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, Prof 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 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, NP-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) 3p5F55p d) F=1.67 in e) All of the above f) None of the above, the correct answer is:

Answer 1

(1) loves diom Solution) briven that - A Gear - pinion groteam of zoo involute full depth = 83.33 R-PM No= 250 R.P.M Pp= 90 HP Ops ruzo Pinioni hear → → Piniom amd gear specified as hardened steel Brinell - Hardness Number B.H. NU=300 Sey flexutal endurance limite Te = 1.75 BHM = 1.75% gora ge = (525) MPa N = 2 factor of safety +8= pressure angle = 20°

(2 gint diametral Pitch al T T Tomodule =10 inch Em nomi (ga). for Pinion and gear (as for Pinion Teeth (Tp) and hear Teeth (Tol at point of contact (Vsside) p = Vslideln → Np Du Ten Te Ney 250 L T Te 73.33 Tes "Те 1250 83.33 - 83.33 = 0.33 >> ITP Tilta 250 Ty

3 univem that centre distance a c = Isinch Det Dor 15 2 2 > (572 =30 m (Ppt To [Те Tpt. To 30 V Tot To 270) - from equation iſ andlig +0.335Tr & Tin = 270 Tr 270 3 202.5 15333 Tu = 20 205 m 202 Similarly from equation ij - Te 0.333x Tv Te 67-4 0.333 x 20 2.5 m 62

A @ for circular Pitch (P) circular Pitch apa sixmodule P- r sex L g = (0.349) g P = 0.349) inch/ tooth do (@) for Da ding of Pinian (ro) and Radius of bear (VG) D شو module = T for pinion m= De a De TP 67 - 43 200 67.43 → re 1 x 67. 43 g 2 = 67.43 10 re - 3074G (3.75) irich - re (3.757 inch &

for hear Du TH mx 202.5 232.5 > Doa » om x 202.5 rui 202.5 2ru 202.5 in [2025] r = (11.25) inch fors n 982 (d) for velocity at com tact point Vp=gal ft/min. Vp= Itlmin We a rp = Terno) x 8p 12 (Ve) piman 22x 250 x 3.75 Iftlimia 12 207 250 3:25] [Velaimir = (490,87) ft / min it in same fur hear

Ab) of calculate face width F Piniom first calculate load factor CK) 17 oalb BHM (no 12 for $2.2017 Q salt a ( Boo 1044 18 Design unidaez & Aatice Awth Load V 2 " (Power) pimian 2rn Torave bo [Torare] pinion Pxoo 2r Me [Tonave] 40% 7 45.7x60 22x250 = (1139. 34 ) No-om > > Te

(7) Su for design Wgs e) by m >> " for static Load > Wg 2:1.25 WD woo Dynannie Load to 25 wo (by m) ba face width=f Ma factor of safety flexural endurancé Limit (1.75 A BHA 525 to lomme Tea 1•25 * 2 Texno < (525XFX Y & LA2S. XR y = form factor 0.912 Ypa urls 4 [for for 9=200 0,154- Те 09.12. 67 0.1403

3 0.1403 1025% 23 1139.34X11D X ? < (525) Fa 0.1403x (254/9 XR 7.5 x 25.4 V (2990 3.93 € (653-067 ) F =) F (face width ) 2 29903.93 653.067 F 2 (45.789) amam » închi F 2 145.789 10 x 280 2:54 V F 1 108 سرا face width