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

.. _RefNumPara__635_2160324876:


Internal forces modeling A
-------------------------------

At node :math:`D`, loading is the twister of efforts :math:`{T}_{D}=({F}_{x},{F}_{y},{F}_{z},{M}_{x},{M}_{y},{M}_{z})`

The torsors of the internal forces, expressed in the local coordinate system of the elements, are as follows:

• For beam 1 going from node :math:`A` to node :math:`B`:

:math:`{\mathit{Fa}}_{1}=[\mathit{Fx},\mathit{Fy},\mathit{Fz}]`

:math:`{\mathit{Ma}}_{1}=[{M}_{x}+3\ast {F}_{z}\ast L,{M}_{y}-2\ast {F}_{z}\ast L,{M}_{z}+2\ast {F}_{y}\ast L-3\ast {F}_{x}\ast L]`

:math:`{\mathit{Fb}}_{1}=[{F}_{x},{F}_{y},{F}_{z}]`

:math:`{\mathit{Mb}}_{1}=[{M}_{x}+3\ast {F}_{z}\ast L,{M}_{y}-{F}_{z}\ast L,{M}_{z}+{F}_{y}\ast L-3\ast {F}_{x}\ast L]`

• For beam 2 going from node :math:`B` to node :math:`C`:

:math:`{\mathit{Fb}}_{2}=[({F}_{y}+{F}_{x})/\sqrt{(2)},({F}_{y}-{F}_{x})/\sqrt{(2)},{F}_{z}]`

:math:`{\mathit{Mb}}_{2}=[({M}_{y}+{M}_{x}+2\ast {F}_{z}\ast L)/\sqrt{(2)},({M}_{y}-{M}_{x}-4\ast {F}_{z}\ast L)/\sqrt{(2)},({M}_{z}+{F}_{y}\ast L-3\ast {F}_{x}\ast L)]`

:math:`{\mathit{Fc}}_{2}=[({F}_{y}+{F}_{x})/\sqrt{(2)},({F}_{y}-{F}_{x})/\sqrt{(2)},{F}_{z}]`

:math:`{\mathit{Mc}}_{2}=[({M}_{y}+{M}_{x}+2\ast {F}_{z}\ast L)/\sqrt{(2)},({M}_{y}-{M}_{x}-2\ast {F}_{z}\ast L)/\sqrt{(2)},{M}_{z}-2\ast {F}_{x}\ast L]`

• For beam 3 going from node :math:`C` to node :math:`D`:

:math:`{\mathit{Fc}}_{3}=[{F}_{y},-{F}_{x},{F}_{z}]`

:math:`{\mathit{Mc}}_{3}=[{M}_{y},(-{M}_{x}-2\ast {F}_{z}\ast L),({M}_{z}-2\ast {F}_{x}\ast L)]`

:math:`{\mathit{Fd}}_{3}=[{F}_{y},-{F}_{x},{F}_{z}]`

:math:`{\mathit{Md}}_{3}=[{M}_{y},-{M}_{x},{M}_{z}]`

The local coordinate system of the elements is the one by default: the local axis :math:`X` is carried by the SEG2, the local axis :math:`Z` is the global :math:`Z`.


.. _RefNumPara__1505_2142876063:


Internal forces (modeling B)
-------------------------------

At node :math:`D`, loading is the twister of efforts :math:`{T}_{\mathit{Dr}}=({F}_{\mathit{xr}},{F}_{\mathit{yr}},{F}_{\mathit{zr}},{M}_{\mathit{xr}},{M}_{\mathit{yr}},{M}_{\mathit{zr}})`

Twister :math:`{T}_{\mathit{DR}}` corresponds to twister :math:`{T}_{D}` to which the same rotations as geometry are applied.

:math:`{F}_{\mathit{xr}}=(0.6942720440\ast {F}_{x})+(-0.6438648260\ast {F}_{y})+(-0.3215966648\ast {F}_{z})`

:math:`{F}_{\mathit{yr}}=(0.3237443710\ast {F}_{x})+(0.6784690681\ast {F}_{y})+(-0.6594462116\ast {F}_{z})`

:math:`{F}_{\mathit{zr}}=(0.6427876097\ast {F}_{x})+(0.3537199593\ast {F}_{y})+(0.6794897197\ast {F}_{z})`

:math:`{M}_{\mathit{xr}}=(0.6942720440\ast {M}_{x})+(-0.6438648260\ast {M}_{y})+(-0.3215966648\ast {M}_{z})`

:math:`{M}_{\mathit{yr}}=(0.3237443710\ast {M}_{x})+(0.6784690681\ast {M}_{y})+(-0.6594462116\ast {M}_{z})`

:math:`{M}_{\mathit{zr}}=(0.6427876097\ast {M}_{x})+(0.3537199593\ast {M}_{y})+(0.6794897197\ast {M}_{z})`

As the force twister has been rotated, the internal forces that are always expressed in the local coordinate system of the elements are the same as those in modeling :math:`A`. On the other hand, it is necessary to define the local axes of the elements of the beams, which have also rotated, using the affe_cara_elem keyword orientation command.

.. csv-table::

    "**GROUP_MA**", "**CARA**", "**VALE**"
    "**PAB**", "v ECT_Y ", "(-0.6438648260, 0.6784690681, 0.3537199593)"
    "**PBC**", "v ECT_Y ", "(-0.9462056549, 0.2508282388, -0.2044016958)"
    "**PCD**", "v ECT_Y ", "(-0.6942720440, -0.3237443710, -0.6427876097)"



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

None.