# HW5 - Problem 3.32 3.32(a For the solid steel shaﬁ...

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Unformatted text preview: Problem 3.32 3.32 (a) For the solid steel shaﬁ shown [G = 77'GPa), determine the angle of twist at A. (b) Solve part 0, assuming that the steel shaﬁ is hollow with a 30-mm-outer diameter and a 20-mm-im1er diameter‘ \ Ct—é-di'. 0.0l5rw) J:TT—2-Cue %S:0,0:Sjw I J = "3?. 52:2 my” m”, L: 1.3», J 6: WHO" Pa mmm WE; 251mm g) : L250)CL3-) = '13.qu!0 rad (77%l0qu'l‘1522wo" q) : s____*’73-‘tq_>‘r10'33‘§°: 4.21" at l. C2: 0.015 w) CI:-édl:o_olohj J=%(CZ‘J_CI4) — 4‘ - TL. = %(o.015“— 0.0m } = gaswnoqmt g; = 6—5 a 2%: qu’led‘md : L359(9l.53¥{o'3)=5:25 9'“ SD (?7>qo")(63.8twlo" Problem 3.35 "WWW-- 300 X - m 3.35 The electric motor exerts a 500 N - m torque on the aluminum shaft ABCD when it is rotating at a constant speed. Knowing that G = 27 GPa and that the torques exerted on pulleys B and C are as Shown, determine the angle of twist between (a) B and C, (b) 3 and D. (<1) Angle é rho-{sf hefweeia 8 and C : Tag: 200 Nun) Lu: L2 M c Plat : 0.022% G : 27x/o‘ Pa ‘1‘}: lie” 3c19‘2xto” m q) 2 E r (200ml) -3 Bit GJ' WT: Zq'm‘uo “‘4 QM? l.38‘-t° GD} Angie B‘F +UES+ Ee'llnueeu B 5.45; D... ’11,, .500 Numba LCD—.ORMJ crwroezqm G:2?w0°10q a, 31:. = 3(o_02\$)’= aardssvto'ﬁm‘“ ,% _ (27V{09)(52l.t\$3 not) ' I! 3!, agenda we (PEA!) : (Peat (pt/n r Q‘LlS—r’xto'34- 3L‘ré3mxo“ = Seawateer F (para 1 «2.on 4% Problem 3.40 3.40 Two shafts, each of g-in. diameter are connected by the gears shown. Knowing that G = l 1.2 x106 psi and that the shaft at F is ﬁxed, determine the angle through which end A rotates when a 1.2 kip ' in. torque is applied at A. Calpculpe‘il'om o‘F 4oviues¢ C in» win audit) cav‘i Mi ‘Pav‘ce Lei-wee». clears B Mai E. : To; ' G E” E 113-1.? kip-in '25-’ch rn Er: [\$30200 " iGOOIPii‘Hr Twis‘i' in sim‘H FIE-Ea L: :2 in.) c _ ‘f - __ FW _ ~1— ' I; C" ‘ 1'36!" F TL 06003013 ‘3 ape = _: _______ -_» 29,78? .Io rad ’F GI {In«:0‘1-557..5"+8v.rc53) x Quin/hm (A E, :95 = QWF = 29.?2ﬁw/0'3ma' Tamievzi’ir) yifsrajacew-wj 64+ 954’“ ’56-“er 5: F“E \$5 = {‘8 {'05 Rowen w‘i‘ B . 30v 33:0; eiizmzwo'ﬁ = sawewo'g m? -——_ Q ~ LLS‘ Tw.‘_=.+ {m sfmff BA, L= 8‘4 :2 = Win- J: 57.542Ho‘3 IN TL -: T 7G_Dé§¥iD-3 V91! so” i 63 (H.2w H3" 3(57-s#~3>:[o'3) Roam“ J A, cpm = 993+ cpM8 = 6.5.7‘EBXSO-3r‘ui (par 3.7? 4 r_ Problem 3 53 3.53 The composite shaﬂ shown consists of a 5-mm-lhjck brassjacket (GmSi = 39 M 2m Brassjacket L : 2M 511:} care: C Lag: 2C3 burr. = 0.02.0 M 5m J - lei-c" - E01020)”: 29.33am” m* 1' 5,33 (77.2a;o")(25I,33ym“") : 19_%25~40‘ N-m‘ Towﬁue aarm‘aﬁ 5?! she—(3 can. T = GKJ 3} Brass Each}? :2, = H. = 20 m. r «3.020 #1 c2: 20 +5: 2.5-qu =o.ozb-m J; : :LCC-f- C.")= %(0-915“- 0-020”) -‘ 332. 245mm“? an" @151 = (392!0")(3£2.2é5wo”j T H. IESSHUE N-M" Torque carriage far-4‘55 Tom? 4mg”, T= T, +71 = {aria 4 613' Q : T Goo = W—hr ‘3 ‘ L. G‘s-I ‘4' Gad-z Iq.q025’{os + Juf‘mgg xiog 17.8 {o Mai/M (a) ﬂwlihﬁuhﬂ SLEQr.‘;%_5iWJSS M EMSS n ed, TM: G “am = {31 C 5Q = (33512041(0'0251Cr7tgqgﬂo.