14_ch 13 Mechanical Design budynas_SM_ch13

14_ch 13 Mechanical Design budynas_SM_ch13 - F r a 2 = W r...

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346 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Gear 2 T a 2 = 63 025(200) / 1800 = 7003 lbf · in W t = 7003 / 4 . 5 = 1556 lbf W r = 1556 tan 20° = 566 lbf Gears 3 and 4 W t (4 . 5) = 1556(8), W t = 2766 lbf W r = 2766 tan 20 = 1007 lbf Ans. 13-31 Given: P = 5 teeth/in, N 2 = 18 T , N 3 = 45 T , φ n = 20°, H = 32 hp, n 2 = 1800 rev/min. Gear 2 T in = 63 025(32) 1800 = 1120 lbf · in d P = 18 5 = 3 . 600 in d G = 45 5 = 9 . 000 in W t 32 = 1120 3 . 6 / 2 = 622 lbf W r 32 = 622 tan 20° = 226 lbf F t a 2 = W t 32 = 622 lbf,
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Unformatted text preview: F r a 2 = W r 32 = 226 lbf F a 2 = (622 2 + 226 2 ) 1 / 2 = 662 lbf Each bearing on shaft a has the same radial load of R A = R B = 662 / 2 = 331 lbf. 2 a T in W t 32 W r 32 F r a 2 F t a 2 b 3 4 y x W r 5 566 lbf W t 5 1556 lbf W t 5 2766 lbf W r 5 1007 lbf 2 a W t 5 1556 lbf W r 5 566 lbf T a 2 5 7003 lbf•in...
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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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