25_ch 20 Mechanical Design budynas_SM_ch20

# 25_ch 20 Mechanical Design budynas_SM_ch20 -...

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Chapter 20 25 20-34 x = S u = W [70 . 3, 84 . 4, 2 . 01] Eq. (20-28) µ x = 70 . 3 + (84 . 4 70 . 3) ± (1 + 1 / 2 . 01) = 70 . 3 + (84 . 4 70 . 3) ± (1 . 498) = 70 . 3 + (84 . 4 70 . 3)0 . 886 17 = 82 . 8 kpsi Ans. Eq. (20-29) ˆ σ x = (84 . 4 70 . 3)[ ± (1 + 2 / 2 . 01) ± 2 (1 + 1 / 2 . 01)] 1 / 2 ˆ σ x = 14 . 1[0 . 997 91 0 . 886 17 2 ] 1 / 2 = 6 . 502 kpsi C x = 6 . 502 82 . 8 = 0 . 079 Ans. 20-35 Take the Weibull equation for the standard deviation ˆ σ x = ( θ x 0 )[ ± (1 + 2 / b ) ± 2 (1 + 1 / b )] 1 / 2 and the mean equation solved for ¯ x x 0 ¯ x x 0 = ( θ x 0 ) ± (1 + 1 / b ) Dividing the ﬁrst by the second,
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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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