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.

Question

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.

Accepted Answer

[latex] ext { Given } F_{r}=8 kN F_{a}=3 kN X=0.56 \quad Y=1.5 [/latex].

n = 1200 rpm d = 75 mm C = 112 000 N
Step I Bearing life with 90% reliability
From Eq. (15.3),

[latex] P=X F_{r}+Y F_{a} [/latex] (15.3).

[latex] P=X F_{r}+Y F_{a}=0.56(8000)+1.5(3000) [/latex].

= 8980 N
From Eq. (15.6),

[latex] L_{10}=\left(\frac{C}{P} ight)^{p} [/latex] (15.6).

[latex] L_{10}=\left(\frac{C}{P} ight)^{3}=\left(\frac{112000}{8980} ight)^{3}=1940.10 ext { million rev. } [/latex].

[latex] L_{10 h }=\frac{L_{10}\left(10^{6} ight)}{60 n}=\frac{1940.10\left(10^{6} ight)}{60(1200)}=26945.83 h [/latex] (i)

Step II Reliability for 20 000 hr life
From Eq. (15.17),

[latex] \left(\frac{L}{L_{10}} ight)=\left[\frac{\log _{e}\left(\frac{1}{R} ight)}{\log _{e}\left(\frac{1}{R_{90}} ight)} ight]^{1 / b} [/latex] (15.17).

[latex] \left(\frac{L}{L_{10}} ight)=\left[\frac{\log _{e}\left(\frac{1}{R} ight)}{\log _{e}\left(\frac{1}{R_{90}} ight)} ight]^{1 / b} [/latex].

or

[latex] \left.\left(\frac{L}{L_{10}} ight)^{b}=\frac{\log _{e}\left(\frac{1}{R} ight)}{\log _{e}\left(\frac{1}{R_{90}} ight)} ight] [/latex].

Substituting the following values,

[latex] L=20000 h \quad L_{10}=26945.83 h \quad R_{90}=0.90 [/latex].

b = 1.17 we get,

[latex] \left(\frac{20000}{26945.83} ight)^{1.17}=\left[\frac{\log _{e}\left(\frac{1}{R} ight)}{\log _{e}\left(\frac{1}{0.90} ight)} ight] [/latex].

[latex] herefore \quad R=0.9283 ext { or } 92.83 \% [/latex] (ii).

Question 15.15: A single-row deep groove ball bearing is subjected to a radi...

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