20_ch 09 Mechanical Design budynas_SM_ch09

# 20_ch 09 Mechanical Design budynas_SM_ch09 - 107 = ² . 537...

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258 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The von Mises stress σ ± is σ ± = ( σ 2 y + 3 τ 2 xy ) 1 / 2 = [2648 2 + 3(1662) 2 ] 1 / 2 = 3912 psi Thus, the factor of safety is, n = S y σ ± = 32 3 . 912 = 8 . 18 Ans. The clip on the mooring line bears against the side of the 1 / 2-in hole. If the clip ﬁlls the hole σ = F td = 1200 0 . 25(0 . 50) =− 9600 psi n =− S y σ ± =− 32(10 3 ) 9600 = 3 . 33 Ans. Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 with S ut = 58 kpsi S ± e = 0 . 5 S ut = 0 . 5(58) = 29 kpsi Table 7-4: k a = 14 . 4(58) 0 . 718 = 0 . 780 For the size factor estimate, we ﬁrst employ Eq. (7-24) for the equivalent diameter. d e = 0 . 808 0 . 707 hb = 0 . 808 ± 0 . 707(2 . 5)(0 . 25) = 0 . 537 in Eq. (7-19) is used next to ﬁnd k b k b = ² d e 0 . 30 ³ 0 .
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Unformatted text preview: 107 = ² . 537 . 30 ³ − . 107 = . 940 The load factor for shear k c , is k c = . 59 The endurance strength in shear is S se = . 780(0 . 940)(0 . 59)(29) = 12 . 5 kpsi From Table 9-5, the shear stress-concentration factor is K f s = 2 . 7 . The loading is repeatedly-applied. τ a = τ m = K f s τ max 2 = 2 . 7 1 . 537 2 = 2 . 07 kpsi Table 7-10: Gerber factor of safety n f , adjusted for shear, with S su = . 67 S ut n f = 1 2 ´ . 67(58) 2 . 07 µ 2 ² 2 . 07 12 . 5 ³ − 1 + ¶ 1 + ´ 2(2 . 07)(12 . 5) . 67(58)(2 . 07) µ 2 = 5 . 52 Ans. Attachment metal should be checked for bending fatigue....
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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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