2. Benchmark solution#

The analytical solution of SIMO_MIEHE makes it possible to define the load to be applied to obtain a Kirchhoff stress of \(\mathrm{1500MPa}\). This load is then applied to the other models.

2.1. Generic result to formalisms#

For a uniaxial tensile test in direction \(x\), the Kirchhoff \(\tau\) and Cauchy \(\sigma\) stress tensors are of the form:

\(\sigma \mathrm{=}\left[\begin{array}{ccc}\sigma & 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],\tau \mathrm{=}J\sigma \mathrm{=}\left[\begin{array}{ccc}\tau & 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\),

The gradient tensors of the transformation

_images/Object_15.svg

and

_images/Object_16.svg

express themselves:

\(F\mathrm{=}\left[\begin{array}{ccc}F& 0& 0\\ 0& {F}_{y}& 0\\ 0& 0& {F}_{y}\end{array}\right],\stackrel{ˉ}{F}\mathrm{=}{J}^{\frac{\mathrm{-}1}{3}}\mathrm{=}\left[\begin{array}{ccc}\stackrel{ˉ}{F}& 0& 0\\ 0& \stackrel{ˉ}{{F}_{y}}& 0\\ 0& 0& \stackrel{ˉ}{{F}_{y}}\end{array}\right]\)

The move verifies:

\(F\mathrm{=}1+\frac{{u}^{\mathit{meca}}}{{l}_{0}}\)

The linear isotropic work hardening function is written as:

\(R(p)\mathrm{=}\frac{E{E}_{T}}{E\mathrm{-}{E}_{T}}\)

2.2. Results for Simo_Miehe#

The isochoric plastic deformation tensor \({G}^{P}\) is of the form:

\({G}^{P}\mathrm{=}\left[\begin{array}{ccc}{G}^{P}& 0& 0\\ 0& {G}_{y}^{P}& 0\\ 0& 0& {G}_{y}^{P}\end{array}\right]\), with \({G}_{y}^{P}\mathrm{=}\frac{1}{\sqrt{{G}^{P}}}\) because \(\mathit{det}({G}^{P})\mathrm{=}1\)

The law of behavior (hydrostatic part) is written as:

\(\mathit{tr}(\tau )\mathrm{=}\frac{\mathrm{3K}}{2}({J}^{2}\mathrm{-}1)\mathrm{\Rightarrow }{J}^{2}\mathrm{=}\frac{2\tau }{\mathrm{3K}}+1\)

The plasticity threshold function is written as:

\(f\mathrm{=}0\mathrm{=}\tau \mathrm{-}R(p)\mathrm{-}{\sigma }_{y}\mathrm{\Rightarrow }p\mathrm{=}\frac{E\mathrm{-}{E}_{T}}{E{E}_{T}}(\tau \mathrm{-}{\sigma }_{y})\)

The law of plastic flow is written as:

\(\stackrel{ˉ}{F}\mathrm{.}\dot{{G}^{P}}\mathrm{.}\stackrel{ˉ}{{F}^{T}}\mathrm{=}\mathrm{-}3\dot{p}\frac{1}{{\tau }_{\mathit{eq}}}\mathit{dev}(\tau )\mathrm{.}(\stackrel{ˉ}{F}\mathrm{.}{G}^{P}\mathrm{.}\stackrel{ˉ}{{F}^{T}})\)

Taking the first component, we get:

\(\frac{\dot{{G}^{P}}}{{G}^{P}}\mathrm{=}\mathrm{-}2\dot{p}\mathrm{\Rightarrow }{G}^{p}\mathrm{=}{e}^{\mathrm{-}\mathrm{2p}}\) because \({G}^{P}(0)\mathrm{=}1\)

To conclude the problem, we use the first component of the deviatory part of the constraint:

\(\mathit{dev}(\tau )\mathrm{=}\mu \mathit{dev}(\stackrel{ˉ}{{b}^{e}})\mathrm{\Rightarrow }\tau \mathrm{=}\mu (\stackrel{ˉ}{{F}^{2}}{G}^{P}\mathrm{-}\stackrel{ˉ}{{F}_{y}^{2}}{G}_{y}^{P})\mathrm{\Rightarrow }\stackrel{ˉ}{{F}^{3}}\mathrm{-}\stackrel{ˉ}{F}\frac{\tau }{\mu {G}^{P}}\mathrm{-}\frac{1}{{({G}^{P})}^{(3\mathrm{/}2)}}\mathrm{=}0\),

Car \(\stackrel{ˉ}{{b}^{e}}\mathrm{=}\stackrel{ˉ}{F}\mathrm{.}{G}^{P}\mathrm{.}\stackrel{ˉ}{{F}^{T}}\)

Elastic energy is then written as:

\({\psi }_{\mathrm{SM}}^{\mathrm{elas}}=\frac{K}{2}\left[\frac{{J}^{2}-1}{2}-\mathrm{ln}J\right]+\frac{\mu }{2}\left[\mathrm{tr}\stackrel{ˉ}{{b}^{e}}-3\right]=\frac{K}{2}\left[\frac{{J}^{2}-1}{2}-\mathrm{ln}J\right]+\frac{\mu }{2}\left[\stackrel{ˉ}{{F}^{2}}{G}^{P}+\frac{2}{\stackrel{ˉ}{F}\sqrt{{G}^{P}}}-3\right]\)

For a Kirchhoff constraint of 1500 MPa, can then successively determine:

  • \(J\mathrm{=}\mathrm{1,003}\)

  • \(\sigma \mathrm{=}1495\mathit{MPa}\)

  • \(p\mathrm{=}\mathrm{0,2475}\)

  • \({G}^{p}\mathrm{=}\mathrm{0,6096}\)

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

  • \(F\mathrm{=}\mathrm{1,290}\)

  • \({u}^{\mathit{meca}}\mathrm{=}290\mathit{mm}\)

  • \({\psi }_{\mathrm{SM}}^{\mathrm{elas}}=\mathrm{5,63}\mathrm{MPa}\) at the material point.

