directions and hence, will have two degrees of freedom; and similarly in a “spatial” system, each mass will have three degrees of freedom As long as the number of lumped inertia elements is finite, then one has a lumpedparameter (or, discrete) system with a finite degree of freedom Note that time is a system variable and not a system parameter Discrete-time models are used in computer analysis and simulation, regardless of whether the system is a discrete-parameter or continuous-parameter one Generally, in practical vibrating systems, inertial, elastic, and dissipative effects are found continuously distributed in one, two, or three dimensions Correspondingly, there are 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 and will require an infinite number of coordinates to represent its motion In other words, when extending the concept of a finite-degree-of-freedom system as analyzed previously, an infinite-dimensional vector is needed to represent the general motion of a continuous system Equivalently, a one-dimensional 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 The present chapter concerns vibration analysis of continuous systems 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 may be various, individual members may be nonuniform, and the material characteristics may be inhomogeneous and anistropic 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 vibration analysis of a few representative continuous members is discussed in this chapter The concepts of modal analysis can be extended from lumped-parameter systems to continuous systems In particular, because the number of principal modes is equal to the number of degrees of freedom of the system, a distributed-parameter system will have an infinite number of natural modes of vibration A particular mode can 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 ©2000 CRC Press 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 But, as in the case of discrete-parameter systems, the general motion can be analyzed and expressed in terms of modal motions, through modal analysis As discussed in Chapter 5, 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, all infinite number of natural frequencies and mode shapes are not significant and typically the very high modes can be neglected Such a modaltruncation procedure, although carried out by continuous-system analysis, is equivalent to approximating the original infinite-degree-of-freedom system by a finite-degree-of-freedom system Vibration analysis of continuous systems can be applied in 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 6.1 TRANSVERSE VIBRATION OF CABLES The first continuous member to be studied in this chapter is a string or cable in tension This is a line structure for which its 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) As usual, some simplifying assumptions will be made for analytical convenience; but 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 to 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 (y) The cable tension (T) remains constant during motion In other words, the initial tension is sufficiently large that the variations during motion are negligible Variations in slope (θ) along the structure are small Hence, for example, θ ≅ sin θ ≅ tan θ = ∂v ∂x A general configuration of a cable (or string) is shown in Figure 6.1(a) Consider a small element of length dx of the cable at location x, as shown in Figure 6.1(b) The equation (Newton’s second law) of motion (transverse) of this element is given by f ( x, t )dx − T sin θ + T sin(θ + dθ) = m( x ) ⋅ dx ©2000 CRC Press ∂ v( x , t ) ∂t (6.1) FIGURE 6.1 (a) Transverse vibration of a cable in tension, and (b) motion of a general element where v(x,t) f(x,t) m(x) T θ = = = = = transverse displacement of the cable lateral force per unit length of the cable mass per unit length of the cable cable tension cable slope at location x Note that the dynamic loading f(x,t) may arise due to such causes as aerodynamic forces, fluid drag, and electromagnetic forces, depending on the specific application Using the small slope assumption, one obtains sinθ ≅ θ and sin(θ + dθ) ≅ θ + dθ with ∂v ∂2v θ= and dθ = dx as dx → ∂x ∂x On substitution of these approximations into equation (6.1) and canceling out dx, one obtains m( x ) ∂ v( x , t ) ∂ v( x , t ) =T + f ( x, t ) ∂t ∂x (6.2) Now consider the case of free vibration, where f(x,t) = Then, ∂ v( x , t ) ∂ v( x , t ) = c ∂t ∂x ©2000 CRC Press (6.3) with c= T m (6.4) Also assume that the cable is uniform so that m is constant 6.1.1 WAVE EQUATION The solution to any equation of the form (6.3) will appear as a wave, traveling either in the forward (+ve x) or in the backward (–ve x) direction at speed c Hence, (6.3) is