Homework_5_Sol - Mechanical Vibrations Homework #5 Due:...

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Unformatted text preview: Mechanical Vibrations Homework #5 Due: 5Nov04 - Start of Class Part 1: Use the following parameters for all problems: m = 3 Kg, IG = 4 Kg metersz, k1 = k2 = 200 N/meter, r = 2 meters, f1 (t) = F0 cos Qt. Also, you should use a coordinate system with x(t) and 0(t). 1. Derive the equations of motion for the system shown in Figure 1. Add in the assumption of proportional damping with modal damping ratios of (1 = 0.01 and (2 = 0.02. Show how to decouple the equations of motion into modal coordinates and solve for the transient motions of the system (in modal coordinates) using the following initial condi,tions_and parameters: F = 0, x(0) = 7r, = 0, 0(0) = fl, = 0. Convert the solutions back into physical coordinates and plot the transient motion of the system (i.e. use x(t) and 0(t) coordinates for graphing). x(t) 1—. Figure 1: A two degree of freedom system with translation and rotation. 2. Find the expressions the transfer functions x(t) / f1 (t) and 0(t) / f1 (t) and plot the real and imaginary portion of these transfer functions for a frequency range of 0 < 9 < 2% (where tag is the highest natural frequency of the system). You can assume F0 = 10 in f1 (t) = F0 cos Qt. Modal coordinates should be used in your solution approach. 3. An accelerometer is placed on the translating mass during a frequency sweep and the accelerance it / f1 (t) is measured, where accelerance is a measure of acceleration per input force. Show your work to relate the accelerance of the system to the desired relationship as/fl 4. Consider that all of the parameters for the system in Figure 1 are fixed, except for Ia. Given . that the input to the system is F0 2 10 in f1 (t) 2 F0 cos Qt, design the [G to provide an amplitude for the translating mass to be zero at a frequency of Q = x/kl/m. You are asked to derive the relationship to obtain a zero response and plot the response amplitude of the combined system receive full credit. 5. Work problem 7.37 from the course text. Notice that you will need matlab or equivalent to complete this analysis. Hint on the last question (a rigid body mode is found when you obtain a natural frequency of zero). 50 SHEETS 22-142 100 SHEETS 22-144- 200 SHEETS 22-141 @fiwmn‘ ’2 T: KE: simg-zA—‘Eiéa 1 v: PE: "fiK‘x‘z’r L2K1CV6_>‘> d 2‘!" _2'T P\m3 wx’na ngvavnan T+C a; 2—11-5- E15 euvcc'N-orfi.‘ F\\'\J E‘Ssz‘L‘k-cg 4‘ o. 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WG-1 — ‘4'“ M12 “‘1 "F v -— u 50 SHEETS 142 100 SHEETS 144 200 SHEETS 141 T mmn- ‘3 { . ‘¢' #5 H “AL/43“ +— ua 'Mz/fi 2i. F\ (32'1— sz ““LH *‘ V12. (737/44 a K I e, a _, “\‘1\/F\ + M'imz/IC-I .' -417.“ 1 TLNS is He X _. So ‘9. 3H. Oesxak {10¢ f A év 0"“ wfl‘wu’r deH-r1"$ 1 MC: -\- KY 5—; '3 3 324* _\ ‘ F0 {flaw}; K-AVEHA: [01¢ W04 9. See “*‘tacL-ec‘ Wa’flkt (‘er 0;ka 3 “"’ . m ' XI: 9"“: I f:co.