Applied Structural and Mechanical Vibrations 2009 Part 8 pdf

8.2.1 which are specified at a given time (usually t=0)—must be satisfied at all times. Let us suppose that our string is attached to a rigid support at x=0 and extends indefinitely in the positive x-direction (semi-infinite string). This is probably the simplest type of boundary condition and it is not difficult to see that such a ‘fixed-end’ situation mathematically translates into (8.23) for all values of t: a condition which must be imposed on the general solution f(x–ct)+g(x+ct). The final result is that the incoming wave g(x+ct) is reflected at the boundary and produces an outgoing wave –g(x–ct) which is an exact replica of the original wave except for being upside down and travelling in the opposite direction. The fact that the original waveform has been reversed is characteristic of the fixed boundary. Another simple boundary condition is the so-called free end which can be achieved, for example, when the end of the string is attached to a slip ring of negligible mass m which, in turn, slides along a frictionless vertical post (for a string this situation is quite artificial, but it is very important in many other cases). In physical terms, we can write Newton’s second law stating that the net transverse force F y (0, t) (due to the string) acting on the ring is equal to Since and m is negligible, the free- end condition is specified by (8.24) which asserts that the slope of string at the free end must be zero at all times. By enforcing the condition (8.24) on the general d’Alembert solution, it is now easy to show that the only difference between the original and the reflected wave is that they travel in opposite directions: that is, the reflected wave has not been inverted as in the fixed-end case. Note that, as expected, in both cases—fixed and free end—the incoming and outgoing waves carry the same amount of energy because neither boundary conditions allow the string to do any work on the support. Other end conditions can be specified, for example, corresponding to an attached end mass, a spring or a dashpot or a combination thereof. Mathematically, all these conditions can be analysed by equating the vertical component of the string tension to the forces on these elements. For instance, if the string has a non-negligible mass m attached at x=0, the boundary condition reads (8.25) Copyright © 2003 Taylor & Francis Group LLC or, say, for a spring with elastic constant k 0 (8.26) Enforcing such boundary conditions on the general d’Alembert solution makes the problem somewhat more complicated. However, on physical grounds, we can infer that the incident wave undergoes considerable distortion during the reflection process. More frequently, the reflection characteristics of boundaries are analysed by considering the incident wave as pure harmonic, thus obtaining a frequency-dependent relationship for the amplitude and phase of the reflected wave. 8.3 Free vibrations of a finite string: standing waves and normal modes Consider now a string of finite length that extends from x=0 to x=L, is fixed at both ends and is subjected to an initial disturbance somewhere along its length. When the string is released, waves will propagate both toward the left and toward the right end. At the boundaries, these waves will be reflected back into the domain [0, L] and this process, if no energy dissipation occurs, will continually repeat itself. In principle, a description of the motion of the string in terms of travelling waves is still possible, but it is not the most helpful. In this circumstance it is more convenient to study standing waves, whose physical meaning can be shown by considering, for example, two sinusoidal waves of equal amplitude travelling in opposite directions, i.e. the waveform (8.27a) which, by means of familiar trigonometric identities, can be written as (8.27b) Two interesting characteristics of the waveform (8.27b) need to be pointed out: 1. All points x j of the string for which sin(kx j )=0 do not move at all times, i.e. y(x j , t)=0 for every t. These points are called nodes of the standing wave and in terms of the waveform (8.27a), we can say that whenever the crest of one travelling wave component arrives there, it is always cancelled out by a trough of the other travelling wave. Copyright © 2003 Taylor & Francis Group LLC 2. At some specified instants of time that satisfy all points x of the string for which reach simultaneously the zero position and their velocity has its greatest value. At other instants of time, when all the above points reach simultaneously their individual maximum amplitude value A sin(kx), and precisely at these times their velocity is zero. Among these points, the ones for which sin(kx)=1 are alternatively crests and troughs of the standing waveform and are called antinodes. In order to progress