## 3 Mathematical Development
### Transient-Dynamic Formulation
\begin{split}\big{(}\rho hLb+4(c)\times\Delta t+0.25\Delta t^{2} (4k)\big{)}\underbrace{\ddot{\mathbf{U}}_{G\ +\Delta t}}_{at\ t\ +\Delta t}=\mathbf{F}_{t+\Delta t}-\\ \underbrace{(4k)\Big{(}\mathbf{U}_{G\ t}+\dot{\mathbf{U}}_{G\ t}\ \Delta t+0.25\Delta t^{2}\ddot{\mathbf{U}}_{G\ t}\ \Big{)}-(4c)\Big{(}\ddot{\mathbf{U}}_{G\ t}\ +0.5 \Delta t\ddot{\mathbf{U}}_{G\ t}\ \Big{)}}_{at\ t},\\ \big{(}\rho hLb^{3}/12+\big{(}4 cb^{2}/4\big{)}\times\Delta t+0.25 \Delta t^{2}\big{(}4 kb^{2}/4\big{)}\big{)}\underbrace{\ddot{\mathbf{\beta}}_{x\ t +\Delta t}}_{at\ t+\Delta t}=\mathbf{M}_{x\ t+\Delta t}-\\ \underbrace{\Big{(}4 kb^{2}/4\Big{)}\Big{(}\theta_{x\ t}+\dot{ \theta}_{x\ t}\ \Delta t+0.25\Delta t^{2}\ddot{\theta}_{x\ t}\ \Big{)}-\Big{(}4 cb^{2}/4\Big{)}\Big{(}\dot{\theta}_{x\ t}\ +0.5 \Delta t\ddot{\theta}_{x\ t}\ \Big{)}}_{at\ t},\\ \big{(}\rho hLb^{3}/12+\big{(}4 cb^{2}/4\big{)}\times\Delta t+0.25 \Delta t^{2}\big{(}4 kb^{2}/4\big{)}\big{)}\underbrace{\ddot{\mathbf{\beta}}_{y\ t +\Delta t}}_{at\ t+\Delta t}=\mathbf{M}_{y\ t+\Delta t}-\\ \underbrace{\Big{(}4 kb^{2}/4\Big{)}\Big{(}\theta_{y\ t}+\dot{ \theta}_{y\ t}\ \Delta t+0.25\Delta t^{2}\ddot{\theta}_{y\ t}\ \Big{)}-\Big{(}4 cb^{2}/4\Big{)}\Big{(}\dot{\theta}_{y\ t}\ +0.5 \Delta t\ddot{\theta}_{y\ t}\ \Big{)}}_{at\ t}.\end{split}  \tag{29}\]  
As an illustration, the analysed plate provided the largest natural frequency \(f_{3}=2.0\ (\mathrm{Hz})\), where \(t_{3}=0.5\ (\mathrm{s})\) and thus, \(\Delta t\leq 0.5\ (\mathrm{s})\). With this value, the solution is warranted for its stability. Hence, the smaller the time step, the better the accuracy, although a value much smaller than the recommended one would be useless since the computational time would increase largely. For instance, if a vehicle accelerates during \(t=0.5\) (s) with no damping, an impulsive force is due to the load on the front axis during \(F=2\times 2.5\) (\(\mathrm{kN}\)) = 5.0 (\(\mathrm{kN}\)) (hubs 1 and 4). Basic equation for centre of mass vibration with no damping is:  
\[4\times 185\times 10^{3}u_{G}+12\times 10^{3}\ddot{u_{G}}=\underbrace{5\times 10^{3 }}_{\left(for\ t\in\ [0,\ 0.5(s)]\right)} \tag{31}\]  
Regarding the natural frequency detected by [PERSON]'s method, one cycle = 1.02 (s); therefore, an analytical evaluation can lead to:  
\[f=1/(2\pi)\times\sqrt{k/m}=1.25\ (\mathrm{Hz}) \tag{32}\]  
Referring to Figure 6, the maximum displacement of the mass centre is \(u_{G}^{Max}=1.23\times 10^{-2}\ (\mathrm{m})\) at the transient step. Concerning the rotation about \(y-axis\) (forces are supposed to act at nodes 1 and 4), 2 nd uncoupled equation: dynamic inertial moment over transverse \(y-axis\). The dimensions of the bus body chassis are listed below. * Rotation stiffness: \(\mathrm{kL}^{2}\);
* Equivalent stiffness: \(k_{Eq}=6439850\ (\mathrm{N/rad})\);
* Equivalent mass: \(J_{y}\ =144\times 10^{3}\ (\mathrm{kg.m}^{2})\);
* Natural frequency of this oscillation: \(f=1/(2\pi)\times\sqrt{k_{Eq}/J_{y}}\ =1.065\ (\mathrm{Hz})\). * Equivalent forces: \[ \begin{cases}F_{u_{G}}(t)\\ M_{\theta_{x}}(t)\\ M_{\theta_{y}}(t)\end{cases} = \begin{bmatrix}1&1&1&1\\ -b/2&-b/2&b/2&b/2\\ -L/2&-L/2&L/2&-L/2\end{bmatrix}\begin{cases}F_{Accel}\\ 0\\ F_{Accel}\end{cases} .\] (33)  
Rotation \(\theta_{y}\) after applying \(F=2\times 2.5\) (\(\mathrm{kN}\)) suddenly (and sustainably) at nodes 1 and 4. Thus, the equivalent moment is \(M=-30.0\) (\(\mathrm{kN.m}\)). From the graph plotted in Figure 7, the maximum rotation is obtained as \(\theta_{y}=-0.00929\ (\mathrm{rad})\). Hence, analytically, true displacement at front of vehicle model (nodes 1 and 4):  
* Maximum rotation obtained as \(\theta_{y}\ =-8.88\times 10^{-3}\ (\mathrm{rad})\);
* Maximum displacement: \(u_{z}^{max}=u_{G}^{max}-\frac{L}{2}\theta_{y}\ =68.0\ (\mathrm{mm})\). Figure 6: Displacement of mass centre, \(u_{G}\). At the rear (nodes 2 and 3), there will be a lowering of the pavement level, thus:  
* Minimum displacement: \(u_{z}^{min}=u_{G}^{max}+\frac{L}{2}\theta_{y}\ =-43.4\ (\mathrm{mm})\).