HW4 - 1 4000 N/m m «0 K7 Ht 400 C05 2041‘ N F0 400 N a w...

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Unformatted text preview: @ 1:: 4000 N/m, m: «0 K7, Ht): 400 C05 2041‘. N F0: 400 N a w: 20" rad/S : 091: 404'0‘ (rad/3),“ .. 'k ' wn’ ‘5," = 20 rad/A SOIHHOn I'S SI‘Ven (3-9): x(‘t)=< 0" COS Uni: + £70— Su'n cant —m092 n . Fa ' . 1 +(m cos (sf (5,, (“3 xo—‘o'l, X'O:o: EZ-(EJ) reduces to I”): {O-I — fl“ } cos 20 .t 4000— lo (404-0!) cos 204i 4000- I0 (404-0!) = 00.075062 C05 201‘ ... 3.975062 cos zo-lt “0:0: i°:|01 E3. (5") reduces fa :— ————————-—- L‘— ' 2:06) «Coo’mway'oofcu 201’: + 20 5m 20* 400 {4000 __ “(QO‘POO cos 20-!”(3 = 3-975062 Cos 20f + 0'5 sin 20": “Ia-975062 cos Zoolt (5'! (C) “9‘0": 7“,= (O: Ez-(E-I) gives x(e)- o-l — 4“ m . _{ HOOO— 50(qoq.o')} COS 20f + --o' Sm 20"; +§ #00 4-000- Io(qol1-ol) } “'3 20""; 138 F 000 8 = A: - a Q st k 400° .. O 025'“ 1 [ (w 2‘ 8“: 0-025 :: 8 . , " m = = = X st (2.); a," X 20x10.a l“ a» w a: = ..25+I =l-5 as“: CO ,.5 = 5(270/6 = 20.944 ra—J/sec m= */63,,"= 400%0-944f = 9-u89 ; w a": ‘154.’ =‘/ 5000710 == 22-3607 ra-J/lec 85*: F9/* .-. zso/sooo = 0.05 m i 6%i-—-‘-£>z} a,“ ’ g Ji . g '2 _ ova (-e., w = 0n <1_ 25:) __. 22-36071] Two (5'- 8H4 rad/83c 140 4‘ é J, *M X a w ' F (1’) Equation of motion for rotational motion about the hinge O: Steady state response (using Eqs. (3.3) and (3.6)): 0p(t) = 9 sin w t (2) F0 5’ where 6 = —————-——-—-—— 3 (k182+k2b2)—(J0+M(2)wz a” e 1 and Jo= “‘12 +m(—2—)2 =5“? (4) For given data, Jo = %- (10) (12) = 3.3333 kg—mz, w = igolség—Wl : 104.72 rad/sec, and 145 .—. _.__________m11________ 5000 (0.252 + 0.52) - (3.3333 + 50 (12)) (104.722) Equation of motion for rotation about 0: 6 e e 6 3 e 3 t’ i 5.; ‘ 'Jo'0'=—k——————k—-— e =—8.5718 (10“) rad ' Ussv 4 4 T i.e., JO'B.+[-58—kt’2 0=Mocoswt k - .. Fe 200 wn=———2aJr 8—’—__———— 3 I"! / , 3* k " 4000—0-05!“ *5: c/cc =(‘/2 m) :(40 z./4ooa(to) ) '-'- °°‘ (I94: 4...?” can -.-.- ‘l1-(oov)" (20) = (3-83974‘3 rad/5 LL IO _ , a where Jo = —1— m 6'2 + In = —7— m 6’2 = L (10) (12) = 1.4583 kg—-m2 _ L 12 4 48 48 _ Far and w = 1000 rpm = 104.72 rad/ sec. Steady state solution is: L 9,,(t) = 8 cos 0) t ‘ where , He n c e=‘5—_M—o—=———_-g—-—}9£————=—0.007772H 3- k t" — J0 w” 5000 (—8-) (12) —- 1.4583 (104.72”) i v 5 k: 4000 N/m, m: IO «7, c: 40 N-s/m, Fog-)3,“ w _ TL“; F0: 200 N, a: to rad/S, xoza-lm, 920:0 ' i 7‘ (’62 : ' F0 r: 09,, = '55 = 0-5 F?" x: 8:+ - 6-—r")2+(2‘sr)2 = 0'” ‘ H a 2 , '- 53 = {0—5; ) Hue-005)) = 0-066082 m 7‘? (0 _ —' arr .. 'l '5 i 55' {1“ (bra) = fat—n. (“aria-2%“) = 0432552 rat SCo “ steady stout-e response, E3. (3.25): X» S" no): Xcos(wt—-¢) x,= =o-06608’2 (.0: (lot — 0432952) m r : ¢°=1 Total resronse, 55.0457: 146 Equation of motion for torsional system: J.,‘é+ct (é—d)+kt (6—a)=0 where 9 = angular displacement of shaft and 0: = angular displacement of base of = a0 sin 0) t. Steady state response of propeller (Eq. (3.67)): k—ZOm—ank— 30112—4; 6pm = a sin (0) t —- 45) 1 , (2) where e = 04, J0 ths = ‘1 _.