6_ch 13 Mechanical Design budynas_SM_ch13

6_ch 13 Mechanical Design budynas_SM_ch13 - (b) n c = n 5 =...

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338 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) m t = 10 . 4 = 3 . 310 mm Ans . φ t = tan 1 tan 20° cos 25° = 21 . 88° Ans . (c) d P = 3 . 310(18) = 59 . 58 mm Ans . d G = 3 . 310(32) = 105 . 92 mm Ans . 13-14 (a) The axial force of 2 on shaft a is in the negative direction. The axial force of 3 on shaft b is in the positive direction of z . Ans. The axial force of gear 4 on shaft b is in the positive z -direction. The axial force of gear 5 on shaft c is in the negative z -direction. Ans.
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Unformatted text preview: (b) n c = n 5 = 14 54 16 36 (900) = + 103 . 7 rev/min ccw Ans . (c) d P 2 = 14 / (10 cos 30) = 1 . 6166 in d G 3 = 54 / (10 cos 30) = 6 . 2354 in C ab = 1 . 6166 + 6 . 2354 2 = 3 . 926 in Ans . d P 4 = 16 / (6 cos 25) = 2 . 9423 in d G 5 = 36 / (6 cos 25) = 6 . 6203 in C bc = 4 . 781 in Ans . 13-15 e = 20 40 8 17 20 60 = 4 51 n d = 4 51 (600) = 47 . 06 rev/min cw Ans . 5 4 c b z a 3 z 2 b...
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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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