HW4ANS - 1W5 Mechanical Vibrations{W Homework#4 v< Q09...

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Unformatted text preview: ._:, 1W5 , Mechanical Vibrations ,/ . {W Homework #4 v\)</ . Q09 Due: 180ct04 - Start of Class (JV Part 1: Work the following problem from the course text Prob: 7.11 {0 Pk Part 2: Work the following problems 1. Derive the equations of motion for the system shown in Figure 1. Use the following parameters: m = 3 Kg, I0 = 4 Kg meters2, k1 = k2 = 200 N/meter, r = 2 meters. x(t) ‘0 pt Figure 1: A two degree of freedom system with translation and rotation; 2. Find the eigenvalues and eigenvectors by hand and check your results with Matlab (or equivalent) for this problem. You are asked to: a) find the eigenvalues and natural frequencies for the system 70,45 shown in Figure 1; b) find eigenvectors for each eigenvalue (V); c) find the “mass-normalized” \ eigenvectors (U); and d) find U TM U and U TK U if the resulting equation of motion was in the form. M5?+KX=6‘ . (1) Example Matlab Matrix Operations: You can check your work, but make sure to turn in your hand calculations, using the following Matlab commands: v first define the A-matrix: type A=[1 2; 3 4] a) transpose of the’matrix'AT, type A’ b) inverse of the matrix A‘l, type inv(A) c) calculate the eigenvalues and‘eigenvectors, type [vector,lamda]=eig(A) d) type “help eig” to learn more 7.11 The experimental arrangement for an airfoil mounted in a wind tunnel is described by the model shown in Figure E7.ll. Determine the equations of motion governing this system when the stiffness of the translation spring is k, the stiffness of the torsion spring is k,, G is the center of the mass of the airfoil located a distance I from the attachment point 0: m is the mass of the airfoil, and la is the mass moment of inertia of the airfoil about the center of mass. Use the 8 53 +m3L [-553 sin (6+03) + 3330 generalized coordinates x and €discussed in Exercise 1.19. Solution 7.11: We assume that a displacement along on the upward vertical direction is positive and a counterclockwise rotation Babout 0’ positive. Then the location of the center of mass from 0 is rm=—-lcos0i+(x—lsin0)j and the velocity of the center of mass is v =i'm =zésin0i+(5c—zécos6)j Then the kinetic energy is T 21160.2 +11% 2 2 =lJGIE}2 +—l—m 2 2 =-1JGI9.z +lm 2 2 and the potential energy is Noting that we do not have any damping and any external forces, the Lagrange equations are On substituting for T and V and evaluating, we obtain m 1 . (vm'vm)=§‘]602+k —-m +63)]+k,3(é+03) (1292 sin2 (9 + (:2 - lécos (9)2) (12492 sin2 0 + x2 — zuécos 0 +1245}2 cos2 0) (1249'2 + x2 - 215:9 cos 0) 1 V=—2-kx2+%k,l92+mg(x—lsin0) d F: 5’. dt ( ( ar 5; g; as l l _a_T+§K=O 3x 3x 3T 3V ‘5???” 146 11—( '—mzécos0)—0+kx+mg =0 dr 3,-(106 +"mlzé -—mb'rcos6) —m&ésin6+k,6-mglcos€ = o 01‘ mSE—mlécos6+mlézsin6+kx=—mg (JG +mlz)§—m&cos6+k,6 = mglcos6 ‘ 7.12 A multi-s ry building is described by the model shown in Figure E7.1. Derive the equations of moti of this system and present them in matrix form. Are the m s and stiffness matrices symmetric. Solution 7.12: Making use of the 'ven coordinates, the kinetic energy and th potential energy are, respectively, ' . 1 . T=-;—mI '12+-;-m25c22+%mgx32+—2-m4xf ; V=k1x12+‘ (xl-x2)2+k3(x2—x3)2+ 4(x3—x4)2 Since there are no external forces, the agrange equations .. e _4_ a 3T+3D .v dt 3%, ax, a 3x1 d[3T]_8T 8D av —— —-+—-=o dt 3x2 ax2 3x2 8x 8x3 3&3 + 3x3 d Myst +a_v_= " 8x4 824 3x4 which on evaluation leads to mp'c'l + 2k1x1 + 21:2(xl x2) = 0 mzjc'2 —2k2 (x1 —x2) +2k3(x2 - 3) = O mSJ'c's —2k3(x2 —x3) +2k4 (x3 —-x4 = 0 mAJt'4 -2k4(x3 -x4) 0 These equat' ans can be written as 147 +/;Ié: + ‘/z KLLKr r93,” 77. K, X1 + 77. K1(\(1-2r‘ex + r291) °__.’i§,fl-‘V‘*7 A T: /2 M, 7L ' PE :7 W 71sz 1 s/Z Y“) X1 V : ° o‘ffi‘r’iffskfl =~o ‘> Pius Lmte ._ all) 5.1 + d 0x Y 1%.: s £91 0% é (59 + Va K1 x2 - Kzrex 4")/1K,‘r‘161 104, Ma) x‘ - KL <32? x311. chie‘ fl” +\4 ~Kro ax ‘(K3 9% ‘L Egupj'fi'hs Tmtlah‘on‘»; mx + LK|+KL5X~ K1Y‘6' :0 Ro‘Q“M~9 ‘9 :43 + Kyle — Elma = G r "H*}+[“‘”‘° “H 6 IC, 9 ”K1? ¥\7_Y‘1 3 0 St Lido 40¢ [a ‘4] 1 + L400 Boo Q} ,Uumm W W cm [WW 1 $346} €4ij 0446/ 6901'; map/x5, {Aymm 5’00 4/00 + 1 O A 0 LI *‘W’ 9’00) A1 3 + <loo 4400 , = 0 ~ 400 ,{1‘4 + 80c» (fife L100)“ ,1" 1.3m) 1- lbOpoo :- o H c), 91% 7C“ V! :Qr 61 £3) 681M MIPS Dar! +K17 = O LDC‘jmueco‘ar 3A} Hoe ~ 400 L - '400 ‘M, +600 We V” \? l. \“rum V2| ”2’ 0.014 4‘00. 5er ””00 \qOO ”3.47.4008 I : UT [‘1 U : 3d) 44.,(0-(05'0 3.9xz Lied—J. 15H) : Sex,” + Lickflowm)‘ obi, a; o( | (0.05143 $44115”) j 0 3*‘9‘z+ Ll ..o(z(o.usw\(-M§1LI) “'0 7. z 7- Sdndma- w—waflomfi 3&1 +4o(,_ L-HSH) ’ 1 L 2 L A? 0“. (5 ”(O-WW) 120MB 1- 4 HM» oksfouum okz=io.3kl7o Rake leufl? U’) MCIWM CW U7 Wan..- U : OJ-HoH 03470 '______w,, « ‘3? \" ‘ 10/14(04 1:41 gM MATLAB Coggiqqmflindow ‘ 1 of 1 >> [U,D] s eig(K,M) U ‘\—o.4614 -o.347o 18* -o.3006 0.3996 46.4816 0 \I’ "rro 286.8517 C; >> ...
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