Question

so 125 mm 100 N.m 250 mm d30 mm - f - 250 mm -- 125 mm a = 150 mm View a-a Figure-1: Shaft-gear arranagement Figure I shows a shaft-gear arrangemenet. The force and moment acting on shaft ABCD is shown in figure-2. 150 h - 450 lbf-in 165.5 the 450 IM.in 180 lbf For the above shaft application, the shaft is driven at a constant speed of 280 rev/min. Obtain a Basic Load Rating for a cylindrical roller bearing at location A for a life of 14 kh with a 98 percent reliability, assuming distribution data from manufacturer 2 in Table 11-6.

Answer 1

C10 = 12121bf

or,

C10 = 5.39KN

The detailed solution is given below:

Given: Mo. 280 revimin I lo= 14,000hr | R = 0.98 #Ro= 50.98 = 0.9899 for each bensing A RAY Isolbf Ak Ain>< B 65.5lbf ARDY Z KAZ loin---> loin Coins Isolbf Dit In x-z plane: - Roz 180 lbf xll X (4 bin D OLL (9) (Maly = 0; [ Mement at A about y-aris] 1804F X[Carto) in ] – Roz x[4+10+6 )in] =0 :: Roz = 180X14 – 12666 → () {Fz=09 → Raz -180 +Rpz = 0 • RAZ = 180–126 = 54lbf 180X14 20

# In X-ypline : LUE MAD2 = 0; [Moment at A about Z-axis] → 4150 lbf) x (fin) -(65.513f)x[(4+10)in] + Roy x[(4+10+6)in] = :: Roy = 150x4 + 65.5X14 – 75.8566f 20 (+1) { Ey= 0; → Ray -150–65.5 + Roy = 0 : Ray = 150+65.5 – 75.85 = 139.65 Obf #. Resultant Force at A: RA= (RAZ ] + (Ray)2 = (54)?+ (139.65)? = 149.7366f # The design life, x = 20 = boxlox no LR Lio = 60x (14000)x (280) X = 235.2

# Fo= RA= 149.73 lbf I s Q = 10/3 , for Gylindrical roller beanings. # The Basic land rating [af=1] | 5o = 2:05 [, +10-17Chelny ] 5 10 Cara Hilamine) Ek are 7 1/1013) 235.2 (). - 149.73 x 8.094 = R ebf 60 = $.39 KN