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.

Neglecting gravity, the equation is written (in total constraints):

\(\text{Div}(\sigma )=0\)

Since the Poisson module \(\nu\) is zero, and being elastic in the case, we have \(\sigma =Eϵ\).

The volume studied is spherically symmetric, made up of a homogeneous and isotropic material; the boundary conditions also have spherical symmetry. We are therefore required to seek a solution to the problem in a spherical coordinate system \((r,\theta ,\varphi )\) such that the displacement fields, stress fields and deformation fields are respectively of the form:

\(\{\begin{array}{c}{u}_{\text{r}}=h(r)\\ {u}_{\theta }={u}_{\varphi }=0\end{array}\) \(\{\begin{array}{c}{\sigma }_{\text{rr}}={f}_{1}(r)\\ \begin{array}{c}{\sigma }_{\theta \theta }={\sigma }_{\varphi \varphi }={g}_{1}(r)\\ {\sigma }_{r\theta }={\sigma }_{r\varphi }={\sigma }_{\theta \varphi }=0\end{array}\end{array}\) \(\{\begin{array}{c}{ϵ}_{\text{rr}}={f}_{2}(r)\\ \begin{array}{c}{ϵ}_{\theta \theta }={ϵ}_{\varphi \varphi }={g}_{2}(r)\\ {ϵ}_{r\theta }={ϵ}_{r\varphi }={ϵ}_{\theta \varphi }=0\end{array}\end{array}\)

The equilibrium equation \(\text{Div}(\sigma )=0\) is then reduced to: \(\frac{d{\sigma }_{\text{rr}}}{\mathit{dr}}+\frac{2}{r}\ast ({\sigma }_{\text{rr}}-{\sigma }_{\theta \theta })=0\)

Static boundary conditions are of the form: \({\sigma }_{\text{rr}}(R)=-p\)

The kinematic equations have the form: \(\{\begin{array}{c}{ϵ}_{\text{rr}}=\frac{{\mathit{du}}_{\text{r}}}{\mathit{dr}}\\ {ϵ}_{\theta \theta }=\frac{{u}_{\text{r}}}{r}\end{array}\) or finally: \(\{\begin{array}{c}{\sigma }_{\text{rr}}=E\ast \frac{{\mathit{du}}_{\text{r}}}{\mathit{dr}}\\ {\sigma }_{\theta \theta }=E\ast \frac{{u}_{\text{r}}}{r}\end{array}\)

Substituting these two relationships into the equilibrium equation we get:

\(\frac{\mathit{d²}{u}_{\text{r}}}{\mathit{dr²}}+\frac{2}{r}\ast (\frac{{\mathit{du}}_{\text{r}}}{\mathit{dr}}-\frac{{u}_{\text{r}}}{r})=0\) or \(\frac{d}{\mathit{dr}}(\frac{1}{\mathit{r²}}\frac{d({\mathit{r²u}}_{r})}{\mathit{dr}})=0\)

The solution to this differential equation is: \({u}_{r}(r)={C}_{1}r+\frac{{C}_{2}}{\mathit{r²}}\). Since the solution sought is discontinuous in \(R\), this equation is solved separately on the two domains \([{R}_{i},R]\) and \([R,{R}_{e}]\).

Finally: \(\{\begin{array}{c}{u}_{\text{r}}(r)={C}_{1}r+\frac{{C}_{2}}{\mathit{r²}}\text{}\mathit{sur}[{R}_{i},R]\\ {u}_{r}(r)={C}_{3}r+\frac{{C}_{4}}{\mathit{r²}}\text{}\mathit{sur}[R,{R}_{e}]\end{array}\)

Based on kinematic boundary conditions, \({u}_{r}({R}_{i})={u}_{r}({R}_{e})=0\) so \(\{\begin{array}{c}{C}_{1}{R}_{i}+\frac{{C}_{2}}{{R}_{i}\mathrm{²}}=0\\ {C}_{3}{R}_{e}+\frac{{C}_{4}}{{R}_{e}\mathrm{²}}=0\end{array}\)

Moreover \({\sigma }_{\text{rr}}=E\ast \frac{{\mathit{du}}_{\text{r}}}{\mathit{dr}}=\{\begin{array}{c}E\ast {C}_{1}-2E\ast \frac{{C}_{2}}{\mathit{r³}}\text{}\mathit{sur}[{R}_{i},R]\\ E\ast {C}_{3}-2E\ast \frac{{C}_{4}}{\mathit{r³}}\text{}\mathit{sur}[R,{R}_{e}]\end{array}\) therefore according to the static boundary conditions: \(\{\begin{array}{c}{C}_{1}-2\ast \frac{{C}_{2}}{\mathit{R³}}=-\frac{p}{E}\\ {C}_{3}-2\ast \frac{{C}_{4}}{\mathit{R³}}=-\frac{p}{E}\end{array}\).

The system resolution is \(\{\begin{array}{c}\begin{array}{c}{C}_{1}=\frac{-p}{E\ast (2\ast \frac{{R}_{i}\mathrm{³}}{R\mathrm{³}}+1)}\\ {C}_{2}=\frac{p}{E\ast (\frac{2}{\mathit{R³}}+\frac{1}{{R}_{i}\mathrm{³}})}\end{array}\\ \begin{array}{c}{C}_{3}=\frac{-p}{E\ast (2\ast \frac{{R}_{e}\mathrm{³}}{R\mathrm{³}}+1)}\\ {C}_{4}=\frac{p}{E\ast (\frac{2}{\mathit{R³}}+\frac{1}{{R}_{e}\mathrm{³}})}\end{array}\end{array}\)

2.2. Reference quantities and results

The value of the radial displacements on both sides of the interface is tested.

Quantities tested	Reference type	Reference value
URL (below)	“ANALYTIQUE”	-0.0001142742582
UR (above)	“ANALYTIQUE”	6.173526141E-05

2.3. Uncertainty about the solution

None, the solution is analytical.