24_ch 08 Mechanical Design budynas_SM_ch08

24_ch 08 Mechanical Design budynas_SM_ch08 -...

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Chapter 8 227 F i = 0 . 75(561)(10 3 )(600) = 252 . 45 kN σ i = 252 . 45(10 3 ) 561 = 450 MPa σ a = CP 2 A t = 0 . 33(80)(10 3 ) 2(561) = 23 . 53 MPa Eq. (8-42): S a = 1 2(129) ± 830 ² 830 2 + 4(129)(129 + 450) 830 2 2(450)(129) ³ = 77 . 0MPa Fatigue factor of safety n f = S a σ a = 77 . 0 23 . 53 = 3 . 27 Ans . Load factor from Eq. (8-28), n = S p A t F i CP = 600(10 3 )(561) 252 . 45 0 . 33(80) = 3 . 19 Ans . Separation load factor from Eq. (8-29), n = F i (1 C ) P = 252 . 45 (1 0 . 33)(80) = 4 . 71 Ans . 8-38 (a) Table 8-2: A t = 0 . 0775 in 2 Table 8-9: S p = 85 kpsi, S ut = 120 kpsi Table 8-17: S e = 18 . 6 kpsi Unthreaded grip k b = A d E l = π (0 . 375) 2 (30) 4(13 . 5) = 0 . 245 Mlbf/in per bolt Ans . A m = π 4 [(
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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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