PSI - Issue 47

Victor Rizov / Procedia Structural Integrity 47 (2023) 3–12 Author name / Structural Integrity Procedia 00 (2019) 000–000

9

7

and (23) in (5), one obtains   1 F G 

  

   M

  



 

u dA ST 0

u dA LN 0





 



3 4 2 3 CD D CD D

3 4 D D D D 2 3

2

R



2

A ( ) 2

(

)

A

LN

  

  

  



 EG A

 BP u dA u dA 0 0 EG



.

(29)

( 3) A

(

)

In order to verify (29), the strain energy release rate is found also by applying (25). For this purpose, 1 a and 1 R are replaced with 2 a and 2 R , respectively. The strain energy release rate derived by substituting of the complementary strain energy in (25) is match of that obtained by (29).

Fig. 2. The strain energy release rate in non-dimensional form plotted against q (curve 1 – at increase of crack 1, curve 2 – at increase of crack 2 and curve 3 – at increase of crack 3). Solution to the strain energy release rate is found also when a small increase of the length of crack 3 is assumed. For this purpose, first, 1 a and 1 R are replaced, respectively, with 3 a and 3 R in (5). Then, by substituting of (6), (7), (18), (19), (21), (22) and (23) in (5), one derives          D D UN CD D CUN M F G      u dA u dA EG BP 0 0 Formula (30) is verified by applying (25). For this purpose, 1 a and 1 R are replaced, respectively, with 3 a and 3 R . The strain energy release rate found by substituting of the complementary strain energy in (25) matches that calculated by (30). 3. Numerical results Calculations of the strain energy release rate are performed by using the solutions derived in the previous section of the paper. The strain energy release rate is presented in non-dimensional form by applying the formula    R 3 2 1  3 4 3 4    A A ( ) 3 EG ) (      ) ( 0 UN u dA . UN A (30)