315 HW_1_Solution

# 315 HW_1_Solution - I Eguivwlence of Pafenh'al energies...

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Unformatted text preview: I Eguivwlence of Pafenh'al energies gives 2 2. z 2- 2 2% “t4 9 +1} *eze +é 19,0911) #51401.) +‘2'Z *a(efz) = {- 1< e2 O ' l 2 @ Ezuivalenl: spring constant: in Jl‘ffent alfredrubns are 1: k k "r “ kc: =(———s——‘—7——) , kc; ‘ —'9_) . k - *1 “2. ‘k _ *3 1‘1, ea - k—k— ’ e4. " _ 1+ 2 k3 + If {he force P moves by x, SPrI'nj loco-fed at e; undergoes an Aisplacement 05- x; = 3: cos a; (JerEVah'an cos .‘n froHem 1.12), 4 Ezw‘vwlenc: of Po+en+chl energu gives i— 1<¢Z x1: 1. z «a 2:: i=1 The steel and aluminum hollow shafts can be treated as two torsional springs in parallel. For a holgw shaft, _ 7" 4 _ 4 he — 32‘, (D d ) For the steel slllgft, G = 80 (109) Pa, 6’ = 5 m, D = 0.25 m, d = 0.15 m, and hence kti = 71\$?)- (0.254 — 0.154) = 5.34072 (10") N—m/rad (00) For the aluminum shaft, G = 26 (10°) Pa, I = 5 m, D = 0.15 m, d = 0.1 m, and hence 9 kt? = W (0.154 — 0.10‘) = 0.207395 (10") N—m/rad keq = kt] + kt: = 5.34072 (10“) + 0.20739 (10") = 5.54811 (10°) N-m/rad a: (’09) ?w, [3 5m, 9:045", an‘! d: O'OSM, 11 q k = 'IT (zsna’) 4 *2 32(5) ~o'os )= a-255‘255ua‘ N'm/‘a-J 6 6 6 key: «hﬂtz = 5.31.072 and + 0-255255 m = 5.595975 m (o-lS N-m/rad ’Let a; : angular velodfy of the mofor (input) Angular veIOcn'h‘es' of dn‘fferenl' gear sets are: - ' ' ' ' ‘ ‘ ' ’ ' "T""’“"I""""""']"""‘T""""""""’“' "‘ ‘Tmeiﬂ" 3-! ' rial, ' J" '75 : I.“ ' J‘IZN’ aluac‘ - - — - - — - ---r—-------p---—---—-,-----o---—-----—----—---- , I'ﬂ I'nn' "_V:_|Y‘3.nz~-1 9* ' “('12) Maﬁa-3;): : a n; r" ) _______ -- ..--- _ -_----._-_-_ -_---_.-.. -_-___»_-__I’M. -_--_ Eguu'vdence cf kineh‘c energies gives .1- o .1 .Z 7- ‘55 a: = .25 Junior 9.: +‘é Z: J 91‘ + J- aicd 91¢..J \/ ® When mass m is displaced by x, the bell crank lever rotates by the angle 01, = This 1 makes the center of the sphere displace by x, = 01, 6’2. Since the sphere rotates with out slip, it rotates by an angle 15 0 __x1_9bl2 = 1:63 ' r, r, (1r, The kinetic energy of the system can be expressed as .2 ’2 .2 I .42 T=tm*+%%0+z"’4°a+imﬁ"* 1 t. u 3 >7‘ 1 1' I '2 i 2' _L 2. = — It 3' ——-— —-m r f 2"" + 0(g,)+2(5 5’5 h": since for asphere, J, = €- 111, r3. Equating this toT = %- meqiz, we obtain 1 7 6’3 ' =m+J —+— — m“ “a: sm'ef Fl '2 /@ F1 t: F; = clamping force of C; = C‘- (1'1;- ,2:); i=l,2,3 . F, c, . pea: Anny-n3 force of ca” 92012- .2.) "I .o #1: C35: Ca+c2+ca e ll (WILEEEFE‘I F . . . e a: cu(7<;— 7“) l—gc, l—j'tz hi, PA, F2: C1013" ’22.) ——E‘?‘ F3=‘3'(iq-"3) C2” C3 C1 ‘ . I ,_ I L L We F=a= Fm Fa= Fa - a + en .2, (c)EZua.ﬂn3 the energies dissipated in cu cycle, 1. z ' 2. 7r Cewa, =7rc,w x. + TrClCJ xf... 1r cam x3 where X,-.-' 91., x2: 912 M14 x3: 9!, .'. cab: cl+<z %z+ C3(_[?,’_>2 Ezua‘h'nj Hag energies dis'ru'patea' in a. cycle, 7" Carla: 9,1 =1r cum 9:4. 11' can} 52+ rceaw a: where 92: 9, omo‘ 93= 9! 2. .'. C = ‘C "I z n ...
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## This note was uploaded on 09/08/2010 for the course MAE 315 taught by Professor Wu during the Summer '08 term at N.C. State.

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315 HW_1_Solution - I Eguivwlence of Pafenh'al energies...

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