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

Let :math:`\mathrm{\Omega }=[-\mathrm{5,}+5]\times [-\mathrm{5,}+5]` be the domain occupied by the solid, in plan :math:`(X,Y)`. Domain :math:`\mathrm{\Omega }` is partitioned in :math:`\mathrm{\Omega }={\mathrm{\Omega }}_{1}\cup {\mathrm{\Omega }}_{2}\cup {\mathrm{\Omega }}_{3}\cup {\mathrm{\Omega }}_{4.}`, where we set:

:math:`\begin{array}{c}{\mathrm{\Omega }}_{1}=\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right],\\ {\mathrm{\Omega }}_{2}=\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[,\\ {\mathrm{\Omega }}_{3}=\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right],\\ {\mathrm{\Omega }}_{4}=\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\mathrm{.}\end{array}`


Contactless case
----------------

Without contact, each zone must undergo a rigid body movement corresponding to the limit condition imposed on its edge (right or left).

The energy of the structure is therefore:

:math:`{E}^{e}=0.`

The analytical solution displacement field is:

:math:`u={u}_{x}(x,y){e}_{x},`

with:

:math:`{u}_{x}(x,y)=\{\begin{array}{c}-\frac{1}{4}\text{pour}(x,y)\in \left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right[,\\ -\frac{1}{2}\text{pour}(x,y)\in \left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[,\\ +\frac{3}{4}\text{pour}(x,y)\in \left]0,+5\right]\times \left]\mathrm{0,}+5\right],\\ +1\text{pour}(x,y)\in \left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[,\end{array}`

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

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

So we have:

:math:`{\Vert u\Vert }_{{L}^{2}}^{2}=25\left(\frac{1}{16}+\frac{1}{4}+\frac{9}{16}+1\right)=25\frac{15}{8}.`

Either:

:math:`{\parallel u\parallel }_{{L}^{2}}=\frac{5}{2}\sqrt{\frac{15}{2}}\approx \mathrm{6,84653196881}{\text{m}}^{2}\mathrm{.}`

 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:

:math:`{\parallel u\parallel }_{{L}^{2}}=\frac{5}{2}\sqrt{\frac{15}{2}}\approx \mathrm{6,84653196881}{\text{m}}^{\frac{5}{2}}\mathrm{.}`


Contact case
----------------

Either:

:math:`{p}_{x}(y)=\{\begin{array}{c}1\text{MPa pour}y\in \left[-\mathrm{5,}0\right[\\ 2\text{MPa pour}y\in \left]\mathrm{0,}+5\right]\end{array}\text{et}{p}_{y}(x)=\{\begin{array}{c}1\text{MPa pour}x\in \left[-\mathrm{5,}0\right[\\ 2\text{MPa pour}x\in \left]\mathrm{0,}+5\right]\end{array}.`

So by definition we have:


.. _RefEquation 2.2-1:

:math:`\begin{array}{c}{p}_{x}(y)=2\text{et}{p}_{y}(x)=1\text{, dans}{\mathrm{\Omega }}_{\mathrm{1,}}\\ {p}_{x}(y)=1\text{et}{p}_{y}(x)=1\text{, dans}{\mathrm{\Omega }}_{\mathrm{2,}}\\ {p}_{x}(y)=2\text{et}{p}_{y}(x)=2\text{, dans}{\mathrm{\Omega }}_{\mathrm{3,}}\\ {p}_{x}(y)=1\text{et}{p}_{y}(x)=2\text{, dans}{\mathrm{\Omega }}_{4.}\end{array}` eq 2.2-1


Case of plane deformations
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The stress tensor for the analytical solution is:

:math:`\mathrm{\sigma }=-{p}_{x}(y){e}_{x}\otimes {e}_{x}-{p}_{y}(x){e}_{y}\otimes {e}_{y}-\mathrm{\nu }({p}_{x}(y)+{p}_{y}(x)){e}_{z}\otimes {e}_{z}`

We have:

:math:`\mathit{tr}(\mathrm{\sigma })=-(1+\mathrm{\nu })({p}_{x}(y)+{p}_{y}(x)).`

The strain tensor is obtained by applying Hooke's law:

:math:`\mathrm{\epsilon }=\frac{1+\mathrm{\nu }}{E}\mathrm{\sigma }-\frac{\mathrm{\nu }}{E}\mathit{tr}(\mathrm{\sigma })I,`

where :math:`I` is the identity tensor.

