Issue 47

V. Rizov, Frattura ed Integrità Strutturale, (2047) 468-481; DOI: 10.3221/IGF-ESIS.47.37

1 2 



m

i m



   

1

i

m  

i

1

2





m

m

2  3

i

i     







6 

i 



i 

1  i

1 

6 

  

y

(28)

2

cos

0

i

i

ai i

i

i

i

2

m

m



i

i

n equations with 6 1 L n 

Eqns. (19), (24) – (28) can be written for each layer of the left-hand crack arm. In this way, 6 L

 , 2 i

 , 3 i

 , 4 i

 , 5 i

 , 6 i

 and 1

 , where



i

1, 2, ..., L n

, can be constructed. Another equation can be written

unknowns, 1 i

by considering the equilibrium of the elementary forces in the left-hand crack arm cross-section

h

L y i n 

2

1 1 i 

1      1 i i 

z dy dz 

M

(29)

i

1 1 1

h y

2

where M is the bending moment in the left-hand crack arm. It is obvious that (Fig. 1)

M Fl 

(30)

1

By substituting of (16) in (29), one derives

i n 

  

  

3

3

h

h

L 







  2    y y



 

 

2









5 

M

y

y

y

y

(31)

3 1 1 i i 

1 1 i 

i

i

ai

i

ai

1

1

12

24



i

1

Eqns. (19), (24) – (28) and (31) should be solved with respect to 1 i  , 2 i  , 3 i  , 4 i  , 5 i  , 6 i  and 1

 by using the MatLab

computer program. Formula (16) is applied also to present

z . For this purpose, 1 i  ,

2 i  , 3 i

 , 4 i

 ,

5 i  , 6 i

 ,

R  as a function of

2 y and 2

i

 , 2 Ri

 , 3 Ri

 , 4 Ri

 , 5 Ri

 , 6 Ri

 , 2

1 y and 1

z are replaced with 1 Ri

y and 2

z , respectively. It should be noted that Eqs. (19),

 , 2 Ri

 , 3 Ri

 , 4 Ri

 , 5 Ri

 , 6 Ri

 and 2

 where 2

 is the curvature of

(24) – (28) and (31) can be used also to determine 1 Ri

3 1 a x l l      , of the un-cracked part of the beam. For this purpose, L n , 1 i y , 1 1 i y  , 1 i  , 2 i  , 3 i  2 2

the portion, 1 2 l l

 , 5 i

 , 6 i

 and 1

 are replaced, respectively, with n , 2 i y ,

 , 2 Ri

 , 3 Ri

 , 4 Ri

 , 5 Ri

 , 6 Ri

 and 2

 in (19),

y 

, 4 i

, 1 Ri

2 1 i

(22) – (28) and (31). By substituting of (4), (5) and (14) in (3), one arrives at

h

h

    

    

y

y

i n 

2

2

i n  

1

1

1 1 i 

2 1 i 

L 

 

 

* 0

*





G

u dy dx

u dy dz

2

(32)

L

0 2 2 R

1 1

h

h

i

i





i

i

1

1

h

h

y

y





i

i

1

2

2

2

where * u x a  . It should be noted that the term in the brackets in (32) is doubled in view of the symmetry (Fig. 1). The integration in (32) should be carried-out by using the MatLab computer program. The delamination fracture behavior is analyzed also by applying the J -integral approach [19] in order to verify the solution to the strain energy release rate (32). The integration of the J -integral is performed along the integration contour,  , showed by a dashed line in Fig. 1. Since the right-hand crack arm is free of stresses, the solution of the J -integral is written as   1 2 2 J J J     (33) where 1 J  and 2 J  are the values of the J -integral, respectively, in segments, 1  and 2  , of the integration contour (segments, 1  and 2  , coincide with the cross-section of the left-hand crack arm and un-cracked beam portion, respectively). The term in brackets in (33) is doubled because of the symmetry. 0 i L and * 0 i R u are obtained by (12), (13), (16), (19), (20), (23), (25) – (28) and (31) at 1

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