Question

9. AGMA Shigley 5th Ed. 14-27 A 17-tooth 20°-pressure angle spur pinion rotates at 1800 rev/min and transmits 4 hp to a 52-tooth gear. The diametral pitch is 10, the face width is 1.25 in, the quality standard is No. 6, and the mounting is with exceptional rigidity. The pinion is steel, hardened to Brinell 240, and the gear is of No. 30 cast iron. Find the factor of safety based on surface fatigue strength if the life is to be no more than 10° cycles corresponding to a reliability of 90 percent 10. AGMA Shigley 5th Ed. 14-29 of 10 cycles and a reliability of Find the factor of safety based on surface fatigue fa 99.9 percent for the following 20° gearset Pinion. 500 rev/min, 19 teeth. 360 Bhn steel, H 6 kW, F 50 mm, m 4 mm, Q = 9, and Cm 1.3 for Gear. 25 teeth, 300 Bhn steel

Answer 1

Solution There will be many terms to obtain so use Figs. 14-17 and 14-18 as needed guides to what is dp = Np/Pa = 17/10 = 1.7 in dG = 52/10 = 5.2 in t (1.7)1800 лdрпр V = = 801.1 ft/min 12 12 33 000(4) 33 000 H W = 164.8 lbf 801.1 Assuming uniform loading, K, = 1. To evaluate Ky, from Eq. (14-28) with a quality number Qu = 6, B = 0.25(12- 6)2/3 = 0.8255 A= 5056(1 -- 0.8255) = 59.77 Then from Eq. (14-27) the dynamic factor is 0.8255 59.77 801.1 = 1.377 Ку 59.77

To determine the size factor, Kg, the Lewis form factor is needed. From Table 14-2 with Np 17 teeth, Yp = 0.303. Interpolation for the gear with Ng = 52 teeth yields YG = 0.412. Thus from Eq. (a) of Sec. 14-10, with F = 1.5 in, 0.0535 1.5 0.303 (Ks)p = 1.192 = 1.043 10 0.0535 1.5 0.412 (Ks)G = 1.192 = 1.052 10

The load distribution factor Km is determined from Eq. (14-30), where five terms are needed. They are, where F = 1.5 in when needed: Uncrowned, Eq. (14-30): Cmc = 1 Eq. (14-32): Cpf = 1.5/[10(1.7)] - 0.03750.0125(1.5) = 0.0695 Bearings immediately adjacent, Eq. (14-33): Cpm = 1 Commercial enclosed gear units (Fig. 14-11): Cma = 0.15 Еq. (14-35): Се 3D1 Thus, K/m = 1Cmc(CpfCpm + CmaC¢) = 1 + (1)[0.0695(1) +0.15(1)] = 1.22 Assuming constant thickness gears, the rim-thickness factor KB = 1. The speed ratio is mG NG/Np ment, with N(pinion) = 108 cycles and N(gear)= 10/mG = 10%/3.059 cycles, are 52/17 3.059. The load cycle factors given in the problem state- (YN)P 1.3558(108)-0.0178 0.977 (YN)G = 1.3558(108/3.059)-0.0178 = 0.996 From Table 14.10, with a reliability of 0.9, Kr = 0.85. From Fig. 14-18, the tempera- ture and surface condition factors are Kī = 1 and Cf = 1. From Eq. (14-23), with mN =1 for spur gears, cos 200 sin 20° I 3.059 = 0.121 3.0591 2 From Table 14-8, Cp = 2300/psi

Next, we need the terms for the gear endurance strength equations. From Table 14-3, for grade 1 steel with HBp 240 and HBG = 200, we use Fig. 14-2, which gives 77.3(240)12 800 31 350 psi (S) P (S G=77.3(200) 12 800 = 28 260 psi Similarly, from Table 14-6, we use Fig. 14-5, which gives 322(240) 29 100 = 106 400 psi (Sc)P (Sc)G 322(200) 29 100 = 93 500 psi From Fig. 14-15 1.4488(108)-0.023 0.948 (ZN)P 1.4488(108/3.059)-0.023 = 0.973 (ZN)G 240/200 1.2. Then, For the hardness ratio factor CH, the hardness ratio is HBp/HBG from Sec. 14-12 A' 8.98(103(Hpp/HBG) 8.29(103) = 8.98(103)(1.2) 8.29(103) = 0.002 49 Thus, from Eq. (14-36) CH=10.002 49(3.059 1) = 1.005

(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into Eq. (14-15) gives -- (w.K.N.) 10 1.22 (1) B WK K,K F 164.8(1)1.377(1.043) J 1.5 0.30 = 6417 psi Substituting the appropriate terms for the pinion into Eq. (1441) gives S.YN/(KT KR) 31 350(0.977)/[1(0.85)| 5.62 Answer (Sf)p = 6417 O Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14-15) gives 10 1.22(1) (o)G = 164.8(1)1.377(1.052) 4854 psi 1.5 0.40 Substituting the appropriate terms for the gear into Eq. (14-41) gives 28 260(0.996)/[1(0.85)]| Answer (SF)G = 6.82 4854 (b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14-16) gives 1/2 Кт Су W'KoK,K, (oc)p = Cp dpF I P 1/2 1.22 2300 164.8(1)1.377(1.043) = 70 360 psi 1.7(1.5) 0.121 Substituting the appropriate terms for the pinion into Eq. (1442) gives

106 400(0.948)/[1(0.85)]| ScZN/(K7KR) (SH)P = 1.69 Answer = 70 360 Oc P Gear tooth wear. The only term in Eq. (14-16) that changes for the gear is Ks. Thus j 1/2 (K:)с 1.052\ 1/2 70 360 = 70 660 psi (oc)G = (oc)p = (К)Р 1.043 Substituting the appropriate terms for the gear into Eq. (14-42) with CH 1.005 gives 93 500(0.973)1.005/[1(0.85)] Answer (SH G 1.52 70 660 or 5.73 with 1.692 = 2.86, so the (c) For the pinion, we compare (Sp)p with (Sh)p threat in the pinion is from wear. For the gear, we compare (Sp)G With (SH) with 1.522 2.31, so the threat in the gear is also from wear or 6.96