Chaper 14 14-1 d N = = = 6.667 in Table 14-: Y = 0.1 Eq. (1-4): πdn π(.667)(100) V = = = 115 f/min 1 1 Eq. (14-4b): 100 + 115 = = 1.96 100 Eq. (1-5) : 15 = 000 = 000 = 49.7 lbf V 115 Eq. (14-7): 1.96(49.7)(6) = = = 76 psi = 7.6 kpsi FY (0.1) Ans. 14- d = N 18 = = 1.8 in 10 Table 14-: Y = 0.09 Eq. (1-4): π dn π (1.8)(600) V = = = 8.7 f/min 1 1 Eq. (14-4b): 100 + 8.7 = = 1.6 100 Eq. (1-5) : = 000 = 000 V 8.7 =.5 lbf Eq. (14-7): 1.6(.5)(10) = = = 940 psi = 9.4 kpsi FY 1.0(0.09) Ans. 14- d = mn = 1.5(18) =.5 mm Table 14-: Y = 0.09 V π dn π (.5)(10 )(1800) = = =.11 m/s 60 60 Eq. (14-6b): 6.1 +.11 = = 1.48 6.1 Eq. (1-6): 60 000 60 000(0.5) 0.58 kn 5.8 N π dn π (.5)(1800) Eq. (14-8): 1.48(5.8) = = = 68.6 Ma FmY 1(1.5)(0.09) Ans. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9
14-4 d = mn = 8(16) = 18 mm Table 14-: Y = 0.96 V π dn π (18)(10 )(150) = = = 1.005 m/s 60 60 Eq. (14-6b): 6.1 + 1.005 = = 1.165 6.1 Eq. (1-6): 60 000 60 000(6) 5.968 kn 5968 N π dn π (18)(150) Eq. (14-8): 1.165(5968) = = =.6 Ma FmY 90(8)(0.96) Ans. 14-5 d = mn = 1(16) = 16 mm Table 14-: Y = 0.96 V π dn π (16)(10 )(400) = = = 0.5 m/s 60 60 Eq. (14-6b): 6.1 + 0.5 = = 1.055 6.1 Eq. (1-6): 60 000 60 000(0.15) 0.4476 kn 447.6 N π dn π (16)(400) Eq. (14-8): F 1.055(447.6) = = = 10.6 mm my 150(1)(0.96) From Table 1-, use F = 11 mm or 1 mm, depending on availabiliy. Ans. 14-6 d = mn = (0) = 40 mm Table 14-: Y = 0. V π dn π (40)(10 )(00) = = = 0.419 m/s 60 60 Eq. (14-6b): 6.1 + 0.419 = = 1.069 6.1 Eq. (1-6): 60 000 60 000(0.5) 1.194 kn 1194 N π dn π (40)(00) Eq. (14-8): F 1.069(1194) = = = my 75(.0)(0.) 6.4 mm From Table 1-, use F = 8 mm. Ans. Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
14-7 d = N 4 = = 4.8 in 5 Table 14-: Y = 0.7 Eq. (1-4): V π dn π (4.8)(50) = = = 6.8 f/min 1 1 Eq. (14-4b): 100 + 6.8 = = 1.05 100 Eq. (1-5) : 6 = 000 = 000 = 151 lbf V 6.8 Eq. (14-7): 1.05(151)(5) F = = =.46 in Y 0(10 )(0.7) Use F =.5 in Ans. 14-8 d = N 16 = = 4.0 in 4 Table 14-: Y = 0.96 Eq. (1-4): π dn π (4.0)(400) V = = = 418.9 f/min 1 1 Eq. (14-4b): 100 + 418.9 = = 1.49 100 Eq. (1-5) : 0 = 000 = 000 = 1575.6 lbf V 418.9 Eq. (14-7): 1.49(1575.6)(4) F = = =.9 in Y 1(10 )(0.96) Use F =.5 in Ans. 14-9 Try = 8 which gives d = 18/8 =.5 in and Y = 0.09. Eq. (1-4): V π dn π (.5)(600) = = = 1 1 5.4 f/min Eq. (14-4b): 100 + 5.4 = = 1.95 100 Eq. (1-5):.5 = 000 = 000 =.4 lbf V 5.4 Eq. (14-7): 1.95(.4)(8) F = = = 0.78 in Y 10(10 )(0.09) Using coarse ineger piches from Table 1-, he following able is formed. Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
d V F 9.000 141.717.178 58.56 0.08 6.000 94.478 1.785 87.55 0.15 4 4.500 706.858 1.589 116.71 0.40 6.000 471.9 1.9 175.069 0.47 8.50 5.49 1.95.46 0.78 10 1.800 8.74 1.6 91.78 1.167 1 1.500 5.619 1.196 50.19 1.67 16 1.15 176.715 1.147 466.85.77 Oher consideraions may dicae he selecion. ood candidaes are = 8 (F = 7/8 in) and =10 (F = 1.5 in). Ans. 14-10 Try m = mm which gives d = (18) = 6 mm and Y = 0.09. π dn π (6)(10 )(900) V = = = 1.696 m/s 60 60 6.1 + 1.696 Eq. (14-6b): = = 1.78 6.1 60 000 60 000(1.5) Eq. (1-6): = 0.884 kn 884 N π dn = π (6)(900) = = 1.78(884) Eq. (14-8): F = = 4.4 mm 75()(0.09) Using he preferred module sizes from Table 1-: m d V K v F 1.00 18.0 0.848 1.19 1768.88 86.917 1.5.5 1.060 1.174 1414.711 57.4 1.50 7.0 1.7 1.09 1178.96 40.987.00 6.0 1.696 1.78 884.194 4.8.00 54.0.545 1.417 589.46 1.015 4.00 7.0.9 1.556 44.097 7.4 5.00 90.0 4.41 1.695 5.678 5.174 6.00 108.0 5.089 1.84 94.71.888 8.00 144.0 6.786.11 1.049.519 10.00 180.0 8.48.91 176.89 1.84 1.00 16.0 10.179.669 147.66 1.414 16.00 88.0 1.57.5 110.54 0.961 0.00 60.0 16.965.781 88.419 0.71 5.00 450.0 1.06 4.476 70.76 0.547.00 576.0 7.14 5.450 55.6 0.406 40.00 70.0.99 6.56 44.10 0.1 50.00 900.0 4.41 7.95 5.68 0.4 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9
Oher design consideraions may dicae he size selecion. For he presen design, m = mm (F = 5 mm) is a good selecion. Ans. 14-11 Eq. (14-4b): Eq. (1-6): N 0 N 50 d = = =.5 in, d = = = 6.5 in 8 8 π (.5)(100) V = = 785.4 f/min 1 100 + 785.4 = = 1.655 100 1 = 000 = 000 = 504. lbf V 785.4 Table 14-8: C = 100 psi [Noe: Using Eq. (14-1) can resul in wide variaion in p C p due o wide variaion in cas iron properies.].5sin 0 6.5 sin 0 Eq. (14-1): r1 = = 0.475 in, r = = 1.069 in Eq. (14-14): C C K 1 1 v = p + F cosφ r r 1 1.655(504.) 1 1 = 100 + 1.5 cos 0 0.475 1.069 = 9.5(10 ) psi = 9.5 kpsi Ans. 1/ 1/ 14-1 d V 16 48 = = 1. in, d = = 4 in 1 1 π(1.)(700) = = 44. f/min 1 Eq. (14-4b): Eq. (1-6): Table 14-8: 100 + 44. = = 1.04 100 1.5 = 000 = 000 = 0.6 lbf V 44. C = 100 psi [Noe: Using Eq. (14-1) can resul in wide variaion p in C p due o wide variaion in cas iron properies.] 1.sin 0 4 sin 0 Eq. (14-1): r1 = = 0.8 in, r = = 0.684 in Eq. (14-14): Shigley s MED, 10 h ediion Chaper 14 Soluions, age 5/9
