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# hw14-10sol - W i\4 ~ 40mm 5—32 The shaft is made from a...

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Unformatted text preview: W i \4 ~ 40mm 5—32. The shaft is made from a solid steel section AB and a tubular portion made— of steel and having a brass core. ‘ If it is ﬁxed to a rigid support at A, and a torque of T = 50 lb - ft is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear - stress and maximum shear strain in the brass and steel. Take Gst = 11.5(103) ksi, Gbr = 5.6(103) ksi. Equilibrium: w.” Tb, + Ts, —- 50 =‘ 0 on TV +T¢ = 1000 us-ia (1) Both the steel tube and brass core undergo the same angle of twist due/B ¢ _ :I;_ T:*(2)(12) = T:;(2)(12) C/B JG 2 (0.54)(5.6)(1o")" 4(14 — 0.54)(11.5)(106) Tb, = 0.032464 7,, (2) Solving Eqs. (1) and (2) yields: T,, = 48.4281b-ft; Tb, = 1.5721b-ft ¢ = Z'E_1.572(12)(2)(12)+ 50(12)(3)(12) : d + (5 C JG 2 (o 54)(5 6x106) :(1‘)(11.5)(106> “‘3 W = 0.002019 rad = 0.116° Ans. T 50 12 1 (Tst)max AB = 336 = “£7129 = 382 psi (13,)mm BC = 23—6 = W - 394. 63 psi = 395 psi (Max) Ans. (mm _3___94. 63 _ = = 4 X b . (7:!)max G 11 5(106) =W 3 b6 \0 Vol Ans ‘ 'T , 1.572 12 0.5 . . (Tbr)max = Sc: ”Hi—((15):?) = 96.07 p51 = 96.1 p51 (Max) Ans. _ (Tbr)max = 96-07 = -6 (mm — G 5.6(106) 17.2(10 )rad Ans, (ﬁe) Tau): 7" Tn \NQYL‘ITLWHAK'L g W3»? Y/W‘J: ”V1.4 m 4% AL? N %0LUT\U~\' \ng' 7‘ a 6 j VJ» 6 {Mi/— 1'3 {,mga‘h. is ”QM. Egg": 1‘ ( \II momm um *5 331% \A + T3 — ‘50 mm A». L10 Q 423 or TV. \000 IVE N T}, “MW“ “910'“ U3?" 4331145 : T‘ao * ‘SO'Lrsqm (1.; {U} 5E :{5 \\ : T - 80x 11 “mm?“ Ugo-n AL: JET—43., " “WWW“: MW '— WM (or We + ¢Hc ““530 8:13” CT‘B ' 50x\ Ax v Us 5") a 14- ' \. 4 1&1 Q 6 12%“ G ’— 7’ \10 I. \ \ \Y)X' 9/87— a v (\OOO—TQ )6) Toy/UN? DI\A\M\$M‘ > 4 ' W M, mum (\Y’LW {3/4 ER ‘ a5 ‘lOTx; _. \opoo : 33000 f 5-73 k «w QM" W T CW ' “@013 “310% + \"L‘soTxa : \ﬂ‘éOIOOO Hw‘ooo -' 440 000 \ :; \‘ ‘ : 0 _ , B \uoo Ck Ow ”1 “A’: WOO u: w “its ~ \00 Lu \V\ ‘ A : TAO = \ODuevquOJlSvn 0.14 ” m A‘ “”“‘ 3’ 1: mm A‘ (’6‘ 7/ VT MB» T’ _ T33} __ w : 43:64 Pg; LAIﬂcL‘M Mkmrum) ”’5‘ g T E {0‘11”“4 _.________ Ti 4‘11wa 5‘3 Ooolboo ”M— E\ ‘ “4 ‘ / (“Wm : \cmw N.“ \Lo ‘ \ H‘W COIOEQ m 2 SM r9“. (_ {ij 01:91:: : w w AA 13 : Tm : \M‘m N-Mﬂﬁm A) = Howls N-«A How. E , v P" M ' buxom 15 [HA wk 3 \ Z; ‘ ”m “M" ‘3 -, wﬂfiégﬁﬂ 51:)? _ 0 0,0373“ w lqb N.“ - C (ALL' QM \,m r ga'i.ué%'c“' Jug ~ ' E \l .. V j- ' if M \W’MJMAY (Via—iv X7) c.0- aml More, vow \ 3., 4%“: ("Pail-9;? \ (Lawn 3w Lug W [email protected] Quote)“ 5 “mi (anti/(H I 9 E 7" ‘30 A £4le +321? 95: Tug, 2 (DOWN (1,4) ‘3‘wa 1%? Y quékwx I'M 03145; (DON-H + 4on ~13 :qs V (,mkm \M} T K00 N'M ’2 ,30M Y ’L‘K (i — EM“ : \»1‘5 “‘1 ”-7 EW 3 0.07 \L 3; \AJS {CW ham me 932) (64ch A'%- W ; M 1' ' (OONW‘LO 0M3 . HIM. K j ' \AJE) 1; LETS—7:3? : KD‘R13[000 N/MQ 3 L20“). N00. (JISAA'Q WM (umw C . -- T“? C’ " \DQN'“ (0.015 '9 '9. 4x 9. v My”! .. “mm W ' T Y; UHZS a“ 3 4.01 Mpa ...
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