28_ch 06 Mechanical Design budynas_SM_ch06

28_ch 06 Mechanical Design budynas_SM_ch06 - = − 85 Table...

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174 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-37 For a non-rotating bar subjected to completely reversed torsion of T a = 2400 lbf · in From Prob. 6-36: S ± e = 29 . 3 LN (1, 0 . 138) kpsi k a = 0 . 782 LN (1, 0 . 11) k b = 0 . 955 For k c use Eq. (6-74): k c = 0 . 328(58) 0 . 125 LN (1, 0 . 125) = 0 . 545 LN (1, 0 . 125) S Se = 0 . 782[ LN (1, 0 . 11)](0 . 955)[0 . 545 LN (1, 0 . 125)][29 . 3 LN (1, 0 . 138)] ¯ S Se = 0 . 782(0 . 955)(0 . 545)(29 . 3) = 11 . 9 kpsi C Se = (0 . 11 2 + 0 . 125 2 + 0 . 138 2 ) 1 / 2 = 0 . 216 Table A-16: d / D = 0, a / D = 0 . 1, A = 0 . 92, K ts = 1 . 68 From Eqs. (6-78), (6-79), Table 6-15 K fs = 1 . 68 LN (1, 0 . 10) 1 + ( 2 / 0 . 125 ) [(1 . 68 1) / 1 . 68](5 / 58) = 1 . 403 LN (1, 0 . 10) Table A-16: J net = π AD 4 32 = π (0 . 92)(1 . 25 4 ) 32 = 0 . 2201 τ a = K fs T a c J net = 1 . 403[ LN (1, 0 . 10)] ± 2 . 4(1 . 25 / 2) 0 . 2201 ² = 9 . 56 LN (1, 0 . 10) kpsi From Eq. (5-43), p. 242: z =− ln ³ (11 . 9 / 9 . 56) ´ (1 + 0 . 10 2 ) / (1 + 0 . 216 2 ) µ ´ ln[(1 + 0 . 10 2 )(1 + 0 . 216 2
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Unformatted text preview: )] = − . 85 Table A-10, p f = . 1977 R = 1 − p f = 1 − . 1977 = . 80 Ans . 6-38 This is a very important task for the student to attempt before starting Part 3. It illustrates the drawback of the deterministic factor of safety method. It also identifies the a priori de-cisions and their consequences. The range of force fluctuation in Prob. 6-23 is − 16 to + 4 kip, or 20 kip. Repeatedly-applied F a is 10 kip. The stochastic properties of this heat of AISI 1018 CD are given....
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