C 1/ 1.04(0.6) 1 1 100 100(10 ) = + = F cos 0 0.8 0.684 100 1.04(0.6) 1 1 F = 0.669 in 100(10 ) + = cos 0 0.8 0.684 Use F = 0.75 in Ans. 14-1 d = 5(4) = 10 mm, d = 5(48) = 40 mm p (10)(10 )(50) V π = = 0.14 m/s 60.05 + 0.14 Eq. (14-6a): = = 1.10.05 60 000 60(10 ) =.18 π dn = π (10)(50) = where is in k and is in kn Table 14-8: C = 16 Ma [Noe: Using Eq. (14-1) can resul in wide variaion in p C p due o wide variaion in cas iron properies]. 10sin 0 40sin 0 Eq. (14-1): r1 = = 0.5 mm, r = = 41.04 mm Eq. (14-14): ( ) 1.10(.18) 10 1 1 690 = 16 + o 60 cos 0 0.5 41.04 =.94 k Ans. 1/ 14-14 Eq. (14-6a): d = 4(0) = 80 mm, d = 4() = 18 mm π (80)(10 )(1000) V = = 4.189 m/s 60.05 + 4.189 = =.7.05 = 60(10)(10 ).87 kn 87 N π (80)(1000) = = Table 14-8: C = 16 Ma [Noe: Using Eq. (14-1) can resul in wide variaion in p C p due o wide variaion in cas iron properies.] Shigley s MED, 10 h ediion Chaper 14 Soluions, age 6/9
80sin 0 18sin 0 Eq. (14-1): r1 = = 1.68 mm, r = = 1.89 mm.7(87) 1 1 Eq. (14-14): C = 16 + = 617 Ma Ans. 50 cos 0 1.68 1.89 14-15 The pinion conrols he design. Bending Y = 0.0, Y = 0.59 17 0 d = = 1.417 in, d = =.500 in 1 1 πdn π(1.417)(55) V = = = 194.8 f/min 1 1 100 + 194.8 Eq. (14-4b): = = 1.16 100 Eq. (6-8), p. 90: S e = 0.5(76) = 8.0 kpsi Eq. (6-19), p. 95: k a =.70(76) 0.65 = 0.857.5.5 l = = = 0.1875 in d 1 Y (0.0) Eq. (14-): x = = = 0.079 in (1) Eq. (b), p. 79: = 4lx = 4(0.1875)(0.079) = 0.1686 in Eq. (6-5), p. 97: d = 0.808 hb = 0.808 0.875(0.1686) = 0.10 in e 0.107 0.10 Eq. (6-0), p. 96: k b = = 0.996 0. k c = k d = k e = 1 Accoun for one-way bending wih k f = 1.66. (See Ex. 14-.) 1/ Eq. (6-18), p. 95: S e = 0.857(0.996)(1)(1)(1)(1.66)(8.0) = 5.84 kpsi For sress concenraion, find he radius of he roo fille (See Ex. 14-). 0.00 0.00 r = f 0.05 in = 1 = From Fig. A-15-6, r r 0.05 = f = = 0.148 d 0.1686 Approximae D/d = wih D/d = ; from Fig. A-15-6, K = 1.68. From Fig. 6-0, wih S u = 76 kpsi and r = 0.05 in, q = 0.6. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 7/9
Eq. (6-): K f = 1 + 0.6 (1.68 1) = 1.4 Se 5.84 all = = = 16.85 psi K n 1.4(.5) f d FY all 0.875(0.0)(16 850) = = = 0.4 lbf d 1.16(1) V 0.4(194.8) = = = 1.89 hp Ans. 000 000 ear ν 1 = ν = 0.9, E 1 = E = 0(10 6 ) psi Eq. (14-1): C p 1/ 1 = = 1 0.9 π 6 0( 10 ) 85 psi d 1.417 Eq. (14-1): r1 = sin φ = sin 0 = 0.4 in d.500 r = sinφ = sin 0 = 0.48 in 1 1 1 1 1 + = + = 6.469 in r r 0.4 0.48 1 Eq. (6-68), p. 7: ( S ) 8 = 0.4 10 kpsi = [0.4(149) 10](10 ) = 49 600 psi C 10 From he discussion and equaion developed on he boom of p. 7, ( SC ) 10 8 49 600 C,all = = = 067 psi n.5 067 0.875cos 0 Eq. (14-14): = =.6 lbf 85 1.16(6.469) V.6(194.8) = = = 0.1 hp Ans. 000 000 Raing power (pinion conrols): 1 = 1.89 hp = 0.1 hp all = (min 1.89, 0.1) = 0.1 hp Ans. B Shigley s MED, 10 h ediion Chaper 14 Soluions, age 8/9
14-16 See rob. 14-15 soluion for equaion numbers. inion conrols: Y = 0., Y = 0.447 Bending d = 0/ = 6.667 in, d = 100/ =. in V = π dn / 1 = π (6.667)(870) / 1 = 1519 f/min = (100 + 1519) / 100 =.66 S e = 0.5(11) = 56.5 kpsi 0.65 ka =.70(11) = 0.771 l =.5 / d =.5 / = 0.75 in x = (0.) / [()] = 0.161 in = 4(0.75)(0.161) = 0.695 in d e = 0.808.5(0.695) = 1.065 in 0.107 kb = (1.065 / 0.0) = 0.87 kc = kd = ke = 1 k f = 1.66 (See Ex. 14-.) S = 0.771(0.87)(1)(1)(1)(1.66)(56.5) = 6.1 kpsi e rf = 0.00 / = 0.100 in r rf 0.100 = = = 0.144 d 0.695 K = 1.75, q = 0.85, K f = 1.64 Se 6.1 all = = = 5.7 kpsi K n 1.64(1.5) f d FY all.5(0.)(5 700) = = = 04 lbf d.66() = V / 000 = 04(1519) / 000 = 140 hp Ans. ear Eq. (14-1): C p 1/ 1 = = 1 0.9 π 6 0( 10 ) 85 psi Eq. (14-1): r 1 = (6.667/) sin 0 = 1.140 in r = (./) sin 0 = 5.700 in Eq. (6-68), p. 7: S C = [0.4(6) 10](10 ) = 94 800 psi C,all = SC / nd = 94 800 / 1.5 = 77 400 psi Shigley s MED, 10 h ediion Chaper 14 Soluions, age 9/9
C,all F cosφ 1 = C p 1 / r1 + 1 / r 77 400.5cos 0 1 = 85.66 1 / 1.140 + 1 / 5.700 = 110 lbf V 110(1519) = = = 5.0 hp Ans. 000 000 For 10 8 cycles (revoluions of he pinion), he power based on wear is 5.0 hp. Raing power (pinion conrols): 1 = 140 hp = 5.0 hp raed = min(140, 5.0) = 5.0 hp Ans. 14-17 See rob. 14-15 soluion for equaion numbers. iven: φ = 0, n = 1145 rev/min, m = 6 mm, F = 75 mm, N = 16 milled eeh, N = 0T, S u = 900 Ma, B = 60, n d =, Y = 0.96, and Y = 0.59. inion bending d = mn = 6(16) = 96 mm d = 6(0) = 180 mm π dn π (96)(10 )(1145) V = = = 5.76 m/s 60 (60) 6.1 + 5.76 = = 1.944 6.1 S e = 0.5(900) = 450 Ma 0.65 ka = 4.51(900) = 0.744 l =.5m =.5(6) = 1.5 mm x = Ym / = (0.96)6 / =.664 mm = 4lx = 4(1.5)(.664) = 1.0 mm d e = 0.808 75(1.0) = 4. mm 0.107 4. kb = = 0.884 7.6 kc = kd = ke = 1 k f = 1.66 (See Ex. 14-) S = 0.744(0.884)(1)(1)(1)(1.66)(450) = 491. Ma r f e = 0.00m = 0.00(6) = 1.8 mm r/d = r f / = 1.8/1 = 0.15, K = 1.68, q = 0.86, K f = 1.58 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 10/9
