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


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

Analytical solution [:ref:`bib1 <bib1>`] and [:ref:`bib2 <bib2>`].


Recessed case-free, unit loads at the end
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Single pull :math:`{u}_{x}=\frac{{F}_{x}L}{ES}`

Simple bending :math:`{u}_{y}=\frac{{F}_{y}{L}^{3}(4+{\phi }_{y})}{12E{I}_{z}}` :math:`{\theta }_{z}=\frac{{L}^{2}{F}_{y}}{2E{I}_{z}}` :math:`{\phi }_{y}=\frac{12E{I}_{y}}{{L}^{2}G{A}_{y}^{\text{'}}}`

Simple bending :math:`{u}_{z}=\frac{{F}_{z}{L}^{3}(4+{\phi }_{z})}{12E{I}_{y}}` :math:`{\theta }_{y}=\frac{-{L}^{2}{F}_{z}}{2E{I}_{y}}` :math:`{\phi }_{z}=\frac{12E{I}_{z}}{{L}^{2}G{A}_{z}^{\text{'}}}`

Twist :math:`{\theta }_{x}=\frac{{M}_{x}L}{G{J}_{x}}`

Pure flex :math:`{u}_{z}=-\frac{{M}_{y}{L}^{2}}{2E{I}_{y}}` :math:`{\theta }_{y}=\frac{{M}_{y}L}{E{I}_{y}}`

Pure flex :math:`{u}_{y}=\frac{{M}_{z}{L}^{2}}{2E{I}_{z}}` :math:`{\theta }_{z}=\frac{{M}_{z}L}{E{I}_{z}}`

**Note 1:**

*For the angle section, as the center of shear is not confused with the center of gravity* :math:`({e}_{y}\ne 0)` *, the torsional moment must be added:* :math:`{M}_{x}={F}_{z}{e}_{y}` *to the load* :math:`{F}_{z}=1` *.*

*This changes the movement:*

:math:`{u}_{z}=\frac{{F}_{z}{L}^{3}(4+{\phi }_{z})}{12E{I}_{y}}+{\theta }_{x}{e}_{y}` :math:`{\theta }_{x}=\frac{{M}_{x}L}{G{J}_{x}}`

*In the same way, loading* :math:`{M}_{x}=1` *results in a move* :math:`{u}_{z}={\theta }_{x}{e}_{y}`.

Linear distributed loading:

:math:`\begin{array}{cc}{u}_{y}(x)=\frac{px}{360LEI}(3{x}^{4}-10{L}^{2}{x}^{2}+7{L}^{4})& {u}_{y}^{\mathrm{max}}=\frac{0.00652p{L}^{4}}{EI}\\ & \mathrm{en}x=0.519L\end{array}`

**Note 2:**

*As far as modeling A is concerned, the beam is carried by the vector* :math:`{e}_{1}=\frac{1}{\sqrt{3}}(\begin{array}{}1\\ 1\\ 1\end{array})` *. The other vectors of the local coordinate system are:* :math:`{e}_{2}=\frac{1}{\sqrt{2}}(\begin{array}{}-1\\ 1\\ 0\end{array})` and :math:`{e}_{3}\mathrm{=}\frac{1}{\sqrt{6}}(\begin{array}{c}\mathrm{-}1\\ \mathrm{-}1\\ 2\end{array})`

*The components of the displacement vector in the global coordinate system are obtained by:*

:math:`{u}_{G}\mathrm{=}(\begin{array}{ccc}\frac{1}{\sqrt{3}}& \frac{\mathrm{-}1}{\sqrt{2}}& \frac{\mathrm{-}1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{2}}& \frac{\mathrm{-}1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}& 0& \frac{2}{\sqrt{6}}\end{array}){u}_{\text{local}}`

Generalized efforts and constraints in :math:`O`:

:math:`N(O)={F}_{x}` :math:`{\sigma }_{\mathrm{xx}}=\frac{N}{S}`

:math:`{M}_{z}(O)={T}_{y}L` :math:`{T}_{y}={F}_{y}` :math:`{\sigma }_{\mathrm{xx}}(y)=\frac{{M}_{z}y}{{I}_{z}}` :math:`{\sigma }_{\mathrm{xy}}=\frac{{T}_{y}}{{k}_{y}S}`

:math:`{M}_{y}(O)=-{T}_{z}L` :math:`{T}_{z}(O)={F}_{z}` :math:`{\sigma }_{\mathrm{xx}}(y)=\frac{-{M}_{y}z}{{I}_{y}}` :math:`{\sigma }_{\mathrm{xz}}=\frac{{T}_{z}}{{k}_{z}S}`

:math:`{M}_{x}(0)={M}_{x}(B)` :math:`{\sigma }_{\mathrm{xy}}={\sigma }_{\mathrm{xz}}=\frac{{M}_{x}{R}_{T}}{{J}_{x}}`

:math:`{M}_{y}(0)={M}_{y}(B)` :math:`{\sigma }_{\mathrm{xx}}(z)=\frac{{M}_{y}z}{{I}_{y}}`

:math:`{M}_{z}(0)={M}_{z}(B)` :math:`{\sigma }_{\mathrm{xx}}(y)=\frac{{M}_{y}y}{{I}_{z}}`


Case: linear distributed transverse loading
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

:math:`y` :math:`{\mathrm{f}}_{y}\mathrm{.}L`

                                                 :math:`x`

:math:`O` :math:`B`

The equilibrium in rotation around :math:`O` gives the value of the reaction, supported by :math:`B`:

         :math:`{R}_{\mathit{By}}=-{\int }_{0}^{L}{\mathrm{f}}_{y}\mathrm{.}{x}^{2}\mathit{dx}=-\frac{1}{3}{\mathrm{f}}_{y}\mathrm{.}{L}^{2}=-12000`; then: :math:`{R}_{\mathit{Oy}}=-6000`.

Then we have:

:math:`\begin{array}{ccc}{M}_{z}(x)=\frac{-1000}{6}\left({L}^{2}x-{x}^{3}\right)& {V}_{y}(x)=\frac{1000{L}^{2}}{6}-\frac{1000{x}^{2}}{2}& {\sigma }_{\mathit{xx}}^{\mathit{max}}=\frac{{M}_{z}^{\mathit{max}}R}{{I}_{z}}\\ \phantom{\rule{2em}{0ex}}& \phantom{\rule{2em}{0ex}}& \mathit{en}x=\frac{L\sqrt{3}}{3}\end{array}`


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


Recessed case-free, unit loads at the end
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Moving point :math:`B`,

Generalized efforts at point :math:`O`,

Point :math:`O` constraints.


Case: linear distributed transverse loading
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Shearing forces and transversal reactions at point :math:`O` and point :math:`B`. Maximum bending moment at abscissa point :math:`x=\frac{L\sqrt{3}}{3}`.


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

Analytical solution.


Bibliographical references
---------------------------


1. J.L. BATOZ, G. DHATT: "Modeling structures by finite elements" - Volume 2 Ed. HERMES.


2. N.D. PIKLEY: "Formulas for Stress, Stain & Structural Matrixes" Ed. John Wiley & Sons.