19_ch 08 Mechanical Design budynas_SM_ch08

19_ch 08 Mechanical Design budynas_SM_ch08 - 7854 in 2 If P...

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222 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design σ a = CP 2 A t = 0 . 239(9 . 72)(10 3 ) 2(58) = 20 MPa n f = S a σ a = 86 . 8 20 = 4 . 34 Ans . (c) ASME elliptic S a = S e S 2 p + S 2 e ± S p ² S 2 p + S 2 e σ 2 i σ i S e ³ = 162 830 2 + 162 2 ´ 830 µ 830 2 + 162 2 622 2 622(162) = 84 . 90 MPa n f = 84 . 90 20 = 4 . 24 Ans . 8-33 Let the repeatedly-applied load be designated as P . From Table A-22, S ut = 93 . 7 kpsi . Referring to the Figure of Prob. 3-74, the following notation will be used for the radii of Section AA. r i = 1in , r o = 2in , r c = 1 . 5in From Table 4-5, with R = 0 . 5in r n = 0 . 5 2 2 ( 1 . 5 1 . 5 2 0 . 5 2 ) = 1 . 457 107 in e = r c r n = 1 . 5 1 . 457 107 = 0 . 042 893 in c o = r o r n = 2 1 . 457 109 = 0 . 542 893 in c i = r n r i = 1 . 457 107 1 = 0 . 457 107 in A = π (1 2 ) /
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Unformatted text preview: . 7854 in 2 If P is the maximum load M = Pr c = 1 . 5 P σ i = P A · 1 + r c c i er i ¸ = P . 7854 · 1 + 1 . 5(0 . 457) . 0429(1) ¸ = 21 . 62 P σ a = σ m = σ i 2 = 21 . 62 P 2 = 10 . 81 P (a) Eye: Section AA k a = 14 . 4(93 . 7) − . 718 = . 553 d e = . 37 d = . 37(1) = . 37 in k b = · . 37 . 30 ¸ − . 107 = . 978 k c = . 85 S ± e = . 5(93 . 7) = 46 . 85 kpsi S e = . 553(0 . 978)(0 . 85)(46 . 85) = 21 . 5 kpsi...
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