## 3 Application of the Proposed Method
From Equations (28) and (29) the \(A\) and \(B\) values to determine \(\Delta\) are:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]0  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]1  
Therefore, from Equation (27) the values are:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]2  
Then, from Equations (14) to (16) the principal contact stresses values are:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]3  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]4  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]5  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]6  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]7  
Additionally, from Equation (17) the maximum shear stress value is:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]8  
Now, based on the above stress values, the Weibull parameters used to determine the actual life and reliability of the ball bearing are determined, as mentioned in Section 2.4. Step 12. Determine the Weibull \(\eta\) and \(\beta\) parameters. By using the maximum contact stress value \(\sigma_{1}\) and minimum contact stress value in Equation (30), the use Weibull scale parameter is [22]:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]9  
Additionally, from Equation (31) the Weibull shape parameter is:  
\[k_{e}\ =\ (22.7)^{2/n}\ =\ 7.3\]0  
Therefore, the Weibull parameters to determine the life of the ball bearing are W (1.28, 910,020,039.1 rev). With these parameters, the actual reliability of the ball bearing is as mentioned in Section 2.5. Step 13a. From Equation (32) using the catalogues \(L_{10}\) life of \(765.77\times 10^{6}\) rev and the Weibull scale parameter, the reliability of the ball bearing is estimated to be:  
\[k_{e}\ =\ (22.7)^{2/n}\ =\ 7.3\]1  
At this point, it is very important to note that the used \(L_{10}\) value does not represent the applied stress, (it represents the expected life at the conditions at which the ball bearing was designed). In addition, the \(\eta_{use}\) value does not represent the strength conditions at which the ball bearing was designed. Therefore, the estimated reliability of \(R(t)=0.4485\) does not represent the life that corresponds to the actual environment at which shaft 2 operates. Therefore, to determine the actual reliability we must proceed with the next steps. Step 13b. By setting Equation (32) at \(R(t)=0.90\), the actual \(L_{10(use)}\) life that represents the actual conditions at which shaft 2 operates is given from Equation (33) as \(L_{10(use)}=\eta_{use}\Big{(}\sqrt[6]{-\ln(0.90)}\Big{)}=156,862,111.04\)_rev_. Now that the \(L_{10(use)}\) value represents the environmental actual stress, it is necessary to determine the Weibull scale parameter that represents the strength of the selected ball bearing. It is determined as follows. Step 13c. The catalogues strength \(\eta_{cat}\) value that represents the strength of the ball bearing is determined from Equation (34) using the catalogue \(L_{10}\) life. It is \(\eta_{cat}=\frac{L_{10}}{\sqrt[6]{-\ln(0.90)}}=4,442,538,996.4\)_rev_. Finally, using the \(L_{10(use)}\) value and the \(\eta_{cat}\) value, the actual reliability is as follows. Step 13d. From Equation (32) the actual reliability of the selected ball bearing is:  
\[k_{e}\ =\ (22.7)^{2/n}\ =\ 7.3\]2  
Finally, in step 13d, the actual reliability of the ball bearing is higher than the designed reliability of \(R(t)=0.90\). In addition, this occurs mainly since the estimated \(L_{10(use)}\) life is lower than the catalogues \(L_{10}\) life (\(L_{10(use)}<L_{10}\)). The actual reliability resulting in the ball bearing is greater than the reliability of the life equation \(L_{10}\) offered by the manufacturer due to the fact that it was analyzed with different loads than those used by the manufacturer.