53_ch 03 Mechanical Design budynas_SM_ch03

53_ch 03 Mechanical Design budynas_SM_ch03 - I r c r o = 20...

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66 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design r c = 1 A ± r o r i br dr = 1 2 . 197 225 ± 6 2 2 r r dr = 2 2 . 197 225 ( 6 2 ) = 3 . 640 957 in r n = A ² r o r i ( b / r ) dr = 2 . 197 225 ² 6 2 (2 / r 2 ) dr = 2 . 197 225 2[1 / 2 1 / 6] = 3 . 295 837 in e = R r n = 3 . 640 957 3 . 295 837 = 0 . 345 12 c i = r n r i = 3 . 2958 2 = 1 . 2958 in c o = r o r n = 6 3 . 2958 = 2 . 7042 in σ i = 20 000 2 . 197 + 20 000(3 . 641)(1 . 2958) 2 . 197(0 . 345 12)(2) = 71 330 psi Ans. σ o = 20 000 2 . 197 20 000(3 . 641)(2 . 7042) 2 . 197(0 . 345 12)(6) =− 34 180 psi Ans. 3-79 r c = 12 in, M = 20(2 + 2) = 80 kip · in From statics book, I = π 4 a 3 b = π 4 (2 3 )1 = 2 π in 4 Inside: σ i = F A + My I r c r i = 20 2 π + 80(2) 2 π 12 10 = 33 . 7 kpsi Ans . Outside: σ o = F A My
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Unformatted text preview: I r c r o = 20 2 80(2) 2 12 14 = 18 . 6 kpsi Ans . Note: A much more accurate solution (see the 7th edition) yields i = 32 . 25 kpsi and o = 19 . 40 kpsi 3-80 For rectangle, d A r = b ln r o / r i For circle, A ( d A / r ) = r 2 2 ( r c r 2 c r 2 ) , A o = r 2 d A r = 2 r c r 2 c r 2 0.4" R 0.4" 0.4" 1" 1"...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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