2. Reference solution#

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

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

2.1. 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:

\({E}^{e}=0.\)

The analytical solution displacement field is:

\(u={u}_{x}(x,y){e}_{x},\)

with:

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

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

\({\Vert u\Vert }_{{L}^{2}}^{2}={\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }.\)

So we have:

math:: `{Green orGreen} _ {{L} ^ {2}}} ^ {2}} ^ {2} =frac {1} {16}left| {mathrm {Omega}} _ {1}right|+frac {1}right|+frac {1}right|+frac {9} {right|+frac {9}} 16}left| {mathrm {Omega}}} _ {3}right|+left| {mathrm {Omega}} _ {4}right|. `

We have:

math:: begin {array} {c}left| {mathrm {Omega}} | {mathrm {Omega}} _ {1}\ right|=20 {text {m}}} ^ {2} {m}}} ^ {2}left| {mathrm {Omega}}} _ {2}\ left| {mathrm {Omega}}} _ {2}\ left| {mathrm {Omega}}} _ {2}\ right|=30 {text {m}}} ^ {2}\ left| {mathrm {Omega}}} _ {2}\ left| {mathrm {Omega}}}} _ {2}\ left| Omega}} _ {3}right|=20 {text {m}}} ^ {2}\left| {mathrm {Omega}} _ {4}right|=30 {text {m}} =30 {text {m}}} ^ {2}end {array}}

From where:

\({\Vert u\Vert }_{{L}^{2}}^{2}=\frac{1}{16}20+\frac{1}{4}30+\frac{9}{16}20+30=50.\)

Either:

\({\Vert u\Vert }_{{L}^{2}}=\sqrt{50}\approx \mathrm{7,071067812}{\text{m}}^{2}.\)

2.2. Contact case#

Either:

\({p}_{x}(y)=\{\begin{array}{c}1\text{MPa pour}y\in \left[-\mathrm{5,}-2\right[\\ 2\text{MPa pour}y\in \left[-\mathrm{2,}+2\right[\\ 3\text{MPa pour}y\in \left]+\mathrm{2,}+5\right]\end{array}\text{et}{p}_{y}=1\text{MPa}.\)

So by definition we have:

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

2.2.1. Case of plane deformations#

The stress tensor for the analytical solution is:

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

We have:

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

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

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

where \(I\) is the identity tensor.

So we have:

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

So we have:

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

From where:

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

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

\(\begin{array}{c}{E}^{e}=\frac{1}{2}\frac{1+\mathrm{\nu }}{E}[{\int }_{{\mathrm{\Omega }}_{1}\cup {\mathrm{\Omega }}_{3}}\left[4(1-\mathrm{\nu })-4\mathrm{\nu }+(1-\mathrm{\nu })\right]d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left[9(1-\mathrm{\nu })-6\mathrm{\nu }+(1-\mathrm{\nu })\right]d\mathrm{\Omega }\\ +{\int }_{{\mathrm{\Omega }}_{4}}\left[(1-\mathrm{\nu })-2\mathrm{\nu }+(1-\mathrm{\nu })\right]d\mathrm{\Omega }].\end{array}\)

Either:

math:: `{E} ^ {e} =frac {1} {2} {2}frac {1+mathrm {nu}} {E}left (5-9mathrm {nu}right)right)left (left| {mathrm {Omega}} _ {1}right|+left| {mathrm {Omega}} _ {1}right|+left| {mathrm {Omega}} _ {1}right|+left| {mathrm {Omega}}}} _ {3}right|right) +left (10-16mathrm {nu}right)left| {mathrm {Omega}} _ {2}right|+left (+left (2-4mathrm {nu}right)left| {mathrm {Omega}}} _ {4}right|+left (2-4left (2-4mathrm {nu}right)right)left| {mathrm {Omega}} _ {4}right|right] . `

So we have:

math:: `{E} ^ {e} =frac {1} {2} {2}frac {1+mathrm {nu}} {E}left [40left (5-9mathrm {nu}right) +30left (10-16mathrm {nu}right) +30left (2-4mathrm {nu}right) +30left (2-4mathrm {nu}right)right)left| {mathrm {Omega}}} _ {4}right|right] . `

And finally:

\({E}^{e}=\frac{40(1+\mathrm{\nu })(7-12\mathrm{\nu })}{E}=\mathrm{1,768}\text{MJ}\times {\text{m}}^{-1}.\)

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

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

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

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

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

We thus have:

