Issue 49

E. Breitbarth et alii, Frattura ed Integrità Strutturale, 49 (2019) 12-25; DOI: 10.3221/IGF-ESIS.49.02

subdomains. But this contour leads to discontinuity in the corners of C L in the derivative functions shown in Fig. 4 (c) and (d). Due to the rectangular shape of the domain these discontinuities are unavoidable. This shape was preferred instead of a circular shape as it is more flexible and much more space of the image section could be used. Furthermore, the q-function should not significantly influence the results of the J or interaction integral. and C R

2

h

         

b        

out 

y

2

   



x

i

2

out,R i



q

C

out h h



b

b

R

 



in

out,R 

in,R



2 2

2

h

b        

out 

y

2

   



x

i

2

out,L

i



q

C

out h h



b

b

L

 



in

out,L 

in,L



2 2

(7)

h

out y q h h    B out 2

i

in

2 2



b

x

out,R i



q

A



b

b

R

out,R in,R



b

x

out,L

i



q

A



b

b

L

out,L in,L

Finally, all values required for solving the J and interaction integral are available (Fig. 1 computation of the interaction integral an auxiliary field is needed (Fig. 1 ⑥ , Auxiliary field As mentioned above, the computation of the interaction integral requires an auxiliary field. All values with the superscript (2) correspond to this field; please see Eqns. (3) and (6). For the determination of K I and K II the first term of the Williams series expansion is utilized. Generally, the Williams field describes the stress field in the vicinity of the crack tip under linear elastic conditions. Here, K I and K II represent the coefficients of the first term. The corresponding stresses and displacements are summarized in Eqn. 8 [26] [26]. The elastic strains can be calculated from the stresses using Hooke’s law (see Eqn. 1). ⑬ ). As mentioned before, for the ⑫ ), as given in the next part (Eq. 8).

  2                2   2 11 22 12      2            2 1 2 u u

        

        

     

     

     

     

I

II

f

f f f

11

11

  2

  2

K

K

I

II

I

II



f f

22

22

2  r 

2  r 

I

II

12

12

(8)

     

     

  2

  2

   

   

   

   

I 1 I 2

II

g g

g g

K

r

K

r

1

I

II



2 2 



2 2 



II

2

17