6. D modeling#

6.1. Characteristics of modeling#

Modeling DKT. We mesh only half of the cylinder (symmetry with respect to plane \(y=0\)) 30 mesh QUAD4 in the height and 60 on the half-circumference.

_images/100000000000021A0000038E7D3C7A498A4E3653.png

6.2. Characteristics of the mesh#

Number of knots: 1894

Number of meshes and types: 1800 QUAD4

Note:

To obtain a precise solution to this problem, it is necessary to use a fairly refined mesh (here 1800 QUAD4).

The following errors are observed as a function of discretization:

Number of items

Maximum displacement error

450 QUAD4

0.5%

1800 QUAD4

0.2%

900 TRIA3

17%

3600 TRIA3

1.4%

It can be seen that for this problem, meshing in quadrangles is preferable.

6.3. Tested values#

Isotropic material

Value

Identification

Reference

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DY}(\mathrm{PM})\)

5.8018E—05

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DX}(\mathrm{A1})\)

5.8018E—05

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DX}(\mathrm{A2})\)

—5.8018E—05

\(\mathrm{Uz}(z=L)\)

\(\mathrm{DZ}(\mathrm{A3})\)

—2.4429E—05

\(\mathrm{Uz}(z=L)\)

\(\mathrm{DZ}(\mathrm{A4})\)

—2.4429E—05

\(\mathrm{SigmaTT}(z=0)\)

\(\mathrm{SIZZ}(\mathrm{PM})\)

2.1375E+06

Orthotropic material

Value

Identification

Reference

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DY}(\mathrm{PM})\)

5.8018E—05

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DX}(\mathrm{A1})\)

5.8018E—05

\(\mathrm{Ur}(z=0)\)

\(\mathrm{DX}(\mathrm{A2})\)

—5.8018E—05

\(\mathrm{Uz}(z=L)\)

\(\mathrm{DZ}(\mathrm{A3})\)

—6.10714E—06

\(\mathrm{Uz}(z=L)\)

\(\mathrm{DZ}(\mathrm{A4})\)

—6.10714E—06

\(\mathrm{SigmaTT}(z=0)\)

\(\mathrm{SIZZ}(\mathrm{PM})\)

2.1375E+06