12 E. Proner and E. Mucchi Fig. 3 Specimen mounted on the shaker. equal to the mean TTF of TE-0. If the specimen survives the first sequence of excitations, the test is performed again until failure. 3. Test Environment 2 (TE-2): the PSDs obtained for TE-1 are applied simultaneously to the specimen. The phase and coherence between each axis is defined according to the Extreme Dynamic Response Method (EDRM) [14]. 4. Test Environment 3 (TE-3): The vibration environment of TE-2 is scaled to have the same MI-FDS, computed according to (6), and the same TTF of TE-0. The SDM of TE-3 is obtained according to the following equation: SinTE−3 = MI-FDSTE−0 MI-FDSTE−2 SinTE−2 (7) whereSinTE−3 andSinTE−2 are the excitation SDM of TE-3 and TE-2 andMI-FDSTE−0 andMI-FDSTE−2 are theMI-FDS of TE-0 and TE-2. The resulting vibration environment is a SDM with the same phase and coherence of TE-2 but with the PSDs that match the MI-FDS of TE-0. Figure 4 shows the SDM of TE-0 (solid blue), TE-2 (dashed red) and TE-3 (dashed green). The plots on the diagonal show the excitation PSD along each-axis. The upper triangular plots show the coherence, the lower triangular plots the phase between each pair of excitations. Figure 5 shows the FDS (left plot) and MI-FDS (right plot) computed for each test environment. The parameters used for the computation of the FDS and the MI-FDS are b = 5.4 , C = 5.1 · 1017 and a damping ratio of ζ = 0.1%. In the plot on the left, the total FDS of each test environment is depicted. TE-0, TE-1 and TE-2 are represented by the same curve (dashed gray). The blue curve depicts the FDS computed for TE-3. The total FDSs are obtained by summing the FDSs of each single-axis excitation. The plot on the right shows the MI-FDS computed for TE-0 (blue curve), TE-1 (gray curve), TE-2 (red curve) and TE-3 (green curve), computed according to Equation (6). Finally, Table 1 shows the TTFs obtained for each specimen in the testing campaign. The TTFs from Table 1 are also depicted in the box plot of Figure 6. In Figure 6, each box describes the first quartile of the TTFs obtained from each test environment, the red line and the black cross are the median and the mean of the TTFs, respectively. Discussion of the results The first result that needs to be addressed is the difference between the traditional FDS and the novel MI-FDS. In Figure 5, in the left plot, the total FDS computed for TE-0, TE-1 and TE-2 is the same. In fact, the traditional FDS considers only the PSD of each excitation. TE-0, TE-1 and TE-2 excite the system with the same PSDs, therefore, the total FDS is the same for these test environments. Moreover, the total FDS of TE-3 resulted lower than the the total FDS of TE-0, as the PSDs of TE-3 are lower. On the other hand, the MI-FDS produces three distinct curves for TE-0, TE-1 and TE-2. In particular TE-1
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