\subsection{The unit cell}
\label{sec:unit_cell}
The area of a hexagonal cell with sides of length $a$, as shown in Figure \ref{fig:hex_unit} is given by
\begin{equation}
    \begin{split}
        A &= 6\frac{a^2}{2}\sin \left( \frac{\sqrt{3}}{2}\right)\\
        A &= \frac{3\sqrt{3}a^2}{2}
    \end{split}
\end{equation}
The area of fibers in the unit cell is given by
\begin{equation}
    A_f = 3\pi r_f^2
\end{equation}
Thereby the volume fraction of fibers in the composite can be related to the unit cell parameter $a$ and the radius of the fibers by
\begin{equation}
    V_f = \frac{A_f}{A} = \frac{2 \pi r_f^2}{\sqrt{3}a^2}.
\end{equation}

The dimensions of the tetragonal unit cell can be related to the dimensions of the hexagonal cell by
\begin{equation}
    \begin{split}
        a_2 &= 2a\cos \left(\frac{\pi}{6}\right) = a\sqrt{3} \\
        a_3 &= a
    \end{split}
\end{equation}

\begin{figure}[htb]
    \centering
    \includegraphics[width=.7\textwidth]{figs/tetragonal_unit_cell.png}
    \caption{Unit cell of a hexagonal fiber composite}
    \label{fig:hex_unit}
\end{figure}

For all simulations, the parameter $a$ was kept constant and equal to unity, while the fiber radius and unit cell depth was varied.

