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

Equations of balance
---------------------

The study is carried out around the initial position of the structure in plane :math:`\mathrm{xy}`. The equations are written at the center of gravity of the beam.

Inertial force:

:math:`M\mathrm{.}{\gamma }_{g}=\left\{\begin{array}{}\mathrm{Mx}\text{'}\text{'}\\ \mathrm{My}\text{'}\text{'}\\ \frac{M{L}^{2}}{12}\mathrm{.}\theta \text{'}\text{'}\end{array}\right\}`

Effort at point :math:`\mathrm{A1}`

:math:`\mathrm{Fa}=\left\{\begin{array}{}-\mathrm{kxa}\mathrm{.}\delta \mathrm{xa}\\ -\mathrm{kya}\mathrm{.}\delta \mathrm{ya}\\ \mathrm{L.}(\delta \mathrm{ya}\mathrm{.}\mathrm{kya}\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )-\delta \mathrm{xa}\mathrm{.}\mathrm{kxa}\mathrm{.}\mathrm{sin}({\theta }_{0}+\theta ))/2\end{array}\right\}\begin{array}{}\text{avec les déplacements du point}\mathrm{A1}\\ \delta \mathrm{xa}=\mathrm{L.cos}({\theta }_{0})/2-\mathrm{L.cos}({\theta }_{0}+\theta )/2+x\\ \delta \mathrm{ya}=\mathrm{L.sin}({\theta }_{0})/2-\mathrm{L.sin}({\theta }_{0}+\theta )/2+y\end{array}`

Effort at point :math:`\mathrm{B1}`

:math:`\mathrm{Fb}=\left\{\begin{array}{}-\mathrm{kxb}\mathrm{.}\delta \mathrm{xb}\\ -\mathrm{kyb}\mathrm{.}\delta \mathrm{yb}\\ \mathrm{L.}(-\delta \mathrm{yb}\mathrm{.}\mathrm{kyb}\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )+\delta \mathrm{xb}\mathrm{.}\mathrm{kxb}\mathrm{.}\mathrm{sin}({\theta }_{0}+\theta ))/2\end{array}\right\}\begin{array}{}\text{avec les déplacements du point}\mathrm{B1}\\ \delta \mathrm{xb}=-\mathrm{L.cos}({\theta }_{0})/2+\mathrm{L.cos}({\theta }_{0}+\theta )/2+x\\ \delta \mathrm{yb}=-\mathrm{L.sin}({\theta }_{0})/2+\mathrm{L.sin}({\theta }_{0}+\theta )/2+y\end{array}`

Effort due to wind

• Relative speed of a :math:`M` point

:math:`{V}_{r}=\left\{\begin{array}{}\mathrm{Vvx}+\mathrm{s.}\mathrm{sin}({\theta }_{0}+\theta )\mathrm{.}\theta \text{'}-x\text{'}\\ \mathrm{Vvy}-\mathrm{s.}\mathrm{cos}({\theta }_{0}+\theta )\mathrm{.}\theta \text{'}-y\text{'}\\ 0\end{array}\right\}`

.. csv-table::

    "with", ":math:`s`: the curvilinear abscissa of the point :math:`M` on the :math:`s\in [–L/\mathrm{2,}L/2]` beam"
    "", ":math:`\mathrm{Vvx}`, :math:`\mathrm{Vvy}`: wind speed along the x axis and the :math:`y` axis."


• Relative speed perpendicular to the bar at point M:

:math:`{V}_{p}=\left\{\begin{array}{}\mathrm{sin}({\theta }_{0}+\theta )\mathrm{.}(-\mathrm{Vvy}\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )+\mathrm{Vvx.}\mathrm{sin}({\theta }_{0}+\theta )+s\mathrm{.}\theta \text{'}-\mathrm{sin}({\theta }_{0}+\theta )\mathrm{.}x\text{'}+\mathrm{cos}({\theta }_{0}+\theta )\mathrm{.}y\text{'})\\ \mathrm{cos}({\theta }_{0}+\theta )\mathrm{.}(\mathrm{Vvy}\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )-\mathrm{Vvx.}\mathrm{sin}({\theta }_{0}+\theta )-s\mathrm{.}\theta \text{'}+\mathrm{sin}({\theta }_{0}+\theta )\mathrm{.}x\text{'}-\mathrm{cos}({\theta }_{0}+\theta )\mathrm{.}y\text{'})\\ 0\end{array}\right\}`

Wind force at one point :math:`M`

:math:`{\mathrm{Fvent}}_{(M)}={\mathrm{Fcx}}_{(M)}\mathrm{.}\frac{{V}_{p}}{∥{V}_{p}∥}` in our case we choose :math:`{\mathrm{Fcx}}_{(M)}=∥{V}_{p}∥`

So we get :math:`{\mathrm{Fvent}}_{(M)}={V}_{p}`

• Resultant of the force due to the wind on the bar

:math:`\mathrm{Fvent}=\left\{\begin{array}{}L\mathrm{.}\mathrm{sin}({\theta }_{0}+\theta )\mathrm{.}((-\mathrm{Vvy}+y\text{'})\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )+(\mathrm{Vvx}-x\text{'})\mathrm{.}\mathrm{sin}({\theta }_{0}+\theta ))\\ L\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )\mathrm{.}((\mathrm{Vvy}-y\text{'})\mathrm{.}\mathrm{cos}({\theta }_{0}+\theta )+(-\mathrm{Vvx}+x\text{'})\mathrm{.}\mathrm{sin}({\theta }_{0}+\theta ))\\ -{L}^{3}\mathrm{.}\theta \text{'}/12\end{array}\right\}`

Final dynamic equation

:math:`M\mathrm{.}{\gamma }_{g}=\mathrm{Fa}+\mathrm{Fb}+\mathrm{Fvent}`


Reference quantities and results
-----------------------------------

Displacements and rotation of the point :math:`G` at the times: :math:`\mathrm{2.0sec}`, :math:`\mathrm{3.0sec}`,, :math:`\mathrm{4.0sec}`, :math:`\mathrm{5.0sec}` and :math:`\mathrm{6.0sec}`.


Uncertainties about the solution
----------------------------

None. The equilibrium equation is solved by a Runge Kutta integration method of order 4.