3_ch 12 Mechanical Design budynas_SM_ch12

3_ch 12 Mechanical Design budynas_SM_ch12 - 7 395 = 169 psi...

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306 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Fig. 12-12: SAE 40, µ ± = 4 . 5 µ reyn S = 0 . 0354 ± 4 . 5 1 . 75 ² = 0 . 0910 h o / c = 0 . 19, P / p max = 0 . 275 h o = 0 . 19(0 . 0025) = 0 . 000 475 in Ans . p max = 177 . 78 / 0 . 275 = 646 psi Ans . 12-4 c min = b min d max 2 = 3 . 006 3 . 000 2 = 0 . 003 r . = 3 . 000 / 2 = 1 . 5in l / d = 1 r / c = 1 . 5 / 0 . 003 = 500 N = 750 / 60 = 12 . 5rev/s P = 600 3(3) = 66 . 7 psi Fig. 12-14: SAE 10W, µ ± = 2 . 1 µ reyn S = (500 2 ) ³ 2 . 1(10 6 )(12 . 5) 66 . 7 ´ = 0 . 0984 From Figs. 12-16 and 12-21: h o / c = 0 . 34, P / p max = 0 . 395 h o = 0 . 34(0 . 003) = 0 . 001 020 in Ans . p max = 66
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Unformatted text preview: . 7 . 395 = 169 psi Ans . Fig. 12-14: SAE 20W-40, µ ± = 5 . 05 µ reyn S = (500 2 ) ³ 5 . 05(10 − 6 )(12 . 5) 66 . 7 ´ = . 237 From Figs. 12-16 and 12-21: h o / c = . 57, P / p max = . 47 h o = . 57(0 . 003) = . 001 71 in Ans . p max = 66 . 7 . 47 = 142 psi Ans ....
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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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