\({u}_{x}=\{\begin{array}{c}{\int }_{-5}^{x}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ {\int }_{-5}^{x}\left[-\frac{(1+\mathrm{\nu })(3-4\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\\ {\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dx}+{\int }_{0}^{x}\left[-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ {\int }_{-5}^{0}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dx},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array},\)

and

\({u}_{y}=\{\begin{array}{c}{\int }_{-5}^{-2}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy}+{\int }_{-2}^{y}\left[-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ \begin{array}{c}{\int }_{-5}^{-2}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy}+{\int }_{-2}^{+2}\left[-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}\right]\mathit{dy}\\ +{\int }_{+2}^{y}\left[-\frac{(1+\mathrm{\nu })(1-4\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\end{array}\\ {\int }_{-5}^{-2}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy}+{\int }_{-2}^{y}\left[-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ {\int }_{-5}^{y}\left[-\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}\right]\mathit{dy},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array}.\)

Either:

\({u}_{x}=\{\begin{array}{c}-\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ -\frac{(1+\mathrm{\nu })(3-4\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(3-4\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\\ -\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(2-3\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ -\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}x-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array},\) eq 2.2-2

and:

\({u}_{y}=\{\begin{array}{c}-\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}y-\frac{(1+\mathrm{\nu })(5-12\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ -\frac{(1+\mathrm{\nu })(1-4\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\\ -\frac{(1+\mathrm{\nu })(1-3\mathrm{\nu })}{E}y-\frac{(1+\mathrm{\nu })(5-12\mathrm{\nu })}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ -\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E}y-5\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array}.\) 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:

\({u}_{y}(y=-{2}^{\text{-}})={u}_{y}(y=-{2}^{\text{+}})=-3\frac{(1+\mathrm{\nu })(1-2\mathrm{\nu })}{E},\)

and:

\({u}_{y}(y=+{2}^{\text{-}})={u}_{y}(y=+{2}^{\text{+}})=-\frac{(1+\mathrm{\nu })(7-18\mathrm{\nu })}{E}.\)

We also have:

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

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

\({u}_{x}(y=-{2}^{\text{+}})-{u}_{x}(y=-{2}^{\text{-}})=-\frac{(x+5)(1+\mathrm{\nu })(1-\mathrm{\nu })}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[,\)

and:

\({u}_{x}(y=+{2}^{\text{+}})-{u}_{x}(y=+{2}^{\text{-}})=-\frac{(x+5)(1+\mathrm{\nu })(1-\mathrm{\nu })}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[,\)

while:

\({u}_{y}(x={0}^{\text{+}})-{u}_{y}(x={0}^{\text{-}})=0\text{, pour}y\in \left]-\mathrm{2,}+2\right[.\)

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 \(y=-2\) and \(y=+2\).

So we have:

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

We finally have:

\({\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{20\left(8436{\mathrm{\nu }}^{4}+7587{\mathrm{\nu }}^{3}-7334{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3{E}^{2}}.\)

From where:

\({\Vert u\Vert }_{{L}^{2}}=\frac{1}{E}\sqrt{\frac{20\left(8436{\mathrm{\nu }}^{4}+7587{\mathrm{\nu }}^{3}-7334{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3}}\approx \mathrm{0,933673961652}{\text{m}}^{2}.\)

2.2.2. Case of plane stresses#

The stress tensor for the analytical solution is:

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

We have:

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

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

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

And finally:

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

The energy of the structure is:

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

We have:

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

Either:

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

So we have:

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

Hence according to equation 2.2-1:

\({E}^{e}=\frac{1}{2}\frac{1}{E}\left[{\int }_{{\mathrm{\Omega }}_{1}\cup {\mathrm{\Omega }}_{3}}\left(4-4\mathrm{\nu }+1\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}}\left(9-6\mathrm{\nu }+1\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}}\left(1-2\mathrm{\nu }+1\right)d\mathrm{\Omega }\right].\)

Either:

math:: `{E} ^ {e} =frac {1} {2} {2}frac {1} {E}left [(5-4mathrm {nu})left (left| {mathrm {Omega}} = {mathrm {Omega}} = {mathrm {Omega}})left (left| {mathrm {omega}})left (left| {mathrm {omega}} | {3}right|}} _ {3}right|right) + (10-6mathrm {nu})left| {mathrm {Omega}} _ {2}right|+ (2-2mathrm {nu})left| {mathrm {omega}}} _ {4}right|right] . `

So we have:

\({E}^{e}=\frac{1}{2}\frac{1}{E}\left[40(5-4\mathrm{\nu })+30(10-6\mathrm{\nu })+30(2-2\mathrm{\nu })\right].\)

And finally:

\({E}^{e}=\frac{1}{E}40(7-5\mathrm{\nu })=\mathrm{2,2}\text{MJ}\times {\text{m}}^{-1}.\) eq 2.2-5

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

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

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

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

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

We thus have:

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

and

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

Either:

\({u}_{x}=\{\begin{array}{c}-\frac{2-\mathrm{\nu }}{E}x-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ -\frac{3-\mathrm{\nu }}{E}x-5\frac{3-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\\ -\frac{2-\mathrm{\nu }}{E}x-5\frac{2-\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ -\frac{1-\mathrm{\nu }}{E}x-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array},\) eq 2.2-7

and:

