In matrix form, this is
{d휀푝}=휆 1 {휕푝
휕휎
}+휆 2 {
휕푞
휕휎
}. (29)
It is obvious that
휆 1 =d휀V푝,휆 2 =d휀푝. (30)According to (27a), (27b), ( 29 ), and ( 30 ),
{d휀푝}=(퐴d푝+퐵d푞){휕푝
휕휎
}+(퐶d푝+퐷d푞){휕푞
휕휎
}.
(31)
Since
d푝={휕푝
휕휎
}
푇
{d휎}, d푞={휕푞
휕휎
}
푇
{d휎}, (32)it follows that
{d휀푝}=[퐶휎푝]{d휎}, (33a)[퐶휎푝]=퐴{
휕푝
휕휎
}{
휕푝
휕휎
}
푇
+퐵{휕푝
휕휎
}{
휕푞
휕휎
}
푇+퐶{
휕푞
휕휎
}{
휕푝
휕휎
}
푇
+퐷{휕푞
휕휎
}{
휕푞
휕휎
}
푇
,(33b)where[퐶휎푝]is the plastic compliance matrix.
Since{d휀} = {d휀푒}+{d휀푝},then{d휀}=[퐶휎푒푝]{d휎}, (34a)where
[퐶휎푒푝]=[퐶푒]+[퐶휎푝]. (34b)[퐶푒]is the elastic compliance matrix and[퐶휎푒푝]is the elasto-
plastic compliance matrix.
Therefore, the elastoplastic model in stress space can
be established once the parameters퐴,퐵, 퐶,and퐷are
determined. Note that퐴, 퐵,퐶,and퐷are not constants,
which evolve with stress and strain, as presented in the
following sections.
6. Elastoplastic Matrix in Strain Space
According to ( 17 ), the plastic stress increment is defined as
{d휎푝}=[퐷푒]{d휀푝},where[퐷푒]is the elastic matrix and{d휀푝}
is the plastic strain increment.
휀V,휀,and휓are three invariants of the strain tensor, where
휓is the strain Lode’s angle. If we ignore the effect of Lode’s
angle and take휀Vand휀as potential functions, that is,퐹 1 =
휀V,퐹 2 =휀in ( 19 ), then
d휎푖푗푝=휆 1휕휀V
휕휀푖푗
+휆 2
휕휀
휕휀푖푗
. (35)
Wr i t t e n i n m at r i x f o r m ,{d휎푝}=휆 1 {휕휀V
휕휀
}+휆 2 {
휕휀
휕휀
}. (36)
Obviously,휆 1 =d푝푝,휆 2 =d푞푝. (37)From the definition of ( 17 ),d푝푝=퐾푒d휀V푝=퐾푒(d휀V−d휀푒V)=퐾푒d휀V−d푝, (38a)d푞푝=3퐺푒d휀푝=3퐺푒(d휀−d휀푒)=3퐺푒d휀−d푞, (38b)where퐾푒,퐺푒are the elastic bulk modulus and shear modulus.
Consideringd휀V푒=1
퐾푒
d푝, (39a)d휀푒=1
3퐺푒
d푞, (39b)substituting (39a)and(39b)into(27a)and(27b) yieldsd휀V=(1
퐾푒
+퐴)d푝+퐵d푞, (40a)d휀=퐶d푝+(1
3퐺푒
+퐷)d푞. (40b)It can now be calculated thatd푝=퐴d휀V+퐵d휀, (41a)d푞=퐶d휀V+퐷d휀, (41b)where퐴=
1
|퐴|
(퐷 +
1
3퐺푒
), 퐵=−
퐵
|퐴|
, 퐶=−
퐶
|퐴|
,
퐷=
1
|퐴|
(퐴 +
1
퐾푒
),
|퐴|=
3퐷퐺푒+퐴퐾푒+1
3퐾푒퐺푒
+(퐴퐷 − 퐵퐶).
(42)
Substituting (41a)and(41b)into( 37 ), (38a), and (38b)gives휆 1 =d푝푝=(퐾푒−퐴)d휀V−퐵d휀, (43a)휆 2 =d푞푝=−퐶d휀V+ (3퐺푒−퐷)d휀. (43b)In addition,d휀V={휕휀V
휕휀
}
푇
{d휀}, d휀={휕휀
휕휀
}
푇
{d휀}. (44)Substituting ( 44 )into(43a), (43b), and ( 36 ) yields{d휎푝}=[퐷휀푝]{d휀}, (45a)