Estimate the remaining life in revolutions of an 02-30 mm angular-contact ball bearing already subjected to 200 000 revolutions with a radial load of 18 kN, if it is now to be subjected to a change in load to 30 kN.

Question

Accepted Answer

Express Eq. (11-1) as

[latex]F_{1}^{a} L_{1}=C_{10}^{a} L_{10}=K[/latex]

For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, [latex]C_{10}=20.3 kN[/latex] .

[latex]K=(20.3)^{3}\left(10^{6} ight)=8.365\left(10^{9} ight)[/latex]

At a load of 18 kN, life [latex]L_{1}[/latex] is given by:

[latex]L_{1}=\frac{K}{F_{1}^{a}}=\frac{8.365\left(10^{9} ight)}{18^{3}}=1.434\left(10^{6} ight) rev[/latex]

For a load of 30 kN, life [latex]L_{2}[/latex] is:

[latex]L_{2}=\frac{8.365\left(10^{9} ight)}{30^{3}}=0.310\left(10^{6} ight) rev[/latex]

In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be expressed as

[latex]\frac{l_{1}}{L_{1}}+\frac{l_{2}}{L_{2}}=1[/latex]

Substituting,

[latex]\begin{aligned}&\frac{200000}{1.434\left(10^{6} ight)}+\frac{l_{2}}{0.310\left(10^{6} ight)}=1 \&l_{2}=0.267\left(10^{6} ight) rev \end{aligned}[/latex]

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Eq. (11-1) : [latex]F L^{1 / a}= ext { constant }[/latex]

Eq. (6-57): [latex]\sum \frac{n_{i}}{N_{i}}=c[/latex]

Question 11.38: Estimate the remaining life in revolutions of an 02-30 mm an...

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