1. Reference problem#

1.1. Description#

This test case comes from sheet 26795 reporting a sudden stop in Hujeux’s law due to a segmentation error in a study of the construction by layers of an embankment dam.

It reproduces on a material point the loading path that caused the crash. This loading path leads to the failure of local Newton iterations and activates a heuristic mechanism for restarting the resolution.

1.2. Material properties#

1.2.1. Elastic properties of the material#

The material is of the type of dense sand. The elastic properties are:

  • Young’s modulus: \(E=2029431300.40069\mathit{Pa}\)

  • Poisson’s ratio: \(\mathrm{\nu }=0.45\)

The anelastic properties (Humjeux) are:

  • power of the nonlinear elastic law: \({n}_{e}=0\)

  • \(\mathrm{\beta }=200\)

  • \(d=3.5\)

  • \(b=0.6\)

  • friction angle: \(\mathrm{\phi }=40°\)

  • angle of expansion: \(\mathrm{\psi }=30°\)

  • critical pressure: \({P}_{c0}=-\mathrm{2,24}\mathit{MPa}\)

  • reference pressure: \({P}_{\mathit{ref}}=-1\mathit{MPa}\)

  • elastic radius of the isotropic mechanism: \({r}_{\mathit{éla}}^{s}=0.01\)

  • elastic radius of the deviatory mechanism: \({r}_{\mathit{éla}}^{d}=0.01\)

  • \({a}_{\mathit{mon}}=0.03\)

  • \({a}_{\mathit{cyc}}=0.00001\)

  • \({c}_{\mathit{mon}}=0.0003\)

  • \({c}_{\mathit{cyc}}=0.0003\)

  • \({r}_{\mathit{hys}}=0.1\)

  • \({r}_{\mathit{mob}}=0.9\)

  • \({x}_{m}=2\)

  • \(\text{dila}=1\)

1.3. Initial conditions and mechanical loading#

1.3.1. Initial conditions#

The initial deformation conditions are as follows:

  • \(\mathit{EPXX}0=-1.350354802792579E-021\)

  • \(\mathit{EPYY}0=-3.980032078861482E-007\)

  • \(\mathit{EPZZ}0=0\)

  • \(\mathit{EPXY}0=\frac{8.492341581286122E-008}{\sqrt{2}}\)

  • \(\mathit{EPXZ}0=0\)

  • \(\mathit{EPYZ}0=0\)

The initial stress conditions are as follows:

  • \(\mathit{SIXX}0=-125\mathit{kPa}\)

  • \(\mathit{SIYY}0=-125\mathit{kPa}\)

  • \(\mathit{SIZZ}0=-125\mathit{kPa}\)

  • \(\mathit{SIXY}0=0\)

  • \(\mathit{SIXZ}0=0\)

  • \(\mathit{SIYZ}0=0\)

The initial internal variables are zero.

1.3.2. Loading#

The deformation increment applied is as follows:

  • \(\mathrm{\Delta }\mathit{EPXX}=7.372770706199615E-006\)

  • \(\mathrm{\Delta }\mathit{EPYY}=4.632919275111915E-005\)

  • \(\mathrm{\Delta }\mathit{EPZZ}=0\)

  • \(\mathrm{\Delta }\mathit{EPXY}=\frac{1.733367998412452E-006}{\sqrt{2}}\)

  • \(\mathrm{\Delta }\mathit{EPXZ}=0\)

  • \(\mathrm{\Delta }\mathit{EPYZ}=0\)