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)
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