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

We assign to the discrete element, via the law of bilinear behavior, two stiffness :math:`{K}_{1}` and :math:`{K}_{2}` functions of the temperature of the form :math:`{K}_{i}={\alpha }_{i}-{\beta }_{i}\dot{T}`.

Here we set arbitrarily:

:math:`{\alpha }_{1}=2\dot{{\alpha }_{2}}={2.10}^{8}{\mathrm{N.m}}^{-1}`

:math:`{\beta }_{1}=2\dot{{\beta }_{2}}={4.10}^{6}{\mathrm{N.m}}^{-1}`

We check that, regardless of the displacement :math:`U` or the stress force :math:`{F}_{P}` that is imposed, we obtain a nodal reaction :math:`F` of the discrete verifying the elastic law :math:`\Delta F={K}_{i}\mathrm{.}\Delta U`, where :math:`i=1` if :math:`F\le {F}_{P}` and :math:`i=2` if :math:`F>{F}_{P}`.


.. image:: images/Shape3.gif


.. _RefSchema_Shape3.gif:

The discrete nodal reactions are recorded for displacements (:math:`{U}_{1}<{U}_{2}`) located on either side of the slope break.

Since the slope break takes place for :math:`{U}_{P}={F}_{P}/{K}_{1}`, the reference solutions are obtained directly:

:math:`{F}_{1}={K}_{1}\mathrm{.}{U}_{1}` and :math:`{F}_{2}={F}_{P}+{K}_{2}\mathrm{.}({U}_{2}-{U}_{P})`

Be :math:`{F}_{2}={F}_{P}\mathrm{.}(1-{K}_{2}/{K}_{1})+{K}_{2}\mathrm{.}{U}_{2}`