Se 491. all = = = K n 1.58 1. f d ( ) 9. Ma FYm all 75(0.96)(6)(9.) Eq. (14-8): = = = 16 90 N K 1.944 Eq. (1-6): ear: inion and gear v π dn 16.9 π (96)(1145) = = = 94. k Ans. 60 000 60 000 Eq. (14-1): r 1 = (96/) sin 0 = 16.4 mm r = (180/) sin 0 = 0.78 mm 1 Eq. (14-1): C p = = 190 Ma 1 0.9 π 07( 10 ) Eq. (6-68), p. 7: S C = 6.89[0.4(60) 10] = 647.7 Ma 647.7 C,all = SC / nd = = 568 Ma 1. Eq. (14-14): Eq. (1-6): 1/ F cosφ 1 C,all = C p K 1 / r1 + 1 / r v o 568 75cos 0 1 = = 190 1.944 1 / 16.4 + 1 / 0.78 π dn.469 π (96)(1145) = = = 0.0 k 60 000 60 000 469 N Thus, wear conrols he gearse power raing; = 0.0 k. Ans. 14-18 N = 17 eeh, N = 51 eeh N 17 d = = =.8 in 6 51 d = = 8.500 in 6 V = π d n / 1 = π (.8)(110) / 1 = 80.7 f/min Eq. (14-4b): = (100 + 80.7)/100 = 1.69 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 11/9
Sy 90 000 all = = = n d 45 000 psi Table 14-: Y = 0.0, Y = 0.410 all (0.0)(45 000) Eq. (14-7): FY = = = 686 lbf K 1.69(6) Eq. (1-5): v V 686(80.7) = = = 67.6 hp 000 000 Based on yielding in bending, he power is 67.6 hp. (a) inion faigue Bending Eq. (-1), p. 5: S u = 0.5 B = 0.5() = 116 kpsi Eq. (6-8), p. 90: S = 0.5S = 0.5(116) = 58 kpsi Eq. (6-19), p. 95: e k a u 0.65.70(116) 0.766 = = Table 1-1, p. 688: l 1 1.5.5.5 = + = = = 6 d d d 0.75 in Eq. (14-): x Y (0.0) = = = 0.0758 in (6) Eq. (b), p. 79: = 4lx = 4(0.75)(0.0758) = 0.7 in Eq. (6-5), p. 97: d = 0.808 F = 0.808 (0.7) = 0.66 in e 0.107 0.66 Eq. (6-0), p. 96: k b = = 0.919 0.0 k c = k d = k e = 1 Accoun for one-way bending wih k f = 1.66. (See Ex. 14-.) Eq. (6-18): S e = 0.766(0.919)(1)(1)(1)(1.66)(58) = 67.8 kpsi For sress concenraion, find he radius of he roo fille (See Ex. 14-). 0.00 0.00 r = f 0.050 in = 6 = r r 0.05 Fig. A-15-6: = f = = 0.148 d 0.8 Esimae D/d = by seing D/d =, K = 1.68. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9
Fig. 6-0, p. 0: q = 0.86 Eq. (6-), p. 0: K f = 1 + (0.86)(1.68 1) = 1.58 Se 67.8 all = = = 1.5 kpsi K n 1.58() (b) inion faigue f d FY all (0.0)(1 500) = = = 18 lbf K 1.69(6) v d V 18(80.7) = = =. hp Ans. 000 000 ear Eq. (14-1): C p 1/ 1 = 6 = π[(1-0.9 ) / 0(10 )] 85 psi Eq. (14-1): d.8 o sin φ sin 0 0.485 in r1 = = = d 8.500 o r = sinφ = sin 0 = 1.454 in 1 1 1 1 + = + =.750 in r1 r 0.485 1.454 Eq. (6-68): ( S ) 8 = 0.4 10 kpsi C 10 In erms of gear noaion B C = [0.4() 10]10 = 8 800 psi e will inroduce he design facor of n d = and because i is a conac sress apply i o he load by dividing by n d =. (See p. 7.) c 8 800 C,all = = = 58 548 psi Solve Eq. (14-14) for : o 58 548 cos 0 = 65 lbf 85 1.69(.750) = V 65(80.7) all = = = 6.67 hp Ans. 000 000 For 10 8 cycles (urns of pinion), he allowable power is 6.67 hp. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9
(c) ear faigue due o bending and wear Bending Eq. (14-): Y (0.410) x = = = 0.106 in (6) Eq. (b), p. 79: = 4lx = 4(0.75)(0.106) = 0.9 in ± Eq. (6-5): d = 0.808 F = 0.808 (0.9) = 0.715 in e 0.107 0.715 Eq. (6-0): k b = = 0.911 0.0 k c = k d = k e = 1 k f = 1.66. (See Ex. 14-.) Eq. (6-18): S e = 0.766(0.911)(1)(1)(1)(1.66)(58) = 67. kpsi 0.050 r d r f = = = 0.18 0.9 Approximae D/d = by seing D/d = for Fig. A-15-6; K = 1.80. Fig. 6-0: q = 0.8 Eq. (6-): K f = 1 + (0.8)(1.80 1) = 1.66 Se 67. all = = = 0. kpsi K n 1.66() f d FY all (0.410)(0 00) = = = 16 lbf K 1.69(6) v d V 16(80.7) all = = = 41.1 hp Ans. 000 000 The gear is hus sronger han he pinion in bending. ear Since he maerial of he pinion and he gear are he same, and he conac sresses are he same, he allowable power ransmission of boh is he same. Thus, all = 6.67 hp for 10 8 revoluions of each. As ye, we have no way o esablish S C for 10 8 / revoluions. (d) inion bending: 1 =. hp inion wear: = 6.67 hp ear bending: = 41.1 hp ear wear: 4 = 6.67 hp ower raing of he gear se is hus raed = min(., 6.67, 41.1, 6.67) = 6.67 hp Ans. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 14/9
14-19 d = 16/6 =.667 in, d = 48/6 = 8 in π (.667)(00) V = = 09.4 f/min 1 000(5) = = 787.8 lbf 09.4 Assuming uniform loading, K o = 1. / Eq. (14-8): Qv = 6, B = 0.5(1 6) = 0.855 A = 50 + 56(1 0.855) = 59.77 0.855 59.77 + 09.4 Eq. (14-7): = = 1.196 59.77 Table 14-: Y = 0.96, Y = 0.4056 From Eq. (a), Sec. 14-10 wih F = in 0.055 0.96 ( Ks) = 1.19 = 1.088 6 0.055 0.4056 ( Ks) = 1.19 = 1.097 6 From Eq. (14-0) wih C mc = 1 C p f = 0.075 + 0.015() = 0.065 10(.667) C = 1, C = 0.09 (Fig. 14-11), C = 1 K p m ma e m = 1 + 1[0.065(1) + 0.09(1)] = 1.156 Assuming consan hickness of he gears K B = 1 m = N /N = 48/16 = ih N (pinion) = 10 8 cycles and N (gear) = 10 8 /, Fig. 14-14 provides he relaions: 8 0.0178 ( YN ) = 1.558(10 ) = 0.977 8 0.0178 ( Y ) = 1.558(10 / ) = 0.996 N Fig. 14-6: J = 0.7, J 0.8 Table 14-10: K R = 0.85 K T = C f = 1 Eq. (14-): Table 14-8: o o cos 0 sin 0 I = = (1) + 1 C p = 00 psi 0.105 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 15/9
