22_ch 08 Mechanical Design budynas_SM_ch08

# 22_ch 08 Mechanical Design budynas_SM_ch08 -...

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Chapter 8 225 8-35 (a) Table 8-2: A t = 0 . 1419 in 2 Table 8-9: S p = 85 kpsi, S ut = 120 kpsi Table 8-17: S e = 18 . 6 kpsi F i = 0 . 75 A t S p = 0 . 75(0 . 1419)(85) = 9 . 046 kip C = 4 . 94 4 . 94 + 15 . 97 = 0 . 236 σ a = CP 2 A t = 0 . 236 P 2(0 . 1419) = 0 . 832 P kpsi Eq. (8-40) for Goodman criterion S a = 18 . 6(120 9 . 046 / 0 . 1419) 120 + 18 . 6 = 7 . 55 kpsi n f = S a σ a = 7 . 55 0 . 832 P = 2 P = 4 . 54 kip Ans . (b) Eq. (8-42) for Gerber criterion S a = 1 2(18 . 6) ± 120 ² 120 2 + 4(18 . 6) ³ 18 . 6 + 9 . 046 0 . 1419 ´ 120 2 2 ³ 9 . 046 0 . 1419 ´ 18 . 6 µ = 11 . 32 kpsi n f = S a σ a = 11 . 32 0 . 832 P = 2 From which P = 11 . 32 2(0 . 832) = 6 . 80 kip Ans . (c) σ a = 0 . 832 P = 0 . 832(6 . 80) = 5 . 66 kpsi σ m = S a + σ a = 11 . 32 + 63 . 75 = 75 . 07 kpsi Load factor, Eq. (8-28) n = S p A t F i
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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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