4_ch 13 Mechanical Design budynas_SM_ch13

4_ch 13 Mechanical Design budynas_SM_ch13 - 2 N P sin 2 13...

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336 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-8 From Ex. 13-1, a 16-tooth spur pinion meshes with a 40-tooth gear, m G = 40 / 16 = 2 . 5 . Equations (13-10) through (13-13) apply. (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10) N P 4 k 6 sin 2 φ µ 1 + q 1 + 3 sin 2 φ 4(1) 6 sin 2 20° ³ 1 + p 1 + 3 sin 2 20° ´ 12 . 32 13 teeth Ans. (b) The smallest pinion that will mesh with a gear ratio of m G = 2 . 5, from Eq. (13-11) is N P 2(1) [1 + 2(2 . 5)] sin 2 20° ½ 2 . 5 + q 2 . 5 2 + [1 + 2(2 . 5)] sin 2 20° ± 14 . 64 15 pinion teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-12) N P 4 k 2 sin 2 φ = 4(1) 2 sin 2 20° 17 . 097 18 teeth Ans. (d) The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-13) is N G N 2 P sin 2 φ 4 k 2 4 k
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Unformatted text preview: 2 N P sin 2 13 2 sin 2 20 4(1) 2 4(1) 2(13) sin 2 20 16 . 45 16 teeth Ans. 13-9 From Ex. 13-2, a 20 pressure angle, 30 helix angle, p t = 6 teeth/in pinion with 18 full depth teeth, and t = 21 . 88 . (a) The smallest tooth count that will mesh with a like gear, from Eq. (13-21), is N P 4 k cos 6 sin 2 t 1 + q 1 + 3 sin 2 t 4(1) cos 30 6 sin 2 21 . 88 1 + p 1 + 3 sin 2 21 . 88 9 . 11 10 teeth Ans. (b) The smallest pinion-tooth count that will run with a rack, from Eq. (13-23), is N P 4 k cos 2 sin 2 t 4(1) cos 30 2 sin 2 21 . 88 12 . 47 13 teeth Ans ....
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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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