PSI - Issue 59

Hryhorii Habrusiev et al. / Procedia Structural Integrity 59 (2024) 494–501 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

496

3

and at the lower boundary plane of layer ( z h  )

0  

 3





  

 

 

, rz r h c   

Ashh A sshh hchh       







 



31

1

2

0

       2 0 1 Bch h Bsch h hsh h J rd              , 1  

 , z urh m Ashh A sshh hchh                            1 2 1 0 Bch h Bsch h hsh h J rd              .         2 1 2 1 0

(6)

1 m , s , 0 s , 1 s are constants depending on the elastic potential ( Guz’ and Rudnitskii , 2006; 1 A , 1 B , 2 A , 2 B are unknown functions that can be found from boundary conditions.

Here, 31 c , 33 c ,

Habrusiev et al, 2022),

We assume that the indenter was formed by the rotation of line

0, 0    

;

r r

a

 

W r

 

1



 2

, r r r r   .



a

a

2

R

about the Oz axis. Here, R is the focal parameter of parabola, and ra is radius of a plane domain at the indenter footing. So we can choose   f r in the form       z f r u a r    . Then the boundary condition (3) will take the form

    



2

a r

, 0

;

r r

 

a

a

2

R

  r

  ,0

  ,0

  r

(7)



 

r a   ,

, 0

u r

u a







z

z

1  



  2   



 

2

,

. r r a  

a r

r r







a

a

a

2

R

3. Solving of the problem Using boundary condition (1), find from (5) the following relation between unknown functions

1 B and 2 B



0;

(8)

1 2 0 B B s  

1 0 2 B s B  .

1 A and 2 A

Substituting (8) into the relations (6) and using conditions (4), obtain a system of equations for

    Ash h Assh h hch h Bhsh h Ashh Asshh hchh B sschh hshh                                                       1 2 0 2 1 2 1 2 1 0 ;

(9)

.

After solving the system (9) we will have       0 1 2 2 h s sh h ch h A B sh h       ;

      2 sh h ch h sh h   

(10)

.

A

B



2

2

Using relations (8) and (10), expressions (5) will become: