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252_Mechanics Homework Mechanics of Materials Solution

252_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 3.113 The solid circular drill rod AB is made ofa steel which is assumed to be PROBLEM 3.117 elastoplastic with r, = 160 MPa and G = 77 GPa. Knowing that a torque T= 5 kN-m is applied to the rod and then removed, determine the maximum residual shearing SOLUTION From +1.: soruh'o». +0 PROBLEM 3.:13 c: 0.025% J J.- 913.57 ~10" qu = 0.5447 3 P" =o.omms m . _ (P. _ T_L7‘r_ L1? A?“ logo’uma Y- %‘-- Q- A}? - W -- Ena- _ Ioheowo‘) ... .. ° C990“: ‘W — 1.9713 “A ~ 89.30 During Uhjouia'vw (P’= L (Eapns'Hc) 4:13 5‘10: N-IM G-J' '- (Extos‘xro) _ _ o ‘53 “(77“0'X61337x10'? ‘ "05 33 H4 - 60.64 Perms-new} +wn‘s‘i' angle qum = (Prod .4) r JAWS—1-0533 .- omaow = 23.7“ 3.114 The solid circular shalt AB is made of a steel which is assumed to be elastoplastic with G = 11.2 X 10 6'psi and ry= 21 ksi. The torque T is increased until the radius of the elastic core is 0.25 in. Detemrine the maximum residual shearing stress in the shafi after the torque T has been removed. SOLUTION 3.118 In Prob. 3.114, determine the permanent angle of twist of the shaft. PROBLEM 3.118 From Hm 5.0)on +° ”PROBLEM 2mm) a =03): o.) J”: 03470: M A-F‘P‘er pot-Jinn T = 18.3823 k-f- I"! ’ Pr 1'- 0.’l5'3'1 J 2 _ ,. -51-...11-2’ Y'PTCP ' (P‘p ”fire- {it} 7 2116:) -Gr=llr7._‘>flo_‘ P55 -. ”-Q‘fIOspsi - (231(21) . 9 _3 _ D _(O.25)Z_Tt.2>e_ae*') - ’30 “to ml - 10.31 t T: 18.383 Mrs. where L: 2'4 in.) -.- TL; _(_is.3331(a~t) .. ends = 5'4” GJ (ll.2¥103)(o.‘ifi7o|) _ ”'2 “J u' Parmanan‘i‘ M39: o?+w15-} (ppm: Q20“. _ (P; Chem : "3° “10's - 7‘7-26 3‘10“ = 100.74 ”6%.! = 5.77” :1 ...
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