Srengh: rade 1 seel wih B = B = 00 Fig. 14-: Fig. 14-5: (S ) = (S ) = 77.(00) + 1 800 = 8 60 psi (S c ) = (S c ) = (00) + 9 100 = 9 500 psi Fig. 14-15: (Z N ) = 1.4488(10 8 ) 0.0 = 0.948 (Z N ) = 1.4488(10 8 /) 0.0 = 0.97 Sec. 14-1: B / B = 1 C = 1 inion ooh bending d K mk B Eq. (14-15): ) = Ko K s F J 6 (1.156)(1) = 787.8(1)(1.196)(1.088) 0.7 = 1 170 psi Ans. Eq. (14-41): SYN / ( KT KR) ( SF ) = 8 60(0.977) / [(1)(0.85)] = = 1 170.47 Ans. ear ooh bending 6 (1.156)(1) Eq. (14-15): ) = 787.8(1)(1.196)(1.097) = 94 psi Ans. 0.8 8 60(0.996) / [(1)(0.85)] Eq. (14-41): ( SF ) = =.51 Ans. 94 inion ooh wear Eq. (14-16): ) K C C K K K d F I m f c = p o v s 1/ 1.156 1 = 00 787.8(1)(1.196)(1.088).667() 0.105 = 98 760 psi Ans. 1/ Eq. (14-4): ScZ N /( KTK ) R 9 500(0.948) /[(1)(0.85)] ( S ) = = = 1.06 Ans. c 98 760 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 16/9
ear ooh wear 1/ 1/ ( K ) s 1.097 c) = c) = (98 760) = 99 170 psi Ans. ( Ks) 1.088 9 500(0.97)(1) /[(1)(0.85)] ( S ) = = 1.08 Ans. 99 170 The hardness of he pinion and he gear should be increased. 14-0 d =.5(0) = 50 mm, d =.5(6) = 90 mm πdn π(50)(10 )(100) V = = = 0.618 m/s 60 60 60(10) = = 458.4 N π(50)(10 )(100) ih no specific informaion given o indicae oherwise, assume K B = K o = Y θ = Z R = 1 Eq. (14-8): Qv = 6, B = 0.5(1 6) / = 0.855 A = 50 + 56(1 0.855) = 59.77 0.855 59.77 + 00(0.618) Eq. (14-7): = = 1.099 59.77 Table 14-: Y = 0., Y = 0.775 Similar o Eq. (a) of Sec. 14-10 bu for SI unis: C ma K s 1 = = 0.84 k b ( mf Y ) 0.055 0.055 ( Ks) = 0.84.5(18) 0. = 1.00 use 1 0.055 ( Ks) = 0.84.5(18) 0.775 =1.007 use 1 Cmc = Ce = Cpm = 1 18 F = 18 / 5.4 = 0.709 in, Cpf = 0.05 = 0.011 10(50) = + = 4 0.47 0.0167(0.709) 0.765(10 )(0.709 ) 0.59 K = 1 + 1[0.011(1) + 0.59(1)] = 1.7 Fig. 14-14: (Y N ) = 1.558(10 8 ) 0.0178 = 0.977 (Y N ) = 1.558(10 8 /1.8) 0.0178 = 0.987 Fig. 14-6: (Y J ) = 0., (Y J ) = 0.8 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 17/9
Eq. (14-8): Y Z = 0.658 0.0759 ln(1 0.95) = 0.885 o o cos 0 sin 0 1.8 Eq. (14-): Z I = = 0.10 (1) 1.8 + 1 Table 14-8: Z E = 191 Ma Srengh rade 1 seel, B = B = 00 Fig. 14-: (S ) = (S ) = 0.5(00) + 88. = 194.9 Ma Fig. 14-5: (S c ) = (S c ) =.(00) + 00 = 644 Ma Fig. 14-15: (Z N ) = 1.4488(10 8 ) 0.0 = 0.948 ( ) 8 0.0 = 1.4488(10 / 1.8) = 0.961 Z N Fig. 14-1: / = 1 Z = C = 1 B B inion ooh bending 1 K K B Eq. (14-15): ) = KoKs bm Y Eq. (14-41) for SI: J 1 1.7(1) = 458.4(1)(1.099)(1) = 4.08 Ma Ans. 18(.5) 0. S Y 194.9 0.977 N ( SF ) = = = 4.99 Ans. Y θ YZ 4.08 1(0.885) ear ooh bending 1 1.7(1) ) = 458.4(1)(1.099)(1) 7.4 Ma. 18(.5) 0.8 = Ans 194.9 0.987 ( SF ) = 5.81 Ans. 7.4 = 1(0.885) inion ooh wear Eq. (14-16): ) Eq. (14-4) for SI: ear ooh wear K Z = Z K K K d b Z R c E o v s w1 I 1.7 1 = 191 458.4(1)(1.099)(1) = 501.8 Ma Ans. 50(18) 0.10 Sc Z NZ 644 0.948(1) ( S ) = = = 1.7 Ans. Y θ Y 501.8 1(0.885) c Z 1/ 1/ ( K ) s 1 c) = c) = (501.8) = 501.8 Ma Ans. ( Ks) 1 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 18/9
644 0.961(1) ( S ) = = 1.9 Ans. 501.8 1(0.885) 14-1 = n cosψ = 6 cos 0 = 5.196 eeh/in 16 48 d = =.079 in, d = (.079) = 9.8 in 5.196 16 π (.079)(00) V = = 41.8 f/min 1 0.855 000(5) 59.77 + 41.8 = = 68. lbf, = = 1.10 41.8 59.77 From rob. 14-19: Y = 0.96, Y = 0.4056 ( Ks) = 1.088, ( Ks) = 1.097, KB = 1 m =, ( YN ) = 0.977, ( YN ) = 0.996, KR = 0.85 ( S ) = ( S ) = 8 60 psi, C = 1, ( S ) = ( S ) = 9 500 psi c c ( Z ) = 0.948, ( Z ) = 0.97, C = 00 psi N N The pressure angle is: 1 an 0 φ = an =.80 cos0.079 ( rb ) = cos.8 = 1.419 in, ( rb ) = ( rb ) = 4.58 in a = 1 / n = 1 / 6 = 0.167 in Eq. (14-5): p 1/ 1/.079 9.8 Z = + 0.167 1.419 + + 0.167 4.58.079 9.8 + sin.8 = 0.9479 +.185.865 = 0.7466 Condiions O. K. for use p π = p cosφ = cos 0 = 0.490 in 6 N n n Eq. (14-1): m N pn 0.49 = = = 0.697 0.95Z 0.95(0.7466) Shigley s MED, 10 h ediion Chaper 14 Soluions, age 19/9
Eq. (14-): I sin.8 cos.8 = (0.697) = + 1 0.19 Fig. 14-7: J 0.45, J 0.54 Fig. 14-8: Correcions are 0.94 and 0.98. J = 0.45(0.94) = 0.4, J = 0.54(0.98) = 0.59 Cmc = 1, C pf = 0.075 + 0.015() = 0.055 10(.079) C = 1, C = 0.09, C = 1 K pm ma e m = 1 + (1)[0.055(1) + 0.09(1)] = 1.146 inion ooh bending 5.196 1.146(1) ) = 68.(1)(1.1)(1.088) = 6 psi Ans. 0.4 8 60(0.977) / [1(0.85)] ( SF ) = = 5.14 Ans. 6 ear ooh bending 5.196 1.146(1) ) = 68.(1)(1.1)(1.097) = 5097 psi Ans. 0.59 8 60(0.996) / [1(0.85)] ( SF ) = = 6.50 Ans. 5097 inion ooh wear 1.146 1 c) = 00 68.(1)(1.1)(1.088) 67 700 psi..078() = Ans 0.19 9 500(0.948) / [(1)(0.85)] ( S ) = = 1.54 Ans. 67 700 ear ooh wear 1/ 1/ 1.097 c) = (67 700) = 67 980 psi Ans. 1.088 9 500(0.97) /[(1)(0.85)] ( S ) = = 1.57 Ans. 67 980 14- iven: R = 0.99 a 10 8 cycles, B = hrough-hardening rade 1, core and case, boh gears. N = 17T, N = 51T, Table 14-: Y = 0.0, Y = 0.410 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 0/9