\\ 1‘ 2-31} “€444: '\ . . [21: Qezn’c = [_n}v\+k_] {F3 (hm: \ 50 SHEETS ¢ 22-142 100 SHEETS @1040 22-144 200 SHEETS 22-141 Sea atiucktd Cod-C auJ pKoQQ, > with (LinearAlgebra) : [> m:=<<—omega‘2*m1+k1+k2l-k2>, <-k2 l—omega‘2*Ig+k2>>; -602 m] + k1 + k2 -k2 m := -k2 -w2 Ig + k2 L I-r> MatrixInverse (m) ; -w2 1g + k2 -a)4m11g+w2m1 k2+k1 wzlg-kl k2+k2w21g, k2 -w4m11g+w2m1k2+k1 (ozlg-kl k2+k2 wzlg] |: k2 -w4m11g+a)2m1k2+k'1 wZIg-kl k2+k2w21g wzmI-kI-kZ -w4m11g+a)2m1k2+k1 wZIg-k1k2+k2co21g _ ‘ < .2 2.3: #4315 jafifigafiii .4 is y. Transient Response 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 time [3] Ir L‘ LI 0 L [l o 2 L E x 6m: 99: TF of mass M TF of IG 0.15 0.15 0.1 - 0.1 A 0.05 A 0.05 g g, E 0 E 0 (E (B a: g -0.05 -0.05 0.1 0.1 0.05 005 5 ° 5 ' f -0.05 55 0 g 601-; £005 ' ' -0.1 -0.2 _0_25 -0.15 ' ‘ -0.2 ' ‘ 0 10 20 30 0 10 20 30 Freq [rad/s] Freq [rad/s] Magnitude Plot for combined system Freq [rad/s] C:\Documents and Settings\mkoplow\My Documents\FALL_O4...\HW5.m Page 1 November 4, 2004. 7:35:43 AM % ________________________________________________ __ % HW 3, PROB 1 SOLUTIONS KEY % KOPLOW, MIKE % _____________________________________________ __s__ close all; clear all; clc; % _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ — _ _ _ _ _ — _ _ _ _ _ . _ _ _.. '% Part 1: Time Series % _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ t = O .001 8, zetal = .01; zeta2 = .02; w1 = 6.818; w2 = 16.94; laml = -zeta1*w1 + w1*sqrt(zeta1A2—l); 1am2 = -zeta1*w1 - w1*sqrt(zeta1A2—1); lam3 = —zeta2*w2 + w2*sqrt(zeta2A2—1); lam4 = —zeta2*w2 - w2*sqrt(zeta2A2-1); Ull = .4614; U12 = .3740; U21 = .3006; U22 = —.3995; cl = 4.06—.0406*i; c2 = 4.06+.O406*i; c3 = -.875+.Ol75*i; c4 = -.875—.Ol75*i; etal = c1*exp(lam1.*t)+c2*exp(lam2.*t); eta2 = c3*exp(lam3.*t)+c4*exp(lam4.*t); X = U11*eta1+U12*eta2; theta = U21*etal+U22*eta2; figure(l) subplot(2,l,l) plot(t,X,'linewidth',2) title(‘Transient Response‘,‘fontsize‘,14) ylabel('X [m]') subplot(2,l,2) plot(t,theta,'linewidth',2) xlabel('time [s]') ylabel('Theta [rad]') % _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ . _ _ __ % Part 2: Transfer Functions % _ _ _ _ _ _ _ _ _ . . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ . _ _ _ _ . _ _ _ _ _ __ freq = .1:.005:10; omega = 2*pi*freq; H1 = Ull./(w1A2-omega.A2+2*i*zeta1*omega.*wl); H2 = U12./(w2A2—omega.A2+2*i*zeta2*omega.*w2); X_F = Ull*Hl+Ul2*H2; TH_F = U21*H1+U22*H2; figure(2) subplot(2,2,1), plot(omega,real(X_F),'linewidth',2); C: \Documents and Settings \mkoplow\My Documents\FALL__O4 . . . \HW5 .m Page 2 November 4, 2004 . 7:35:43 AM title('TF of mass M','fontsize',14) ylabel('Real H(\Omega)') grid on; axis([0 30 —.15 .15]) - subplot(2,2,3), plot(Omega/imag(X_F),'linewidth',2) ylabel('Imag H(\Omega)') xlabel('Freq [rad/s]') grid on; axis([0 30 —.3 .15]) subplot(2,2,2), plot(omega,real(TH_F),'linewidth',2); title('TF of IG','fontsize',14) ylabel('Real H(\Omega)') grid on; axis([0 3O —.15 .15]) subplot(2,2,4), plot(omega,imag(TH_F),'linewidth',2); ylabel('Imag H(\Omega)') xlabel('Freq [rad/s]') grid on; axis([0 30 —.2 .15]) % _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ % Part 4 New System Response % _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ — . _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ __ Ig = 3; k1 = 200; k2 = 200; m1 = 3; fo = 10; X1 = (omega.A2*Ig—k2)*fo./(—omega.A4*m1*Ig+omega.A2*m1*k2+k1*omega.A2*Ig-k1*k2+k2*om K ega.A2*Ig); figure(3) plot(omega,abs(X1),'1inewidth',2) axis([0 20 0 8]) ylabel('Mag(X)') xlabel('Freq [rad/51') title('Magnitude Plot for combined