Reference problem
=====================


Geometry
---------

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

The thickness of the plate is :math:`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 :math:`{S}_{a}\mathrm{=}\mathrm{1,5}{.10}^{\mathrm{-}4}{m}^{2}`.


.. image:: images/Object_1.svg
    :width: 296
    :height: 250


.. _RefImage_Object_1.svg:


Material properties
------------------------


.. csv-table::

    "Concrete material constituting the plate:", "Young's Module :math:`{E}_{b}\mathrm{=}{3.10}^{10}\mathit{Pa}`"
    "Steel material constituting the cable:", "Young's Module :math:`{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 :math:`x`).

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


Boundary conditions and loads
-------------------------------------

The lower vertex of the left edge of the plate, that is, the origin node :math:`(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 :math:`(0;2)`, is supported bi-laterally: the blocked degrees of translational freedom are :math:`\mathit{DX}` and :math:`\mathit{DZ}`.

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