2. Benchmark solution

The imposed loading allows us to obtain a homogeneous solution, equivalent to a bar of length \(L\) subjected to uniaxial tensile loading. We can then express elastic energy in the following way:

\({W}_{\mathrm{el}}=\frac{E}{2}{(\frac{\mathrm{dx}}{L})}^{2}\)

The damage values associated with the moments \({t}_{1}\), \({t}_{2}\) and \({t}_{3}\) are deduced analytically from the formula:

\(d=1-{(\frac{{\sigma }_{y}L}{E\mathrm{dx}})}^{2}\)

That is: \({d}_{1}=0.\), \({d}_{2}=0.36\) and \({d}_{3}=0.75\). We then consider that the test is verified if Newton returns the same damage values to us, at a precision of \({10}^{-6}\).