2. Reference solution

2.1. Calculation method used for the reference solution

Modeling A:

The solution is analytical. It is calculated in the local coordinate system.

either:

\(U=(u,v,w,{\theta }_{x},{\theta }_{y},{\theta }_{z})\) the displacement of the beam and \(E=({\epsilon }_{x},{\chi }_{y},{\chi }_{z},{\gamma }_{\mathrm{xy}},{\gamma }_{\mathrm{xz}})\) the generalized deformation.

Let’s be the solution:

\(u=\alpha x\) \(v=\gamma \frac{{x}^{2}}{2}\) \(w=-\beta \frac{{x}^{2}}{2}\) \({\theta }_{x}=0\) \({\theta }_{y}=\beta x\) \({\theta }_{z}=\gamma x\)

so:

\({\epsilon }_{x}={u}_{,x}=\alpha\) \({\chi }_{y}={\theta }_{y,x}=\beta\) \({\chi }_{z}={\theta }_{z,x}=\gamma\) \({\gamma }_{\mathrm{xy}}={v}_{,x}-{\theta }_{z}=0\) \({\gamma }_{\mathrm{xz}}={w}_{,x}+{\theta }_{y}=0\)

If we choose \(\alpha ={\epsilon }_{X}(\text{=}0.001)\), \(\beta ={\chi }_{y}^{0}(\text{=}0.002)\), \(\gamma ={\chi }_{z}^{0}(\text{=}0.003)\) then \(E-{E}_{\mathrm{init}}=0\) and the efforts are zero: the balance is therefore verified. In addition, the solution verifies the conditions at the embedment limits in \(A\). It is therefore the solution to the problem posed.

Modeling B:

The functions used to impose the pre-deformations are of the « staircase » function type depending on the coordinates of the nodes of the cells. To obtain the reference solution, the values of the pre-deformations are assigned cell by cell.

C modeling:

In C modeling, the beam cells are duplicated as bar cells.

To validate the pre-deformations by value, a first calculation is made in which the pre-deformations are imposed on the beam cells, it serves as a reference for the second calculation in which the pre-deformations are imposed on the bar cells.

For the validation of pre-deformations by function, the reference solution is obtained by assigning to each mesh the value of the function depending on the geometry (coordinates of the Gauss point).

2.2. Benchmark results

Modeling A:

The results expressed in the local coordinate system are:

In \(B\):

\(\mathrm{Dx}=0.10\mathrm{mm};\mathrm{Dy}=15.0\mathrm{mm};\mathrm{Dz}=–10.0\mathrm{mm};\mathrm{DRx}=0.0\mathrm{rd};\mathrm{DRy}=0.2\mathrm{rd};\mathrm{DRz}=0.30\mathrm{rd}\)

In \(C\):

\(\mathrm{Dx}=0.05\mathrm{mm};\mathrm{Dy}=3.75\mathrm{mm};\mathrm{Dz}=–2.50\mathrm{mm};\mathrm{DRx}=0.0\mathrm{rd};\mathrm{DRy}=0.1\mathrm{rd};\mathrm{DRz}=0.15\mathrm{rd}\)

In the global frame of reference, we find at points \(B\) and \(C\):

\(\begin{array}{}\mathrm{DX}(B)=\frac{\sqrt{3}}{30}+5\frac{\sqrt{3}}{6}(-3\sqrt{6}+2\sqrt{2})[\mathrm{mm}]\\ \mathrm{DY}(B)=\frac{\sqrt{3}}{30}+5\frac{\sqrt{3}}{6}(3\sqrt{6}+2\sqrt{2})[\mathrm{mm}]\\ \mathrm{DZ}(B)=\frac{\sqrt{3}}{30}+5\frac{\sqrt{3}}{6}(-4\sqrt{2})[\mathrm{mm}]\\ \mathrm{DRX}(B)=\frac{1}{20}(-\sqrt{6}-2\sqrt{2})[\mathrm{rd}]\\ \mathrm{DRY}(B)=\frac{1}{20}(-\sqrt{6}+2\sqrt{2})[\mathrm{rd}]\\ \mathrm{DRZ}(B)=\frac{1}{20}(2\sqrt{6})[\mathrm{rd}]\end{array}\) \(\begin{array}{}\mathrm{DX}(C)=\frac{\sqrt{3}}{60}+5\frac{\sqrt{3}}{24}(-3\sqrt{6}+2\sqrt{2})[\mathrm{mm}]\\ \mathrm{DY}(C)=\frac{\sqrt{3}}{60}+5\frac{\sqrt{3}}{24}(3\sqrt{6}+2\sqrt{2})[\mathrm{mm}]\\ \mathrm{DZ}(C)=\frac{\sqrt{3}}{60}+5\frac{\sqrt{3}}{24}(-4\sqrt{2})[\mathrm{mm}]\\ \mathrm{DRX}(C)=\frac{1}{40}(-\sqrt{6}-2\sqrt{2})[\mathrm{rd}]\\ \mathrm{DRY}(C)=\frac{1}{40}(-\sqrt{6}+2\sqrt{2})[\mathrm{rd}]\\ \mathrm{DRZ}(C)=\frac{1}{40}(2\sqrt{6})[\mathrm{rd}]\end{array}\)

B and C modeling:

Results from other calculations performed in the same test.

2.3. Uncertainty about the solution

Modeling A:

The solution is exact for the theory of Euler beams (or Timoshenko beams because there is no shear). Since twisting does not occur, the solution is also valid for elements POU_D_TG and POU_D_TGM.