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Code_Aster
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Version
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
1/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Organization (S)
: EDF/AMA














Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
Document
V
V
6
6
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0
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2
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2
2



SSNL112 - Bar subjected has a loading
cyclic thermics



Summary:

This case test enters within the framework of the validation of the relations of behavior in elastoplasticity of the elements
bar for the quasi-static mechanics of the structures.
An embedded bar has these two ends undergoes a cyclic thermal loading inducing efforts of
traction and compression.

Each modeling makes it possible to validate one of the relations of non-linear behavior introduced:
Linear isotropic work hardening with criterion of Von Mises (modeling A), linear kinematic work hardening
with criterion of Von Mises (modeling B), as well as a model known as of Pinto-Menegotto, representing it
cyclic behavior of the steel reinforcements in the reinforced concrete (modelings C and D).
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Code_Aster
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Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
2/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
1
Problem of reference
1.1 Geometry
N1
N2
y
X
Z


Length of the bar: 1 m
Section of the bar
: 5 cm
2

1.2
Properties of materials
1.2.1 Linear work hardenings isotropic and kinematics
E
E
y
T
Young modulus:
E = 2. 10
11
AP
Slope D `work hardening:
E
T
= 2.10
9
AP
Elastic limit:
= 2.10
8
AP
Poisson's ratio:
= 0,3
Thermal expansion factor:
= 1.10
- 5
K
- 1
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Code_Aster
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5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
3/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
1.2.2 Model of Pinto-Menegotto
E
U
H
U
y
0
y
0
Young modulus:
E
=
2. 10
11
AP
Elastic limit:
y
0
=
2.10
8
AP
Poisson's ratio:
=
0,3
Thermal expansion factor:
=
1.10
- 5
K
- 1
Deformation of work hardening:
H
=
2.3 10
- 3
Ultimate stress:
U
=
2.58 10
8
AP
Ultimate deformation:
U
=
3.10
- 2
Coefficient defining the curve
:
R
0
=
20
Coefficient defining the curve
:
With
1
=
18.5
Coefficient defining the curve
:
With
2
=
0.15
Coefficient of buckling:
C
=
0.5
Coefficient of buckling:
With
=
0.008
1.3
Boundary conditions and loading
Boundary conditions:
The bar is embedded. Displacements are thus locked in the three directions.
In N1 and N2: DX = DY = DZ = 0
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Code_Aster
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
4/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Loading:
The way of loading is described by the change of the temperature, uniform in the bar:


T 0
1 2 3 4 5 6 7
T (°C)
50
­ 50
­ 300
­ 100 50 ­ 150 ­ 350 ­ 200


















The temperature of reference is taken to 0
O
C
Temperatures
- 400
- 350
- 300
- 250
- 200
- 150
- 100
- 500
50
0
1
2
3
4
5
6
7
Time
T (degrees
C)
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Code_Aster
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Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
5/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
2
Reference solutions
2.1
Method of calculation used for the reference solutions
2.1.1 Work hardenings
linear
Isotropic work hardening
For a uniaxial traction, the criterion of plasticity is written:
L
R p
-
() 0
where p is the cumulated plastic deformation
R p
R p
y
()
= +
and
=
-
R
E E
E E
T
T
The criterion is written then:
L
y
R p
- -
0
The tensor of the stresses is obtained by:
(
)
(
)
=
-
-
-
With
U
Id
. ()
p
ref.
K T T
3
One thus deduces the expression from it from
L
:
(
)
L
p
ref.
E
T
E
T
=
-
-
=
(
)
0
In our case,
= 0
thus:
L
L
p
L
E
E
T
=
-
= -
with
Thus:
·
If
L
R p
-
<
() 0
:
p=0
and
L
L
E
=
·
If
L
R p
-
=
() 0
:
p
R
y
=
-




L
L
p
E
E
=
-
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Code_Aster
®
Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
6/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Application to the way of loading
Moment 1:
el
E
MPa
=
= 200
and
R p
R p
MPa
y
()
= +
=
100
because
p=0
.
One has well
L
R p
-
() 0
.
The criterion is not crossed, the evolution is elastic:
L
MPa
= 100
and
NR
kN
= 100
Moment 2:
The criterion is reached:
(
)
(
)
L
L
y
E
E R R
MPa
NR
kN
=
+
+
=
+
×
+
=
=
-
2 10
2 10
2 02 10 2 02 10
35 10
2 10
205
102 5
11
11
9
9
3
8
.
.
.
.
and
p
=
-
2.475 10
3
.
Moment 3:
One discharges elastically:
L
L
p
E
E
MPa
NR
kN
=
-
=
-
= -
= -
-
-
2 10 15 10
2.475 10
195
97 5
11
3
3
(.
.
)
.
Moment 4:
One plasticizes again:
The criterion is written:
- -
=
R p
y
0
with
p
p
p
=
+
1
2
where
p
1
3
2.475 10
=
-
.
One thus obtains:
(
)
p
R
p
E
E p
p
R
R
E
E p
E
R
MPa
y
p
y
2
1
1
2
1
2
207 9
=
-
-
= -
= -
-
=
+




