2. Benchmark solution

2.1. Calculation method used for the reference solution

For a circular crack of radius \(a\) in an infinite medium, subjected to a uniform traction \(\mathrm{\sigma }\) following the normal to the plane of the lips, the local energy restoration rate \(G(s)\) is independent of the curvilinear abscissa \(s\) and is equal to [bib1]:

\(G(s)\mathrm{=}\frac{(1\mathrm{-}{\nu }^{2})}{\pi E}4{\sigma }^{2}a\)

then the stress intensity coefficient \(\mathrm{K1}\) is given by Irwin’s formula:

Be \({K}_{I}\mathrm{=}\frac{2\sigma \sqrt{a}}{\sqrt{\pi }}\)

If this crack is subject to shear spread over the lips: \({\sigma }_{\theta z}\mathrm{=}\tau \frac{r}{a}\)

(which is equivalent to twisting to infinity), then we are in pure 3 mode and the corresponding stress intensity factor is:

\({K}_{3}\mathrm{=}\frac{4\tau \sqrt{a}}{3\sqrt{\pi }}\) so by Irwin’s formula \(G(s)\mathrm{=}\frac{(1+\nu )}{E}{K}_{3}^{2}\)

In the presence of the two combined modes, we will have:

\(G(s)\mathrm{=}\frac{(1\mathrm{-}{\nu }^{2})}{E}{K}_{1}^{2}+\frac{(1+\nu )}{E}{K}_{3}^{2}\)

2.2. Benchmark results

Digital application (case with traction load only):

For the load in question and \(a=2m\), we then get:

\(G(s)\mathrm{=}11.586J\mathrm{/}{m}^{2}\)

\(\mathit{K1}\mathrm{=}1.5958E6J\mathrm{/}{m}^{2}\)

For modeling \(G\) (3 different crack backgrounds) with the same load, we obtain:

For \(a=2m\)

\(G(s)=10.586J/{m}^{2}\)

\(\mathrm{K1}=1.5958E6J/{m}^{2}\)

For \(a=1.88m\)

\(G(s)=10.891J/{m}^{2}\)

\(\mathrm{K1}=1.5472E6J/{m}^{2}\)

For \(a=1.76m\)

\(G(s)=10.196J/{m}^{2}\)

\(\mathrm{K1}=1.4969E6J/{m}^{2}\)

Digital application (cases with torsional loading only):

\(G(s)=7.3565J/{m}^{2}\)

\(\mathrm{K1}=1.0638E6J/{m}^{2}\)

2.3. Bibliographical references

1. Solution by Sneddon (1946) in G.C. SIH: Handbook of stress-intensity factors Institute of Fracture and Solid Mechanics - Lehigh University Bethlehem, Pennsylvania