PSI - Issue 41

Efstathios E. Theotokoglou et al. / Procedia Structural Integrity 41 (2022) 361–371 Efstathios E.Theotokoglou/ Structural Integrity Procedia 00 (2022) 000–000

369

9

p=0.5

p=1

p=5

p=10

discon

‐5E+09 ‐4E+09 ‐3E+09 ‐2E+09 ‐1E+09 0 1E+09 2E+09 3E+09 4E+09 5E+09

‐0,11

‐0,08

‐0,05

‐0,02

0,01

0,04

0,07

0,1

σyy(Pa)

y(m)

Figure 11 Results for σ yy for the case with the metal core

p=0.5

p=1

p=5

p=10

discon

3E+09

2E+09

1E+09

0

0

0,5

1

1,5

2

‐1E+09 σxy(Pa)

‐2E+09

‐3E+09

x(m)

Figure 12 Results for σ xy for the case with the metal core

The normal stresses are calculated at the left fixed support. The shear stress for the case of the ceramic core is calculated at the middle of the height of the beam. In the case with the metal core the shear stress is calculated in the upper surface of the top face sheet. We also give a table (Table 2) with the maximum (tensile) stresses for the two cases. We can observe from Figures 7and 8 that the stress distributions for the normal stresses of the sandwich beam with the ceramic core and for the volume fraction indexes 0.5,1, are different from the stress distributions for the volume fraction indexes 5,10. We can also observe from Figure 9 that the stress distributions of the shear stress for the different volume fraction indexes are similar. Similarly from Figures 10-12 for the sandwich beam with the metal core, the stress distributions for both normal and shear stresses for the different volume fraction indexes are similar. We can see from Figures 7-12 that for the different volume fraction indexes the highest maximum stress for the ceramic core is less than the lowest maximum stress for the metal core.