Issue 48

F. V. Antunes et alii, Frattura ed Integrità Strutturale, 48 (2019) 676-692; DOI: 10.3221/IGF-ESIS.48.64 680 I loading. Different values of  and  were considered for each of these 24 crack tip positions. The values considered for  were 0, 10, 20, 30, 40, 50 and 60º, while for  were 0, 10, 20, 30, 40, therefore a total number of 1120 numerical analysis were performed. J integral values (J I and J II ) were obtained for each analysis, from which K I and K II were calculated according: E.J I K I 2 1 υ   E.J II K II 2 1 υ   (5) Finally, Y I and Y II were obtained using relations 3 and 4. Tabs. B1 and B2 in Appendix B present results obtained for Y I and Y II , respectively, with CosmosM software. Typical variations of geometric factor, Y I , with (  -  ),  , y and (W/(W-x)) 1.5 can be seen in Figs. 5a, 5b, 5c and 5d, respectively. Maximum values can be observed in Fig. 5a for (  -  ) close to zero, which could be expected as zero correspond to pure mode I loading. However, the maximum value is not always zero, which indicates that there are others aspects influencing the geometric factors apart from loading direction. A good fitting is obtained with second order polynomial or sinusoidal function. A second order polynomial also fits well to results in Fig. 5b. It can be seen that the increase of slope at crack tip (  ), keeping all the others parameters fixed, reduces Y I . The typical results presented in Fig. 5c show a linear variation with y. Finally, Fig. 5d shows a linear variation of Y I with (W/(W-x)) 1.5 . This variation is related with the asymmetric geometry of the specimen, which imposes flexure to non-cracked section. The increase of crack length decreases the height of rectangular non-cracked section, and increases flexure moment. The comparison between Figs. 5 indicates that highest variation of Y I occurs with x and then with (  -  ). Figure 5 : Variation of Y I with (a) (  ) (for x=52.5 mm; y=0). (b) crack tip slope, (  ) (for x=52.5 mm; y=0). (c) y (for x=37.5 mm;  =0º). (d) (W/(W-x)) 1. 5 (for y=0;  =10º). A numerical solution with 39 constants was fitted to the numerical results presented in Tab. 4 (Appendix B) and illustrated in Figs. 5: Y I = m. 1.5 (W / (W x))  +b (6) (a) (b) (c) (d)