2. 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 \(X\) and it depends on the Pressure which varies according to \(Y\). The displacement along \(Y\) only depends linearly on \(Y\) because the imposed pressure \({\mathit{Press}}_{y}\) is constant.

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

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

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

2.1. Calculation of the energy of the structure

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

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

where we put:

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

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

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

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

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

So we have:

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

That is:

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

So for \({p}_{x}\) and \({p}_{x}\) expressed in MPa, we have:

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

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:

\({E}^{e}=\mathrm{2,6}\times {10}^{7}\text{MJ}\times {\text{m}}^{-1}\),

in 3D with contact:

\({E}^{e}=\mathrm{2,6}\times {10}^{7}\text{MJ}\),

in plane stresses and in plane deformations, without contact:

\({E}^{e}=\mathrm{2,2}\times {10}^{7}\text{MJ}\times {\text{m}}^{-1}\),

in 3D, without contact:

\({E}^{e}=\mathrm{2,2}\times {10}^{7}\text{MJ}\).

2.2. Calculation of the L2 norm of displacement

The analytical solution displacement field is

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

The \({L}^{2}\) travel standard is defined by:

\({\Vert u\Vert }_{{L}^{2}}^{2}={\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}\mathit{dS}\)

We have:

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

So we have:

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

That is:

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

From where:

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

And finally:

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

That is:

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

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 \({L}^{2}\) displacement norm is the same, but the units are changed.

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

\({\Vert u\Vert }_{{L}^{2}}\approx \mathrm{1,00664459137}{\text{m}}^{2}\)

in 3D with contact:

\({\Vert u\Vert }_{{L}^{2}}\approx \mathrm{1,00664459137}{\text{m}}^{\frac{5}{2}}\)

in plane stresses and in plane deformations, without contact:

\({\Vert u\Vert }_{{L}^{2}}\approx \mathrm{0,765941686205}{\text{m}}^{2}\)

in 3D, without contact:

\({\Vert u\Vert }_{{L}^{2}}\approx \mathrm{0,765941686205}{\text{m}}^{\frac{5}{2}}\)