2. Benchmark solution

The solution to the problem posed can be determined analytically.

We note:

• \({\varepsilon }_{0}\) the deformation applied in the \(z\) direction,

• \({\varepsilon }_{1}\), \({\varepsilon }_{2}\), and \({\varepsilon }_{3}\) the main deformations

2.1. First loading step: simple compression

• The deformation tensor is equal to:

  

with \({\varepsilon }_{0}<0\)

• The equivalent deformation is therefore equal to:

  

• As soon as

  

, there is an evolution of the damage which is equal to:

• Finally, constraint \({\sigma }_{\mathit{zz}}\) is equal to:

2.2. Second loading stage: thermal expansion in plane deformations

• The tensor of the total deformations is equal to:

with fixed \({\varepsilon }_{0}<0\)

• The elastic deformation equal

  

, the equivalent deformation is equal to:

• The damage is worth:

• Finally, constraint \({\sigma }_{\mathit{zz}}\) is equal to:

Note:

• • In a given state, the material parameters used are those defined at the maximum temperature seen by the material and not at the current temperature.

  • The assessment of damage \(D\) involves the concept of maximum reached during the history of the load; the solution is therefore not completely analytical but involves discretization. In the case where there is no thermal influence, simply take \(\stackrel{~}{\mathrm{\epsilon }}\) equivalent to the maximum equivalent deformation reached. When the thermal aspect is taken into account, heating can contribute to « reducing » or « delaying » damage at a given deformation; this is the case with the evolution of \({B}_{c}\) retained. In this case, it is in fact necessary to discretize the load quite finely to have the right damage value \(D\) (which in fact has a maximum in our case) .