Chapter in. π(3.667)(1200) W = = = lbf W P 1.96(429.7)(6) FY 2(0.331) 2 V 282.7

Transcription

1 Chaper d N = = = in Table 14-: Y = 0.1 Eq. (1-4): πdn π(.667)(100) V = = = 115 f/min 1 1 Eq. (14-4b): = = 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): = = 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 Eq. (14-6b): = = Eq. (1-6): (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

2 14-4 d = mn = 8(16) = 18 mm Table 14-: Y = 0.96 V π dn π (18)(10 )(150) = = = m/s Eq. (14-6b): = = Eq. (1-6): (6) kn 5968 N π dn π (18)(150) Eq. (14-8): 1.165(5968) = = =.6 Ma FmY 90(8)(0.96) Ans d = mn = 1(16) = 16 mm Table 14-: Y = 0.96 V π dn π (16)(10 )(400) = = = 0.5 m/s Eq. (14-6b): = = Eq. (1-6): (0.15) kn 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 d = mn = (0) = 40 mm Table 14-: Y = 0. V π dn π (40)(10 )(00) = = = m/s Eq. (14-6b): = = Eq. (1-6): (0.5) 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

3 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): = = 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 d = N 16 = = 4.0 in 4 Table 14-: Y = 0.96 Eq. (1-4): π dn π (4.0)(400) V = = = f/min 1 1 Eq. (14-4b): = = Eq. (1-5) : 0 = 000 = 000 = lbf V Eq. (14-7): 1.49(1575.6)(4) F = = =.9 in Y 1(10 )(0.96) Use F =.5 in Ans Try = 8 which gives d = 18/8 =.5 in and Y = Eq. (1-4): V π dn π (.5)(600) = = = f/min Eq. (14-4b): = = 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

4 d V F Oher consideraions may dicae he selecion. ood candidaes are = 8 (F = 7/8 in) and =10 (F = 1.5 in). Ans Try m = mm which gives d = (18) = 6 mm and Y = π dn π (6)(10 )(900) V = = = m/s Eq. (14-6b): = = (1.5) Eq. (1-6): = 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 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9

5 Oher design consideraions may dicae he size selecion. For he presen design, m = mm (F = 5 mm) is a good selecion. Ans Eq. (14-4b): Eq. (1-6): N 0 N 50 d = = =.5 in, d = = = 6.5 in 8 8 π (.5)(100) V = = f/min = = = 000 = 000 = 504. lbf V 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 sin 0 Eq. (14-1): r1 = = in, r = = in Eq. (14-14): C C K 1 1 v = p + F cosφ r r (504.) 1 1 = cos = 9.5(10 ) psi = 9.5 kpsi Ans. 1/ 1/ 14-1 d V = = 1. in, d = = 4 in 1 1 π(1.)(700) = = 44. f/min 1 Eq. (14-4b): Eq. (1-6): Table 14-8: = = = 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 = = in Eq. (14-14): Shigley s MED, 10 h ediion Chaper 14 Soluions, age 5/9

6 C 1/ 1.04(0.6) (10 ) = + = F cos (0.6) 1 1 F = in 100(10 ) + = cos Use F = 0.75 in Ans d = 5(4) = 10 mm, d = 5(48) = 40 mm p (10)(10 )(50) V π = = 0.14 m/s Eq. (14-6a): = = (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 = = mm Eq. (14-14): ( ) 1.10(.18) = 16 + o 60 cos =.94 k Ans. 1/ Eq. (14-6a): d = 4(0) = 80 mm, d = 4() = 18 mm π (80)(10 )(1000) V = = m/s = =.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

7 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 The pinion conrols he design. Bending Y = 0.0, Y = d = = in, d = =.500 in 1 1 πdn π(1.417)(55) V = = = f/min Eq. (14-4b): = = Eq. (6-8), p. 90: S e = 0.5(76) = 8.0 kpsi Eq. (6-19), p. 95: k a =.70(76) 0.65 = l = = = in d 1 Y (0.0) Eq. (14-): x = = = in (1) Eq. (b), p. 79: = 4lx = 4(0.1875)(0.079) = in Eq. (6-5), p. 97: d = hb = (0.1686) = 0.10 in e Eq. (6-0), p. 96: k b = = k c = k d = k e = 1 Accoun for one-way bending wih k f = (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-) r = f 0.05 in = 1 = From Fig. A-15-6, r r 0.05 = f = = d Approximae D/d = wih D/d = ; from Fig. A-15-6, K = 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

