\section{Theoretical equations}
\label{sec:theory}

\subsection{Loads}
\label{load_explanation}
The beam needs to support its own weight and the weight of the car. The load from the vehicle can be computed as $P = mg$, with m = 2000 kg and g = 9,81 $m.s^{-2}$, we obtain that $P_{tot} = 19620$ N. This load will be assumed subdivided equally between the 4 wheels so for each beam we can consider the application of two forces $P = 4905N$ that have a distance from the centre of the beam of $b = 2.5\si{\meter}$ as can be represented by the figure  \ref{fig:load_simplify}. In the same way, this study will analyse the worse scenario for the beam. Its mean that $a = 3,75\si{\meter}$. The gravity load is distributed along the beam. The constant used in this formula are explained in Figure \ref{fig:load_simplify}.

\begin{figure}[h!]
    \centering
    \includegraphics[width=.5\textwidth]{Figures/Beam_formula.PNG}
    \caption{Parameters used in computation of beam loading.}
    \label{fig:load_simplify}
\end{figure}

The maximum momentum at the center can be computed using this equation given by the literature \cite{matte3}
\begin{equation}
M_{max}=Pa
\label{max_load}
\end{equation}

In the previous computation, the weight of the beam is neglected. The following computation proves that its influence can be neglected. Considering that the weight of the beam is 60kg and that it is evenly distributed as a pressure Q, we obtain that the maximum moment in the beam at $x=\frac{L}{2}$ is

\begin{equation}
    \begin{split}
        M_{max} &= Pa+Q(\frac{L}{2}x-L)\\
        M_{max} &= \SI{15480}{\newton\meter}.
    \end{split}
\end{equation}

Without the weight, the maximum moment is,

\begin{equation}
    \begin{split}
        M_{max} &= Pa\\
        M_{max} &= \SI{15450}{\newton\meter}.
    \end{split}
\end{equation}

The weight influence is equal to 0,2\% and can be thereby neglected, even for weights far exceeding \SI{60}{\kilo\gram}.

\subsection{2D failure criterion}
\label{Hashin_2D_part}
Here, in this case the Hashin 2D criterion is possible to use, in fact, the force applied on the beam is only following the x and y axis. It is noted that the resulting 2D-criterion may be erroneous at edges, as the plane stress assumption is generally not valid at free edges. 

\begin{equation}
f_{FF} = \max{\left( \frac{\sigma_1}{X_T}, -\frac{\sigma_1}{X_C}\right)}
\end{equation}

for fiber failure criterion.

\begin{equation}
f_{IFF} = \left(\frac{\sigma_2}{Y_T}\right)^2 + \left(\frac{\tau_{12}}{S_{12}}\right)^2, \hspace{.5cm} \sigma_2 > 0,
\end{equation}

for tensile loads, and

\begin{equation}
f_{IFF} = \left(\frac{\sigma_2}{2 S_{12}}\right)^2+  \left(\frac{\tau_{12} }{S_{12}}\right)^2 +
\left[ \left(\frac{Y_C}{2S_{12}}\right)^2 - 1 \right]\frac{\sigma_2}{Y_C}, \hspace{.5cm} \sigma_2 < 0
\end{equation}

for compressive loads. In this analysis, both inter-fiber failure and fiber-failure will be regarded as ultimate failure of the material.

\subsection{Stiffness Matrix}

\label{st_matrix}

For computing the stiffness matrix of the laminate layup, the following computations are used. The compliance matrix is calculable with the relation

\begin{equation}
\mathbf{S}=\begin{bmatrix} 1/E_1 & -\nu_{12}/E_1 & 0 \\ -\nu_{12}/E_1 & 1/E_{2} & 0 \\ 0 & 0 & 1/G_{12} \end{bmatrix}.
\end{equation}

The 2D stiffness matrix is the inverse of the compliance matrix.

