24_ch 06 Mechanical Design budynas_SM_ch06

24_ch 06 Mechanical Design budynas_SM_ch06 - k a = 4 ....

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170 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Now C Se . = (0 . 12 2 + 0 . 139 2 ) 1 / 2 = 0 . 184 S e = 91 . 1 LN (1, 0 . 184) kpsi Ans . 6-33 A Priori Decisions: • Material and condition: 1018 CD, S ut = 440 LN (1, 0 . 03), and S y = 370 LN (1, 0 . 061) MPa • Reliability goal: R = 0 . 999 ( z =− 3 . 09) • Function: Critical location—hole •Variabilities: C ka = 0 . 058 C kc = 0 . 125 C φ = 0 . 138 C Se = ( C 2 ka + C 2 kc + C 2 φ ) 1 / 2 = (0 . 058 2 + 0 . 125 2 + 0 . 138 2 ) 1 / 2 = 0 . 195 C kc = 0 . 10 C Fa = 0 . 20 C σ a = (0 . 10 2 + 0 . 20 2 ) 1 / 2 = 0 . 234 C n = ± C 2 Se + C 2 σ a 1 + C 2 σ a = ± 0 . 195 2 + 0 . 234 2 1 + 0 . 234 2 = 0 . 297 Resulting in a design factor n f of, Eq. (6-88): n f = exp[ ( 3 . 09) ² ln(1 + 0 . 297 2 ) + ln ² 1 + 0 . 297 2 ] = 2 . 56 • Decision: Set n f = 2 . 56 Now proceed deterministically using the mean values: Table 6-10:
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Unformatted text preview: k a = 4 . 45(440) . 265 = . 887 k b = 1 Table 6-11: k c = 1 . 43(440) . 0778 = . 891 Eq. (6-70): S e = . 506(440) = 222 . 6 MPa Eq. (6-71): S e = . 887(1)0 . 891(222 . 6) = 175 . 9 MPa From Prob. 6-10, K f = 2 . 23 . Thus, a = K f F a A = K f F a t (60 12) = S e n f and, t = n f K f F a 48 S e = 2 . 56(2 . 23)15(10 3 ) 48(175 . 9) = 10 . 14 mm...
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