PSI - Issue 40

A.I. Chanyshev et al. / Procedia Structural Integrity 40 (2022) 97–104 Chanyshev A.I. at al. / Structural Integrity Procedia 00 (2022) 000 – 000

100

4

will be fulfilled. By virtue of (5), the obtained solution (11) may not necessarily be analytic inside the circle, that is, have poles. The task is to find them inside the circle by solving (11) and to establish the singularity of the solution at 0 z  . Let us consider the numerical implementation of this problem setting. Laplace's equation (1) in cylindrical coordinates is represented as 2 2

1 1 

d u u

u



0

,

 



2

2

2

r r 

dr

r





boundary conditions (2), (3)

2 ( ) a   

u r

 

,

.

2 ( )  



r a u

 

r a 

Let us write down the difference schemes of these equations

2  

2  

u

, i j u u

1, u u   i j

u

, i j u u

1

1

1,

1,

1,

, 1 

, 1 

i

j

i

j

i

j

i j

i j







0  ,

(12)





2

2

2

2

r

h

h

r



i

i

2 ( ) a    j

1, u u h   2 i j

1,

i

j



(13)

2 ( )    ,



u

.

, i j

j

Substituting boundary conditions (13) into (12), with the second order of accuracy on layer 1 ( . i j u . Changing j we find . i j u for the entire layer 1. Then we go to the second layer ( the value or integrating (12) are known. A similar idea can be obtained by considering the problems of the theory of elasticity. Let the following conditions be given on the surface of the half-plane 0 y  : 1 ( ) y f x   , 2 ( ) xy f x   , 3 ( ) x u f x  , 4 ( ) y u f x  , where functions 1 ( ) f x , 2 ( ) f x , 3 ( ) f x , 4 ( ) f x are not related to each other. The task is to find distribution of stresses, strains, and displacements inside the half-plane 0 y  . In this area, the Kolosov-Muskhelishvili formula is valid [5] 0 i  ), we find 1 i  ), all the values

( ) ( ) z z z z z                , ( ) ( ) xy i y

(14)

0 )          . ( ) [ ( ) ( )] z z z x y u i u z C 2 ( 

(15)

The system (14), (15) is a system of two equations for determining two complex functions ( ) z  , ( ) z  .Similarly to finding solution (11), we solve the following problem:



[ ( ) f z if z dz  ( )]

2

[ ( ) f z if z C    ( )]

1

2

( ) z

,

(16)







3

4

1

1

1

 

2

z f z if z

  

[ ( ) f z if z z f z if z C        ( ) ( ( ) ( ))]

[ ( )

( )]  

( ) z

[ ( ) f z if z dz  ( )]









, (17)

1

2

3

4

3

4

2

1

2

1

1

1





3 4    ,  is Poisson's ratio.

where