10_ch 15 Mechanical Design budynas_SM_ch15

10_ch 15 Mechanical Design budynas_SM_ch15 - v t = d P n P...

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388 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Pinion case ( s ac ) P = 341(373) + 23 620 = 150 813 psi ( σ c ,all ) P = 150 813(1) 1(1)(1 . 118) = 134 895 psi W t = ± 134 895 2290 ² 2 ³ 1 . 25(3 . 333)(0 . 086) 1(1 . 374)(1 . 106)(0 . 593 75)(2) ´ = 689 . 0 lbf Gear case ( s ac ) G = 341(345) + 23 620 = 141 265 psi ( σ c ,all ) G = 141 265(1 . 0685)(1) 1(1)(1 . 118) = 135 010 psi W t = ± 135 010 2290 ² 2 ³ 1 . 25(3 . 333)(0 . 086) 1(1 . 1374)(1 . 106)(0 . 593 75)(2) ´ = 690 . 1 lbf The equations developed within Prob. 15-7 are effective. 15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: N P = 20 teeth, N G = 40 teeth, φ n = 20 , F = 0 . 71 in, J P = 0 . 241, J G = 0 . 201, P d = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Q v = 5 uncrowned. Mesh d P = 20 / 10 = 2 . 000 in, d G = 40 / 10 = 4 . 000 in
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Unformatted text preview: v t = d P n P 12 = (2)(1200) 12 = 628 . 3 ft/min K o = 1, S F = 1, S H = 1 Eq. (15-6): B = . 25(12 5) 2 / 3 = . 9148 A = 50 + 56(1 . 9148) = 54 . 77 Eq. (15-5): K v = 54 . 77 + 628 . 3 54 . 77 . 9148 = 1 . 412 Eq. (15-10): K s = . 4867 + . 2132 / 10 = . 508 Eq. (15-11): K m = 1 . 25 + . 0036(0 . 71) 2 = 1 . 252 where K mb = 1 . 25 Eq. (15-15): ( K L ) P = 1 . 6831(10 9 ) . 0323 = . 862 ( K L ) G = 1 . 6831(10 9 / 2) . 0323 = . 881 Eq. (15-14): ( C L ) P = 3 . 4822(10 9 ) . 0602 = 1 . 000 ( C L ) G = 3 . 4822(10 9 / 2) . 0602 = 1 . 043...
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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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