5. E modeling

5.1. Geometry

It is a parallelepiped cut into 9 hexahedra.

5.2. Boundary conditions and loading

Side \(x=0\) is embedded.

Press, inwards, on the \(x=6\) side.

We are in the presence of gravity.

Note :

Gravity is tilted with respect to the axes to break the symmetry of the problem. This ensures a single extremum for the error indicator and therefore an identical selection of the mesh to be refined, regardless of the execution machine.

5.3. Characteristics of the mesh

The domain is meshed with 9 HEXA8 hexahedra.

The edges of the domain are meshed in QUAD4 quadrangles.

Number of node groups: 2

\(B\): \(\mathit{NO4}(6.\mathrm{,0}\mathrm{.}\mathrm{,0}\mathrm{.})\)

\(A\): \(\mathit{NO32}(6.\mathrm{,6}\mathrm{.}\mathrm{,2}\mathrm{.})\)

Number of mesh groups: 5

\(\text{X\_0}\): the quadrangles of the face x=0

\(\text{X\_MAX}\): the quadrangles of the face x=6

\(\text{Z\_MI\_MA}\): the quadrangles of the lower side, z=0, and the upper side, z=2

\(\text{Y\_MI\_MA}\): the quadrangles of the front side, y=0, and the back side, y=6

\(\mathit{VOLUME}\): the hexahedra of volume

5.4. Benchmark results

On the result of the calculation:

	After adaptation 2
DZ	-17.3329132
ERREST	51818.753

On the displacement field interpolated with the second adaptation; the two nodes are created in the center of the cut hexahedra. This makes it possible to avoid possible changes in the numbering of the nodes in HOMARD and to perpetuate the test case.

	Interpolated values
X	6.03638
DZ	-14.0615426