2. Benchmark solution#

2.1. Internal forces modeling A#

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

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

For beam 1 going from node \(A\) to node \(B\):

\({\mathit{Fa}}_{1}=[\mathit{Fx},\mathit{Fy},\mathit{Fz}]\)

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

\({\mathit{Fb}}_{1}=[{F}_{x},{F}_{y},{F}_{z}]\)

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

For beam 2 going from node \(B\) to node \(C\):

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

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

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

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

For beam 3 going from node \(C\) to node \(D\):

\({\mathit{Fc}}_{3}=[{F}_{y},-{F}_{x},{F}_{z}]\)

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

\({\mathit{Fd}}_{3}=[{F}_{y},-{F}_{x},{F}_{z}]\)

\({\mathit{Md}}_{3}=[{M}_{y},-{M}_{x},{M}_{z}]\)

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

2.2. Internal forces (modeling B)#

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

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

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

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

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

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

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

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

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 \(A\). 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.

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)

2.3. Uncertainties about the solution#

None.