“ r Imawxo‘ Pa ~ ££M:=§7.LFS'MP:~‘Q (U Maw—imam sketch}; sci-mas in 5.429,? core, w“ ’L’m= GYM = G, a = C72.zxro?)(o-ozo)(r7.8w mil? C t 27. c xiv Pa lgwczzs Ma in Ed} AMgfe o‘P ‘HJI'S'fi. cp= Lf‘? - mommwo") v 3.5—.7‘3‘29W0-3HJ g = 2.05” A '— Prob'em 3-57 3.5? and 3.58 Two Solid steel shafts are ﬁtted with ﬂanges that are then . bolts are slightly undersized and permit a 1.5“ rotation of one flange with respect to the other before the ﬂanges begin to rotate as a single unit. Knowmg that G = 11.2 x 106 psi, determine the maximum 3.57 The torque T is applied 10 ﬂange B. T _ fAeJ£g5=2JI1t Tszva-) C 1'- : O_51_S'.'«1. cc) Lag: :36r-W-)C:%OiTO-7S\$ﬂ, = gwﬂsv -‘ ammo: [V1 " 63in GJcn 1- ‘1‘ G -' ' T”: _:ip£ : chc : f_v;L,I.(;,2§,\qr0s CRL Cpéqrance. r‘d'I'a'tfow have #jamﬁe E 1' " L50 -‘ 26J3>WOA mud naive +0 we‘moua aﬁeamma: 7;; '=_(m-253~:o‘)(26. lauds]: H22.3}é. .'n Equal— : ._ . ' 'f's: cause «Jam‘an V‘c‘z‘hﬂi'fon (PH: T” _ H ._ T” r In + H 5 F u: =5040— 2‘?28,3= 2H :,7 is”; 7mm =(HL353*IoatﬁPWUSLLGZErIoUQD‘ q)”: 7_q23m—'3mp 73;.- (:;3,353 xgog‘jﬁ'fﬁzsxlo'S) : 236.2! 154»: Tc; = 059-92930310. 513* to“) = new. as; 49th.}. Maxr‘mUM sheet-"I'm; stress in H g . I) . ' - LE: Téac r 29,223+ 32.5.2 (0 e25): W60 Fe“. GAE, .23 96% Maximum shew-tn s+v~ess in CD. _—-—‘4________ _ T c U225.oqt(0;75) _ . T" 33:: : TW— = "3"“? re ‘3‘“?“5' “‘ ; 15.3: g-ﬁfks‘. 4 Asimmﬂ-L - — ' 0 rpm. Knowing that G = 3. I A t e1 shaft must transmit 150 kW at speed of36 I ‘ Problem 3'81 7752 GPS: design a solid shaft so that the maximum shearing stress Will not exceed 50 MPa and the angle of twist in a 2.5m length must not exceed 3°. 3&0 _ r, 1 PHSokwmgovsoaw ¥= go - art: "— P = -——"5°"l°1 = 3_¢i789><!o3 N-w l ' QTT'f 7.1T (a) Tc 2T S‘l‘ress r: utII/‘emewfi. T? SOVI’O‘P = Hf _ T‘C.3 ______‘L.—l 3 _ lg/(2\(3‘ﬁ-lgﬁallow5 ‘-_ sloo‘ﬂ‘lodim = 3?-00 run 5: W h 'lTCSvaO“): a An )5 as". ‘l'wl's‘ll M Ottawa-Jill. CPrgo=52-gé’lo V‘M’l q; _ IL? = ﬂue—sz}, 6:77.22rloqua} L= 25w " G5 'n’ r. - rm- ‘f/KTL— _ “ (xwtseisewoﬁMZ‘Ei = 3332,40“ m: {be m, a: TTGC? h ‘rr(77.2xlo“)(52.scxlo"-‘j 5?: 2c =?L}.omm 423 Use. Jam-:36.“ Vﬂ/pUE_ a: 37100 mm Problem 3 88 3.88 The Stcppcd shaﬁ shown must transmit 45 kW. Knowing that the allowable shearing stress in the shaft is 40 MP3 and that the radius of the ﬁllet is r = 6mm. determine the smallest permissible speed of the shaft. 2 _ E _ 2- £19 e d‘so'o'z cl'30'2 “V‘DM 3.32 LZG T For shqaallller Ej'rlﬁ‘ a: 410% lSmM r 0.0l5m 7;]: m z ZKT' 3 ( 3 J) 77:23 “rt: ’3.” 7.? 0.061(40 not . :5 T7 "-' ____———————-—————— : lé8,30x10 -m 2% (210.20 N P = '45 w = 45mm P = 2m?!— ¥ __ P I __l “SHOE __ 2w we Hz -l‘-'LlQ.éHz-e¢ ...
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## This note was uploaded on 02/25/2010 for the course ENGR 2530 taught by Professor Lindaschadlerfeist during the Spring '08 term at Rensselaer Polytechnic Institute.

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HW5 - Problem 3.32 3.32(a For the solid steel shaﬁ...

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