Tài liệu shi20396 chương 13 doc

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 = πm = 3π 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 = 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 = 7in d 1b = 7 cos 20° = 6.578 in Ans. Gear Base-Circle: d 2 = N 2 /P = 28/3 = 9.333 in d 2b = 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. shi20396_ch13.qxd 8/29/03 12:16 PM Page 333 334 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-5 (a) A O =   2.333 2  2 +  5.333 2  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, 0.3A 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 30Њ P G 2 1 3 " 5 1 3 " A O ⌫ ␥ 10.5Њ Arc of approach ϭ 0.87 in Ans. Arc of recess ϭ 0.77 in Ans. Arc of action ϭ 1.64 in Ans. L ab ϭ 1.53 in 10Њ O 2 O 1 14Њ 12.6Њ P B A shi20396_ch13.qxd 8/29/03 12:16 PM Page 334 Chapter 13 335 (b) 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 ◦ ) = 22.8° 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 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 φ t = tan −1  tan 14.5° cos 20°  = 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 20Њ P shi20396_ch13.qxd 8/29/03 12:16 PM Page 335 336 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 ≥ 4k 6 sin 2 φ  1 +  1 + 3 sin 2 φ  ≥ 4(1) 6 sin 2 20°  1 +  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 +  2.5 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 ≥ 4k 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 − 4k 2 4k − 2N P sin 2 φ ≤ 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 φ t = 21.88°. (a) The smallest tooth count that will mesh with a like gear, from Eq. (13-21), is N P ≥ 4k cos ψ 6 sin 2 φ t  1 +  1 + 3 sin 2 φ t  ≥ 4(1) cos 30° 6 sin 2 21.88°  1 +  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 ≥ 4k cos ψ 2 sin 2 φ t ≥ 4(1) cos 30 ◦ 2 sin 2 21.88° ≥ 12.47 → 13 teeth Ans. shi20396_ch13.qxd 8/29/03 12:16 PM Page 336 Chapter 13 337 (c) The largest gear tooth possible, from Eq. (13-24) is N G ≤ N 2 P sin 2 φ t − 4k 2 cos 2 ψ 4k cos ψ − 2N 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  tan 20° cos 30°  = 22.796° Program Eq. (13-24) on a computer using a spreadsheet or code and increment N P . The first 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 first 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. 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 φ t − 4 cos 2 ψ 4 cos ψ − 2N P sin 2 φ t Set the denominator to zero 4 cos ψ − 2N P sin 2 φ t = 0 From which sin φ t =  2 cos ψ N P φ t = sin −1  2 cos ψ N P For N P = 9 teeth and cos ψ = 1 φ t = sin −1  2(1) 9 = 28.126° Ans. 13-13 (a) p n = πm n = 3π mm Ans. p t = 3π/cos 25° = 10.4mm Ans. p x = 10.4/tan 25° = 22.3mm Ans. 18T 32 T ␺ ϭ 25Њ, ␾ n ϭ 20Њ, m ϭ 3 mm shi20396_ch13.qxd 8/29/03 12:16 PM Page 337 338 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) m t = 10.4/π = 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. (b) n c = n 5 = 14 54  16 36  (900) =+103.7rev/min ccw