.18Modeling for StructuralVibrations: FEM Models,Damping, Similarity Laws18–1Chapter 18: MODELING FOR STRUCTURAL VIBRATIONS: FEM MODELS, DAMPING, SIMILARITY LAWS18–2§18.1 FINITE ELEMENT MODELING OF VIBRATION PROBLEMSIn modeling of cable vibration problems by linear elements, it was observed that one needs to model the cableby more than 50 elements if the fundamental frequency is to be computed within 4 digit accuracy. What wasnot discussed therein is the accuracy of mode shapes. To gain further insight into finite element modeling ofvibration problems, let us consider the modeling of plane beam vibration problems. For simplicity, a beamwith simple-simple supports is used to model 5th modes. Classical theory tells us that we have the followingsolution:k-th mode: ωk=kπLEIρk-th mode shape: W(x) = sinkπ xL(18.1)where EI is the bending rigidity, ρ is the mass per unit beam length, and L is the beam span.5 10152025303540455010-410-310-210-1100101Number of Beam ElementsFrequency Error (%) Frequency error vs. elements for simple support beam6-th mode beta*L = 18.8497Figure 18.1 Frequency error vs. number of elements for simply-simply supported beam18–218–3 §18.1 FINITE ELEMENT MODELING OF VIBRATION PROBLEMS5 10152025303540455010-310-210-1100101Number of Beam ElementsFrequency Error (%) Frequency error vs. elements for fixed-simple support beam6-th mode beta*L = 19.6351Figure 18.2 Frequency error vs. number of elements for fixed-simply supported beamIn using the finite element method for modeling of beam vibrations, the question arises: how many elementsdoes one need for an accurate computation of its k-th mode and mode shape. While there exists a vastamount of literature on the accuracy and convergence properties of various elements which one may employto answer the question, we will adopt a posteriori assessment approach. To this end, let us take a simplysupported beam and concentrate on its 6th modes. The frequency error vs. the number of beam elementsused are shown in fig. 18.1. As can be seen, five-digit accuracy is achieved with about 45 elements. If allone needs is the frequency with one percent accuracy, one could use only 10 elements. In Figure 18.2 thefrequency error of a fixed-simply supported beam is illustrated. Although the error for this case is a littlehigher than that of the simply supported case, the frequency error converges with the same trend.let’s now focus on the mode shape accuracy.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam5 elements 6-th mode beta*L = 19.8826 Freq(Hertz) = 918.1601computed mode shapespline fitconverged mode shape0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam10 elements 6-th mode beta*L = 18.9243 Freq(Hertz) = 831.7825Figure 18.3 Frequency error vs. number of elements for fixed-simply supported beam18–3Chapter 18: MODELING FOR STRUCTURAL VIBRATIONS: FEM MODELS, DAMPING, SIMILARITY LAWS18–40 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam15 elements 6-th mode beta*L = 18.8652 Freq(Hertz) = 826.59610 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam30 elements 6-th mode beta*L = 18.8506 Freq(Hertz) = 825.3172Figure 18.4 Frequency error vs. number of elements for fixed-simply supported beam0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam40 elements 6-th mode beta*L = 18.8499 Freq(Hertz) = 825.2570 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1-1.5-1-0.500.511.5Beam spanMode shape Simple - Simple Support Beam50 elements 6-th mode beta*L = 18.8497 Freq(Hertz) = 825.2404Figure 18.5 Frequency error vs. number of elements for fixed-simply supported beamFigures 18.3-5 illustrate how the mode shapes converge to the exact solution as the number of elementsincreases. Also plotted is spline-fitted curves that utilize only the sample points (or computed discrete modeshape points). It is clear that curve fitting in general enhances the discrete raw data points, especially for thecase of crude models. Scanning over the six mode shape plots (Fig. 18.3-5) vs. the increasing number ofelements, an acceptable number of element for capturing the sixth mode shape appears to be 30. Note thatthere are six half sine waves in the 6th mode shape. Hence, for the 30-element model, each half sine waveis sampled by five elements or six nodal points, meaning that ten elements span one full sine wave.To conclude, ten elements captures the sixth mode frequency with less than one percent error, whereas for anadequate mode shape capturing three time of that elements (30 elements) are needed. This is often referredto as ”three-to-one rule.” Finally, the case of ten-element model satisfies the so-called Shannon’s samplingcriterion which state that for the minimum sampling number for a sinusoidal signal is three.18–418–5 §18.2 CHARACTERIZATION OF LINEAR STRUCTURAL DYNAMICS EQUATIONS§18.2 CHARACTERIZATION OF LINEAR STRUCTURAL DYNAMICS EQUATIONSLet us now study the characteristics of the second-order damped systemM¨u + D˙u + Ku = f(t), D = (αM + βK)(18.2)where α and β are constants. The damping matrix, D = αM + βK, is called a Rayleigh damping as it isproportional both to mass and stiffness of the system.The coupled equations of motion for a linear structure(18.2) can be decoupled by the (n × n) eigenvectormatrix T, which relates the displacement vector u to a generalized solution vector q viau = Tq (18.3)Substituting (18.3) into (18.2) and premultiplying the resulting equations by TTone obtainsI¨q + (αI + β)˙q +  q = fq, fq= TTf( f )(18.4)in whichTTMT = I (18.5)TTKT =  = diag(ω21, ..., ω2N)(18.6)where ωjis the j-th undamped frequency component of (18.4).The solution of (18.4) for its homegeneous part can be expressed asq = cest(18.7)where s is in general complex constants. Substitution of (18.7) into (18.4) with fq= 0 yields{s2I + (α I + β)s + }c = 0 (18.8)Equation (18.8) has a nontrivial solution only ifdet |s2I + (α I + β)s + |=0 (18.9)from which the k-th solution component can be expressed asqk= akeskt+¯ake¯skt(18.10)where skand¯skare the k-th complex conjugate pairs of (18.9) and akand¯akare arbitrary constants.Two