PSI - Issue 47

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

8

6

  

   .







 EG A

U

a a

 u dA u dA

 

(22)

3

2

0

0

D D

BP

EG

3 4

( ) 3 A

(

)





BP u 0 and

EG u 0 , are found by replacing of

EG  in (20). In

The strain energy densities,

with

and

1 2 DD

3 4 D D

the un-cracked beam portion, the strain energy is found as      ) ( 0 3 UN A UN UN U l a u dA ,

(23)



UN u 0 is obtained by replacing of

UN  in formula (20).

where the strain energy density,

with

1 2 DD

By substituting of (6), (7), (18), (19), (21), (22) and (23) in (5), one derives          D D D D CDD CD D M F G     u dA 01

1 2 1 R 

  

 ( ) 1 A

 

u dA RQ 0



1 2

2 3

1 2

2 3

(

)

A

RQ

  

  

  



 LN A

 ST u dA u dA 0 0 LN



.

(24)

( ) 2 A

(

)

The integration in (24) is performed by the MatLab computer program. In order to verify (24), the strain energy release rate is derived also by using the formula (Rizov (2021))

1 1 2 Rda dU  *

G



,

(25)

* U is the complementary strain energy,

1 da is an elementary increase of crack 1. The complementary strain

where

1 2 DD U

2 3 D D U

3 4 D D U

UN U , are

energy is calculated by (18). For this purpose, the strain energies,

,

,

and

replaced, respectively, with the complementary strain energies, * 1 2 D D U , * 2 3 D D U , * 3 4 D D U and * UN U . The complementary strain energy, * 1 2 D D U , is found by replacing of 01 u and RQ u 0 with the complementary strain energy densities, * 01 u and * 0 RQ u , in (19). The complementary strain energy density is equal to the area that supplements the area enclosed by the stress-strain curve to a rectangle. Thus, * 01 u , is written as

* 01 u   

u

.

(26)

01

By substituting of (1) and (20) in (26) and replacing of  with



, one obtains

1 2 DD

1

p

p







  

     p BH

  

     p BH

   

  

D D

D D

*

1

1

u

B









 

.

(27)

1 2

1 2

0

D D

1

1

H

H

1 2

Formula (27) is applied also to calculate the complementary strain energy densities in the other beam portions by replacing of 1 2 DD  with the corresponding strains. By substituting of the complementary strain energy in (25), one obtains

  

1 2 1 R 

  

 ( ) 1 A



  A ST u dA u dA .  ( ) 2 ) ( * 0 * 0 LN A LN

u dA * 01

u dA RQ * 0





G



(28)

(

)

A

RQ

The integration in (28) is carried-out by the MatLab computer program. It should be noted that the strain energy release rate obtained by (28) is match of that found by (24). This fact is a verification of the analysis developed in the present paper. The strain energy release rate is derived also assuming a small increase of the length of crack 2. For this purpose, 1 a and 1 R are replaced, respectively, with 2 a and 2 R in (5). By substituting of (6), (7), (18), (19), (21), (22)