Ans. (c) d P2 = 14/(10 cos 30°) = 1.6166 in d G3 = 54/(10 cos 30°) = 6.2354 in C ab = 1.6166 + 6.2354 2 = 3.926 in Ans. d P4 = 16/(6 cos 25°) = 2.9423 in d G5 = 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 shi20396_ch13.qxd 8/29/03 12:16 PM Page 338 Chapter 13 339 13-16 e = 6 10  18 38  20 48  3 36  = 3 304 n a = 3 304 (1200) = 11.84 rev/min cw Ans. 13-17 (a) n c = 12 40 · 1 1 (540) = 162 rev/min cw about x. Ans. (b) d P = 12/(8 cos 23°) = 1.630 in d G = 40/(8 cos 23°) = 5.432 in d P + d G 2 = 3.531 in Ans. (c) d = 32 4 = 8inatthe large end of the teeth. Ans. 13-18 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus n A = n 3 = 1200(17/54) = 377.8rev/min Ans. (b) n F = n 5 = 0, n L = n 6 , e =−1 −1 = n 6 − 377.8 0 − 377.8 377.8 = n 6 − 377.8 n 6 = 755.6rev/min Ans. Alternatively, the velocity of the center of gear 4 is v 4c ∝ N 6 n 3 . The velocity of the left edge of gear 4 is zero since the left wheel is resting on the ground. Thus, the ve- locity of the right edge of gear 4 is 2v 4c ∝ 2N 6 n 3 . This velocity, divided by the radius of gear 6 ∝ N 6 , is angular velocity of gear 6–the speed of wheel 6. ∴ n 6 = 2N 6 n 3 N 6 = 2n 3 = 2(377.8) = 755.6rev/min Ans. (c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. 13-19 (a) The motive power is divided equally among four wheels instead of two. (b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels, rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50. shi20396_ch13.qxd 8/29/03 12:16 PM Page 339 340 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. Alternatively, since N ∝ r , let v = Nn (crazy units). v = N 6 n 6 N 6 = 20 + 30 − 16 = 34 teeth v A N 4 = v N 4 − N 5 ⇒ v A = N 4 N 6 n 6 N 4 − N 5 n A = v A N 2 + N 4 = N 4 N 6 n 6 (N 2 + N 4 )(N 4 − N 5 ) = 30(34)(12) (20 + 30)(30 − 16) = 17.49 rev/min 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. Alternatively, since N ∝ r , let v = Nn (crazy units). v A N 5 = v N 4 − N 5 = N 2 n 2 N 4 − N 5 v A = N 5 N 2 n 2 N 4 − N 5 n A = v A N 2 + N 4 = N 5 N 2 n 2 (N 2 + N 4 )(N 4 − N 5 ) = 16(20)(180) (20 + 30)(30 − 16) = 82.29 rev/min cw Ans. 45 v ϭ 0 v ϭ N 2 n 2 N 2 v A 2 4 5 v v ϭ 0 v A 2 shi20396_ch13.qxd 8/29/03 12:16 PM Page 340 Chapter 13 341 13-22 N 5 = 12 + 2(16) + 2(12) = 68 teeth Ans. Let gear 2 be first, n F = n 2 = 320 rev/min . Let gear 5 be last, n L = n 5 = 0 e = 12 16  16 12  12 68  = 3 17 , e = n L − n A n F − n A 320 − n A = 17 3 (0 − n A ) n A =− 3 14 (320) =−68.57 rev/min The negative sign indicates opposite of n 2 ∴ n A = 68.57 rev/min cw Ans. Alternatively, n A = n 2 N 2 2(N 3 + N 4 ) = 320(12) 2(16 + 12) = 68.57 rev/min cw Ans. 13-23 Let n F = n 2 then n L = n 7 = 0. e =− 24 18  18 30  36 54  =− 8 15 e = n L − n 5 n F − n 5 =− 8 15 0 − 5 n 2 − 5 =− 8 15 ⇒ n 2 = 5 + 15 8 (5) = 14.375 turns in same direction 13-24 (a) Let n F = n 2 = 0, then n L = n 5 . e = 99 100  101 100  = 9999 10 000 , e = n L − n A n F − n A = n L − n A 0 − n A n L − n A =−en A n L = n