Hutton - Fundamentals of Finite Element Analysis (523155), страница 79
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Forexample, a bar element with damping is mathematically modeled as a linearspring and a dashpot connected in parallel to the element nodes as in Figure 10.15.The element damping matrix is (e) c −cc=(10.133)−c cand the element equations of motion are (e) m {ü} + c (e) {u̇} + k (e) {u} =f (e)!(10.134)The element damping matrix is symmetric and singular, and the individual termsare assigned to the global damping matrix in the same manner as the mass andstiffness matrices. Assembly of the global equations of motion for a finite element model of a damped structure is simple.
Determination of the effective viscous damping coefficients for structural elements is not so simple.Damping due to internal friction is known as structural damping, and experiments on many different elastic materials have shown that the energy loss permotion cycle in structural damping is proportional to the material stiffness andthe square of displacement amplitude [2].
That is,U cycle = k X 2(10.135)where is a dimensionless structural damping coefficient, k is the material stiffness, and X is the displacement amplitude. By equating the energy loss per cycleto the energy loss per cycle in viscous damping, an equivalent viscous dampingcoefficient is obtained:ceq =k(10.136)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.9 Energy Dissipation: Structural Dampingwhere is circular frequency of oscillation.
That the equivalent damping coefficient depends on frequency is somewhat troublesome, since the implication isthat different coefficients are required for different√ frequencies. If we consider asingle degree-of-freedom system for which = k/m, the equivalent dampingcoefficient given by Equation 10.136 becomes√kk= kmceq == √(10.137)k/mindicating that the damping coefficient is proportional, at least in a general sense,to both stiffness and mass.
We return to this observation shortly.Next we consider the application of the transformation using the normalized matrix as described in Section 10.7. Applying the transformation to Equation 10.132 results in{ p̈} + [ A] T [C ][ A]{ ṗ} + [ 2 ]{ p} = [ A] T {F (t )}(10.138)The transformed damping matrix[C ] = [ A] T [C ][ A](10.139)is easily shown to be a symmetric matrix, but the matrix is not necessarily diagonal. The transformation does not necessarily result in decoupling the equationsof motion, and the simplification of the mode superposition method is not necessarily available. If, however, the damping matrix is such that[C ] = ␣[M ] + [K ](10.140)where ␣ and  are constants, then[C ] = ␣[ A] T [M ][ A] + [ A] T [K ][ A] = ␣[I ] + [ 2 ](10.141)is a diagonal matrix and the differential equations of motion are decoupled.
Notethat the assertion of Equation 10.140 leads directly to the diagonalization of thedamping matrix as given by Equation 10.141. Hence, Equation 10.138 becomes{ p̈} + (␣ + [ 2 ]){ ṗ} + [ 2 ]{ p} = [ A] T {F (t )}(10.142)As the differential equations represented by Equation 10.142 are decoupled, letus now examine the solution of one such equationP (i)p̈i + ␣ + i2 ṗi + i2 pi =A j Fj (t )(10.143)j=1where P is the total number of degrees of freedom.
Without loss of generality andfor convenience of illustration, we consider Equation 10.143 for only one of theterms on the right-hand side, assumed to be a harmonic force such that p̈i + ␣ + i2 ṗi + i2 pi = F0 sin f t(10.144)and assume that the solution ispi (t ) = X i sin f t + Yi cos f t(10.145)429Hutton: Fundamentals ofFinite Element Analysis43010. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsSubstitution of the assumed solution into the governing equation yields − X i 2f sin f t − Yi 2f cos f t + ␣ + i2 f ( X i cos f t − Yi sin f t )+ i2 X i sin f t + i2 Yi cos f t = F0 sin f t(10.146)Equating coefficients of sine and cosine terms yields the algebraic equations& ' i2 − 2f− f ␣ + i2F0Xi =(10.147)Yi0 f ␣ + i2i2 − 2ffor determination of the forced amplitudes X i and Yi .