So we have:

:math:`\begin{array}{c}\mathrm{\epsilon }=-\left(\frac{(1+\mathrm{\nu })(1-\mathrm{\nu }){p}_{x}(y)}{E}-\frac{\mathrm{\nu }(1+\mathrm{\nu }){p}_{y}(x)}{E}\right){e}_{x}\otimes {e}_{x}\\ -\left(\frac{(1+\mathrm{\nu })(1-\mathrm{\nu }){p}_{y}(x)}{E}-\frac{\mathrm{\nu }(1+\mathrm{\nu }){p}_{x}(y)}{E}\right){e}_{y}\otimes {e}_{y}\end{array}`

So we have:

:math:`\mathrm{\sigma }\mathrm{:}\mathrm{\epsilon }=\frac{(1+\mathrm{\nu })(1-\mathrm{\nu })}{E}{\left({p}_{x}(y)\right)}^{2}-2\frac{\mathrm{\nu }(1+\mathrm{\nu })}{E}{p}_{x}(y){p}_{y}(x)+\frac{(1+\mathrm{\nu })(1-\mathrm{\nu })}{E}{\left({p}_{y}(x)\right)}^{2}.`

From where:

:math:`{E}^{e}=\frac{1}{2}\frac{1+\mathrm{\nu }}{E}{\int }_{\mathrm{\Omega }}\left[(1-\mathrm{\nu }){\left({p}_{x}(y)\right)}^{2}-2\mathrm{\nu }{p}_{x}(y){p}_{y}(x)+(1-\mathrm{\nu }){\left({p}_{y}(x)\right)}^{2}\right]d\mathrm{\Omega }`

Therefore, according to equation 2.2-1, we have:

:math:`\begin{array}{c}{E}^{e}=\frac{1}{2}\frac{1+\mathrm{\nu }}{E}[{\int }_{{\mathrm{\Omega }}_{1}}\left[4(1-\mathrm{\nu })-4\mathrm{\nu }+(1-\mathrm{\nu })\right]d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left[(1-\mathrm{\nu })-2\mathrm{\nu }+(1-\mathrm{\nu })\right]d\mathrm{\Omega }\\ +{\int }_{{\mathrm{\Omega }}_{3}}\left[4(1-\mathrm{\nu })-8\mathrm{\nu }+4(1-\mathrm{\nu })\right]d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}}\left[(1-\mathrm{\nu })-4\mathrm{\nu }+4(1-\mathrm{\nu })\right]d\mathrm{\Omega }].\end{array}`

Either:

:math: `{E} ^ {e} =\ frac {1} {2} {2}\ frac {1+\ mathrm {\nu}} {E}\ left [\ left (5-9\ mathrm {\nu}\ right)\ right)\ right)\ left| {\ mathrm {\ omega}}} _ {1}\ right|+\ left (2-4\ mathrm {\nu}\ right)\ left| {\ mathrm {\ Omega}} _ {2}\ right|+\ left (8-16\ mathrm {\nu}\ right)\ left| {\ mathrm {\ Omega}} _ {3}\ right|+\ right|+\ left (5-9\ mathrm {\nu}\ right)\ left| {\ mathrm {\ Omega}}} _ {4}\ right|+\ left (5-9\ mathrm {\nu}\ right)\ left| {\ mathrm {\ Omega}}} _ {Omega}} _ {4}\ right|+\ left (5-9\ mathrm {\ Omega}}} right|\ right] . `

We have:

:math: `\ left| {\ mathrm {\ Omega}} _ {1}\ right|=\ left| {\ mathrm {\ Omega}} _ {2}\ right|=\ left| {\ mathrm {\ Omega}}} _ {\ mathrm {\ Omega}}} _ {\ mathrm {\ Omega}} | {4}\ right|=5\ times 5=25. `