Fig. 14-6: J = 0.9, J = 0.96 d = N / = 17 / 6 =.8 in, d = 51 / 6 = 8.500 in. inion bending From Fig. 14-: ( S ) = 77. + 1 800 0.99 10 7 B = 77.() + 1 800 = 0 74 psi Fig. 14-14: Y N = 1.681(10 8 ) 0.0 = 0.98 Eq. (14-15): V = π d n / 1 = π (.8)(110 / 1) = 80.7 f/min KT = KR = 1, SF =, S = 0 74(0.98) all = = 14 61 psi (1)(1) Qv = = = A = 50 + 56(1 0.9148) = 54.77 K v / 5, B 0.5(1 5) 0.9148 0.9148 54.77 + 80.7 = = 1.47 54.77 0.055 0.0 K s = 1.19 = 1.089 use 1 6 K = C = 1 + C ( C C + C C ) C C C C m m f mc p f p m ma e mc pf pm ma C = 1 e = 1 F = 0.075 + 0.015F 10d = 0.075 + 0.015() = 0.0581 10(.8) = 1 = + = 4 0.17 0.0158() 0.09(10 )( ) 0.1586 Km = 1 + 1[0.0581(1) + 0.1586(1)] = 1.17 KB = 1 FJ all = KoKsd KmKB (0.9)(14 61) = = 775 lbf 1(1.47)(1)(6)(1.17)(1) V 775(80.7) = = = 19.5 hp 000 000 inion wear Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9
Fig. 14-15: Z N =.466N 0.056 =.466(10 8 ) 0.056 = 0.879 m = 51 / 17 = Eq. (14-): o o cos 0 sin 0 I = = 1.05, C = 1 + 1 Fig. 14-5: ( S ) 7 = + 9 100 Eq. (14-16): 0.99 c 10 B = () + 9 100 = 10 804 psi 10 804(0.879) c,all = = 64 519 psi (1)(1) FdI C K K K C c,all = p o s m f 64 519 (.8)(0.105) = 00 1(1.47)(1)(1.167)(1) = 00 lbf V 00(80.7) = = = 7.55 hp 000 000 The pinion conrols, herefore raed = 7.55 hp Ans. 14- l =.5/ d, x = Y / d.5 Y.674 = 4lx = 4 = Y d d d.674 de = 0.808 F = 0.808 F Y = 1.5487 d k b 0.107 1.5487 F Y / d F Y = = 0.889 0.0 d 0.055 0.055 F Y 1 F Y Ks = = 1.19 Ans. k b d 14-4 Y = 0.1, Y = 0.4, J = 0.45, J = 0.410, K o = 1.5. The service condiions are adequaely described by K o. Se S F = S = 1. d = / 4 = 5.500 in d = 60 / 4 = 15.000 in d Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
π (5.5)(1145) V = = 1649 f/min 1 inion bending ( S ) = 77. + 1 800 = 77.(50) + 1 800 = 15 psi Eq. (14-17): ( ) 0.99 10 7 B 9 0.0 YN = 1.681[(10 )] = 0.8 15(0.8) all = = 678 psi 1(1)(1) / 0.5(1 6) 0.855 B = = A = 50 + 56(1 0.855) = 59.77 K K C v s mc C ma 0.855 59.77 + 1649 = = 1.54 59.77 = 1, C = 1 m F = 0.075 + 0.015F 10d.5 = 0.075 + 0.015(.5) = 0.06 10(5.5) = + = 4 0.17 0.0158(.5) 0.09(10 )(.5 ) 0.178 C e = 1 K = C = 1 + (1)[0.06(1) + 0.178(1)] = 1.40 K m B m f = 1, K = 1 T 6 78(.5)(0.45) Eq. (14-15): 1 = = 1.5(1.54)(1)(4)(1.40) 151(1649) 1 = = 157.5 hp 000 151 lbf ear bending By similar reasoning, = 861 lbf and = 19.9 hp inion wear m = 60 / =.77 o o cos 0 sin 0.77 I = = 0.1176 1 +.77 ( S ) = (50) + 9 100 = 109 600 psi 0.99 c 10 7 9 0.056 ( Z N ) =.466[(10 )] = 0.77 9 0.056 ( Z N ) =.466[(10 ) /.77] = 0.769 109 600(0.77) c,all) = = 79 679 psi 1(1)(1) Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
c,all = FdI C K K K C p o s m f 79 679.5(5.5)(0.1176) = = 1061 lbf 00 1.5(1.54)(1)(1.4)(1) 1061(1649) = = 5.0 hp 000 ear wear Similarly, 4 = 118 lbf, 4 = 59.0 hp Raing raed = min( 1,,, 4) = min(157.5, 19.9, 5, 59) = 5 hp Ans. Noe differing capaciies. Can hese be equalized? 14-5 From rob. 14-4: = 151 lbf, = 861 lbf, = 1061 lbf, = 118 lbf 000Ko 000(1.5)(40) = = = 1000 lbf V 1649 1 4 inion bending: The facor of safey, based on load and sress, is 1 151 ( S F) = = =.15 Ans. 1000 1000 ear bending based on load and sress 861 ( S F) = = =.86 Ans. 1000 1000 inion wear 1061 based on load: n = = = 1.06 1000 1000 based on sress: ( S ) = 1.06 = 1.0 Ans. ear wear 4 118 based on load: n 4 = = = 1.18 1000 1000 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9
based on sress: ( S ) = 1.18 = 1.09 Ans. Facors of safey are used o assess he relaive hrea of loss of funcion.15,.86, 1.06, 1.18 where he hrea is from pinion wear. By comparison, he AMA safey facors (S F ), (S F ), (S ), (S ) are.15,.86, 1.0, 1.09 or.15,.86, 1.06 1/, 1.18 1/ and he hrea is again from pinion wear. Depending on he magniude of he numbers, using S F and S as defined by AMA, does no necessarily lead o he same conclusion concerning hrea. Therefore be cauious. 14-6 Soluion summary from rob. 14-4: n = 1145 rev/min, K o = 1.5, rade 1 maerials, N = T, N = 60T, m =.77, Y = 0.1,Y = 0.4, J = 0.45, J = 0.410, d = 4T /in, F =.5 in, Qv = 6, (N c ) = (10 9 ), R = 0.99, K m = 1.40, K T = 1, K B = 1, d = 5.500 in, d = 15.000 in, V = 1649 f/min, = 1.54, (K s ) = (K s ) = 1, (Y N ) = 0.8, (Y N ) = 0.859, K R = 1 inion B : 50 core, 90 case ear B : 50 core, 90 case Bending = 6 78 psi ( S ) all) = 15 psi all) = 7 546 psi ( ) S = 15 psi = 151 lbf, = 157.5 hp = 861 lbf, = 19.9 hp 1 1 ear φ = I = Z N = o 0, 0.1176, ( ) 0.77 ( Z ) = 0.769, C = 00 psi N ( S ) = S = (90) + 9 100 = 154 680 psi c c 154 680(0.77) c,all) = = 11 450 psi 1(1)(1) 154 680(0.769) c,all) = = 118 950 psi 1(1)(1) 11 450 11(1649) = (1061) = 11 lbf, = = 105.6 hp 79 679 000 118 950 54(1649) = (118) = 54 lbf, = = 117.6 hp 109 600(0.769) 000 4 4 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 5/9
