# Damage Analysis of Rolling Stock Automatic Coupler under Cyclic Loads
## 3 Results
### Damage Summation under Stationary Load
In the case of stress-controlled loading, the number of cycles of the calculated low-cycle fatigue curve \(k_{N}\) is required to estimate the damage \(d_{F}\), which reflects only fatigue failure. For the purpose of determination of this curve, the stress-controlled loading is further analyzed as a non-stationary strain-controlled loading, with the value of fatigue damage per half-cycle expressed by Equation (12). Then, for the whole strain process to crack initiation, the following could be written [22]:  
\[\frac{\overline{\delta}_{1}}{k_{f1}}+\frac{\overline{\delta}_{2}}{k_{f2}}+ \frac{\overline{\delta}_{3}}{k_{f3}}+\dots+\frac{\overline{\delta}_{f}}{k_{f 4}}=1 \tag{12}\]  
\(\sum\limits_{1}^{k_{f1}}\overline{\delta}_{k}\)--the accumulated fatigue damage to crack initiation at the loading level corresponding to loop width \(\overline{\delta}_{1}\); \(\sum\limits_{1}^{k_{f2}}\overline{\delta}_{k}\)--the accumulated fatigue damage to crack initiation at the loading level corresponding to loop width \(\overline{\delta}_{2}\), etc. Following the series of mathematical actions, the following equation was obtained:  
\[\frac{\overline{\delta}_{1}\overline{\epsilon}_{1}^{\overline{\delta}_{1}}}{ C_{2}C_{3}^{\overline{\delta}_{3}}}+\frac{\overline{\delta}_{2}\overline{ \epsilon}_{3}^{\overline{\delta}_{3}}}{C_{2}C_{3}^{\overline{\delta}_{3}}}+ \frac{\overline{\delta}_{3}\overline{\epsilon}_{3}^{\overline{\delta}_{3}}}{ C_{2}C_{3}^{\overline{\delta}_{3}}}+\dots+\frac{\overline{\delta}_{k_{f}}\overline{ \epsilon}_{k_{f}}^{\overline{\epsilon}_{3}}}{C_{2}C_{3}^{\overline{\delta}_{3} }}=1 \tag{13}\]  
Or  
\[\sum\limits_{1}^{k_{f}}\overline{\delta}_{k}\overline{\epsilon}_{k}^{\overline {\epsilon}_{3}}=C_{2}C_{3}^{\overline{\delta}_{3}} \tag{14}\]  
where \(C_{2}=\frac{\overline{\delta}_{max}}{k_{f}^{\overline{\epsilon}_{2}}}\), \(C_{3}=2^{\overline{\epsilon}_{1}}C_{Figure 11 provides the quasi-static damage \(d_{Q}\) for the symmetric cycle calculated according to the actual loop width (dotted curve) and the mean loop width (dashed curve). Figure 11 proposes that the discrepancy between the curves is larger in the case of small numbers of half-cycles, and the final damage is almost identical. The total quasi-static damage calculated according to the actual loop width is \(d_{Q}=0.425\), and according to the mean loop width--\(d_{Q}=0.393\). Figure 12 represents the fatigue damage \(d_{F}\) for the \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline \multirow{2}{*}{**Damage**} & \multicolumn{8}{c}{\(\sigma_{\
u}\)**MPa**} \\ \cline{2-10}  & **410** & **385** & **362** & **335** & **315** & **295** & **275** & **255** \\ \hline \(d_{F}\) & 0.405 & 0.510 & 0.608 & 0.612 & 0.632 & 0.643 & 0.650 & 0.660 \\ \(d_{Q}\) & 0.689 & 0.637 & 0.443 & 0.388 & 0.356 & 0.337 & 0.288 & 0.263 \\ \(d_{total}\) & 1.090 & 1.148 & 1.051 & 1.000 & 0.988 & 0.980 & 0.938 & 0.923 \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Damage accumulation in case of symmetric loading. Figure 10: Symmetric cycle damage (continuous lines—total damage, dotted lines—fatigue damage, dashed lines—quasi-static damage). symmetric cycle calculated according to the actual loop width (dotted curve) and the mean loop width (dashed curve). The total fatigue damage calculated on the basis of the actual loop width is \(d_{F}=0.590\), and on the basis of the mean loop width--\(d_{F}=0.579\). As can be seen from Figures 11 and 12, the discrepancy is greater for fatigue damage than for quasi-static damage. Table 7 presented the fatigue \(d_{F}\), quasi-static \(d_{Q}\), and total damage \(d_{total}\) for asymmetric loading. It could be noted that the quasi-static damage is higher at higher load levels, while the fatigue damage is higher at lower load levels. The mean damage for asymmetric load \(d_{total}=0.96876\). Figure 13 shows the variation of the asymmetric cycle fatigue \(d_{F}\), quasi-static \(d_{Q}\), and total damage \(d_{total}\) depending on the number of half-cycles from 1 to a failure half-cycle \(k_{f}\).