10_ch 06 Mechanical Design budynas_SM_ch06

10_ch 06 Mechanical Design budynas_SM_ch06 -...

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Unformatted text preview: budynas_SM_ch06.qxd 156 6-17 11/29/2006 17:40 FIRST PAGES Page 156 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Table A-20: Sut = 64 kpsi, S y = 54 kpsi A = 0.375(1 − 0.25) = 0.2813 in2 Fmax 3000 = (10−3 ) = 10.67 kpsi A 0.2813 σmax = ny = 54 = 5.06 Ans. 10.67 Se = 0.5(64) = 32 kpsi ka = 2.70(64) −0.265 = 0.897 kb = 1, kc = 0.85 Se = 0.897(1)(0.85)(32) = 24.4 kpsi Table A-15-1: w = 1 in, d = 1/4 in, d /w = 0.25 K t = 2.45. ˙ Fig. 6-20, with r = 0.125 in, q = 0.8 Eq. (6-32): K f = 1 + 0.8(2.45 − 1) = 2.16 Fmax − Fmin 2A σa = K f = 2.16 σm = K f 3.000 − 0.800 = 8.45 kpsi 2(0.2813) Fmax + Fmin 2A = 2.16 3.000 + 0.800 = 14.6 kpsi 2(0.2813) (a) Gerber, Table 6-7 nf = 1 2 64 14.6 2 8.45 −1 + 24.4 1+ 2(14.6)(24.4) 8.45(64) 2 = 2.17 Ans. (b) ASME-Elliptic, Table 6-8 nf = 6-18 (8.45/24.4) 2 1 = 2.28 Ans. + (14.6/54) 2 Referring to the solution of Prob. 6-17, for load fluctuations of −800 to 3000 lbf σa = 2.16 3.000 − ( −0.800) = 14.59 kpsi 2(0.2813) σm = 2.16 3.000 + ( −0.800) = 8.45 kpsi 2(0.2813) ...
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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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