further along this line of reasoning, we must investigate the possibility of motions satisfying the wave equation in which all parts of the string oscillate in phase with simple harmonic motion of the same frequency. From the discussions of previous chapters, we recognize this statement as the definition of normal modes. The mathematical form of eq (8.27b) suggests that the widely adopted approach of separation of variables can be used in order to find standing- wave, or normal-mode, solutions of the one-dimensional wave equation. So, let us assume that a solution exists in the form y(x, t)=u(x)z(t), where u is a function of x alone and z is a function of t alone. On substituting this solution in the wave equation we arrive at which requires that a function of x be equal to a function of t for all x and t. This is possible only if both sides of the equation are equal to the same constant (the separation constant), which we call – ω 2 . Thus (8.28) The resulting solution for y(x, t) is then (8.29) where and it is easy to verify that the product (8.29) results in a series of terms of the form (8.27b). The time dependent part of the solution represents a simple harmonic motion at the frequency ω , whereas for the space dependent part we must require that (8.30) Copyright © 2003 Taylor & Francis Group LLC because we assumed the string fixed at both ends. Imposing the boundary conditions (8.30) poses a serious limitation to the possible harmonic motions because we get A=0 and the frequency equation (8.31) which implies (n integer) and is satisfied only by those values of frequency ω n for which (8.32) These are the natural frequencies or eigenvalues of our system (the flexible string of length L with fixed ends) and, as for the MDOF case, represent the frequencies at which the system is capable of undergoing harmonic motion. Qualitatively, an educated guess about the effect of boundary conditions could have led us to argue that, when both ends of the string are fixed, only those wavelengths for which the ‘matching condition’ (where n is an integer) applies can satisfy the requirements of no motion at x=0 and x=L. This is indeed the case and the allowed wavelengths satisfy etc. The first four patterns of motion (eigenfunctions) are shown in Fig 8.2: the motion for n=1, 3, 5,…result in symmetrical (with respect to the point x=L/2) modes, while antisymmetrical modes are obtained for n=2, 4, 6,… So, for a given value of n, we can write the solution as (8.33) where, for convenience, the constant of the space part has been absorbed in the constants A n and B n . Then, given the linearity of the wave equation, the general solution is obtained by the superposition of modes: (8.34) where the (infinite) sets of constants A n and B n represent the amplitudes of the standing waves of frequency ω n . The latter quantities, in turn, are related to the allowed wavenumbers by the equation On physical grounds, since we observed that there is no motion at the nodes and hence no energy flow between neighbouring parts of the string, one could ask at this point how a standing wave gets established and how it is maintained. To answer this question we must remember that a standing wave represents a steady-state situation; during the previous transient state (which, broadly speaking, we may call the ‘travelling wave era’) the nodes move and allow the transmission of energy along the string. Moreover, it should also be noted that nodes are not perfect in real strings where friction is present; they are only points of minimum amplitude of vibration. Copyright © 2003 Taylor & Francis Group LLC Fig. 8.2 Vibration of a string with fixed ends: (c) third and (d) fourth modes. which we recognize as Fourier series with coefficients A n and ω n B n , respectively. Following the standard methods of Fourier analysis, we multiply both sides of eqs (8.35) by sin k m x and integrate over the interval [0, L] in order to obtain, by virtue of (8.36) Copyright © 2003 Taylor & Francis Group LLC the expressions (8.37) which establish the motion of our system. Note that eq (8.34) emphasizes the fact that the string is a system with an infinite number of degrees of freedom, where, in the normal mode representation, each mode represents a single degree of freedom; furthermore, from the discussion above it is clear that the boundary conditions determine the mode shapes and the natural frequencies, while the initial conditions determine the contribution of each mode to the total response (or, in other words, the contribution of each mode to the total response depends on how the system has been started into motion). If, for example, we set the string into motion by pulling it aside at its centre and then letting it go, the ensuing free motion will comprise only the odd (symmetrical) modes; even modes, which have a node at the centre, will not contribute to the