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Solutions-HW6-Chapter13-14

# Solutions-HW6-Chapter13-14 - budynas_SM_ch13.qxd 7:06 PM...

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Chapter 13 13-1 d P = 17 / 8 = 2 . 125 in d G = N 2 N 3 d P = 1120 544 (2 . 125) = 4 . 375 in N G = Pd G = 8(4 . 375) = 35 teeth Ans. C = (2 . 125 + 4 . 375) / 2 = 3 . 25 in Ans. 13-2 n G = 1600(15 / 60) = 400 rev/min Ans. p = π m = 3 π mm Ans. C = [3(15 + 60)] / 2 = 112 . 5 mm Ans. 13-3 N G = 20(2 . 80) = 56 teeth Ans. d G = N G m = 56(4) = 224 mm Ans. d P = N P m = 20(4) = 80 mm Ans. C = (224 + 80) / 2 = 152 mm Ans. 13-4 Mesh : a = 1 / P = 1 / 3 = 0 . 3333 in Ans. b = 1 . 25 / P = 1 . 25 / 3 = 0 . 4167 in Ans. c = b a = 0 . 0834 in Ans. p = π/ P = π/ 3 = 1 . 047 in Ans. t = p / 2 = 1 . 047 / 2 = 0 . 523 in Ans. Pinion Base-Circle : d 1 = N 1 / P = 21 / 3 = 7 in d 1 b = 7 cos 20° = 6 . 578 in Ans . Gear Base-Circle : d 2 = N 2 / P = 28 / 3 = 9 . 333 in d 2 b = 9 . 333 cos 20° = 8 . 770 in Ans. Base pitch : p b = p c cos φ = ( π/ 3) cos 20° = 0 . 984 in Ans. Contact Ratio : m c = L ab / p b = 1 . 53 / 0 . 984 = 1 . 55 Ans. See the next page for a drawing of the gears and the arc lengths. budynas_SM_ch13.qxd 1/29/07 7:06 PM Page 325

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332 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-20 Let gear 2 be first, then n F = n 2 = 0. Let gear 6 be last, then n L = n 6 = − 12 rev/min. e = 20 30 16 34 = 16 51 , e = n L n A n F n A (0 n A ) 16 51 = − 12 n A n A = 12 35 / 51 = − 17 . 49 rev/min (negative indicates cw) Ans . 13-21 Let gear 2 be first, then n F = n 2 = 180 rev/min. Let gear 6 be last, then n L = n 6 = 0 . e = 20 30 16 34 = 16 51 , e = n L n A n F n A (180 n A ) 16 51 = (0 n A ) n A = 16 35 180 = − 82 . 29 rev/min The negative sign indicates opposite n 2 n A = 82 . 29 rev/min cw Ans .
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