Discrete Modeling of Dynamical Systems | ASEN 5022, Study notes of Aerospace Engineering

Download Discrete Modeling of Dynamical Systems | ASEN 5022 and more Study notes Aerospace Engineering in PDF only on Docsity! Lecture 15 (09 March 2004): Discrete Modeling of Dynamical Systems 1. Introduction With a plethora of engineering analysis software packages available, the dynam- ics 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 dy- namical 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 cul- tivate 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. 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 ] (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 1(a). Since node 1 is fixed, the resulting one-DOF model can be obtained from the elemental mass and stiffness matrices given by (1) as ρL 3 q̈2(t) + T L q2(t) = f (t) (2) which yields its frequency to be ωn = √√√√ TL ρL 3 = √ 3 L √ T ρ = 1.7321 L √ T ρ (3) Comparing this with the analytical solution, ωn = πL √ T ρ , we find a frequency error of about 45% by this one degree of freedom simple model. one linear element model one quadratic element model two linear element model  L=  L= /2  L= /2 fixed boundaries (a) one dof model from one element (b) one dof model from two elements (c) one-dof model from one quadratic element 1 1 2 2 3 2 2  L= /2  L= /2 1 32 1 32 Figure 1 Finite element modeling of a cable 2. 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  (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) (11) one dof model by an assumed mode L w(x,t) = q(t) sin( x/L) q(t) x π Figure 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) (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) (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) (14) Using the assumed displacement given by (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 (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 (16) Hence, we observe that the assumed displacement (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! 3. 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 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 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 assem- ble the middle mass-spring model, the resulting equation before applying boundary condition is given as ( see equation(2)) as shown in Figure 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 } (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 (18) As in the homework problem, 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) The above equation gives two frequencies: ωmax = √ k m = π  √ T ρ , ωmin = √ a √ k m = √a π  √ T ρ (25) Observe that ωmax is the same as the half-length cable. Hence, it is the lowest frequency, ωmin has to be half of π √ T ρ . This determines the stiffness distribution factor a to be a = 1 4 (26) Let us now assemble the flexible middle support as shown in Figure 6. Km O k m k m Mm (1-a)(1-a) k ka a R ig h t ca b le Le ft c ab le xx x 1 2 m (1-b) Kmb_2 m = ρ 2 k = T π 2 2  = L/2 µm = Mm/(ρ) κm = Km T Model parameters used Figure 6 A simple equivalent model of a cable with a middle support By adhering to the stiffness redistribution approach, the flexible support Km is redistributed as shown in Figure 6. The equation of motion for the system shown in Figure 6 can be derived as mẍ1 + (k + b 2 Km)x1 − [3k 4 + b 2 Km] xm = f1 mẍ2 + (k + b 2 Km)x2 − [3k 4 + b 2 Km] xm = f2 Mm ẍm + (Km + 3k 2 )xm − [3k 4 + b 2 Km] (x1 + x2) = fm (27) whose characteristic equation is given by det   (k + b 2 Km − m ω2) 0 −[ 3k4 + b2 Km] 0 (k + b2 Km − m ω2) −[ 3k4 + b2 Km] −[ 3k4 + b2 Km] −[ 3k4 + b2 Km] (Km + 3k2 − Mm ω2)   = 0 (28) For computational expediency, divide each of the three rows of the above charac- teristic matrix by m and using ωn = √ k/m, we have det   (1 + b 2 κm)ω 2 n − ω2 0 −[ 34 + b2κm]ω2n 0 (1 + b2 κm)ω2n − ω2 −[ 34 + b2κm]ω2n −[ 34 + b2κm]ω2n −[ 34 + b2 κm]ω2n (κm + 32 )ω2n − 2µm ω2)   = 0 (29) where κm and µm are defined as κm = Km T 2 π2 , µm = Mm ρ (30) For this characteristic equation, by varying b, κm was computed. The result is that the best value of κm is obtained when b = 0. When b = 0 is substituted with ω = 1.5π and µm = 1, κm is determined to be κm = 2.2 = Km T 2 π2 ⇒ Km T = 10.856 (31) which, when compared with the exact solution given by Km/T = 10.264, yields about 5% error. There are several additional aspects in modeling of simple dynamical systems, which include the use of lumped mass matrix (diagonal mass matrix as opposed to coupled mass matrix), redistribution of kinetic energy (or mass distribution), tran- sient response analysis, reduction of large-order models to smaller-order models, incorporation of experimental observations into finite element models, nonlinear phenomena, among others. Some of these aspects (not all) will be discussed in the remainder of the course.