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exam1soln - flehwmwwa CE 573 Structural Dynamics Exam#1...

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Unformatted text preview: flehwmwwa CE 573: Structural Dynamics Exam #1 Problem #1 (25 pts.) 5 ft 5 ft Given: A compressor unit weighing 3000 lb is supported by two parallel, simply supported steel beams (E = 30, 000 ksi for each beam). The motor in the unit runs at 300 revolutions per minute and has an unbalance W'e = 180 lb-in. You may assume that the mass of the beams is negligible, the supports share all loads equally, and that the damping ratio 5 for the system as a whole is negligible. Reguired: (a) You wish to choose a value for] such that the force that the beams exert on the supports in the steady state is always downward. There are two possible ranges of I that will accomplish this. Detetmine each range. (Hint: there is more than one contribution to the force exerted on a support!) (b) There is a nonzero value for 6 Such that any choice of I will lead to an always downward force on the supports in the steady state. What is this value of 6 ? (You may assume that this value is small.) _ PL’ 48E] ymax rél — Uh... Soiu‘l‘lcfiA; (a) Em Mam ‘ {flfi wand (liq Mad PTA) = EwTr~M(a++¢we),w/zw Randal LA. MM) FWVt—t MQWM we Em) . ammo AW WMwflwJ'l ) W \U 7 z \ Fn;O,Tv . HUM [I-rll>fi;3=o.)533 E l L0 1.0 > glfizw = :7 k >(3L‘HS-fi) ’0 #1201 M ‘ 14 3 330$H‘517Q1V3H-’§%)L= 9052??? = 9.032 - 17,:‘(3El_ _ 59651 kbw‘ Y“ La "’> --/L’3 ‘ : 46 301105le . 7 W>q052£§bxa m) m 1-94): £6 $470.63 => 9%le W‘i’lm‘q 3HZ’%L ’ o < ’ =zq_zs%i :7 k < (igi‘fflgawafi = éew ‘5/0; = 4M “4%; =7 £6 (LOWE < 4.6qu ‘El’m I< 346i ( ((wm” (by F0 f M, 114,! W uni/(L2 067} WJWWZV4 , f ‘ l4 (2 V‘ w («\W~ m9 .‘ =\)(’Lwl MMWM \] u-nm (2y 25) \fi <é-SZB :7 (a?)sz (4255(6) (ijzozLMS W/MW Problem #2 (25 pts.) Ri id bas’ g L \ NJ» / Rigid base Given: You are asked to analyze a landing platform used to cushion supplies being drOpped from airplanes into remote regions. The platform consists of a two rigid bases of negligible mass, two springs having identical stiffnesses kl 2 = 80 kN/m, and a viscous damper having a damping coefficient c = 6.4 kN-sec/m. The distance between the top and bottom base is initially h = 026 m, which you may assume is the unstretched length of each spring. The platform is designed to carry a mass m = 400 kg. When it lands, you may assume that the downward speed of the combined system is 4.5 m/s and that the platform does not bounce (tie, the bottom base remains in contact with the surface). Reguired: At what time afler impact does the top base come closest to the bottom base, and what is the value of this minimum separation? 55%;“ 1 Max tfle m MM 1%: Malawi New astut- m “I T mamafiimwewra - g 2 ,(Ht 9 : its: («walt‘w‘ui (£1 {7% ) a A K HOHO’UIu / ( I : Z‘l-S‘IO’D’M f. TM u“: -A=-Z‘1_S'r/O—sm Tumumaftiehfiawaurqsm. _ TM 4' u(t\;Ce'TwLA/wdi+oci c: = W mm (i ‘ Z 035?: iKO‘EIO-“J/K : I M 906143 C : [(4484635 E L“‘OHOMzogx-mvm 1 ‘L (sq-3% gimme“ 04 5M [CW ° =~O.Io%~.(. (29%) Gunsmm 4 45% Wm, M WW finwm m =0 0 , w+ . u [H = 6 Y [vgw «Mn/\{wfiffl * wdmfwd++0013 0 ’7 gm M(Wd++o‘\) ‘ wt; mflofiu), (n 3 $_ 12.33%: wwém w - (OWWJNZZW =7 wfiuw HS‘M :7 1L : HS‘MJwt _ l.|§‘Zw+O./9M 4 WW grit; ooqu _’ 2 - (. .0 q “[‘L JR») 0135M 9 WW 6 Mfll‘833)(.om~o.10fl = OJZ‘fém Mum mmm 1mm W70wa 4 A+ UH=+WX = 0.2qu W ‘A‘ Wm M5 Problem #3 (30 pts.) Given: The dynamic system shown consists of two masses (ml and m2) connected to each other by a spring of stiffness k and a viscous damper with a damping coefficient 0. Each mass is subjected to an external force, respectivelyF, (t) and FZU) . The displacements of each mass are ua (t) and 112(1) , and the system is in static equilibrium when 141 = u2 = 0. Define u,(t) =u2 (t) —uI (t) as the relative . _ 1 l 1 displacement between the two masses, and define meq (the “equwalent mass”) Via — = — + — . m mm eq 1 2 Reguired: (a) Show