2. Benchmark solution

2.1. Equations of balance

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

Inertial force:

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

Effort at point \(\mathrm{A1}\)

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

Effort at point \(\mathrm{B1}\)

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

Effort due to wind

• Relative speed of a \(M\) point

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

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

• Relative speed perpendicular to the bar at point M:

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

Wind force at one point \(M\)

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

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

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

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

Final dynamic equation

\(M\mathrm{.}{\gamma }_{g}=\mathrm{Fa}+\mathrm{Fb}+\mathrm{Fvent}\)

2.2. Reference quantities and results

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

2.3. Uncertainties about the solution

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