system','fontsize',14) % % ———————————————————————————————————————————————— —- % % Prob 7.37: Mode Shapes % % ———————————————————————————————————————————————— —- % % M = [21 0 O 0 0 O 0 0; % 0 21 0 0 O 0 0 0; % 0 O 21 0 0 0 O 0; % O 0 0 21 0 O O 0; % 0 0 O 0 21 O 0 0; % 0 0 0 0 O 21 0 0; % 0 0 0 O 0 0 98 0; % O 0 0 0 0 O 0 49}; % % K = [51 —51 0 O O O 0 0; % —51 102 —51 0 0 0 0 0; % 0 -51 102 -51 0 0 O O; % 0 O —51 102 —51 0 0 0; % O 0 O -51 102 -51 0 0; C:\Documents and Settings\mkoplow\My Documents\FALL_04...\HW5.m Page 3 November 4, 2004 . 7:35:43 AM 0 O O O -51 117 -66 O; O O 0 O O -66 81 -15; O O O O O O —15 15]*1OA6; [VIN eig(K,M); RMK = rank(K); w = sqrt(D); for n=1:8 plot([l 8],[2*n—1 2*n—1],'k—-') plot(1:8,2*n—1+V(:,n)/max(abs(V(:,n))),'sk-') hold on end figure(10) subplot(2,1,1) plot(V(:,2)/max(abs(V(:,2)))) subplot(2,1,2) plot(V(:,2)) figure(11) plot(V) Y = V(:,2); Y2 = V(:,2)/max(abs(V(:,2))); Z = Y—Y2; w w w w w w w w w w w w w w w w w w w w'w w w w w w w w 51 -51 0 0 0 —51 102 —51 0 0 0 —51 102 -51 0 0 O —51 102 —51. 0 0 -51 102 —51 0 0 0 —51 117 —66 O 0 0 0 —66 81 —15 0 O 0 0 0 0 —15 15 0 0 0 . 0 x106 Nm/rad OOOOO OOOOOO ogo Determine the natural frequencies and mode shapes associated with the system and plot the mode shapes. Does the system have any rigid-body modes? Solution 7.37: a) Using the MATLAB function eig, we find the natural frequencies in rad/s and mode shapes, which are given below. b) The rank of the stiffness matrix is 7; therefore, there is one rigid-body mode, which in the figure is the mode shape corresponding to (01 = 0. Figure E737 MATLAB script that generatg the numericg results and figure for Exercise 7.37 , M=diag([2121212121219849]); K=[51-51000000 -51102-5100000 0-51102“-510000 00-51102-51000 ' 000-51102-5100 0000-51117-660 00000-6681-15 174 0 0 0 0 0 0 -1515]*10"6; m1k=rank(K) [modes E]=eig(K,M); w=sqrt(E) hold on w(l,l)=0; for n=1 :8 « plot([l 8],[2*n-1 2*n-1],'k--') . ‘ plot(1:8.2*n-1+modes(:,n)/max(éibs(modes(:,n))),'sk-') text(-.4.2*n-l, [\omega_(' num25tr(n) '}=' num2str(w(n,n).4)],'fontsize',l4,'HorizontalA1ignment',‘left') end v=axis; v(l)=0.5; axis v ; axis off 7.38 The eigenvalues determined for the two different three-degree-of-freedom systems are as follows: ‘ (a) id,” = —a1 $jb], al > 0, br>0 Adz”: 02$]b2, az>O, b2>0 2”” = a3$jb3, a3 >0, b3 >0 (b) 1d,”: «a] 1 jbl; al > O, b] > 0 Ads” = a3,-b3; a3 > 0, b3 > 0 Determine which of these systems is stable. Solution 7.38: In case (a), since the real part of each eigenvalue is negative, the system is stable. However, since the real part of one of the eigenvalues in the third pair in case (b) is positive, the system is unstable. 7.39 (a) Determine the equations of motion for the system shown in Figure E739 and put them in matrix form. (b) From the results in (a) determine the determinant form of the characteristic equation. Solution 7.39: a) The kinetic energy and the potential energy are, respectively, 1 - . 1 - T =i-2-J1612 +5.56: 1 1 1 V =§k1 (r4902 +'2'k2(’l€1—r262)2 +5k3(r292)2 Since the system is undamped and free of external forces, the Lagrange equations are 175 ...
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Homework_5_Sol - Mechanical Vibrations Homework #5 Due:...

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