Question 11.17: Different bearing metallurgy affects bearing life. A manufac......
Different bearing metallurgy affects bearing life. A manufacturer reports that a particular heat treatment increases bearing life at least threefold. A bearing identical to that of Prob. 11–15 except for the heat treatment, loaded to 18 kN and run at 2000 rev/min, revealed an L_{{{10}}} life of 360 h and an L_{{{80}}} life of 2000 h. Do you agree with the manufacturer’s assertion concerning increased life?
| 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. | ||||
The horizontal separation of the R = 0.90 loci in a log F-log x plot such as Fig. 11-5 will be demonstrated. We refer to the solution of Prob. 11-15 to plot point G (F = 18 kN, x_{G}=13.8). We know that (C_{10})_{1}=39.6\,\mathrm{kN},x_{1}=1. This establishes the unimproved steel R = 0.90 locus, line AG. For the improved steel
(x_{m})_{1}={\frac{360(2000)(60)}{10^{6}}}=43.2We plot point G^{\prime}(F=18\;\mathrm{kN},x_{G^{\prime}}=43.2), and draw the R = 0.90 locus A_{m}G^{\prime} parallel to AG
We can calculate (C_{10})_{m} by similar triangles.
{\frac{\log(C_{10})_{m}-\log18}{\log43.2-\log1}}={\frac{\log39.6-\log18}{\log13.8-\log1}}\log(C_{10})_{m}={\frac{\log43.2}{\log13.8}}\log\left({\frac{39.6}{18}}\right)+\log18
(C_{10})_{m}=55.8\,\mathrm{kN}
The usefulness of this plot is evident. The improvement is 43.2/13.8 = 3.13 fold in life. This result is also available by (L_{10})_{m}/(L_{10})_{1} as 360/115 or 3.13 fold, but the plot shows the improvement is for all loading. Thus, the manufacturer’s assertion that there is at least a 3-fold increase in life has been demonstrated by the sample data given.
