## 5 Load Factor of the Most Loaded Rolling Element of Bearing
### BSO Position of Support  
The equation of equilibrium for the BSO support position can be written as (Figure 2a)  
\[Q=F_{0}+2\cdot\sum_{i=1}^{\frac{q-1}{2}}F_{i}\cdot\cos{(i\gamma)}, \tag{10}\]  
where \(Q\) is the external radial load, \(F_{0}\) is the load of the most loaded rolling element, \(F_{i}\) is the load of the \(i\)-th rolling element, \(\gamma\) is the angular spacing between rolling bearing elements, and \(q\) is the number of active rolling elements.  
The load transferred by the \(i\)-th rolling elements is equal to  
\[F_{i}=K\cdot\delta_{i}{}^{n}, \tag{11}\]  
where \(\delta_{i}\) is the contact deformation at the position of the \(i\)-th rolling element.  
Relative to Equation (11) the load of the 0-th rolling element is equal to:  
\[F_{0}=K\cdot\delta_{0}{}^{n}. \tag{12}\]  
By dividing Equation (11) by Equation (12), the following can be obtained:  
\[\frac{F_{i}}{F_{0}}=\left(\frac{\delta_{i}}{\delta_{0}}\right)^{n}, \tag{13}\]  
Figure 2: The load distribution for two boundary positions of bearing support.  
According to [26], the contact deformation \(\delta_{i}\) is obtainable as  
\[\delta_{i}=(\delta_{o}-a_{2:i+1})\cdot\cos\left(i\gamma\right), \tag{14}\]  
where \(a_{2:i+1}\) is the boundary-bearing deflection until entering of \(i\)-th rolling element in contact with bearing rings.  
Furthermore, if Equation (14) is included in Equation (13), the relationship between the load of the most loaded and the arbitrary rolling element is obtained:  
\[F_{i}=F_{0}\left[\frac{(\delta_{o}-a_{2:i+1})\cdot\cos\left(i\gamma\right)}{ \delta_{0}}\right]^{n}, \tag{15}\]  
Since this is for the BSO support position, the contact deformation of the most loaded rolling element is equal to total deflection of the bearing, i.e., \(\delta_{0}=w_{o}\); thus, Equation (15) can be written as  
\[F_{i}=F_{0}\left[\frac{(w_{o}-a_{2:i+1})\cdot\cos\left(i\gamma\right)}{w_{o}} \right]^{n}, \tag{16}\]  
If the above equation is inserted in Equation (10), after arranging, the relationship between the external load and the load of the most loaded rolling element was obtained, is  
\[F_{0}=\frac{Q}{1+2\cdot\sum\limits_{i=1}^{\frac{q-1}{2}}\left[\left(1-\frac{a _{2:i+1}}{w_{o}}\right)^{n}\cdot\cos^{n+1}\left(i\gamma\right)\right]}, \tag{17}\]  
If it is further taken that  
\[k_{max,o}=\frac{1}{1+2\cdot\sum\limits_{i=1}^{\frac{q-1}{2}}\left[\left(1- \frac{a_{2:i+1}}{w_{o}}\right)^{n}\cdot\cos^{n+1}\left(i\gamma\right)\right]}, \tag{18}\]  
The force which is transferred by the most loaded rolling element is obtained, as:  
\[F_{0}=k_{max,o}\cdot Q, \tag{19}\]  
In the equation above \(k_{max,o}\) represents a load factor of the most loaded rolling element. This factor is, in fact, a ratio of the size of the external load and the load transferred by the most loaded rolling element of bearing.  
In the boundary case, when switching from one support system to the other [25], the bearing rings' relative movement is equal to the boundary deflection, which is required for supporting the inner ring on the \(q\) of the rolling elements, i.e., \(w_{o}=a_{q}\). Then the expression for the load factor of the most loaded rolling element is obtained as  
\[k_{maxq}=\frac{1}{1+2\cdot\sum\limits_{i=1}^{\frac{q-1}{2}}\left[\left(1- \frac{a_{2:i+1}}{a_{q}}\right)^{n}\cdot\cos^{n+1}\left(i\gamma\right)\right]}, \tag{20}\]  
Further taking Equation (4) into account, after arranging, the above equation is obtained in its final form:  
\[k_{maxq}=\frac{1}{1+2\cdot\sum\limits_{i=1}^{\frac{q-1}{2}}\left[\left(1- \frac{t_{2:i+1}}{t_{q}}\right)^{n}\cdot\cos^{n+1}\left(i\gamma\right)\right]}, \tag{21}\]Respectively, if the expressions for \(t_{q}\) from Equation (5) are taken into account, then, after arranging, we can obtain  
\[k_{omaxq}=\frac{1}{1+2\cdot\sum\limits_{i=1}^{\frac{q-1}{2}}\left\{\left[\cos i \gamma-\frac{(1-\cos i\gamma)\cdot\cos\frac{q-1}{2}\gamma}{1-\cos\frac{q-1}{2} \gamma}\right]^{n}\cdot\cos\left(i\gamma\right)\right\}}. \tag{22}\]