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Sol HW5 - 7 3.82 A 1.5-m-long mum steel shaft of38—mm...

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Unformatted text preview: 7 3.82' A 1.5-m-long mum steel shaft of38—mm outer diameterdi is to be made of aeteelforwhich (“=65 MPa andG=77.2 GPa Knowingthatthe angle oft-wist must ‘ ’ not- exceed 4° when the shaft is subjected to agerque of§00 N; m, determine the large-‘15 , inner diametm' d2 which can be specified in the design. ' - 7 Problem 3.82 151.5”. :1: ac); Hm». = 0-0mm '7.’= eswo‘ Pa". 99: 5‘":- 6?-813210'5Vw! .3" =— lgkjecf) Tcz 7 7..- 2ch S'h'f-SS \(‘eqfifrewxefljl' = "—37—- " FLCLHI'CI‘i) , - “f ‘f ' 7 _- , _ . f ‘1. .QTCz / q _ (ZMGOOKQOI‘H = :- D 3 =m[ g3 M ,c'. = C1. — TT’L” r 0.9!? ————-——-7-, 1T(é.5’xl(34‘) H.435!!! . M liar-1‘ ‘ » 7 ‘ ‘ I _ TL -_ 2T1; - - ngsfil WJé requ-uremew‘f . 99 ~ —G;T “ MTGLCfi—Cfi) __q q“ 2T1. _: * ODHQ‘ (21(6093053 Ct— 9 "”5? ' , TF(7‘7-2>¢'ID")(G?.8£3*(O'5 c, = mew-«1055 = 12.49: My , Use smaller Vague elf-c. . C, m1 7‘ , A 2-,” Inescr'mm . aiz-zcezeAh,‘ Problem 4-5. - " 4-5 A beam ofthe cross sectienshown is extraded from an almm'm' um alloy for which ay= 250 MPa and av :45!) MPa. Using a factor of samy of 3 00 determm‘ e 7 ' V I I the largest couple that can be applied to the beam when it is bent about the taxis. Aflflan-fig 5M9. = g- =‘ Egg ='135'O MPG. = Isoxio‘ Pa. _Mbme-n+. A? mart; “w 2 Ma's. ‘ ,= fidekao'V: Q32.é7¥103 Mm»- Ia: £01: “32?: ' 4376'? x 103 WW4- ' I37= I. = £82._£.? xlog “.4“ I. + I1 4» I3 = Lemozxm‘ m“: 130902210"; ‘m't M I ' _ 7- > = "is" wL'H-p C: “i(30) =. 405...... 2 04??? w. ' .-_' I + 4:.4Loqozxtc9“)(*§0vt ‘) - r M — 736i - 0.04;: O 25.227“? Nam M razsz-m Emblem 4.18 418 The_ beam shown is made of a nylon for which the allowable stress is 24 MP2 ' fiztension and 30 MPaincom pression: Determine the largest couple M that can be applied to the beam. Tine he-Ji‘V‘c-fi ayr‘s pics ,Illfmm mind/e 'HLZ bo'i‘f‘om. m ,-= 3047.5 =12; m»; 1 0.0125». g . ym = 47.: man = —o.O.!7e"m~ I - fib$3+ {ME = 6440305?+_(.'6oa)(s“=)i = 25.25 w-osmm’ \ I 2 [2 =' fiblhfr+ Ag; = 7':- (2o-¥1533+ (5350400)" = 35.51.0105 ”M ‘I = I, + I2 = GILS‘ISXIOJ w.“ : €1.37: x ’0’? m ‘* _ ' El; , ‘ _ _ 13F: +mm Shie M: nghoggmma :- nsis Mm .' ' I . , 6‘! 5 , ' ‘ BGHOM: coupmh‘ou M =,§fllgg—T:;———W~—Ji 105-" NEW-- Claws:- sm-erw Vaguer ‘ Problem 4.21 4.21 Knowing that a,“ P» 24 ksi for the steeistrip AB, determine (a) the largest coug‘le M that can be applied, (1;) the corresponding radius of curvature. Use E = 29 '->< 1 psi. ‘ i ' . k ’ I= T'ib‘hi‘ ¥Cfiiilifiis = 1.36208 um" M" 6‘ —'. %E , c: (“,lzifii : 0475‘.“ w M= g; : W ,C I I 0.115 M: 250 jé‘fn 4 c 6: , EC (zqwo‘ )(0 £15) I - in 2 ”a _ = 2 . 0 r 5_0 I . I‘ ( P 'E‘ f 6.1., 24 x103 I I h ...
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