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


Calculation method used for the reference solution
--------------------------------------------------------


Case 1: free end
~~~~~~~~~~~~~~~~~~~~~~~~~

The equation for the beam is written as:


.. math::
 :label: eq-1

    \ frac {{d} ^ {4} v} {{\ mathit {xx}}} ^ {4}} =\ frac {-\ rho S} {\ mathit {EI}}}\ frac {{d}} ^ {d} ^ {d} ^ {2} v} {{2} v} {{\ mathit {dt}}} ^ {2}}

We are looking for a solution in the form of:


.. math::
 :label: eq-2

    v (x) =\ mathrm {sin} (\ omega t)\ left [A\ mathrm {cos} (\ mathit {kx}) +B\ mathrm {sin} (\ mathit {kx}) +C\ mathrm {kx}) +C\ mathrm {kx}) +C\ mathrm {kx}) +C\ mathrm {kx}) +C\ mathrm {kx}) +C\ mathrm {kx}) +C\ mathrm {kx})\ right]

with :math:`\omega =\sqrt{\frac{\mathit{EI}}{\rho S}}{k}^{2}`

The boundary conditions are as follows:

In :math:`x=0`, :math:`v=0` and :math:`\frac{{d}^{2}v}{{\mathit{dx}}^{2}}=0`

In :math:`x=L`, :math:`\frac{{d}^{2}v}{{\mathit{dx}}^{2}}=0` and :math:`\frac{{d}^{3}v}{{\mathit{dx}}^{3}}=0`

Replacing :math:`v` with the expression (), we get:


* In :math:`x=0`, :math:`A=C=0`;


* In :math:`x=L`, :math:`\mathrm{sin}(\mathit{kL})\mathrm{cosh}(\mathit{kL})=\mathrm{cos}(\mathit{kL})\mathrm{sinh}(\mathit{kL})` (3)

The first six roots of () are given in the following table:


.. csv-table::

    ":math:`\mathit{kL}`"
    ":math:`0`"
    ":math:`3.9266`"
    ":math:`7.06858`"
    ":math:`10.2102`"
    ":math:`13.3518`"
    ":math:`16.4934`"


From this we deduce the values of :math:`{\omega }_{i}`, :math:`i=\mathrm{1,}\mathrm{...},6`.


Case 2: elastic end
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The eigenvalue equation is as follows:


.. math::
: label: eq-4

    \ mathrm {sin} (\ mathit {kL})\ left ({k}} ^ {3} {L} ^ {3}\ frac {\ mathit {EI}} {{L} ^ {3}}}\ mathrm {kL}}}\ mathrm {kL})\ mathrm {kL})\ right)\ mathrm {3}}}\ mathrm {3}}}\ mathrm {3}}}\ mathrm {cosh} (\ mathit {kL})\ right) -\ mathrm {cosh} (\ mathit {kL}) sinh} (\ mathit {kL})\ left ({k}} ^ {3} {3} {L} ^ {3}\ frac {\ mathit {EI}} {{L} ^ {3}}\ mathrm {cos}} (\ mathrm {cos}} (\ mathit {cos}} (\ mathit {cos}}) (\ mathit {cos}} (\ mathit {cos}} (\ mathit {kL}}) (\ mathit {kL})\ right) (\ mathit {kL})\ right) =0

The own pulsations are given by:


.. math::
: label: eq-5

    \ omega =\ frac {1} {{L} ^ {2}}}\ sqrt {\ frac {\ mathit {EI}} {\ rho S}} {k}} {k} ^ {2} ^ {2} {2} {L} ^ {2}

The first six roots of () are given in the following table:


.. csv-table::

    ":math:`\mathit{kL}`"
    ":math:`2.7880886`"
    ":math:`4.5619625`"
    ":math:`7.1895724`"
    ":math:`10.249031`"
    ":math:`13.368916`"
    ":math:`16.502406`"


We then deduce the values of :math:`{\omega }_{i}`, :math:`i=\mathrm{1,}\mathrm{...},6`.


Benchmark results
----------------------


Case 1: free end
~~~~~~~~~~~~~~~~~~~~~~~~~

The structure has a rigid mode at zero frequency. The following table shows the non-zero frequencies.


.. csv-table::

    "Mode", "Frequency (:math:`\mathit{Hz}`)"
    "1"," :math:`85.5`"
    "2"," :math:`277`"
    "3"," :math:`577.9`"
    "4"," :math:`988.2`"
    "5"," :math:`1507.9`"



Case 2: elastic end
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


.. csv-table::

    "Mode", "Frequency (:math:`\mathit{Hz}`)"
    "1"," :math:`43.1`"
    "2"," :math:`115.4`"
    "3"," :math:`286.5`"
    "4"," :math:`582.3`"
    "5"," :math:`990.7`"
    "6"," :math:`1509.6`"



Uncertainty about the solution
---------------------------

Analytical solution.