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# homework5 - Prob. 4.6-8. Solve Prob. 4.6-? if. instead of...

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Unformatted text preview: Prob. 4.6-8. Solve Prob. 4.6-? if. instead of being a solid core with outer diamater d; = 2.0 in., the inn: element is tubular with an outer diameter d: = 2.0 in. and an innar diameter (d1); = 1.0 in. fr'o‘ n .' (-4) Maximum Maw r/reﬂcm ff a/A‘énktm ,' [Mt-'0: 7,775: 7; M 55mm: [Z fa» £143 — [la/Zr/ 584:? afar .' 5,21 ‘ {55/ 21'. 4mm Wm 7%“K6MVMWV—MJN WWW-m s _ 4; ___.3;4____ 9% 74;: 742 («91,»): zrﬁfﬂﬂf—MW1.0/3353 ’7’” two f‘ﬂ. ﬂame/g of o/c/ﬂ/ma/zbm' 4? ’45,: :gﬁﬂ {TH ‘42». M m (z) nJ-rJ. ’ I " K M far/gig, gag—z) ﬂ zrﬁlﬂrwwm 7; r Z; -7,7-‘ m: Mférgwa, {ﬁrmer .' ,w MUM/2L xa mm, L (mu)! : If“ _ [[Kuéij‘foéjrf] = n lament/z] Liege/«FL _ , (W6 rm = Fmam ' " 7’7 4* 932, = 0.1’245'm/ I’l’mb. 4.3-8. Solvc Prob. 4.3-? with d. - 4.0 in. and T = 60 kip‘ in. for the following values of ran: (a) rm... - 12 ksi, and (b) r... = 20m. r’ ﬁrm ﬂ: J‘ﬂ/u/ﬁ'ﬂn 24/ 4;}‘7; aé‘Z/¢/%)4- 7:" } m i7 77“!va Z MwKi'f. 1002 f?) (Q) 4/0 = 4’051 J 7: ‘50 é’é‘ﬁﬂ' ,« Kile“; : {'2 £37: 4 . f f . .e 4 z /(¢frm/4 _ diam Wf/Zkﬁ'] ZWM = [ﬂy-wok": = 725% 22 .OM. Wig-w 7.7,éZ i Pub. 4.3-1.0. The solid circular shaft of diameter d in Fig. P43-lﬂo has a twist rate of dﬂd; 9, under the action of the torqin 7, 1m: solid m is repucea'isy'mawsm withauﬁootouuide diametertoimide diameterofde; - 1.2 but weighing the same as the solid shaft (Fig. P4.3—10b), by what percentage would the torque have to be increased inordertoproduoeﬂiesametudstrmainthetubularahaﬂ? 4 Via: Wb 7rd 2 16.1 Ema: 2% zt(zﬁr}=xt(—ai~7') 431(4- .4) d2: doz_di2 (Tab 524d: q = \$41-\$92] =15d2'9‘D-(313H _dfﬂ_ do = l-(‘i‘icibz _d""_ 3 “(L/[.232 5% gig—1: ear : 9L: g T [6(ng dui -l %‘%Ei[l-{aﬂ] ' = (I.%o=aﬂq[1—(T.L2)4]Ta. = 5.54 553 ‘7on91 = loo 7:. ' ‘ look: " 354,55 ‘2 Zinnia '5 455 92 Prob. 43-14. Solve Prob. 43-13 if we stated extemally-ep- plied torque distribution is replaced by the cubﬁnﬂy-vuying torque shown in Fig. P4344. 53445—190! ﬂfj “(HE-«m Wé/ (ﬁrst? . [EI‘A‘JFJTQ’M If 5/1}; = 0: Wm? 'EI L :[f/rm = 474:4;th L 943-1 4 =5, H Ti: Jﬁ= 5% {av—4%) 5H] Mex/mam Nita: r/rcrr.’ ' 75M“, /6 77:: fwd/K}: 31*? = [0/3 ‘- (5) 5517/.- m‘z/Zw/i/f Qty, ‘ 0 L ﬁne/x raw “5%) =0 _ é’ziéz x x L I x r ' gin/4 [f/r)—Z/Z'} +?/z)] 5 45:35 féﬂd‘ 4%.! é! 4w; [Is—Mg) +/£-J‘]=«¢< 535:400/y12 L 0' Em ' 7?:(1 l—C I '< ‘\ 1—x —+—H'L-x —>{ ' Problem! 44-4 though 4.4-7. For these four pmbkmi. m the genetic torsion member 18mm above. For each problem (a) determine the maximum tensile stress, the maximum compnztstve meat, and the maximum shear stress- tmtl' (b) Show due mute: dew-mined in PM (a) on property Mimi 3m elements (as, see Hg. (.14). _ In “Mug ﬁlm problem“, make the team of the same: comment with the sense qf the torque T shown on the generic Wm! mauberﬁgm Ike remaining probfmsfor Section 4.4. 