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Question 11.18: Estimate the remaining life in revolutions of an 02-30 mm an......

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.

Manufacturer Rating Life, revolutions Weibull Parameters Rating Lives
{x}_{0} \theta b
1 90(10^{6}) 0 4.48 1.5
2 1(10^{6}) 0.02 4.459 1.483
Tables 11–2 and 11–3 are based on manufacturer 2.
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Express Eq. (11-1) as

F L^{1/a}={\mathrm{constant}}                        (11-1)

F_{1}^{a}L_{1}=C_{10}^{a}L_{10}=K

For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C_{10}=20.3\,\mathrm{kN}.

K=(20.3)^{3}(10^{6})=8.365(10^{9})

At a load of 18 kN, life L_{\mathrm{1}} is given by:

L_{1}={\frac{K}{F_{1}^{a}}}={\frac{8.365(10^{9})}{18^{3}}}=1.434(10^{6})\,\mathrm{rev}

For a load of 30 kN, life {L}_{2} is:

L_{2}={\frac{8.365(10^{9})}{30^{3}}}=0.310(10^{6})\,\mathrm{rev}

In this case, Eq. (7-57) – the Palmgren-Miner cycle ratio summation rule – can be expressed as

{\frac{l_{1}}{L_{1}}}+{\frac{l_{2}}{L_{2}}}=1

Substituting,

{\frac{200\,000}{1.434(10^{6})}}+{\frac{l_{2}}{0.310(10^{6})}}=1

 

l_{2}=0.267(10^{6})\;\mathrm{rev}

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