So we have:

:math:`{E}^{e}=\frac{1}{2}\frac{1+\mathrm{\nu }}{E}25\left[(5-9\mathrm{\nu })+(2-4\mathrm{\nu })+(8-16\mathrm{\nu })+(5-9\mathrm{\nu })\mathrm{.}\right]`

And finally:

:math:`{E}^{e}=\frac{25(1+\mathrm{\nu })(10-19\mathrm{\nu })}{E}=\mathrm{1,3975}\text{MJ}\times {\text{m}}^{-1}.`

The analytical displacement field :math:`u={u}_{x}{e}_{x}+{u}_{y}{e}_{y}` is obtained by integrating the deformations:

:math:`\begin{array}{c}{u}_{x}={\int }_{-5}^{x}{\mathrm{\epsilon }}_{\mathit{xx}}\mathit{dx},\\ {u}_{y}={\int }_{-5}^{y}{\mathrm{\epsilon }}_{\mathit{yy}}\mathit{dy},\end{array}`

because the boundary conditions applied are :math:`{u}_{x}(x=-5)=0` and :math:`{u}_{y}(y=-5)=0`.

It should be noted that the strain tensor is discontinuous. In fact, we have:

:math:`\mathrm{\epsilon }=\{\begin{array}{c}-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}{e}_{x}\otimes {e}_{x}-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}{e}_{y}\otimes {e}_{y},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}{e}_{x}\otimes {e}_{x}-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}{e}_{y}\otimes {e}_{y},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}{e}_{x}\otimes {e}_{x}-\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}{e}_{y}\otimes {e}_{y},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}{e}_{x}\otimes {e}_{x}-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}{e}_{y}\otimes {e}_{y},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}`

Note that the deformation field is discontinuous through the lines of equations :math:`x=0` and :math:`y=0`. It is therefore necessary to distinguish cases according to the sign of the coordinate on which one integrates in order to explain the value of integrals.

We thus have:

:math:`{u}_{x}=\{\begin{array}{c}{\int }_{-5}^{x}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{x}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ {\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dx}+{\int }_{0}^{x}\left[-\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dx}+{\int }_{0}^{x}\left[-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array},`

and

:math:`{u}_{y}=\{\begin{array}{c}{\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy}+{\int }_{0}^{y}\left[-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{y}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ {\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dy}+{\int }_{0}^{y}\left[-\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{y}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}.`

Either:

:math:`{u}_{x}=\{\begin{array}{c}-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array},` eq 2.2-2

and:

:math:`{u}_{y}=\{\begin{array}{c}-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}.` eq 2.2-3

It should be noted that the field of movement is not continuous. Since the field is the solution of a contact problem, the part normal to the interfaces of the movement is continuous and we have:

:math:`{u}_{x}(x={0}^{\text{-}})={u}_{x}(x={0}^{\text{+}})=\{\begin{array}{c}-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\text{, pour}y\in \left[-\mathrm{5,}0\right[\\ -5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\text{, pour}y\in \left]\mathrm{0,}+5\right]\end{array},`

and

:math:`{u}_{y}(y={0}^{\text{-}})={u}_{y}(y={0}^{\text{+}})=\{\begin{array}{c}-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[\\ -5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\text{, pour}x\in \left]\mathrm{0,}+5\right]\end{array}.`

On the other hand, the tangential part of the displacement is discontinuous and we have:

:math:`{u}_{y}(x={0}^{\text{+}})-{u}_{y}(x={0}^{\text{-}})=-\frac{(y+5)(1+\mathrm{\nu })(1-\mathrm{\nu })}{E},`

and

:math:`{u}_{x}(y={0}^{\text{+}})-{u}_{x}(y={0}^{\text{-}})=-\frac{(x+5)(1+\mathrm{\nu })(1-\mathrm{\nu })}{E}.`

The calculation of the integral of the square of the displacement norm must therefore once again use a domain partition in accordance with the equation interfaces :math:`x=0` and :math:`y=0`.

