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29 211, 1000
525
84
1900
2005 by
by
1865 843830,
1865 853333,
222 7504744, 90
01923,
7508400, 20 7631
no by
by
2005
20 7631 5500.
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1
p
,
no 0,
,,,
by
21.3.
on
by
.p
and
and a and
by
2 .p
0
0.
593
2
. ......
0 and
0
0
0
10
0. 2,
10
2 by 0
k,,,
by 0 p by 0 qb 0
o
0 and o,
g and 0
,
0
a
0
by
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594 0
g,
g
g 10
on g 2 ___ on 2 2
0
0
2
0
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a 2h 0 b,
a hq a by
0,
0. 0
0 2
0
0.
0
a
.
?
10,
p
0g
0
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595
10.2.
1 0
0. 10
22.3,
22.2
21.3
body
no
no 0 body
596
a
,, -
-- f((l - a)y (1) + ay (2))
I',
1
y(1) l
y(2)
=y
1, 2 , . . . ,
a 0
1 _
+ (I -ct)y~2) e C
and 0
1.
by by
a
a
597
. . . . . .
by
f(ory:I)+ (I -- a)y:2)) <
otf(y~I))+ (I -- ot)f(y~ 2))
From the definitionof the set C, we have f(y:l)) < K and f(y:2)) < K and _
by 22.3. a and
and 0
,
0
0
and
0 ,
u,
598
y
by
on
on *
on
,
,
by
0 ,
,
,
0
0
599
and
and 0
and
0
2
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do
a
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a.
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by
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no
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by o
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2 p.
von
_
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.
603
9.6
p.. p
p
p
p o
p
1
v
2
eeff =
vlvP(ePeff ) "--"~H(Eef f ) ,
~fP(eP/f)is made
If a Taylor expansion of p
1
IC
(22.26)
about the point
0
0
no
a
n
0
= ka.(e~:/)"
_.
0, w e obtain
0
0
p
p
ee/fv=
2
1
p
n
1
604
p
p
Eeff
p
,
von
2
o
o
g
2
0
0
b
-
2
3
_
-
3
o
605
von a
0
l(o'q
4, _
0
0 0
,1
,
,91
ag
p
1
2 n
1
. p p
von
606
1
2
by
g
dg =
=
_ _
=
12,,,:: by
p
elf
2
g*
=
-
.p
/~ = c%//(I - ~-~-=)
(22.35)
607 0
by
by by
2
2
by
yon
3
d
,
0 .p
2
608
1 , 0
0 no
1
1
3
by by
609
von 3 d
d
d.
.p 28yo
.p . fcij = - :2~i ) = ~.
28yo
=
E ij '
2 fc =
- i~ - ~
-
E ef f
From (22.1) and as p~t'(tcij, ic)
,c)
p
,c)
p
p =
+ P
021p
k
2
p =
3
3 =
0
0 1
1
3
2
610 by
02111
021# p
2
0 .p
25yo
2#yo 2 0
von
3
, 1
p
q
o
3
2
eff
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0
3 3
611
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d
d
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x
h
aoo oo
h
.
by
2
2
.p
p
021p
,
612
02q1
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p
.p
p p
1
3
by by by
p
1
d
3
d
3 0 3
d 0
2a,,o
d d
0
613
3
o
=
3
g=
p
d
h
=;of
.,
og
g
., _
aoo
3
0
2h
h
2 p
1
d
g
and d
..
;.,
614 p* p*
.p
2
p
p.
2
0
gp 0
0 0
22.5
0
0 p
0
0.
615
g
11
0
0
0, 22.6
0.
.p
o
0.
22.6.
0,
0, 0
0
by
616
.p
0
p ___0 on
p
p
2
617
g
a
0,
and a*
., x
0
a
a*
and
and and p
,
p
2
22.4
von
9.6,
von
p. p. von
618
p p
do by
x
.
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n.
nq
q 14
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619
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1
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620
by
1, 2 . . . . .
on 1, 2 . . . . .
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by
0
0
and
by dp
1, 2 .....
621
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1
p
_
p
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a
622
a
a,
n 0 0
by
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AA=I
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1
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624
by
1,
0
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0,
0
625
no 0
10.3.
0. 0
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22.8 on 22.9. by k k
1
626
1
1
2
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1
2
1
S
0
by by
02
,,
S
3
627
22.6 15
.
on by
by .vp
.vp
0
0 0 by
.vp
..
