HW3_Solution - Problem 3.4(d Summarizing these results together with given data:a Machine running at 20 rpm i” 2 39— 0.1 can 200

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Unformatted text preview: Problem 3.4 (d) Summarizing these results together with given data: :a) Machine running at 20 rpm. i”. 2 39— : 0.1 can 200 no (‘55:)0 (a) [1"(0/6911 )- 3; The isoiator is effective at (IO/£0“ = 0.9; it reduces the (a J _ _ deformation amplitude to 39% 0f the response without 0.2 5 W => (“500 = 0-1930 15- isolators. At 03/0),I : 0.} or 3, the isolator has eSSentially \ ' ) no influence on reducing the defamation. For f = 0.25 and 60/50“ = 0.1, l = 0.}. , Cb) 2 « 2 [1 ~— {cu/wn) ] + [Qéw/mn] 3r 1 up = 0.1980 2 a a w/[l - (0.1) ]“ + [2 (0.25)(0.1)]‘ = 0.399701. 13) Machine running at 180 rpm. 3 = 3&9. a 03 can 200 gram Eq. (0), 1.042 = “35539-1— 2; (Mo = 0.198001. |1 — (0.9)~ For 4' z 0.25 and (0/60,, = 0.9,Eq. (b) reads Mo = 0.1980 1 1 [t — (0.9fl” + [2(0.25)(0.9)}2 : 0.4053 in. 0) Machine running at 600 rpm. _w___6_9.9_3 OJ” 200 trom Eq’ (a)! 0.0248 = 43¢; = (me = 0.1980111. j} m (3)- 30! 5 = 0.25 and (0/0)“ = 3,Eq.(b} reads : 1 1:0 = 0.1930 2 e — (3)3? + [201206)] = 0.024301. 'EProbIem 3.11 _ The given data is pierced in the farm of the frequency i’esponse curve shown ii: the accompanying figure: H O Accelerafiun anlplimdc, “1" x g Frequency. Hz (a) Natural frequency The frequency response carve peaks at f” a 1. 487 Hz fissuméng small damping, this value is the natural :frequency of the system. {13) Damping ratio I The acceéemtion at the peak is rim“,c = 8.6 x10‘3g. brew a horizontal line at I}, —:- J5 a 6.08 x 10-3 g to ebtain fa and fb in Hz: mk- i 131 = 3.47GHz fl, = 1.507Hz Then, gm 1?, -F fa = 3.507 -— 1.473 : 0-0114 2}; 2(}.487) = 1.14% 11 iProbIem 3.14 ugh) = ago Sinai: => fig (1*) = ~59qu since): pefifi) = —miig(t) = mwgugo sinmr a pa sin a): yummy a) a 2 no = (u3,)oRd : k R, = uga[;--) Rd = ugoRa n Spring force: f5 = [my 1 The resonant frequency for f5 is {1:12 same as that for Eta, which is the resonsmt frequency f0; Ra {fiom section j3.2.5): can (are! = .7 lWZC‘ 8.786 «ll—2&4): Resonance occurs at {his forcing frequency, which implies a speed of cuL (13.655)}00 v:-—-——-= 291' 27: «1 13655 rads/ soc =169.6 ft/soc = 316 mph 14 ...
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This note was uploaded on 01/04/2010 for the course CE 225 taught by Professor Chopra during the Spring '09 term at University of California, Berkeley.

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HW3_Solution - Problem 3.4(d Summarizing these results together with given data:a Machine running at 20 rpm i” 2 39— 0.1 can 200

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