8 Eq. (6-): K f = (1.68 1) = 1.4 Se 5.84 all = = = 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 ear ν 1 = ν = 0.9, E 1 = E = 0(10 6 ) psi Eq. (14-1): C p 1/ 1 = = π 6 0( 10 ) 85 psi d Eq. (14-1): r1 = sin φ = sin 0 = 0.4 in d.500 r = sinφ = sin 0 = 0.48 in = + = in r r Eq. (6-68), p. 7: ( S ) 8 = kpsi = [0.4(149) 10](10 ) = psi C 10 From he discussion and equaion developed on he boom of p. 7, ( SC ) C,all = = = 067 psi n cos 0 Eq. (14-14): = =.6 lbf (6.469) V.6(194.8) = = = 0.1 hp Ans 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

9 14-16 See rob soluion for equaion numbers. inion conrols: Y = 0., Y = Bending d = 0/ = in, d = 100/ =. in V = π dn / 1 = π (6.667)(870) / 1 = 1519 f/min = ( ) / 100 =.66 S e = 0.5(11) = 56.5 kpsi 0.65 ka =.70(11) = l =.5 / d =.5 / = 0.75 in x = (0.) / [()] = in = 4(0.75)(0.161) = in d e = (0.695) = in 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 / = in r rf = = = d 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 = = π 6 0( 10 ) 85 psi Eq. (14-1): r 1 = (6.667/) sin 0 = in r = (./) sin 0 = in Eq. (6-68), p. 7: S C = [0.4(6) 10](10 ) = psi C,all = SC / nd = / 1.5 = psi Shigley s MED, 10 h ediion Chaper 14 Soluions, age 9/9

10 C,all F cosφ 1 = C p 1 / r1 + 1 / r cos 0 1 = / / = 110 lbf V 110(1519) = = = 5.0 hp Ans 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 See rob 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 = 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) = = S e = 0.5(900) = 450 Ma 0.65 ka = 4.51(900) = l =.5m =.5(6) = 1.5 mm x = Ym / = (0.96)6 / =.664 mm = 4lx = 4(1.5)(.664) = 1.0 mm d e = (1.0) = 4. mm kb = = 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

11 Se 491. all = = = K n f d ( ) 9. Ma FYm all 75(0.96)(6)(9.) Eq. (14-8): = = = N K Eq. (1-6): ear: inion and gear v π dn 16.9 π (96)(1145) = = = 94. k Ans 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 π 07( 10 ) Eq. (6-68), p. 7: S C = 6.89[0.4(60) 10] = Ma 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 cos 0 1 = = / / 0.78 π dn.469 π (96)(1145) = = = 0.0 k N Thus, wear conrols he gearse power raing; = 0.0 k. Ans N = 17 eeh, N = 51 eeh N 17 d = = =.8 in 6 51 d = = in 6 V = π d n / 1 = π (.8)(110) / 1 = 80.7 f/min Eq. (14-4b): = ( )/100 = 1.69 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 11/9

12 Sy all = = = n d psi Table 14-: Y = 0.0, Y = 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 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 (116) = = Table 1-1, p. 688: l = + = = = 6 d d d 0.75 in Eq. (14-): x Y (0.0) = = = in (6) Eq. (b), p. 79: = 4lx = 4(0.75)(0.0758) = 0.7 in Eq. (6-5), p. 97: d = F = (0.7) = 0.66 in e Eq. (6-0), p. 96: k b = = k c = k d = k e = 1 Accoun for one-way bending wih k f = (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-) r = f in = 6 = r r 0.05 Fig. A-15-6: = f = = d 0.8 Esimae D/d = by seing D/d =, K = Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9

13 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 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 in r1 = = = d o r = sinφ = sin 0 = in = + =.750 in r1 r Eq. (6-68): ( S ) 8 = kpsi C 10 In erms of gear noaion B C = [0.4() 10]10 = 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 C,all = = = psi Solve Eq. (14-14) for : o cos 0 = 65 lbf (.750) = V 65(80.7) all = = = 6.67 hp Ans 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

