2. Benchmark solution#
2.1. Calculation method#
The analytic expressions of the stress intensity factors \({K}_{I}\) and \({K}_{\mathrm{II}}\) are functions of the distributed force \(p\), of the length of the crack a, of the width of the plate \(\mathrm{Lx}\):
\(\begin{array}{}{K}_{I}=p\sqrt{\pi a}f(\frac{a}{\mathrm{Lx}})\\ {K}_{\mathrm{II}}=0\end{array}\)
where function \(f\) can be determined in several different ways. We choose the one obtained by [1], and that is true for \(\begin{array}{}\frac{a}{\mathrm{Lx}}<\mathrm{0,6}\end{array}\):
\(\begin{array}{}f(\frac{a}{\mathrm{Lx}})=\mathrm{1,12}-\mathrm{0,231}(\frac{a}{\mathrm{Lx}})+\mathrm{10,55}{(\frac{a}{\mathrm{Lx}})}^{2}-\mathrm{21,72}{(\frac{a}{\mathrm{Lx}})}^{3}+\mathrm{30,39}{(\frac{a}{\mathrm{Lx}})}^{4}\end{array}\)
We are moving the crack forward thanks to the Paris law:
\(\begin{array}{}\frac{\mathrm{da}}{\mathrm{dN}}=C\Delta {K}^{m}\end{array}\) where a is the crack length, \(C\) and \(m\) are material constants, \(\begin{array}{}\Delta K\end{array}\) is the difference between two consecutive FICs’s, and \(N\) is the number of cycles.
With the numerical values of the test:
No spread: \(\mathrm{0,25}m\)
\(\mathrm{Lx}\): \(10m\)
2.2. Reference quantities and results#
Reference |
||
\(a(m)\) |
\({K}_{I}({\mathrm{Pa.m}}^{\mathrm{0,5}})\) |
\({K}_{\mathrm{II}}({\mathrm{Pa.m}}^{\mathrm{0,5}})\) |
2.5 |
4,205998 106 |
0 |
2.75 |
4,63286 106 |
0 |
3 |
5,09492 106 |
0 |
3.25 |
5,59908 106 |
0 |
3.5 |
6,15349 106 |
0 |
3.75 |
6,76776 106 |
0 |
4 |
7,4531 106 |
0 |
4.25 |
8,2224 106 |
0 |
4.5 |
9,0905 106 |
0 |
4.75 |
1,0074 107 |
0 |
5 |
1,1192 107 |
0 |
5.25 |
1,2465 107 |
0 |
5.5 |
1,3916 107 |
0 |
5.75 |
1,55716 107 |
0 |
6 |
1,74586 107 |
0 |
Table 2.2-1 : reference values for \({K}_{I}\) and \({K}_{\mathrm{II}}\)
2.3. Uncertainties about the solution#
None, analytical solution.
2.4. Bibliographical references#
TADA H., PARIS P., IRWIN G.:The stress analysis of cracks, Handbook. Del Research Corporation, Hellertown, PA, 1973.