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


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

The problem to be solved is:

:math:`{\mathrm{\int }}_{\Omega }\sigma (\frac{\mathrm{\partial }{u}^{\to }}{\mathrm{\partial }x})+\mathrm{\sum }(\frac{\mathrm{\partial }{\nu }^{\to }}{\mathrm{\partial }x})\mathrm{-}\lambda (\frac{\mathrm{\partial }{u}^{\to }}{\mathrm{\partial }x}\mathrm{-}{\nu }^{\to })+{\lambda }^{\to }(\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\mathrm{-}\nu )+r(\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\mathrm{-}\nu )(\frac{\mathrm{\partial }{u}^{\to }}{\mathrm{\partial }x}\mathrm{-}{\nu }^{\to })\mathrm{=}0`

with :math:`\forall {u}^{},{\nu }^{},{\lambda }^{}` cinematically eligible.

We have:

:math:`\mathrm{\{}\begin{array}{c}\sigma \mathrm{=}E\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\\ \Sigma \mathrm{=}F\frac{\mathrm{\partial }\nu }{\mathrm{\partial }x}\end{array}`

Let's note :math:`\omega` the second-order form functions and :math:`\text{N}` the first-order form functions on linear elements,

At the same time, let's recall the definitions of SEG2 and SEG3:

SEG2: 2-node segment

number of knots: 2

number of vertex nodes: 2


.. image:: images/10000000000001510000003C177B13C850F9EF4B.png
    :width: 2.6126in
    :height: 0.4764in


.. _RefImage_10000000000001510000003C177B13C850F9EF4B.png:

**Figure 2.1-a**

SEG3: 3-node segment

number of knots: 3

number of vertex nodes: 2


.. image:: images/100000000000016B0000006C23D2AE35EF82DE67.png
    :width: 2.8138in
    :height: 0.8571in


.. _RefImage_100000000000016B0000006C23D2AE35EF82DE67.png:

**Figure 2.1-b**

:math:`{N}_{1}=\frac{1-x}{2}` and :math:`{N}_{2}=\frac{1+x}{2}` on the reference element :math:`x\in \left[-\mathrm{1,}+1\right]` with :math:`{N}_{1}` the form function

at the first node.

The shape functions of the segment at the 3 nodes are then:

:math:`{\omega }_{1}=-\frac{x(1-x)}{2}` :math:`{\omega }_{2}=\frac{x(1+x)}{2}` and :math:`{\omega }_{3}=1-{x}^{2}`

By setting :math:`{u}^{}={\nu }^{}=0` we show that:

:math:`{\int }_{-1}^{+1}{\lambda }^{}(\frac{\partial u}{\partial x}-\nu )\mathrm{dx}=0` or :math:`u={\omega }_{1}\mathrm{.}{u}_{1}+{\omega }_{2}\mathrm{.}{u}_{2}+{\omega }_{3}\mathrm{.}{u}_{3}` with :math:`{u}_{2}=0` (:math:`\text{C.L.}`)

Hence :math:`\frac{\partial u}{\partial x}=(x-\frac{1}{2}){u}_{1}-2x{u}_{3}` and :math:`v={N}_{1}\mathrm{.}{\nu }_{1}+{N}_{2}\mathrm{.}{\nu }_{2}` with :math:`{\nu }_{2}=0` (:math:`\text{C.L.}`)

So we have :math:`\nu =\frac{(1-x)}{2}{\nu }_{1}` but it should be noted after integration that: :math:`{\nu }_{1}=-{u}_{1}` (1) :math:`\forall {u}^{},{\nu }^{},{\lambda }^{}` kinematically eligible.

Proceeding in the same way and asking: :math:`{\nu }^{}={\lambda }^{}=0` we find that:

:math:`{u}_{3}=\frac{2(E-r)}{4(E+r)}\mathrm{.}{u}_{1}\approx -\frac{{u}_{1}}{4}` (2)

Hence the general formula :math:`{S}_{1}={\mathrm{3.a}}^{1}\mathrm{.}\frac{\partial \nu }{\partial x}=-\frac{3}{2}\mathrm{.}{a}^{1}\mathrm{.}{\nu }_{1}` and after simplification :math:`{S}_{1}=\frac{3}{2}\mathrm{.}{a}^{1}\mathrm{.}{u}_{1}`.


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

The fundamental and very general formula for solid is :math:`{S}_{1}=\frac{3}{2}\mathrm{.}{a}^{1}\mathrm{.}{u}_{1}`.


Uncertainties about the solution
----------------------------

Analytical solution.