Budynas ch13

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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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 = . 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 . 5mm 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 = 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 = 7in 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 = ( 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.
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326 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-5 (a) A O = ± ² 14 / 6 2 ³ 2 + ² 32 / 6 2 ³ 2 ´ 1 / 2 = 2 . 910 in Ans. (b) γ = tan 1 (14 / 32) = 23 . 63° Ans. = tan 1 (32 / 14) = 66 . 37° Ans. (c) d P = 14 / 6 = 2 . 333 in, d G = 32 / 6 = 5 . 333 in Ans. (d) From Table 13-3, 0 . 3 A O = 0 . 873 in and 10 / P = 10 / 6 = 1 . 67 0 . 873 < 1 . 67 F = 0 . 873 in Ans. 13-6 (a) p n = π/ 5 = 0 . 6283 in p t = p n / cos ψ = 0 . 6283 / cos 30° = 0 . 7255 in p x = p t / tan ψ = 0 . 7255 / tan 30° = 1 . 25 in 30 ± P G 2 1 3 " 5 1 3 " A O ² ³ 10.5 ± Arc of approach ´ 0.87 in Ans. Arc of recess ´ 0.77 in Ans. Arc of action ´ 1.64 in Ans. L ab ´ 1.53 in 10 ± O 2 O 1 14 ± 12.6 ± P B A
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Chapter 13 327 (b) Eq. (13-7): p nb = p n cos φ n = 0 . 6283 cos 20° = 0 . 590 in Ans. (c) P t = P n cos ψ = 5 cos 30° = 4 . 33 teeth/in φ t = tan 1 (tan φ n / cos ψ ) = tan 1 (tan 20° / cos 30 ) = 22 . Ans. (d) Table 13-4: a = 1 / 5 = 0 . 200 in Ans. b = 1 . 25 / 5 = 0 . 250 in Ans. d P = 17 5 cos 30° = 3 . 926 in Ans. d G = 34 5 cos 30° = 7 . 852 in Ans. 13-7 N P = 19 teeth, N G = 57 teeth, φ n = 14 . 5°, P n = 10 teeth/in (a) p n = π/ 10 = 0 . 3142 in Ans. p t = p n cos ψ = 0 . 3142 cos 20° = 0 . 3343 in Ans. p x = p t tan ψ = 0 . 3343 tan 20° = 0 . 9185 in Ans. (b) P t = P n cos ψ = 10 cos 20° = 9 . 397 teeth/in Ans . φ t = tan 1 ± tan 14 . cos 20° ² = 15 . 39° Ans. (c) a = 1 / 10 = 0 . 100 in Ans. b = 1 . 25 / 10 = 0 . 125 in Ans. d P = 19 10 cos 20° = 2 . 022 in Ans. d G = 57 10 cos 20° = 6 . 066 in Ans. G 20 ± P
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328 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-8 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10) N P 2 k 3 sin 2 φ ± 1 + ² 1 + 3 sin 2 φ ³ 2(1) 3 sin 2 20° ´ 1 + µ 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° · 2 . 5 + ² 2 . 5 2 + [1 + 2(2 . 5)] sin 2 20° ¸ 14 . 64 15 pinion teeth Ans. The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is N G N 2 P sin 2 φ 4 k 2 4 k 2 N P sin 2 φ 15 2 sin 2 20° 4(1) 2 4(1) 2(15) sin 2 20° 45 . 49 45 teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-13) N P 2 k sin 2 φ = 2(1) sin 2 20° 17 . 097 18 teeth Ans. 13-9 φ n = 20°, ψ = 30°, φ t = tan 1 (tan 20° / cos 30°) = 22 . 80° (a) The smallest pinion tooth count that will run itself is found from Eq. (13-21) N P 2 k cos ψ 3 sin 2 φ t ± 1 + ² 1 + 3 sin 2 φ t ³ 2(1) cos 30° 3 sin 2 22 . 80° ´ 1 + µ 1 + 3 sin 2 22 . 80° 8 . 48 9 teeth Ans.
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Budynas ch13 - Chapter 13 13-1 d P = 17/8 = 2.125 in dG =...

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