So we have:

:math:`{\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }={\int }_{{\mathrm{\Omega }}_{1}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{3}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }.`

We finally have:

:math:`{\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{1250\left(131{\mathrm{\nu }}^{4}+119{\mathrm{\nu }}^{3}-115{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+40\right)}{3{E}^{2}}.`

From where:

:math:`{\Vert u\Vert }_{{L}^{2}}=\frac{25}{E}\sqrt{\frac{2\left(131{\mathrm{\nu }}^{4}+119{\mathrm{\nu }}^{3}-115{\mathrm{\nu }}^{2}-63\mathrm{\nu }+40\right)}{3}}\approx \mathrm{0,791204250915}{\text{m}}^{2}.`


Case of plane stresses
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The stress tensor for the analytical solution is:


.. _RefEquation 2.2-4:

:math:`\mathrm{\sigma }=-{p}_{x}(y){e}_{x}\otimes {e}_{x}-{p}_{y}(x){e}_{y}\otimes {e}_{y}` eq 2.2-4

We have:

:math:`\mathit{tr}(\mathrm{\sigma })=-({p}_{x}(y)+{p}_{y}(x)).`

The strain tensor is obtained by applying Hooke's law and we have:

:math:`\begin{array}{c}\mathrm{\epsilon }=\left[-\frac{1+\mathrm{\nu }}{E}{p}_{x}(y)+\frac{\mathrm{\nu }}{E}({p}_{x}(y)+{p}_{y}(x))\right]{e}_{x}\otimes {e}_{x}+\left[-\frac{1+\mathrm{\nu }}{E}{p}_{y}(x)+\frac{\mathrm{\nu }}{E}({p}_{x}(y)+{p}_{y}(x))\right]{e}_{y}\otimes {e}_{y}\\ +\frac{\mathrm{\nu }}{E}({p}_{x}(y)+{p}_{y}(x)){e}_{z}\otimes {e}_{z}\end{array}`

And finally:

:math:`\mathrm{\epsilon }=-\left(\frac{{p}_{x}(y)}{E}-\mathrm{\nu }\frac{{p}_{y}(x)}{E}\right){e}_{x}\otimes {e}_{x}-\left(\frac{{p}_{y}(x)}{E}-\mathrm{\nu }\frac{{p}_{x}(y)}{E}\right){e}_{y}\otimes {e}_{y}+\frac{\mathrm{\nu }}{E}\left({p}_{x}(y)+{p}_{y}(x)\right){e}_{z}\otimes {e}_{z}`

The energy of the structure is:

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

We have:

:math:`\mathrm{\sigma }\mathrm{:}\mathrm{\epsilon }={p}_{x}(y)\left(\frac{{p}_{x}(y)}{E}-\mathrm{\nu }\frac{{p}_{y}(x)}{E}\right)+{p}_{y}(x)\left(\frac{{p}_{y}(x)}{E}-\mathrm{\nu }\frac{{p}_{x}(y)}{E}\right)`

Either:

:math:`\mathrm{\sigma }\mathrm{:}\mathrm{\epsilon }=\frac{{({p}_{x}(y))}^{2}}{E}-2\mathrm{\nu }\frac{{p}_{x}(y){p}_{y}(x)}{E}+\frac{{({p}_{y}(x))}^{2}}{E}`

So we have:

:math:`{E}^{e}=\frac{1}{2}\frac{1}{E}{\int }_{\mathrm{\Omega }}\left[{({p}_{x}(y))}^{2}-2\mathrm{\nu }{p}_{x}(y){p}_{y}(x)+{({p}_{y}(x))}^{2}\right]d\mathrm{\Omega }`

Hence according to equation 2.2-1:

:math:`{E}^{e}=\frac{1}{2}\frac{1}{E}\left[{\int }_{{\mathrm{\Omega }}_{1}}\left(4-4\mathrm{\nu }+1\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left(1-2\mathrm{\nu }+1\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{3}}\left(4-8\mathrm{\nu }+4\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}}\left(1-4\mathrm{\nu }+4\right)d\mathrm{\Omega }\right].`