called the wave equation and c is the wave speed To prove this fact, first show that a solution to equation (6.3) can take the form v( x, t ) = v1 ( x − ct ) (6.5) To this end, let x – ct = z Hence, v1(x – ct) = v1(z) Then, ∂v1 dv1 ∂z = ⋅ ∂x dz ∂x and ∂v1 dv1 ∂z = ⋅ ∂t dz ∂t with ∂z =1 ∂x and ∂z = −c ∂t ∂ v1 = v1′′ ∂x and ∂ v1 = c v1′′ ∂t It follows that where v1′′ = d v1 ∂z Clearly then v1 satisfies equation (6.3) Now examine the nature of the solution v1(x – ct) It is clear that v1 will be constant when x – ct = constant But, the equation x – ct = 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 = will “appear” to travel along the cable at speed c This is analogous to the waves one observes in a pond when excited by dropping a stone Note that the particles of the cable not travel along x, and it is the deformation “shape” (the wave) that travels Similarly, it can be shown that v( x, t ) = v2 ( x + ct ) ©2000 CRC Press (6.6) is also a solution to equation (6.3), and this corresponds to a wave that travels backward (–ve x direction) at speed c The general solution, of course, will be of the form v( x, t ) = v1 ( x − ct ) + v2 ( x + ct ) (6.7) which represents two waves, one traveling forward and the other backward 6.1.2 GENERAL (MODAL) SOLUTION As usual in modal analysis, one looks for a separable solution of the form v( x , t ) = Y ( x ) ⋅ q ( t ) (6.8) for the cable/string vibration problem given by the wave equation (6.3) If a solution of the form of equation (6.8) is obtained, it will be essentially a modal solution This should be clear from the separability of the solution Specifically, at any given time t, the time function q(t) will be fixed, and the structure will have a shape give by Y(x) Hence, at all times, the structure will maintain a particular “shape” Y(x) and this will be a mode shape Also, at a given point x of the structure, the value of the space function Y(x) will be fixed, and the structure will vibrate according to the time response q(t) It will be shown that q(t) 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 (6.8), with different natural frequencies The corresponding functions Y(x) 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 of the form (6.8) will be a complete set With this qualitative understanding, one can now seek a solution of the form of equation (6.8) for the system equation (6.3) Substitute equation (6.8) in (6.3) to obtain Y ( x) d q (t ) d Y (α ) = c q (t ) dt dx or d 2Y ( x ) d q (t ) = = −λ2 Y ( x ) dx c q(t ) dt (6.9) In equation (6.9), because 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 –λ2, which is called the separation constant and is designated to be negative There are two good reasons for that If this common constant were positive, the function q(t) would be nonoscillatory and transient, which is contrary to the nature of undamped vibration Furthermore, it can be shown that a nontrivial solution for Y(x) would not be possible if the common constant were positive The unknown constant λ is determined by solving the space equation (mode shape equation) of (6.9); specifically, ©2000 CRC Press d 2Y ( x ) + λ2 Y ( x ) = dx (6.10) and then applying the boundary conditions of the problem There will be an infinite number of solutions for λ, with corresponding natural frequencies ω and mode shapes Y(x) The characteristic equation of (6.10) is p + λ2 = (6.11) which has the characteristic roots (or eigenvalues) p = ± jλ (6.12) The general solution is Y ( x ) = A1e jλx + A2 e − jλx = C1 cos λx + C2 sin λx (6.13) Note that, since Y(x) 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 λx = e jλx + e − jλx e jλx − e − jλx and sin λx = , one can show that 2j A1 = (C − jC2 ) and A2 = (C + jC2 ) For analytical convenience, use the real-parameter form of equation (6.13) Note that one cannot determine both constants C1 and C2 using boundary conditions Only their ratio is determined, and the constant multiplier is absorbed into q(t) in equation (6.8) and then determined using the appropriate initial conditions (at t = 0) It follows that the ratio of C1 and C2 and the value of λ are determined using the boundary conditions Two boundary conditions will be needed Some useful situations and appropriate relations are given in Table 6.1 6.1.3 CABLE WITH FIXED ENDS One can obtain the complete solution for free vibration of a taut cable that is fixed at both ends The applicable boundary conditions are Y ( ) = Y (l ) = where l is the length of the cable Substitution into equation (6.13) gives C1 × + C2 × = C1 cos λl + C2 sin λl = Hence, ©2000 CRC Press (6.14) TABLE 6.1 Some Useful Boundary Conditions for the Cable Vibration Problem Type of End Condition Nature of End x = xo Boundary Condition Modal Boundary Condition