-
-




= -
.
And thus
NR
kN
= - 103 95
.
Moment 5:
One discharges elastically:
L
L
p
E
E
MPa
NR
kN
=
-
=
-
=
=
-
-
2 10 2 10
10395 10
192 1
96 05
11
3
3
(
.
)
.
.
Moments 6 and 7:
The reasoning is identical
One finds:
NR
kN
NR
kN
inst
inst
(
. )
(
. )
.
.
6
7
10587
44 13
=
= -
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Code_Aster
®
Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
7/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Kinematic work hardening
The method of calculation is identical, but in this case, the criterion of plasticity is written:
(
)
-
-
X
eq
y
0
with
X
C
C
E E
E E
eq
p
eq
p
T
T
p
=
=
=
-
()
3
2
With the preceding notations, the criterion is written:
L
p
y
R
-
-
0
And
L
p
y
R
=
±
(according to the direction of the flow).
Application to the way of loading
Moment 1:
The criterion is not crossed, the evolution is elastic:
L
MPa
= 100
and
NR
kN
= 100
Moment 2:
The criterion is reached:
L
p
y
R
-
-
= 0
L
p
y
R
MPa
=
+
=
×
+
=
-
2 02 10
2.475 10
2 10
205
9
3
8
.
.
Moment 3:
One discharges elastically:
L
L
p
E
E
MPa
NR
kN
=
-
=
-
= -
= -
-
-
2 10 15 10
2.475 10
195
97 5
11
3
3
(.
.
)
.
Moment 4:
One a:
-
-
=
=
-
R
p
p
p
y
p
0
1
2
with
p
1
3
2.475 10
=
-
.
(
)
p
p
R
E
E p
p
E
R
MPa
NR
kN
y
p
y
2
1
1
2
198
99
=
-
+
= -
= -
-
= -
+


= -
= -
Moment 5:
One discharges elastically:
L
L
p
E
E
MPa
NR
kN
=
-
=
-
=
=
-
-
2 10 2 10
9 9 10
202
101
11
3
4
(
.
)
Moments 6 and 7:
The reasoning is identical
One finds:
NR
kN
NR
kN
inst
inst
(
. )
(
. )
6
7
103
47
=
= -
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Code_Aster
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
8/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
2.1.2 Model of Pinto-Menegotto
This model is described in the Manual of Reference of Code_Aster [R5.03.09] [bib1]. The constitutive law
steels is made up of two distinct parts: the monotonous loading composed of three areas
successive (linear elasticity, plastic bearing and work hardening) and the cyclic loading where the way
between two points of inversion (semi-cycle) is described by an analytical curve of expression of the type
= F ()
.
As previously the imposed deformations are thermal deformations:
= - T

2.1.2.1 Case without buckling
First loading
·
Linear elasticity:
= E
Moment 1:
NR
E S
kN
=
=
×
×
=
-
-
2 10
1 10
5 10
100
11
3
4
·
Plastic bearing:
=
y
·
Polynomial of degree 4:
=
-
-
-
-




U
known
U
U
H
y
(
)
0
4
Moment 2:
=
>
=
-
-
3510
2 310
3
3
.
.
H
, one uses the polynomial of degree 4:
= 209 416
.
MPa
and
NR
kN
= 104 708
.
Cycles
Semi-cycle 1:
One determines
p
0
:
p
R
y
0
0
0
3
3
3
3510
1 10
2 510
=
-
=
-
=
-
-
-
.
.
.
because
R
inst
0
2
=
(
. )
.
Then
0
:
0
0
9
3
2 10
2 5 10
5
=
=
×
=
-
E
MPa
H
p
.
From where
y
y
p
sign
MPa
1
0
0
0
200 5
195
=
-
+
= -
+ = -
(
)
·
One calculates then
y
1
:
y
R
y
R
E
1
0
1
0
3
6
11
3
3510
195.209.416 10
2 0 10
1477 10
=
+
-
=
+ -
-
=
-
-
.
(
.
)
.
.
One determines thus
= F ()
, defined by:
(
)
=
+
-
+


