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


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

For the problem studied, the displacement :math:`u` is radial is therefore of
shape :math:`u=[u\mathrm{,0}\mathrm{,0}]`.

From this we deduce the general shape of the strain tensor in
large deformations:

.. math::
 b\ mathrm {=} F {F} ^ {T}\ mathrm {=}\ left [\ begin {array} {ccc}
    {(1+u\ text {'})}} ^ {2} & 0& 0
                    \\ 0& {(1+\ frac {u} {r})} ^ {2} & 0
                    \\ 0& 0& 1\ end {array}\ right]


as well as the expression for the stress tensor, which is simply written
if we take into account the fact that :math:`J=\text{det}F=1`
for an incompressible problem:

.. math::
\ sigma =-p {I} _ {d} _ {d} +\ mu {b} ^ {d}

, that is:

.. math::
\ left\ {\ begin {array} {c} {
    \ sigma} _ {\ mathit {rr}}\ mathrm {=}\ mathrm {=}\ mathrm {-} p+\ mu (\ frac {2} {3} {(1+u\ text {'})}} ^ {2}}\ mathrm {-}\ mathrm {-}}\ mathrm {-}}\ frac {1} {3} {3} {(1+\ frac {u} {r})} ^ {2}\ mathrm {-}\ frac {1} {3})\\
    {\ sigma} _ {\ theta\ theta}\ mathrm {=}\ mathrm {=}\ mathrm {-}\ frac {1} {3} {(1+u\ text {'})}} ^ {2})} ^ {2})} ^ {2}} +\ frac {2} {3} {3} {r})} ^ {2} {r})} ^ {2}\ mama thrm {-}\ frac {1} {3})\\
    {\ sigma} _ {\ mathit {zz}}\ mathrm {=}\ mathrm {=}\ mathrm {-}\ frac {1} {3} {{3} {(1+u\ text {'})}}\ mathrm {'})}} ^ {2}\ mathrm {-}\ frac {1} {3} {(1+\ frac {u} {r})} ^ {2} +\ frac {2} {3})\\
    {\ sigma} _ {r\ theta}\ mathrm {=} {\ sigma} _ {\ mathit {riz}}\ mathrm {=} {\ sigma} _ {\ sigma} _ {\ theta z} _ {\ theta z}}\ mathrm {=} 0\ end {array} _ {\ theta z}}\ mathrm {=}
    \ right.

Writing balance equations leads to the verification of a single equation:

.. math::
\ sigma {\ text {'}} _ {\ mathit {rr}}} +\ frac {{\ sigma} _ {\ mathit {rr}}\ mathrm {-} {\ sigma}} _ {\ sigma} _ {\ theta\ theta}}} {r}\ mathrm {=} 0

which makes it possible to determine the pressure :math:`p` knowing the radial displacement field :math:`u`:

.. math::
 p\ text {'}\ mathrm {=}\ mu\ left (\ frac {4} {3} (1+u\ text {'}) u\ text {'}\ text {'}\ mathrm {-}\ mathrm {-}\ frac {-}\ frac {2} {3} (1+\ frac {u} {r} {r}) (\ frac {u\ text {'}} {'}} {r}} {r} {3} (1+\ frac {u} {r} {r}) (\ frac {u\ text {'}} {'}} {r}} {r} {r} {r}) thrm {-}\ frac {u} {{r} {{r} ^ {2}}) +\ frac {{(1+u\ text {'})} ^ {2}} {r}\ mathrm {-}\ frac {-}\ frac {{}}\ frac {{2}} {r}}\ right)


Particularization of the solution
--------------------------------

The incompressibility condition is written as :math:`\text{det}F=1`
with :math:`F=\left[\begin{array}{ccc}1+u\text{'}& 0& 0\\ 0& 1+\frac{u}{r}& 0\\ 0& 0& 1\end{array}\right]`. The displacement :math:`u` therefore verifies the following differential equation:


.. math::
 :label: v6-04-112-eqdiff

    \ mathrm {ru}\ text {'} +u+u\ text {'} u=0

The imposed load is as follows :math:`u={U}_{0}` in :math:`r=a`.