v( x o , t ) = Yi ( x o ) = ∂v( x o , t ) dYi ( x o ) Fixed Free Flexible Flexible and inertial T T ∂x ∂v( x o , t ) ∂x T C1 = dx − kv( x o , t ) = ∂v( x o , t ) =M =0 ∂x − kv( x o , t ) ∂ v( x o , t ) ∂t T T dYi ( x o ) dx dYi ( x o ) dx ( =0 − kYi ( x o ) = ) − k − ω i2 M Yi ( x o ) = and C2 sin λl = (6.15) A possible solution for equation (6.15) is C2 = But this is the trivial solution, which corresponds to Y(x) = 0; that is, a stationary cable with no vibration It follows that the applicable, nontrivial solution is sin λl = ©2000 CRC Press which produces an infinite number of solutions for λ given by λi = iπ l with, i = 1, 2, K, ∞ (6.16) As mentioned earlier, the corresponding infinite number of mode shapes is given by Yi ( x ) = Ci sin iπx l (6.17) for i = 1, 2, , ∞ Note: If one had used a positive constant λ2 instead of –λ2 in equation (6.9), only a trivial solution (with C1 = and C2 = 0) would be possible for Y(x) This further justifies the decision to use –λ2 Substitute equation (6.16) into (6.9) to determine the corresponding time response (generalized coordinates) qi(t); thus, d qi ( t ) + ω i2 qi (t ) = dt (6.18) iπ l (6.19) in which ωi = λic = T m for i = 1, 2, K, ∞ Equation (6.18) represents a simple harmonic motion with the modal natural frequencies ωi given by equation (6.19) It follows that there are an infinite number of natural frequencies, as mentioned earlier The general solution of equation (6.18) is given by qi (t ) = ci sin(ω i t + φ i ) (6.20) where the amplitude parameter ci and the phase parameter φi are determined using two initial conditions of the system It should be clear that it is redundant to use a separate constant Ci for Yi(x) in equation (6.17) as it can be absorbed into the amplitude constant in equation (6.20), to express the general free response of the cable as v( x , t ) = ∑ c sin i iπx sin(ω i t + φ i ) l (6.21) 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 was obtained above is a complete set of normal modes (normal in the sense of perpendicular or orthogonal) The system is able to move independently in each mode, with a unique spatial shape, at the corresponding natural frequency, because each modal solution is separable into a space function Yi(x) and a time function (generalized coordinate) qi(t) Of course, the system will be able to simultaneously move in a linear combination of two modes (say, C1Y1(x)q1(t) + C2Y2(x)q2(t)) because this combination satisfies the original system equation (6.3), in view of its linearity, and because each modal component satisfies the equation But, clearly, this solution (with two modes) is not separable into a product of a space ©2000 CRC Press function and a time function Hence, it is not a modal solution In this manner, it can be argued ∑ that the infinite sum of modal solutions ci Yi ( x )qi (t ) is the most general solution to system (6.3) Orthogonality of mode shapes plays a key role in this argument and, furthermore, it is useful in the analysis of the system In particular, in equation (6.21), the unknown constants ci and φi are determined using the system initial conditions, and the orthogonality property of modes is useful in that procedure This will be addressed next 6.1.4 ORTHOGONALITY OF NATURAL MODES A cable can vibrate at frequency ωi called natural frequency while maintaining a unique natural shape Yi(x) called mode shape of the cable It has been shown that for the fixed-ended cable, the iπx natural mode shapes are given by sin , with the corresponding natural frequencies ωi given by l equation (6.19) It can be easily verified that l ∫ sin iπx jπx sin dx = l l for i ≠ j l = for i = j (6.22) In other words, the natural modes are orthogonal Equation (6.22) 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 The orthogonality holds for other types of boundary conditions as well To show this, one sees from equation (6.9) that d Yi ( x ) + λ2i Yi ( x ) = dx d Yj ( x ) dx + λ2j Yj ( x ) = for mode i (6.23) for mode j (6.24) Multiply equation (6.23) by Yj(x), equation (6.24) by Yi(x), subtract that second result from the first and integrate with respect to x along the cable length from x = to l One obtains l ∫ d Yi d Yj − Yi dx + λ2i − λ2j Yj 2 dx dx ( l )∫ Y Y dx = i j Integrating by parts, one obtains the results l ∫ ©2000 CRC Press Yj d Yi dY dx = Yj i dx dx l l − ∫ dYi dYj dx dx dx (6.25) Z j ( y) = sin β j y [ ω ij = c α + β i j ] 12 i2 j2 = πc + b a with β j = jπ b (6.243) 12 for i = 1, 2, 3, K and j = 1, 2, 3, K (6.244) Note that the spatial mode shapes are given by Yj ( x ) Z j ( y) = sin 6.6.3 TRANSVERSE VIBRATION OF iπx jπy sin a b (6.245) THIN PLATES Consider a thin plate of thickness h in a Cartesian coordinate system as shown in Figure 6.19 The usual assumptions