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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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 $5 = {‘8 {'05 Rowen w‘i‘ B . 30v 33:0; eiizmzwo'fi = sawewo'g m? -——_ Q ~ LLS‘ Tw.‘_=.+ {m sfmff BA, L= 8‘4 :2 = Win- J: 57.542Ho‘3 IN TL -: T 7G_D駥iD-3 V91! so” i 63 (H.2w H3" 3(57-s#~3>:[o'3) Roam“ J A, cpm = 993+ cpM8 = 6.5.7‘EBXSO-3r‘ui (par 3.7? 4 r_ Problem 3 53 3.53 The composite shafl shown consists of a 5-mm-lhjck brassjacket (GmSi = 39 M 2m Brassjacket L : 2M 511:} care: C Lag: 2C3 burr. = 0.02.0 M 5m J - lei-c" - E01020)”: 29.33am” m* 1' 5,33 (77.2a;o")(25I,33ym“") : 19_%25~40‘ N-m‘ Towfiue aarm‘afi 5?! she—(3 can. T = GKJ 3} Brass Each}? :2, = H. = 20 m. r «3.020 #1 c2: 20 +5: 2.5-qu =o.ozb-m J; : :LCC-f- C.")= %(0-915“- 0-020”) -‘ 332. 245mm“? an" @151 = (392!0")(3£2.2é5wo”j T H. IESSHUE N-M" Torque carriage far-4‘55 Tom? 4mg”, T= T, +71 = {aria 4 613' Q : T Goo = W—hr ‘3 ‘ L. G‘s-I ‘4' Gad-z Iq.q025’{os + Juf‘mgg xiog 17.8 {o Mai/M (a) flwlihfiuhfl SLEQr.‘;%_5iWJSS M EMSS n ed, TM: G “am = {31 C 5Q = (33512041(0'0251Cr7tgqgflo.“ r Imawxo‘ Pa ~ ££M:=§7.LFS'MP:~‘Q (U Maw—imam sketch}; sci-mas in 5.429,? core, w“ ’L’m= GYM = G, a = C72.zxro?)(o-ozo)(r7.8w mil? C t 27. c xiv Pa lgwczzs Ma in Ed} AMgfe o‘P ‘HJI'S'fi. cp= Lf‘? - mommwo") v 3.5—.7‘3‘29W0-3HJ g = 2.05” A '— Prob'em 3-57 3.5? and 3.58 Two Solid steel shafts are fitted with flanges that are then . bolts are slightly undersized and permit a 1.5“ rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowmg that G = 11.2 x 106 psi, determine the maximum 3.57 The torque T is applied 10 flange B. T _ fAeJ£g5=2JI1t Tszva-) C 1'- : O_51_S'.'«1. cc) Lag: :36r-W-)C:%OiTO-7S$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¢ ...
View Full Document

  • Spring '08
  • LINDASCHADLERFEIST
  • Trigraph, aluminum shaft ABCD, solid steel shafi, B‘F +UES+ Ee'llnueeu, Numba LCD—.ORMJ crwroezqm, Torque carriage far-4‘55

{[ snackBarMessage ]}

Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online