The solutions are F0 i2 − 2fXi = 2 2i2 − 2f + 2f ␣ + i2(10.148) − F0 f ␣ + i2Yi = 2 2i2 − 2f + 2f ␣ + i2To examine the character of the solution represented by Equation 10.145, weconvert the solution to the formpi (t ) = Z i sin( f t + i )withZi =(X i2 + Yi2and i = tan−1(10.149)YiXito obtainF0pi (t) = )sin( f t + i )*2 2i2 − 2f + 2f ␣ + i2$i = tan−1 %−2f ␣ + i2i2 − 2f(10.150)(10.151)Again, the mathematics required to obtain these solutions are algebraicallytedious; however, Equations 10.150 and 10.151 are perfectly general, in that theequations give the solution for every equation in 10.142, provided the appliednodal forces are harmonic.
Such solutions are easily generated via digital computer software. The actual displacements are then obtained by application ofEquation 10.112, as in the case of undamped systems.The equivalent viscous damping described in Equation 10.140 is known asRayleigh damping [6] and used very often in structural analysis. It can be shown,by comparison to a damped single degree-of-freedom system that␣ + i2 = 2i i(10.152)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.9 Energy Dissipation: Structural Damping431where i is the damping ratio corresponding to the i th mode of vibration, that is,i =␣i+2i2(10.153)represents the degree of damping for the i th mode.
Equation 10.153 provides ameans of estimating ␣ and  if realistic estimates of the degree of damping fortwo modes are known. The realistic estimates are most generally obtained experimentally or may be applied by rule of thumb. The following example illustrates the computations and the effect on other modes.EXAMPLE 10.9Experiments on a prototype structure indicate that the effective viscous damping ratiois = 0.03 (3 percent) when the oscillation frequency is = 5 rad/sec and = 0.1(10 percent) for frequency = 15 rad/sec. Determine the Rayleigh damping factors ␣and  for these known conditions.■ SolutionApplying Equation 10.153 to each of the known conditions yields␣5+2(5)2␣150.1 =+2(15)20.03 =Simultaneous solution provides the Rayleigh coefficients as␣ = −0.0375 = 0.01350.250.20.15i0.10.050⫺0.051357911 13 15 17 19 21 23 25 27 29iFigure 10.16 Equivalent damping factor versus frequency forExample 10.9.Hutton: Fundamentals ofFinite Element Analysis43210.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsIf we were to apply the equivalent damping given by these values to the entire frequencyspectrum of a structure, the effective damping ratio for any mode would be given byi =−0.0375 + 0.0135 i22iIf the values of ␣ and  are applied to a multiple degrees-of-freedom system, the damping ratio for each frequency is different. To illustrate the variation, Figure 10.16 depictsthe modal damping ratio as a function of frequency. The plot shows that, of course, theratios for the specified frequencies are exact and the damping ratios vary significantly forother frequencies.Rayleigh damping as just described is not the only approach to structuraldamping used in finite element analysis.
Finite element software packages alsoinclude options for specifying damping as a material-dependent property, asopposed to a property of the structure, as well as defining specific dampingelements (finite elements) that may be added at any geometric location in the structure. The last capability allows the finite element analyst to examine the effects ofenergy dissipation elements as applied to specific locations.10.10 TRANSIENT DYNAMIC RESPONSEIn Chapter 7, finite difference methods for direct numerical integration of finiteelement models of heat transfer problems are introduced. In those applications,we deal with a scalar field variable, temperature, and first-order governing equations.
Therefore, we need only to develop finite difference approximations to firstderivatives. For structural dynamic systems, we have a set of second-order differential equations¨ + [C ]{␦}˙ + [K ]{␦} = {F (t )}[M ]{␦}(10.154)representing the assembled finite element model of a structure subjected to general (nonharmonic) forcing functions. In applying finite difference methods toEquation 10.154, we assume that the state of the system is known at time t andwe wish to compute the displacements at time t + t ; that is, we wish to solve¨ + t )} + [C ]{␦(t˙ + t )} + [K ]{␦(t + t )} = {F (t + t )}[M ]{␦(t(10.155)for {␦(t + t )}.Many finite difference techniques exist for solving the system of equationsrepresented by Equation 10.155.
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