=
B
B
B
E
E
R
R
H
1
1
1
()
,
with
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Code_Aster
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
9/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
=
-
-
R
y
R
0
1
0
=
-
-
R
y
R
0
1
0
p
p
y
R
0
0
1
0
=
-
and
R
R
With
With
p
p
1
0
1
0
2
0
=
-
+
One obtains
p
0
123
= -.
and
R
1
351
=.
One can then calculate the value of
at moments 3 and 4:
Moment 3:
-
-
-
-
=
-
-
=
-
-
=
(
. )
.
.
.
.
.
inst
R
y
R
3
0
1
0
3
3
3
3
15 10
35 10
1477 10
35 10
0 988
(
)
=
+
-
+


=
×
+
-
+




=
B
B
R
R
1
1
0 01 0 988
1 0 01
1
0 988
0 82
1
3 51 1 3 51
()
.
.
.
(
(.
)
)
.
.
/.
and
=
-
+
=
× -
-
+
= -
(
)
.
(
.
)
.
y
R
R
MPa
1
0
0
0 82
195.209.416
209 416
122
from where
NR
kN
= - 61
Moment 4:
One uses the same method, with
= 0
.

=
=
= -
= -
173
0 56
20
10
.
.
MPa
NR
kN
Semi-cycle 2:
Moment 5 and 6:
The method of calculation is identical, one determines:
p
1
,
y
2
,
y
2
,
p
1
,
R
2
, then
(
. )
(
. )
(
)
inst
inst
F
5
5
=
and
(
. )
(
. )
(
)
inst
inst
F
6
6
=
and finally
(
. )
inst
5
and
(
. )
inst
6
.
Semi-cycle 3:
Moment 7: Idem
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Code_Aster
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
10/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A

2.1.2.2 Case with buckling
First loading
Identical to the preceding case.
Cycles
Semi-cycle 1 (compression):
The method of calculation is identical, but the value of the slope of the asymptote is modified:
A new coefficient is calculated
B
C
:
B
has
L D E
E
C
B
E
y
=
-
=
×
-
= -
-




×
-






-
(.
/)
.
(.
. )
.
.
.
.
5 0
0 006
5 0 5 9
0 0057
0 01 1.477 10
2 10
2 10 1 36 10
3
11
8
8
It is necessary then, as in the model without buckling, to determine
y
N
. The reasoning is identical,
but a complementary stress is added
S
in order to position the curve correctly by
report/ratio with the asymptote.
S
S
C
C
B E B B
B
MPa
=
-
-
=
×
×
×
+
+
=
1
0.028 0 01 2 10
0 01 0 0057
1 0 0057
0 87
11
.
.
.
.
.
.
where
S
is given by:
(
)
S
C L D
L D
E
=
-
-
=
110
10
10
0 028
.
/
.
.
(/)
Semi-cycle 2 (traction):
·
In traction, one adopts a reduced Young modulus:
(
)
(
)
E
E has
E has
E
MPa
R
has
=
+
-




=
×
+ -
=
-
-
×
-
5
5
11
620 1.473 10
11
10
2 10
0 88
1 0 88
199 10
6
2
6
(.
)
(.
(
. )
)
.
.
with
has
L D
5
10
5 0
7 5 0 88
=
+
-
=
.
(.
/)/.
.
The remainder of the method is identical.
2.2
Results of reference
Normal effort NR constant on the bar
2.3
Uncertainty on the solution
No, the solution is analytical
2.4 References
bibliographical
[1]
Manual of reference of Code_Aster [R5.03.09].
[2]
S. ANDRIEUX: Thermoelastoplastic TD 1 Three bars perfect Von Mises. In “Initiation with
thermoplasticity in Code_Aster “, HI-74/November 96,/013 1996 (manual of reference
course).
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Code_Aster
®
Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
11/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
3 Modeling
With
3.1
Characteristics of modeling
The model is made up D `an element of bar (
BAR
).
Law of behavior: elastoplasticity with linear isotropic work hardening - Criterion of Von Mises
3.2
Characteristics of the mesh
2 nodes.
1 mesh SEG2
3.3
Functionalities tested
Controls
DEFI_MATERIAU ECRO_LINE
STAT_NON_LINE COMP_INCR
VMIS_ISOT_LINE
OPTION
SIEF_ELNO_ELGA