as for the derivation of the Bernoulli-Euler beam equation are used In particular, h is assumed small compared to the surface dimensions (a and b for a rectangular plate) Then, shear deformation and rotatory inertia can be neglected, and normal stresses in the transverse direction (z) can also be neglected Furthermore, any end forces in the planar directions (x and y) are neglected The governing equation is ∂ v( x, y, t ) ∂2 ∂2 + c + v( x, y, t ) = ∂t ∂y ∂x (6.246) with c2 = E ′I ′ Eh = ρA′ 12 − υ ρ ( ) (6.247) where E′ = I′ = A′ ρ E υ = = = = E − υ2 ( (6.248) ) h3 = second moment of area per unit length of section 12 area per unit length of section mass density of material Young’s modulus of elasticity of the plate material Poisson’s ratio of the plate material (6.249) If one attempts modal analysis by assuming a completely separable solution of the form v(x,y,t) = Y(x)Z(y)q(t) in equation (6.246), a separable grouping of functions of x and y will not be achieved in general But, the space and the time will be separable in modal motions Hence, seek a solution of the form v( x, y, t ) = Y ( x, y)q(t ) ©2000 CRC Press (6.250) One will get q˙˙(t ) + ω q(t ) = (6.251) ∇ ∇ Y ( x, y) − λ4 Y ( x, y) = (6.252) and with the natural frequencies ω given by ω = λ2 c (6.253) and ∇2 is the Laplace operator given by ∇2 = ∂2 ∂2 + 2 ∂x ∂y (6.254) and, hence, ∇2∇2 is the biharmonic operator ∇4 given by ∇4 = ∂4 ∂4 ∂4 + + ∂x ∂x ∂y ∂y (6.255) Solution of equation (6.252) will require two sets of boundary conditions 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 (6.252) and the boundary conditions is employed next for a simply supported rectangular plate The solution tried is in fact the correct solution for the particular problem 6.6.4 RECTANGULAR PLATE WITH SIMPLY SUPPORTED EDGES As a special case, now consider a thin rectangular plate of length a, width b, and thickness h, as shown in Figure 6.19, whose edges are simply supported For each edge, the boundary conditions are that the displacement is zero and the bending moment about the edge is zero Specifically, v( x, y, t ) = v( x, y, t ) = and ∂2v ∂2v M x = E ′I ′ + v = ∂y ∂x and ∂2v ∂2v M y = E ′I ′ + v = ∂x ∂y for x = and a; ≤ y ≤ b (6.256) for y = and b; ≤ x ≤ a where E′ and I′ are given by equations (6.248) and (6.249), respectively In this case, the mode shapes are found to be Yij ( x, y) = sin iπx jπy sin a b for i = 1, 2, K and j = 1, 2, K (6.257) which clearly satisfy the boundary conditions (6.256) and the governing model equation (6.252) There exists an infinite set of solutions for λ given by ©2000 CRC Press FIGURE 6.20 Mode shapes of transverse vibration of a simply supported rectangular plate i2 j2 λ2ij = π + b a (6.258) and, hence, from equation (6.253), the natural frequencies are i2 j2 ω ij = π c + b a (6.259) where c is given by equation (6.247) The overall response, then, is given by v( x, y, t ) = ∞ ∞ ∑ ∑ [ A sin ω t + B cos ω t] sin ij i =1 j =1 ij ij ij iπx iπy sin a b (6.260) The unknown constants Aij and Bij are determined by the system initial conditions v(x,y,0) and v˙(x,y,0) The first six mode shapes of transverse vibration of a rectangular plate are sketched in Figure 6.20 ©2000 CRC Press PROBLEMS 6.1 Consider the traveling wave solution given by v( x, t ) = v1 ( x − ct ) + v2 ( x + ct ) to the problem of free transverse vibration of a taut cable (wave equation): ∂2v ∂ v = c ∂t ∂x 6.2 6.3 6.4 Suppose that the system is excited by an initial displacement d(x) and an initial speed s(x) Show that the functions v1(x – ct) and v2(x + ct) can be completely determined by these two initial conditions This also means that the traveling wave solution does not depend on the boundary conditions of the system Under what conditions would this assumption be satisfied and, hence, the solution be valid? Provide some justification for the solution a List several differences between lumped-parameter (discrete) and distributed-parameter (continuous) vibratory systems b Consider a cable of length l that is stretched across a flexible pole and a rigid pole, modeled as shown in Figure P6.2 The flexible pole is represented by a spring of stiffness k The cable is assumed to have a uniform cross section, with a mass of m per unit length The cable tension T is assumed constant Analyze the problem of free transverse vibration of the cable, giving appropriate boundary actions Show that the orthogonality property is satisfied by the mode shapes of transverse vibration of a taut cable under a flexibly supported end with stiffness k The cord of a musical instrument is mounted horizontally with fixed ends and maintained in tension T The length of the cord is l and mass per unit length is m An impulsive speed of aδ x − is applied