29_ch 06 Mechanical Design budynas_SM_ch06

# 29_ch 06 Mechanical Design budynas_SM_ch06 - σ a = S e n...

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Chapter 6 175 Function Consequences Axial F a = 10 kip Fatigue load C Fa = 0 C kc = 0 . 125 Overall reliability R 0 . 998; z =− 3 . 09 with twin ﬁllets C Kf = 0 . 11 R 0 . 998 0 . 999 Cold rolled or machined C ka = 0 . 058 surfaces Ambient temperature C kd = 0 Use correlation method C φ = 0 . 138 Stress amplitude C Kf = 0 . 11 C σ a = 0 . 11 Signiﬁcant strength S e C Se = (0 . 058 2 + 0 . 125 2 + 0 . 138 2 ) 1 / 2 = 0 . 195 Choose the mean design factor which will meet the reliability goal C n = ± 0 . 195 2 + 0 . 11 2 1 + 0 . 11 2 = 0 . 223 ¯ n = exp ² ( 3 . 09) ³ ln(1 + 0 . 223 2 ) + ln ³ 1 + 0 . 223 2 ´ ¯ n = 2 . 02 Review the number and quantitative consequences of the designer’s a priori decisions to accomplish this. The operative equation is the deﬁnition of the design factor
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Unformatted text preview: σ a = S e n ¯ σ a = ¯ S e ¯ n ⇒ ¯ K f F a w 2 h = ¯ S e ¯ n Solve for thickness h . To do so we need ¯ k a = 2 . 67 ¯ S − . 265 ut = 2 . 67(64) − . 265 = . 887 k b = 1 ¯ k c = 1 . 23 ¯ S − . 078 ut = 1 . 23(64) − . 078 = . 889 ¯ k d = ¯ k e = 1 ¯ S e = . 887(1)(0 . 889)(1)(1)(0 . 506)(64) = 25 . 5 kpsi Fig. A-15-5: D = 3 . 75 in, d = 2 . 5 in, D / d = 3 . 75 / 2 . 5 = 1 . 5, r / d = . 25 / 2 . 5 = . 10...
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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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