Benchmark solution
=====================


Calculation method used for the reference solution
--------------------------------------------------------

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

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

then the stress intensity coefficient :math:`\mathrm{K1}` is given by Irwin's formula:


.. image:: images/Object_7.svg
    :width: 127
    :height: 49


.. _RefImage_Object_7.svg:

Be :math:`{K}_{I}\mathrm{=}\frac{2\sigma \sqrt{a}}{\sqrt{\pi }}`

If this crack is subject to shear spread over the lips: :math:`{\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:

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

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

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


Benchmark results
----------------------

Digital application (case with traction load only):

For the load in question and :math:`a=2m`, we then get:


.. csv-table::

    ":math:`G(s)\mathrm{=}11.586J\mathrm{/}{m}^{2}`"
    ":math:`\mathit{K1}\mathrm{=}1.5958E6J\mathrm{/}{m}^{2}`"


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

For :math:`a=2m`


.. csv-table::

    ":math:`G(s)=10.586J/{m}^{2}`"
    ":math:`\mathrm{K1}=1.5958E6J/{m}^{2}`"


For :math:`a=1.88m`


.. csv-table::

    ":math:`G(s)=10.891J/{m}^{2}`"
    ":math:`\mathrm{K1}=1.5472E6J/{m}^{2}`"


For :math:`a=1.76m`


.. csv-table::

    ":math:`G(s)=10.196J/{m}^{2}`"
    ":math:`\mathrm{K1}=1.4969E6J/{m}^{2}`"


Digital application (cases with torsional loading only):


.. csv-table::

    ":math:`G(s)=7.3565J/{m}^{2}`"
    ":math:`\mathrm{K1}=1.0638E6J/{m}^{2}`"



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