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


Formal solution
-----------------

The movements in the cable and in the concrete are continuous and homogeneous. So the horizontal displacement on the interval :math:`\mathrm{[}0;L\mathrm{]}` is equal to :math:`u(x)\mathrm{=}\frac{{u}_{L}}{L}x`

The normal stresses in the concrete plate and the steel cable are written respectively (elasticity hypothesis):

:math:`\mathrm{\{}\begin{array}{c}{\sigma }_{b}\mathrm{=}{E}_{b}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\mathrm{=}{E}_{b}\frac{{u}_{L}}{L}\\ {\sigma }_{a}\mathrm{=}{E}_{a}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}+{\sigma }_{0}\mathrm{=}{E}_{a}\frac{{u}_{L}}{L}+{\sigma }_{0}\end{array}`

where :math:`{\sigma }_{0}\mathrm{=}\frac{{F}_{0}}{{S}_{a}}` is the initial prestress in the cable

and :math:`{u}_{L}` is the horizontal displacement at abscissa :math:`L`.

The balance of the plate and cable assembly is written as: :math:`{\sigma }_{b}{S}_{b}+{\sigma }_{a}{S}_{a}\mathrm{=}0\mathrm{\Rightarrow }{u}_{L}\mathrm{=}\frac{\mathrm{-}L{F}_{0}}{{E}_{b}eH+{E}_{a}{S}_{a}}`

The normal effort in the cable is worth: :math:`{N}_{a}\mathrm{=}{E}_{a}{S}_{a}\frac{{u}_{L}}{F}+{F}_{0}\mathrm{=}{F}_{0}\frac{{E}_{b}eH}{{E}_{b}eH+{E}_{a}{S}_{a}}`

The total effort on the vertical section of the concrete plate is equal to:

:math:`{N}_{b}\mathrm{=}{E}_{b}eH\frac{{u}_{L}}{L}\mathrm{=}\mathrm{-}{F}_{0}\frac{{E}_{b}eH}{{E}_{b}eH+{E}_{a}{S}_{a}}`

The linear density of normal force on the concrete plate is deduced.

:math:`{N}_{\mathit{xx}}\mathrm{=}\mathrm{-}{F}_{0}\frac{{E}_{b}e}{{E}_{b}eH+{E}_{a}{S}_{a}}`


Numeric reference values
-------------------------------

The numerical reference values are:


.. csv-table::

    ":math:`{u}_{L}\mathrm{=}–\mathrm{1,11013974}{.10}^{\mathrm{-}5}m`"
    ":math:`{N}_{a}\mathrm{=}\mathrm{1,99825153}{.10}^{5}N`"
    ":math:`{N}_{b}\mathrm{=}–\mathrm{1,99825153}{.10}^{5}N`"
    ":math:`{N}_{\mathit{xx}}\mathrm{=}–\mathrm{9,99125765}{.10}^{4}N\mathrm{/}m`"