Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
1/10
Organization (S): EDF/IMA/MN
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
V7.90.02 document
HSNV100 - Elasto-plasticity under thermal load
Summary:
To build the reference solution to test the processing of the relation of behavior on a situation 0D
(uniform field), in elastoplasticity under thermal load, with imposed displacement and temperature
increasing. The elastic limit depends on the temperature. The modeling of the geometry can be:
· axisymmetric 2D or plane constraints,
· 3D.
This solution corresponds to test HSNV100 [V7.22.100].
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
2/10
Contents
1 Presentation ........................................................................................................................................... 3
2 Kinematics, balances ........................................................................................................................... 4
2.1 Axisymmetric case (2D) ................................................................................................................... 4
2.2 Parallelepipedic case .................................................................................................................... 4
3 Relation of behavior ..................................................................................................................... 5
4 thermal Loading .......................................................................................................................... 6
5 Solution .................................................................................................................................................. 7
5.1 Elastic phase ............................................................................................................................... 8
5.2 Elastoplastic phase ..................................................................................................................... 9
6 numerical Application .......................................................................................................................... 10
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
3/10
1 Presentation
The studied model problem is such as the solution is uniform in space, without any external effort
given, so as to test only the processing of the relation of behavior.
The following solid thus is considered:
· height
H,
· axisymmetric (of radii has and b),
· or parallelepipedic (thickness B - has).
Z
H
0
R
has
B
It is placed between two lubricated rigid plates.
The material is thermoelastoplastic homogeneous (see hereafter) with isotropic work hardening and criterion of
Von Mises.
One supposes the uniform temperature spaces some, and increasing.
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
4/10
2 Kinematics,
balance
2.1
Axisymmetric case (2D)
Fields of displacement:
U = ur (R) er
(blocking in Z)
U '0
0
R
R
Fields of deforamation: (U) = 0
0
0
according to Z
ur
0 0
R
0 0
0
R
Stress fields:
= 0 1 0 (cf boundary conditions)
L
according to
Z
0 0 0
2.2 Case
parallelepipedic
Fields of displacement:
U = ux (X) ex + uy (y) ey
(blocking in Z)
U '0 0
X
X
Fields of deforamation: (U) = 0
0
0
according to Z
0 0 uy '
y
0 0
0
X
Stress fields:
= 0 1 0 (cf boundary conditions)
L
according to
Z
0 0 0
y
The case could be studied in plane constraints and 3D.
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
5/10
3
Relation of behavior
Isotropic, linear work hardening (tangent AND constant module).
Criterion of Von Mises.
The elastic coefficients, E and, as well as the tangent module AND are invariants according to
temperature.
The elastic limit varies there according to the temperature T:
O
O
y (T)
= y (1 - (St - T)
(for the temperature range studied, is positive there!).
The thermal dilation coefficient is constant.
AND
y
2µ =
E
1 +
E
3K =
E
1 - 2
The law of behavior is written (variable scalar intern p):
1
1
=
tr Id +
D + p +
O
(T - T) Id
9K
2µ
1
with: D
= - tr Id
(diverter of the constraints)
3
3
D
P
3
!
=
p!
, with
=
D D
2
éq
2
éq
p! =
0 if F (, p) =
-
éq
R (p) < 0
p!
0 if F (, p) = 0
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
6/10
R (p) indicates the function of work hardening:
E AND
R (p) = +
p
y
E - AND
The rate!p can be expressed, when F (, p) = 0. Indeed, of!p F identically no one, one draws:
!p!f+!p F = 0. Thus, when one is on the criterion (F =)
0, necessarily!F = 0. I.e.:
3D!
D
- R T!
,
T - R, p!p = 0
2
éq
3D!
D
E E
+ O S!
T
y
T -
!p = 0
2
E - E
éq
T
From where:
E - E
D
D
T
3!
!p
O
=
+ S!T
y
if!p 0,
for
R
2
=
éq
(p)
E AND
éq
(criterion reached, in “load”)
4 Loading
thermics
Uniform temperature in space
T (T) = T + To, > 0
1
T [
,
0 T fine]; with tfin < S
T
To
T
Virgin initial State: L = 0; p = 0
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
7/10
5 Solution
The stress field being uniaxial, one a:
- 1 0 0
D
L
=
0 2 0
3 0 0 - 1
As follows:
=
éq
L
and:
- 1 0 0
P
!p
!
