PSI - Issue 2_B

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

3444

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

6

The last formula is similar to the one given by Hutchinson and Suo (1992) however in that work the influence of the transverse force was not considered. The displacements far from the crack tip is polynomial with the two leading terms (3-d and 2-nd power of x ) corresponding to deflection of the equivalent beam under the applied force and moment. The term of tangential displacement proportional to x corresponds to tension of a road. The other terms corresponding to the constant relative displacements and rotations proportional to the applied total force and moment with the help of (1) may be written as follows (expressions for 11 33 , i i a a are too awkward to be written here).

          

          

   

   





2 12 9 3  



2

2   12

a

4 3



 





11

2

8





   

   

(1) E E 

(2)

2

2

12 9 

3 2



2    2 6

A

h

3

6

, for

1













(1) (2)

E E

2





3 2

  

2

2  

2    2 6

a

4 3

12

33

8

0  

d

ds

4log 2

3

1

  

  









   

2

2        2 2 4 3 ,

a

A s B s

4 3

2 3

log



1 

 

  



11

1

ds

s

2





 

    

s ds   

  

0        

24 3 

3

d

A s 

3 4log 2 

1

1

coth

  

 

3   2 

2

AB

a

3

log

log





1    





  

 

3

3

3

3

ds

B

s

2







2 2 s

2 2 s   

1

1

1





     

i

a

3

1 3 



11 3

1

2

A

12

3 / 2 6 







for h 

(20)

(2)

E

i

2

a

1 3 3 / 2 6   



33

1   ) value of  becomes very close to the value obtained by

For rather rigid semi-infinite substrates (

considering the problem of a beam half-plane, see Salganik and Ustinov (2011).

2/3 5/6 2 3 0,636  

3     d d ,

(21)

0

0

3. Solution of the scalar problems

On assuming the possibility to neglect the cross-terms in (5), the matrix Riemann problem is reduced to two scalar ones. For normal separation we have       F p K p F p    , p L  ,       1 sin cos / sin cos / K p p p p d h p h p h p d      (22)

(2)

 

0

  ,0 v x v x e dx x           ,0 (2) (1) 0 px

 

  ,0 x e dx   px yy

  F p E  

F p 

 

(23)

2



The energy release rate and coefficients of elastic compliance for which are (here  is Riemann zeta function)