14 (c) ear faigue due o bending and wear Bending Eq. (14-): Y (0.410) x = = = in (6) Eq. (b), p. 79: = 4lx = 4(0.75)(0.106) = 0.9 in ± Eq. (6-5): d = F = (0.9) = in e Eq. (6-0): k b = = k c = k d = k e = 1 k f = (See Ex. 14-.) Eq. (6-18): S e = 0.766(0.911)(1)(1)(1)(1.66)(58) = 67. kpsi r d r f = = = Approximae D/d = by seing D/d = for Fig. A-15-6; K = 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 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

15 14-19 d = 16/6 =.667 in, d = 48/6 = 8 in π (.667)(00) V = = 09.4 f/min 1 000(5) = = lbf 09.4 Assuming uniform loading, K o = 1. / Eq. (14-8): Qv = 6, B = 0.5(1 6) = A = ( ) = Eq. (14-7): = = Table 14-: Y = 0.96, Y = From Eq. (a), Sec wih F = in ( Ks) = 1.19 = ( Ks) = 1.19 = From Eq. (14-0) wih C mc = 1 C p f = () = (.667) C = 1, C = 0.09 (Fig ), C = 1 K p m ma e m = 1 + 1[0.065(1) (1)] = 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 provides he relaions: ( YN ) = 1.558(10 ) = ( Y ) = 1.558(10 / ) = 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 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 15/9

16 Srengh: rade 1 seel wih B = B = 00 Fig. 14-: Fig. 14-5: (S ) = (S ) = 77.(00) = 8 60 psi (S c ) = (S c ) = (00) = psi Fig : (Z N ) = (10 8 ) 0.0 = (Z N ) = (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 = psi Ans. Eq. (14-41): SYN / ( KT KR) ( SF ) = 8 60(0.977) / [(1)(0.85)] = = Ans. ear ooh bending 6 (1.156)(1) Eq. (14-15): ) = 787.8(1)(1.196)(1.097) = 94 psi Ans (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)(1.196)(1.088).667() = psi Ans. 1/ Eq. (14-4): ScZ N /( KTK ) R 9 500(0.948) /[(1)(0.85)] ( S ) = = = 1.06 Ans. c Shigley s MED, 10 h ediion Chaper 14 Soluions, age 16/9

17 ear ooh wear 1/ 1/ ( K ) s c) = c) = (98 760) = psi Ans. ( Ks) (0.97)(1) /[(1)(0.85)] ( S ) = = 1.08 Ans The hardness of he pinion and he gear should be increased d =.5(0) = 50 mm, d =.5(6) = 90 mm πdn π(50)(10 )(100) V = = = m/s (10) = = 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) / = A = ( ) = (0.618) Eq. (14-7): = = Table 14-: Y = 0., Y = Similar o Eq. (a) of Sec bu for SI unis: C ma K s 1 = = 0.84 k b ( mf Y ) ( Ks) = (18) 0. = 1.00 use ( Ks) = (18) =1.007 use 1 Cmc = Ce = Cpm = 1 18 F = 18 / 5.4 = in, Cpf = 0.05 = (50) = + = (0.709) 0.765(10 )(0.709 ) 0.59 K = 1 + 1[0.011(1) (1)] = 1.7 Fig : (Y N ) = 1.558(10 8 ) = (Y N ) = 1.558(10 8 /1.8) = Fig. 14-6: (Y J ) = 0., (Y J ) = 0.8 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 17/9

18 Eq. (14-8): Y Z = ln(1 0.95) = o o cos 0 sin Eq. (14-): Z I = = 0.10 (1) Table 14-8: Z E = 191 Ma Srengh rade 1 seel, B = B = 00 Fig. 14-: (S ) = (S ) = 0.5(00) = Ma Fig. 14-5: (S c ) = (S c ) =.(00) + 00 = 644 Ma Fig : (Z N ) = (10 8 ) 0.0 = ( ) = (10 / 1.8) = 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 N ( SF ) = = = 4.99 Ans. Y θ YZ (0.885) ear ooh bending 1 1.7(1) ) = 458.4(1)(1.099)(1) 7.4 Ma. 18(.5) 0.8 = Ans ( 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)(1.099)(1) = Ma Ans. 50(18) 0.10 Sc Z NZ (1) ( S ) = = = 1.7 Ans. Y θ Y (0.885) c Z 1/ 1/ ( K ) s 1 c) = c) = (501.8) = Ma Ans. ( Ks) 1 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 18/9

