3. Modeling A#

3.1. Characteristics of modeling#

2D axisymmetric finite elements

_images/Object_12.svg

The discretized geometry is shown above:

\(\mathrm{Bord}\)

\(\mathrm{group}\text{\_}\mathrm{no}\)

\(\mathrm{BC}\)

\(\mathrm{BC}\)

\(\mathrm{DA}\)

\(\mathrm{DA}\)

\(\mathrm{AB}\)

\(\mathrm{BAS}\)

\(\mathrm{CD}\)

\(\mathrm{HAUT}\)

3.2. Characteristics of the mesh#

The mesh is regular: 4 elements in the height, 8 in the thickness.

Number of knots: 121

Number of meshes and type: 32 QUAD8

3.3. Boundary conditions while on the go#

3.3.1. Gravity#

Move \(\mathrm{DY}\) is stuck at point \(F\) alone.

3.3.2. Rotation#

Move \(\mathrm{DY}\) is stuck on sides \([\mathrm{AB}]\) (GROUP_NO = “BAS”) and on \([\mathrm{CD}]\) (GROUP_NO = “HAUT”).

3.3.3. Thermal expansion case no. 1#

Move \(\mathrm{DY}\) is stuck all over the structure.

3.3.4. Thermal expansion case no. 2#

Move \(\mathrm{DY}\) is stuck on sides \([\mathrm{AB}]\) (GROUP_NO = “BAS”) and \([\mathrm{CD}]\) (GROUP_NO = “HAUT”).

3.4. Model A results#

Identification

Knot (Knit)

Value tested

Reference

Gravity

\(\mathrm{N78}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

—2.34000 10—8

\(\mathrm{N120}\)

\(\mathrm{DY}\) \((\mathrm{mm})\)

—1.185 10—9

\(\mathrm{N13}\)

\(\mathrm{DY}\) \((\mathrm{mm})\)

1.2150 10—9

\(\mathrm{N78}(\mathrm{M13})\)

\(\mathrm{SIYY}\) \((\mathrm{MPa})\)

8.0000 10—4

Uniform rotation - centrifugal force

\(\mathrm{N120}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

—2.94240 10—7

\(\mathrm{N13}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

2.88010 10—7

\(\mathrm{N120}(\mathrm{M1})\)

\(\mathrm{SIYY}\) \((\mathrm{MPa})\)

9.94880 10—4

\(\mathrm{N13}(\mathrm{M32})\)

\(\mathrm{SIYY}\) \((\mathrm{MPa})\)

9.26310 10—4

Dilation case 1

\(\mathrm{N120}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

1.056145 10—6

\(\mathrm{N13}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

1.110317 10—6

\(\mathrm{N120}(\mathrm{M1})\)

\(\mathrm{SIYY}\) \((\mathrm{MPa})\)

1.4321427

Dilation case 2

\(\mathrm{N120}\)

\(\mathrm{DX}\) \((\mathrm{mm})\)

2.53500 10—5

\(\mathrm{N120}(\mathrm{M1})\)

\(\mathrm{SIYY}\) \((\mathrm{MPa})\)

—2.00000 10—1