Benchmark solution
=====================

Calculation method
-----------------

Reminders
~~~~~~

Modeling BARRE does not transmit any shear force or bending moment. Therefore, if we note :math:`E` the Young modulus of the element, :math:`A` the area of its section, the area of its section and :math:`L` its length, the elementary stiffness matrix :math:`{K}^{\mathit{elem}}` of a bar is as follows (with the components in the order :math:`({\mathrm{DX}}_{1}{\mathrm{DY}}_{1}{\mathrm{DZ}}_{1}{\mathrm{DX}}_{2}{\mathrm{DY}}_{2}{\mathrm{DZ}}_{2})`):

:math:`{K}^{\mathit{elem}}\mathrm{=}(\begin{array}{cccccc}\frac{\mathit{EA}}{L}& 0& 0& \frac{\mathrm{-}\mathit{EA}}{L}& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ \frac{\mathrm{-}\mathit{EA}}{L}& 0& 0& \frac{\mathit{EA}}{L}& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array})`

The profile of the mass matrix :math:`{M}^{\mathit{elem}}` is different from the stiffness matrix because the mass must be taken into account in all directions of space. Thus, if we write :math:`\rho` the density of the element, the elementary mass matrix, with :math:`m\mathrm{=}\rho AL`, is as follows:

:math:`{M}^{\mathit{elem}}\mathrm{=}(\begin{array}{cccccc}m\mathrm{/}3& 0& 0& m\mathrm{/}6& 0& 0\\ 0& m\mathrm{/}3& 0& 0& m\mathrm{/}6& 0\\ 0& 0& m\mathrm{/}3& 0& 0& m\mathrm{/}6\\ m\mathrm{/}6& 0& 0& m\mathrm{/}3& 0& 0\\ 0& m\mathrm{/}6& 0& 0& m\mathrm{/}3& 0\\ 0& 0& m\mathrm{/}6& 0& 0& m\mathrm{/}3\end{array})`


Calculation of internal forces and support reactions
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let :math:`m` be the mass of the element and :math:`U` the displacement, so if we choose a gravity field according to :math:`X`, the external forces at each node are as follows: :math:`(\mathrm{-}m\mathrm{/}\mathrm{2,0}\mathrm{,0})`. For elastic stiffness behavior :math:`K`, the only non-imposed displacement :math:`{U}_{{B}_{X}}` is equal to :math:`\frac{{F}_{{B}_{X}}^{\mathit{ext}}}{K}\mathrm{=}\frac{\mathrm{-}m}{\mathrm{2K}}`.

If we write the elementary stiffness matrix :math:`{K}^{\mathit{elem}}`, we have the relationship :math:`{F}^{\mathit{int}}\mathrm{=}{K}^{\mathit{elem}}U`. The following results are obtained for internal forces: :math:`{F}_{{O}_{X}}^{\mathit{int}}\mathrm{=}\frac{m}{2}`, :math:`{F}_{{O}_{Y}}^{\mathit{int}}\mathrm{=}0`,,, :math:`{F}_{{O}_{Z}}^{\mathit{int}}\mathrm{=}0`, :math:`{F}_{{B}_{X}}^{\mathit{int}}\mathrm{=}\frac{\mathrm{-}m}{2}`, :math:`{F}_{{B}_{Y}}^{\mathit{int}}\mathrm{=}0`, and :math:`{F}_{{B}_{Z}}^{\mathit{int}}\mathrm{=}0`.

By noting :math:`{R}^{\mathit{ap}}` the supporting reactions, we have the :math:`{R}^{\mathit{ap}}\mathrm{=}{F}^{\mathit{int}}\mathrm{-}{F}^{\mathit{ext}}` relationship. So we easily have: :math:`{R}_{{O}_{X}}^{\mathit{ap}}\mathrm{=}m` and all the other zero components.

*Note: for the loading directions* :math:`Y` *and* :math:`Z` *, there is no displacement so the internal forces are zero and the support reactions are equal to the external forces.*


Projection of the mass matrix on a modal basis
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The purpose of this calculation is to verify that for a unitary mode of movement :math:`\phi` in a given direction, we have equality:

:math:`{\phi }^{T}M\phi \mathrm{=}m`


Calculation of kinetic energy
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The purpose of this calculation is to verify that for a unit speed mode :math:`\phi` in a given direction, we have equality:

:math:`\frac{1}{2}{\phi }^{T}M\phi \mathrm{=}\frac{\mathit{mv}}{2}` with :math:`v\mathrm{=}1`