So the solution on the go is:

.. math::
\ left\ {\ begin {array} {c} {c} {u} _ {u} _ {r} =-r+r\ sqrt {{r} ^ {2} + {U} _ {0} ({U} _ {0}} +\ mathrm {2a})}
    \\ {u} _ {\ theta} = {u} _ {z} =0\ end {array}\ right.

The deformation tensor therefore has the following expression:

.. math::
\ left\ {\ begin {array} {c} {b} {b} _ {\ mathrm {rr}}} =\ frac {{r} ^ {2}} {{r} ^ {2}} + {U} _ {2} + {U} _ {0}}}\\
     {b} _ {\ theta\ theta} =\ frac {{r} ^ {2} + {U} _ {0} ({U} _ {0} +\ mathrm {2a})} {{r} ^ {2}}\\
      {b} _ {\ mathrm {zz}} =1\\
       {b} _ {r\ theta} = {b} _ {z\ theta} = {b} _ {\ theta z} =0\ end {array}\ right.

And the constraints apply to:

.. math::
\ left\ {\ begin {array} {c}
    {\ sigma} _ {\ mathit {rr}}\ mathrm {=}\ mathrm {=}\ mathrm {-} p+\ mu (\ frac {2} {3}}\ frac {{r} ^ {2}}} {{r}}}} {{r}}} {r} ^ {2}}} {{r}}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {{r}}} {r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}}} {{r}}} {r} ^ {2}} frac {1} {3}\ frac {{r} ^ {2} ^ {2} + {U} _ {0} ({U} _ {0} +\ mathrm {2a})} {{r} ^ {2}}}\ mathrm {2}}}\ mathrm {-}\ frac {1} {3})\\
    {\ sigma} _ {\ theta\ theta}\ mathrm {=}\ mathrm {=}\ mathrm {-}\ frac {1} {3}\ frac {{r} ^ {2}} ^ {2}}} {{r}}} {{r}}} {{r} ^ {2}}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}} {{r}}} {r} ^ {2}} + {{r}} {2}}} {{r}} {2}} + {{r}} {2}} {{r}} {2}} + {{r}\ frac {2} {3}\ frac {{r} ^ {2} ^ {2} + {U} _ {0} ({U} _ {0} +\ mathrm {2a})} {{r} ^ {2}}}\ mathrm {2}}}\ mathrm {-}\ frac {1} {3})} {{r} ^ {2}}} {{r} ^ {2}}}\ mathrm {2}}}\ mathrm {2}}}\
    {\ sigma} _ {\ mathit {zz}}\ mathrm {=}\ mathrm {=}\ mathrm {-}\ frac {1} {3}\ frac {{r} ^ {2}} ^ {2}} {{2}}} {{r}}} {{r}} {2}} +\ mathrm {2}}}\ mathrm {-}\ frac {1} {3} {3}\ frac {{r} ^ {2} + {U} _ {0} ({U} _ {0} +\ mathrm {2a})}} {{r} {2}})} {{r} {3})} {{r})} {{r}})} {{r} ^ {2}}} +\ frac {2} {3})\\
    {\ sigma} _ {r\ theta}\ mathrm {=} {\ sigma} _ {\ sigma} _ {z\ theta}\ mathrm {=} {\ sigma} _ {\ theta z}\ mathrm {=}} 0\ end {array}\ right.

with :math:`p` obtained by integrating :eq:`v6-04-112-eqdiff` which is equal to:

