PSI - Issue 13

Ivan Shatskyi et al. / Procedia Structural Integrity 13 (2018) 1482–1487 Author name / Structural Integrity Procedia 00 (2018) 000–000

1484

3

0 E , 0  are Young's modulus and Poisson's ratio for the filler material. 2.2. Integrodifferential equations

To solve the problem (1)–(3) we used the method of singular integral equations. The integral expressions of the forces and moments through derivatives from jumps of the displacements and the rotation angles of the normal look in the following way (Osadchuk (1985), Savruk (1981)):

y u   

  x u      l

 

 

d x  

l

d x  

B

B

  , 0

  , 0



1

2

N x

n  

,

,

xy N x

s  

y

4

4



  



l

l

  



  x     y d

  x

  x     d

l

l

B

B

  , 0 M x m

  , 0 M x C p 



3



4

,

,

(4)

 

  

y

xy

4

4

l   

 

l



    3 1 3       , E and  are Young's modulus and Poisson's

3 3 4 2 B B Eh    ,

where 1 2 2 B B Eh   ,

ratio for the plate material, C is arbitrary constant. By placing (4) into the respective expressions (2), we received singular integrodifferential equations referring to the jumps functions:       0 1 1 4 2 l y y l u d B B u x n x b x                  ,           0 2 2 4 2 l x x l u d B B u x s x b x            ,       0 3 3 4 2 l y y l d B B x m x b x                   ,           0 4 4 4 2 l x x l d B B x p C x b x              ,   , x l l   . (5) These equations must be solved according to additional conditions on the edges of the cut:     0 x u l   ,   0 y u l       ,     0 x l    ,   0 y l        . (6) In case of arbitrary shape of the slit ( ) b x the solution to the problem (5)–(6) is constructed only by numerical methods. In this article we will find an analytical solution to the case of a slit of a special shape, namely, of elliptical one. Let us assume 2 2 1/2 ( ) ( ) b x l x    , where (0) / b l   , 2 (0) b is maximum width of the crack. Then, in case of uniform loads the solution will be as follows:   2 2 1 1 4 1 y n u x l x B         ,     2 2 2 2 4 1 x s u x l x B     ,   2 2 3 3 4 1 y m x l x B           ,     0 x x   , C p  . (7) where key dimensionless parameters of the problem are: 3. Results and discussions 3.1. Solution

1 2    

2

0 E E



0 (1 )  

0 3 (1 )    

, 2 

, 3 

, 4 

,

.







 

3



 

Forces and moments intensity factors near the tip of the slit, normalized as in Osadchuk (1985), have been found by the formulae: