Issue 68

S. Kotrechko et alii, Frattura ed Integrità Strutturale, 68 (2024) 410-421; DOI: 10.3221/IGF-ESIS.68.27

For a transverse shear crack in the first approximation (see Appendix A):   2 4 2 3.75 1 2 4 f Y f Y S l                    

(9)

Substituting (9) to (8) and taking into account (6) and (7):

2      

  





l

4

4

f f

E



T  

(10)

4

f      





 



2

C   

   

l

3.75 1 2

Y      

f

In our case (see Appendix B):

E 

2 f f l



4  

(11)

whence

2



4

f

Y T  

(12)

4

f   





 

2      

C E 

3.75 12

Y      

Note that Y  is the yield strength of metal at strain rate ahead of dynamic crack tip. For bcc metals and alloys, this value can significantly exceed the value st Y  under quasi-static load. The data given in [13, 15,16] enable to estimate Y  :   1.5 2 st Y Y     (13)

In our case, the critical stress applied to the transversal shear crack, f  , can be estimated as:





1

f 





(14)

f

2

where 0.3   is the friction coefficient [17]. This dependence takes into account the fact that the sides of a shear crack are pressed against each other by a force of 2 f  . At 2000MPa f   and 0.3   one has 700 f MPa   . Dependence (9) for the area of local yield region and, accordingly, relation (12) for the increase in the average temperature in this region, are obtained in the approximation of an ideally plastic material. It means that the effect of strain hardening isn’t taken into account. This is due to the fact that above effect, when determining the sizes of local yield region ahead of a crack, considerably complicates the problem and requires to employ numerical methods. So, the effective value is usually applied to estimate the yield stress st Yef  . According to [18], in our case: f  is the macroscopic fracture stress of a specimen;

st Y f  

2 

st Yef





(15)

417