{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

14_ch 13 Mechanical Design budynas_SM_ch13

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}