13_ch 13 Mechanical Design budynas_SM_ch13

13_ch 13 Mechanical - d 3 = 32 2 = 16 in d 4 = 18 2 = 9 in d 5 = 48 2 = 24 in 4" 5" 1576 lbf 1576 lbf T out 1 1 1 5 W t 5 788 lbf F r 5 287

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Chapter 13 345 Input torque: T 2 = 63 025 H n T 2 = 63 025(25) 1000 = 1576 lbf · in For 100 percent gear ef±ciency T arm = 63 025(25) 200 = 7878 lbf · in Gear 2 W t = 1576 2 = 788 lbf F r 32 = 788 tan 20° = 287 lbf Gear 4 F A 4 = 2 W t = 2(788) = 1576 lbf Gear 5 Arm T out = 1576(9) 1576(4) = 7880 lbf · in Ans. 13-30 Given: P = 2 teeth/in, n P = 1800 rev/min cw, N 2 = 18 T , N 3 = 32 T , N 4 = 18 T , N 5 = 48 T . Pitch Diameters: d 2 = 18 / 2 = 9in
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Unformatted text preview: ; d 3 = 32 / 2 = 16 in ; d 4 = 18 / 2 = 9 in ; d 5 = 48 / 2 = 24 in . 4" 5" 1576 lbf 1576 lbf T out 1 1 1 5 W t 5 788 lbf F r 5 287 lbf 2 W t 5 1576 lbf W t F r 4 n 4 F A 4 W t W t F r F r 2 T 2 5 1576 lbf • in n 2 F t a 2 W t F r a 2 F r 42...
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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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