Question

fast please

CHAPTER 9 9.1. A ball bearing is subjected to a radial load of 2884pounds for 40% of the time and 1884pounds for the remainder of the load cycle. Speed is 1884rpm. Expected life is to be 10 years at 250 days per year and 6 hours per day. Calculate the minimum value of the basic load rating that the bearing must have. All loads are steady.

Answer 1

Given in Radial Load Free 2884 eb for 40% time Fr2=1884 lb Gay time for 1884 rpm Speed Expected life = 10 years at 2.50 days per year and 6 hours per day Tox 250 X 6 X 60 X1884 So Number revolutions Lo q year days minutes days hours No. of revolution in one minute Lio z 1695.6 million revolutions -10

Basic Load Rating ©) - It is that load which a group of apparently identical bearing can withstand for a rating life of one million revolutions, Here the work cy de is divided into 2 elements Let P, and P2 be the loads and n, and n₂ be the speed during these element- According to load life relationship, during first element the life L, corresponding to load Pi is Li 00 million rev. N, = 0.4 x 1884: 7536 pi N23 0.6x1884: 1130.4 3 given by LIE C D xro 3 Pi CI en one revolution liße concumet ijs (6) 10" 2 Nipi so life Let us assume first dement consists of N, revolutions, consumed by first element is 106 CS life bed by second element is N, P, 3 consurnbed 106 c? Life consumed by complete work cy de = NIP, NP? 10 10fc of Pe is equivalent load for complete work cycle, the

3 life consumed By coork cycle is given by N Pe? 10°C N: Ni Na : 1884 them. equating N, 0,3 + N₂ P² . Nie 3 = 1884 Pe 74536) (2884)+ (1884(+132.4) Pe : 2 387-39 lb Also Lo: 1695.6 million revolution Basic load rating - C= P(410) 's C = (2387.39)(1695.6) Y 28468.49 lb