11_ch 06 Mechanical Design budynas_SM_ch06

# 11_ch 06 Mechanical Design budynas_SM_ch06 -...

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Unformatted text preview: budynas_SM_ch06.qxd 11/29/2006 17:40 FIRST PAGES Page 157 157 Chapter 6 (a) Table 6-7, DE-Gerber nf = 1 2 64 8.45 2 14.59 −1 + 24.4 1+ 2(8.45)(24.4) 64(14.59) 2 = 1.60 Ans. (b) Table 6-8, DE-Elliptic nf = 6-19 1 = 1.62 Ans. (14.59/24.4) 2 + (8.45/54) 2 Referring to the solution of Prob. 6-17, for load ﬂuctuations of 800 to −3000 lbf σa = 2.16 σm = 2.16 0.800 − ( −3.000) = 14.59 kpsi 2(0.2813) 0.800 + ( −3.000) = −8.45 kpsi 2(0.2813) (a) We have a compressive midrange stress for which the failure locus is horizontal at the Se level. nf = Se 24.4 = = 1.67 Ans. σa 14.59 nf = Se 24.4 = = 1.67 Ans. σa 14.59 (b) Same as (a) 6-20 Sut = 0.495(380) = 188.1 kpsi Se = 0.5(188.1) = 94.05 kpsi ka = 14.4(188.1) −0.718 = 0.335 For a non-rotating round bar in bending, Eq. (6-24) gives: de = 0.370d = 0.370(3/8) = 0.1388 in kb = 0.1388 0.3 −0.107 = 1.086 Se = 0.335(1.086)(94.05) = 34.22 kpsi 30 − 15 30 + 15 = 7.5 lbf, Fm = = 22.5 lbf Fa = 2 2 32 Mm 32(22.5)(16) σm = = (10−3 ) = 69.54 kpsi 3 πd π (0.3753 ) ...
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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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