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16_ch 14 Mechanical Design budynas_SM_ch14

# 16_ch 14 Mechanical Design budynas_SM_ch14 -...

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Gear tooth wear ( σ c ) G = ( K s ) G ( K s ) P 1 / 2 ( σ c ) P = 1 . 097 1 . 088 1 / 2 (98 760) = 99 170 psi Ans. ( S H ) G = 93 500(0 . 973)(1) / [(1)(0 . 85)] 99 170 = 1 . 08 Ans. The hardness of the pinion and the gear should be increased. 14-20 d P = 2 . 5(20) = 50 mm, d G = 2 . 5(36) = 90 mm V = π d P n P 60 = π (50)(10 3 )(100) 60 = 0 . 2618 m/s W t = 60(120) π (50)(10 3 )(100) = 458 . 4 N Eq. (14-28): K o = 1, Q v = 6, B = 0 . 25(12 6) 2 / 3 = 0 . 8255 A = 50 + 56(1 0 . 8255) = 59 . 77 Eq. (14-27): K v = 59 . 77 + 200(0 . 2618) 59 . 77 0 . 8255 = 1 . 099 Table 14-2: Y P = 0 . 322, Y G = 0 . 3775 Similar to Eq. ( a ) of Sec. 14-10 but for SI units: K s = 1 k b = 0 . 8433 ( mF Y ) 0 . 0535 ( K s ) P = 0 . 8433 2 . 5(18) 0 . 322 0 . 0535 = 1 . 003 use 1 ( K s ) G = 0 . 8433 2 . 5(18) 0 . 3775 0 . 0535 > 1 use 1 C mc = 1, F = 18 / 25 . 4 = 0 . 709 in, C pf = 18 10(50) 0 . 025 = 0 . 011 C pm = 1, C ma = 0 . 247 + 0 . 0167(0 . 709) 0 . 765(10 4 )(0 . 709 2 ) = 0 . 259 C e = 1 K H = 1 + 1[0 . 011(1) + 0 . 259(1)] = 1 . 27 Eq. (14-40):
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