5_ch 13 Mechanical Design budynas_SM_ch13

5_ch 13 Mechanical Design budynas_SM_ch13 - Use 11:66 Ans ....

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Chapter 13 337 (c) The largest gear tooth possible, from Eq. (13-24) is N G N 2 P sin 2 φ t 4 k 2 cos 2 ψ 4 k cos ψ 2 N P sin 2 φ t 10 2 sin 2 21 . 88° 4(1 2 ) cos 2 30° 4(1) cos 30° 2(10) sin 2 21 . 88° 15 . 86 15 teeth Ans . 13-10 Pressure Angle: φ t = tan 1 µ tan 20° cos 30° = 22 . 796° Program Eq. (13-24) on a computer using a spreadsheet or code and increment N P . The Frst value of N P that can be doubled is N P = 10 teeth, where N G 26 . 01 teeth. So N G = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use 10:20 Ans . 13-11 Refer to Prob. 13-10 solution. The Frst value of N P that can be multiplied by 6 is N P = 11 teeth where N G 93 . 6 teeth. So N G = 66 teeth.
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Unformatted text preview: Use 11:66 Ans . 13-12 Begin with the more general relation, Eq. (13-24), for full depth teeth. N G = N 2 P sin 2 t 4 cos 2 4 cos 2 N P sin 2 t Set the denominator to zero 4 cos 2 N P sin 2 t = rom which sin t = s 2 cos N P t = sin 1 s 2 cos N P or N P = 9 teeth and cos = 1 t = sin 1 r 2(1) 9 = 28 . 126 Ans . 13-13 (a) p n = m n = 3 mm Ans . p t = 3 / cos 25 = 10 . 4 mm Ans . p x = 10 . 4 / tan 25 = 22 . 3 mm Ans . 18 T 32 T c 5 25 8 , f n 5 20 8 , m 5 3 mm...
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