A (−e + 1) n L n A = 1 − e = 1 − 9999 10 000 = 1 10 000 = 0.0001 Ans. (b) d 4 = N 4 P = 101 10 = 10.1in d 5 = 100 10 = 10 in d housing > 2  d 4 + d 5 2  = 2  10.1 + 10 2  = 30.2in Ans. v ϭ 0 n A (N 2 ϩ N 3 ) v ϭ n 2 N 2 2n A (N 2 ϩ 2N 3 ϩ N 4 ) ϭ n 2 N 2 ϩ 2n A (N 2 ϩ N 3 ) 2n A (N 2 ϩ 2N 3 ϩ N 4 ) Ϫ 2n A (N 2 ϩ N 3 ) ϭ n 2 N 2 n A (N 2 ϩ 2N 3 ϩ N 4 ) shi20396_ch13.qxd 8/29/03 12:16 PM Page 341 342 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 13-25 n 2 = n b = n F , n A = n a , n L = n 5 = 0 e =− 21 444 = n L − n A n F − n A − 21 444 (n F − n A ) = 0 − n A With shaft b as input − 21 444 n F + 21 444 n A + 444 444 n A = 0 n A n F = n a n b = 21 465 n a = 21 465 n b ,inthe same direction as shaft b, the input. Ans. Alternatively, v A N 4 = n 2 N 2 N 3 + N 4 v A = n 2 N 2 N 4 N 3 + N 4 n a = n A = v A N 2 + N 3 = n 2 N 2 N 4 (N 2 + N 3 )(N 3 + N 4 ) = 18(21)(n b ) (18 + 72)(72 + 21) = 21 465 n b in the same direction as b Ans. 13-26 n F = n 2 = n a , n L = n 6 = 0 e =− 24 18  22 64  =− 11 24 , e = n L − n A n F − n A = 0 − n b n a − n b − 11 24 = 0 − n b n a − n b ⇒ n b n a = 11 35 Ans. Yes, both shafts rotate in the same direction. Ans. Alternatively, v A N 5 = n 2 N 2 N 3 + N 5 = N 2 N 3 + N 5 n a , v A = N 2 N 5 N 3 + N 5 n a n A = n b = v A N 2 + N 3 = N 2 N 5 (N 2 + N 3 )(N 3 + N 5 ) n a n b n a = 24(22) (24 + 18)(22 + 18) = 11 35 Ans. n b rotates ccw ∴ Yes Ans. 13-27 n 2 = n F = 0, n L = n 5 = n b , n A = n a e =+ 20 24  20 24  = 25 36 3 5 v ϭ 0 v A n 2 N 2 2 3 4 2 v ϭ 0 v A n 2 N 2 shi20396_ch13.qxd 8/29/03 12:16 PM Page 342 [...]... So y FB = 613 N F B = 318.5i + 613j N Ans shi20396_ ch13.qxd 8/29/03 12:16 PM Page 357 357 Chapter 13 Or FB = [( 613) 2 + (318.5) 2 ]1/2 = 691 N radial F = F A + W + RB = 0 Radial F A = −(W + F B ) = −(−637i + 133 3j + 5119k + 318.5i + 613j) = 318.5i − 1946j − 5119k Ans Fr = 318.5i − 1946j N, A r FA = [(318.5) 2 + (−1946) 2 ]1/2 = 1972 N Thrust 13- 42 a FA = −5119 N From Prob 13- 41 WG = 637i − 133 3j − 5119k... n2 = shi20396_ ch13.qxd 8/29/03 12:16 PM Page 347 347 Chapter 13 Gear 3 r Fb 3 3 Tout ϭ W t23 r3 ϭ 2799 lbf • in t t W23 = W32 = 622 lbf r r W23 = W32 = 226 lbf t Fb3 b Fb3 = Fb2 = 662 lbf RC = R D = 662/2 = 331 lbf W t23 Wr 23 Each bearing on shaft b has the same radial load which is equal to the radial load of bearings, A and B Thus, all four bearings have the same radial load of 331 lbf Ans 13- 32... 179(5.625/8.98)]2 + [179(7/8.98)]2 = 145 lbf shi20396_ ch13.qxd 8/29/03 12:16 PM Page 349 349 Chapter 13 The shear forces are independent of the rotational sense The bolt tensions and the shear forces for cw rotation are, Tension (lbf) 0 0 1109 1109 145 145 300 300 Tension (lbf) A B C D Shear (lbf) Shear (lbf) 182 182 0 0 145 145 300 300 For ccw rotation, A B C D 13- 33 Tin = 63 025H/n = 63 025(2.5)/240 =... FC = (11.42 + 39.32 ) 1/2 = 40.9 lbf Ans F=0 F D = −5.78i + 9.363j − 69.3k lbf FD (radial) = [(−5.78) 2 + (−69.3) 2 ]1/2 = 69.5 lbf Ans FD (thrust) = W a = 9.363 lbf Ans shi20396_ ch13.qxd 8/29/03 12:16 PM Page 351 351 Chapter 13 13-35 Sketch gear 2 pictorially z y Pt = Pn cos ψ = 4 cos 30° = 3.464 teeth/in Wa 2 W Wr t φt = tan−1 x tan φn tan 20° = tan−1 = 22.80° cos ψ cos 30° T Sketch gear 3 pictorially,... 