Either:

:math: `{E} ^ {e} =\ frac {1} {2} {2}\ frac {1} {E}\ left [(5-4\ mathrm {\nu})\ left| {\ mathrm {\ Omega}} =\ frac {\ Omega}}} _ {Omega}} _ {2}\ right|+ (8-8\ mathrm {\nu})\ left| {\ mathrm {\ Omega}} _ {3}\ right|+ (5-4\ mathrm {\nu})\ left| {\ mathrm {\nu})\ left| {\ mathrm {\nu})\ left| {\ right] . `

So we have:

:math:`{E}^{e}=\frac{1}{2}\frac{1}{E}25\left[(5-4\mathrm{\nu })+(2-2\mathrm{\nu })+(8-8\mathrm{\nu })+(5-4\mathrm{\nu })\mathrm{.}\right]`

And finally:


.. _RefEquation 2.2-5:

:math:`{E}^{e}=\frac{1}{E}25(10-9\mathrm{\nu })=\mathrm{1,825}\text{MJ}\times {\text{m}}^{-1}.` eq 2.2-5

The analytical displacement field :math:`u={u}_{x}{e}_{x}+{u}_{y}{e}_{y}` is obtained by integrating the deformations:

:math:`\begin{array}{c}{u}_{x}={\int }_{-5}^{x}{\mathrm{\epsilon }}_{\mathit{xx}}\mathit{dx},\\ {u}_{y}={\int }_{-5}^{y}{\mathrm{\epsilon }}_{\mathit{yy}}\mathit{dy},\end{array}`

because the boundary conditions applied are :math:`{u}_{x}(x=-5)=0` and :math:`{u}_{y}(y=-5)=0`.

It should be noted that the strain tensor is discontinuous. In fact, we have:

:math:`\mathrm{\epsilon }=\{\begin{array}{c}-\frac{2-\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{1-2\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+3\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{1-\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{1-\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+2\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2-2\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{2-2\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+4\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{1-2\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{2-\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+3\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}.` eq 2.2-6

Note that the deformation field is discontinuous through the lines of equations :math:`x=0` and :math:`y=0`. It is therefore necessary to distinguish cases according to the sign of the coordinate on which one integrates in order to explain the value of integrals. We thus have:

:math:`{u}_{x}=\{\begin{array}{c}{\int }_{-5}^{x}\left(-\frac{2-\mathrm{\nu }}{E}\right)\mathit{dx},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{x}\left(-\frac{1-\mathrm{\nu }}{E}\right)\mathit{dx},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ {\int }_{-5}^{0}\left(-\frac{2-\mathrm{\nu }}{E}\right)\mathit{dx}+{\int }_{0}^{x}\left(-\frac{2-2\mathrm{\nu }}{E}\right)\mathit{dx},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{0}\left(-\frac{1-\mathrm{\nu }}{E}\right)\mathit{dx}+{\int }_{0}^{x}\left(-\frac{1-2\mathrm{\nu }}{E}\right)\mathit{dx},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array},`

and

:math:`{u}_{y}=\{\begin{array}{c}{\int }_{-5}^{0}\left(-\frac{1-\mathrm{\nu }}{E}\right)\mathit{dy}+{\int }_{0}^{y}\left(-\frac{1-2\mathrm{\nu }}{E}\right)\mathit{dy},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{y}\left(-\frac{1-\mathrm{\nu }}{E}\right)\mathit{dy},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ {\int }_{-5}^{0}\left(-\frac{2-\mathrm{\nu }}{E}\right)\mathit{dy}+{\int }_{0}^{y}\left(-\frac{2-2\mathrm{\nu }}{E}\right)\mathit{dy},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ {\int }_{-5}^{y}\left(-\frac{2-\mathrm{\nu }}{E}\right)\mathit{dy},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}.`

Either:

:math:`{u}_{x}=\{\begin{array}{c}-\frac{2-\mathrm{\nu }}{E}x-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{1-\mathrm{\nu }}{E}x-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2-2\mathrm{\nu }}{E}x-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{1-2\mathrm{\nu }}{E}x-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array},` eq 2.2-7

and:

:math:`{u}_{y}=\{\begin{array}{c}-\frac{1-2\mathrm{\nu }}{E}y-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\\ -\frac{1-\mathrm{\nu }}{E}y-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\\ -\frac{2-2\mathrm{\nu }}{E}y-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\\ -\frac{2-\mathrm{\nu }}{E}y-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\end{array}.` eq 2.2-8

It should be noted that the field of movement is not continuous. Since the field is the solution of a contact problem, the part normal to the interfaces of the movement is continuous and we have:

:math:`{u}_{x}(x={0}^{\text{-}})={u}_{x}(x={0}^{\text{+}})=\{\begin{array}{c}-5\frac{1-\mathrm{\nu }}{E}\text{, pour}y\in \left[-\mathrm{5,}0\right[\\ -5\frac{2-\mathrm{\nu }}{E}\text{, pour}y\in \left]\mathrm{0,}+5\right]\end{array},`

and

:math:`{u}_{y}(y={0}^{\text{-}})={u}_{y}(y={0}^{\text{+}})=\{\begin{array}{c}-5\frac{2-\mathrm{\nu }}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[\\ -5\frac{1-\mathrm{\nu }}{E}\text{, pour}x\in \left]\mathrm{0,}+5\right]\end{array}.`

On the other hand, the tangential part of the displacement is discontinuous and we have:

:math:`{u}_{y}(x={0}^{\text{+}})-{u}_{y}(x={0}^{\text{-}})=-\frac{y+5}{E},`

and

:math:`{u}_{x}(y={0}^{\text{+}})-{u}_{x}(y={0}^{\text{-}})=-\frac{x+5}{E}.`

The calculation of the integral of the square of the displacement norm must therefore once again use a domain partition in accordance with the equation interfaces :math:`x=0` and :math:`y=0`.

So we have:

:math:`{\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }={\int }_{{\mathrm{\Omega }}_{1}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{3}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}}\left({u}_{x}^{2}+{u}_{y}^{2}\right)d\mathrm{\Omega }.`

We finally have:

:math:`{\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{1250\left(28{\mathrm{\nu }}^{2}-63\mathrm{\nu }+40\right)}{3{E}^{2}}.`

From where:

:math:`{\Vert u\Vert }_{{L}^{2}}=\frac{25}{E}\sqrt{\frac{2\left(28{\mathrm{\nu }}^{2}-63\mathrm{\nu }+40\right)}{3}}\approx \mathrm{0,992051745962}{\text{m}}^{2}.`


