A single row deep-groove ball bearing No. 6403 is used to support the lay shaft of a four speed automobile gear box. It is subjected to the following loads:

The lay shift is connected to the engine shaft and rotates at 1750 nr.p.m If the bearing is expected to be in use for 4000 hours, determine the reliability of bearing.

Mumbai University > Mechanical Engineering > Sem 7 > Machine Design 2

Marks: 10M

Year: Dec 2016

Calculation of revolution for each phase

$n_1 = 0.01 \times N = 17.5\\ n_2 = 0.03 \times N = 52.5\\ n_3 = 0.21 \times N = 367.5 \\ n_4 = 0.75 \times N = 1312.5$

Mean revolution $\sum n = n_1 + n_2 + n_3 + n_4 = 1750 rpm$

calculation of mean load from variable load

$F_{rn} = \sqrt[3]{Fr_1^3n_1 + \frac{Fr_2^3n_2+Fr_3^3}{\sum n}+ Fr_4^3n_4}\\ \hspace{0.5cm} = 1754.122 N/m\\ 175.4 kgf$

$F_{am} = 702.749 N\\ \hspace{0.5cm} = 70.2749 kgf$

Calculation of life induced in bearing

$40' = (\frac{c}{p})^k 40 --- 1$

Brg No = 6403 d = 17mm c = 1800 kgf $c_0$ = 1280 kgf N = 10000

P = equivalent load

P = (XVFr + YFa)kt

V = 1 for inner race rotating.

s = 1.3 medium shock

K = 1 temp not given

$\frac{F_a}{C_0} = 0.054 \hspace{0.5cm} \therefore e = 0.27\\ \frac{F_a}{F_r} = 0.4\\ \frac{F_a}{F_r} \gt e$

X = 0.56 Y = 1.6

P = 273.871 kgf

40' = 283.906 mr

Probably for 4000 hr

$L_x = \frac{4000 \times 1750 \times 60}{106}\\ L_x = 420 mr\\ \frac{L_x}{40'} = \bigg[ \frac{l_n (1/p_x)}{l_n (1/0.9)}\bigg]^1/1.34\\ p.x = 0.8368$

Probability 83.68%