A ball bearing, subjected to a radial load of 5 kN, is expected to have a life of 8000 h at 1450 rpm with a reliability of 99%. Calculate the dynamic load capacity of the bearing, so that it can be selected from the manufacturer’s catalogue based on a reliability of 90%.

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Accepted Answer

[latex] ext { Given } F_{r}=5 kN \quad n=1450 rpm \quad L_{99 h }=8000 h [/latex].

Step I Bearing life with 99% reliability

[latex] L_{99}=\frac{60 n L_{99 h }}{10^{6}}=\frac{60(1450)(8000)}{10^{6}} [/latex].

[latex] =696 ext { million rev } [/latex].

Step II Bearing life with 90% reliability
From Eq. (15.17),

[latex] \left(\frac{L_{99}}{L_{10}} ight)=\left[\frac{\log _{e}\left(\frac{1}{R_{99}} ight)}{\log _{e}\left(\frac{1}{R_{90}} ight)} ight]^{1 / 1.17} [/latex] (15.17).

[latex] \left(\frac{L_{99}}{L_{10}} ight)=\left[\frac{\log _{e}\left(\frac{1}{R_{99}} ight)}{\log _{e}\left(\frac{1}{R_{90}} ight)} ight]^{1 / 1.17}=\left[\frac{\log _{e}\left(\frac{1}{0.99} ight)}{\log _{e}\left(\frac{1}{0.90} ight)} ight]^{1 / 1.17} [/latex].

=0.1342.

Therefore,

[latex] L_{10}=\frac{L_{99}}{0.1342}=\frac{696}{0.1342}=5186.29 ext { million rev. } [/latex]

Step III Dynamic load carrying capacity of bearing

[latex] C=P\left(L_{10} ight)^{1 / 3}=5000(5186.29)^{1 / 3}=86547.7 N [/latex].

Question 15.16: A ball bearing, subjected to a radial load of 5 kN, is expec...

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