## 3 Mathematical Development
### Matrix Formulation of the Model
\begin{array}{c}k_{11}=k_{1}+k_{2}+k_{3}+k_{4}\\ k_{12}=k_{21}=\frac{b}{2}(-k_{1}+k_{2}+k_{3}-k_{4})\\ k_{13}=k_{31}=\frac{t}{2}(-k_{1}+k_{2}+k_{3}-k_{4})\\ k_{22}=\frac{b^{2}}{4}(k_{1}+k_{2}+k_{3}+k_{4})\\ k_{23}=k_{32}=\frac{tL}{2}(-k_{1}+k_{2}+k_{3}-k_{4})\\ k_{33}=\frac{t^{2}}{4}(k_{1}+k_{2}+k_{3}+k_{4})\end{array} \right.. \tag{5}\]  
It must be mentioned that using springs with stiffness, \(k\), at the four corner nodes of the plate contributes to rendering the previous stiffness matrix diagonal, which led to significantly simplifying the problem complexity once a set of uncoupled systems of equations is attained:  
\[\left[ \begin{matrix}4k&0&0\\ 0&\frac{b^{2}}{4}(4k)&0\\ 0&0&\frac{L^{2}}{4}(4k)\end{matrix} \right]\left\{ \begin{matrix}u_{G}\\ \theta_{x}\\ \theta_{y}\end{matrix} \right\}=\left\{ \begin{matrix}F_{U_{G}}\\ M\theta_{x}\\ M\theta_{y}\end{matrix} \right\}. \tag{6}\]  
As an illustration, if a static concentrated transverse force \(F_{1}\) at node 1 is considered prior to evaluating the _RHS_ (Right Hand Side) of equation system (6), it is important to recover the transverse displacement at node 1 presented in Equation (1). Hence, the work \(W\) released by a force term of \(F_{1}\) is:  
\[W(F_{1})=F_{1}\times u_{1}=F_{1}\times[u_{G}-(b/2)\times\theta_{x}-(L/2) \times\theta_{y}]. \tag{7}\]  
Applying [PERSON]'s theorem [25], the external force system, performing the work equivalent to the one due to force \(F_{1}\), can be formulated as:  
\[\left\{\begin{matrix}\frac{\partial W(F_{1})}{\partial U_{G}}=\frac{\partial[ F_{1}\times[u_{G}-(b/2)\times\theta_{x}-(L/2)\times\theta_{y}]]}{\partial U _{G}}=F_{1}=F_{u_{G}}\\ \frac{\partial W(F_{1})}{\partial\theta  
A dynamic analysis of structures requires not only the definition of their internal reactions due to external force systems but also the computation of inertial forces as a consequence of the acceleration field. To achieve this goal, it is necessary to evaluate the structure mass matrix, a task that can be solved by variational techniques on evaluating the work performed by inertial forces (involving accelerations) on the respective displacement field. Considering that the bus platform behaves as a plane, any displacement at a point can be interpolated from the 4 nodal displacements using shape or interpolation polynomial functions. Then, in a matrix, the displacement vector of any internal point can take the following form:  
\[u(x,y)=\{N_{1}\quad N_{2}\quad N_{3}\quad N_{4}\} \begin{cases}u_{1}\\ u_{2}\\ u_{3}\\ u_{4}\end{cases} , \tag{12}\]  
where \(N_{1}\dots N_{4}\) and \(u_{1}\dots u_{4}\) are accordingly the nodal shape functions and the displacement. Indeed, a similar interpolation applies to the acceleration field, which is calculated in a similar mode to the displacements:  
\[\ddot{u}(x,y)=\{N_{1}\quad N_{2}\quad N_{3}\quad N_{4}\} \begin{cases}\ddot{u}_ {1}\\ \ddot{u}_{2}\\ \ddot{u}_{3}\\ \ddot{u}_{4}\end{cases}  \tag{13}\]  
Table 2 reports the corresponding interpolation functions. Figure 4 represents the finite element with four nodes. Assuming that each variable is attributed to the relation of \(x=\frac{L}{2}(1+\xi);y=\frac{b}{2}(1+\eta)\), the corresponding differential relation is described as \(dx=\frac{L}{2}d\xi;dy=\frac{b}{2}d\eta\). Hence, the work performed by inertial forces, assuming a consistent mass distribution, takes the following form:  
\[U_{Inertia}=\left\{\delta_{1}\quad\delta_{2}\quad\delta_{3}\quad(\delta_{1}+ \delta_{3}-\delta_{2})\right\}\rho h\left(\int_{-1}^{1}\int_{-1}^{1} \begin{pmatrix} N_{1}\\ N_{2}\\ N_{3}\\ N_{4}\end{pmatrix} \{N_{1}\quad N_{2}\quad N_{3}\quad N_{4}\}d\xi d\eta\right) \begin{pmatrix}\ddot{u}_{1}\\ \ddot{u}_{2}\\ \ddot{u}_{3}\\ \ddot{u}_{4}\end{pmatrix} . \tag{14}\] \begin{table}
\begin{tabular}{c c} \hline \hline
**Shape \(N_{i}\)** & **Product Shape** \\ \hline
1 & \(N_{1}=(1-\xi)(1-\eta)/4\) \\
2 & \(N_{2}=(1+\xi)(1-\eta)/4\) \\
3 & \(N_{3}=(1+\xi)(1+\eta)/4\) \\
4 & \(N_{4}=(1-\xi)(1+\eta)/4\) \\ \hline \hline \end{tabular}
\end{table}