2. Benchmark solution#
2.1. Calculation method used for the reference solution#
1 The reference solution is an analytical solution resulting from, for a circular crack with radius \(a\) in an infinite medium, subject to a uniform surface force \(\sigma\) inclined at an angle \(\alpha\) to the plane of the crack, the stress intensity factors for a point \(A\) placed on the crack front are equal to: —-
\({K}_{I}\mathrm{=}\frac{2}{\pi }\sigma ({\mathrm{sin}}^{2}\alpha )\sqrt{\pi a}\)
\({K}_{\mathit{II}}\mathrm{=}\frac{4}{\pi (2\mathrm{-}\nu )}\sigma (\mathrm{sin}\alpha \mathrm{cos}\alpha )\mathrm{cos}\omega \sqrt{\pi a}\)
\({K}_{\mathit{III}}\mathrm{=}\frac{4(1\mathrm{-}\nu )}{\pi (2\mathrm{-}\nu )}\sigma (\mathrm{sin}\alpha \mathrm{cos}\alpha )\mathrm{sin}\omega \sqrt{\pi a}\)
\(\omega\) being the angle characterizing the position of the point \(A\) on the circular background (see).
2.2. Benchmark results#
For the load under consideration and \(a=\mathrm{0,2}m\), the values are shown in the figure and the table.
Figure 2.2-1 : Reference values for SIFs
Angle \(\mathrm{\omega }\) (rad) |
\({K}_{I}\) \((\mathit{MPa}\mathrm{.}\sqrt{M})\) |
|
|
|
0 |
0, 252 |
0, 297 |
0 |
|
\(\mathrm{\pi }/4\) |
0, 252 |
0, 2099 |
0, 147 |
|
\(\mathrm{\pi }/2\) |
0, 252 |
0 |
0 |
208 » |
\(3\mathrm{\pi }/4\) |
0, 252 |
-0, 2099 |
0, 147 |
|
\(\mathrm{\pi }\) |
0, 252 |
-0, 297 |
0 |
2.3. Bibliographical references#
TADA H., PARIS P., IRWIND G.: The stress analysis of cracks handbook, 3rd ed., 2000