Benchmark solution ===================== Calculation method used for the reference solution -------------------------------------------------------- We rely on the methodology already described in fascicle [:ref:`V6.04.263 `] in which a stress state with axial symmetry is imposed and whose components :math:`{\sigma }_{\mathit{rr}}={\sigma }_{\theta \theta }`, :math:`{\sigma }_{\mathit{zz}}` and :math:`{\sigma }_{\mathit{rz}}` do not depend on the position :math:`(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: :math:`f=-{\sigma }_{\mathit{rz}}/r{e}_{z}` * top edge :math:`z=H`: :math:`T={\sigma }_{\mathit{zz}}{e}_{z}+{\sigma }_{\mathit{rz}}{e}_{r}` * bottom edge :math:`z=0`: :math:`T=-{\sigma }_{\mathit{zz}}{e}_{z}-{\sigma }_{\mathit{rz}}{e}_{r}` * inner wall :math:`r={R}_{i}`: :math:`T=-{\sigma }_{\mathit{rr}}{e}_{r}-{\sigma }_{\mathit{rz}}{e}_{z}` * outer wall :math:`r={R}_{e}`: :math:`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 :math:`{\epsilon }_{\mathit{rr}}={\epsilon }_{\theta \theta }`, :math:`{\epsilon }_{\mathit{zz}}` and :math:`{\epsilon }_{\mathit{rz}}`. This deformation field is geometrically compatible with the following field of displacement (with axial symmetry): :math:`{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 :math:`{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: :math:`\sigma (x,t)=q(t){\mathrm{\Sigma }}^{0}(x)` :math:`{\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, :math:`{\sigma }_{\mathit{eq}}=q`. The equation for the evolution of plastic deformation is easy to integrate since the direction is fixed, by noting :math:`\kappa` the work-hardening variable: :math:`{\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 :math:`\epsilon ={\epsilon }^{e}+{\epsilon }^{p}` and to the displacements: :math:`{\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 :math:`\kappa` according to the loading level :math:`q(t)`. Note :math:`R(\kappa )` the work hardening function: :math:`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: :math:`\dot{\kappa }={\left(\frac{⟨q-R(\kappa )⟩}{K}\right)}^{n}` Instead of solving this differential equation explicitly, we can start by inverting it: :math:`q=R(\kappa )+V(\dot{\kappa })\text{;}V(\dot{\kappa })=K{\dot{\kappa }}^{\frac{1}{n}}` By choosing a linear evolution in time :math:`\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 :math:`{\dot{ϵ}}_{0}=0.05{s}^{-1}`. It remains to consider the initial state in :math:`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 :math:`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 :math:`{q}_{0}`. In the case of a formulation with a plasticity gradient, the results remain unchanged. In fact, the work-hardening variable :math:`\kappa` is homogeneous, so that its gradient is zero: non-local effects do not appear in this problem (except for spatial discretization errors). 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 (:math:`{\sigma }_{\mathit{rr}}`), displacements and plastic deformations (:math:`{\epsilon }_{\mathit{rz}}^{p}`). Uncertainties about the solution ---------------------------- Nil. Bibliographical references --------------------------- Néant