PSI - Issue 2_B

K.B. Ustinov / Procedia Structural Integrity 2 (2016) 3439–3446 K.B. Ustinov / Structural Integrity Procedia 00 (2016) 000 – 000

3443

5

 

  

ds

1



 

   

J p 

A s B s

exp   

log

 





 

is p

4





(15)

   

   

   

A s B s

s

s

ds

1

coth









1/ 2



  p

  p

  p

2 2 s

1

ln

 



S 

T 

 













is p 

2



2 2 s   

1

1







2 2 p

p i

log

1  





2 2 p   

1

1

1 log



  p

  p

,

,

,

0

T 



S 

 

S 

S     

T 











p



2 2 p

2 2 p

1

2 1









  p  X , the final solution is obtained with the help of Liouville's theorem       1 p p p     F X Π

On finding

(16)

Here   p Π is the vector function to be determined from (8). The most interesting results are related to the stress field near the crack tip, (SIF), and the displacement field far from the crack tip, which relates to coefficients of elastic clamping. According to the obtained solution the SIF is .

I     II   K K



  

  

M    k

k k

T N  k

,

(17)

M

T

M

N

2

sin cos

cos







  

  

  

  

3



2

k

k

k k

k

a

h

12

,

4 3

,

, for

1



 

   M



M

T

N

T

sin





2

a

4 3 

sin cos

cos

 





  

  

  

  

12

1

k

k

k k

3 , for k

h

,

,

   M







M

T

N

T

sin



1

1









         log A s B s    

   

s

s

coth

 









1/ 2



2 2

s

ds

1





 





4 2



2 2   s 

1

1



(18)

    

   

   

   

   

A s B s

d

s

s ds 

4ln 2 1

coth







1/ 2



2 2

A s B s

s

log

1  

log





 

0 

ds

s

 

2 2   s 

1

1

According to the generalization of Irwin’s formula, Salganik (1963), the energy release rate is

   

    

2

2

  

   

(1) E E J 

(2)  

3





  

  





2 a T 

M N T   

N

h

12

4 3

, for

1

 





(1) (2)

E E

2

2

2

a

4 3 

(19)

    

   

2

T

1

1 2





  

  

2

J

M N T N     

h

6

3 , for 





(2)

E

2