.. math::
 p\ mathrm {=}\ mu\ left (\ frac {{U}} _ {0} _ {0} +\ mathrm {2a})} {6 {r} ^ {2}}\ mathrm {-}}\ mathrm {-}}\ mathrm {-}}\ mathrm {-}} -}\ mathrm {-}} -}\ mathrm {-}}\ mathrm {-} -}\ frac {2} -}\ frac {2} _ {U} _ {0})} {6 {r} ^ {2}}}}\ mathrm {2}}}\ mathrm {2}}}\ mathrm {-}} -}\ mathrm {-}}\ mathrm {-}}} _ {0} ({U} _ {0} +\ mathrm {2a}) + {r} ^ {2})}\ mathrm {-}\ mathrm {log} (r) +\ frac {1} {2} +\ frac {1} {{2} +\ frac {1} {2} {2} +\ frac {1} {2} {2}) +\ frac {1} {2} {2}}\ frac {1} {2}} +\ frac {1} {2} {2}}\ frac {1} {2}} +\ frac {1} {2} {2}) +\ frac {1} {2} {2}) + {r} ^ {2})\ right) +C

where :math:`C` is a constant

The following numerical values are finally obtained:


.. csv-table::

    "in :math:`r\mathrm{=}0.1`:", "in :math:`r\mathrm{=}0.2`:"
    ":math:`\begin{array}{c}{u}_{r}={6.10}^{-5}\\ {\mathrm{\sigma }}_{\mathit{rr}}=-59.9955\\ {\mathrm{\sigma }}_{\mathrm{\theta }\mathrm{\theta }}=99.9566\\ {\mathrm{\sigma }}_{\mathit{zz}}=19.9326\\ {E}_{\mathit{rr}}=\mathrm{0,0005994604316761909}\\ {E}_{\mathrm{\theta }\mathrm{\theta }}=-0.0006001799999999502\end{array}` "," :math:`\begin{array}{c}{u}_{r}\mathrm{=}3.0067{10}^{\mathrm{-}5}\\ {\sigma }_{\mathit{rr}}\mathrm{=}0.\\ {\sigma }_{\theta \theta }\mathrm{=}40.006\\ {\sigma }_{\mathit{zz}}\mathrm{=}20.\end{array}`"


The transition through the Cartesian system is made using the following relationships:

.. math::
\ begin {array} {c}
    {\ sigma} _ {\ mathit {xx}}}\ mathrm {=} {\ sigma} _ {\ mathit {rr}} {\ mathrm {cos}} ^ {2}\ theta + {\ sigma} _ {\ sigma} _ {\ sigma} _ {\ theta\ theta}} {\ theta\ theta\ theta} theta\ theta {theta} theta\ theta} _ {r theta}\ mathrm {sin}\ theta\ mathrm {cos}\ theta\\
     {\ sigma} _ {\ theta\ theta}\ mathrm {=} {\ sigma} _ {\ mathit {rr}} {\ mathrm {sin}} ^ {2}\ theta + {\ sigma} _ {\ sigma} _ {\ theta\ theta} _ {\ theta\ theta} _ {\ theta} _ {\ theta} _ {\ theta} _ {\ theta}} _ {\ theta}\ theta} _ {\ theta}\ theta} _ {\ theta}\ theta}\ mathrm {sin}\ theta\ mathrm {cos}\ theta\\
      {\ sigma} _ {\ mathit {zz}}}\ mathrm {=} {\ sigma} _ {\ mathit {rr}}\ mathrm {sin}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos}\ theta\ mathrm {cos} thrm {-} 2 {\ sigma} _ {r\ theta} ({\ mathrm {cos}}} ^ {2}\ theta\ mathrm {-} {\ mathrm {sin}}}} ^ {2}\ theta)\ end {array}


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

The following are compared to the reference values:


* trips :math:`(u,v)` to points :math:`A` and :math:`F`,


* constraints :math:`({\sigma }_{\mathit{xx}},{\sigma }_{\mathit{yy}},{\sigma }_{\mathit{zz}},{\sigma }_{\mathit{xy}})` at points :math:`A` and :math:`F`,


* the Von Mises and Tresca constraints as well as the eigenvalues of the stress tensor at point :math:`A`.