20_ch 14 Mechanical Design budynas_SM_ch14

20_ch 14 Mechanical Design budynas_SM_ch14 -...

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Fig. 14-14: Y N = 1 . 6831(10 8 ) 0 . 0323 = 0 . 928 V = π d P n / 12 = π (2 . 833)(1120 / 12) = 830 . 7ft/min K T = K R = 1, S F = 2, S H = 2 σ all = 30 734(0 . 928) 2(1)(1) = 14 261 psi Q v = 5, B = 0 . 25(12 5) 2 / 3 = 0 . 9148 A = 50 + 56(1 0 . 9148) = 54 . 77 K v = ± 54 . 77 + 830 . 7 54 . 77 ² 0 . 9148 = 1 . 472 K s = 1 . 192 ± 2 0 . 303 6 ² 0 . 0535 = 1 . 089 use 1 K m = C mf = 1 + C mc ( C pf C pm + C ma C e ) C mc = 1 C pf = F 10 d 0 . 0375 + 0 . 0125 F = 2 10(2 . 833) 0 . 0375 + 0 . 0125(2) = 0 . 0581 C pm = 1 C ma = 0 . 127 + 0 . 0158(2) 0 . 093(10 4 )(2 2 ) = 0 . 1586 C e = 1 K m = 1 + 1[0 . 0581(1) + 0 . 1586(1)] = 1 . 2167 Eq. (14-15): K β = 1 W t = FJ P σ all K o K v K s P d K m K B = 2(0 . 292)(14 261) 1(1 . 472)(1)(6)(1 . 2167)(1) = 775 lbf H = W t V 33 000 = 775(830 . 7) 33 000 = 19 . 5hp Pinion wear
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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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