Issue 41

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

where Γ is a contour of integration going from the lower crack face to the upper crack face in the counter clockwise direction, 0 u is the strain energy density, α is the angle between the outwards normal vector to the contour of integration and the crack direction, x p and y p are the components of stress vector, u and v are the components of displacement vector with respect to the crack tip coordinate system xy ( x is directed along the crack), ds is a differential element along the contour. The J -integral is solved by using an integration contour, Γ , that coincides with the beam contour (Fig. 1). It is obvious that the J -integral value is non-zero only in segments 1  and 2  of the integration contour. Therefore, the J -integral solution is written as

 

J J

J  

(25)

1

2

J 

J 

are the J -integral values in segments 1

where

and

 and

2  , respectively ( 1

 and

2  coincide with the free

1

2

end of lower crack arm and the clamping, respectively). The components of J -integral in segment, 1

 , of the integration contour are written as (Fig. 3)

m B 

y p 

0

 

 

,

(26)

p

x

  

1 ds dz  , cos

(27)

1

where 1

z varies in the interval

.

h

h

[

/ 2,

/ 2]

1

1

/ u x   , in (24) is found as

Partial derivative,

u x

   

z z

(28)

1 1 1 1 n

where 1 1 n z

and

1  are obtained from Eqs. (18) and (19). The strain energy density is calculated by substituting of (1) in

(10)

1 1 m B u m   

(29)

0

By substituting of (1), (12), (16), (26), (27), (28) and (29) in (24) and integrating in boundaries from 1 / 2 h  to 1 / 2 h , one derives

  

   

 m J m   1 1 m  1

2 y B

2

B

2 B r

4

m

m

1

1

0

1 1

u

u

1 

2 

u     

 h m 2

2

1

1 m b m

1

1

u

u

z

       

  

B r

2

1

n

1

m

m

m

m

2

2

1

1

2 2

1

u

u

u

u

1 

2 

1 

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 q

1

n

1

q

q

q

q

u

1

u

u

u

u

1 

2 

1 

2 

(30)

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 



f

q

 

u u

498

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