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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 1/14
Organization (S): EDF/AMA
Handbook of Validation
V6.02 booklet: Nonlinear statics of the linear structures
V6.02.112 document
SSNL112 - Subjected Barre 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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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 2/14
1
Problem of reference
1.1 Geometry
N1
y
X
N2
Z
Length of the bar: 1 m
Section of the bar
: 5 cm2
1.2
Properties of materials
1.2.1 Linear work hardenings isotropic and kinematics
E
T
y
E
Young modulus:
E = 2. 1011 Pa
Slope D `work hardening:
And = 2.109 Pa
Elastic limit:
= 2.108 Pa
Poisson's ratio:
= 0,3
Thermal dilation coefficient:
= 1.10-5 K-1
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Code_Aster ®
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 3/14
1.2.2 Model of Pinto-Menegotto
U
0y
E
0y
H
U
Young modulus:
E
=
2. 1011 Pa
Elastic limit:
0
=
2.108 Pa
y
Poisson's ratio:
=
0,3
Thermal dilation coefficient:
=
1.10-5 K-1
Deformation of work hardening:
H
=
2.3 10-3
Ultimate constraint:
U
=
2.58 108 Pa
Ultimate deformation:
U
=
3.10-2
Coefficient defining the curve:
R0
=
20
Coefficient defining the curve:
A1
=
18.5
Coefficient defining the curve:
A2
=
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 blocked in the three directions.
In N1 and N2: DX = DY = DZ = 0
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 4/14
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
Temperatures
0
1
2
3
4
5
6
7
50
0
- 50
C) - 100
- 150
- 200
- 250
T (degrees - 300
- 350
- 400
Time
The temperature of reference is taken with 0oC
Handbook of Validation
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Code_Aster ®
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 5/14
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
E E
R p
R p
y
() = + and R =
T
E - And
The criterion is written then:
- R p
- y
L
0
The tensor of the constraints is obtained by:
= A. ((U) - p) - 3 (- ref.
K T T) Id
One thus deduces the expression from it from:
L
= E (- T) - E p
ref.
L
(T =) 0
In our case, = 0 thus:
= E - E p
with =
L
L
L
- T
Thus:
·
If L - R (p) < 0:
p=0 and = E
L
L
·
If L - R (p) = 0:
y
-
p =
R
= - p
E
E
L
L
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Code_Aster ®
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 6/14
Application to the way of loading
Moment 1:
= E = 200MPa and R p
R p
y
() = + = 100MPa bus p=0.
el
One has well - R (p) 0.
L
The criterion is not crossed, the evolution is elastic: = 100 MPa and NR = 100kN
L
Moment 2:
The criterion is reached:
11
E
2
=
(
10
R
+ y =
2 0
. 2.109 × 35
. -
10 3 + 2.108 = 205MPa
L
L
)
11
9 (
)
E + R
2 10 + 2 02
. 10
NR = 102 5
. kN
and p =
-
2.475 10 3
.
Moment 3:
One discharges elastically:
= - p
E
E
= 11
- 3 -
-
2 10 15 10
2.475 10 3
(.
.
) = - 195MPa
L
L
NR = - 97. K
5 NR
Moment 4:
One plasticizes again:
The criterion is written: - RP - y = 0 with p = p + p where p
- 3
= 2 475
.
10
1
2
1
One thus obtains:
y
-
p =
- p
2
R
1
= - E p
= - E (p - p
1
2)
R
E y
=
2nd p
= - 207 9
. MPa
R + E
1
-
-
R
And thus NR = - 103
K
95
.
NR
Moment 5:
One discharges elastically:
= - p
E
E
= 11 - 3 -
-
2 10 2 10
10395 10 3
(
.
) = 192 1
. MPa
L
L
NR = 96. K
05 NR
Moments 6 and 7:
The reasoning is identical
One finds:
NR
=
.
10587kN
(inst.6)
NR
= -.
44
K
13 NR
(inst.7)
Handbook of Validation
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Code_Aster ®
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 7/14
Kinematic work hardening
The method of calculation is identical, but in this case, the criterion of plasticity is written:
(- X) - y 0
eq
3
p
p
E E
with X
C ()
C
T
p
=
=
=
eq
eq
2
E - And
With the preceding notations, the criterion is written:
- p
- y
R
0
L
And = p ± y
R
(according to the direction of the flow).
L
Application to the way of loading
Moment 1:
The criterion is not crossed, the evolution is elastic: = 100 MPa and NR = 100kN
L
Moment 2:
The criterion is reached: - p - y
R
= 0
L
= p + y
R
=
9 ×
-
2 02 10
2.475 10 3 + 2 108
.
.
= 205MPa
L
Moment 3:
One discharges elastically:
= - p
E
E
= 11
- 3 -
-
2 10 15 10
2.475 10 3
(.
.
) = - 195MPa
L
L
NR = - 97. K
5 NR
Moment 4:
One a:
- R
p - y = 0 with
p
- 3
= 2 475
.
