33_ch 17 Mechanical Design budynas_SM_ch17

33_ch 17 Mechanical Design budynas_SM_ch17 - 1 = Pl AE + (...

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452 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From which d * = a 2 cn 1 Ans. Substituting this result for d in Eq. (1) gives m * = 4 bcn 1 a 2 Ans. 17-34 A m = 0 . 40 d 2 = 0 . 40(2 2 ) = 1 . 6in 2 E r = 12 Mpsi, w = 1 . 6 d 2 = 1 . 6(2 2 ) = 6 . 4 lbf/ft w l = 6 . 4(480) = 3072 lbf Treat the rest of the system as rigid, so that all of the stretch is due to the cage weighing 1000 lbf and the wire’s weight. From Prob. 4-6
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Unformatted text preview: 1 = Pl AE + ( w l ) l 2 AE = 1000(480)(12) 1 . 6(12)(10 6 ) + 3072(480)(12) 2(1 . 6)(12)(10 6 ) = . 3 + . 461 = . 761 in due to cage and wire. The stretch due to the wire, the cart and the cage is 2 = 9000(480)(12) 1 . 6(12)(10 6 ) + . 761 = 3 . 461 in Ans. 17-35 to 17-38 Computer programs will vary....
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