2. Benchmark solution#

2.1. Equations of balance#

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

\(\mathit{Fa}\mathrm{=}\)

_images/Object_3.svg

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

\(\mathit{Fb}\mathrm{=}\)

_images/Object_4.svg

Effort due to the wind

  • Wind speed at one point \(M\mathrm{\in }\mathit{barre}\)

\({V}_{r}\mathrm{=}\)

_images/Object_5.svg

with \(\mathit{Vvx}\), \(\mathit{Vvy}\): wind speed along the \(x\) axis and the \(y\) axis.

  • Relative speed perpendicular to the bar at point \(M\):

\({V}_{p}\mathrm{=}\)

_images/Object_6.svg
  • Force due to wind at one point \(M\)

\({\mathit{Fvent}}_{(M)}\mathrm{=}{\mathit{Fcx}}_{(M)}\)

_images/Object_7.svg

In our case we choose \({\mathit{Fcx}}_{(M)}\mathrm{=}\mathrm{\parallel }{V}_{p}\mathrm{\parallel }\)

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

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

Wind =

_images/Object_8.svg

Equation of balance: \(\mathit{Fa}+\mathit{Fb}+\mathit{Fvent}\mathrm{=}0\)

2.2. Reference quantities and results#

Displacement of points \(\mathit{A1}\) and \(\mathit{B1}\) at the times: \(1.s\), \(1.05s\) and \(2.s\). These times correspond to wind speeds of \(10\), \(15\) and \(\mathrm{20m}\mathrm{/}s\) respectively.

The resolution of the 3 equilibrium equations, projection of \(\mathit{Fa}+\mathit{Fb}+\mathit{Fvent}\mathrm{=}0\), is done by iterations. The 3 unknowns of the problem are the position of the center of gravity of the bar \(G\): \((x,y)\) and the variation of the angle: \(\theta\).

In Code_Aster, the wind effect is taken into account by a force distributed along the line element. The expression for the modulus of this distributed force is as follows:

\({\mathit{Fcx}}_{(v)}\mathrm{=}\frac{1}{2}\mathrm{.}\rho \mathrm{.}{V}^{2}\mathrm{.}\mathit{Cx}(v)\mathrm{.}{D}_{h}\)

where

\({\mathit{Fcx}}_{(v)}\): is the modulus of the force distributed along the cable in \(N\mathrm{/}m\), depending on the speed.

\(\rho\): is the density of the air in \(\mathit{kg}\mathrm{/}{m}^{3}\). \(V\): is the relative speed of the cable in \(m\mathrm{/}s\). \(\mathit{Cx}(v)\): is the drag coefficient of the cable, depending on the relative speed. \({D}_{h}\): is the hydraulic diameter of the cable in \(m\).

To obtain a simple analytical reference solution, the function \({\mathit{Fcx}}_{(\mathit{Vp})}\) is taken to be equal to \(\mathrm{\parallel }{V}_{p}\mathrm{\parallel }\). In the Code_Aster command file the \(\mathit{Fcxv}\) function is therefore defined as follows:

FCXV = DEFI_FONCTION (

NOM_PARA =' VITE ',

VALE =( 0.0, 0.0,

10.0, 10.0),

PROL_GAUCHE =' LINEAIRE ',

PROL_DROITE =' LINEAIRE ',

)

2.3. Uncertainties about the solution#

None. The balance equation is solved by iterations with an error less than \(1.0E\mathrm{-}09\).

2.4. Bibliographical reference#

HM77 /01/046/A. « Project M7-01-70. Evolution of the*Code_Aster to better take into account dynamic wind loads on line elements ».