2. Benchmark solution#

2.1. Calculation method used for the reference solution#

We rely on the methodology already described in fascicle [V6.04.263] in which a stress state with axial symmetry is imposed 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. The equations of equilibrium in the volume are verified by imposing volume forces in the cylinder and surface forces on the edges of the cylinder:

    • in volume: \(f=-{\sigma }_{\mathit{rz}}/r{e}_{z}\)

    • 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 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.

The path of the transient in the monotonic estradial stress space:

\(\sigma (x,t)=q(t){\mathrm{\Sigma }}^{0}(x)\)

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

And in this case, \({\sigma }_{\mathit{eq}}=q\). The equation for the evolution of plastic deformation is easy to integrate since the direction is fixed, by noting \(\kappa\) the work-hardening variable:

\({\epsilon }^{p}=\frac{3}{2}\kappa \mathit{dev}({\mathrm{\Sigma }}^{0})\)

Finally, the elastic deformation is deduced directly from the stress and makes it possible to go back to deformation \(\epsilon ={\epsilon }^{e}+{\epsilon }^{p}\) and to the displacements:

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

It therefore remains to determine the evolution of \(\kappa\) according to the loading level \(q(t)\). Note \(R(\kappa )\) the work hardening function:

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

Norton’s law provides the following expression that links the load to the work-hardening variable:

\(\dot{\kappa }={\left(\frac{⟨q-R(\kappa )⟩}{K}\right)}^{n}\)

Instead of solving this differential equation explicitly, we can start by inverting it:

\(q=R(\kappa )+V(\dot{\kappa })\text{;}V(\dot{\kappa })=K{\dot{\kappa }}^{\frac{1}{n}}\)

By choosing a linear evolution in time \(\kappa (t)={\dot{ϵ}}_{0}t\), we deduce the load to be applied by the previous relationship. In practice, we consider a transient lasting 1 s, with \({\dot{ϵ}}_{0}=0.05{s}^{-1}\).

It remains to consider the initial state in \(t=0\). The work-hardening variable is zero as is the plastic deformation, taking into account the preceding relationships. On the other hand, the effort applied is not zero since \(q(0)=V({\dot{ϵ}}_{0})={q}_{0}\). The initial mechanical state is therefore characterized by a non-zero deformation, a solution to the elastic problem under load \({q}_{0}\).

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).

2.2. Benchmark results#

We will ensure that by targeting an cumulative plastic deformation of 5% reached in one second at a constant rate, 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