# Influence of Structural Stiffness and Loss Factor on Railroad Vehicle Comfort
## 2 Formulation  
The model in Figure 1 is a uniform Euler-Bernoulli beam on two spring-damper sets. Let \(l\) be the length of the beam (equal, in our case, to the distance between axles, as mentioned in the introduction); \(l\), its cross section area moment of inertia; \(m\), its uniformly distributed mass (with \(\mu=m/l\), the mass per unit length); \(E_{r}\), the material storage modulus; and \(\eta\), its loss factor (so that \(E=E_{r}(1+i\eta)\)). The suspension is simplified as linear springs and dampers at each end of the beam with no unsprung mass considered. As shown in Figure 1, \(k\) and \(c\) are the suspension stiffness and damping constants, respectively; \(z_{1}\) and \(z_{2}\), the front and rear wheel vertical displacements, respectively; \(x\), the coordinate along the beam measured from the front axle; and \(w(x,t)\), the vertical displacement of the beam. The equation of motion is  
\[EIw^{IV}+\mu\bar{w}=f \tag{1}\]where Roman numerals indicate derivation with respect to \(x\) and dots derivation with respect to time. The right-hand side \(f\) (\(f(x,t)\), that is) is the load distribution on the beam (gravitational forces not included since displacements are measured from the static equilibrium position). In our case, \(f\) is concentrated at each end of the beam and corresponds to the spring and damper forces that can be written as  
\[f=(k(z_{1}-w_{0})+c(\dot{z}_{1}-\dot{w}_{0}))\delta(0)+(k(z_{2}-w_{l})+c(\dot{ z}_{2}-\dot{w}_{l}))\delta(l) \tag{2}\]  
where \(w_{0}=w(0,t)\), \(w_{l}=w(l,t)\), and \(\delta\) is [PERSON]'s delta. Displacements are expressed as the following superposition of \(n\) modes:  
\[w(x,t)=\sum_{i=1}^{n}W_{i}(x)\,q_{i}(t) \tag{3}\]  
where  
\[ \begin{array}{rcl}W_{1}(x)&=&1\\ W_{2}(x)&=&x-l/2\\ W_{i}(x)&=&\cos\beta_{i}x+\cosh\beta_{i}x-\frac{\cos\beta_{i}l-\cosh\beta_{i}l }{\sin\beta_{i}l-\sinh\beta_{i}l}\left(\sin\beta_{i}x+\sinh\beta_{i}x\right) \qquad i\geq 3\end{array}  \tag{4}\]  
The first two functions being the rigid body modes, while the rest are flexible modes for the free-free boundary conditions, with \(\beta_{i}l\approx(2i-3)\pi/2\) for \(i\geq 3\). In principle, the set of modes for the case of a simply supported beam could also be used. Nevertheless, in that case, all resulting equations except for the two involving rigid body modes become homogeneous, and, therefore, amplitudes obtained from those equations could be arbitrarily scaled. Introducing Expressions (2) and (3) in Equation (1), multiplying by mode \(W_{i}\), integrating along the length of the beam, and verifying that all modes (including rigid body modes) are orthogonal, the following equation is obtained:  
\[ \begin{array}{l}EI\delta_{1}^{4}\frac{m_{i}}{\dot{\mu}}\,q_{i}+m_{i}\dot{q}_ {i}=\\ \left(k(z_{1}-w_{0})+c(\dot{z}_{1}-w_{0})\right)W_{i}(0)+\left(k(z_{2}-w_{l})+ c(\dot{z}_{2}-\dot{w}_{l})\right)W_{i}(l)\end{array}  \tag{5}\]  
where mode \(W_{i}\) could be any of the \(n\) modes included (making Equation (5), in fact, a system of \(n\) ordinary differential equations), and where \(m_{i}=m\) for \(i=1,3,\ldots n\), and \(m_{2}=l_{g}\), the rigid beam moment of inertia with respect to its center of mass. Using Equation (3) to express displacements and velocities at each end of the beam (\(w_{0}\), \(\dot{w}_{0}\), \(w_{l}\), and \(\dot{w}_{l}\)) as a superposition of modes, the set of equations can be written as  
\[ \begin{array}{l}EI\delta_{1}^{4}\frac{m_{i}}{\dot{\mu}}q_{i}+m_{i}\dot{q}_{i }=(kz_{1}+cz_{1})\,W_{i}(0)+(kz_{2}+cz_{2})\,W_{i}(l)-\\ k\sum_{j=1}^{n}\left(W_{j}(0)W_{i}(0)+W_{j}(l)W_{i}(l)\right)q_{j}-c\sum_{j=1} ^{n}\left(W_{j}(0)W_{i}(0)+W_{j}(l)W_{i}(l)\right)\,q_{j}\qquad i=1,\ldots n \end{array}  \tag{6}\]  
Figure 1: Railway vehicle car model. Defining vectors \(\mathbf{q}\), \(\mathbf{W}\), \(\mathbf{W}_{0}\), and \(\mathbf{W}_{l}\) as  
\[\mathbf{q}=\left\{ \begin{aligned} & q_{1}\\ &\vdots\\ & q_{n}\end{aligned} \right\}\quad\mathbf{W}=\left\{ \begin{aligned} & W_{1}(x)\\ &\vdots\\ & W_{n}(x)\end{aligned} \right\}\quad\mathbf{W}_{0}=\left\{