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

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

The problem consists in analyzing the response of a mobile mass subjected to a non-zero initial speed and coming to shock against a wall that follows a buckling type of behavior law.

The equilibrium equation for the system is written as:

.. math::
 :label: eq-1

    M\ frac {{\ partial} ^ {2} x} {\ partial {t} ^ {2}} +\ mathrm {Kx} = {F} _ {\ mathrm {ext}}

The resolution of takes place in several phases: before buckling of the wall, loading after buckling, unloading after buckling and free flight.


First phase
~~~~~~~~~~~~~~~

Before buckling, the reaction force is equal: :math:`{F}_{\mathrm{ext}}=-{K}_{1}x`

The solution to the equation is in the following form: :math:`x(t)={A}_{1}\mathrm{sin}{\omega }_{1}t+{B}_{1}\mathrm{cos}{\omega }_{1}t`

With: :math:`{\omega }_{1}^{2}=\frac{K+{K}_{1}}{M}`

Taking into account the initial conditions: :math:`x(0)=0` and :math:`\frac{\partial x}{\partial t}(0)={v}_{0}`

We identify: :math:`{A}_{1}=\frac{{v}_{0}}{{\omega }_{1}}` and :math:`{B}_{1}=0`

That is: :math:`x(t)=\frac{{v}_{0}}{{\omega }_{1}}\mathrm{sin}{\omega }_{1}t`

The buckling limit threshold :math:`{F}_{\mathrm{fl}}` is reached at time :math:`{t}_{\mathrm{fl}}`, such as: :math:`x({t}_{\mathrm{fl}})={d}_{\mathrm{fl}}=\frac{{F}_{\mathrm{fl}}}{{K}_{1}}`.

From where: :math:`{t}_{\mathrm{fl}}=\frac{1}{{\omega }_{1}}\mathrm{Arc}\mathrm{sin}(\frac{{F}_{\mathrm{fl}}{\omega }_{1}}{{K}_{1}{v}_{0}})`

and :math:`\frac{\partial x}{\partial t}({t}_{\mathrm{fl}})={v}_{\mathrm{fl}}`


Second phase
~~~~~~~~~~~~~~~

The next phase is the loading phase after buckling; the reaction force is constant as long as the speed remains positive. This reaction force is equal to: :math:`{F}_{\mathrm{ext}}=-{F}_{\mathrm{seuil}}` and the solution of the equation is written as: :math:`x(t)={A}_{2}\mathrm{sin}{\omega }_{0}t+{B}_{2}\mathrm{cos}{\omega }_{0}t-\frac{{F}_{\mathrm{seuil}}}{M{\omega }_{0}^{2}}`

With: :math:`{\omega }_{0}^{2}=\frac{K}{M}`

Like :math:`{\omega }_{0}^{2}\ll 1`, we can do a limited expansion of trigonometric functions.

That is: :math:`x(t)={A}_{21}{t}^{2}+{B}_{21}t+{C}_{21}`.

Taking into account the initial conditions :math:`x({t}_{\mathrm{fl}})={d}_{\mathrm{fl}}` and :math:`\frac{\partial x}{\partial t}({t}_{\mathrm{fl}})={v}_{\mathrm{fl}}`, we obtain:

 :math:`{A}_{21}=(\frac{{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}}{2}-\frac{{d}_{\mathrm{fl}}}{2}-\frac{{F}_{\mathrm{seuil}}^{2}}{4M}){\omega }_{0}^{2}-\frac{{F}_{\mathrm{seuil}}}{2M}`

 :math:`{B}_{21}={v}_{\mathrm{fl}}+\frac{{F}_{\mathrm{seuil}}{t}_{\mathrm{fl}}}{M}+({d}_{\mathrm{fl}}{t}_{\mathrm{fl}}-\frac{{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}^{2}}{2}){\omega }_{0}^{2}`

 :math:`{C}_{21}={d}_{\mathrm{fl}}-\frac{{F}_{\mathrm{seuil}}{t}_{\mathrm{fl}}^{2}}{2M}-{v}_{\mathrm{fl}}{t}_{\mathrm{fl}}-\frac{{d}_{\mathrm{fl}}{t}_{\mathrm{fl}}^{2}}{2}{\omega }_{0}^{2}`

The moment at which the speed is cancelled is: :math:`{t}_{m}=-\frac{{B}_{21}}{2{A}_{21}}`

By neglecting the terms in :math:`{\omega }_{0}^{2}`, we deduce: :math:`{t}_{m}={t}_{\mathrm{fl}}+\frac{M{v}_{\mathrm{fl}}}{{F}_{\mathrm{seuil}}}`,

and the maximum displacement is equal to: :math:`x({t}_{m})={d}_{m}`

This makes it possible to obtain the cumulative plastic deformation: :math:`{d}_{p}={d}_{m}-\frac{{F}_{\mathrm{seuil}}}{{K}_{2}}`