to the stationary cord in the transverse direction; for 2 FIGURE P6.2 Cable with flexible and rigid supports ©2000 CRC Press example, by gently hitting the mid-span with a hammer Determine the resulting vibration of the cord Note: δ(x – x0) is the Dirac delta function such that b ∫ f ( x)δ( x − x )dx = f ( x ) 0 a 6.5 where x0 is within the integration interval [a,b] and f(x) is an arbitrary function a Discuss why boundary conditions have to be known explicitly in solving a continuous vibrating system, but not for solving a discrete vibrating system b Determine the modal boundary condition at x = l for a rod carrying a mass M and restrained by a fixture of stiffness k, as shown in Figure P6.5 The area of cross section of the rod is A, and the Young’s modulus is E Show that the orthogonality property, in the conventional form, is not satisfied by the mode shapes of the rod, and that the following modified orthogonality condition is satisfied: l ∫ ρI Xi X j dx + MXi (l ) X j (l ) = for i ≠ j FIGURE P6.5 A rod with a dynamic restraint FIGURE P6.6 Longitudinal vibration of a rod with an end mass FIGURE P6.7 A helical spring with dominant axial deformation in the coil wire ©2000 CRC Press 6.6 6.7 How would the result change if I = I(x) and if M is located at x = 0? a What factors govern the use of the discrete-parameter assumption for a continuousparameter vibrating system? b A uniform rod of length l, cross-sectional area A, mass density ρ, and Young’s modulus E is fixed horizontally to a rigid wall at one end and carries a mass M at the other end, as shown in Figure P6.6 Analyze the system to determine the natural frequencies of longitudinal vibration When a helical (coil) spring is stretched, the coil wire will be subjected to bending and torsion at each cross section, as well as tension along the axis of the wire The former effects can dominate unless the spring is almost stretched or loosely wound so that the coil pitch (distance between adjacent coils turns) is large compared to the coil diameter and that the coil diameter is not large compared to the wire diameter Assume that for a (somewhat unusual) helical spring, the bending and torsion effects are negligible compared to the longitudinal deflection of the wire For such a spring with one end fixed and the other end free, as shown in Figure P6.7, determine the stiffness k and the natural frequencies of vibration The following parameters are given: M l A E 6.8 6.9 = = = = mass of the spring nominal, relaxed length of the coil wire area of cross section of the coil wire Young’s modulus of the coil material Consider a uniform structural column of height l, mass density ρ, and Young’s modulus E, and mounted on a rigid base, as schematically shown in Figure P6.8 Two types of initial conditions are considered: a An impulsive impact is made at the top end of the column so as to impart an instantaneous velocity of v0δ(x – l) at that point b The column is pressed down at the top through a displacement u0 and released suddenly from rest In each case, determine the subsequent longitudinal vibration of the column A uniform metal post of height l, mass density ρ, and Young’s modulus E is fixed vertically on a rigid floor as schematically shown in Figure P6.9 The top end of the post is harmonically excited using a shaker device in the following two ways: FIGURE P6.8 A structural column in longitudinal vibration: (a) initial impulsive impact at the top end, and (b) releasing from rest with initial elastic displacement ©2000 CRC Press FIGURE P6.9 A post mounted on a rigid floor and excited at the top by a shaker a The shaker head is displacement-feedback controlled so that a sinusoidal displacement of u0sinωt is generated b The shaker head is force-feedback controlled so that a sinusoidal force of f0sinωt is generated In each case, determine the longitudinal vibratory displacement of the post, under steadystate conditions 6.10 Consider a uniform shaft, not necessarily of circular cross section, in torsional vibration The following shaft parameters are known: J Jt ρ G = = = = polar moment of area of the shaft cross section (about the axis of rotation) Saint Venant torsional parameter (equal to J for a circular cross section) mass density shear modulus Six sets of boundary conditions, as shown in Figure P6.10, are studied In cases (e) and (f), there is an element of moment of inertia Ie, about the common axis of rotation, attached to one end of the shaft For each case of boundary conditions, determine the natural frequencies and mode shapes of torsional vibration In cases (e) and (f), show that the orthogonality property, in the conventional form, is not satisfied, but the following modified orthogonality condition governs: l ∫ ρJΘ ( x)Θ ( x)dx + I Θ (l)Θ (l) = i j e i j for i ≠ j 6.11 A mounted drill can slide along its guidepost so as to engage the drill bit with a workpiece A schematic diagram is given in Figure P6.11(a) Suppose that the drill bit rotates at a constant angular speed ω0 prior to engagement At the instant of engagement, the power ©2000 CRC Press FIGURE P6.10 Some boundary conditions for torsional vibration of shafts is lost and the motor drive torque become zero Assuming a large workpiece, the drill bit can be assumed fixed at the engagement end, under these conditions, as represented in Figure P6.11(b) Analyze the resulting torsional vibrations in the drill bit The following parameters are known: Length of the drill bit = l Moment of inertia of the drive rotor = Ir The usual parameters G, ρ, J, and Jt of the drill bit 6.12 A vibration engineer proceeds to estimate the shear modulus G of an unknown material She prepares a uniform shaft of circular cross section and length 50.0 cm of the material, rigidly mounts one end on a heavy fixture, and leaves the other end free She excites the shaft and measures the fundamental natural frequency of torsion It is found to be 1.6 kHz Also, the density of the material is measured to be 7.8 × 103 kg m–3 a Indicate a possible method of exciting the shaft and measuring the frequency in this experiment b Estimate the value of G c Guess the material type 6.13 A grinding tool is modeled as in Figure P6.13 The drive torque that is applied at one end of the tool, through electromagnetic means, is given by T0 + Ta sinωt, where T0 is a constant denoting the steady torque There is a torque ripple of amplitude Ta and frequency ω The grinding process is represented by an energy dissipation as in a viscous torsional damper with damping constant b The tool cross section is circular and has a polar moment of area J about its axis of rotation The tool length is l, the mass density is ρ, and the shear modulus is G Determine the rotational motion θ(x,t) ©2000 CRC Press FIGURE P6.11 The problem of a failed drill FIGURE P6.13 Torsional vibration of a grinding tool of the tool at steady state, where x is measured from the driven end, along the tool axis, as shown 6.14 Consider the torsional guideway transit problem shown in Figure 6.8 and analyzed in Example 6.5 a If the crossing frequency ratio vc is small compared to 1, give an expression for the angle of twist θ(t) of the guideway at the kth vehicle suspension, when there are Nv vehicle suspensions moving along the particular guideway span Use only the first Nt modes in the modal summation b Neglecting the dynamics of the guideway, what is the angle of twist at a vehicle suspension, when only that suspension is on the guideway span? What is its maximum value θmax? c For a single suspension, determine an expression for e = (θ s − θ) × 100%, where θ is θ max the angle of twist obtained in (a), with Nt modes, when there is only one suspension vt on the guideway span Plot e versus the fractional vehicle location for the five l cases of Nt = 1, 2, 3, 4, and How many modes would be adequate for a “good” approximation in the present application? ©2000 CRC Press 6.15 An AC induction motor drives a pump through a shaft The free torsional vibration of the system is to be analyzed A schematic representation of this free system (i.e., with zero motor torque) is shown in Figure P6.15 It is not necessary to assume a circular cross section for the shaft The usual parameters G, Jt, ρ, J, and l are known for the shaft Also, given are: Im = moment of inertia of the motor rotor Ip = moment of inertia of the pump impeller a Neglecting bearing friction, formulate the problem and identify the boundary conditions b What are the modal boundary conditions? c Giving all the steps and the necessary equations, describe how the natural frequencies and mode shapes of torsional vibration can be determined for this system d If the shaft is massless, what are the natural frequencies of the system? Show how these frequencies could be derived from the general solution given in (c), as the shaft inertia approaches zero 6.16 a A beam in transverse vibration is represented by the equation ∂ v( x , t ) ∂ v( x , t ) ∂2 EI + ρ A = f ( x, t ) ∂x ∂x ∂t i Define the parameters E, I, ρ, A, and the variables t, x, f, and v in this equation ii What are the assumptions made in deriving this equation? b Prove that the mode shape functions Yi(x) of the beam in part (a) are orthogonal c For a simply supported beam with uniform cross section, it can be shown that the mode shape functions are given by: Yi ( x ) = sin iπx l i = 1, 2, K Express the natural frequencies ωi corresponding to these modes, in terms of E, I, ρ, A, l, and the mode number i 6.17 a