.19Modeling of DynamicalSystems: Discrete Models19–1Chapter 19: MODELING OF DYNAMICAL SYSTEMS: DISCRETE MODELS 19–2§19.1 INTRODUCTIONWith a plethora of engineering analysis software packages available, the dynamics specialist canperformavarietyofcomputerizedmodelingandsimulationofcomplexsystems. Itisnotuncommonthese days that a typical structural dynamics model may consist of several thousands to severalmillionsofdegreesoffreedom, oftenemployingthefiniteelement methodasthedominantmodelingtool. As a result, what used to be a challenge to the dynamics specialist several decades ago, viz.,more elaborate models that result in a large number of degrees of freedom, has become routinepractices.At present, the most time-consuming part of engineering analysis including dynamical systemsis model development tasks. While still significant, the core of computational effort has becomerelatively insignificant thanks to the advances in computational power known in the computer worldas Gordon Moor’s law. To the engineer, a new challenge has emerged: how to interpret a vast arrayof analysis results in terms of plots, tables and other graphical data representations. A first stepto a meaningful interpretation of computer-generated analysis results is to cultivate an ability toreduce the complex systems to a set of simplified models, from which one can relate the observedphenomena to the complex models. This lecture is aimed at introducing to the students how toconstruct simple models, how to interpret the simple model results, and how to relate the resultsobtained by simple models to complex simulation models.§19.2 ONE-DOF MODELING OF CABLE VIA THE FINITE ELEMENT METHODSuppose we do not know how to obtain a sensible simple discrete model and we are asked toconstruct a single, two or at most three degrees of freedom cable model by the finite elementmethod. A quick reference to a standard finite element text provides the following two-nodedelemental mass and stiffness matrices:mel=ρ 62112 kel=T 1 −1−11 (19.1)where T is the tension of the cable (N), is the elemental length (not to be confused with the totalcable length!) (m), ρ is the cable mass per unit length (kg/m).As shown in the beam finite element modeling, one can construct a single, two and three degreesof freedom model as follows.One-DOF Model: A finite element-based construction of simple models can be constructed in twoways:(1)From onelinearelementmodelby constrainingeitherof thetwonodalpoint. Thisapproximationis shown in Figure 19.1(a). Since node 1 is fixed, the resulting one-DOF model can be obtainedfrom the elemental mass and stiffness matrices given by (19.1) asρL3¨q2(t) +TLq2(t) = f (t)(19.2)19–219–3 §19.2 ONE-DOF MODELING OF CABLE VIA THE FINITE ELEMENT METHODwhich yields its frequency to beωn=TLρL3=√3LTρ=1.7321LTρ(19.3)Comparing this with the analytical solution, ωn=πLTρ,wefind a frequency error of about 45%by this one degree of freedom simple model.one linear element modelone quadraticelement modeltwo linear element model L= L=/2 L=/2fixed boundaries(a) one dof model from one element(b) one dof modelfrom two elements(c) one-dof modelfrom one quadraticelement1122322 L=/2 L=/2132132Figure 19.1 Finite element modeling of a cable(2) From two linear element model by constraining the two ends. This approximation is shown inFigure 19.1(b). First, the free-free two-element model can be written asρ 6210141012¨q1(t)¨q2(t)¨q3(t)+T 1 −10−12−10 −11q1(t)q2(t)q3(t)=f1(t)f2(t)f3(t)(19.4)Constraining q1(t) and q3(t) from the preceding equation, which is equivalent to eliminating thefirst and third rows and columns, yields4ρ 6¨q2(t) +2T q2(t) = f2(t), = L/2 (19.5)which yields its frequency to beωn=2T 4ρ 6=√12LTρ=3.4641LTρ(19.6)We thus conclude that a finite-element based one degree of freedom approximation yields about 10% frequency error.(3) From one quadratic element model by constraining the two ends. This approximation is shownin Figure 19.1(c). First, the free-free one quadratic element model can be written asρL3042−12162−12 4¨q1(t)¨q2(t)¨q3(t)+2T3L2 −2 −1−28−2−1 −22q1(t)q2(t)q3(t)=f1(t)f2(t)f3(t)(19.7)19–3Chapter 19: MODELING OF DYNAMICAL SYSTEMS: DISCRETE MODELS 19–4Constraining w1(t) and w3(t) from the preceding equation, which is equivalent to eliminating thefirst and third rows and columns, yields16ρL30¨q2(t) +16T3q2(t) = f2(t)(19.8)which yields its frequency to beωn=16T3L16ρL30=√10LTρ=3.16227LTρ(19.9)We thus conclude that a quaadratic-element based one degree of freedom approximation yieldsabout less than one percent in frequency error.Remark: It is worth noting that one quadratic element model captures the fundamental frequencyof cable with less than 1.0% error, whereas one and two linear element models yield about 45% and10% in frequency error, respectively. Note from Figure 19.1 that cable modeling by linear elementsin general causes thecable slopediscontinuous, which isa poorapproximation ofcontinuous cables.On the other hand, modeling by quadratic elements satisfies the inter-element slope continuity. Itis this inter-element slope continuity that plays a key role for simple cable modeling. Of course, ifone employs a large number of linear elements, linear element models will capture the analyticalfrequencies. This is illustrated in Figure 19.2.0 50 100 150 200 250 30010-410-310-210-1100101102Number of linear cable elementsFrequency error (%) of first cable mode Cable model by linear fem methodFigure 19.2 Assumed Mode-based simple models of a cableThe error analysis of linear elements for cable vibration analysis indicates that about 40 linearelements are needed to capture the first cable mode for three digit accuracy, and about 200 elementsfor five digit accuracy. Also, note that the frequency error for models of in excess of 150 elementsdecreases linearly, a classical convergence rate result in the finite element method.19–419–5 §19.3 MODELING OF CABLE VIA ASSUMED MODE APPROXIMATIONS§19.3 MODELING OF CABLE VIA ASSUMED MODE APPROXIMATIONSIn a broadest sense, the finite element method is a polynomial-based assumed mode approximation.For example, a linear element assumesw(ξ , t) =12(1 − ξ)q1(t) +12(1 + ξ)q2(t), −1 ≤ ξ ≤ 1,ξ=2x (19.10)where the element coordinate origin is located at the center of the element.