3.464 TG = W t r = 800(9.238) = 7390 lbf · in 13- 36 From Prob 13- 35 solution, 3 800 462 800 4 462 336 336 336 336 2 462 800 800 462 Notice that the idler shaft reaction contains a couple tending to turn the shaft end-overend Also the idler teeth are bent both ways Idlers are more severely loaded than other gears, belying their name Thus be cautious shi20396_ ch13.qxd 352 8/29/03 12:16 PM Page 352 Solutions... 1114j − 1929k lbf Ans 13- 38 Pt = 6 cos 30° = 5.196 teeth/in y 3 d3 = C D 42 = 8.083 in 5.196 T3 φt = 22.8° T2 d2 = x B 2 A 16 = 3.079 in 5.196 63 025(25) = 916 lbf · in 1720 916 T = 595 lbf Wt = = r 3.079/2 T2 = W a = 595 tan 30° = 344 lbf W r = 595 tan 22.8° = 250 lbf W = 344i + 250j + 595k lbf R DC = 6i, R DG = 3i − 4.04j shi20396_ ch13.qxd 8/29/03 12:16 PM Page 353 353 Chapter 13 y y FC x x FC C 3"... 43 b 3 Fx b3 Fa 43 Fb3 Fy b3 Fa 23 z Fr 23 F t23 y 13- 40 y FD D x FD 2.6"R b z FD 2" Fr 54 1.55"R 3 Fa 54 G F t54 1" 2 Fa 23 3" x FC 3 4 x H Fr 23 F t23 C z y FC 14 N 36 = = 2.021 in, d3 = = 5.196 in Pn cos ψ 8 cos 30° 8 cos 30° 15 45 d4 = = 3.106 in, d5 = = 9.317 in 5 cos 15° 5 cos 15° d2 = shi20396_ ch13.qxd 8/29/03 12:16 PM Page 355 355 Chapter 13 For gears 2 and 3: φt = tan−1 (tan φn /cos ψ) = tan−1.. .shi20396_ ch13.qxd 8/29/03 12:16 PM Page 343 343 Chapter 13 nb − na 25 = 36 0 − na 11 nb = Ans na 36 Same sense, therefore shaft b rotates in the same direction as a Ans Alternatively, ( N2 + N3 )n a v5 = N3 − N4 N3 4 (N2 ϩ N3)na v5 = 3 v5 vϭ0 nb = ( N2 + N3 )( N3 − N4 )n N3 v5 ( N2 + N3 )( N3 − N4 )n a = N5 N3 N5 (20 + 24)(24 − 20) nb = na 24(24) = 13- 28 11 36 same sense Ans... 65.8 lbf W z = 263[cos 14.5◦ (cos 7.59◦ ) − 0.05 sin 7.59◦ ] = 251 lbf So W = 46.7i + 65.8j + 251k lbf Ans T = 46.7(0.75) = 35 lbf · in Ans 13- 44 100:101 Mesh dP = 100 = 2.083 33 in 48 dG = 101 = 2.104 17 in 48 shi20396_ ch13.qxd 8/29/03 12:16 PM Page 359 359 Chapter 13 Proper center-to-center distance: 2.083 33 + 2.104 17 d P + dG = = 2.093 75 in 2 2 2.0833 cos 20◦ = 0.9788 in = r cos φ = 2 C= rb P 99:100... (b) on the following page shi20396_ ch13.qxd 344 8/29/03 12:16 PM Page 344 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design r4 = (b) 5(51) m N4 = = 127.5 mm 2 2 9.36 F t43 3 O b Fb3 18.73 F t23 9.36 Tc4 = 9.36(127.5) = 1193 N · m ccw ∴ T4c = 1193 N · m cw Ans 4 Note: The solution is independent of the pressure angle c Tc4 ϭ 1193 9.36 13- 29 4 2 5 6 N 6 d2 = 4 . Eq. (13- 13) is N G ≤ N 2 P sin 2 φ − 4k 2 4k − 2N P sin 2 φ ≤ 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,. Eq. (13- 23), is N P ≥ 4k cos ψ 2 sin 2 φ t ≥ 4(1) cos 30 ◦ 2 sin 2 21.88° ≥ 12.47 → 13 teeth Ans. shi20396_ ch13.qxd 8/29/03 12:16 PM Page 336 Chapter 13