Compare and contrast linear, lumped-parameter systems and distributed-parameter systems, considering for example the nature of their i equations of motion ii natural frequencies iii mode shapes FIGURE P6.15 A pump driven by an induction motor ©2000 CRC Press What are possible practical problems that might arise when a distributed-parameter system is approximated by a lumped-parameter system? b Consider the Bernoulli-Euler beam equation given by ∂2 ∂2v ∂2v + = f ( x, t ) ρ A EI ∂x ∂x ∂t 6.18 6.19 i Define all the parameters and variables in this equation ii What are the assumptions made in deriving this equation? iii What are the analytical steps involved in solving for natural frequencies and mode shapes of this system? iv What analytical steps could be followed in obtaining the response of the beam to a specified forcing function f(x,t)? v Suppose that the beam is simply supported and a constant force F0 moves along the beam at constant speed u0 from one end to the other What is f(x,t) in this case? What are the modal boundary conditions of a thin beam in transverse vibration, with the end x = fixed (clamped) and the other end x = l sliding? Giving all the necessary steps, derive an equation for which the solutions provide the natural frequencies of transverse vibration of this cantilever What are the corresponding mode shape functions? Give three different forms of these eigenfunctions For mode shapes of a thin and non-uniform beam in bending vibration, show that the orthogonality condition l ∫ EI ( x ) 6.20 d Yi ( x ) d Yj ( x ) dx = dx dx holds under common boundary conditions The parameter EI represents flexural stiffness of the beam One way of normalizing the mode shapes Yi(x) of a beam in transverse vibration is according to l l ∫ ρAY dx = ∫ ρAdx i 6.21 for i ≠ j for all i Note that the right-hand-side integral is equal to the beam mass Determine the normalized mode shape functions according to this approach for a simply supported beam of uniform cross section A non-uniform beam is excited in the transverse direction by a distributed harmonic force of f ( x, t ) = f ( x ) sin ωt as sketched in Figure P6.21 Determine the resulting transverse response v(x,t) of the beam for initial conditions ©2000 CRC Press FIGURE P6.21 A beam excited by a transverse, distributed, harmonic force v( x , ) = d ( x ) 6.22 ∂v ( x , ) = s( x ) ∂t What is the steady-state response, assuming very light damping? Specialize your results to the case of a uniform simply supported beam Assume that the usual parameters E, I, ρ, A, and l of the beam are known Beams on elastic foundations are useful in many applications such as railroad tracks, engine baseblocks and mounts, and seismic motions of structures For a beam resting on an elastic foundation, the following parameters are defined: kf = = bf = = 6.23 and foundation elastic modulus force per unit length of the beam that causes a unit deflection in the foundation foundation damping modulus force per unit length of the beam that causes a unit velocity in the foundation Indicate how the transverse vibration of a beam can be modified to include kf and bf A schematic diagram of the system is shown in Figure P6.22 For the undamped case (bf = 0), explain how modal analysis could be performed for this system In particular, solve the case of a uniform, simply supported beam Consider a thin beam with a constant axial force P Perform a modal analysis for transverse vibration For the special case of a uniform, simply supported beam (with the usual parameters EI, ρ, A, l), obtain the complete solution giving all the natural frequencies and mode shapes Show that the natural frequencies increase due to the tensile force FIGURE P6.22 A beam on an elastic and dissipative foundation ©2000 CRC Press a Show that each natural frequency has two contributions: one for a beam without an axial force and the other for a cable that cannot support a bending moment In particular, show that when P = 0, one gets the former component; and when the beam cannot support a bending moment (i.e., EI = 0), one gets the latter component b Show that the compressive force Pcr at which the fundamental natural frequency of transverse vibration becomes zero is the first Euler buckling load for the beam 6.24 Consider a non-uniform and thin beam in bending vibration The left end of the beam has a translatory-dynamic boundary condition consisting of a mass m, a spring of stiffness k, and a linear viscous damper of damping constant b The right end of the beam has a rotatory-dynamic boundary condition consisting of a moment of inertia J, a spring of torsional stiffness K, and a linear viscous damper with rotatory damping constant B A schematic representation of the system is given in