____...._—___.__. and ¢ tan kt (kt " J0 W2) + (W “)2 (4) Here Jo = 104 kg—m2, g = 0.1, and o.) = 314.16 rad/sec. Torsional stifl'nesses of shafts: (so (109)) (0.0‘ — 0.44)] G J 0:.)1 = —;,—l = ——-——30—————— = 27.2272 (10°) N—m/rad l (80 (109)) 1r— (0.4‘ — 0.24) G2 J2 a (kt), = 6,2 = —————é(—)————————— = 9.4248 (10 )N—m/rad Series springs give: _ _ _ _m m t _ (kt): +(ke)z — 27.2272 (10°) +9.424s (106) “’7‘0013 (10°) N / d et = g (2 ’\/Jo kt = 0.1 (2) 'V (104) (7.0013 (10°) = 52,919.8624 N—m—s/rad From Eq. (3), 1 2 (7.0013 (105))2 + {5.2920 (104) (314.1152)} 8 = 0.05 7.0013 (105) — (104) (314.16”) + 5.2920 (10‘) (314.10) = 9.2028 (10“) tad (104) (5.2920 (104)) (314.103) 05 = tan"1 7.0013 (10“) 7.0013 (10°) —- (104) (314.162) + (5.2920 (10‘) (314.1(1))2 = tan—1 (59.3664) = 89.0350° = 1.5540 rad ——___—_________________—_______—__——_— 151 @ C°"'leé‘ “W” “ “W Xoéw"tcos(w4*+¢.)+x cd(w*-¢) ca: zTr(3-5) = 2b33lz mA/m , wn=Jg= [25:30 = Sam, rag/m 8*: —%—= égrOOB-z 0072». _ c - +5 _ _ (.3 _ 21.79”. T “ zmcan ‘ 2('Io)(l5c8|i4) ' “+23 ’ r - 7.97. ‘ m‘ l 3908 jun-.- 1.7.5J (dd = Ji-T‘ can: IS-GSOS x : 89¢ _ 0-072 J(1"")z+ (1"): [(1- havoc"): + (“o-m: x 1-3908)1] h = 49-07015 m 4958 =ML (“zrr)= '4 0 >2 __ . o #5 1J2. fan ( 4H,“ 22. 9591 -2451: x(t) = X, e M (IS-6505f+ ¢.) + 0.07095 cad (zwrmt + 22.- 757I°) ' -- I I — ' f 7‘“): - 2'15 X0 3 2 1 «5 (’5'6505*+ 3‘0) - 5.6505 x, e1 25 4;,Q5.esast+g) - 2149:2(0-07095) man-Wu t +u- 75W) 1(a) = cools = X 06!. + 0'0709'5 ‘9‘ “'95?” X (15¢. = .. 0-0503; 0-- (51) Ho): 5 = -2-25 x. as 93, .45.ng x.4€..' ¢._u.s€os Ac—zz- 957? a—Ou — o 5 x M “ 5 x Mn 94° = 6°“ 2 2 ° ¢° = .. o-asn ----(Ez) cS-esos E%$.(Eg) axial (E1) give 1/ x, = {Go-05033)1+(_o.35ll)1} z = 0-3547 -1 -35n _ . go: far. (333-555) = thm'(G-37GO) = 31.8423 155 F; sin at Fx, reaction-7 3 at O ‘m 1‘ It Equation of motion for rotation of pulley about 0: —k2 (0r)r'—Jo.9.—k1 x(2 r)—ci(2 r)+Fo sinwt(2 r)-—mii(2 r)=0 (l) where 0 = x/(2r). Equation (1) can be rearranged as: J . ~2i+2mr ii+2crx+l2k1r+}ik2r]x=2rFosinwt (2) :- For given data, Eq. (2) becomes 115i+50ic+112.5x=5sin20t (3) : Steady state response is given by Eq. (3.25): xp(t) =X cos (w t —- gb) where X = 5 l = 0.001136 m 2 2 ; {112.5 — 11 (202)} + {50 (20)} so 20 and ¢ = tan“1 ——-——(—l-;- = — 0.2291 rad = — 13.12s7° 112.5 — 11 (2o ) . (a) 2 Mo = 0 (about hinge): 103+ k914£]§;£+(c€d)f=%Fosinwt Foe sinwt or Io§+clzd+—126-k(20= Magnitude of steady state response: 1 e,=[F°€]/{—9—ke’240ml}2 +(c6’2w)2 ‘2— 2 16 158 ...
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This note was uploaded on 03/26/2008 for the course MAE 315 taught by Professor Wu during the Spring '08 term at N.C. State.

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HW4 - 1 4000 N/m m «0 K7 Ht 400 C05 2041‘ N F0 400 N a w...

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