9_ch 05 Mechanical Design budynas_SM_ch05

9_ch 05 Mechanical Design budynas_SM_ch05 - ductile ε f>...

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Chapter 5 123 For graphical solution, plot load lines on DE envelope as shown. (a) σ A = 9, σ B =− 5 kpsi n = OB OA = 3 . 5 1 = 3 . 5 Ans. (b) σ A , σ B = 12 2 ± ± ² 12 2 ³ 2 + 3 2 = 12 . 7, 0 . 708 kpsi n = OD OC = 4 . 2 1 . 3 = 3 . 23 (c) σ A , σ B = 4 9 2 ± ± ² 4 9 2 ³ 2 + 5 2 =− 0 . 910, 12 . 09 kpsi n = OF OE = 4 . 5 1 . 25 = 3 . 6 Ans. (d) σ A , σ B = 11 + 4 2 ± ± ² 11 4 2 ³ 2 + 1 2 = 11 . 14, 3 . 86 kpsi n = OH OG = 5 . 0 1 . 15 = 4 . 35 Ans. 5-10 This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0 . 06 . The steel is
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Unformatted text preview: ductile ( ε f > . 05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis (DCM) of Fig. 5-19 applies — confine its use to first and fourth quadrants. ( c ) ( a ) ( b ) ( d ) E C G H D B A O F 1 cm ± 10 kpsi ± B ± A...
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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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