Question 11.15: In bearings tested at 2000 rev/min with a steady radial load......
In bearings tested at 2000 rev/min with a steady radial load of 18 kN, a set of bearings showed an L_{{{10}}} life of 115 h and an L_{{{80}}} 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.
| 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. | ||||
Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken: b = 1.5, θ = 4.48. 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, (x)_{1}={\frac{115(2000)(16)}{10^{6}}}=13.8
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, x_{0}=0 and points A and B are related by [see Eq. (20-25)]:
R(x)=\exp\biggl[-\left(\frac{x}{\theta}\right)^{b}\biggr]\quad\quad x\ge0 (20-25)
x_{A}=\theta[\mathrm{ln}(1/0.90)]^{1/b} (1)
x_{B}=\theta[{\mathrm{ln}}(1/0.20)]^{1/b}and x_{B}/x_{A} is in the same ratio as 600/115. Eliminating θ
b={\frac{\ln[\ln(1/0.20)/\ln(1/0.90)]}{\ln(600/115)}}=1.65Solving for θ in Eq. (1)
\theta={\frac{x_{A}}{[\ln(1/R_{A})]^{1/1.65}}}={\frac{1}{[\ln(1/0.90)]^{1/1.65}}}=3.91Therefore, for the data at hand,
R=\exp\biggl[-\left(\frac{x}{3.91}\right)^{1.65}\biggr]Check R at point B: x_{B}=(600/115)=5.217
R=\exp\biggl[-\left(\frac{5.217}{3.91}\right)^{1.65}\biggr]=0.20
Note also, for point 2 on the R = 0.20 line.
\log(5.217)-\log(1)=\log(x_{m})_{2}-\log(13.8)(x_{m})_{2}=72
