2. Benchmark solution#

2.1. Calculation method#

The deformation field is obtained from the displacements and the stress field is obtained using Hooke’s law. The elastic energy \({E}_{\mathit{elas}}\) is then calculated by the following formula:

\({E}_{\mathit{elas}}\mathrm{=}{\mathrm{\int }}_{V}\frac{1}{2}\sigma \mathrm{:}\epsilon \mathit{dV}\) equation 2.1-1

2.1.1. Modeling A#

The homogeneous deformation and stress fields on the element are shown in the.

Component

\(X\)

\(Y\)

\(Z\)

\(\mathit{XY}\)

\(\mathit{YZ}\)

\(\mathit{ZX}\)

Deformation

0.001

-0.002

0.002

0.003

0.003

0.005

-0.0001

-0.0003

Constraint (\(\mathit{Pa}\))

1.3888889E6

-1.111111E6

3.0555556E6

4.1166667E5

-8.3333333E5

-8.3333333E4

-2.5E5

Table 2.1.1-1 : Strain and stress field

2.1.2. Modeling B: plane deformations#

The homogeneous deformation and stress fields on the element are shown in the.

Component

\(X\)

\(Y\)

\(Z\)

\(\mathit{XY}\)

\(\mathit{YZ}\)

\(\mathit{ZX}\)

Deformation

0.001

-0.002

0.002

0.0

0.0

0.0

0.0

Constraint (\(\mathit{Pa}\))

5,5555556 E 5

-1, 9444444 E6

-2,7777778 E 5

4,1666667E5

-8,3333333E5

-8,3333333E4

-2,5E5

Table 2.1.2-1 : Strain and stress field

2.1.3. C (plane stresses) and D (shell and plate elements) modeling#

The homogeneous deformation and stress fields on the element are shown in the. For shell and plate elements, this is membrane deformation, with the curvature being zero.

Component

\(X\)

\(Y\)

\(Z\)

\(\mathit{XY}\)

\(\mathit{YZ}\)

\(\mathit{ZX}\)

Deformation

0.001

-0.002

-0.002

-0.00025

0.0005

0.0

0.0

Constraint (\(\mathit{Pa}\))

6.25 E 5

-1, 875 E5

-1, 875 E6

0.0

0.0

0.0

0.0

Table 2.1.3-1 : Strain and stress field

2.1.4. E modeling: axisymmetric.#

The homogeneous deformation and stress fields on the element are shown in the.

Component

\(X\)

\(Y\)

\(Z\)

\(\mathit{XY}\)

Deformation

0.0

0.0001

0.0

0.0

Constraint (\(\mathit{Pa}\))

0.0

1.0

1.0E5

0.0

0.0

Table 2.1.4-1: Deformation and Stress Field

The elastic energy is calculated according to the by expressing the elementary volume \(\mathit{dV}\) in the cylindrical coordinate system for a slice of infinitesimal thickness \(d\theta\):

\({E}_{\mathit{elas}}\mathrm{=}{\mathrm{\int }}_{V}\frac{1}{2}\sigma \mathrm{:}\epsilon \mathit{dV}\mathrm{=}\frac{1}{2}\sigma \mathrm{:}\epsilon {\mathrm{\int }}_{{R}_{1}}^{{R}_{2}}\mathit{rdr}{\mathrm{\int }}_{{H}_{1}}^{{H}_{2}}\mathit{dz}\)

The terminals of the integral are: \({R}_{1}\mathrm{=}\mathrm{1m}\), \({R}_{2}\mathrm{=}\mathrm{2m}\),, \({H}_{1}\mathrm{=}\mathrm{0m}\), and \({H}_{2}\mathrm{=}\mathrm{1m}\).

2.2. Reference quantities and results#

The elastic energy calculated analytically for each of the models is shown in the.

Modeling

\(A\)

\(B\)

\(C\)

\(D\)

\(E\)

Elastic energy

6680,555556 J

2430.555556 J

2395,833333 J

2395,833333 J

2395,833333 J

\(\mathrm{7,5}{\mathit{J.rad}}^{1}\)

Table 2.2-1: Elastic energy

2.3. Uncertainties about the solution#

None. It is an analytical solution.

2.4. Bibliographical references#