Raed power raed = min(157.5, 19.9, 105.6, 117.6) = 105.6 hp Ans. rob. 14-4: raed = min(157.5, 19.9, 5.0, 59.0) = 5 hp The raed power approximaely doubled. 14-7 The gear and he pinion are 910 grade 1, carburized and case-hardened o obain Brinell 85 core and Brinell 580 600 case. Table 14-: ( S ) 7 = 55 000 psi 0.99 10 Modificaion of S by (Y N ) = 0.8 produces ) = 45 657 psi, all Similarly for (Y N ) = 0.859 ) = 47 161 psi, and all = 4569 lbf, = 8 hp = 5668 lbf, = 8 hp 1 1 From Table 14-8, C = 00 psi. Also, from Table 14-6: p ( S ) = 180 000 psi 0.99 c 10 7 Modificaion of S c by Y N produces and c,all) = 10 55 psi ) = 18 069 psi c,all = 489 lbf, = 14. hp = 767 lbf, = 18. hp 4 4 Raing raed = min(8, 8, 14, 18) = 14 hp Ans. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 6/9
14-8 rade, 910 carburized and case-hardened o 85 core and 580 case in rob. 14-7. Summary: Table 14-: ( S ) 7 = 65 000 psi and i follows ha 0.99 10 all) = 5 959 psi ) = 55 76 psi all = 5400 lbf, = 70 hp = 6699 lbf, = 5 hp 1 1 From Table 14-8, C = 00 psi. Also, from Table 14-6: p Consequenly, S c = 5 000 psi ) = 181 85 psi c,all ) = 191 76 psi c,all = 4801 lbf, = 40 hp = 57 lbf, = 67 hp 4 4 Raing raed = min(70, 5, 40, 67) = 40 hp. Ans. 14-9 iven: n = 1145 rev/min, K o = 1.5, N = T, N = 60T, m =.77, d =.75 in, d = 7.5 in, Y = 0.1,Y = 0.4, J = 0.5, J = 0.405, = 8T /in, F = 1.65 in, B = 50, case and core, boh gears. C m = 1, F/d = 0.0591, C f = 0.0419, C pm = 1, C ma = 0.15, C e = 1, K m = 1.194, K T = 1, K B = 1, K s = 1,V = 84 f/min, (Y N ) = 0.818, (Y N ) = 0.859, K R = 1, I = 0.117 58 ( S ) = 15 psi 0.99 10 7 all) = 6 668 psi ) = 7 546 psi all and i follows ha = 879. lbf, = 1.97 hp = 1098 lbf, = 7.4 hp 1 1 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 7/9
For wear = 04 lbf, = 7.59 hp = 40 lbf, = 8.50 hp 4 4 Raing raed = min(1.97, 7.4, 7.59, 8.50) = 7.59 hp In rob. 14-4, raed = 5 hp. Thus, The ransmied load raing is In rob. 14-4 7.59 1 1 = 0.14 =, no Ans. 5.0 6.98 8 raed = min(879., 1098, 04, 40) = 04 lbf raed = 1061 lbf Thus 04 1 1 = 0.865 =, no Ans. 1061.49 4 14-0 S = S = 1, d = 4, J = 0.45, J = 0.410, K o = 1.5 Bending Table 14-4: ( S ) 7 = 1 000 psi 0.99 10 1 000(1) all) = all) = = 1 000 psi 1(1)(1) allfj 1 000(.5)(0.45) 1 = = = 15 lbf Ko Ksd KmKB 1.5(1.54)(1)(4)(1.4)(1) 15(1649) 1 = = 76.6 hp 000 = 1 J / J = 15(0.410) / 0.45 = 18 lbf = J / J = 76.6(0.410) / 0.45 = 91.0 hp 1 ear Table 14-8: C p = 1960 psi Table 14-7: ( S ) 7 = 75 000 psi = ) = ) 0.99 c 10 c,all c,all Shigley s MED, 10 h ediion Chaper 14 Soluions, age 8/9
c,all p = Cp KoKsKmC f 4 4 ) Fd I 75 000.5(5.5)(0.1176) = = 195 lbf 1960 1.5(1.54)(1)(1.4)(1) = = 195 lbf = 195(1649) = = 64.7 hp 000 Raing raed = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp Ans. Noice ha he balance beween bending and wear power is improved due o CI s more favorable S c /S raio. Also noe ha he life is 10 7 pinion revoluions which is (1/00) of (10 9 ). Longer life goals require power de-raing. 14-1 From Table A-4a, E av = 11.8(10 6 ) Mpsi For φ = 14.5 and B = 156 For φ = 0 S C 1.4(81) S = = 51 69 psi C 6 sin14.5 / [11.8(10 )] 1.4(11) = = 5 008 psi 6 sin 0 / [11.8(10 )] SC = 0.(156) = 49.9 kpsi The firs wo calculaions were approximaely 4 percen higher. 14- rograms will vary. 14- ( Y ) = 0.977, ( Y ) = 0.996 N N ( S ) = ( S ) = 8.(50) + 1 150 = 75 psi 75(0.977) all) = = 7 615 psi 1(0.85) 7 615(1.5)(0.4) 1 = = 1558 lbf 1(1.404)(1.04)(8.66)(1.08)(1) 1558(95) 1 = = 4.7 hp 000 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 9/9
75(0.996) all) = = 8 46 psi 1(0.85) 8 46(1.5)(0.546) = = 1(1.404)(1.04)(8.66)(1.08)(1) 007(95) = = 56. hp 000 ( Z ) = 0.948, ( Z ) = 0.97 N N 007 lbf Table 14-6: ( S ) 7 = 150 000 psi 0.99 c 10 0.948(1) c,allow) = 150 000 = 167 94 psi 1(0.85) 167 94 1.96(1.5)(0.195) = = 074 lbf 00 1(1.404)(1.04) 074(95) = = 58.1 hp 000 0.97 c,allow) = (167 94) = 171 706 psi 0.948 171 706 1.96(1.5)(0.195) 4 = = 167 lbf 00 1(1.404)(1.05) 167(95) 4 = = 60.7 hp 000 raed = min(4.7, 56., 58.1, 60.7) = 4.7 hp Ans. inion bending is conrolling. 14-4 8 0.0 ( Y N ) = 1.681(10 ) = 0.98 8 0.0 ( Y N ) = 1.681(10 /.059) = 0.96 Table 14-: S = 55 000 psi 55 000(0.98) all) = = 60 047 psi 1(0.85) 60 047(1.5)(0.4) 1 = = 1(1.404)(1.04)(8.66)(1.08)(1) 487(95) 1 = = 69.7 hp 000 0.96 all) = (60 047) = 6 47 psi 0.98 487 lbf Shigley s MED, 10 h ediion Chaper 14 Soluions, age 0/9
