2. Benchmark solution#
2.1. Calculation method used for the reference solution#
Case without fluid:
Let us consider the heterogeneous domain described in [§1] in the absence of the fluid. It is assumed that the beams respect the modeling assumptions of a straight Euler Bernoulli beam. Since the boundary conditions applied to all the beams are the same as for each of them, we can reduce the search for the natural frequencies of the set to that of a single beam.
The following problem is therefore studied:
Or a bi-embedded beam [fig 2.1-a] with the same geometric and material characteristics as the beams of the heterogeneous environment. We note \(A\) the area of the section, \(L\) its length, and \(I\) the flexural moment of inertia.
Using the dynamic stiffness method, it is shown that such a beam admits frequencies that are double the same as:
\({f}_{i}\mathrm{=}\frac{{\lambda }_{i}^{2}}{2\pi {L}^{2}}{\left[\frac{\mathit{EI}}{\rho A}\right]}^{\frac{1}{2}}\)
\({\lambda }_{i}=(2i+1)\pi /2i=\mathrm{1,2},\mathrm{...}\) for the second case of boundary conditions: [fig 2.1-a].
Figure 2.1-a
The domain contains \(N\) beams that are independent of each other (no boundary conditions that couple the movements of two different beams). As a result, the multiplicity of frequencies is equal to \(\mathrm{2N}\) (2 modes of flexure per beam).
For the homogenized medium discretized by the finite elements hexahedron with 8 knots or hexahedron with 20 nodes, the number \(N\) must be replaced by the number of lines parallel to the axis of the beams.
Case with fluid:
The case with fluid is more difficult to solve analytically: no analytical results have been found until this test case was written. The reference results that have been established therefore come from a numerical resolution by finite elements of the complete heterogeneous problem. For this reason, version 3.6.2 of*Code_Aster* was used.
Each beam is represented in the mesh by its average fiber modeled by POU_D_E (Euler’s right beam). For all the beams, each node of the middle fiber is linked to the nodes of the lateral surface, located in the same cross section as the node in question, by LIAISON_SOLIDE. The fluid-beam interface is modelled by FLUI_STRU, which reflects the continuity of the normal velocities at the walls. The fluid, had to be perfect incompressible, we deduced its modeling from that of the perfect compressible fluid 3D_ FLUIDE by removing the contribution of pressure.
The boundary conditions imposed on the domains [§1.3], and especially the relationship that couples the displacement of all the beams at the level of \(\mathrm{Sh}\), reveal two types of specific modes of the structure:
Assembly modes: all the beams are deformed in the same way and the top surface admits a non-zero displacement.
Local modes: they correspond to modes of embedded and embedded beams. The top surface therefore admits zero displacement. None of these modes can correspond to an overall mode.
The action of the fluid results in an added mass effect and therefore a reduction in frequencies compared to the case without fluid. In the case of local modes, it also has the effect of spreading the spectrum of associated frequencies. In the case without fluid, we saw that this spectrum was concentrated in a single vibration frequency.
2.2. Benchmark results#
Value of natural frequencies.
For mass and inertia at the center of gravity:
In the absence of fluid, the mass is determined by the product of the volume occupied by the beams and the density of these elements: \({\mathrm{MASSE}}_{\mathrm{solide}}={\rho }_{\mathrm{poutre}}\times {\mathrm{Vol}}_{\mathrm{poutres}}\) where \({\mathrm{Vol}}_{\mathrm{poutres}}\) is determined by the product of the number of beams, the area of the section and the length of the beam. From the data defined above, we can calculate \({\mathrm{MASSE}}_{\mathrm{solide}}\): we thus obtain \({\mathrm{MASSE}}_{\mathrm{poutres}}=7641\times 16\times (1.5/10)\mathrm{²}\times 4.1=11278.116\mathrm{kg}\).
Since the mass is distributed uniformly in the volume (due to the position of the beams in the volume), the center of gravity is therefore located at its center, namely at the coordinate point \((0.42,0.42,2.05)\). This result is confirmed analytically by calculating the following integrals:
\({X}_{G}={Y}_{G}=\frac{1}{\text{Vol}}{\int }_{V}x\text{.}\rho \text{dV}\) and \({Z}_{G}=\frac{1}{\text{Vol}}{\int }_{V}z\text{.}\rho \text{dV}\)
The inertias at the center of gravity \(G\) are calculated by: .Analytically, we obtain:
\({I}_{\mathrm{xx}}(G)={\int }_{V}({(y-{y}_{G})}^{2}+{(z-{z}_{G})}^{2})\rho \text{dV}\); \({I}_{\mathrm{yy}}(G)={\int }_{V}({(x-{x}_{G})}^{2}+{(z-{z}_{G})}^{2})\rho \text{dV}\); \({I}_{\mathrm{zz}}(G)={\int }_{V}({(y-{y}_{G})}^{2}+{(x-{x}_{G})}^{2})\rho \text{dV}\)
Analytically, we obtain:
\({I}_{\text{xx}}(G)={I}_{\text{YY}}(G)=\frac{8\rho }{3}(4\times {0.15}^{2}\times {2.05}^{2}+4.1\times 0.15\times ({0.39}^{3}-{0.24}^{3}+{0.18}^{3}-{0.03}^{3}))=16441.61\); \({I}_{\text{zz}}(G)=\frac{16\rho \times 4.1\times 0\text{.}\text{15}}{3}({0.39}^{3}-{0.24}^{3}+{0.18}^{3}-{0.03}^{3})=1285.71\)
In the presence of fluid, the total mass corresponds to the sum of the solid mass and the fluid mass. Knowing the total volume, we determine the volume occupied by the fluid:
\({\text{Vol}}_{\text{fluide}}=\text{Vol}-{\text{Vol}}_{\text{poutres}}={0.84}^{2}\times 4.1–16\times {(1.5/10)}^{2}\times 4.1=1.41696{m}^{3}\); \({\text{MASSE}}_{\text{fluide}}=1000\times 1.41696=1416.96\mathrm{kg}\); \({\text{MASSE}}_{\text{totale}}={\text{MASSE}}_{\text{solide}}+{\text{MASSE}}_{\text{fluide}}=12695.076\mathrm{kg}\)
From this we deduce the density of the element (with fluid):
\(\rho =\frac{{\text{MASSE}}_{\text{totale}}}{\text{Vol}}=4388.265\mathrm{kg}/{m}^{3}\)
The center of gravity remains the same with or without fluid.
The inertias at the center of gravity \(G\) are:
\({I}_{\text{xx}}(G)={I}_{\text{YY}}(G)=\frac{2\rho \times 0.84(4.1\times {0.42}^{3}+0.84\times {2.05}^{3})}{3}=18530.155\); \({I}_{\text{zz}}(G)=\frac{2\rho \times 0.84\times 4.1\times 2\times {0.42}^{3}}{3}=1492.941\)
2.3. Bibliographical references#
Walter D. Pilkey: « Formulas for Stress, Strain and Structural Matrixes », A Wiley-Interscience Publication JOHN WILEY & SONS, INC. 1994 edition.