14_ch 11 Mechanical Design budynas_SM_ch11

14_ch 11 Mechanical Design budynas_SM_ch11 - log basic load...

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302 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design From the solution of Prob. 11-18, L 1 = 1 . 434(10 6 )rev and L 2 = 0 . 310(10 6 )rev . Palmgren-Miner rule: l 1 L 1 + l 2 L 2 = f 1 l L 1 + f 2 l L 2 = 1 from which l = 1 f 1 / L 1 + f 2 / L 2 l = 1 { 0 . 40 / [1 . 434(10 6 )] }+{ 0 . 60 / [0 . 310(10 6 )] } = 451 585 rev Ans. Total life in loading cycles 4 min at 2000 rev/min = 8000 rev 6 min 10 min/cycle at 2000 rev/min = 12 000 rev 20 000 rev/cycle 451 585 rev 20 000 rev/cycle = 22 . 58 cycles Ans. Total life in hours ± 10 min cycle ²± 22 . 58 cycles 60 min/h ² = 3 . 76 h Ans. 11-20 While we made some use of the log F -log x plot in Probs. 11-15 and 11-17, the principal
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Unformatted text preview: log basic load rating for a case at hand. Point D F D = 495 . 6 lbf log F D = log 495 . 6 = 2 . 70 x D = 30 000(300)(60) 10 6 = 540 log x D = log 540 = 2 . 73 K D = F 3 D x D = (495 . 6) 3 (540) = 65 . 7(10 9 ) lbf 3 turns log K D = log[65 . 7(10 9 )] = 10 . 82 F D has the following uses: F design , F desired , F e when a thrust load is present. It can include application factor a f , or not. It depends on context....
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