# Study on Load Distribution and Fatigue Elastic Life of Ball Screw under Ultimate Conditions
## 2 Theoretical Analysis
### The Ball Screw's Static Load Analysis Considering Nut Position and Screw Length
\[\omega_{M\eta i}=\sum\limits_{j=1}^{i-1}\frac{M_{M\eta i}}{6\varepsilon_{i}TL} \Big{\{}-L_{i}^{3}+3 LL_{2}{}^{2}+\Big{[}L^{2}-3L_{1}{}^{2}\Big{]}L_{i}\Big{\}}+ \sum\limits_{j=i}^{N}\frac{M_{M\eta i}L_{i}}{6\varepsilon_{i}TL}\Big{[}L^{2}-3 L_{1}{}^{2}-L_{i}^{2}\Big{]}. \tag{5}\]  
Taking the nut raceway as a reference, the equivalent axial deformation and radial deformation \(\delta_{ri}\) can be calculated by Equations (6) and (7), where \(\varepsilon_{i}\) is positive towards the positive side of the \(Z_{2}\) axis, and \(\delta_{ri}\) is positive in the positive direction of the \(y_{2}\) axis [20]. \[\varepsilon_{i}-\varepsilon_{i-1}=(\Delta_{si}-\Delta_{ni})\cos\lambda \tag{6}\]  
\[\delta_{ri}=-\big{(}\omega_{Qxi}+\omega_{Myi}\big{)}\cos\theta_{i}-\big{(} \omega_{Qyi}+\omega_{Mxi}\big{)}\sin\theta_{i}. \tag{7}\]  
Figure 2 represents the relationship between the equivalent deformation in \(y_{2}\) and \(z_{2}\) directions and the deformation of the ball. \(O_{ni}\) and \(O_{si}\) are the centers of the nut and screw raceways, respectively. The deformation of the screw along the \(z_{2}\) axis increases the ball's elastic deformation, while deformation along the \(y_{2}\) axis decreases the elastic deformation of the ball, and Equation (8) can be obtained. \(A_{0}\), \(B_{0}\), and \(C_{0}\) are the components of the distance between the screw and nut raceways' centers along the normal direction of the raceway, the \(y_{2}\) axis, and the \(y_{2}\) axis, respectively. \[\left\{ \begin{array}{l}A_{0}=2R-d_{b}\\ B_{0}=A_{0}\cos\alpha_{0}\\ C_{0}=A_{0}\sin\alpha_{0}.\end{array} \right. \tag{8}\]  
When the nut is subjected to a load, the relationship between \(A_{0}\), \(B_{0}\), \(C_{0}\), and the deformation and equivalent deformation can be calculated through Figure 2 by Equation (9). In addition, it can also be derived that \(C_{i}=\sqrt{A_{i}^{2}-B_{i}^{2}}\). Because ball screws usually use Gothic raceway, and the elastic deformation of the screw and the nut is much smaller than their structural parameters, the contact angle does not change much before and after the load is applied [20]. \[\left\{ \begin{array}{l}A_{i}=A_{0}+KQ_{i}^{2/3}\\ B_{i}=B_{0}-\delta_{ri}\\ C_{i}=C_{0}+\varepsilon_{i}.\end{array} \right. \tag{9}\]  
From Equations (8) and (9), the relationship between the equivalent screw axial deformation and other deformations can be further derived. Here, the equivalent axial deformation is expressed as Equation (10) and then substituted into Equation (6) to obtain Equation (11). \[\varepsilon_{i}=\sqrt{A_{i}^{2}-B_{i}^{2}}-C_{0} \tag{10}\]  
\[\sqrt{A_{i}^{2}-B_{i}^{2}}-\sqrt{A_{i-1}^{2}-B_{i-1}^{2}}=(\Delta_{si}-\Delta _{ni})\cos\lambda. \tag{11}\]  
By substituting Equations (2) and (9) into Equation (11), Equation (12) is obtained, where \(i=2\sim N\). Equation (12) consists of (\(N-1\)) equations. We also need to introduce the axial force balance Equation (13) of the nut to solve for the contact loads of \(N\) balls, where  
\[S_{1}=\sqrt{\big{(}A_{0}+KQ_{i}^{2/3}\big{)}^{2}-(B_{0}-\delta_{ri})^{2}},\,S_ {2}=\sqrt{\big{(}A_{0}+KQ_{i-1}^{2/3}\big{)}^{2}-\Big{(}B_{0}-\delta_{r(i-1)} \big{)}^{2}}.\]  
\[S_{1}-S_{2}=\bigg{(}-\frac{\Delta LF_{si}}{E_{s}S_{s}}-\frac{\Delta LF_{ni}}{E _{n}S_{n}}\bigg{)}\cos\lambda \tag{12}\]  
\[F_{a}=\sum\limits_{i=1}^{N}Q_{i}\sin\alpha_{0}\cos\lambda. \tag{13}\]  
Equations (12) and (13) together form a set of nonlinear equations \(f\), and can be numerically solved by [PERSON]'s iteration using Equation (14) to obtain the contact loads of each ball, where \(Q^{(r+1)}\) is the solution of the contact load at the \(r\)-th iteration, \(J\) is the Jacobian matrix of the nonlinear function \(f\), and the iteration ends when \(\|Q^{(r+1)}-Q^{(r)}\|\leq 10^{-3}\). \[Q^{(r+1)}=Q^{(r)}-J\Big{(}Q^{(r)}\Big{)}^{-1}\cdot f\Big{(}Q^{(r)}\Big{)}. \tag{14}\]