2. Benchmark solution

2.1. Calculation method used for the reference solution

The aim is to impose a stress state with axial symmetry and whose components \({\sigma }_{\mathit{rr}}\), \({\sigma }_{\theta \theta }\), \({\sigma }_{\mathit{zz}}\) and \({\sigma }_{\mathit{rz}}\) do not depend on the position \((r,z)\) step. We are also restricted to the situation where \({\sigma }_{\mathit{rr}}={\sigma }_{\theta \theta }\). The equations of equilibrium in the volume are verified as long as a vertical volume force is imposed:

\({f}_{z}=-\frac{{\sigma }_{\mathit{rz}}}{r}\)

The equilibrium equations at the edges of the cylinder require the imposition of surface forces \(T\) equal to \(\sigma \cdot n\) where \(n\) designates the outgoing normal to the surface. They are written:

• • top edge \(z=H\): \(T={\sigma }_{\mathit{zz}}{e}_{z}+{\sigma }_{\mathit{rz}}{e}_{r}\)

  • bottom edge \(z=0\): \(T=-{\sigma }_{\mathit{zz}}{e}_{z}-{\sigma }_{\mathit{rz}}{e}_{r}\)

  • inner wall \(r={R}_{i}\): \(T=-{\sigma }_{\mathit{rr}}{e}_{r}-{\sigma }_{\mathit{rz}}{e}_{z}\)

  • outer wall \(r={R}_{e}\): \(T={\sigma }_{\mathit{rr}}{e}_{r}+{\sigma }_{\mathit{rz}}{e}_{z}\)

The constraint field is then in balance with the imposed load.

Given the form of the behavioral equations, in particular their isotropy, the components of the deformation field in the cylindrical coordinate system are also constant (in space) and are limited to \({\epsilon }_{\mathit{rr}}={\epsilon }_{\theta \theta }\), \({\epsilon }_{\mathit{zz}}\) and \({\epsilon }_{\mathit{rz}}\). This deformation field is geometrically compatible with the following field of displacement (with axial symmetry):

\({u}_{r}={\epsilon }_{\mathit{rr}}r\text{;}{u}_{z}={\epsilon }_{\mathit{zz}}z+2{\epsilon }_{\mathit{rz}}(r-{R}_{i})\)

where we set the integration constant so that \({u}_{z}({R}_{i}\mathrm{,0})=0\), a boundary condition that blocks the movement of rigid bodies.

It now remains to fix the path of the transient in the space of the constraints. To simplify the integration of the behavior, we opt for a monotonic radial path. The constraint is thus imposed in the form \(\sigma =q{\mathrm{\Sigma }}^{0}\), where \(q\ge 0\) refers to the intensity of the stresses. The following components are chosen for \({\mathrm{\Sigma }}^{0}\) in the cylindrical coordinate system:

\({\mathrm{\Sigma }}_{\mathit{rr}}^{0}={\mathrm{\Sigma }}_{\theta \theta }^{0}=\frac{1}{6}\text{;}{\mathrm{\Sigma }}_{\mathit{zz}}^{0}=\frac{2}{3}\text{;}{\mathrm{\Sigma }}_{\mathit{rz}}^{0}=\frac{1}{2}\)

It is a tensor whose VonMises equivalent stress is equal to 1, so that \({\sigma }_{\mathit{eq}}=q\). We note \({N}^{0}\) the deviator sound. The equation for the evolution of plastic deformation is easily integrated since the direction is fixed and equal to \({N}^{0}\):

\({\epsilon }^{p}=\frac{3}{2}\kappa {N}^{0}\)

The work hardening variable \(\kappa\) depends (in plastic load) only on the equivalent stress \(q\), via the coherence equation:

\(q=R(\kappa )={R}_{0}+{R}_{H}\kappa +{R}_{1}(1-{e}^{-{\gamma }_{1}\kappa })+{R}_{2}(1-{e}^{-{\gamma }_{2}\kappa })+{R}_{K}{(\kappa +{p}_{0})}^{{\gamma }_{M}}\)

It is obviously easier to give yourself a target value for \(\kappa\) (5%, for example) and to derive from it the required \(q\) load (830.364 MPa).

Finally, the elastic deformation is deduced directly from the stress:

\({\epsilon }^{e}=q\left[\frac{1-\nu }{E}{\mathrm{\Sigma }}^{0}-\frac{\nu }{E}\mathit{tr}{\mathrm{\Sigma }}^{0}\mathit{Id}\right]\)

It makes it possible to go back to deformation \(\epsilon ={\epsilon }^{e}+{\epsilon }^{p}\) and to displacements.

In the case of a formulation with a plasticity gradient, the results remain unchanged. In fact, the work-hardening variable \(\kappa\) is homogeneous, so that its gradient is zero: non-local effects do not appear in this problem (except for spatial discretization errors).

Finally, we can also apply the history of movements at any point in the structure to find the stress response. We take advantage of this approach to also test the case with Lüders plate in work hardening (a constrained approach is then not possible due to the perfect plastic nature of the tray). The stresses are unchanged beyond the plateau and locked at the equivalent stress of the plateau for deformations less than the limit of the plateau.

2.2. Benchmark results

We will ensure that by targeting an cumulative plastic deformation of 5%, the model regains the expected levels of stresses (\({\sigma }_{\mathit{rr}}\)), displacements and plastic deformations (\({\epsilon }_{\mathit{rz}}^{p}\)).

2.3. Uncertainties about the solution

Nil.

2.4. Bibliographical references

Néant