12_ch 13 Mechanical Design budynas_SM_ch13

12_ch 13 Mechanical - 36 ¶µ 36 144 = 1 6 n P = n 2 = 1000 rev/min n L = n 6 = e = n L − n A n F − n A = − n A 1000 − n A n A = − 200

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344 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) r 4 = mN 4 2 = 5(51) 2 = 127 . 5 mm T c 4 = 9 . 36(127 . 5) = 1193 N · m ccw T 4 c = 1193 N · m cw Ans. Note: The solution is independent of the pressure angle. 13-29 d = N 6 d 2 = 4 in, d 4 = 4 in, d 5 = 6 in, d 6 = 24 in e = 24 24 µ 24
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Unformatted text preview: 36 ¶µ 36 144 ¶ = 1 / 6, n P = n 2 = 1000 rev/min n L = n 6 = e = n L − n A n F − n A = − n A 1000 − n A n A = − 200 rev/min 2 4 5 6 9.36 4 c T c 4 5 1193 b 9.36 O 3 F t 43 9.36 18.73 F t 23 F b 3...
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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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