1. Reference problem

1.1. Geometry

The concrete plate is square; the sides have the same length \(L\mathrm{=}H\mathrm{=}2m\).

The thickness of the plate is \(e\mathrm{=}\mathrm{0,6}m\).

The cable crosses the plate horizontally, halfway up, without eccentricity in the thickness. The cross-sectional area of the cable is \({S}_{a}\mathrm{=}\mathrm{1,5}{.10}^{\mathrm{-}4}{m}^{2}\).

1.2. Material properties

Concrete material constituting the plate:	Young’s Module \({E}_{b}\mathrm{=}{3.10}^{10}\mathit{Pa}\)
Steel material constituting the cable:	Young’s Module \({E}_{a}\mathrm{=}\mathrm{2,1}{.10}^{11}\mathit{Pa}\)

The Poisson’s ratio is taken to be equal to 0 for both materials; the direction of application of the normal force is thus preferred (direction \(x\)).

Since voltage losses are neglected, the various parameters used to estimate them are set to 0.

1.3. Boundary conditions and loads

The lower vertex of the left edge of the plate, that is, the origin node \((0;0)\), is embedded: all degrees of freedom of translation and rotation are blocked.

The upper vertex of this same left edge, i.e. node \((0;2)\), is supported bi-laterally: the blocked degrees of translational freedom are \(\mathit{DX}\) and \(\mathit{DZ}\).

A normal tensile force is applied to both ends of the cable (which are fixed to the concrete in \(A\) and \(B\)): \((–{F}_{0};0)\) at the node \(A\) \((0;1)\) and \(({F}_{0};0)\) at the node \(B\) \((2;1)\), with \({F}_{0}\mathrm{=}{2.10}^{5}N\).