2. Reference solution

2.1. Calculation method

It is an analytical solution. Taking into account the boundary conditions, displacements can be obtained from the analytical resolution of the equation for the conservation of momentum.

Since Poisson’s Ratio \(\nu\) is zero, the problem is one-dimensional according to \(y\). The stress tensor is uniform across the domain: \(\sigma =Eϵ=E\ast {ϵ}_{\mathit{yy}}{e}_{y}\otimes {e}_{y}\).

Or \(\text{Div}(\sigma )=0\) so \(\frac{\partial {ϵ}_{\mathit{yy}}}{\partial y}=0\). Based on the boundary conditions applied \({ϵ}_{\mathit{yy}}=\frac{{u}_{\text{y, impo}}}{L}\)

Finally, \(\sigma =E\ast \frac{{u}_{\text{y, impo}}}{L}{e}_{y}\otimes {e}_{y}={\sigma }_{\text{yy}}{e}_{y}\otimes {e}_{y}\).

At the crack level, \({e}_{r}=-\mathrm{sin}(\theta ){e}_{\text{x}}+\mathrm{cos}(\theta ){e}_{\text{y}}\) and \({e}_{\theta }=-\mathrm{cos}(\theta ){e}_{\text{x}}-\mathrm{sin}(\theta ){e}_{\text{y}}\).

At a point in the crack with coordinates \((R,\theta )\), if there was no crack, we would have:

\(\sigma \mathrm{.}{e}_{\text{r}}=({e}_{\text{r}}\mathrm{.}\sigma \mathrm{.}{e}_{\text{r}}){e}_{\text{r}}+({e}_{\text{r}}\mathrm{.}\sigma \mathrm{.}{e}_{\theta }){e}_{\theta }={\sigma }_{\text{yy}}\ast [{({e}_{y}\mathrm{.}{e}_{r})}^{2}{e}_{\text{r}}+({e}_{y}\mathrm{.}{e}_{r})x({e}_{y}\mathrm{.}{e}_{\theta }){e}_{\theta }]\)

\(\sigma \mathrm{.}{e}_{\text{r}}={\sigma }_{\text{yy}}\ast (\mathit{cos²}(\theta ){e}_{\text{r}}-\mathrm{sin}(\theta )\mathrm{cos}(\theta ){e}_{\theta })\)

Finally the solution to the problem is:

\(\sigma ={\sigma }_{\text{yy}}{e}_{y}\otimes {e}_{y}\)

\(u(x,y)=\frac{{u}_{\text{y, impo}}}{L}\ast (\frac{L}{2}+y)\)

2.2. Reference quantities and results

We test the value of the constraints SIGMAXX and SIGMAYY on the whole column.

Quantities tested	Reference type	Reference value
SIGMAXX (MPa)	“ANALYTIQUE”	0.0
SIGMAYY (MPa)	“ANALYTIQUE”	-5800.0

2.3. Uncertainty about the solution

None, the solution is analytical.