2. Benchmark solution

2.1. Calculation method

2.1.1. Reminders

Modeling BARRE does not transmit any shear force or bending moment. Therefore, if we note \(E\) the Young modulus of the element, \(A\) the area of its section, the area of its section and \(L\) its length, the elementary stiffness matrix \({K}^{\mathit{elem}}\) of a bar is as follows (with the components in the order \(({\mathrm{DX}}_{1}{\mathrm{DY}}_{1}{\mathrm{DZ}}_{1}{\mathrm{DX}}_{2}{\mathrm{DY}}_{2}{\mathrm{DZ}}_{2})\)):

\({K}^{\mathit{elem}}\mathrm{=}(\begin{array}{cccccc}\frac{\mathit{EA}}{L}& 0& 0& \frac{\mathrm{-}\mathit{EA}}{L}& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ \frac{\mathrm{-}\mathit{EA}}{L}& 0& 0& \frac{\mathit{EA}}{L}& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array})\)

The profile of the mass matrix \({M}^{\mathit{elem}}\) is different from the stiffness matrix because the mass must be taken into account in all directions of space. Thus, if we write \(\rho\) the density of the element, the elementary mass matrix, with \(m\mathrm{=}\rho AL\), is as follows:

\({M}^{\mathit{elem}}\mathrm{=}(\begin{array}{cccccc}m\mathrm{/}3& 0& 0& m\mathrm{/}6& 0& 0\\ 0& m\mathrm{/}3& 0& 0& m\mathrm{/}6& 0\\ 0& 0& m\mathrm{/}3& 0& 0& m\mathrm{/}6\\ m\mathrm{/}6& 0& 0& m\mathrm{/}3& 0& 0\\ 0& m\mathrm{/}6& 0& 0& m\mathrm{/}3& 0\\ 0& 0& m\mathrm{/}6& 0& 0& m\mathrm{/}3\end{array})\)

2.1.2. Calculation of internal forces and support reactions

Let \(m\) be the mass of the element and \(U\) the displacement, so if we choose a gravity field according to \(X\), the external forces at each node are as follows: \((\mathrm{-}m\mathrm{/}\mathrm{2,0}\mathrm{,0})\). For elastic stiffness behavior \(K\), the only non-imposed displacement \({U}_{{B}_{X}}\) is equal to \(\frac{{F}_{{B}_{X}}^{\mathit{ext}}}{K}\mathrm{=}\frac{\mathrm{-}m}{\mathrm{2K}}\).

If we write the elementary stiffness matrix \({K}^{\mathit{elem}}\), we have the relationship \({F}^{\mathit{int}}\mathrm{=}{K}^{\mathit{elem}}U\). The following results are obtained for internal forces: \({F}_{{O}_{X}}^{\mathit{int}}\mathrm{=}\frac{m}{2}\), \({F}_{{O}_{Y}}^{\mathit{int}}\mathrm{=}0\),,, \({F}_{{O}_{Z}}^{\mathit{int}}\mathrm{=}0\), \({F}_{{B}_{X}}^{\mathit{int}}\mathrm{=}\frac{\mathrm{-}m}{2}\), \({F}_{{B}_{Y}}^{\mathit{int}}\mathrm{=}0\), and \({F}_{{B}_{Z}}^{\mathit{int}}\mathrm{=}0\).

By noting \({R}^{\mathit{ap}}\) the supporting reactions, we have the \({R}^{\mathit{ap}}\mathrm{=}{F}^{\mathit{int}}\mathrm{-}{F}^{\mathit{ext}}\) relationship. So we easily have: \({R}_{{O}_{X}}^{\mathit{ap}}\mathrm{=}m\) and all the other zero components.

Note: for the loading directions \(Y\) and \(Z\) , there is no displacement so the internal forces are zero and the support reactions are equal to the external forces.

2.1.3. Projection of the mass matrix on a modal basis

The purpose of this calculation is to verify that for a unitary mode of movement \(\phi\) in a given direction, we have equality:

\({\phi }^{T}M\phi \mathrm{=}m\)

2.1.4. Calculation of kinetic energy

The purpose of this calculation is to verify that for a unit speed mode \(\phi\) in a given direction, we have equality:

\(\frac{1}{2}{\phi }^{T}M\phi \mathrm{=}\frac{\mathit{mv}}{2}\) with \(v\mathrm{=}1\)