19 (1) ( S ) = = 1.9 Ans (0.885) 14-1 = n cosψ = 6 cos 0 = eeh/in d = =.079 in, d = (.079) = 9.8 in π (.079)(00) V = = 41.8 f/min (5) = = 68. lbf, = = From rob : Y = 0.96, Y = ( 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 ) = 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 = in, ( rb ) = ( rb ) = 4.58 in a = 1 / n = 1 / 6 = in Eq. (14-5): p 1/ 1/ Z = sin.8 = = Condiions O. K. for use p π = p cosφ = cos 0 = in 6 N n n Eq. (14-1): m N pn 0.49 = = = Z 0.95(0.7466) Shigley s MED, 10 h ediion Chaper 14 Soluions, age 19/9

20 Eq. (14-): I sin.8 cos.8 = (0.697) = Fig. 14-7: J 0.45, J 0.54 Fig. 14-8: Correcions are 0.94 and J = 0.45(0.94) = 0.4, J = 0.54(0.98) = 0.59 Cmc = 1, C pf = () = (.079) C = 1, C = 0.09, C = 1 K pm ma e m = 1 + (1)[0.055(1) (1)] = inion ooh bending (1) ) = 68.(1)(1.1)(1.088) = 6 psi Ans (0.977) / [1(0.85)] ( SF ) = = 5.14 Ans. 6 ear ooh bending (1) ) = 68.(1)(1.1)(1.097) = 5097 psi Ans (0.996) / [1(0.85)] ( SF ) = = 6.50 Ans inion ooh wear c) = (1)(1.1)(1.088) psi..078() = Ans (0.948) / [(1)(0.85)] ( S ) = = 1.54 Ans ear ooh wear 1/ 1/ c) = (67 700) = psi Ans (0.97) /[(1)(0.85)] ( S ) = = 1.57 Ans 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 = Shigley s MED, 10 h ediion Chaper 14 Soluions, age 0/9

21 Fig. 14-6: J = 0.9, J = 0.96 d = N / = 17 / 6 =.8 in, d = 51 / 6 = in. inion bending From Fig. 14-: ( S ) = B = 77.() = 0 74 psi Fig : 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 = = psi (1)(1) Qv = = = A = ( ) = K v / 5, B 0.5(1 5) = = K s = 1.19 = 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 = F 10d = () = (.8) = 1 = + = () 0.09(10 )( ) Km = 1 + 1[0.0581(1) (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 inion wear Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9

22 Fig : Z N =.466N =.466(10 8 ) = m = 51 / 17 = Eq. (14-): o o cos 0 sin 0 I = = 1.05, C = Fig. 14-5: ( S ) 7 = Eq. (14-16): 0.99 c 10 B = () = psi (0.879) c,all = = psi (1)(1) FdI C K K K C c,all = p o s m f (.8)(0.105) = 00 1(1.47)(1)(1.167)(1) = 00 lbf V 00(80.7) = = = 7.55 hp 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 = F = F Y = d k b F Y / d F Y = = d 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 = in d = 60 / 4 = in d Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9

23 π (5.5)(1145) V = = 1649 f/min 1 inion bending ( S ) = = 77.(50) = 15 psi Eq. (14-17): ( ) B YN = 1.681[(10 )] = (0.8) all = = 678 psi 1(1)(1) / 0.5(1 6) B = = A = ( ) = K K C v s mc C ma = = = 1, C = 1 m F = F 10d.5 = (.5) = (5.5) = + = (.5) 0.09(10 )(.5 ) C e = 1 K = C = 1 + (1)[0.06(1) (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 = = hp 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 = = ( S ) = (50) = psi 0.99 c ( Z N ) =.466[(10 )] = ( Z N ) =.466[(10 ) /.77] = (0.77) c,all) = = psi 1(1)(1) Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9

24 c,all = FdI C K K K C p o s m f (5.5)(0.1176) = = 1061 lbf (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 inion bending: The facor of safey, based on load and sress, is ( S F) = = =.15 Ans ear bending based on load and sress 861 ( S F) = = =.86 Ans inion wear 1061 based on load: n = = = based on sress: ( S ) = 1.06 = 1.0 Ans. ear wear based on load: n 4 = = = Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9

