7_ch 06 Mechanical Design budynas_SM_ch06

# 7_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 153 153 Chapter 6 −0.107 2.00 0.30 kb = = 0.816 Se = 0.759(0.816)(60) = 37.2 kpsi [0.9(120)]2 = 313.5 37.2 0.9(120) 1 = −0.15429 b = − log 3 37.2 a= S f = 313.5(3450) −0.15429 = 89.2 kpsi M 305.6 30 = = σ0 = I /c 0.098 17d 3 d3 305.6 = 38.2 kpsi 23 d 2 r= = = 0.2 10 10 ˙ q = 0.87 = Fig. 6-20: K f = 1 + 0.87(1.68 − 1) = 1.59 ˙ Eq. (6-32): σa = K f σ0 = 1.59(38.2) = 60.7 kpsi Sf 89.2 = = 1.47 nf = σa 60.7 Design is adequate unless more uncertainty prevails. Choose d = 2.00 in Ans. 6-12 σmax = [1722 + 3(1032 )]1/2 = 247.8 kpsi Yield: n y = S y /σmax = 413/247.8 = 1.67 Ans. √ √ σa = 172 MPa σm = 3τm = 3(103) = 178.4 MPa (a) Modiﬁed Goodman, Table 6-6 nf = (b) Gerber, Table 6-7 nf = 1 2 551 178.4 2 1 = 1.06 Ans. (172/276) + (178.4/551) 172 −1 + 276 1+ 2(178.4)(276) 551(172) 2 = 1.31 Ans. (c) ASME-Elliptic, Table 6-8 1 nf = 2 + (178.4/413) 2 (172/276) 1/2 = 1.32 Ans. ...
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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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