6 47 0.546 = (487) = 58 lbf 60 047 0.4 58 = (69.7) = 91. hp 487 Table 14-6: S c = 180 000 psi 8 0.056 ( ) =.466(10 ) = 0.8790 Z N 8 0.056 ( Z N ) =.466(10 /.059) = 0.958 180 000(0.8790) c,all) = = 186 141 psi 1(0.85) 186 141 1.96(1.5)(0.195) = = 568 lbf 00 1(1.404)(1.04) 568(95) = = 7.0 hp 000 0.958 c,all) = (186 141) = 198 169 psi 0.8790 198 169 1.04 4 = (568) = 886 lbf 186 141 1.05 886(95) 4 = = 80.9 hp 000 raed = min(69.7, 91., 7, 80.9) = 69.7 hp Ans. inion bending conrolling 14-5 (Y N ) = 0.98, (Y N ) = 0.96 (See rob. 14-4) Table 14-: S = 65 000 psi 65 000(0.98) all) = = 70 965 psi 1(0.85) 70 965(1.5)(0.4) 1 = = 99 lbf 1(1.404)(1.04)(8.66)(1.08) 99(95) 1 = = 8.4 hp 000 65 000(0.96) all) = = 7 565 psi 1(0.85) 7 565 0.546 = (99) = 850 lbf 70 965 0.4 850 = (8.4) = 108 hp 99 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9
Table 14-6: S c = 5 000 psi ( Z ) = 0.8790, ( Z ) = 0.958 N N 5 000(0.879) c,all) = = 676 psi 1(0.85) 676 1.96(1.5)(0.195) = = 401 lbf 00 1(1.404)(1.04) 401(95) = = 11.5 hp 000 0.958 c,all) = ( 676) = 47 711 psi 0.8790 47 711 1.04 4 = (401) = 4509 lbf 676 1.05 4509(95) 4 = = 16 hp 000 raed = min(8.4, 108, 11.5, 16) = 8.4 hp Ans. The bending of he pinion is he conrolling facor. 14-6 = eeh/in, d = 8 in, N = d = 8 () = 16 eeh π π F = 4 p = 4 = 4 = π o o M = 0 = 10(00) cos 0 4F cos0 x F B = 750 lbf o o = F B cos 0 = 750cos 0 = 705 lbf n = 400 / = 100 rev/min π dn π (8)(100) V = = = 51 f/min 1 1 e will obain all of he needed facors, roughly in he order presened in he exbook. B Fig. 14-: S = 10(00) + 16 400 = 47 000 psi Fig. 14-5: S c = 49(00) + 4 00 = 19 000 psi Fig. 14-6: J = 0.7 Eq. (14-): o o cos 0 sin 0 I = = 0.107 (1) + 1 Table 14-8: C p = 00 psi Assume a ypical qualiy number of 6. / / Eq. (14-8): B = 0.5(1 Q ) = 0.5(1 6) = 0.855 v Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
Eq. (14-7): A = 50 + 56(1 B) = 50 + 56(1 0.855) = 59.77 K v B 0.855 A + V 59.77 + 51 = = = 1.65 A 59.77 To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = 0.96. From Eq. (a), Sec. 14-10, 0.055 0.055 F Y π 0.96 Ks = 1.19 = 1.19 = 1. The load disribuion facor is applicable for sraddle-mouned gears, which is no he case here since he gear is mouned ouboard of he bearings. Lacking anyhing beer, we will use he load disribuion facor as a rough esimae. Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): π C p f = 0.075 + 0.015( π ) = 0.1196 10(8) Eq. (14-): C pm = 1.1 Fig. 14-11: C ma = 0. (commercial enclosed gear uni) Eq. (14-5): C e = 1 Eq. (14-0): K m = 1 + 1[0.1196(1.1) + 0.(1)] = 1.6 For he sress-cycle facors, we need he desired number of load cycles. N = 15 000 h (100 rev/min)(60 min/h) = 1.1 (10 9 ) rev Fig. 14-14: Y N = 0.9 Fig. 14-15: Z N = 0.8 K = 0.658 0.0759 ln 1 R = 0.658 0.0759ln 1 0.95 = 0.885 Eq. 14-8: ( ) ( ) R ih no specific informaion given o indicae oherwise, assume K o = K B = K T = C f = 1 Tooh bending d KmK B Eq. (14-15): = KoK s F J (1.6)(1) = 705(1)(1.65)(1.) = 94 psi π 0.7 Eq. (14-41): S F SYN / ( KT KR) = 47 000(0.9) / [(1)(0.885)] = = 0.8 Ans. 94 Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9
Tooh wear Eq. (14-16): K C m f c = C p Ko Ks d F I 00 705(1)(1.65)(1.) 1.6 1 = 8( π ) 0.107 = 4 750 psi Since gear B is a pinion, C is no used in Eq. (14-4) (see p. 75), where S = S Z / ( K K ) c N T R c 19 000(0.8) / [(1)(0.885)] = =.9 Ans 4 750 14-7 m = 18.75 mm/ooh, d = 00 mm N = d/m = 00 / 18.75 = 16 eeh F = b = 4 p = 4 π m = 4π 18.75 = 6 mm M x 1/ ( ) ( ) o = 0 = 00(11) cos 0 150F cos5 F B =.81 kn o o = F B cos 5 =.81cos 5 = 0.67 kn n = 1800 / = 900 rev/min π dn π (0.00)(900) V = = = 14.14 m/s 60 60 e will obain all of he needed facors, roughly in he order presened in he exbook. B o 1/ Fig. 14-: S = 0.70(00) + 11 = 4 Ma Fig. 14-5: S c =.41(00) + 7 = 960 Ma Fig. 14-6: J = Y J = 0.7 Eq. (14-): o o cos 0 sin 0 5 I = Z I = = (1) 5 + 1 Table 14-8: Z E = 191 Ma 0.14 Assume a ypical qualiy number of 6. / / Eq. (14-8): B = 0.5(1 Qv ) = 0.5(1 6) = 0.855 A = 50 + 56(1 B) = 50 + 56(1 0.855) = 59.77 Eq. (14-7): K v B 0.855 A + 00V 59.77 + 00(14.14) = = = 1.69 A 59.77 To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = 0.96. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9
