1. Reference problem#

1.1. Geometry#

The crack is circular (penny shaped crack) with radius \(a=\mathrm{0,2}m\) in the \(\mathit{OXY}\) plane. The side of the cube is \(L=2m=10a\) long. Thus, it is considered that the crack is in an infinite medium.

_images/100000000000034E000002C1E48DED210B8710B1.png

Figure 1.1-1 : geometry of the cracked cube 2222

1.2. Material properties#

We consider a homogeneous isotropic elastic linear material whose characteristics are the following:

  • Young’s module \(E=200000\mathit{MPa}\),

  • Poisson’s ratio \(\nu \mathrm{=}\mathrm{0,3}\).

1.3. Boundary conditions and loads#

Given the symmetries of the structure, the crack, and the loads, only half of the structure is modeled: half-space such as \(Y>0\). Symmetry conditions are therefore applied to the face in \(Y\mathrm{=}0\): on this face, the following movement \(Y\) is blocked.

The structure is subject to a \({\sigma }_{\mathit{xx}}\mathrm{=}1\mathit{MPa}\) constraint. The local coordinate system \((x,y,z)\) is obtained by rotating an angle \(\alpha \mathrm{=}\pi \mathrm{/}4\) around the axis \(\mathit{OY}\).

_images/10000000000002B20000022EE428EC6E1D6BBC34.png

Figure 1.3-1 : Constraints in the global coordinate system

_images/100000000000015B000000E8C71967D18D36A085.png

Figure 1.3-2 : Local location

So, we have:

on the upper side:

  • \({\sigma }_{\mathit{ZZ}}=\sigma {\mathrm{sin}}^{2}(\alpha )\)

  • \({\sigma }_{\mathit{ZX}}=\sigma \mathrm{cos}(\alpha )\mathrm{sin}(\alpha )\)

on the underside:

  • \({\sigma }_{\mathit{ZZ}}\mathrm{=}\mathrm{-}\sigma {\mathrm{sin}}^{2}(\alpha )\)

  • \({\sigma }_{\mathit{ZX}}\mathrm{=}\mathrm{-}\sigma \mathrm{sin}(\alpha )\mathrm{cos}(\alpha )\)

on the right side face:

  • \({\sigma }_{\mathit{XX}}\mathrm{=}\sigma {\mathrm{cos}}^{2}(\alpha )\)

  • \({\sigma }_{\mathit{XZ}}\mathrm{=}\sigma \mathrm{cos}(\alpha )\mathrm{sin}(\alpha )\)

on the left side face:

  • \({\sigma }_{\mathit{XX}}\mathrm{=}\mathrm{-}\sigma {\mathrm{cos}}^{2}(\alpha )\)

  • \({\sigma }_{\mathit{XZ}}=-\sigma \mathrm{cos}(\alpha )\mathrm{sin}(\alpha )\)

In order to block rigid body modes, the point \(\mathit{D1}\) \((L\mathrm{/}\mathrm{2,0}\mathrm{,0})\) is blocked following \(X\) and \(Z\) and the point \(\mathit{D2}\) \((\mathrm{-}L\mathrm{/}\mathrm{2,0}\mathrm{,0})\) is blocked following \(Z\).

_images/10000000000002C200000249CE148B0A9575D322.png

Figure 1.3-3 : Blocking rigid modes