The displacement applied for the two models and the 3 formalisms will therefore be \({u}^{\mathit{meca}}\mathrm{=}290\mathit{mm}\).

2.3. Results for GDEF_LOG#

Logarithmic deformation is written as:

\({E}_{\mathrm{log}}\mathrm{=}\frac{1}{2}\mathrm{ln}\mathrm{[}C\mathrm{]}\mathrm{=}\frac{1}{2}\mathrm{ln}\mathrm{[}{F}^{T}\mathrm{.}F\mathrm{]}\mathrm{=}\left[\begin{array}{ccc}\mathrm{ln}F& 0& 0\\ 0& \mathrm{ln}{F}_{y}& 0\\ 0& 0& \mathrm{ln}{F}_{y}\end{array}\right]\)

The upper left quarter of the Lagragian projector is deduced from this, in Voigt notation:

\(P\mathrm{=}2\frac{\mathrm{\partial }{E}_{\mathrm{log}}}{\mathrm{\partial }C}\mathrm{=}\frac{1}{2}\left[\begin{array}{ccc}\frac{2}{{F}^{2}}& 0& 0\\ 0& \frac{2}{{F}_{y}^{2}}& 0\\ 0& 0& \frac{2}{{F}_{y}^{2}}\end{array}\right]\)

And because of the expression of the second Piola-Kirchhoff constraint:

\(S\mathrm{=}T\mathrm{:}P\mathrm{=}{F}^{\mathrm{-}1}\mathrm{.}\tau \mathrm{.}{F}^{\mathrm{-}T}\mathrm{\Rightarrow }T\mathrm{=}\tau\)

The law of behavior is written as:

\(T\mathrm{=}E(\mathrm{ln}F\mathrm{-}p)\)

Because of the plasticity threshold:

\(f\mathrm{=}0\mathrm{=}T\mathrm{-}R(p)\mathrm{-}{\sigma }_{y}\mathrm{\Rightarrow }p\mathrm{=}\frac{E\mathrm{-}{E}_{T}}{E{E}_{T}}(T\mathrm{-}{\sigma }_{y})\mathrm{=}\frac{E\mathrm{ln}F\mathrm{-}{\sigma }_{y}}{E+\frac{E{E}_{T}}{E\mathrm{-}{E}_{T}}}\),

The elastic energy of this formalism is therefore written as:

\({\psi }_{\mathrm{log}}^{\mathit{elas}}\mathrm{=}\frac{1}{2}\frac{{T}^{2}}{E}\)

From the imposed displacement \({u}^{\mathit{meca}}\mathrm{=}290\mathit{mm}\), we deduce:

  • \(F\mathrm{=}\mathrm{1,290}\)

  • \(\mathrm{ln}F\mathrm{=}\mathrm{0,255}\)

  • \(p\mathrm{=}\mathrm{0,2475}\)

  • \(T\mathrm{=}\mathrm{1500MPa}\)

  • \(\sigma \mathrm{=}\mathrm{1495MPa}\)

  • \({\psi }_{\mathrm{log}}^{\mathit{elas}}\mathrm{=}\mathrm{5,625}\mathit{MPa}\) at the material point.

2.4. Results in small deformations#

In small deformations, the result is classical.

Axial deformation:

\({\varepsilon }_{x}\mathrm{=}\frac{{u}^{\mathit{meca}}}{{l}_{0}}\)

Behaviour:

\(\sigma \mathrm{=}E({\varepsilon }_{x}\mathrm{-}p)\)

Threshold function:

\(\sigma \mathrm{-}R(p)\mathrm{-}{\sigma }_{y}\mathrm{=}0\)

From where:

\(\begin{array}{c}p\mathrm{=}\frac{E{\varepsilon }_{x}\mathrm{-}{\sigma }_{y}}{E+\frac{E{E}_{T}}{E\mathrm{-}{E}_{T}}}\\ \sigma \mathrm{=}E({\varepsilon }_{x}\mathrm{-}p)\end{array}\)

Elastic energy:

\({\psi }_{\mathit{HPP}}^{\mathit{elas}}\mathrm{=}\frac{{\sigma }^{2}}{2E}\)

From the imposed displacement \({u}^{\mathit{meca}}\mathrm{=}290\mathit{mm}\), we deduce:

  • \({\varepsilon }_{x}\mathrm{=}\mathrm{0,029}\)

  • \(p\mathrm{=}\mathrm{0,281}\)

  • \(\sigma \mathrm{=}\mathrm{1570MPa}\)

  • \({\psi }_{\mathrm{log}}^{\mathit{elas}}\mathrm{=}\mathrm{6,16}\mathit{MPa}\) at the material point.

2.5. Tests carried out#

For each of the formalisms and each of the models, we test the values of the imposed displacement, of the Cauchy stress \(\sigma\), of the cumulative plastic deformation \(p\) and of the elastic energy.

Attention, the elastic energy tested is the value for the bar (not the material point). For axisymmetric modeling, it is therefore necessary to multiply the value of the elastic energy of the material point by \(\frac{\pi {R}^{2}}{2\pi }\).