## 3 Application of the Proposed Method
Similarly, as shown in Figure 7b, the forces and reactions that are acting in the x-z plane are: \(+\ \bigcirc\ \sum M_{D}\ =\ 0\ =\ (R_{Az})(330.2\ \mathrm{mm})\ -\ (1500\ \mathrm{N})(254\ \mathrm{mm})\ -\ (2500\ \mathrm{N})(76.2\ \mathrm{mm})\); \(R_{Az}\ =\ 1730\ \mathrm{N}\); \(\sum F_{z}\ =\ 0\ =\ -1730\ \mathrm{N}+1500\ \mathrm{N}+2500\ \mathrm{N}-R_{Dz}\); \(R_{Dz}\ =\ 2270\ \mathrm{N}\). Therefore, the reaction forces in points \(A\) and \(D\) are:  
Figure 5.1  
Figure 5.2  
Additionally, since \(R_{D}>R_{A}\), then \(R_{D}\) represents the reaction force value base on which the ball bearing is selected. Step 1. Since we do not have an axial force for shaft 2, then from Equation (1) the design load is directly given by the radial force represented by \(R_{D}\). Therefore, \(P=R_{D}=2415.6\ \mathrm{N}\). Additionally, since the inner race is the one that is rotating, then \(V=I\), and consequently, from Equation (2) the designed load is:  
Figure 5.3  
Step 2. Since from the design shaft phase, the diameter of shaft 2 is 45 mm, and from step 1 \(P_{d}=2415.6\ \mathrm{N}\), then from the SKF catalogue the selected ball bearing is the 6009 SKF type ball bearing with a bore diameter of 45 mm. The related ball bearing characteristics are given in Figure 8 and in Figure 5.4 \begin{table}
\begin{tabular}{l l} \hline \hline \multicolumn{1}{c}{**Bearing Type**} & \multicolumn{1}{c}{**Capped Single Row Deep Groove Ball Bearings**} \\ \hline Bore Diameter (\(d\)) & 45 mm \\ \hline Outer Diameter (\(D\)) & 75 mm \\ \hline Width (\(B\)) & 16 mm \\ \hline Dynamic Load Rating (C) & 22,100 N \\ \hline Static Load Rating (\(C_{o}\)) & 14,600 N \\ \hline Max Speed & 1047 rad/s \\ \hline Max. Shaft Shoulder Dia. Inner (\(U_{i}\)) & 67.8 mm \\ \hline Min. Housing Shoulder Dia., Outer (\(U_{o}\)) & 54.7 mm \\ \hline Chamfer radius (\(r\)) & 1 mm \\ \hline Ball Quantity & 13 \\ \hline Ball Diameter (\(d_{b}\)) & 8.731 mm \\ \hline Material & 52,100 Chrome Steel \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Ball bearing specifications. Figure 8: Ball bearing dimensions. Step 3. According to the catalogues, the designed \(L_{10}\) life of the ball bearing is:  
\[L_{10}\ =\ 10^{6}\bigg{(}\frac{22,100\ \mathrm{N}}{2415.6\ \mathrm{N}}\bigg{)}^{3}\ =\ 765.77\ \times\ 10^{6}\ \mathrm{rev}\]  
With this data, the contact area generated in the ball bearing mentioned in Section 2.1 is determined as follows. Step 4. Since the curvature generated in the x direction is \(r_{ax}\ =\ r_{ay}\ =\ 4.365\ \mathrm{mm}\) and \(r_{bx}\ =\ 35.003125\ \mathrm{mm}\), then \(\frac{1}{R_{x}}\ =\ \frac{1}{4.365\ \mathrm{mm}}\ -\ \frac{1}{35.003125\ \mathrm{mm}}\ =\ 0.20052\ \mathrm{mm}^{-1}\), and therefore, \(R_{x}\ =\ 4.986\ \mathrm{mm}\). Additionally, since the outer ratio \(r_{by}\) is unknown, then from Equation (7) by using the compliance value of 0.52, the \(r_{by}\) value is \(r_{by}\ =\ dR_{r}\ =\ (8.731\ \mathrm{mm})(0.52)\ =\ 4.54\ \mathrm{mm}\), therefore, \(\frac{1}{R_{y}}\ =\ \frac{1}{4.365\ \mathrm{mm}}\ -\ \frac{1}{4.54\ \mathrm{mm}}\ =\ 0.00883\ \mathrm{mm}^{-1}\) implying \(R_{y}\ =\ 113.24\ \mathrm{mm}\). Moreover, since \(\frac{1}{R}\ =\ 0.20052\ \mathrm{mm}^{-1}+0.00883\ \mathrm{mm}^{-1}\ =\ 0.20935\ \mathrm{mm}^{-1}\), then the total curvature is \(R\ =\ 4.77\ \mathrm{mm}\). Step 5. From Equation (8), the curvature's index \(\alpha_{r}\) is:  
\[\alpha_{r}\ =\ \frac{113.24\ \mathrm{mm}}{4.986\ \mathrm{mm}}\ =\ 22.7\]  
Step 6. From Equation (9), the elliptical parameter \(k_{e}\) is:  
\[k_{e}\ =\ (22.7)^{2/n}\ =\ 7.3\]  
Step 7. Since \(k_{e}<100\), the equations of Table 1 are selected. Step 7.1. The simplified values of the first and second order elliptical equations \(\mathcal{F}\ \mathrm{y}\ \mathcal{E}\) are:  
\[\mathcal{F}\ =\ \frac{\pi}{2}+\bigg{(}\frac{\pi}{2}-1\bigg{)}\ln(22.7)\ =\ 3.353\]  
\[\mathcal{E}\ =\ 1+\frac{\bigg{(}\frac{\pi}{2}-1\bigg{)}}{22.7}\ =\ 1.025\]  
Step 8.