Third phase
~~~~~~~~~~~~~~~~

This phase corresponds to unloading, the reaction force is equal to: :math:`{F}_{\mathrm{ext}}=-{K}_{2}(x-{d}_{p})`

The solution to the equation is written as: :math:`x(t)={A}_{3}\mathrm{sin}{\omega }_{2}t+{B}_{3}\mathrm{cos}{\omega }_{2}t+\frac{{K}_{2}{d}_{p}}{K+{K}_{2}}`

With: :math:`{\omega }_{2}^{2}=\frac{K+{K}_{2}}{M}`

Taking into account initial conditions :math:`x({t}_{m})={d}_{m}` and :math:`\frac{\partial x}{\partial t}({t}_{m})=0`,

we get: :math:`{A}_{3}=({d}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{sin}{\omega }_{2}{t}_{m}` and :math:`{B}_{3}=({d}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{cos}{\omega }_{2}{t}_{m}`.

The reaction force is cancelled out when the displacement of the material point reaches the cumulative plastic deformation value :math:`{d}_{p}`.

Like :math:`K\ll {K}_{2}`, we make the following approximation: :math:`\frac{{K}_{2}{d}_{p}}{K+{K}_{2}}\approx {d}_{p}`.

Thus, the moment :math:`{t}_{d}` which corresponds to the cancellation of the reaction force is such that:

:math:`x({t}_{d})={d}_{p}={A}_{3}\mathrm{sin}{\omega }_{2}{t}_{d}+{B}_{3}\mathrm{cos}{\omega }_{2}{t}_{d}+{d}_{p}`

That is: :math:`\mathrm{sin}{\omega }_{2}{t}_{m}\mathrm{sin}{\omega }_{2}{t}_{d}+\mathrm{cos}{\omega }_{2}{t}_{m}\mathrm{cos}{\omega }_{2}{t}_{d}=\mathrm{cos}{\omega }_{2}({t}_{d}-{t}_{m})=0`.

From where: :math:`{t}_{d}={t}_{m}+\frac{\pi }{2{\omega }_{2}}`.


Fourth phase
~~~~~~~~~~~~~~~~

The next phase corresponds to free flight phase :math:`{F}_{\mathrm{ext}}=0`.

The solution to the equation is written as: :math:`x(t)={A}_{4}\mathrm{sin}{\omega }_{0}t+{B}_{4}\mathrm{cos}{\omega }_{0}t`

The initial conditions are:

:math:`x({t}_{d})={d}_{p}` and :math:`\frac{\partial x}{\partial t}({t}_{d})={v}_{d}={\omega }_{2}({x}_{m}-\frac{{K}_{2}{d}_{p}}{K+{K}_{2}})\mathrm{sin}{\omega }_{2}({t}_{m}-{t}_{d})`

This results in:

:math:`{A}_{4}={d}_{p}\mathrm{sin}{\omega }_{0}{t}_{d}+\frac{{v}_{d}}{{\omega }_{0}}\mathrm{cos}{\omega }_{0}{t}_{d}`

:math:`{B}_{4}={d}_{p}\mathrm{cos}{\omega }_{0}{t}_{d}-\frac{{v}_{d}}{{\omega }_{0}}\mathrm{sin}{\omega }_{0}{t}_{d}`

Like :math:`{\omega }_{0}^{2}\ll 1`, by doing a limited expansion of trigonometric functions up to order 2, the solution is in the following form:

:math:`x(t)={A}_{41}{t}^{2}+{B}_{41}t+{C}_{41}`

With:

:math:`{A}_{41}=\frac{{\omega }_{0}^{2}}{2}[{v}_{d}{t}_{d}-{d}_{p}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})]`

:math:`{B}_{41}={d}_{p}{t}_{d}{\omega }_{0}^{2}+{v}_{d}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})`

:math:`{C}_{41}={d}_{p}(1-\frac{{\omega }_{0}^{2}{t}_{d}^{2}}{2})-{v}_{d}{t}_{d}`

By neglecting the terms in :math:`{\omega }_{0}^{2}`, we get:

:math:`x(t)={v}_{d}t+{d}_{p}-{v}_{d}{t}_{d}`.

And we deduce the instant :math:`{t}_{0}` of transition to the initial position (:math:`x=0`).

That is: :math:`{t}_{0}={t}_{d}-\frac{{d}_{p}}{{v}_{d}}`.


C modeling reference solution
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The initial conditions chosen make it possible to have a kinetic energy of :math:`4J`, which should lead to a total compression of :math:`6.5m` and a residual compression of :math:`3.1m`.

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

*For A and B models*

It is proposed to test the following quantities:

:math:`{t}_{\mathrm{fl}}`: moment of onset of buckling;

:math:`{d}_{p}`: cumulative plastic deformation;

:math:`{t}_{0}`: moment of ironing at the initial position (after buckling and unloading).

Taking into account the numerical values of the input data, we obtain:

:math:`{t}_{\mathrm{fl}}=\frac{\pi }{6}` (expressed in seconds)

:math:`{d}_{p}=3` (expressed in meters)

:math:`{t}_{0}=\frac{\pi }{6}+2\sqrt{3}+\frac{\pi +6}{\sqrt{2}}` (expressed in seconds)

*For C* modeling

Maximum compression :math:`\mathit{max}(d)=6.5` (expressed in meters)

Residual compression :math:`{d}_{p}=3.1` (expressed in meters)

Rebound speed after shock :math:`{V}_{\mathit{rebond}}=-1.303840481` (expressed in meters per second)

*For modelling*D*

The non-regression is tested without an analytical reference value.


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

The reference solution is analytical (to the second order).