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

Since the Poisson ratio is zero, uniform compression must be restored at each of the 4 floors of the structure. In other words, for a given floor, the displacement of the structure is proportional to the position according to :math:`X` and it depends on the Pressure which varies according to :math:`Y`. The displacement along :math:`Y` only depends linearly on :math:`Y` because the imposed pressure :math:`{\mathit{Press}}_{y}` is constant.

We therefore have for all the points of the structure the displacement which is equal to:


.. _RefEquation 2.1-1:

:math:`{\mathit{Depl}}_{X}(X,Y)\mathrm{=}(2\mathrm{-}X)\frac{{\mathit{Press}}_{x}(Y)}{E}` eq 2.1-1


.. _RefEquation 2.1-2:

:math:`{\mathit{Depl}}_{Y}(Y)=-Y\frac{{\mathit{Press}}_{y}}{E}` eq 2.1-2


Calculation of the energy of the structure
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's say :math:`{p}_{x}=10\text{MPa}` and :math:`{p}_{y}=10\text{MPa}` in cases with contact, :math:`{p}_{y}=0\text{MPa}` in cases without contact. The stress tensor for the analytical solution is:

:math:`\mathrm{\sigma }={\mathrm{\sigma }}_{\mathit{xx}}(y){e}_{x}\otimes {e}_{x}-{p}_{y}{e}_{y}\otimes {e}_{y}`

where we put:

:math:`{\sigma }_{\mathit{xx}}(y)=\{\begin{array}{c}0\text{pour}y<\frac{1}{2},\\ -{p}_{x}\text{pour}y\in \left[\frac{1}{2},\frac{3}{2}\right[,\\ -2{p}_{x}\text{pour}y\in \left[\frac{3}{2},\frac{5}{2}\right[,\\ -3{p}_{x}\text{pour}y\in \left[\frac{5}{2},\frac{7}{2}\right[,\\ -4{p}_{x}\text{pour}y\ge \frac{7}{2},\end{array}`

Let :math:`\Omega =[\mathrm{0,}2]\times [\mathrm{0,}4]` be the domain occupied by the solid. The energy of the structure is defined by:

:math:`{E}^{e}=\frac{1}{2}{\int }_{\mathrm{\Omega }}\mathrm{\sigma }\mathrm{:}\mathrm{\epsilon }\mathit{dS}`

Since :math:`\nu =0`, we have :math:`\epsilon =\frac{1}{E}\sigma`. From where:

:math:`\mathrm{\sigma }\mathrm{:}\mathrm{\epsilon }=\frac{1}{E}\left({\mathrm{\sigma }}_{\mathit{xx}}{(y)}^{2}+{p}_{y}^{2}\right)`

So we have:

:math:`{E}^{e}=\frac{1}{2}\frac{1}{E}\underset{0}{\overset{2}{\int }}\left(\underset{0}{\overset{4}{\int }}\left({\mathrm{\sigma }}_{\mathit{xx}}{(y)}^{2}+{p}_{y}^{2}\right)\mathit{dy}\right)\mathit{dx}`

That is:

:math:`{E}^{e}=\frac{1}{E}\left(\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}{p}_{x}^{2}\mathit{dy}+\underset{\frac{3}{2}}{\overset{\frac{5}{2}}{\int }}4{p}_{x}^{2}\mathit{dy}+\underset{\frac{5}{2}}{\overset{\frac{7}{2}}{\int }}9{p}_{x}^{2}\mathit{dy}+\underset{\frac{7}{2}}{\overset{4}{\int }}16{p}_{x}^{2}\mathit{dy}+\underset{0}{\overset{4}{\int }}{p}_{y}^{2}\mathit{dy}\right)`

So for :math:`{p}_{x}` and :math:`{p}_{x}` expressed in MPa, we have:

:math:`{E}^{e}=\frac{1}{E}\left(22{p}_{x}^{2}+4{p}_{y}^{2}\right)`

This result is valid in the case of plane stresses and plane deformations. In the 3D case, the thickness chosen is 1 m. The expression for energy is the same, but the units are changed. We then have, in plane stresses and plane deformations, with contact:

:math:`{E}^{e}=\mathrm{2,6}\times {10}^{7}\text{MJ}\times {\text{m}}^{-1}`,

in 3D with contact:

:math:`{E}^{e}=\mathrm{2,6}\times {10}^{7}\text{MJ}`,

in plane stresses and in plane deformations, without contact:

:math:`{E}^{e}=\mathrm{2,2}\times {10}^{7}\text{MJ}\times {\text{m}}^{-1}`,

in 3D, without contact:

:math:`{E}^{e}=\mathrm{2,2}\times {10}^{7}\text{MJ}`.


