Reference solution
=====================


Calculation method
-----------------

*Ratings:*

 :math:`u`: displacement;

 :math:`\sigma`: constraint;

 :math:`\epsilon`: deformation;

 :math:`\delta =⟦u⟧`: displacement jump through cohesive discontinuity;

 :math:`{\delta }_{c}=\frac{2{G}_{c}}{{\sigma }_{c}}`: displacement jump corresponding to the rupture of the bar (zero stress).

It is an analytical solution. Since the problem is one-dimensional, all quantities are scalar and the constraint is constant in space. The problem is symmetric and can therefore be restricted to interval :math:`[\mathrm{0,}L]`. We then have the following set of equations:

In the bar:

         :math:`\sigma =E\epsilon` (eq1)

         :math:`\frac{\mathit{du}}{d\widehat{x}}=\epsilon` (eq2)

At the level of discontinuity (cohesive law refines in a softening regime):

         :math:`\delta ={\delta }_{c}(1-\frac{\sigma }{{\sigma }_{c}})` (eq3)

By integrating (eq2) over the interval :math:`[\mathrm{0,}L]` and using (eq1), we obtain the following relationship:

         :math:`u(\widehat{x}=L)-u(\widehat{x}=0)=L\epsilon`, or :math:`U-\frac{\delta }{2}=L\frac{\sigma }{E}` (eq4)

The imposed displacements that correspond to stress levels :math:`{\sigma }_{c}` (threshold for opening the discontinuity) and 0 (broken material) respectively are denoted :math:`{U}_{c}` and :math:`{U}_{f}`. By applying (eq4), it comes:

         :math:`{U}_{c}=L\frac{{\sigma }_{c}}{E}` and :math:`{U}_{f}=\frac{{\delta }_{c}}{2}` (eq5)

It is assumed that the values chosen for the material parameters :math:`E`, :math:`{\sigma }_{c}` and :math:`{G}_{c}`, and for the length of the bar :math:`L`, lead to a stable response of the bar (absence of "snap-backs"). We also assume increasing monotonic loading, so we have :math:`{U}_{c}⩽U⩽{U}_{f}` in a non-linear regime. The values of the material and geometric parameters must therefore verify the following inequality:

         :math:`L\frac{{\sigma }_{c}}{E}<\frac{{\delta }_{c}}{2}`, or :math:`\frac{L}{E}<\frac{{G}_{c}}{{\sigma }_{c}^{2}}` (eq6)

When this inequality is true, we can then express the constraint as a function of the imposed displacement:

         :math:`\sigma =\frac{\frac{{\delta }_{c}}{2}-U}{\frac{{\delta }_{c}}{2{\sigma }_{c}}-\frac{L}{E}},\forall U\in [:ref:`\frac{L{\sigma }_{c}}{E},\frac{{\delta }_{c}}{2} <\frac{L{\sigma }_{c}}{E},\frac{{\delta }_{c}}{2}>`] `(eq7)


Reference quantities and results
-----------------------------------

At the instant corresponding to the imposed displacement :math:`{U}_{\mathit{test}}=0.0199\mathit{mm}`, we test the constraint :math:`\sigma` (constant in space) as well as the displacement :math:`u` at the point :math:`\widehat{x}={0}^{\text{+}}` (i.e. the half-jump in displacement) are tested.

 :math:`\sigma =\frac{\frac{{\delta }_{c}}{2}-\mathrm{2L}\frac{{\sigma }_{c}}{E}}{\frac{{\delta }_{c}}{2{\sigma }_{c}}-\frac{L}{E}}` digital application: :math:`\sigma =1.72344975053\mathit{MPa}`

 :math:`u(\widehat{x}={0}^{\text{+}})=2{\delta }_{c}(1-\frac{\sigma }{{\sigma }_{c}})` digital application: :math:`u(\widehat{x}={0}^{\text{+}})=0.0141838916607\mathit{mm}`


.. csv-table::

    "**Identification**", "**Reference type**", "**Reference value**"
    "Under load :math:`{U}_{\mathit{test}}=0.0199\mathit{mm}`, at any point: :math:`\sigma` ", "'ANALYTIQUE'", ":math:`1.72344975053\mathit{MPa}`"
    "Under load :math:`{U}_{\mathit{test}}=0.0199\mathit{mm}`, in :math:`\widehat{x}={0}^{\text{+}}`: :math:`u` ", "'ANALYTIQUE'", ":math:`0.0141838916607\mathit{mm}`"