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8_ch 05 Mechanical Design budynas_SM_ch05

8_ch 05 Mechanical Design budynas_SM_ch05 - (a σ ± =[9 2...

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122 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-8 See Prob. 5-7 for plot. (a) For all methods: n = OB OA = 1 . 55 1 . 03 = 1 . 5 (b) BCM: n = OD OC = 1 . 4 0 . 8 = 1 . 75 All other methods: n = OE OC = 1 . 55 0 . 8 = 1 . 9 (c) For all methods: n = OL OK = 5 . 2 0 . 68 = 7 . 6 (d) MNS: n = OJ OF = 5 . 12 0 . 82 = 6 . 2 BCM: n = OG OF = 2 . 85 0 . 82 = 3 . 5 MM: n = OH OF = 3 . 3 0 . 82 = 4 . 0 5-9 Given: S y = 42 kpsi, S ut = 66 . 2 kpsi, ε f = 0 . 90 . Since ε f > 0 . 05, the material is ductile and thus we may follow convention by setting S yc = S yt . Use DE theory for analytical solution. For σ ± , use Eq. (5-13) or (5-15) for plane stress and Eq. (5-12) or (5-14) for general 3-D.
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Unformatted text preview: (a) σ ± = [9 2 − 9( − 5) + ( − 5) 2 ] 1 / 2 = 12 . 29 kpsi n = 42 12 . 29 = 3 . 42 Ans. (b) σ ± = [12 2 + 3(3 2 )] 1 / 2 = 13 . 08 kpsi n = 42 13 . 08 = 3 . 21 Ans. (c) σ ± = [( − 4) 2 − ( − 4)( − 9) + ( − 9) 2 + 3(5 2 )] 1 / 2 = 11 . 66 kpsi n = 42 11 . 66 = 3 . 60 Ans. (d) σ ± = [11 2 − (11)(4) + 4 2 + 3(1 2 )] 1 / 2 = 9 . 798 n = 42 9 . 798 = 4 . 29 Ans....
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