that the equation of motion for the relative displacement ur (t) is 2 m mm, d Md”r maniac). dt dt m1 m2 (b) Let m1 = 40 kg, m2 = 50 kg, k = 200 N/m, (2 = 4N-s/m. F10) = 0 and F2(t) = 135 cos 2t N. Find the steady-state solution for u, (t). meg +ku, =— (0) Suppose that both mass #1 and mass #2 are initially at rest when the force F20) in part (b) is applied. Would you expect the M solution for ur(t) (i. e., transient plus steady-state solution) to be the same as the steady-state solution found in part (b)? Explain why or Why not. (Note: you do n_ot need to solve for the full solution to answer this part.) Soil/ATM: (4i ‘ (Fm LIZ‘M/l. w CIZN‘M :7 “r70 m ‘19“) F, {A _> 7 k (UI’U‘\ : Lah- , - w - 7i 5 WI 7 C(Ql'u|\: Cur D ~ r:v\7(:\ 4’ % “r % [A l l l kw . W\L ——>E_({-i Em k v Q th— Z 11‘ El " wt r I. ~- ~- # AM Mr: “J Uh /W 579‘” %‘M E?" Z ~ _Em Em Lg , 9+9 $1-)!“ Me ’ W1“ “V: “\(w MASH" Wt :L‘l) ‘ " - ll' _, 9 Mr - FLGA/vnl " F[ W” Ur C/Me+ MT ) ZF'WWI :3 3 [UM-"f Car) (5pm :F: Wq '37 Wlaz: [4“1'1 Ca" / (b) Fmfigfiwwm) MQ$=[@L( 4 J_ —(= 22.22 kg 3 9013 L0». g: Zoo“/.._30-o‘1,(. FL:__"_{_:*7_W_,_ “a; 22ml ' ' “J E’wa 2(22‘22w3m) : 0.03 _ K“. r 2122 M“) H” “MW 3:? (ISszr) = gen 0H ‘21) M T?” -’- ’4 W1 ,rzyl: :éao 7 F2 gown it» SIFW w 391%" MSW: '1‘: mm: 0.300w) W W 9: 2;”: Z r = -( 2*,0'5M67 “(5”, P Lu; [hr-‘1 {M ptmv} 0'07)“ .: 9qu «m >4 mm MMWWWJ’Q L W DMT: ~ I .. [l]: s (4023‘ f ggjjfl‘k‘ : :2???“ “lama: [0-3w1.799)w2+:a+,2 ,_ 07 la“? ‘— O‘SMMQHLWN M (CB M M Mr({'): ETUfE/q‘mwdf +U4awwéfl + U‘JH SMT 53.11 ' ‘TMMJM&%WM%M.MM%M=O M541 Lab? W “buoy :0 (4+ mm = A + Merkwg-sm .7 A: -uJMM‘oH‘M: —_§7¢(M{).‘14?M3=—.338:¢O WWM /MMM-M\WM%W~ Problem #4 (20 pts.) Given: A free vibration test is conducted on an elevated water tank as shown. A cable attached to the tank is pulled (by a very strong technician!) such that it applies a force of 20.02 kips at an angle 9 =35°. At this time, you measure an initial horizontal displacement for the tank of 0.20 in. The cable is then released and the subsequent free vibration of the tank is recorded. It is noted that it takes 2.0 seconds to complete four full cycles of vibration and that the measured amplitude of the vibration at that time is 0.10 in. Reguired: Estimate the following quantities: (a) the damping factor 4‘ , (b) the undamped natural frequency a) , (c) the mass, stiffness, and viscous damping coefficient for the tank, (d) the number of m cycles it would take for the vibration amplitude to go below 0.02 inches. (6) The test is now repeated, but this time the initial horizontal displacement of the tank is 1% larger. The initial displacement and amplitude of vibration after four full cycles are again measured, and it is found that it still takes 2.0 seconds to complete four cycles. Which test (small initial displacement or large initial displacement) would you expect to give the best estimates for the quantities in parts (a)-(c)? Give at least one reason to explain your choice. (Assume that you can make your measurements with the same level of precision in both tests.) Scum: (a3 Amawmwiflewwtm/ “WW' g“- ‘ifim “l = $13"[9£51=O.l733 out“ 0.1;. AWL? :1 S My} 6(737 I r ' In .‘2 LO 7 047F142 T \iL/n‘«.l733‘ i? O Zjé (to) Ema. my Witt WWI = E—(‘Ofiflamc 11 . , 4—144 .: wA= =12.S7*—;{‘= writ/:1 flu): I257“ mmmg \il~..¢27tz (MWMM,%£QMM¢Q§/Mmmwmwd) (it) d1 k_ Eagle, (20.0mmm390 :FZ A O. my - 41¢ / {43 AWWQ 7% 0.20 w E \ J— ; —' S "‘ \A fl/M mm» 3 o l E 3 LIA A1 0.: - VL aft/a 0’ L ‘ 7 — l 3' meawfié/fifififlfmow‘ [,1 MM WWW/424.) (e, W \ A4fmfirwfifl W, W \ flag/Mk [figmmmgflwyim ‘Wdf Wéd/MdMZ-é we m . m L/ mflffifigyflg ...
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This note was uploaded on 05/09/2008 for the course CE 573 taught by Professor Whalen during the Fall '05 term at Purdue.

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exam1soln - flehwmwwa CE 573 Structural Dynamics Exam#1...

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