PSI - Issue 33

K. Mysov et al. / Procedia Structural Integrity 33 (2021) 365–370 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

368

4

and a variation of the Hankel integral transform by standard scheme ( ) ( ) w rI r w r dr   =

0 

1 2

k

k



k

+

(2.5)

0  

1 2

( )    I r w d

( ) ( ) 1 = −

k

+

w r

1 2

k

k



k

+

to correspondence (1.7) and receive the solution in transform domain ( ) ( ) ( ) 1 1 sin , cos I w P        • − 

 



( ) ,  

(

) d   

0 

1

cos

w

P

−

 

1 2

k

k

k

k





k

+

,

(2.6)

w

= −

k



2

2

q

+



( ) 1/ 2 ,

where are Bessel’s modified functions. After applying the inverse integral transforms (2.5) and (2.4) to the expression (2.6) and considering ( ) ( ) ( ) ( ) 1 , , 0 , 0 cot w w w         • • • = − − + = one receive the discontinuous solution in the original domain ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 1 2 1 , 1 1 2 , sin , cos cos 1 b q k k k k a k w r r P P d k k          = + = −  +   , (2.7) where ( ) ( ) ( ) ( ) ( ) 1 1 2 2 , 1 1 2 2 , , , , 2 k k q k k k J q H rq r i r q c J rq H q r r         + + + +    = =    (2.8) ( ) , 1,2 i k P x i = are Legendre polynomials and ( ) k H x is the Hankel function of the first kind. The general solution of the initial problem can be written as the superposition ( ) ( ) ( ) 1 2 , , , w r w r w r    = + (2.9) To determine the unknown function ( ) 1   which has next form ( ) ( ) 1 1 /      = (2.10) one must substitute equality (2.9) in condition (1.5) for 0   = + and solve the equation ( ) 2 1 0 1 1 , r G G        = + = − (2.11) 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). Knowing the displacement one can derive stress ( ) ( ) ( ) ( ) 1 , , , r r G w r r w r      −  = − and rewrite the singular integral equation (2.11) in form ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 1 , 1 1 2 , cos , 1 sin b q k k r k a k r r P d r k k G           = +  = +   , (2.12) where ( ) ( ) 1 1 , , / r r R R        = ( ( ) 1 , r R    is found knowing (2.2)). This singular integral equation is solved approximately with the help of orthogonal polynomial method Popov (1982). Realizing the standard scheme of orthogonal polynomial method leads to the singular integral equation 1, 2,... k I x k + =

  

   

2 1





2   n 

( ) y U y dy A x + 2

( ) , 

( ) , 

1 ln 1 1 x y −

F x

,

(2.13)

−

=

n

n

x

 

1

n

=

−

where

3 2

(

)

1 1 a b x b a + + −  1 1

  

2



 

(

)

2

1 1 a b x b a + + −  1 1

  

( ) , 

1

,

F x

=





,

(2.14)

 



2

2

1 1 a b b a +   −  1 1 x    +

(

)

1 1 G b a −