PSI - Issue 28

K. Mysov et al. / Procedia Structural Integrity 28 (2020) 352–357 Author name / Structural Integrity Procedia 00 (2019) 000–000

355

4

where   1/2 k I x  is Bessel’s modified functions. After applying the inverse integral transforms (2.5) and (2.4) to the expression (2.6) and considering         1 1 , 0, 0, w R w R w R R               one receive the discontinuous solution in the original domain        2 2 0 1 , , sin , ; , q w r R X R K r R d R R                   , (2.7) where







   

1 sin



  , 

(2.8)

X R

 

sin



 1 !         1 2 k



k



   

  q K r R   , ; ,



   1 , r R P

  0 k P



cos

cos ,

(2.9)





 



 1 !

, k q k

k

k

k



0

k



    1 J rq H Rq r R rR      1 2 2 k k J Rq H rq r R   i , ,    



  , r R

,

(2.10)

q







, q k

c

2



1

1

k

k





2

2

  , 1, 2 i

  k H x is the Hankel function of the first kind.

k P x i 

are Legendre polynomials and

The general solution of the initial problem can be written as the superposition       1 2 , , , w r w r w r     

(2.11)

To determine the unknown function 

 , X R  one must substitute equality (2.11) in condition (1.5) for

0 r R  

and solve the equation

1

1





2

1

,

(2.12)

R

r 

 

r 







G

G

0

r R

 

as well as equation (1.2) to derive the unknown rotation angle. For further calculations it is required to select the arbitrary function from (1.1) which is   2

F    . Knowing the





  ,    r r

  , 

  , 

displacement one can receive stress

and rewrite the singular integral equation

G w r 

1 r w r 





(2.12)



1



 

  1 r 

   X R

   , q k  



  0 k P





  

0

, sin

cos

cos

0, R R d

,

,

(2.13)

P

R

 



  



 

k

k

G

0

k



0

   , , X R X R     ,   1 

where  

 1    r



 , /

and

,

R

R

 

r





    J Rq H rq J rq H Rq           1 1 2 2 k k

    J Rq H rq J rq H Rq         1 1 2 2 k k

    

    

    

2

3

iq

iq





   , , q k 

2

r R R  



2

4

rR

r rR

1

1

1

1

k

k

k

k









2

2

2

2

(2.14)

    J Rq H rq J rq H Rq         1 1 2 2 k k

    J Rq H rq J rq H Rq       1 1 2 2 k k

    

    

    

3

9

iq

i 







4

8

R rR

rR rR

1

1

1

1

k

k

k

k









2

2

2

2

This singular integral equation is solved approximately with the help of orthogonal polynomial method Popov (1992). According to this method the singular part of the kernel must be separated. It was done using the formula from Popov (1982)       0 0 cos cos 2 , P P S      



0

k

k

0

k



(2.15)

1 sec sec 2 2 2  

2 2  

  

  

  ,  

, W tg tg

S



0

0

with changing variables