Modeling of Dynamical Systems: Discrete Models - One-DOF Cable Modeling, Study notes of Aerospace Engineering

Download Modeling of Dynamical Systems: Discrete Models - One-DOF Cable Modeling and more Study notes Aerospace Engineering in PDF only on Docsity! . 19 Modeling of Dynamical Systems: Discrete Models 19–1 Chapter 19: MODELING OF DYNAMICAL SYSTEMS: DISCRETE MODELS 19–2 §19.1 INTRODUCTION With a plethora of engineering analysis software packages available, the dynamics specialist can perform a variety of computerized modeling and simulation of complex systems. It is not uncommon these days that a typical structural dynamics model may consist of several thousands to several millions of degrees of freedom, often employing the finite element method as the dominant modeling tool. 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 routine practices. At present, the most time-consuming part of engineering analysis including dynamical systems is model development tasks. While still significant, the core of computational effort has become relatively insignificant thanks to the advances in computational power known in the computer world as Gordon Moor’s law. To the engineer, a new challenge has emerged: how to interpret a vast array of analysis results in terms of plots, tables and other graphical data representations. A first step to a meaningful interpretation of computer-generated analysis results is to cultivate an ability to reduce the complex systems to a set of simplified models, from which one can relate the observed phenomena to the complex models. This lecture is aimed at introducing to the students how to construct simple models, how to interpret the simple model results, and how to relate the results obtained by simple models to complex simulation models. §19.2 ONE-DOF MODELING OF CABLE VIA THE FINITE ELEMENT METHOD Suppose we do not know how to obtain a sensible simple discrete model and we are asked to construct a single, two or at most three degrees of freedom cable model by the finite element method. A quick reference to a standard finite element text provides the following two-noded elemental mass and stiffness matrices: mel = ρ 6 [ 2 1 1 2 ] kel = T  [ 1 −1 −1 1 ] (19.1) where T is the tension of the cable (N ),  is the elemental length (not to be confused with the total cable 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 degrees of freedom model as follows. One-DOF Model: A finite element-based construction of simple models can be constructed in two ways: (1) From one linear element model by constraining either of the two nodal point. This approximation is shown in Figure 19.1(a). Since node 1 is fixed, the resulting one-DOF model can be obtained from the elemental mass and stiffness matrices given by (19.1) as ρL 3 q̈2(t) + T L q2(t) = f (t) (19.2) 19–2 19–5 §19.3 MODELING OF CABLE VIA ASSUMED MODE APPROXIMATIONS §19.3 MODELING OF CABLE VIA ASSUMED MODE APPROXIMATIONS In a broadest sense, the finite element method is a polynomial-based assumed mode approximation. For example, a linear element assumes w(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. Note that the cable slope for the linear element is given by w(x, t)x = 2  x(ξ, t)ξ = q2(t) − q1(t)  which suggests that the cable slope is discontinuous (constant!). The only exception is when the cable forms a straight line, a trivial state. On the other hand, the quadratic element approximates the displacement by w(ξ, t) = −ξ 2 (1 − ξ)q1(t) + (1 − ξ 2)q2(t) + ξ 2 (1 + ξ)q3(t) (19.11) one dof model by an assumed mode L w(x,t) = q(t) sin( x/L) q(t) x π Figure 19.3 Assumed mode based modeling of a cable Note that the cable slope for the quadratic element, w(x, t)x = 2 x(ξ, t)ξ , is continuous. A classical one-DOF mass-spring modeling is to introduce the following form: w(x, t) = H(x)q(t) (19.12) where q(t) is the displacement where the mass is located, and H(x) is the assumed deformation shape. Thus, for one-DOF approximation of a cable with two ends fixed the following displacement is assumed: w(x, t) = sin(πx  ) q(t) (19.13) If one needs more than one displacement approximation, then depending on the analyst’s experience and the nature of problem to be modeled, w(x, t) may be modeled as w(x, t) = sin(πx  ) q1(t) + sin(2πx  ) q2(t) (19.14) 19–5 Chapter 19: MODELING OF DYNAMICAL SYSTEMS: DISCRETE MODELS 19–6 Using the assumed displacement given by (19.13), we have T = ∫  0 1 2ρ(x)ẇ 2(x, t) dx = ∫  0 1 2ρ(x){sin( πx  )}2 q̇2(t) dx = 12 [ ρ 2 ] q̇2(t) V = ∫  0 1 2 T (x) w 2 x (x, t) dx = ∫  0 1 2 T (x){ π  cos( πx  )}2 q2(t) dx = 12 [ T π2 2 ] q2(t) δW̄noncons = 0 due to free vibration modeling (19.15) Th discrete Euler-Lagrange’s equation from the above energy expressions yields the following one-DOF equation m q̈(t) + k q(t) = 0, m = ρ 2 , k = T π 2 2 ⇓ ωn = √ k m = √√√√ T π22 ρ 2 = π  √ T ρ (19.16) Hence, we observe that the assumed displacement (19.13) yields the exact fundamental frequency of a cable whose ends are fixed. This si expected because the shape function, sin(πx/) , is the exact mode shape! §19.4 FINITE ELEMENT MODELING OF CABLE WITH FLEXIBLE MIDDLE SUPPORT With the preceding background, let’s revisit the task of modeling of a cable with two fixed end and 19–6 19–7§19.4 FINITE ELEMENT MODELING OF CABLE WITH FLEXIBLE MIDDLE SUPPORT a middle flexible support as shown below. O Km Mm L /2 L /2 O Km Mm L /2 L /2 1 2 3 1 2 3 4 5 (a) Two linear element model of cable (b) Four linear element modeling of cable Figure 19.4 A simple equivalent model of a cable with a middle support A Single Degree of Freedom Model: If one utilizes two linear elements and assemble the middle mass-spring model, the resulting equation before applying boundary condition is given as ( see equation(19.2)) as shown in Figure 19.4(a): ρ 6 [ 2 1 0 1 4 + 6µm 1 0 1 2 ][ q̈1(t) q̈2(t) q̈3(t) ] + T  [ 1 −1 0 −1 2 + κm −1 0 −1 1 ][ q1(t) q2(t) q3(t) ] = [ 0 0 0 ] (19.17) Constraining out the two support conditions, viz., q1 = q3 = 0, we obtain ρ 6 (4 + 6µm) q̈2(t) + T  (2 + κm) q2(t) = 0  = L/2, µm = Mm/(ρ), κm = Km T (19.18) If the required frequency is to be ωn = 1.5πL √ T ρ with µm = 1.0, we obtain κm = Km T = Km L 2T = 7.25275 (19.19) This incurs about 30% error when compared with the analytical solution. A Three Degrees of Freedom Model: Let us utilize four linear elements and assemble the middle mass-spring model as shown in Figure 19.4(b). Following a similar assembling and constraining processes taken for the case of a single degree of freedom case, one arrives at the following three degrees of freedom model equation: 19–7