PSI - Issue 81

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

164

2. Problem Statement and Analytical Approach The plate is modeled as an elastic isotropic layer of h thickness. A cylindrical coordinate system r ,  , z is adopted with the origin at the top surface of the plate. The basic relations for the components of stress and displacement tensors are obtained by expressing the components of the infinitesimal strain tensor as follows:

0 ij ij    = + , , e ij

, , i j r z  = ,

(1)

ij  – the components of the elastic strain tensor, and 0

where ij  – the components of the total strain tensor, e of the free (plastic) strain tensor. Having assumed that the components of 0

ij  – the components

ij  are independent of z , the equilibrium equations for

the isotropic layer is simplified as follows:

1

(

)

( ) 0 

( ) ( )

( ) r

( ) rr r    = + , 0 0 0

(

)

0 

0   − 0 rr

,

dr

2     + + u

=

0 = ,

(2)

2 −  

grad div u

grad



r

where u – a displacement vector. A partial solution can be represented as follows:

r  

z  

r u

, z u

=

=

,

(3)

where  – a scalar function satisfying

1 2 −

.

m

=

2 1 , m   = 0

(4)

1

1

In the general case, the field of residual plastic strains 0 ij  is described by a tensor function. For an axisymmetric problem, having assumed that the tensor is symmetric and the sum of normal components is zero, the nonzero components is reduced to: ( ) 0 0 , rr rr r z   = , ( ) 0 0 0 zz rr     =− + , ( ) 0 0 , r z     = . (5)

ij  , corresponds to the partial solution of

The stress state in the layer is represented as the sum of two stress fields. The first,

the equilibrium equations and, in this case, they will look like

( ) 0 r     +  1

1

 

  

(

)

( ) r

0   + 0 rr

0

2

2

2

,

,

(6)

G

G

G

=

=

+

rr





1  −

r r

r r 

 

1

  

  

( ) r

0   rr

0 + + 

0

2

0 rz  = .

,

G

=

zz

1

1

ij  , corresponds to the general solution of the equilibrium equations and, in the axisymmetric case, is expressed

The second,

using the Love function

( ) ( ) ( ) ( ) A sh z B ch z zC shz D chz J rd         −   = + + +    ( ) ( ) ( ) ( ) ( ) ( ) 0 2

L

(7)

0

by the formulas:

2 1 2 G

1

L

 

2   L

2 1 2 G −

 

 

,

,

2

(8)

L

2

=

L

=

  − 

  − 



rr

z

r r 

2

 − 

r

 

 

 

 

2

2

2

2

G

L

G

L

,

.

(

)

(

)

2

2

2

1

L

L

=

=

 −  − 

 −  − 

zz

rz

2

2

1 2

1 2

z

r

z

z

 − 

 − 

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