3D case
~~~~~

The structure occupies domain :math:`{\mathrm{\Omega }}_{3D}=\mathrm{\Omega }\times [\mathrm{0,}1]`. The boundary conditions of the 3D case are the same as those of the 2D cases. In plane :math:`(X,Y),`, a roller condition is imposed in :math:`Z=0` and the edge :math:`Z=1` is free of constraints. The stress tensor analytical solution is therefore identical to the case of plane stresses (*cf.* eq 2.2-4):

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


.. _RefEquation 2.2-5)::

The elastic energy density is therefore identical to that of the 3D case. The solid is of unit thickness in the :math:`Z` direction. The expression of the energy of the structure is therefore identical to the case of plane stresses, but the units are modified. We then have (*cf.* eq 2.2-5):

:math:`{E}^{e}=\frac{1}{E}25(10-9\mathrm{\nu })=\mathrm{1,825}\text{MJ}.`

The analytical displacement field :math:`u={u}_{x}{e}_{x}+{u}_{y}{e}_{y}+{u}_{z}{e}_{z}` is obtained by integrating the deformations:

:math:`\begin{array}{c}{u}_{x}={\int }_{-5}^{x}{\mathrm{\epsilon }}_{\mathit{xx}}\mathit{dx},\\ {u}_{y}={\int }_{-5}^{y}{\mathrm{\epsilon }}_{\mathit{yy}}\mathit{dy},\\ {u}_{z}={\int }_{0}^{z}{\mathrm{\epsilon }}_{\mathit{zz}}\mathit{dz},\end{array}`

because the boundary conditions applied are :math:`{u}_{x}(x=-5)=0`, :math:`{u}_{y}(y=-5)=0`, and :math:`{u}_{z}(z=0)=0`.


.. _RefEquation 2.2-6)::

It should be noted that the strain tensor is discontinuous. In fact, we have (*cf.* eq 2.2-6):

:math:`\mathrm{\epsilon }=\{\begin{array}{c}-\frac{2-\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{1-2\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+3\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ -\frac{1-\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{1-\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+2\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\\ -\frac{2-2\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{2-2\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+4\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ -\frac{1-2\mathrm{\nu }}{E}{e}_{x}\otimes {e}_{x}-\frac{2-\mathrm{\nu }}{E}{e}_{y}\otimes {e}_{y}+3\frac{\mathrm{\nu }}{E}{e}_{z}\otimes {e}_{z},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\end{array}.`


.. _RefEquationuation :math:`x=0` et :math:`y=0` . Il est donc nécessaire de distinguer les cas selon le signe de la coordonnée sur laquelle on intègre pour expliciter la valeur des intégrales. Les expressions des composantes :math:`{u}_{x}` et :math:`{u}_{y}` sont identiques au cas des contraintes planes ( *cf.* éq 2.2-7et 2.2-8) et on a pour :math:`{u}_{z}` ::

Note that the deformation field is discontinuous through the lines of equations :math:`x=0` and :math:`y=0`. It is therefore necessary to distinguish cases according to the sign of the coordinate on which one integrates in order to explain the value of integrals. The expressions for components :math:`{u}_{x}` and :math:`{u}_{y}` are identical to the case of plane constraints (*cf.* eq 2.2-7 and 2.2-8) and for :math:`{u}_{z}` we have:

:math:`{u}_{z}=\{\begin{array}{c}{\int }_{0}^{1}\frac{3\mathrm{\nu }}{E}\mathit{dz},\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ {\int }_{0}^{1}\frac{2\mathrm{\nu }}{E}\mathit{dz},\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\\ {\int }_{0}^{1}\frac{4\mathrm{\nu }}{E}\mathit{dz},\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ {\int }_{0}^{1}\frac{3\mathrm{\nu }}{E}\mathit{dz},\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\end{array}.`

Either:

:math:`{u}_{z}=\{\begin{array}{c}\frac{3\mathrm{\nu }}{E}z,\text{dans}\left[-\mathrm{5,}0\right[\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ \frac{2\mathrm{\nu }}{E}z,\text{dans}\left[-\mathrm{5,}0\right[\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\\ \frac{4\mathrm{\nu }}{E}z,\text{dans}\left]\mathrm{0,}+5\right]\times \left]\mathrm{0,}+5\right]\times [\mathrm{0,}1]\\ \frac{3\mathrm{\nu }}{E}z,\text{dans}\left]\mathrm{0,}+5\right]\times \left[-\mathrm{5,}0\right[\times [\mathrm{0,}1]\end{array}.` eq 2.2-9

The calculation of the integral of the square of the displacement norm must therefore once again use a domain partition in accordance with the equation interfaces :math:`x=0` and :math:`y=0`. So we have:

:math:`\begin{array}{c}{\int }_{{\mathrm{\Omega }}_{3D}}{\Vert u\Vert }^{2}d\mathrm{\Omega }={\int }_{{\mathrm{\Omega }}_{1}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }\\ +{\int }_{{\mathrm{\Omega }}_{3}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }.\end{array}`

We finally have:

:math:`{\int }_{{\mathrm{\Omega }}_{3D}}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{50(719{\mathrm{\nu }}^{2}-1575\mathrm{\nu }+1000)}{3{E}^{2}}.`

From where:

:math:`{\Vert u\Vert }_{{L}^{2}}=\frac{5}{E}\sqrt{\frac{2(719{\mathrm{\nu }}^{2}-1575\mathrm{\nu }+1000)}{3}}\approx \mathrm{0,99348712456}{\text{m}}^{\frac{5}{2}}.`