17_ch 20 Mechanical Design budynas_SM_ch20

# 17_ch 20 Mechanical Design budynas_SM_ch20 -...

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Chapter 20 17 For no yield, m = S y σ 0 z = m µ m ˆ σ m = 0 µ m ˆ σ m =− µ m ˆ σ m µ m = ¯ S y −¯ σ = 27 . 47 kpsi, ˆ σ m = ± ˆ σ 2 σ σ 2 S y ² 1 / 2 = 12 . 32 kpsi z = 27 . 47 12 . 32 =− 2 . 230 From Table A-10, p f = 0 . 0129 R = 1 p f = 1 0 . 0129 = 0 . 987 Ans. 20-24 For a lognormal distribution, Eq. (20-18) µ y = ln µ x ln ³ 1 + C 2 x Eq. (20-19) ˆ σ y = ³ ln ( 1 + C 2 x ) From Prob. (20-23) µ m = ¯ S y −¯ σ = µ x µ y = ± ln ¯ S y ln ³ 1 + C 2 S y ² ´ ln ¯ σ ln ³ 1 + C 2 σ µ = ln ¯ S y ¯ σ · 1 + C 2 σ 1 + C 2 S y ¸ ˆ σ y = ¹ ln ± 1 + C 2 S y ² + ln ( 1 + C 2 σ ) º 1 / 2 = » ln ¹± 1 + C 2 S y ² ( 1 + C 2 σ ) º z =− µ ˆ σ =− ln ¼ ¯ S y ¯ σ ·
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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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