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Unformatted text preview: Problem 3.1 From the given data, «51 = (0,2, = (27:13:)2 a (37:)2 = 649:2
m k HDMELWOKK. jag/4‘ eggshi e Lymieaa (11) h r 2 :2 r 2 2 2 =
m _ (can) (27:12) (67:) 367:2 Dividing Eq. (a) by Eq. (b) gives Am 16
I+——= m 3
m = 2Am = 23 = 6‘43 Ibs/g
7 7 g g From Eq. (a), (b) (C) k = 54an = 647933 = 10.52lbs/in. (d) 3 Problem 3.3 Assuming that damping is small enough to justify the
approximation that the resonant frequency is (on and the resonant ampiitude of R, is 1/2; then the given data
implies: (mania =(aao4L (@ 29" =("n—«——~——«——~——~ (m
u Ju~aﬁf+mamm Combining Eq. (:1) and Eq. (b): 2 2
l (”a )a.v=1.2£u,I = I
(29’)2 main," (—0.44% +(2.4§)2 ’ For . (“0)m=1.2m" __
m _ 1
(%n=% 3
Eq. (c) gives
64472 = 0.1935 + 5.76;2 => ; = 0.0576 Assumption of small damping implied in Eq'(
reasonable; otherwise we would have to use the] he 1' amplitude = 01“)” /[2{ W] . resonant frequency = COM/l — 2g,"2 and exact resonant) Problem 3.5 Given: . . Natural frequency: w = 1200 lbs, E : 30 x 10‘ ', {k f [32,552
p51 a)" = ~— 2 ——k— = ——«— = 102.3 rads/sec
I = 10in.4. L = 8ft; 4,: 1% m w/g 1200/386 300 Steady state response: p0 = 60 lbs; to :66?) 27: =1023rads/sec 1 Rd 2
2 2
Stiffness of two beams: 5 1/[1 _ (cu/a)" )2] + [ng/wn] where «cu/a)" : 10711023 = 0.3071.Therefore, 1c = 2[i§:£] = 32,552113/in. L _
_. ; R5 = ———_.I_.~__ 2 L104
5 ' 1/[1 — 0.0943]2 + [2 x 0.01 x 0.3071]2 Displacement: 25 =_ (:55, R5 = $15 60
32, 552 x 1.104 = 2.035 x 10‘3 in. II Acceleration amplitude:
ii a (02550 = (107:)2 2.035 x 10—3
= zooms/sec2 = 0.0052g Problem 3.7
(a) The displacement amplitude in given by Eq. (3.2.11): 4/2
an = (M40 ~ [3232 + (241532] (a) where ﬂ 2 (0/61)". Resonance occurs at ﬂ when rigis
maximum, i.e., duo/d1? = 0. Differentiating Eq. (a) With respect to ﬂ gives
[(1 — 5252 + (2:522
x [2(1 — ﬂan—25) + 2(2973) 2;] = o —3/2 01' —4(1 — 563)!” 862ﬁ= 0 1—ﬂ2m252=¢ﬁ=\’1—2§2 (h) Resonant frequency: 605 = (tin“'1' 2:2 (b) Substituting Eq. (1:) in Eq. (:1) gives the resonant;
amplitude: or 1 : . _1__ (c) = _ _ ”g = (us{)9
Llo (Hst)0 (2:1): + 4:3(1... 2:2) 2(f1— {2 or Problem 3.11 The given data is plotted in the form of the frequency
response curve shown in the accompanying ﬁgure: 5 Acceleration amplitude. 10‘3 x g Frequency. Hz (a) Natural frequency
The frequency response curve peaks at f" = 1.487 Hz Assuming small damping, this value is the natural ‘
frequency of the system. i (b) Damping ratio The acceleration at the peak is 'beak = 8.6 X104 g. Drawahorizontal line at rpm + J5 = 6.08 x 10*3 g to
obtain fa and fl, in Hz: i f,. = 1.473Hz fb a 1.507Hz Then,
fb — fa 1.507 —« 1.473 = 0.0114 21; 2 (1.487)
= 1.14% ...
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 Spring '11
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