motion. A final important result must be pointed out: when the motion is written as the summation of modes (8.34), the total energy E of the string—i.e. the integral of the energy density over the length of the string—is given by (8.38) where it is evident that each mode contributes independently to the total energy, without any interaction with other modes (recall Parseval’s theorem stated in Chapter 2). The explicit calculation of (8.38), which exploits the relation (8.36) together with its cosine counterpart (8.39) is left to the reader. Copyright © 2003 Taylor & Francis Group LLC We close this section with a word of caution. Traditionally, cable vibration observations of natural frequencies and mode shapes are compared to those of the taut string model. However, a more rigorous approach must take into account the axial elasticity and the curvature of the cable (for example, power- line cables hang in a shape called ‘catenary’ and generally have a sag-to- span ratio between 0.02 and 0.05) and may show considerable discrepancies compared to the string model. In particular, the natural frequencies and mode shapes depend on a cable parameter (E=Young’s modulus, A=cross-sectional area, ρ g=cable weight per unit volume, L 0 =half-span length) and on the sag-to-span ratio. The interested reader may refer, for example, to Nariboli and McConnell [3] and Irvine [4]. 8.4 Axial and torsional vibrations of rods In the preceding sections we considered in some detail a simple case of continuous system—i.e. the flexible string. However, in the light of the fact that our interest lies mainly in natural frequencies and mode shapes, we note that we can explore the existence of solutions in which the system executes synchronous motions just by assuming a simple harmonic motion in time and asking what kind of shape the string has in this circumstance. This amounts to setting and substituting it into the wave equation to arrive directly at the first of eqs (8.28) so that, by imposing the appropriate boundary conditions (fixed ends), we arrive at the eigenvalues (8.32) and the eigenfunctions (8.40) where C n are arbitrary constants which, a priori, may depend on the index n. If now we consider the axial vibration of a slender rod of uniform density ρ and cross-sectional area A in presence of a dynamically varying stress field σ (x, t), we can isolate a rod element as in Fig. 8.3 and write Newton’s second law as (8.41) where y(x, t) is the longitudinal displacement of the rod in the x-direction. If we assume the rod to behave elastically, Hooke’s law requires that where is the axial strain, and upon substitution in eq (8.41) we get (8.42) Copyright © 2003 Taylor & Francis Group LLC and physical dimensions, the discussions of the preceding sections apply also for the cases above. If now we separate the variables and assume a harmonic solution in time, we arrive at the ordinary differential equations (8.46) where, for convenience, we called u(x) the spatial part of the solution in both cases (8.42) and (8.45) and the parameter γ 2 is equal to ω 2 ρ /E for the first of eqs (8.46) and to ω 2 ρ /G for the second. Again, in order to obtain the natural frequencies and modes of vibrations we must enforce the boundary conditions on the spatial solution (8.47) One of the most common cases of boundary conditions is the clamped-free (cantilever) rod where we have (8.48) so that substitution in (8.47) leads to which, in turn, translates into meaning that the natural frequencies are given by (8.49) for the two cases of axial and torsional vibrations, respectively. Like the Copyright © 2003 Taylor & Francis Group LLC eigenvectors of a finite DOF system, the eigenfunctions are determined to within a constant. In our present situation (8.50) where u n (x) must be interpreted as an axial displacement or an angle, depending on the case we are considering. If, on the other hand, the rod is free at both ends, the boundary conditions (8.51) lead to B=0 and so that and (8.52) are the eigenvalues for the two cases, respectively, while (8.53) (where γ n is as appropriate) are the eigenfunctions. Note that in the case of the free-free rod the solution with n=0 is a perfectly acceptable root and does not correspond to no motion at all (see the taut string for comparison). In fact, for n=0 we get and, from eq (8.46), so that where C 1 and C 2 are two constants whose value is irrelevant for our present purposes. Enforcing the boundary conditions (8.51) gives which corresponds to a rigid body mode at zero frequency. As for the discrete case, rigid-body modes are characteristic of unrestrained systems. For the time being, we do not consider other types of boundary conditions and we turn to the analysis of a more complex one-dimensional system—the beam in flexural vibration. This will help us generalize the discussion on continuous systems by arriving at a systematic approach in which the similarities with discrete (MDOF) systems will be more evident. Copyright © 2003 Taylor & Francis Group LLC [...]