25 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, /, / 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 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 = in, d = 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) = psi ( ) S = 15 psi = 151 lbf, = hp = 861 lbf, = 19.9 hp 1 1 ear φ = I = Z N = o 0, , ( ) 0.77 ( Z ) = 0.769, C = 00 psi N ( S ) = S = (90) = psi c c (0.77) c,all) = = psi 1(1)(1) (0.769) c,all) = = psi 1(1)(1) (1649) = (1061) = 11 lbf, = = hp (1649) = (118) = 54 lbf, = = hp (0.769) Shigley s MED, 10 h ediion Chaper 14 Soluions, age 5/9

26 Raed power raed = min(157.5, 19.9, 105.6, 117.6) = hp Ans. rob. 14-4: raed = min(157.5, 19.9, 5.0, 59.0) = 5 hp The raed power approximaely doubled The gear and he pinion are 910 grade 1, carburized and case-hardened o obain Brinell 85 core and Brinell case. Table 14-: ( S ) 7 = psi Modificaion of S by (Y N ) = 0.8 produces ) = psi, all Similarly for (Y N ) = ) = 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 ) = psi 0.99 c 10 7 Modificaion of S c by Y N produces and c,all) = psi ) = 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

27 14-8 rade, 910 carburized and case-hardened o 85 core and 580 case in rob Summary: Table 14-: ( S ) 7 = psi and i follows ha all) = psi ) = 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 = psi ) = psi c,all ) = psi c,all = 4801 lbf, = 40 hp = 57 lbf, = 67 hp 4 4 Raing raed = min(70, 5, 40, 67) = 40 hp. Ans 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 = , C f = , 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 = ( S ) = 15 psi all) = psi ) = 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

28 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 = 0.14 =, no Ans raed = min(879., 1098, 04, 40) = 04 lbf raed = 1061 lbf Thus = =, no Ans S = S = 1, d = 4, J = 0.45, J = 0.410, K o = 1.5 Bending Table 14-4: ( S ) 7 = psi (1) all) = all) = = 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 = psi = ) = ) 0.99 c 10 c,all c,all Shigley s MED, 10 h ediion Chaper 14 Soluions, age 8/9

29 c,all p = Cp KoKsKmC f 4 4 ) Fd I (5.5)(0.1176) = = 195 lbf (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 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 = = psi C 6 sin14.5 / [11.8(10 )] 1.4(11) = = 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 ) = N N ( S ) = ( S ) = 8.(50) = 75 psi 75(0.977) all) = = 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

30 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 = psi 0.99 c (1) c,allow) = = psi 1(0.85) (1.5)(0.195) = = 074 lbf 00 1(1.404)(1.04) 074(95) = = 58.1 hp c,allow) = (167 94) = psi (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 ( Y N ) = 1.681(10 ) = ( Y N ) = 1.681(10 /.059) = 0.96 Table 14-: S = psi (0.98) all) = = psi 1(0.85) (1.5)(0.4) 1 = = 1(1.404)(1.04)(8.66)(1.08)(1) 487(95) 1 = = 69.7 hp all) = (60 047) = 6 47 psi lbf Shigley s MED, 10 h ediion Chaper 14 Soluions, age 0/9

31 = (487) = 58 lbf = (69.7) = 91. hp 487 Table 14-6: S c = psi ( ) =.466(10 ) = Z N ( Z N ) =.466(10 /.059) = (0.8790) c,all) = = psi 1(0.85) (1.5)(0.195) = = 568 lbf 00 1(1.404)(1.04) 568(95) = = 7.0 hp c,all) = ( ) = psi = (568) = 886 lbf (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 = psi (0.98) all) = = psi 1(0.85) (1.5)(0.4) 1 = = 99 lbf 1(1.404)(1.04)(8.66)(1.08) 99(95) 1 = = 8.4 hp (0.96) all) = = psi 1(0.85) = (99) = 850 lbf = (8.4) = 108 hp 99 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 1/9