\({u}_{y}=\{\begin{array}{c}-\frac{1-2\mathrm{\nu }}{E}y-\frac{5-7\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\\ -\frac{1-3\mathrm{\nu }}{E}y-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\\ -\frac{1-2\mathrm{\nu }}{E}y-\frac{5-7\mathrm{\nu }}{E},\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\\ -\frac{1-\mathrm{\nu }}{E}y-5\frac{1-\mathrm{\nu }}{E},\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\end{array}.\) 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:

\({u}_{y}(y=-{2}^{\text{-}})={u}_{y}(y=-{2}^{\text{+}})=-3\frac{1-\mathrm{\nu }}{E},\)

and:

\({u}_{y}(y=+{2}^{\text{-}})={u}_{y}(y=+{2}^{\text{+}})=-\frac{7-11\mathrm{\nu }}{E}.\)

We also have:

\({u}_{x}(x={0}^{\text{-}})={u}_{x}(x={0}^{\text{+}})=-5\frac{2-\mathrm{\nu }}{E}\text{, pour}y\in \left]-\mathrm{2,}+2\right[.\)

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

\({u}_{x}(y=-{2}^{\text{+}})-{u}_{x}(y=-{2}^{\text{-}})=-\frac{x+5}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[,\)

and:

\({u}_{x}(y=+{2}^{\text{+}})-{u}_{x}(y=+{2}^{\text{-}})=-\frac{x+5}{E}\text{, pour}x\in \left[-\mathrm{5,}0\right[,\)

while:

\({u}_{y}(x={0}^{\text{+}})-{u}_{y}(x={0}^{\text{-}})=0\text{, pour}y\in \left]-\mathrm{2,}+2\right[.\)

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 \(y=-2\) and \(y=+2\).

So we have:

We finally have:

\({\int }_{\mathrm{\Omega }}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{20\left(1951{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3{E}^{2}}.\)

From where:

\({\Vert u\Vert }_{{L}^{2}}=\frac{1}{E}\sqrt{\frac{20\left(1951{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3}}\approx \mathrm{1,11656914997}{\text{m}}^{2}.\)

2.2.3. 3D case#

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

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

The elastic energy density is therefore identical to that of the 3D case. The solid is of unit thickness in the \(Z\) 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):

\({E}^{e}=\frac{1}{E}40(7-5\mathrm{\nu })=\mathrm{2,2}\text{MJ}\times {\text{m}}^{-1}.\)

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

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

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

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

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

Note that the deformation field is discontinuous through the lines of equations \(y=-2\) and \(y=+2\). It is therefore necessary to distinguish cases according to the domain in which one integrates in order to explain the value of integrals. The expressions for components \({u}_{x}\) and \({u}_{y}\) are identical to the case of plane constraints (cf. eq 2.2-7 and 2.2-8) and for \({u}_{z}\) we have:

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

Either:

\({u}_{z}=\{\begin{array}{c}\frac{3\mathrm{\nu }}{E}z,\text{dans}\left[-\mathrm{5,}0\right[\times \left]-\mathrm{2,}+2\right[\times [\mathrm{0,}1]\\ \frac{4\mathrm{\nu }}{E}z,\text{dans}\left[-\mathrm{5,}+5\right]\times \left]+\mathrm{2,}+5\right]\times [\mathrm{0,}1]\\ \frac{3\mathrm{\nu }}{E}z,\text{dans}\left]\mathrm{0,}+5\right]\times \left]-\mathrm{2,}+2\right[\times [\mathrm{0,}1]\\ \frac{2\mathrm{\nu }}{E}z,\text{dans}\left[-\mathrm{5,}+5\right]\times \left[-\mathrm{5,}-2\right[\times [\mathrm{0,}1]\end{array}.\) 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 \(y=-2\) and \(y=+2\).

So we have:

\({\int }_{{\mathrm{\Omega }}_{3D}}{\Vert u\Vert }^{2}d\mathrm{\Omega }={\int }_{{\mathrm{\Omega }}_{1}\times [\mathrm{0,}1]\cup {\mathrm{\Omega }}_{3}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{2}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }+{\int }_{{\mathrm{\Omega }}_{4}\times [\mathrm{0,}1]}\left({u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}\right)d\mathrm{\Omega }.\)

We finally have:

\({\int }_{{\mathrm{\Omega }}_{3D}}{\Vert u\Vert }^{2}d\mathrm{\Omega }=\frac{20\left(1999{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3{E}^{2}}.\)

From where:

\({\Vert u\Vert }_{{L}^{2}}=\frac{1}{E}\sqrt{\frac{20\left(1999{\mathrm{\nu }}^{2}-3685\mathrm{\nu }+2800\right)}{3}}\approx \mathrm{1,11785807089}{\text{m}}^{\frac{5}{2}}.\)