Issue 41

V. Rizov, Frattura ed Integrità Strutturale, 41 (2017) 491-503; DOI: 10.3221/IGF-ESIS.41.61

Formula (21) can also be applied to obtain the distribution of complimentary strain energy density, * 0 U

u , in the un-cracked

z , 1 1 n z

z , 2 2 n z

 have to be replaced with 0 , 2

 , respectively

beam portion. For this purpose, r , 1

y , 1

and 1

y , 2

and 2

( 2 2 n z is the neutral axis coordinate of the cross-section of un-cracked beam portion, 2

 is the curvature of un-cracked

beam portion). The quantities, 2 2 n z

2  , can be determined from Eqs. (18) and (19). For this purpose, r , 1 1 n z , 1 h

and

, 2 h and

 have to be replaced with 0 , 2 2 n z

2  , respectively.

and 1

Finally, by substituting of * 0 L

u , * u

and (8) in (7), one derives

0 U

  

   

1 m m G  1 

2







0 B b

1 B b

2 B br





m

m

1

1

u

u

1 

2 











u m     





 b m



 h m 2





m

1

1 3

1



1

u

u

z

       

  









2 B br

2

1

n

1









m

m

m

m

2

2

1

1

1

u

u

u

u











1 

2 

1 

2 

2





m

m

2

1

h

u

u

3 u u  q

3 u u  q

2 u u  q

2 u u  q

f

f

f

f

   

   

   

   

z

2

2 B bq

1

n

1

q

q

q

q

u

1

u

u

u

u



1 



2 



1 



2 



2





f

q

f

q

3

2

h

u

u

u

u

f q 

f q 

   

  

u u

u u

2 1 1 n

z

  

q

q

u

1 

2 







(22)



f

q

 

u u



m

1



  







2 

0 B b

1 B b

 m b m





m

m

1

1

u

u

1 

2 









   

u

u













m

m

1

1 3

1

u

u

    

3 u u  q

3 u u  q

2 u u  q

2 u u  q

f

f

f

f

   

   

   

   

z

2

2 B bq

1

n

2

q

q

q

q

u

2

u

u

u

u



1 



2 



1 



2 



u

u

u

u

2





f

q

f

q

3

2

h

u

u

u

u

f q 

f q 

   

  

u u

u u

2 2 2 n

z

  

q

q

u





1 

2 



u

u



f

q

 

u u

where . Formula (22) calculates the strain energy release rate in the beam configuration shown in Fig. 1 when the beam mechanical behavior and the material gradient are described by formulae (1) and (2), respectively. It should be noted that at m =1, 0 B E  , 1 2 0 B B   and 1 h h  formula (22) yields 1 u m m   , / u u u q m  ( u f f and u q are positive integers), 1 2 2 n u h z    and 2 2 2 n u h z    

2 2 3 y

M

21



(23)

G

Eb h

4

which is exact match of the expression for the strain energy release rate when the beam considered is linear-elastic and homogeneous [26]. In order to verify (22), the fracture is analyzed also by using the J -integral written as [27]





u x

v

  

  

  

  

 

(24)









J

u

p

p

ds

cos

x

y

0





x

497