8_ch 14 Mechanical Design budynas_SM_ch14

8_ch 14 Mechanical Design budynas_SM_ch14 -...

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From Eq. (6-68), ( S C ) 10 8 = 0 . 4 H B 10 kpsi = [0 . 4(149) 10](10 3 ) = 49 600 psi Eq. (14-14): σ C ,all =− ( S C ) 10 8 n = 49 600 2 . 25 =− 33 067 psi W t = ± 33 067 2285 ² 2 ³ 0 . 875 cos 20° 1 . 162(6 . 469) ´ = 22 . 6 lbf H = 22 . 6(194 . 8) 33 000 = 0 . 133 hp Ans. Rating power (pinion controls): H 1 = 2 . 04 hp H 2 = 0 . 133 hp H all = (min 2 . 04, 0 . 133) = 0 . 133 hp Ans . See Prob. 14-15 solution for equation numbers. Pinion controls: Y P = 0 . 322, Y G = 0 . 447 Bending d P = 20 / 3 = 6 . 667 in, d G = 33 . 333 in V = π d P n / 12 = π (6 . 667)(870) / 12 = 1519 ft/min K v = (1200 + 1519) / 1200 = 2 . 266 S ± e = 0 . 5(113) = 56 . 5 kpsi k a = 2 . 70(113) 0 . 265 = 0 . 771 l = 2 . 25 / P d = 2 . 25 / 3 = 0 . 75 in x = 3(0 . 322) / [2(3)] = 0 . 161 in t = µ 4(0 . 75)(0 . 161) = 0 . 695 in d e = 0 . 808 µ 2 . 5(0 . 695) = 1 . 065 in k b = (1 . 065 / 0 . 30) 0 . 107 = 0 . 873 k c = k d = k e = 1 r f = 0 . 300 / 3 = 0 . 100 in r d = r f t = 0 . 100 0 . 695 = 0 . 144 From Table A-15-6, K t = 1 . 75 ; Fig . 6-20, q = 0 . 85 ; Eq . (6-32), K f = 1 .
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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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