32 Table 14-6: S c = psi ( Z ) = , ( Z ) = N N 5 000(0.879) c,all) = = 676 psi 1(0.85) (1.5)(0.195) = = 401 lbf 00 1(1.404)(1.04) 401(95) = = 11.5 hp c,all) = ( 676) = psi = (401) = 4509 lbf (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 = 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) = psi Fig. 14-5: S c = 49(00) = psi Fig. 14-6: J = 0.7 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. / / Eq. (14-8): B = 0.5(1 Q ) = 0.5(1 6) = v Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9

33 Eq. (14-7): A = (1 B) = ( ) = K v B A + V = = = 1.65 A To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = From Eq. (a), Sec , 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 = ( π ) = (8) Eq. (14-): C pm = 1.1 Fig : 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 = h (100 rev/min)(60 min/h) = 1.1 (10 9 ) rev Fig : Y N = 0.9 Fig : Z N = 0.8 K = ln 1 R = ln = 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) = (0.9) / [(1)(0.885)] = = 0.8 Ans. 94 Shigley s MED, 10 h ediion Chaper 14 Soluions, age /9

34 Tooh wear Eq. (14-16): K C m f c = C p Ko Ks d F I (1)(1.65)(1.) = 8( π ) = 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 (0.8) / [(1)(0.885)] = =.9 Ans m = mm/ooh, d = 00 mm N = d/m = 00 / = 16 eeh F = b = 4 p = 4 π m = 4π = 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 = = = m/s 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) 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) = A = (1 B) = ( ) = Eq. (14-7): K v B A + 00V (14.14) = = = 1.69 A To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = Shigley s MED, 10 h ediion Chaper 14 Soluions, age 4/9

35 Similar o Eq. (a) of Sec bu for SI unis: K s 1 = = 0.84 k b ( mf Y ) Ks = (6) 0.96 = 1.8 Conver he diameer and facewidh o inches for use in he load-disribuion facor equaions. d = 00/5.4 = in, F = 6/5.4 = 9.9 in Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): 9.9 C pf = (9.9) = (11.81) Eq. (14-): C pm = 1.1 Fig : 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 = h (900 rev/min)(60 min/h) = 6.48 (10 8 ) rev Fig : Y N = 0.9 Fig : Z N = 0.85 K = ln 1 R = ln = 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)] = = 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

36 = (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 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 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 Eq. (14-8): Eq. (14-7): / / B = 0.5(1 Qv ) = 0.5(1 6) = A = (1 B) = ( ) = K v B A + 00V (1.5) = = = 1.1 A To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = Similar o Eq. (a) of Sec bu for SI unis: K s 1 = = 0.84 k b ( mf Y ) Ks = (105) 0.90 = 1.17 Conver he diameer and facewidh o inches for use in he load-disribuion facor Shigley s MED, 10 h ediion Chaper 14 Soluions, age 6/9

37 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 = (4.1) = (4.9) Eq. (14-): C pm = 1 Fig : 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 = h (191 rev/min)(60 min/h) = 1.4 (10 8 ) rev Fig : Y N = 0.95 Fig : Z N = 0.88 K = ln 1 R = ln = 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 K Z R c = ZE KoKs d w1 b Z I = (1)(1.1)(1.17) 15(105) = 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

38 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 Eq. (14-8): Eq. (14-7): / / B = 0.5(1 Qv ) = 0.5(1 6) = A = (1 B) = ( ) = K v B A + V = = = 1.18 A To esimae a size facor, ge he Lewis Form Facor from Table 14-, Y = From Eq. (a), Sec , F Y Ks = 1.19 = 1.19 = 1.17 Eq. (14-1): C mc = 1 (uncrowned eeh) Eq. (14-): 4. C pf = (4.) = (5) Eq. (14-): C pm = 1 Fig : 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 = h (140 rev/min)(60 min/h) = 1. (10 8 ) rev Fig : Y N = 0.95 Fig : Z N = 0.88 K = ln 1 R = ln = 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

39 Tooh bending d KmK B Eq. (14-15): = KoK s F J (1.4)(1) = 180(1)(1.18)(1.17) = 1010 psi Eq. (14-41): S F SYN / ( KT KR) = (0.95) / [(1)(0.955)] = = Ans. Tooh wear Eq. (14-16): K C m f c = C p Ko Ks d F I = (1)(1.18)(1.17) 5(4.) = 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 Shigley s MED, 10 h ediion Chaper 14 Soluions, age 9/9