# Railway Bridge Condition Monitoring Using Numerically Calculated Responses from Batches of Trains
## Appendix A Inverse Newmark-\(\beta\) Method of 4-Axle Railway Carriage Model  
Hence, for the inverse problem, \(\ddot{u}_{b1},\ddot{u}_{b2},\dot{\theta}_{b1}\),and \(\dot{\theta}_{b2}\) are known inputs. Consistent with Equation (10), other values can be obtained by integration of the inputs:  
\[\dot{u}_{b1,t+\Delta t}=\dot{u}_{b1,t}+a_{6}\ddot{u}_{b1,t}+a_{7}\ddot{u}_{b1,t+\Delta t} \tag{11}\]  
\[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]  
\[\dot{u}_{b2,t+\Delta t}=\dot{u}_{b2,t}+a_{6}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t} \tag{13}\]  
\[u_{b2,t+\Delta t}=\left(\ddot{u}_{b2,t+\Delta t}+a_{2}\dot{u}_{b2,t}+a_{3} \ddot{u}_{b2,t}\right)/a_{0}+u_{b2,t} \tag{14}\]  
\[\ddot{\theta}_{b1,t+\Delta t}=\left(\dot{\theta}_{b1,t+\Delta t}-\dot{\theta} _{b1,t}-a_{6}\ddot{\theta}_{b1,t}\right)/a_{7} \tag{15}\]  
\[\theta_{b1,t+\Delta t}=\left(\dot{\theta}_{b1,t+\Delta t}+a_{4}\dot{\theta}_{ b1,t}+a_{5}\ddot{\theta}_{b1,t}\right)/a_{1}+\theta_{b1,t} \tag{16}\]  
\[\ddot{\theta}_{b2,t+\Delta t}=\left(\dot{\theta}_{b2,t+\Delta t}-\dot{\theta} _{b2,t}-a_{6}\ddot{\theta}_{b2,t}\right)/a_{7} \tag{17}\]  
\[\theta_{b2,t+\Delta t}=\left(\dot{\theta}_{b2,t+\Delta t}+a_{4}\dot{\theta}_{ b2,t}+a_{5}\ddot{\theta}_{b2,t}\right)/a_{1}+\theta_{b2,t} \tag{18}\]  
According to Equations (11)-(18), the equations of motion of the main body can be expressed:  
\[m_{p}\ddot{u}_{v,t+\Delta t}+(c_{s1}+c_{s2})\dot{u}_{v,t+\Delta t }-c_{s1}\dot{u}_{b1,t+\Delta t}-c_{s2}\dot{u}_{b2,t+\Delta t}+(L_{v1}c_{s1}-L _{v2}c_{s2})\dot{\theta}_{v,t+\Delta t}+(k_{s1}+k_{s2})u_{v,t+\Delta t} \tag{19}\] \[-k_{s1}u_{b1,t+\Delta t}-k_{s2}u_{b2,t+\Delta t}+(L_{v1}s_{1}-L _{v2}k_{s2})\theta_{v,t+\Delta t}=m_{v}g\] \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]0 (20) \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]1  
In the Newmark-\(\beta\) method,  
\[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]2  
\[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]3\[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]4 (A13) \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]5 (A14) Substituting (A11)-(A14) into (A9) and (A10) \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]6 \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]7 (A15) \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]8 \[u_{b1,t+\Delta t}=\left(\ddot{u}_{b1,t+\Delta t}+a_{2}\dot{u}_{b1,t}+a_{3} \ddot{u}_{b1,t}\right)/a_{0}+u_{b1,t} \tag{12}\]9 \[\dot{u}_{b2,t+\Delta t}=\dot{u}_{b2,t}+a_{6}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t} \tag{13}\]0 (A16) \[\dot{u}_{b2,t+\Delta t}=\dot{u}_{b2,t}+a_{6}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t}+a_{7}\ddot{u}_{b2,t} \tag{13}\]1  
In Equations (A15) and (A16), all symbols are given except \(u_{v,t+\Delta t}\) and \(\theta_{v,t+\Delta t}\). So \(u_{v,t+\Delta t}\) and \(\theta_{v,t+\Delta t}\) could be solved now and \(\ddot{u}_{v,t+\Delta t}\), \(\dot{u}_{v,t+\Delta t}\), \(\dot{v}_{v,t+\Delta t}\), \(\ddot{\theta}_{v,t+\Delta t}\) can be calculated from \(u_{v,t+\Delta t}\) and \(\theta_{v,t+\Delta t}\) using Equations (A11)-(A14). Therefore, all the symbols on the left side of Equation (1) are known and the right side of Equation (1), \(F\) can be found, say:  
\[F=\left\{ \begin{array}{c}F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\end{array}