2. Reference solution#

2.1. Calculation method used for the reference solution#

If the beam is modelled by an Euler-Bernoulli beam and by a single finite element, the harmonic problem can be written as follows:

traction problem:

\((1+i\mathrm{\alpha }\mathrm{\omega })\frac{\mathit{ES}}{L}u(B)-{\mathrm{\omega }}^{2}\frac{\mathrm{\rho }SL}{6}u(B)={F}_{x}(B)\)

Hence \(u(B)=\frac{F(B)}{\frac{\mathit{ES}}{L}-{\mathrm{\omega }}^{2}\frac{\mathrm{\rho }SL}{6}+i\mathrm{\alpha }\mathrm{\omega }\frac{\mathit{ES}}{L}}\)

flexure problem:

\(\left[-{\mathrm{\omega }}^{2}\left[\begin{array}{cc}\frac{13L}{35}& \frac{-11{L}^{2}}{210}\\ \frac{-11{L}^{2}}{210}& \frac{{L}^{3}}{105}\end{array}\right]+(1+i\mathrm{\alpha }\mathrm{\omega })\frac{12{\mathit{EI}}_{y}}{{L}^{3}}\left[\begin{array}{cc}1& \frac{-L}{2}\\ \frac{-L}{2}& \frac{{L}^{2}}{3}\end{array}\right]\right]\left[\begin{array}{c}v(B)\\ \mathrm{\theta }(B)\end{array}\right]=\left[\begin{array}{c}{F}_{y}(B)\\ 0\end{array}\right]\)

Note:

If the material has no damping, we then have: AMOR_ALPHA = \(\mathrm{\alpha }=0\) .

The efforts at point \(B\) are calculated as follows:

traction problem:

\(N(B)=\left(\frac{\mathit{ES}}{L}-{\mathrm{\omega }}^{2}\frac{\mathrm{\rho }SL}{6}\right)u(B)\)

flexure problem:

\(\left[\begin{array}{c}\mathit{VY}(B)\\ \mathit{MFZ}(B)\end{array}\right]=\left[-{\mathrm{\omega }}^{2}\left[\begin{array}{cc}\frac{13L}{35}& \frac{-11{L}^{2}}{210}\\ \frac{-11{L}^{2}}{210}& \frac{{L}^{3}}{105}\end{array}\right]+\frac{12{\mathit{EI}}_{y}}{{L}^{3}}\left[\begin{array}{cc}1& \frac{-L}{2}\\ \frac{-L}{2}& \frac{{L}^{2}}{3}\end{array}\right]\left[\begin{array}{c}v(B)\\ \mathrm{\theta }(B)\end{array}\right]\right]\)

Systems \(2\times 2\) are solved analytically to obtain the solution.

2.2. Benchmark results#

The reference results are the displacements, velocities, accelerations, and generalized efforts obtained at point \(B\) during harmonic analysis.

2.3. Note for modeling B#

For B modeling, we want to test the FORCE_POUTRE keyword in the case of the traction problem, which makes it possible to apply distributed forces. To obtain the same solution as the beam subjected to nodal force at its end, the relationship between the constant distributed force and the nodal force is:

\({F}_{x}(B)=\frac{fL}{2}\)

With the values given in paragraph 1.3, we have: \(f=600N/m\)

2.4. Uncertainty about the solution#

If the hypotheses are verified (Euler-Bernoulli beam), the solution is analytical.

2.5. Bibliographical references#

  1. [R3.08.01] « Exact » beam elements (straight and curved).