Question

5) Rayleigh's dissipation function for a stand-alone viscous damper element with damping coefficient of c is defined as D = {ci?. In the presence of viscous damping, the system's EOMs can be found by a modified Lagrange's equations as (0) - d al al ad + dt ağK дак дäk Where L=T-V, T and V are the total kinetic and potential energies of the system, respectively, and Qx is the generalized force due to enternal forcing projected to k-th DOF. Using those definitions, obtain the EOMs of the following system: Xy(t) X2(1) X3(t) C1 C2 C3 -] C4 M M W K K₂ 1 M2 W K₃ f2(0) C5 719 Ww Ko

Answer 1

Given X(t) X200) X3(t) C1 C2 C3 C4 MA WM M K K₂ M2 K3 2(0) C5 f(t) Kg From the System, given that xi(t), xz(t) and x3(t) are the displacements for masses M1, M2, M3. The Lagrange's Equation of the general system: aT av aD d/aT dtax, + + Qi дхі дх; ox,

Where, • T = Total Kinetic Energy of the system • V = Total potential energy of the system • D = Total dissipation energy of the system •j=1, 2, 3, ...... n, when, n= Number of degrees of freedom • Q; = Generalized External forces • X; = Generalized coordinates >> The kinetic energy of the system comes from the motion of the block and the swing of the bob of the pendulum. Total Kinetic Energy, 1 1 T = M 812 + M2x2+ M2:32 2 The potential energy comes from the deformation of the spring and the position of the bob. Total Potential Energy, ŽM, X 1 1 V K1x12 + K(– x3)2 + 2K(x2 – x)? + Kg(x3 – x3)2 +3K, (x2 – xv)2 + 3K; (x3 – x3)2 + 2Kg(x3 – x3)2

Total Damping Energy, 1 (X2 – )2 + ż €5(*; – x3) 1 1 o - zake?+ 30.66 – 8,99 +3.com +żcz(*– x;)2 +żcz(*3 – x3)2 + 2C3(*3 – )2 +244xz? L=1-v=[M&+ +M.&*+¿Miksi) Lagrangian of the system, 1 1 1 1 1 1 1 1x12 + 1 +-K2(x2 - x)2 +5K3(x3 – x)2 + K5(x3 - x1 2 2 Equation of motion for displacement xi(t): dat aT av ad + Q1 dt laxi дх, дx, (5) + дх Mx(t) - 0+KX + K2X + KsX - K2X - KỆX3 + CX + C2X + CsX - CzX2 – Csåg = f (t) Muxi(t) +K_(x1) + K2(x1 - x2)+K5(x1 - x3) + (x) + c2(x1 - x2) + C5X1 – X3) = fi(t)

M,х(t) + x, (К, +К, + K5) — Кух, — K5х2 + x,(с + c, + cs) — сух, — cs X3 = f (t) --- (1) Equation of motion for displacement x2(t): daT ат dt \дх, дх, ду + (8) + aD + дх, дx, Q, Mї,(t) — 0 + Кух, + Кух, — Кух— Кахз – сәх, + сэх, + сәХ, — саха f(t) М.Х.(t) + К.(х2 – х.) + K2 (x, — х3) + с.(x, — х1) + c (х2 – х3) = f(t) Мәх, (t) + x, (К, + K2) — Кух, — K2X3 + ,(с. + с) — сухі — саха f, (t) ––– (2) Equation of motion for displacement x3(t): dат от ду др + дх2 дх, дха + Q dt \дх3

M283 (t) - 0 + K3X3 + K3X3 – K3X2 – K5X1 – Cziz + cziz + C483 + C583 - Cså4 = 0 M383 (t) + K3 (X3 – x2) + K5(X3 – xy) + C483 + c3(X3 – 81) + c3(83 - X2) = 0 Mz8z(t) + x3(K3 + K3) – K3X2 – K3X1 + 83(C4 + 25 + c3) - C581 – Cziz 0---(3) In matrix form, TM 0 0 0 M2 0 --C2 -C5 01X(t) 0 ||X(t) M3] X(t) ( + C2 + C5) [x1(0)] + (C2 + c3) -C3 X2(t) -C3 (C4 + C5 + X(t) (K1 +K2 +K) -K2 -K, xi(t) + -K2 (K2 + K3) -K3 x2(t) -K5 -K3 (K3 +K)][x3(t)) -C2 -C5 [f(t)] [2(t) 0