## 5 Examples
### Appropriate Location of RER
One can conclude that for the continuous beam with the shorter length, RER should be arranged in the middle of the beam, while for the longer length, RER should be arranged at the two ends of the beam, which is conducive to reducing the maximum additional force of rail. As shown in Figure 7, with the length of 120 m, the values in the curves of \"TF Left\" and \"CF Left\" are \(2.2595\times 10^{5}\) and \(-1.2195\times 10^{5}\) N, respectively, and those in the curves of \"TF Mid\" and \"CF Mid\" are \(1.2242\times 10^{5}\) and \(-1.8952\times 10^{5}\) N, respectively. The large value (absolute value) of \(2.2595\times 10^{5}\) N of the former is greater than that of \(-1.8952\times 10^{5}\) N of the latter. That is to say, it is advantageous for RER to be laid above the middle of the beam. With the length of 240 m, the values in the curves of \"TF Left\" and \"CF Left\" are \(2.4378\times 10^{5}\) and \(-1.2195\times 10^{5}\) N, respectively, and those in the curves of \"TF Mid\" and \"CF Mid\" are \(2.2113\times 10^{5}\) and \(-3.2543\times 10^{5}\) N, respectively. The large value (absolute value) of \(2.4378\times 10^{5}\) N of the former is less than that of \(-3.2543\times 10^{5}\) N of the latter. In other words, it is advantageous for RER to be laid above the ends of the beam. With the length of 164 m, the calculated values of \"TF Left\" and \"CF Left\" are \(2.4029\times 10^{5}\) and \(-1.2195\times 10^{5}\) N, respectively; those of \"TF Mid\" and \"CF Mid\" are \(1.6991\times 10^{5}\) and \(-2.3998\times 10^{5}\) N, respectively; the respective large values (absolute values) are approximately equal, with a difference of only 0.13%. Consequently, the critical length of the continuous beam, where RER is arranged at above the middle and the ends of the beam, is 164 m. When the length of the continuous beam is shorter than the critical length, RER should be arranged above the middle of the beam; otherwise, RER should be arranged above the two ends of the beam. In addition, Figure 7 shows that the maximum additional tension force of rail will converge to a certain value, which is 243,967.5 N. The explanation is as follows. A mini segment of rail with the length of \(\mathrm{d}x\) is taken, as shown in Figure 8. Based on the equilibrium condition of longitudinal forces acting on the mini segment, the equation can be written as  
\[P+q\cdot\mathrm{d}x =P+\mathrm{d}P \tag{6}\] \[\frac{\mathrm{d}P}{\mathrm{d}x} =q \tag{7}\]  
in which \(P\) represents the additional force of rail. From Equation (7), if \(q\) is equal to 0 at a certain section of rail, then \(P\) is the extreme value. In fact, it can be judged by the condition that the direction of \(q\) is reversed. Where the direction of \(q\) is reversed, one position is at the same displacements of beam  
Figure 7: The curves of the minimum values of the maximum additional forces of rail changing with the length of the continuous beam for RER located at three positions. In the legend, CF and TF represent the compression and tension forces, respectively, and Left, Mid and Right represent RER located the above the left end, the middle, and the right end of the continuous beam, respectively. and rail, i.e., \(u_{\rm R}=u_{\rm B}\), and the other position is at the junction of beams or at the junction between beam and embankment. The maximum additional tension force of rail must occur at the section with \(u_{\rm R}=u_{\rm B}\), and it must be above the beam.