The tri-axial stress state has been analyzed both theoretically using Lame’s equations and computationally using ANSYS. Figure 6a and Figure 6b depicts the theoretical contours of hoop stress distribution obtained from Lam’s equations with the help of MATLAB whereas Figure 6c represents the computational results using ANSYS. Figure 6 is the testimony of the accurate computational results for hoop stress.
In the same fashion for radial stresses, Figure 7a and Figure 7b represents the theoretical results. While utilizing ANSYS Figure 7c depicts the computational results. Figure 8a shows the computational results of longitudinal stress. Figure 8b is the graphical representation of the Lame’s equations which is in scrutiny with the stress distributions in Figure 6 to Figure 8.
The stress analysis of the tube reveals that the maximum principle stress is the hoop stress which is higher at the inner surface of the tube. The radial stress is compressive while the longitudinal stress is uniform across the wall thickness. The comparison of maximum theoretical and computational stresses in the tube wall has been done in Table 2.
The growth of a semi-elliptical crack in a cube under uniform stress distribution had been analyzed with ten steps. The SIF in the first and last steps have been shown in Figure 9a and Figure 9b, in which the minimum value of SIF occurred at the minor axis which is the deepest point on the crack tip in the simple cubic structure, and the greater on the major axis. As the crack was grown the value of SIF increased as well.
The comparison between crack growth results obtained from ANSYS SMART and MATLAB, depicted in Figure 9c, reveals notable discrepancies. MATLAB employs double 1/3 Simpson’s integration of Paris’s law as in equations 12 and 13 to calculate ∆c and the number of loading cycles, resulting in larger values compared to ANSYS SMART. This deviation could be reduced by increasing the number of steps in both analyses, contours at the crack tip, and the crack front elements in ANSYS.
Figure 10 and Figure 11 show the SIF for the inner and outer surface semi-elliptical crack respectively, likewise the SIF was solved using equations 15 and 16 for the inner and outer surface. The FE results obtained are compared in Table 3.
As for the stress ratio R = 0, ∆K = K-0, Which was estimated using equations 15 for the inner and 16 for the outer surface and plugged in equations 12 and 13 to find the number of cycles and crack growth along the major axis iteratively. The number of cycles for both inner and outer surface cracks using Simpson’s integration and ANSYS SMART are compared in Figure 12a. Using MATLAB for Simpson’s integration the number of cycles is lower for the inner and higher for the outer surface crack. Conversely, the results from ANSYS SMART are the same for cracks on inner and outer surfaces as shown in Figure 12b and Table 4. The similarity in ANSYS results for both cracks is due to the higher a/t ratio for both cracks. Hence this shows faster crack growth for cracks on the inner surface of the tube as compared to an outer surface crack using Simpson’s integration of Paris’s Law, which elucidates that cracks on the inner surface of tubes are critical to supervise.
The higher values of SIFs at each step of solution on the major and minor axis for the inner surface crack as compared to that of the outer surface crack as shown in Figure 13a justify the lower cycles for the inner surface crack. Despite maintaining a constant crack growth per step (∆a) along the minor axis for both cracks, they exhibited growth along their respective major axes with an incremental increase in ∆c in each step, as shown in Figure 13b which is greater for the inner surface as compared to the outer surface which is due to the hoop stress across the thickness as shown in Figure 6 which is greater on the inner as compared to the outer surface of the tube.
The cycles for crack growth have been shown as curves in Figure 14 obtained from ANSYS SMART crack growth simulations for different orientations. These results state that the crack in H orientation exhibited the lowest cycles because the affective stress in H orientation is hoop which is the highest stress as compared to the other two and conversely, the number of cycles is highest for R orientation which is a dead orientation. After all, the solution stops in the first step even at a very fine mesh justifying no crack growth in plane perpendicular to radial stresses as it is compressive stress as shown in Figure 7 leading to no crack growth when the tube is pressurized internally. The crack growth is along the x-axis of each crack coordinate and thus the solution for a number of cycles has been shown till the final leak in Figure 14b. It shows that as the influence of hoop stress increases across a crack plane the number of cycles decreases.
Figure 15 shows the SIFs in modes 1, 2, and 3 respectively, and the equivalent SIF is shown in Figure 16 in the first step. These figures show that mode 1 SIF is the dominant one and has the higher value for (H). The elevated value observed can be attributed to the predominance of hoop stress, which surpasses other stress components, as previously outlined. By comparing Figure 14 and Figure 16 we can say that the higher the equivalent stress intensity factor lower the cycles to failure.
Figure 17 shows the SIFs for (H) in step ten just before the final leak where the SIFs have been increased as compared to step one. This increase is due to the change in crack dimensions which increases in each step and so as an equivalent stress intensity factor K_eq.
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