... from the combination and odd functions, which come from the combination In either case, if we fit the boundary conditions at x=L/2, they will also fit at x=–L/2 For the even functions the boundary conditions Copyright © 2003 Taylor & Francis Group LLC lead to the equation (8. 88a) and for the odd functions we obtain (8. 88b) Both eqs (8. 88a and b) must be solved numerically: from eq (8. 88a) we obtain the... in the previous cases, we called u(x) the spatial part of the solution If now we let eq (8. 81) yields where is positive and is negative It follows that we have the four roots ±η and where we defined (8. 82) The solution of eq (8. 81) can then be written as (8. 83) which is formally similar to eq (8. 58) but it must be noted that the hyperbolic functions and the trigonometric functions have different arguments... u(x) and slope du/dx both vanish at the clamped end, i.e (8. 63a) Copyright © 2003 Taylor & Francis Group LLC and that the bending moment and shear force the free end, i.e both vanish at (8. 63b) We recognize eqs (8. 63a) as geometric boundary conditions and eqs. (8. 63b) as natural boundary conditions Substitution of eqs (8. 63a and b) into (8. 58) gives (8. 64a) which can be arranged in matrix form as (8. 64b)... y From eq (8. 96) we get (8. 98) and by differentiating eq (8. 97) we obtain (8. 99) Copyright © 2003 Taylor & Francis Group LLC Substitution of eq (8. 98) into eq (8. 99) yields the desired result, i.e (8. 100) Now, a closer look at eq (8. 100) shows that: 1 The term arises from shear deformation and vanishes when the beam is very rigid in shear, i.e when 2 The term is due to rotatory inertia and vanishes... conditions are geometrical and read (8. 68) We can follow a procedure similar to the previous case to arrive at and to the frequency equation (8. 69) The first six roots of eq (8. 69) are (8. 70) Copyright © 2003 Taylor & Francis Group LLC which, for can be approximated by Note that the root of eq (8. 69) implies no motion at all, as the reader can verify by solving eq (8. 57) with and enforcing the boundary... Substitution of the four boundary conditions in eq (8. 58) leads to and to the frequency equation (8. 60) which implies and hence (8. 61) The eigenfunctions are then given by (8. 62) and have the same shape as the eigenfunctions of a fixed-fixed string Case 2 One end clamped and one end free (cantilever configuration) Suppose that the end at x=0 is rigidly fixed (clamped) and the end at x=L is free; then the boundary... from Fig 8. 5(b) and from a similar figure in the y–z plane (8. 107a) where s is the membrane mass per unit area and the angles θx and θy are given by ∂w/∂x and ∂w/∂y respectively Note that small deflections have been assumed, so that etc and the area of the deflected element can still be written as dxdy Equation (8. 107a) results in the equation of motion (8. 107b) Obviously, we can arrive at eq (8. 107b)... among the constants Cj and are given by (8. 71) where now (8. 72) Case 4 Both ends free (free-free configuration) The boundary conditions are now all of the force type, requiring that bending moment and shear force both vanish at x=0 and x=L, i.e (8. 73) which, upon substitution into eq (8. 58) give Equating the determinant of the 4×4 matrix to zero yields the frequency equation (8. 74) which is the same... solution of the form which gives and and solve the characteristic equation so that (8. 58) where the arbitrary constants Aj or Cj (j=1, 2, 3, 4) are determined from the boundary and initial conditions The calculation of natural frequencies and eigenfunctions is just a matter of substituting the appropriate boundary conditions in eq (8. 58) ; we consider now some simple and common cases Copyright © 2003... deflection alone and of rotatory inertia alone on the eigenvalues of a pinned-pinned beam Case 1 Shear deflection alone We neglect rotatory inertia in eq (8. 101) and obtain (8. 102) The solution of eq (8. 102) is formally analogous to the case of the beam with axial force: we obtain four roots ±η and where now we define (8. 103) and the allowed frequencies are obtained from the condition as (8. 104a) where . spatial part of the solution. If now we let eq (8. 81) yields where is positive and is negative. It follows that we have the four roots ± η and where we defined (8. 82) The solution of eq (8. 81) can. i.e. (8. 63b) We recognize eqs (8. 63a) as geometric boundary conditions and eqs. (8. 63b) as natural boundary conditions. Substitution of eqs (8. 63a and b) into (8. 58) gives (8. 64a) which can be arranged. four boundary conditions in eq (8. 58) leads to and to the frequency equation (8. 60) which implies and hence (8. 61) The eigenfunctions are then given by (8. 62) and have the same shape as the