Similar o Eq. (a) of Sec. 14-10 bu for SI unis: K s 1 = = 0.84 k b ( mf Y ) 0.055 0.055 Ks = 0.84 18.75(6) 0.96 = 1.8 Conver he diameer and facewidh o inches for use in he load-disribuion facor equaions. d = 00/5.4 = 11.81 in, F = 6/5.4 = 9.9 in Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): 9.9 C pf = 0.075 + 0.015(9.9) = 0.157 10(11.81) Eq. (14-): C pm = 1.1 Fig. 14-11: C ma = 0.7 (commercial enclosed gear uni) Eq. (14-5): C e = 1 Eq. (14-0): K = K = 1 + 1[0.157(1.1) + 0.7(1)] = 1.44 m For he sress-cycle facors, we need he desired number of load cycles. N = 1 000 h (900 rev/min)(60 min/h) = 6.48 (10 8 ) rev Fig. 14-14: Y N = 0.9 Fig. 14-15: Z N = 0.85 K = 0.658 0.0759 ln 1 R = 0.658 0.0759ln 1 0.98 = 0.955 Eq. 14-8: ( ) ( ) R ih no specific informaion given o indicae oherwise, assume K o = K B = K T = Z R = 1. Tooh bending Eq. (14-15): 1 = KoKs bm B K K Y J 1 (1.44)(1) = 0 670(1)(1.69)(1.8) = 5.9 Ma 6(18.75) 0.7 Eq. (14-41): S F SYN / ( KT KR) = 4(0.9) / [(1)(0.955)] = = 5.9 5.66 Ans. Tooh wear Eq. (14-16): K Z R c = ZE KoKs d w1 b Z I 1/ Shigley s MED, 10 h ediion Chaper 14 Soluions, age 5/9
1.44 1 = 191 0 670(1)(1.69)(1.8) 00(6) 0.14 = 498 Ma Since gear B is a pinion, C is no used in Eq. (14-4) (see p. 757), where ScZ N / ( KT KR) S = c 960(0.85) / [(1)(0.955)] = = 1.7 Ans 498 14-8 From he soluion o rob. 1-40, n = 191 rev/min, = 1600 N, d = 15 mm, N = 15 eeh, m = 8. mm/ooh. F = b = 4 p = 4 π m = 4π 8. = 105 mm ( ) ( ) π dn π (0.15)(191) V = = = 1.5 m/s 60 60 e will obain all of he needed facors, roughly in he order presened in he exbook. 1/ Table 14-: S = 65 kpsi = 448 Ma Table 14-6: S c = 5 kpsi = 1550 Ma Fig. 14-6: J = Y J = 0.5 Eq. (14-): o o cos 0 sin 0 I = Z I = = (1) + 1 Table 14-8: Z E = 191 Ma Assume a ypical qualiy number of 6. 0.107 Eq. (14-8): Eq. (14-7): / / B = 0.5(1 Qv ) = 0.5(1 6) = 0.855 A = 50 + 56(1 B) = 50 + 56(1 0.855) = 59.77 K v B 0.855 A + 00V 59.77 + 00(1.5) = = = 1.1 A 59.77 To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = 0.90. Similar o Eq. (a) of Sec. 14-10 bu for SI unis: K s 1 = = 0.84 k b ( mf Y ) 0.055 Ks = 0.84 8.(105) 0.90 = 1.17 Conver he diameer and facewidh o inches for use in he load-disribuion facor 0.055 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 6/9
equaions. d = 15/5.4 = 4.9 in, F = 105/5.4 = 4.1 in Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): 4.1 C pf = 0.075 + 0.015(4.1) = 0.0981 10(4.9) Eq. (14-): C pm = 1 Fig. 14-11: C ma = 0. (open gearing) Eq. (14-5): C e = 1 Eq. (14-0): K = K = 1 + 1[0.0981(1) + 0.(1)] = 1.4 m For he sress-cycle facors, we need he desired number of load cycles. N = 1 000 h (191 rev/min)(60 min/h) = 1.4 (10 8 ) rev Fig. 14-14: Y N = 0.95 Fig. 14-15: Z N = 0.88 K = 0.658 0.0759 ln 1 R = 0.658 0.0759ln 1 0.95 = 0.885 Eq. 14-8: ( ) ( ) R ih no specific informaion given o indicae oherwise, assume K o = K B = K T = Z R = 1. Tooh bending Eq. (14-15): 1 = KoKs bm B K K Y 1 (1.4)(1) = 1600(1)(1.1)(1.17) = 14.7 Ma 105(8.) 0.5 Since gear is a pinion, C is no used in Eq. (14-4) (see p. 75), where J Tooh wear Eq. (14-16): S F SYN / ( KT KR) = 448(0.95) / [(1)(0.885)] = =.7 Ans. 14.7 K Z R c = ZE KoKs d w1 b Z I 1.4 1 = 191 1600(1)(1.1)(1.17) 15(105) 0.107 = 89 Ma 1/ 1/ ScZN / ( KTKR) Eq. (14-4): S = c 1550(0.88) / [(1)(0.885)] = = 5. Ans 89 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 7/9
14-9 From he soluion o rob. 1-41, n = (70) = 140 rev/min, = 180 lbf, d = 5 in N = 15 eeh, = eeh/in. π π F = 4 p = 4 = 4 = 4. in π dn π (5)(140) V = = = 18. f/min 1 1 e will obain all of he needed facors, roughly in he order presened in he exbook. Table 14-: S = 65 kpsi Table 14-6: S c = 5 kpsi Fig. 14-6: J = 0.5 Eq. (14-): o o cos 0 sin 0 I = = (1) + 1 Table 14-8: C p = 00 psi Assume a ypical qualiy number of 6. 0.107 Eq. (14-8): Eq. (14-7): / / B = 0.5(1 Qv ) = 0.5(1 6) = 0.855 A = 50 + 56(1 B) = 50 + 56(1 0.855) = 59.77 K v B 0.855 A + V 59.77 + 18. = = = 1.18 A 59.77 To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = 0.90. From Eq. (a), Sec. 14-10, 0.055 0.055 F Y 4. 0.90 Ks = 1.19 = 1.19 = 1.17 Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): 4. C pf = 0.075 + 0.015(4.) = 0.099 10(5) Eq. (14-): C pm = 1 Fig. 14-11: C ma = 0. (Open gearing) Eq. (14-5): C e = 1 Eq. (14-0): K m = 1 + 1[0.099(1) + 0.(1)] = 1.4 For he sress-cycle facors, we need he desired number of load cycles. N = 14 000 h (140 rev/min)(60 min/h) = 1. (10 8 ) rev Fig. 14-14: Y N = 0.95 Fig. 14-15: Z N = 0.88 K = 0.658 0.0759 ln 1 R = 0.658 0.0759ln 1 0.98 = 0.955 Eq. 14-8: ( ) ( ) R ih no specific informaion given o indicae oherwise, assume K o = K B = K T = C f = 1. Shigley s MED, 10 h ediion Chaper 14 Soluions, age 8/9
Tooh bending d KmK B Eq. (14-15): = KoK s F J (1.4)(1) = 180(1)(1.18)(1.17) = 1010 psi 4. 0.5 Eq. (14-41): S F SYN / ( KT KR) = 65 000(0.95) / [(1)(0.955)] = = 1010 64.0 Ans. Tooh wear Eq. (14-16): K C m f c = C p Ko Ks d F I 1.4 1 = 00 180(1)(1.18)(1.17) 5(4.) 0.107 = 8 800 psi 1/ 1/ Since gear B is a pinion, C is no used in Eq. (14-4) (see p. 75), where S ScZN / ( KTKR) = c 5 000(0.88) / [(1)(0.955)] = = 7.8 Ans 8 800 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 9/9