PSI - Issue 32

Daria Dolgikh et al. / Procedia Structural Integrity 32 (2021) 246–252 D. Dolgikh, M. Tashkinov / Structural Integrity Procedia 00 (2019) 000–000

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Fig.7 Graphs of the dependence of reaction forces on displacement for models of cellular structures with different volume fractions

Dependence of reaction forces on displacement in CAD-models before the optimization process is represented by a solid line, after the optimization process - by a dashed line. The area under the graphs corresponds to the internal energy of deformation. When displacements reach 0.005 mm, in all cases curves for unoptimized structures are higher than curves after optimization, and hence the area under the plot before the optimization process is larger. Thus, according to the given target function, the strain energy of the random structures was minimized. From the analysis of the data presented in the graphs it is evident that minimization of the matrix volume down to 50% of the initial may lead to different mechanical behavior even for the structures with the same initial pores volume fraction. The random structure optimization process depends only on the initial morphology of the material model and is unique for any structure. It can be seen in figure 6 that curves for the structures with a volume fraction of 0.512 and 0.514 are close to each other, whereas the curves for structures with a volume fraction of 0.502 and 0.504 have a significant divergence. This phenomenon can be explained by the stochastic arrangement of the material over the volume of the initial cell structure, i.e. the optimization algorithm is affected by the geometric features of the structure. The revealed dependence allows to state that the topology optimization strongly depends on the initial morphology of the structures. Table 1 shows the change in weight of the structure after optimization.

Table 1. Change in weight of the structures after optimization

Volume fraction of pores

Weight change, %

p = 0.5

32,88 34,30 32,95 30,60 33,53 21,20 22,71 25,00

p = 0.502 p = 0.504 p = 0.512 p = 0.514 p = 0.612 p = 0.619 p = 0.702