## 3 Application of the Proposed Method  
The application of the proposed method is based on the traditional design methodology. Therefore, the application is as shown in Figure 4. As an application, the intermediate shaft of the speed reducer is used to determine the principal stresses values that are acting on the ball bearing. Here, the speed is reduced from the initial 188 rad/s to a final speed of 47 rad/s. Initially, the motor transmits a constant power of \(8.95\times 10^{6}\) N mm/s. Additionally, in the intermediate shaft and through the gears, the initial speed of 188 rad/s is reduced to 94 rad/s, while the power of \(8.95\times 10^{6}\) N mm/s remains constant. The intermediate shaft is shown in Figure 5. This shaft is made of AISI 1020 steel with a shaft's diameter (\(d_{S}\)) of 45 mm. The designed shaft will rotate with an angular speed of 94 rad/s. In addition, while the pass diameter of gear B is 127 mm, for gear C it is 76.2 mm. Moreover, the pression angle between these gears is 20\({}^{\circ}\), as shown in Figure 6.  
Figure 4: Speed reducer design connected to a motor. Figure 5: Top view of shaft 2. By having the dimensions of the selected bearing, the application of the Hertz theory is carried out to obtain the main stresses that are acting in the bearing outer race. This is done in order to use them for determining the Weibull distribution parameters that will allow us to determine the actual reliability of the ball bearing. The step-by-step analysis given in Section 2.1 is as follows. Step 0. The loads which are acting in shaft 2 are first determined. Since the torque generated on axis 2 depends on the angular speed (\(\omega\)) of rotation and the power (\(P\)) that the axis transmits, then the constant torque generated on shaft 2 is:  
\[T\ =\ \frac{P}{\omega}\ =\ \frac{8.95\ \times\ 10^{6}\ \mathrm{N\cdot mm/s}}{94\ \mathrm{rad/s}}\ =\ 95,212.76\ \mathrm{N\cdot mm} \tag{36}\]  
Additionally, since shaft 2 is in the equilibrium, then the generated torque in gears \(B\) and \(C\) is equal but in an opposite direction. Thus, torque in \(C\) is \(T_{c}=-95,212.76\ \mathrm{N\ mm}\). On the other hand, as shown in Figure 6, the radial and tangential forces that are acting on the gears depend on the pression angle of \(\Phi=20^{\circ}\). Therefore, in the function of the torque and convergence ratio, the tangential force is given by:  
\[F_{t}\ =\ \frac{T}{r} \tag{37}\]  
Then, the corresponding radial force is given by:  
\[F_{r}\ =\ (F_{t})\tan\varnothing \tag{38}\]  
Additionally, since for gears \(B\) and \(C\)\(\Phi=20^{\circ}\), and \(r_{B}=63.5\ \mathrm{mm}\) and \(r_{C}=38.1\ \mathrm{mm}\), then numerically: \(F_{tB}\ =\ \frac{T}{r_{B}}\ =\ \frac{95,212.76\ \mathrm{N\cdot mm}}{63.5\ \mathrm{mm}}\ =\ 1500\ \mathrm{N}\), and \(F_{rB}\ =\ \frac{F_{tB}tan\varnothing}{(1500\ \mathrm{N})\tan 20^{\circ}}\ =\ 546 \mathrm{N}\)  
\(F_{tC}\ =\ \frac{T}{r_{C}}\ =\ \frac{95,212.76\ \mathrm{N\cdot mm}}{38.1\ \mathrm{mm}}\ =\ 2500\ \mathrm{N}\), and \(F_{rC}\ =\ \frac{F_{rC}tan\varnothing}{(2500\ \mathrm{N})\tan 20^{\circ}}\ =\ 910\ \mathrm{N}\). The radial and tangential forces, as well as the corresponding reactions that are acting in the x-y and in the x-z plane, are shown in Figure 7.  
Figure 6: Gear’s pressure angle. Figure 5.0 Forces and reactions in the x-y and x-z planes. Thus, as shown in Figure 7a, the forces and reactions that are acting in the x-y plane are: \(+\ \bigcirc\ \sum M_{D}\ =\ 0\ =\ (910\ \mathrm{N})(76.2\ \mathrm{mm})\ +\ (546\ \mathrm{N})(254\ \mathrm{mm})\ -\ (330.2\mathrm{m})\Big{(}R_{Ay}\Big{)}\); \(R_{Ay}\ =\ 630\ \mathrm{N}\); \(\sum F_{y}\ =\ 0\ =\ 630\ \mathrm{N}-546\ \mathrm{N}-910\ \mathrm{N}+R_{Dy}\); \(R_{Dy}\ =\ 826\ \mathrm{N}\).