2. Benchmark solution

2.1. Calculation method used for the reference solution

The analytical solution on the go, in canonical basis \((\overrightarrow{{e}_{x}},\overrightarrow{{e}_{y}},\overrightarrow{{e}_{z}})\), is worth:

\(U(x,y,z)\mathrm{=}k\left\{\begin{array}{c}x\\ y\\ z\end{array}\right\}\mathrm{\pm }2\) where, \(k\mathrm{=}\mathrm{-}\frac{p(1\mathrm{-}2\nu )}{E}\mathrm{=}{4.10}^{\mathrm{-}3}\).

Subsequently, it is verified that this expression of the displacement field is the only solution of the balance problem on both sides of the interface XFEM, that is to say, that the solution proposed § 2.1, respects the limit conditions and the hypotheses of the problem HPP.

By construction, the on-the-go solution verifies the Dirichlet conditions of § 1.3.

The strain tensor is set to: \(\underline{\underline{\epsilon }}\mathrm{=}\frac{1}{2}(\mathrm{\nabla }U+\mathrm{\nabla }{U}^{T})\mathrm{=}\left[\begin{array}{ccc}k& 0& 0\\ 0& k& 0\\ 0& 0& k\end{array}\right]\mathrm{=}kI\).

By taking \(k\mathrm{=}\mathrm{0,4}\text{\%}\), we respect the hypothesis of small disturbances.

Hooke’s law, for an isotropic material, is equal to:

\(\underline{\underline{\sigma }}\mathrm{=}\frac{E}{1+\nu }(\underline{\underline{\epsilon }}+\frac{\nu }{1\mathrm{-}2\nu }\mathit{Tr}(\underline{\underline{\epsilon }})I)\mathrm{=}k\frac{E}{1\mathrm{-}2\nu }I\)

Now, by design, we have: \(k\mathrm{=}\mathrm{-}\frac{p(1\mathrm{-}2\nu )}{E}\)

with, \(\underline{\underline{\sigma }}\mathrm{=}\mathrm{-}pI\).

Finally, the Neumann limit conditions to be imposed on the edges of domain \(\mathrm{-}\underline{\underline{\sigma }}\mathrm{.}\overrightarrow{n}\mathrm{=}p\) correspond well to the loading of § 1.3.

Since the stress tensor is uniform, it respects the static equilibrium equation, \(\text{div}\underline{\underline{\sigma }}\mathrm{=}0\).

2.2. Benchmark results

We test the maximum error on the movements at a few points located on the interface \(x+y+z+\mathit{cte}\mathrm{=}0\). The value of the displacement at these points corresponds to the interpolation of the displacement field calculated by Aster. The displacement calculated by Aster is then compared to the analytical value of the displacement given in § 2.1:

\(U(x,y,z)\mathrm{=}{4.10}^{\mathrm{-}3}\left\{\begin{array}{c}x\\ y\\ z\end{array}\right\}+\mathit{sign}(x+y+z+\mathit{cte})\mathrm{\times }2\)