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26_ch 14 Mechanical Design budynas_SM_ch14

# 26_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 12/05/2006 17:39 Page 374 FIRST PAGES 374 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 14-29 n = 1145 rev/min, K o = 1.25, N P = 22T , NG = 60T , m G = 2.727, d P = 2.75 in, dG = 7.5 in, Y P = 0.331, YG = 0.422, J P = 0.335, JG = 0.405 , P = 8T /in, F = 1.625 in, H B = 250, case and core, both gears. Cm = 1, F /d P = 0.0591, C pm = 1, Cma = 0.152, Ce = 1, K m = 1.1942, K T = 1, C f = 0.0419, K β = 1, K s = 1, V = 824 ft/min, ( Y N ) P = 0.8318, ( Y N ) G = 0.859, K R = 1, I = 0.117 58 0.99 ( St ) 107 = 32 125 psi ( σall ) P = 26 668 psi ( σall ) G = 27 546 psi and it follows that t W1 = 879.3 lbf, H1 = 21.97 hp W2 = 1098 lbf, H2 = 27.4 hp t W3 = 304 lbf, H3 = 7.59 hp t W4 = 340 lbf, H4 = 8.50 hp t For wear Rating Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp In Prob. 14-24, Hrated = 53 hp Thus 7.59 1 1 = 0.1432 = , not Ans . 53.0 6.98 8 The transmitted load rating is t Wrated = min(879.3, 1098, 304, 340) = 304 lbf In Prob. 14-24 t Wrated = 1061 lbf Thus 304 1 = 0.2865 = , 1061 3.49 14-30 S P = S H = 1, Bending Table 14-4: Pd = 4, 0.99 ( St ) 107 J P = 0.345, 1 not , 4 JG = 0.410, Ans . K o = 1.25 = 13 000 psi 13 000(1) = 13 000 psi 1(1)(1) σall F J P 13 000(3.25)(0.345) t W1 = = = 1533 lbf K o K v K s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1) ( σall ) P = ( σall ) G = 1533(1649) = 76.6 hp 33 000 t t W2 = W1 JG / J P = 1533(0.410) /0.345 = 1822 lbf H2 = H1 JG / J P = 76.6(0.410) /0.345 = 91.0 hp H1 = ...
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