4
Results of modeling A
4.1 Values
tested
Identification Moments
Reference
Aster Variation
%
normal effort NR
1
1.0000 10
5
1.0000 10
5
0
normal effort NR
2
1.0250 10
5
1.0250 10
5
0
normal effort NR
3
­ 9.7500 10
4
­ 9.7500 10
4
0
normal effort NR
4
­ 1.0395 10
5
­ 1.0395 10
5
0
normal effort NR
5
9.6050 10
4
9.6050 10
4
0
normal effort NR
6
1.0587 10
5
1.0587 10
5
0
normal effort NR
7
­ 4.4129 10
4
­ 4.4129 10
4
0
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Code_Aster
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Version
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
12/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
5 Modeling
B
5.1
Characteristics of modeling
The model is made up D `an element of bar (
BAR
).
Law of behavior: elastoplasticity with linear kinematic work hardening - Criterion of Von Mises
5.2
Characteristics of the mesh
2 nodes.
1 mesh SEG2
5.3
Functionalities tested
Controls
DEFI_MATERIAU ECRO_LINE
STAT_NON_LINE COMP_INCR
VMIS_CINE_LINE
OPTION
SIEF_ELNO_ELGA


6
Results of modeling B
6.1 Values
tested
Identification Moments
Reference
Aster Variation
%
normal effort NR
1
1.0000 10
5
1.0000 10
5
0
normal effort NR
2
1.0250 10
5
1.0250 10
5
0
normal effort NR
3
­ 9.7500 10
4
­ 9.7500 10
4
0
normal effort NR
4
­ 9.9000 10
4
­ 9.9000 10
4
0
normal effort NR
5
1.0100 10
5
1.0100 10
5
0
normal effort NR
6
1.0300 10
5
1.0300 10
5
0
normal effort NR
7
­ 4.7000 10
4
­ 4.7000 10
4
0
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Code_Aster
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Version
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Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
13/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
7 Modeling
C
7.1
Characteristics of modeling
The model is made up D `an element of bar (
BAR
).
Law of behavior: model of Pinto-Menegotto without buckling (value of
DASH
lower than 5).
7.2
Characteristics of the mesh
2 nodes.
1 mesh SEG2
7.3
Functionalities tested
Controls
DEFI_MATERIAU PINTO_MENEGOTTO
DASH
:
4.9
STAT_NON_LINE COMP_INCR
PINTO_MENEGOTTO
OPTION
SIEF_ELNO_ELGA


8
Results of modeling C
8.1 Values
tested
Identification Moments
Reference
Aster Variation
%
normal effort NR
1
1.0000 10
5
1.0000 10
5
0
normal effort NR
2
1.0470 10
5
1.0470 10
5
0
normal effort NR
3
­ 6.0777 10
4
­ 6.0777 10
4
0
normal effort NR
4
­ 9.1430 10
4
­ 9.1430 10
4
0
normal effort NR
5
7.6082 10
4
7.6082 10
4
0
normal effort NR
6
1.0125 10
5
1.0125 10
5
0
normal effort NR
7
­ 3.7965 10
4
­ 3.7965 10
4
0
background image
Code_Aster
®
Version
5.0
Titrate:
SSNL112 - Bar subjected has a cyclic thermal loading
Date:
09/04/02
Author (S)
:
C. CHAVANT
Key:
V6.02.112-A
Page:
14/14
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
9 Modeling
D
9.1
Characteristics of modeling
The model is made up D `1 element of bar (
BAR
).
Law of behavior: model of Pinto-Menegotto with buckling (value of
DASH
higher than 5).
9.2
Characteristics of the mesh
2 nodes.
1 mesh SEG2
9.3
Functionalities tested
Controls
DEFI_MATERIAU PINTO_MENEGOTTO
DASH
:
5.9
STAT_NON_LINE COMP_INCR
PINTO_MENEGOTTO
OPTION
SIEF_ELNO_ELGA


10 Results of modeling D
10.1 Values
tested
Identification Moments
Reference
Aster Variation
%
normal effort NR
1
1.0000 10
5
1.0000 10
5
0
normal effort NR
2
1.0470 10
5
1.0470 10
5
0
normal effort NR
3
­ 6.0556 10
4
­ 6.0556 10
4
0
normal effort NR
4
­ 8.9078 10
5
­ 8.9078 10
5
0
normal effort NR
5
7.6905 10
5
7.6905 10
5
0
normal effort NR
6
1.0125 10
5
1.0125 10
5
0
normal effort NR
7
­ 3.8119 10
4
­ 3.8119 10
4
0


11 Summary of the results
The results calculated by Code_Aster are in excellent agreement with the analytical solutions.