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27_ch 06 Mechanical Design budynas_SM_ch06

# 27_ch 06 Mechanical Design budynas_SM_ch06 - 9 17 95 ´(1...

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Chapter 6 173 6-36 S ut = 58 kpsi S e = 0 . 506(58) LN (1, 0 . 138) = 29 . 3 LN (1, 0 . 138) kpsi Table 6-10: k a = 14 . 5(58) 0 . 719 LN (1, 0 . 11) = 0 . 782 LN (1, 0 . 11) Eq. (6-24): d e = 0 . 37(1 . 25) = 0 . 463 in k b = 0 . 463 0 . 30 0 . 107 = 0 . 955 S e = 0 . 782[ LN (1, 0 . 11)](0 . 955)[29 . 3 LN (1, 0 . 138)] ¯ S e = 0 . 782(0 . 955)(29 . 3) = 21 . 9 kpsi C Se = (0 . 11 2 + 0 . 138 2 ) 1 / 2 = 0 . 150 Table A-16: d / D = 0, a / D = 0 . 1, A = 0 . 83 K t = 2 . 27. From Eqs. (6-78) and (6-79) and Table 6-15 K f = 2 . 27 LN (1, 0 . 10) 1 + ( 2 / 0 . 125 ) [(2 . 27 1) / 2 . 27](5 / 58) = 1 . 783 LN (1, 0 . 10) Table A-16: Z = π AD 3 3 2 = π (0 . 83)(1 . 25 3 ) 32 = 0 . 159 in 3 σ = K f M Z = 1 . 783 LN (1, 0 . 10) 1 . 6 0 . 159 = 17 . 95 LN (1, 0 . 10) kpsi ¯ σ = 17 . 95 kpsi C σ = 0 . 10 Eq. (5-43), p. 242: z = − ln (21
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Unformatted text preview: . 9 / 17 . 95) ´ (1 + . 10 2 ) / (1 + . 15 2 ) µ ´ ln[(1 + . 15 2 )(1 + . 10 2 )] = − 1 . 07 Table A-10: p f = . 1423 R = 1 − p f = 1 − . 1423 = . 858 Ans . For a completely-reversed design load M a of 1400 lbf · in, the reliability estimate is 0.858. M M D 1 4 1 " D Non-rotating 1 8 "...
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