PSI - Issue 81

Borys Shelestovskyi et al. / Procedia Structural Integrity 81 (2026) 162–169

165

The functions ( ) A  , ( ) B  , ( ) C  , ( ) D  are determined from the boundary conditions of the problem. Let us consider the plate free of external loads, in which a field of residual plastic strains arises due to localized heating during welding. This field can be described analytically as follows: ( ) ( ) 0 * 2 2 2 2 0 1 exp , rr p r p r    =− − − ( ) ( ) 0 * 2 2 2 2 0 1 exp , p r p r     =− + − ( ) 0 0 0 , zz rr     =− + (9) 0  ,  , p are obtained from experimental data (Nedoseka, 2008). The scalar function  defining the partial solution of the three-dimensional elasticity problem with eigenstrains satisfies the modified equation ( ) ( ) ( ) 2 * 2 2 2 2 0 1 0 1 1 exp m p r p r m r     =− − − + , (10) where parameters *

where

(

)

( ) r

0 

* 0 exp

2 2

.

p r

  =−

−

(11)

Since the right part (10) does not depend on z , then

2 2 r              1 1 r r r r r r

2

 = + = 

(12)

the equation (10) looks like

1

r r      r    

r r         

(

) ( exp

)

 1 0 1  ( *

) ( exp

)



*    + − 1

2 2

2 2

2 3

2 2

.

or

m

r p r

p r

m

p r

p r

=−

+ −

−

=−

−

 

1 0

r r

(13)

By integrating the last expression on r we get

  

  

С

1 2

1

  

  



(

)

*

2 2 − + p r

exp

.

m

r





=−

−

1

1 0

2

r

r

rp



(14)

0 r = should be limited due to the limited movement and that

The right part of the last expression at

r r u   = . As the result

we have

*

1

1 2

1

1 2 −

0 

2 2 p r  −





 

  

( ) 2 2 1 exp  − −  p r  

(

)

( )

( )

2 2

.

,

exp

,

F r

С

F r

p r

=

=



=

−

−

1

2

2

1  −  

2

p

r r

(15)

ij  corresponding to the partial solution of equilibrium equations, taking into account the

Formulas (6) for stress components expression for ( ) F r , are written as follows:

*  −   −  0  1 G

1 2

*

  

G

0 

  

  

(

)

(

)

(

)

( )

( )

2 2

2 2

2 1 2 − − − +

exp

,

F r

p r

p r

2 2

exp

,



=

  

−

F r

p r

=

−

−







rr

1

−



(16)

* 0 2 

G

−

(

) ( exp

)

1 − + −   

 

2 2

2 2

2

,

0.

p r

p r

=

−



rz  =



zz

1

−



We will satisfy the boundary conditions of the task:

0 z = , z h = .

0 zz zz   + = ,

(17)

ij  the components of the tensor will be presented as the integral of Hankel

To do this,