2. Benchmark solution#

2.1. Calculation method used for the reference solution#

Cumulative plastic deformation \(P\) is equal to:

\(P\mathrm{=}\frac{{\sigma }_{L}\mathrm{-}{\sigma }^{y}}{C}\)

with: \({\sigma }_{L}\): constraint to the node in question

\({\sigma }^{y}\): elastic limit

\(C\): slope of the traction curve

The constraints are given by:

\({\sigma }_{\mathit{xx}}({N}_{i})\mathrm{=}\mathrm{-}{P}_{\mathit{res}}({N}_{i})\)

The plastic deformation is given by:

\(\mathrm{\mid }{\varepsilon }_{\mathit{xx}}^{p}({N}_{i})\mathrm{=}P({N}_{i})\mathrm{\mid }\)

2.2. Benchmark results#

Uniaxial stress, plastic deformation, and cumulative plastic deformation are calculated at nodes \({N}_{2}\) and \({N}_{3}\).

For the problem in question:

\({N}_{2}\)

\({N}_{3}\)

\({\sigma }_{\mathit{xx}}\)

0

—300

\({\varepsilon }_{\mathit{xx}}\)

0

—6.1658 10—2

\({\varepsilon }_{\mathit{xx}}^{p}\)

0

6.1658 10—2

2.3. Uncertainty about the solution#

Analytical solution.

2.4. Bibliographical references#

  1. LORENTZ E., PROIX J.M., VAUTIER I., VOLDOIRE F., F., WAECKEL F.: Introduction to thermo-plasticity in the Aster code. Course Reference Manual. EDF - DER, SCE IMA, Dept. Mechanics and Numerical Models, HI-74/96/013/0