2. Reference solution#

2.1. Result of the reference solution#

For a tensile test in direction \(x\), the Kirchhoff tensor \(\tau\) is of the form:

_images/Object_14.svg

The gradient tensors of the transformation

_images/Object_15.svg

and

_images/Object_16.svg

and the isochoric plastic deformation tensor

_images/Object_17.svg

are of the form:

_images/Object_18.svg

Through the law of behavior, we obtain the following relationship:

_images/Object_19.svg

either

_images/Object_20.svg

The Cauchy constraint is written as:

_images/Object_21.svg

In plastic filler for isotropic work hardening

_images/Object_22.svg

linear, such as:

_images/Object_23.svg

we have:

_images/Object_24.svg

Integrating the flow law of plastic deformation \({G}^{P}\) gives (knowing that \({G}^{P}(p\mathrm{=}0)\mathrm{=}1\)):

\({G}^{P}\mathrm{=}{e}^{\mathrm{-}2p}\)

The \(\stackrel{ˉ}{F}\) component of the transformation gradient is given by the resolution of:

_images/Object_29.svg

The displacement field \(u\) (in the initial configuration) is of the form \(u\mathrm{=}{u}_{x}X+{u}_{y}Y+{u}_{z}Z\). The components are given by:

_images/Object_32.svg

2.2. Benchmark results#

Displacements, Cauchy stress \(\sigma\) and cumulative plastic deformation \(p\) will be adopted as reference results.

At time \(t\mathrm{=}\mathrm{2 }s\) (\(\Delta T\mathrm{=}100°C\), pull \(u\))

We are looking for the total displacement (thermal + mechanical) such that the stress \(\tau\) is equal to:

\(\tau \mathrm{=}1500\mathit{Mpa}\) (to \(T\mathrm{=}120°C\))

  • \(\mathrm{3K}\mathrm{=}500000\mathit{MPa}\) \(\mu \mathrm{=}76923\mathit{MPa}\)

  • _images/Object_42.svg

  • \(\sigma \mathrm{=}1453\mathit{MPa}\)

  • \(p\mathrm{=}\mathrm{0,2475}\)

  • \({G}^{p}\mathrm{=}\mathrm{0,609}\)

  • \(\stackrel{ˉ}{F}\mathrm{=}\mathrm{1,289}\)

  • \(F\mathrm{=}\mathrm{1,303}\)

  • \(\tilde{u}\mathrm{=}303\mathit{mm}\)

  • \(\tilde{v}\mathrm{=}–110\mathit{mm}\)

With these quantities, it is possible to determine the elastic energy of the bar. Attention, the presence of thermal generates a high Jacobian, requiring a specific correction as described in R5.03.21. In the end, we get at the material point: \({\Psi }_{\mathit{elas}}\mathrm{=}\mathrm{5,6}\mathit{MPa}\)

2.3. Uncertainty about the solution#

The solution is analytical. Except for rounding errors, it can be considered accurate.

2.4. Bibliographical references#

Reference may be made to:

    1. CANO, E. LORENTZ: Introduction to the Code_Aster of a model of behavior in large elastoplastic deformations with isotropic work hardening - Internal note EDF DER HI‑74/98/006/0