# 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  
The ball screw is typically installed with one end fixed and the other supported, while the nut is mounted on a workbench supported by parallel rails [23]. As shown in Figure 1, multiple coordinate systems are established to simplify the calculation. The global coordinate system \(CS_{0}(x_{0},y_{0},z_{0})\) is fixed on the base, its origin is positioned at the central point of the left of the screw, and its \(y_{0}\) axis overlaps with the screw axis. The mobile coordinate system \(CS_{1}(x_{1},y_{1},z_{1})\) is fixed on the nut, its axes are parallel to \(CS_{0}\), and the balls are numbered along the helix, with the leftmost ball numbered the first. The position of each ball in \(CS_{1}\) can be determined by its azimuth angle \(\theta_{i}\). Here, it is assumed that the first ball is on the \(x_{1}\) axis and \(\theta_{1}=0\). The mobile coordinate system \(CS_{2}\) is anchored at the \(i\)-th ball, its origin is situated at the ball's center, the \(x_{2}\) axis is tangent to the helix and points in the positive direction, and the \(y_{2}\) axis points to the screw center along the radial direction. The force of the ball on the screw is a spatially distributed force system and the mobile coordinate system \(CS_{3}(x_{3},y_{3},z_{3})\) is established to simplify the equivalent forces. Its axis is parallel to the global coordinate system \(CS_{0}\), and the origin is on the screw axis and aligns with the contact point of the \(i\)-th ball with the screw. As shown in Figure 1b, the load of the \(i\)-th ball on the screw is decomposed into the \(CS_{3}\) coordinate system, as expressed in Equation (1). \[\left\{ \begin{array}{l}Q_{sxi}=-Q_{i}\cos\alpha_{0}\cos\theta_{i}-Q_{i}\sin \alpha_{0}\sin\theta_{i}\sin\lambda\\ Q_{syi}=-Q_{i}\cos\alpha_{0}\sin\theta_{i}+Q_{i}\sin\alpha_{0}\cos\theta_{i} \sin\lambda\\ Q_{szi}=-Q_{i}\sin\alpha_{0}\cos\lambda\\ M_{sxi}=Q_{szi}\cdot y_{si}\\ M_{syi}=Q_{szi}\cdot x_{si}.\end{array} \right. \tag{1}\]  
When the left side of the nut bears a leftward pulling force, the nut is stretched and the screw is compressed. In the \(CS_{0}\) coordinate system, the axial elastic deformation of the screw and nut is given by Equation (2) [20]. \[\left\{ \begin{array}{l}\Delta_{ni}=\frac{\Delta LF_{ni}}{E_{ss}}\\ \Delta_{si}=-\frac{\Delta LF_{ni}}{E_{s}S_{s}}.\end{array} \right. \tag{2}\]  
Here, \(F_{si}\) and \(F_{ni}\) are the internal forces experienced by the nut and the screw on the left side of the \(i\)-th ball's contact point, as expressed in Equation (3). \[F_{ci}=F_{a}-\sum\limits_{j=1}^{i-1}Q_{j}\sin\alpha_{0}\cos\lambda\quad(\xi=n,s). \tag{3}\]  
Given that the nut's length is significantly shorter compared to the screw's length, and it is usually supported by two parallel linear guides, its elastic deformation is far less than the deflection of the screw. Hence, the radial elastic deformation of the nut is ignored. The deflection caused by the equivalent force and torque at each contact point of the screw can be represented by Equations (4) and (5), where \(L_{1}=L-L_{j}\), \(L_{2}=L_{i}-L_{j}\)[20]. \[\omega_{Qqi}=\sum\limits_{j=1}^{i-1}\frac{Q_{sqi}L_{i}}{6k_{s}IL}\Big{\{}\frac {L_{1}}{L_{2}}L_{3}^{3}+\Big{[}L^{2}-{L_{1}}^{2}\Big{]}L_{i}-L_{i}^{3}\Big{\}} +\sum\limits_{j=i}^{N}\frac{Q_{sqi}L_{i}L_{i}}{6k_{s}IL}\Big{[}L^{2}-{L_{1}}^{ 2}-L_{i}^{2}\Big{]} \tag{4}\]  
Figure 1: (**a**) Coordinate systems on nut and screw, (**b**) resolution and simplification of contact load.