\begin{equation}
\mathbf{Q} = \mathbf{S}^{-1}
\end{equation}

Now it is possible to include the orientation of the matrix with the transformation,

\begin{equation}
\mathbf{Q{'}} = \mathbf{T}_{\sigma}^{-1}\mathbf{Q} \mathbf{T}_{\epsilon}.
\end{equation}

With the two transformation matrix expressed as

\begin{equation}
\mathbf{T}_{\sigma}(\theta)=\begin{bmatrix} c^2 & s^2 & 2cs \\ s^2 & c^2 & -2cs \\ -cs & cs & c^2-s^2 \end{bmatrix} ; \mathbf{T}_{\varepsilon} (\theta)=\begin{bmatrix} c^2 & s^2 & cs \\ s^2 & c^2 & -cs \\ -2cs & 2cs & c^2-s^2 \end{bmatrix}
\end{equation}

Where $c = \cos \theta$ and $s = \sin \theta$. The Q matrix components are described as

\begin{equation} \mathbf{Q}'=
\begin{bmatrix} Q_{xx} & Q_{xy}   & Q_{xs}     \\
                Q_{xy} & Q_{yy}   & Q_{ys}     \\
                Q_{xs} & Q_{ys}   & Q_{ss}     
\end{bmatrix}.              
\end{equation}

To compute $ \mathbf{A}$ from begin $\mathbf{Q}$, laminates loads and constitutive relations are used

\begin{equation}
\mathbf{A} = \sum_k \mathbf{Q}'_k (h_k - h_{k-1})
\label{Get_A}
\end{equation}

where $h_i$ is the distance from the center of the laminate to the top of layer $i$.

\subsection{Linear Buckling Analysis}
\label{Linear_Buckling_Analysis}
Buckling is one the main issue that the analytical approach cannot face but that is possible to verify through the use of FEA. 
Linear buckling is the most common type of analysis and is easy to execute, but it is limited in the results it can provide.

Linear-buckling analysis calculates buckling load magnitudes that cause buckling and associated buckling modes. Abaqus can provide calculations of a large number of buckling modes and the associated buckling-load factors (BLF). The BLF is expressed by a number which the applied load must be multiplied by to obtain the buckling-load magnitude.

The buckling mode presents the shape the structure assumes when it buckles in a particular mode, but says nothing about the numerical values of the displacements or stresses. The numerical values can be displayed, but are merely relative. This is in close analogy to modal analysis, which calculates the natural frequency and provides qualitative information on the modes of vibration (modal shapes), but not on the actual magnitude of displacements.

Theoretically, it is possible to calculate as many buckling modes as the number of degrees of freedom in the FEA model. In this case we looked only the first positive buckling mode and its associated BLF. This is because higher buckling modes have no chance of taking place — buckling most often causes catastrophic failure or renders the structure unusable.

The nomenclature is “the first positive buckling mode” because buckling modes are reported in the ascending order according to their numerical values. A buckling mode with a negative BLF means the load direction must be reversed (in addition to multiplying by the BLF magnitude) for buckling to happen.


\subsection{Safety factor}

A safety factor must be chosen for the design of the beam. This factor will be used to intentionally build the beam stronger than required for normal use, thereby giving enough resistance to the beam for emergency situations, unexpected loads, misuse, or degradation. Usually, safety factors are determined by experimentation. This is obviously impossible in the scope of this project. The safety factor has to be chosen with the parameter given.

Here, the problem is a static problem. The main problem here is that we have no information on how the condition will be, wind, bad supports, bad installation. The probability of a misuse are numerous, and a failure is likely to happen if we dimension our system for a vehicle of 2000kg only. 

It is commonly accepted that, for a static problem, the safety factor should be included between 2 and 4. In this case, a safety factor of 3 is chosen as a compromise. A too low safety factor will include some risk for the user, and one too big will mean that the system will be over designed. This factor mean that the beam should in principle be able to handle a car of 6000kg. It is estimated that this is enough for all situations that are likely to happen.