Question

7) [30 pts] Factor of Safety for Contact Stress A steel spur pinion has 16 teeth cut on the 20 full-depth system with a module of 8 mm and a face width of 90 mm. The pinion rotates at 150 rev/min and transmits 3 kW to the mating steel gear. This pinion is to mesh with a steel gear with a gear ratio of 4:1. The Brinell hardness of the teeth is 200, and the tangential load transmitted by the gears is 6 kN. If the contact fatigue strength of the steel can be estimated from the AGMA formula of Sc = 2.22 H8+ 200 MPa, estimate the factor of safety of the drive based on a surface fatigue failure. The factor of safety of the drive is

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Solution: Pinion: No. of Teeth (20) = 16 Pressure Angle ($) = 20° Fill depth modyle (m) = 8 mm Face Width (b) = 90 mm Np = 150 rpm Power transmitted = 3 kW Gear ratio (6) = 4 Brinell Hardness Number CHB) = 200 Tangention load (P+) ? 6kN Fatigue strength (6c) = (2.22 HB + 200) MP4 (2.22x 200)/ 200 = 644 MPa Pt # Cs here, Effective load Cv of pinion av == π dp Np 60 velocity de where, dp= diameter of pinion ТР dp= 8x 16 = 128 mm = 0.128 m V- 3:14+ 150X0-128 = 1.0048 m/s 4.58 5.58 0.821 ( carefully cost Cv= 4.58 4.58 tv gears) Service Factor Cs = 1,25 6 x 1.25 - 9.135 KN Peff = PtXcs 0-821 CV

" wear loud Fw= dp q kb where, de diameter of Pinion a = Ratio Factor k= load stress Fuctor b= face width From design dute book, When 6c = 644 mp4 k= 1.02 Q = 2G (For Externus genring) +) = 1.6 = 264) 4+1 This, wear loud, Fw = [0.128 * 1.6x 1.02 x 90 1000 18.8 EN Factor of safety based on surface Fatigue Failure FW/ Feff FOS : FOS & 18.8 9.135 Fos: 2.058 Thank you.