PSI - Issue 37

L.V. Stepanova et al. / Procedia Structural Integrity 37 (2022) 920–925 Author name / Structural Integrity Procedia 00 (2019) 000 – 000

923

4

F

F

F

F

F

F

 

,

rr a F

a F 

r 

r 

= −

= −

rr

rr

rr

,







rr

rr

rr







r 

 F

r  F

 

rr

rr





F   

F

F

F

 

,

rrrr b

=

rr

,

, rr

, rr

rr b

b

a F

r 

=

=

= −

r

r

r

(12)

r

r

rr

rrr





r 

 

r  

rr



rr

F

F

F

F

F

, F   

 

 

,

, r 

,

, r 

b

b

b

b

b

b

=

=

=

=

=

=

r





rr

r  

r

r rr 

r r







 

r  

 

r  

rr

rr

( ) ( , k rr r      . Thus, by quasilinearisation the constitutive equations are ( ) k ( ) k , )

are evaluated for the current estimate

replaced by its linearised form. As an example, this problem has been solved for the special case of a creep power law . n B   = Consequently, equations (1)-(5) with linearized creep strain rates (9)-(11) is a system of linear partial differential equations, where stress tensor components for every estimate can be determined in the following form

20  = + + + + + + + + + 40 60 80 62 k k 82 22 k k 42 k k k k

   

( , ) r   rr

cos 2

k

k

+

    

00

02

2 r r r r 4 6

8

2 r r r r k k k k r r r r + + + + +   4 6 26 46 66 06 2 4 6 k

8

64 k k

24 k k

  

  

,

cos 4

cos6

44 + + + + + 2 r r r r 4 6 k 04

84

86

8

8

l

l

l

l

l

l

l

l

62 82  = + + + + + + + + + 22 42

  

(13)

( , ) r   

cos 2

l

l

+

20 40 60 80 r r r r 2 4 6 8

 

00

02

2 r r r r 4 6

8

l

l

l

l

l

l

l

l

  

  

  

  

,

cos 4

cos6

64 84 24 44 + + + + + 2 r r r r 4 6 l 04 8

l

+ + + + +

26 46 66 86 r r r r 2 4 6 8

06

62 m m

64 m m

22 m m

24 m m

  

  

  

  

( , ) r    r

sin 2

sin 4

42 = + + + + m

44 + + + + + m 

+

82

84

02

04

2

4

6

8

2

4

6

8

r

r

r   

r

r

r

r

r

m m m m

 

26 + + + + +  . where results of linear problem (Kirsch problem) used as zero approximation 2 4 2 (0) 2 4 2 1 3 4 ( , ) 1 1 cos 2 2 rr a a a r r r r            = − − + −             , (0) 1 ( , ) 2 r      = 46 66 86 8 sin 6 06 2 4 6 m r r r r  Third approximation obtained by the quasilinearisation method is (3) 2 4 6 8 0.20819 0.31247 0.10884 0.12951 ( , ) 0.5 rr r r r r r   = − − − + + 8 1.74046 0.96566 0.13228 0.40709 0.5 cos 2 r r r r    + − + − + − +     2 4 6 2 8 2.11747 4.30736 1.63138 1.05849 cos 4 r r r r    + − + + +     8 0.213459 0.13539 0.99656 0.64771 cos 6 r r r r    + + − +     , 8 0.20819 0.93742 0.5442 0.90657 ( , ) 0.5 r r r r r    = + + − + + 8 0.96566 0.3307 1.70978 0.5 cos 2 r r r    + − + − + +     4 4 6 2 4 6 (3) 2 4 6 4 6 4 3 2 2 (0)     = ( , ) r 4 − +  1 2 1 2 sin 2 a a r r        .

  

 

   

  

2

4

3

a

a

,

1 + + +     1

cos 2

2

4

r

r

(14)

6 8 1.4357 1.63138 1.06622 cos 4 r r r    − − +    

(15)

Made with FlippingBook Ebook Creator