=
sgn (L) 0 2 0
2
0 0 - 1
The relation of behavior leads to:
!p
!rr =! = -
!
- sgn
L
(L) +!T = =
for the case of the parallelepiped
E
2
(!xx!yy
)
1
!zz = 0 =
!
L +
!p sgn (L) +!T
E
From where:
3
1 -
2
!rr =! =!T +
!L
2
2nd
!p = sgn (
L
L) -!
!
T -
0
if
L R (p)
E =
<
D
D
E - E
T
3!
= max 0;
+ O!
y St
if not
E E
2
T
éq
I.e., in the case L = R (p) (criterion reached):
E - E
!p
Max 0;
T
O
=
+ S T
E E
(sgn (L)!
!
L
y
)
T
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
8/10
5.1 Phase
rubber band
At the beginning of the thermal loading, L being lower than y!p is null.
From where:
!
L = - E!T; !rr =! =!T (1+).
As follows:
L = - E T
(compressionL <) 0
rr = = (1+) T
Validity of the elastic solution
The criterion is:
O
L (T) - Y (T)
= E = T - y (1 - S T) 0
The criterion is not crossed for T = [0, ty], with:
O
T
y
y
= (E O
+ S
y
)
y - L
OY
T
T y
At the moment ty:
E O
y
L (ty) = - E + OY S
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
9/10
5.2 Phase
elastoplastic
T ty. One is on the criterion. Then:
E - E
!p
Max 0;
T
O
=
+ S T
E E
(! sgn
L
(L)
!
y
)
T
By admitting that one is “charges some” (!p >)
0, then one eliminate!p to have:
E - E
!
= - E!
T
O
L
T T + sgn (L)
S y
E AND
then:
E - E
S O
y
!p
T
=
!T sgn
(L) +
E
E
With T = ty, L = - E ty < 0; one integrates then these expressions for T T (T
y! =):
E - E
T
O
L (T) = - AND (T - ty)
-
S y - L T
E E
(y)
T
(
E - E
p T)
T
=
2
[E+soy] (t-ty)
E
Maybe, after rearrangement, (T ty):
E
T
O
T
1
1
L (T)
= y S T - +
-
E
T
y
O
y (E - AND)
T
p (T) =
-
1
E2
T
y
Validity of this elastoplastic solution
It should be made sure that L (T) remains negative. Knowing that S T < 1, and that T > ty, the preceding result
confirm that L (T) < 0.
Lastly, it is noticed that:
1 -
2
sgn (L)
!p +!rr = (1+)!T
2
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
Code_Aster ®
Version
3
Titrate:
HSNV100 - Elasto-plasticity under thermal load
Date:
15/03/96
Author (S):
F. VOLDOIRE
Key:
V7.90.02-A
Page:
10/10
from where:
1 -
2
rr (T) = (T)
= (1+) T +
(
p T),
T
,
2
[ty tfin]
(since L (T) < 0).
6 Application
numerical
E = 200.000 MPa; = 0 3; = -
10 5 °C-1; = 10
. s-1
O
= 400 MPa; To = 0 °C; S = -
10 2 °C-1; T
< 100s
y
end
E
= 50.000 MPa
T
From where:
T
= 66 6666
.
S
y
133 333
.
L
(ty) = -
MPa
elastic phase
-
rr
(ty) = (ty) = 0866666
.
10 3
.
Then, elastoplastic phase:
with T =
S
80:
L ()
80
= - 100 0
. MPa
(
p
)
80
=
-
0 3000
.
10 3
.
3
rr (
)
80
= ()
80
=
-
1100
.
10
.
with T =
S
90:
L ()
90
= - 7500
.
MPa
(
p
)
90
=
-
0 5250
.
10 3
.
3
rr (
)
90
= ()
90
= 1275
.
10
.
Handbook of Validation
V7.90 booklet: Theoretical references of tests into thermomechanical
HI-75/96/014/A
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