In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an L10 life of 115 h and an L80 life of 600 h. The basic load rating of this bearing is 39.6 kN. Estimate the Weibull shape factor b and the characteristic life θ for a two-parameter model. This

Question

In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an [latex]L_{10}[/latex] life of 115 h and an [latex]L_{80}[/latex] life of 600 h. The basic load rating of this bearing is 39.6 kN. Estimate the Weibull shape factor b and the characteristic life θ for a two-parameter model. This manufacturer rates ball bearings at 1 million revolutions.

Accepted Answer

We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB. In this case, B is to the right of A.

For F = 18 kN, [latex](x)_{1}=\frac{115(2000)(60)}{10^{6}}=13.8[/latex]

This establishes point 1 on the R = 0.90 line.

The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, [latex]x_{0}=0[/latex] and points A and B are related by [see Eq. (20-25)]:

[latex]\begin{aligned}&x_{A}= heta[\ln (1 / 0.90)]^{1 / b} (1) \&x_{B}= heta[\ln (1 / 0.20)]^{1 / b}\end{aligned}[/latex]

and [latex]x_{B} / x_{A}[/latex] is in the same ratio as 600/115. Eliminating θ,

[latex]b=\frac{\ln [\ln (1 / 0.20) / \ln (1 / 0.90)]}{\ln (600 / 115)}=1.65[/latex]

Solving for θ in Eq. (1),

[latex] heta=\frac{x_{A}}{\left[\ln \left(1 / R_{A} ight) ight]^{1 / 1.65}}=\frac{1}{[\ln (1 / 0.90)]^{1 / 1.65}}=3.91[/latex]

Therefore, for the data at hand,

[latex]R=\exp \left[-\left(\frac{x}{3.91} ight)^{1.65} ight][/latex]

Check R at point B: [latex]x_{B}=(600 / 115)=5.217[/latex]

[latex]R=\exp \left[-\left(\frac{5.217}{3.91} ight)^{1.65} ight]=0.20[/latex]

Note also, for point 2 on the R = 0.20 line,

[latex]\begin{aligned}&\log (5.217)-\log (1)=\log \left(x_{m} ight)_{2}-\log (13.8) \&\left(x_{m} ight)_{2}=72\end{aligned}[/latex]

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Eq. (20-25) : [latex]R(x)=\exp \left[-\left(\frac{x}{ heta} ight)^{b} ight] \quad x \geq 0[/latex]

Question 11.36: In bearings tested at 2000 rev/min with a steady radial load...

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