16_ch 14 Mechanical Design budynas_SM_ch14

16_ch 14 Mechanical Design budynas_SM_ch14 -...

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Unformatted text preview: budynas_SM_ch14.qxd 364 12/05/2006 17:39 Page 364 FIRST PAGES Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Gear tooth wear ( Ks )G ( σc ) G = ( Ks ) P ( SH )G = 1/2 ( σc ) P = 1.097 1.088 1/2 (98 760) = 99 170 psi Ans. 93 500(0.973)(1) /[(1)(0.85)] = 1.08 Ans. 99 170 The hardness of the pinion and the gear should be increased. 14-20 d P = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm π dP n P π (50)(10−3 )(100) = = 0.2618 m/s 60 60 60(120) = 458.4 N Wt = π (50)(10−3 )(100) V= K o = 1, Eq. (14-28): Q v = 6, B = 0.25(12 − 6) 2/3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 √ 0.8255 59.77 + 200(0.2618) = 1.099 Kv = 59.77 Eq. (14-27): Y P = 0.322, Table 14-2: YG = 0.3775 Similar to Eq. (a) of Sec. 14-10 but for SI units: √ 0.0535 1 = 0.8433 m F Y kb √ 0.0535 ( K s ) P = 0.8433 2.5(18) 0.322 = 1.003 use 1 √ 0.0535 > 1 use 1 ( K s ) G = 0.8433 2.5(18) 0.3775 Ks = 18 − 0.025 = 0.011 10(50) = 0.247 + 0.0167(0.709) − 0.765(10−4 )(0.7092 ) = 0.259 Cmc = 1, F = 18/25.4 = 0.709 in, C pm = 1, Cma Cp f = Ce = 1 K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27 Eq. (14-40): Fig. 14-14: K B = 1, m G = NG / N P = 36/20 = 1.8 ( Y N ) P = 1.3558(108 ) −0.0178 = 0.977 ( Y N ) G = 1.3558(108 /1.8) −0.0178 = 0.987 Fig. 14-6: Eq. (14-38): Sec. 14-15: ( Y J ) P = 0.33, ( Y J ) G = 0.38 Y Z = 0.658 − 0.0759 ln(1 − 0.95) = 0.885 Yθ = Z R = 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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