Calculation of the L2 norm of displacement
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The analytical solution displacement field is

:math:`u=\frac{{\sigma }_{\mathit{xx}}(y)}{E}(2-x){e}_{x}-\frac{{p}_{y}}{E}y{e}_{y},`

The :math:`{L}^{2}` travel standard is defined by:

:math:`{\Vert u\Vert }_{{L}^{2}}^{2}={\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}\mathit{dS}`

We have:

:math:`{\Vert u\Vert }^{2}=\frac{{{\mathrm{\sigma }}_{\mathit{xx}}(y)}^{2}}{{E}^{2}}{(2-x)}^{2}-\frac{{p}_{y}^{2}}{{E}^{2}}{y}^{2}`

So we have:

:math:`{\Vert u\Vert }_{{L}^{2}}^{2}=\frac{1}{{E}^{2}}\underset{0}{\overset{4}{\int }}\left(\underset{0}{\overset{2}{\int }}\left[{{\mathrm{\sigma }}_{\mathit{xx}}(y)}^{2}{(2-x)}^{2}-{p}_{y}^{2}{y}^{2}\right]\mathit{dx}\right)\mathit{dy}`

That is:

:math:`{\parallel u\parallel }_{{L}^{2}}^{2}=\frac{1}{{E}^{2}}\underset{0}{\overset{4}{\int }}\left(\frac{8}{3}{{\sigma }_{\mathit{xx}}(y)}^{2}-2{p}_{y}^{2}{y}^{2}\right)\mathit{dy}\mathrm{.}`

From where:

:math:`{\parallel u\parallel }_{{L}^{2}}^{2}=\frac{1}{{E}^{2}}\left[\frac{8}{3}\left(\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}{p}_{x}^{2}\mathit{dy}+\underset{\frac{3}{2}}{\overset{\frac{5}{2}}{\int }}4{p}_{x}^{2}\mathit{dy}+\underset{\frac{5}{2}}{\overset{\frac{7}{2}}{\int }}9{p}_{x}^{2}\mathit{dy}+\underset{\frac{7}{2}}{\overset{4}{\int }}16{p}_{x}^{2}\mathit{dy}\right)+2\underset{0}{\overset{4}{\int }}{p}_{y}^{2}{y}^{2}\mathit{dy}\right]\mathrm{.}`

And finally:

:math:`{\parallel u\parallel }_{{L}^{2}}^{2}=\frac{1}{{E}^{2}}\left(\frac{176}{3}{p}_{x}^{2}+\frac{128}{3}{p}_{y}^{2}\right)\mathrm{.}`

That is:

:math:`{\Vert u\Vert }_{{L}^{2}}=\frac{1}{E}\sqrt{\frac{176}{3}{p}_{x}^{2}+\frac{128}{3}{p}_{y}^{2}}.`

This result is valid in the case of plane stresses and plane deformations. In the 3D case, the thickness chosen is 1 m. The expression for the :math:`{L}^{2}` displacement norm is the same, but the units are changed.

We then have, in plane stresses and plane deformations, with contact:

:math:`{\Vert u\Vert }_{{L}^{2}}\approx \mathrm{1,00664459137}{\text{m}}^{2}`

in 3D with contact:

:math:`{\Vert u\Vert }_{{L}^{2}}\approx \mathrm{1,00664459137}{\text{m}}^{\frac{5}{2}}`

in plane stresses and in plane deformations, without contact:

:math:`{\Vert u\Vert }_{{L}^{2}}\approx \mathrm{0,765941686205}{\text{m}}^{2}`

in 3D, without contact:

:math:`{\Vert u\Vert }_{{L}^{2}}\approx \mathrm{0,765941686205}{\text{m}}^{\frac{5}{2}}`