# Influence of Non-Uniform Rail Loads on the Rotation of Railway Sleepers
## 3 Calculation Examples
### Introductory Notes  
A simplified model of vehicle movement in a curve was analysed (Figure 3, based on [22]). The simplification is based on the fact that we treated the vehicle as a stiff element, and, additionally, the rotation resulting from the deformation of the suspension springs was ignored. The reactions transferred through the vehicle wheels to both rails constituted--for a given value of the cant h, curve radius R, centre of mass position in relation to the rail heads h\({}_{c}\), and the vehicle speed v--constant rail loads different for both rails (symbols in Figure 3, P\({}_{z}\) and P\({}_{w}\)). Figure 3: Vehicle movement as a stiff element in a curve [22]. Symbols used: m—vehicle mass [kg]; g—standard gravity [m/s\({}^{2}\)]; v—train speed [m/s]; R—curve radius [m]; h\({}_{c}\)—vehicle centre of mass position over rail head [m]; h—track cant [m]; s—track width (rail gauge) [m]. The vertical reactions \(\mathrm{P_{z}}\) and \(\mathrm{P_{w}}\) and a horizontal reaction \(\mathrm{H_{z}}\) were determined using system equilibrium equations. These are expressed by the following formulas:  
\[\mathrm{H_{z}}=\mathrm{m}\frac{\mathrm{v}^{2}}{\mathrm{R}}\mathrm{*cos}\, \alpha-\mathrm{mg}\mathrm{*sin}\,\alpha\]  
\[ \begin{split} P_{z}&=m\cdot\left[g\cdot\left(\frac{1} {2}\cos\alpha-\frac{h_{\mathrm{c}}}{s}\sin\alpha\right)+\frac{\mathrm{v}^{2}}{ \mathrm{R}}\cdot\left(\frac{1}{2}\sin\alpha+\frac{h_{\mathrm{c}}}{s}\cos \alpha\right)\right]\\ P_{w}&=m\cdot\left[g\cdot\left(\frac{1}{2}\cos \alpha+\frac{h_{\mathrm{c}}}{s}\sin\alpha\right)+\frac{\mathrm{v}^{2}}{ \mathrm{R}}\cdot\left(\frac{1}{2}\sin\alpha-\frac{h_{\mathrm{c}}}{s}\cos \alpha\right)\right]\\ &\hskip 56.905512 ptH_{z}=m\cdot\frac{\mathrm{v}^{2}}{\mathrm{R}} \cdot\cos\alpha-mg\cdot\sin\alpha\end{split}  \tag{29}\]  
The trigonometric functions present in the formulas were determined from the following geometrical relationships:  
\[\sin\alpha=\frac{h}{s},\ cos\alpha=\frac{b}{s},\ h^{2}+b^{2}=s^{2},\ b=\sqrt{s ^{2}-h^{2}},\ \cos\alpha=\sqrt{1-\left(\frac{h}{b}\right)^{2}} \tag{30}\]  
In reference to the time-variable load--for the given vehicle axis configuration--multiple instances of geometrical unevenness of the rail were assumed. It was assumed--similarly to article [11]--that rail vibrations are driven by forces resulting from vertical displacements of the wheel along the unevenness for the given stiffness of the wheel-rail contact. This means that the force amplitude will be equal to the amplitude of the unevenness \(\mathrm{r_{0}}\) multiplied by the stiffness of the contact between the wheel and rail \(\mathrm{k_{c}}\), while the frequency of force changes will be based on the length of the wave of the unevenness \(\mathrm{L_{r}}\) and the given train speed v. If a single wave of unevenness described by the equation:  
\[r(x)=r_{0}\cdot\cos\frac{2\pi x}{L_{r}} \tag{31}\]  
is analysed, the driving force for the first wheel of the vehicle will have the following form:  
\[P(t)=r_{0}\cdot k_{c}\cos\frac{2\pi vt}{L_{s}}=\Delta P\cdot\cos\omega t \tag{32}\]  
where \(\Delta P=\mathrm{r_{0}}\mathrm{k_{c}}\), \(\omega=2\pi\mathrm{v}/\mathrm{L_{r}}\). The driving force for the following wheels will depend on the configuration of the train axes in reference to the given length \(\mathrm{L_{r}  
\(\mathrm{P_{w}}=64.1\) kN and \(\mathrm{P_{z}}=94.4\) kN. Examples of the calculations for the rails and sleepers for the speed \(\mathrm{v}=170\) km/h are presented in the following chapters. After determining the forces, the axle spacing corresponding to the actual spacing of the middle bogies EMU 250 Pendolino (Alstom, Warsaw, Poland) was assumed. Additionally, 8-m sections (the distance for which the track response is different than 0; 25.9 m in total) as well as the number of terms of the Fourier series expansion \(\mathrm{n}=3000\) were adopted.