. by p
0
0
0
628
0,
0
d
0.
by
.vp =
3
0 0
f (~Yij,K~,) = qg(nEeff )
~
~(f) = rle~ff
where 9 is the inverse function of q~. A comparison of (22.80) and (22.81b) shows that the non-negative function A is given by
von
by
on by
0 15.4.2 by
8q,
0 o
on
629
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1
by
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by
0
n
0
p
n
0
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k
2
1, 2,
....
n
635
0
0
by
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by by by by
by
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by
23.1
by
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638
30 25 20
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3.8.
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q2 q3
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654
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ao
by
q v
o,
qn dS
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0
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19,
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y
1.
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1
7
1,
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1
1
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n
1,
0
n
1
0
o ao,n
o
ao, ao
17. 1, 2 . . . .
ao ao ao
0, 0
ao
658
0
by
1 ,
1 O[C(ao - ao,.)] = C +
1 C
(23.5I)
k dk
_
k oao
iCo Ko
Ko =
B o -~VONo
= C + At
~o
1
= ~
dV
+ Ko + Ko
1(ao_l - ao.~) + K o ao-I - f o,~+1
ao,n
17 n,
ao,n
659
17.7
a
23.6
23.7,
k 1
17.8,
by
23.5.2
on ao.
an
n n
on
on
0 by
n
1.
17 by a
,
a
on ao,n
660
The starting vector for a ~-I for i = I is taken as a ~ = a., that is the value at the last accepted state n. It appears that the only way in which the temperature field influences the mechanical field equations is through the stresses which, in turn, determine the internal forces ~v BTtrdV; this is the topic of the next section.
23.5.3
18 18.1.1, no 2 1
2
1
.
tr~J) = .-. . (i) -- ekl _p(1). Z)Okttekt ) -- riO(0(i) -- 00) where for simplicity it was assurncd that Dijkl and to are constant. Subtracting the above equations
p
Aait; -~ DijklAEkl --
_
0
no
=
0
p,jA01
661
0 0
18.1.4.
0_.gg
0 2 by
a
0 von 18.1.4 18.1.5
23.6
by 23.8 by
ao
662 by _ 1, 2 . . . .
on by
on
23.6.1
1
1 by n
n
1 on
on
0
o
an, ao,n
n
1.
p
.
k n
k
1
1
1
1
1
1,
663
p
1, 2,..
ao
~to(ai,aio)=llto(a i-I,a i-I) o + ('~a
(ai -
) + ("~ao )
- a~
According to the Newton-Raphson method, we require ~to(a~,a 0) = O, i.e. ,
i-i +(O~o)i-1(ai ai-l)+.O~o,i-1(ao i-i 23.5.1,
by u
--gZ-a o aV
V
_ 1__
on
2 on
23.2.1
664
0
Tia-l(ai - d-l) + Tio-l(ao- ao ) = -~o
1
23.6.2
0
ao
n , a0
1
,
ll/(ai-Ii-I)
v(a t,a~) =
,a 0
0~oo
o
i-I
ao
0 0
,
ao
o
da on
665
dg ,
0
a0
d V - f n+l
~r(a i-I, a o ) =
The starting vectors for a t-1 and a o are taken as a 0 = a~ and a~ = ao,,, i.e.the values at the last accepted state n. From (23.73) and (23.76) it is apparent that the Newton-Raphson formulation of the coupled problem becomes
Tol
aoaoii1
L
1
, ao (a0 -I - ao.n) + K 0 a 0
= C
_
_
k
1
1
,. ao
ao ao.n
1, 2 . . . . . n
by
666
0 23.5.3 on
a
a a
,00
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1
a
pa a
a
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ad 11
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on 23.10.
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Is (il2)_ill) ) IvVi(bl2).(1)
.(I)
b Su
.(2) (I) v, j(:#~> - :,:)))dv =
Suv,(il~> - t,.(I))dS u
0
o Introduce the notation
0
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by
1
1,..
0,
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by on on
by y,
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1.
698
d2 .9 y
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1, 2 , . . . ,
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y
699
,
(.~yi))otAy,(.9(a 2)
_
+
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_ y,
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700
,
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1, 2 , . . . , 1, 2 , . . . , and a
0
2
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704
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and
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0
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37,
and
44,
and
and
34,
69,
4,
35,
and
98,
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44, 1530,
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