PSI - Issue 33

Anna Fesenko et al. / Procedia Structural Integrity 33 (2021) 509–527 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

512

4

2

2

      

  





2

, ( , )

U

   

1

2





1 2             ( , , ) ( , , ) U U

2      ( , , )

( , , )

U

W

1 1

   

 



1

1

2



 

2





2

      

  

      

  



2 , ( , )     W

1

2

1





( , , )        W

2            ( , , ) ( , , ) W U

1 1

 

(5)

1

2



2





1 , 0 1, / a h          .

Boundary conditions (1), taking into account the replacement (4), are transformed into form



( ,0, ) 0,   

( ,1, ) 0, ( ,0, ) 0, U W      

( ,1, ) 0   

U

W

(6)



as the boundary conditions (2) take the form

  

  





1 W aG P      (1, , )

(1, , )

(1, , )

( , )  

3 1

U

U

1 1

  

   

 

(7)











(1, , )     U

(1, , ) 0   

W

(8)





3. Solving the vector boundary problem in transform domain. In order to reduce the problem to the one-dimensional problem, the finite sin- and cos- Fourier integral transforms with regard of the variable  and Laplace integral transform with regard of the variable  (Sneddon, 1955) are applied successively to the differential equations (5) and boundary conditions (6)-(8)

( , ) ( , )    

( , , ) cos ( , , )sin           n

0,1, 2....

U W

U W

n

1        0 



  

  



,

d

n     



n

1, 2,...

n



n



( ) ( )  

U W

( , ) ( , )    

U W

  

     

  

0 

p





p e d  







p





As a result, equations (5) can be written

      

  



1   

2 ( ) W U 

2

2 p U

2    ( )

( )         ( ) U

( ) ( )    

U

1 1

1 1

 

 

*

*

p

p

p

p

p











1





      

  

 

1   

1

2

2



 

( )

( )

( ) W p W

( ) 0  

W

U

2          

1 1

 

(9)



*

*

p

p

p

p









1







( )   

( ,0), p U

, 1    

   

*



During this operation the boundary conditions (6) are automatically satisfied except the first condition ( ,0, ) 0 U    , which derivative goes to the right part of the equations (9). Boundary conditions (7), (8) take the form