2. Benchmark solution#

2.1. Calculation method used for the reference solution#

Analytical solution:

In short trips:

  • in elastic regime (for \(F<{\sigma }_{Y}\)) the theoretical critical value corresponds to the Euler load. In the context of beam kinematics, the critical load is equal to:

_images/Object_1.svg

, so the critical pressure:

_images/Object_2.svg

with

_images/Object_3.svg

and

_images/Object_4.svg

either

_images/Object_5.svg
  • under elastoplastic conditions, as uniform compression without elastic discharge is considered and due to the law of behavior, the critical buckling load is equal to:

_images/Object_6.svg

or a critical pressure of:

_images/Object_7.svg

2.2. Benchmark results#

Critical load values for both load cases.

In elastic mode, for \(F<4\mathrm{MPa}\), which is \(t<0.61538462\), we have to get: \({P}_{\mathrm{cr}}=12.95\mathrm{MPa}\).

Under plastic conditions the critical buckling pressure value is: \(\mathrm{4,32}\mathrm{MPa}\).

The critical coefficients as a function of loading are:

No time

Surface force (en \(\mathrm{MPa}\) )

Critical Coefficient

Critical Load (in \(\mathrm{MPa}\) )

1

0.65

19.9290

12.9539

2

1.3

9.9645

12.9539

3

1.95

6.6430

12.9539

4

2.6

4.9823

12.9539

5

3.25

3.9858

3.9858

12.9539

6

3.9

3.3215

12.9539

7

4.55

0.9490

4.3180

8

5.2

0.8304

4.3180

9

5.85

0.7381

4.3180

10

6.5

0.6643

4.3180