A single-row deep groove ball bearing is subjected to a radial force of 8 kN and a thrust force of 3 kN. The values of X and Y factors are 0.56 and 1.5 respectively. The shaft rotates at 1200 rpm. The diameter of the shaft is 75 mm and Bearing No. 6315 (C =112 000 N) is selected for this application.
(i) Estimate the life of this bearing, with 90% reliability.
(ii) Estimate the reliability for 20 000 h life.
\text { Given } F_{r}=8 kN F_{a}=3 kN X=0.56 \quad Y=1.5 .
n = 1200 rpm d = 75 mm C = 112 000 N
Step I Bearing life with 90% reliability
From Eq. (15.3),
P=X F_{r}+Y F_{a} (15.3).
P=X F_{r}+Y F_{a}=0.56(8000)+1.5(3000) .
= 8980 N
From Eq. (15.6),
L_{10}=\left(\frac{C}{P}\right)^{p} (15.6).
L_{10}=\left(\frac{C}{P}\right)^{3}=\left(\frac{112000}{8980}\right)^{3}=1940.10 \text { million rev. } .
L_{10 h }=\frac{L_{10}\left(10^{6}\right)}{60 n}=\frac{1940.10\left(10^{6}\right)}{60(1200)}=26945.83 h (i)
Step II Reliability for 20 000 hr life
From Eq. (15.17),
\left(\frac{L}{L_{10}}\right)=\left[\frac{\log _{e}\left(\frac{1}{R}\right)}{\log _{e}\left(\frac{1}{R_{90}}\right)}\right]^{1 / b} (15.17).
\left(\frac{L}{L_{10}}\right)=\left[\frac{\log _{e}\left(\frac{1}{R}\right)}{\log _{e}\left(\frac{1}{R_{90}}\right)}\right]^{1 / b} .
or
\left.\left(\frac{L}{L_{10}}\right)^{b}=\frac{\log _{e}\left(\frac{1}{R}\right)}{\log _{e}\left(\frac{1}{R_{90}}\right)}\right] .
Substituting the following values,
L=20000 h \quad L_{10}=26945.83 h \quad R_{90}=0.90 .
b = 1.17 we get,
\left(\frac{20000}{26945.83}\right)^{1.17}=\left[\frac{\log _{e}\left(\frac{1}{R}\right)}{\log _{e}\left(\frac{1}{0.90}\right)}\right] .
\therefore \quad R=0.9283 \text { or } 92.83 \% (ii).