41 = mlaljvc Prob. 4.44. (See the “generic torsion mam " ﬁgure.) T‘Skip'in..d.-1.2in.,andd;-0in. a) mfmum etgeggee ’ ' r' m Egg : Wedge element: T _ ,2 2H = 0'- mA» ‘r QTAnemewse = o “’ O‘n = '2’Dsinecase £\ TWA" 4 = "Islnze "C(Ancosa \<'e ' x Kr 2Fb=ot TmAn "' ’EAH {Coezs- 6in28§=o "I T“: :Lfcos 29 ‘ 6m ‘93 :_TCO‘329 TJAnsineE So (Urnij :- c- 1: (5mm); 7 Gr: (45") =- ‘1: Tm.” “ ’Ilnt{‘io‘] = "L ﬂora mfg-3r yd - 615/ T ‘= lgE; Maximum at ochrewji-‘ace, 80 '1 ' 3} "EH = WT J47 . = (szi— {oinrrﬂ ' “JﬁélkjL n {OMB-r : ﬂu?" (Snug: a “I‘LW Tm =14.74 K5] Hal-.me mximm normal Q-EFZQSES maximum Shem" sireaa ‘ij Kcmm ‘ ‘ \ / I: HEP-I kﬁl ._._._.r.. ’[Lmax '= H .14 k5! ___ 1 m] I: __ __ _ _ x \ "Em ’Em (GMMJJT M “OWE = VH4 L6; This ﬁgure ofa generic roman member applies to a“ of twist angle Prob. 4.5-6. A torque is applied to gearA of a twoshaft system and is transmitted through gears at B and C to a ﬁxed end at D. The shafts are made of steel (G = 80 GPa). Each shaft has a diameterd - 32 mm, and they are supported by frictionless bearings as shown in Fig. P4.5-6. Ifthe tortu applied to gear A is 400 N -m. and D is rtrained, (a) determine the maximum shear stress in each shaft, and (b) determine the angle of rotation of the gear A relative to its no—load position. P456 QW EBDI: shafbeedjon GE £321: member (I) EEzD 16?]de 9x2,ch of member (I) wemlxr‘ (2) A ( A H 5 ' c ) . I] I (2 X1“ {F K— }- x—Q@:r3i ‘lCD N 4‘11 TI LlCO 'm. N Fra FF; ,0. éMﬁo: T, F" I n; Tz”‘Fr¢.‘ﬁm_-m_ ® She ITMd/g] Hem Istqooszt (Imam: I : Ta]; = 71152 I‘l’ll’l’n3 f; r F ET; Kai/21 IQQI'H “£22151 (Lula = :1; ‘ (was " "Hfb'meF =m rat—max}. "g [92.2, (M); 55.5MPa, aw T _ J: ) 2- % (b1 ‘; TitI—rl‘ l; -— 93,. = Wzogsargzmml“ ‘- ldﬂLﬂoiLblm % ® 0 fad: it; = (532,31“ 'lrﬁoePatﬁgmmP = be‘HBllo‘ - (3)2 :ETZ CPI '1 \$s:_ CPe. {Dc CbBrB = _d>crc (132—:d34. r : CD3: -(%é)¢3c r3 (1).: = R. (“'00 NM) + (ll-{o mmﬁmﬁ 28%. ‘5?- M' ml = W cm- oouez rad. 1“? ¢d = Ct): + ® Prob. 4.6-2. 1\ solid circular shun AC of total length L - L; + LgisﬂxedegeitmmteﬂonetendsAdemdls loaded by a torque T. st joint 3, as shown in Fig. PMS-2. If the diameters of the two segments of the shaft are in the ratio 112%, what ratio of lengths, balls, will muse the maxi- mum shear stress to be the same in the two segments, that is. will make (7...). - (r_.),? In your answer, express ML. in terms of dga'dl. TB iMx'O1Tz'TH TB Xe mm s T: B 1; TQEqFAg‘Tﬂ'jgt L ﬁtl = (611:3? 1: \$4 (bl =- itITI :tz: (Eff z: 'TTigsizq 02 ‘ f1;sz (bl = “CD I __. ._ \$3: CD: }% (bl CD2 2mm :u—TI = _5E;2:T2 ; 1:; d x : "Em ‘2 TI- Tz 2* = {Tagging In In 1 1T ﬂ Jéafa 2'5 rm = @5le \$2ng (2’13 :14 ‘5. = | L2. Cl: ...
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## This note was uploaded on 06/11/2008 for the course AME 20241 taught by Professor Wagner during the Spring '08 term at Notre Dame.

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homework5 - Prob. 4.6-8. Solve Prob. 4.6-? if. instead of...

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