14_ch 14 Mechanical Design budynas_SM_ch14

# 14_ch 14 Mechanical Design budynas_SM_ch14 - = 1 156...

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(d) Pinion bending: H 1 = 32 . 3hp Pinion wear: H 2 = 6 . 67 hp Gear bending: H 3 = 38 . 0hp Gear wear: H 4 = 6 . 67 hp Power rating of the gear set is thus H rated = min(32 . 3, 6 . 67, 38 . 0, 6 . 67) = 6 . 67 hp Ans. 14-19 d P = 16 / 6 = 2 . 667 in, d G = 48 / 6 = 8in V = π (2 . 667)(300) 12 = 209 . 4 ft/min W t = 33 000(5) 209 . 4 = 787 . 8 lbf Assuming uniform loading, K o = 1 . From Eq. (14-28), 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 + 209 . 4 59 . 77 ² 0 . 8255 = 1 . 196 From Table 14-2, N P = 16 T , Y P = 0 . 296 N G = 48 T , Y G = 0 . 4056 From Eq. ( a ), Sec. 14-10 with F = 2in ( K s ) P = 1 . 192 ± 2 0 . 296 6 ² 0 . 0535 = 1 . 088 ( K s ) G = 1 . 192 ± 2 0 . 4056 6 ² 0 . 0535 = 1 . 097 From Eq. (14-30) with C mc = 1 C pf = 2 10(2 . 667) 0 . 0375 + 0 . 0125(2) = 0 . 0625 C pm = 1, C ma = 0 . 093 (Fig. 14-11), C e = 1 K m = 1 + 1[0 . 0625(1) + 0 . 093(1)]
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Unformatted text preview: = 1 . 156 Assuming constant thickness of the gears → K B = 1 m G = N G / N P = 48 / 16 = 3 With N (pinion) = 10 8 cycles and N (gear) = 10 8 / 3, Fig. 14-14 provides the relations: ( Y N ) P = 1 . 3558(10 8 ) − . 0178 = . 977 ( Y N ) G = 1 . 3558(10 8 / 3) − . 0178 = . 996 362 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design...
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