Question

Two 20° involute full depth spur gears are used to transmit the power and speeds shown. The material for both pinion and gear is specified as through- hardened steel tempered to 300 BHN surface hardness. Both pinion and gear are precision cut for "very accurate” gearing. For a factor of safety of 2 and assuming a diametral pitch, Pof 9. specify face widths of the pinion and gear using only the AGMA bending fatigue analysis. State all assumptions used to determine the factors. (Be sure to apply the rule of thumb for face width in making your specification) ng=83.33 RPM np=250 RPM @40 HP PT C=15" Note, choose one or more than one of the following MC answers and then type the letter (s) corresponding to the question number in the essay section of this problem with supporting calculations contained on the scratch sheet. If the answer requires "the correct answer is ___"type that in the essay section as well. 4a) For the pinion and gear: a) Number of pinion teeth, N=202 and number of gear teeth, Ng=67 b) Circular pitch, p, is 0.394 in/tooth c) rp=3.75 in and r=11.25 in d) Velocity at contact point P=941 ft/min e) All of the above f) None of the above, the correct answer is: 4b) Face width, F, of pinion should be: a) F=2.76 in b) F=5 in c) 3pxFs5p d) F=1.67 in e) All of the above f) None of the above, the correct answer is:

Answer 1

Solution: Given that, 20° full depth involute. Material for both dame ** 300 BHN 1 storength) facto or of Safety: 2 diametllal pitch (Pd) = 9 ng = 83.33 RPM MP = 250 HP Те rg C:15 - - 381 mm calculating modele fout the maty gears: i Pd I 2.8 mm → → X 25.47 m; 2.8 mm Now, calculating the no. of teeths on Pinion & geam & diameter of Pitch cihicle fog pinion & gear Np NG TG Te forom olelation gear 250 = 3 83.33 Up : 30 G TG 3 Tp 0 u 23 alon TP

and acation we gel forom centre distance: 2 Пх (р 2 m. Tg t 381 mm (m: Module) 2 2.8* Tp + 2.8 2 hou 381 a TP + Tg 2121-4 from 0 Ę 4 Tp = 272 Tp = 68 teeths (3 Tq - 204 teeths Now, calculating 2.8 X 68 95.2 mm 3.74 inch Ip = mx Tp. 2 a eg mx Tg 2.88204 : 285.6 mm >> 11.244 inch 2 2 calculating ciMcular pitch: Ciocular pitch = Mx module Tx module = 1TX 2.8 mm = 0.3463 inch) tooth 25.4 velocity at contant point a DN Tx 95.2 x 2 x 250 25.4 x 12 - 490.616 ft Imin

Answers 3 4b option (f) none of the above C0391ect answer is 0.847 inch 40). options a, b, c ale connect option d, velocity at contact Point P =490.616 ft/min
please give some positive rating.. I have less cf score.. thank you so much..