## 6 Dynamic Effect of Vehicle Suspensions  
Significant differences between the Olson model and the dynamic simulation occurred in the average load of 54.6% and the maximum load of 37.2%, as shown in Table 4. These differences of impact forces transmitted to the DCP are caused due to the dynamic motions of the masses of the wheelsets, the bogie frame, and the vehicle body. In the case of the Olson model, it is assumed that all the masses of the vehicle are transmitted to the DCP all at once. However, in the case of the DCP model, because the masses of the vehicle bodies are transmitted through the suspensions after impact, the dynamic effects of suspensions and masses should be considered to obtain an actually loaded force to the DCP in time. By the way, the dynamic effects can be considered by a simplified spring-mass model using the lateral stiffnesses of the suspensions calculating how much inertia force of the masses is gradually transmitted to the DCP. When a vehicle impacts with a DCP, the masses are transmitted in the sequential order of wheelset, bogie frame and car body. An impact between the DCP and the vehicle can be simplified to a spring-mass model as shown in Figure 11. The left side is a simplified free body diagram with 1 degree of freedom, and the right side is a simplified free body diagram with 3 degrees of freedom. \(m_{t}\) is the mass of the vehicle. \(m_{1}\) is the mass of the wheelset and \(k_{1}\) is the contact stiffness between the wheelset and the DCP. \(m_{2}\) is the mass of the bogie excluding the wheelset, and \(k_{2}\) is the sum of the lateral stiffnesses of the primary suspensions connected to the wheelset and the bogie frame. \(m_{3}\) is the mass of the car body excluding the bogie, and \(k_{3}\) is the sum of the lateral stiffnesses of the secondary suspensions connected to the car body and the bogie frame. The impact force applied to the DCP can be calculated as \(k_{1}x_{t}\) and \(k_{1}x_{1}\). x(t) means the displacement of a mass. The equations of motion are given as follows. 1 Degree of Freedom  
\[\sum F_{mt}=-k_{1}x_{t}=m_{t}\ddot{x_{t}}\to m_{1}\ddot{x_{1}}+k_{1}x_{t}=0\]  
The solution can be assumed to \(x(t)=ae^{i\omega t}\), and by substituting the above equation and putting initial conditions, the following Equation (5) can be finally obtained. \[x(t)=\frac{\sqrt{\omega^{2}x_{0}^{2}+v_{0}^{2}}}{w}sin(\omega t+\phi),\ \phi=tan^{-1}(\frac{\omega x_{0}}{v_{0}}) \tag{5}\]  
2 3 Degrees of Freedom  
\[\sum F_{m1}=-k_{1}x_{1}+k_{2}(x_{2}-x_{1})=m_{1}\ddot{x_{1}}\to m_{1} \ddot{x_{1}}+(k_{1}+k_{2})x_{1}-k_{2}x_{2}=0\]  
\[\sum F_{m2}=k_{2}(x_{2}-x_{1})-k_{3}(x_{3}-x_{2})=m_{2}\ddot{x_{2}}\to m_{2} \ddot{x_{2}}-k_{2}x_{1}+(k_{2}+k_{3})x_{2}-k_{3}x_{3}=0\]  
\[\sum F_{m3}=k_{3}(x_{3}-x_{2})=m_{3}\ddot{x_{3}}\to m_{3}\ddot{x_{3}}+k_{3}(x_ {3}-x_{2})x_{1}=0\]  
Figure 11: Simplified spring–mass model for a train vehicle. The three equations can be written in the form of a matrix equation. \[ \begin{bmatrix}m_{1}&0&0\\ 0&m_{2}&0\\ 0&0&m_{3}\end{bmatrix}\begin{vmatrix}\ddot{x_{1}}\\ \ddot{x_{2}}\\ \ddot{x_{3}}\end{vmatrix} + \begin{bmatrix}k_{1}+k_{2}&-k_{2}&0\\ -k_{2}&k_{2}+k_{3}&-k_{3}\\ 0&-k_{3}&k_{3}\end{bmatrix}\begin{vmatrix}x_{1}\\ x_{2}\\ x_{3}\end{vmatrix} =0=>\text{M}\ddot{\mathbf{x}}+\text{K}\mathbf{x}=0\]  
The solution can be assumed to \(\mathbf{x}(t)=\mathbf{u}e^{i\omega t}(j=\sqrt{-1},\mathbf{u}\
eq 0,\big{)}\) in a vector form and by substituting the equation, the following equation can be obtained. \[(-\omega^{2}\text{M}+\text{K})\mathbf{u}=0,\text{ for motion to occur }\mathbf{u}\
eq 0=>\det(-\omega^{2}\text{M}+\text{K})=0\]  
\[\det \begin{bmatrix}(k_{1}+k_{2})-m_{1}\omega^{2}&-k_{2}&0\\ -k_{2}&(k_{2}+k_{3})-m_{2}\omega^{2}&-k_{3}\\ 0&-k_{3}&k_{3}-m_{3}\omega^{2}\end{bmatrix} =0\]  
The characteristic equation is defined from the \(\det(-\omega^{2}\text{M}+\text{K})=0\) and once \(\omega\) is known from the equation, we can calculate the corresponding vector \(\mathbf{u}(u_{1},u_{2},u_{3})\).