Figure P6.24 Assume that there are no translatory-dynamic effects (mass, translatory stiffness, translatory damping) due to the rotatory dynamic element at the right-hand end (which is not usually the case) a Express the boundary conditions of the beam b For the undamped case (b = 0, B = 0), determine the modal boundary conditions 6.25 Consider a circular shaft of length l, mass density ρ, shear modulus G, and polar moment of area J(x) carrying a circular disk of moment of inertia I at the far end Using the energy variational approach, determine the equation of torsional vibration of the system, including the boundary conditions for the case where the near end of the shaft is a fixed b free Now consider the modal analysis of a uniform circular shaft with its near end fixed and the far end carrying a disk The analysis was done in Problem 6.10(e) For the special case of I = lρJ, determine the fundamental natural frequency (ω1) and the corresponding normalized mode shape (Θ1(x)) 6.26 Using the energy variational approach, derive the terms that should be added to the governing equation and boundary conditions of a circular shaft with an end disk, in torsional vibration, as in Problem 6.25, for the following situations: a A distributed torque τ(x,t) per unit length along the shaft length b A point torque Te at x = l Consider a uniform circular shaft (ρ, l, G, J) with one end (x = 0) fixed and the other end (x = l) carrying a circular disk of moment of inertia I A harmonic point torque T0 sinωt is applied to the disk at x = l Determine the torsional response of the shaft at steady state A schematic diagram of the system is shown in Figure P6.26 i What are the critical excitation frequencies at which the steady-state response of the system would become very high? ii What is the amplitude of the angular response of the end disk at steady state? FIGURE P6.24 A beam with translatory-dynamic and rotatory-dynamic boundary conditions ©2000 CRC Press FIGURE P6.26 A shaft with an end disk excited by a harmonic torque 6.27 6.28 In thin beam theory of transverse vibration (Bernoulli-Euler beam), does the rotation of a beam element vary in phase with the displacement? In thick beam theory (Timoshenko beam), show that if one assumes that displacement and rotation of a beam element are in phase, it will lead to erroneous results Perform a modal analysis for transverse vibration of a thin damped beam given by ∂2 ∂2v ∂v ∂2v + + =0 EI x L ρ A x ( ) ( ) ∂x ∂x ∂t ∂t for the two damping models L a L = ∂v given by: ∂t ∂2 ∂2 * E I( x) 2 ∂x ∂x b L = c ∂ ∂x In (a), determine the damped natural frequencies and modal damping ratios if the energy dissipation in the beam material is represented by the hysteretic damping model: E* = 6.29 g1 + g2 ω Consider a single vehicle moving at constant speed p along a span of an elevated guideway Assume that the guideway is uniform with parameters EI and ρA; the span is simply supported at the support piers; and the vehicle is a point suspension with constant load W, inclusive of the vehicle weight A schematic diagram is given in Figure P6.29 FIGURE P6.29 A vehicle on an elevated guideway ©2000 CRC Press a Determine the transverse motion of the guideway as the vehicle travels along the span b What are the critical speeds that should be avoided by the vehicle? c If initially the span is at rest, what is the deflection of the guideway just underneath the vehicle, measured from the equilibrium configuration of the guideway? 6.30 An elevated guideway of a transit system consists of two-span single beam segments on three support piers, as shown in Figure P6.30 The span lengths are equal at l, and the ends of each two-span beam are simply supported In order to determine the guideway response to vehicles moving on it, first the natural frequencies and mode shapes of each guideway beam must be determined Clearly state the steps that need to be carried out in accomplishing this, giving the equations that need to be solved, along with appropriate boundary conditions Assume that the piers not receive bending moments from the guideway and that the guideway is always attached to the piers FIGURE P6.30 A single-beam, two-span elevated guideway segment ©2000 CRC Press ... 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 λx = e jλx + e − jλx e jλx − e − jλx and sin λx = , one can... determined, and the constant multiplier is absorbed into q(t) in equation (6.8) and then determined using the appropriate initial conditions (at t = 0) It follows that the ratio of C1 and C2 and the... for longitudinal and transverse vibrations of rods and beams, in the case of torsional vibration of shafts, the same equation of motion for circular shafts (equations (6.95) and (6.96)) cannot