10
1
p = p - p
1
2
y
+
p = p -
2
1
R
= - E p
= - E (p - p
1
2)
y
+
= - E
198MPa
R
= -
NR = 99
- kN
Moment 5:
One discharges elastically:
= - p
E
E
= 11 - 3 -
-
2 10 2 10
9 9 10 4
(
.
) = 202 MPa
L
L
NR =
K
101 NR
Moments 6 and 7:
The reasoning is identical
One finds:
NR
=
K
103 NR
(inst.6)
NR
= - 47kN
(inst.7)
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 8/14
2.1.2 Model of Pinto-Menegotto
This model is described in Manuel de Référence of Code_Aster [R5.03.09] [bib1]. The constitutive law
steels is made up of two distinct parts: the monotonous loading composed of three zones
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 = 11 × - 3 × -
2 10
1 10
5 10 4 = 100kN
·
Plastic bearing: = y
- 4
·
Polynomial of degree 4: = -
(
- 0) U
U
known
y -
U
H
Moment 2:
=
- 3
> =
-
3510
2.310 3
.
.
, one uses the polynomial of degree 4:
H
= 209 416
.
MPa
and NR = 104 708
.
kN
Cycles
Semi-cycle 1:
0 are determined:
p
0 = 0 - 0
- 3
- 3
- 3
= 351
. 0 - 1.10 = 2 51
. 0 bus 0 =
.
p
R
y
R
(inst.2)
Then 0:
0 = E 0
9
- 3
= 210 × 2 5
. 10 = 5MPa
H
p
From where 1 = 0 sign (0
-) + 0
= - 200 + 5 = - 195MPa
y
y
p
·
One calculates then 1:
y
1 - 0
6
-
(- 195 - 209 416
.
) 10
1 = 0
y
R
3
- 3
+
= 351
. 0 +
= 1477
.
10
y
R
E
11
2 0
. 10
One determines thus
= F (), defined by:
1 - B
E
B
=
+
H
,
1 R
with B =
(
1 + (R
))
E
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 9/14
- 0
=
R
1 - 0
y
R
- 0
=
R
1 - 0
y
R
0
With
0
p
0
1
p
1
p =
and R = R -
1 - 0
0
With
0
+
y
R
2
p
One obtains 0 = - 12
. 3 and R1 = 35
. 1
p
One can then calculate the value of at moments 3 and 4:
Moment 3:
- 0
- 3
- 3
(inst. )
3
R
.
1510
-.
3510
=
=
=.
0 988
1 - 0
- 3
-
.
1477 10
-.
35
3
y
R
10
1 - B
1 - 0 0
. 1
B
=
+
0 0
. 1 0 988
.
0 8
. 2
1 R
=
×
+
=
3 51
.
1/3 51
.
(
1 + (R
))
1
(+ (0 988
.
)
)
and =
(1 - 0) + 0 =.
0 82 × (- 195 -
.
209
)
416 +
.
209.416 = -
y
R
R
122 MPa
from where NR = - K
61 NR
Moment 4:
One uses the same method, with = 0.
= 17
. 3
= 05
. 6
= - 20MPa
NR = - 10kN
Semi-cycle 2:
Moment 5 and 6:
The method of calculation is identical, one determines:
1, 2, 2, 1, R2, then
F (
=
) and
F (
=
)
p
y
y
p
(inst. )
5
(inst. )
5
(inst.6)
(inst.6)
and finally (inst. )
5 and (inst.6).
Semi-cycle 3:
Moment 7: Idem
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
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Key: V6.02.112-A Page: 10/14
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 bc is calculated:
11
E
210
- 3
B
.
0 01.
1.477 10
8
8
B
has (.
5 0 L/D) E
y
=
-
-
×
=.
0.006 × (.
5 0 -.
5 9) E
2 10 -.
1
36 10
C
= -.
0 0057
It is necessary then, as in the model without buckling, to determine N. The reasoning is identical,
y
but one adds a complementary constraint in order to position the curve correctly by
S
report/ratio with the asymptote.
B - B
0 0
. 1 +
0 0057
.
=
C
11
S
Sb E
= 0 028
.
× 0 0
. 1× 2 10 ×
= 0 8
. 7 MPa
1 - C
B
1 + 0 0057
.
110
. - L/D
where is given by: =
= 0 028
.
S
S
10 (C (L/D
E
) - 1. )
0
Semi-cycle 2 (traction):
·
In traction, one adopts a reduced Young modulus:
(- has 2
6
6
)
11
(- 620×1.47 -
3 10)
E = E has + (. -) E has
(.
(
. ) E
11
10
2 10
0 88
1 0 88
) 19
. 9 10 MPa
R
5
5
=
×
+ -
=
with A = 10
. + 5
(0
. - L/D)/7 5
. = 0 8
. 8
5
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]
Handbook of reference of Code_Aster [R5.03.09].
[2]
S. ANDRIEUX: TD 1 Trois thermoelastoplastic bars perfect Von Mises. In “Initiation with
thermoplasticity in Code_Aster “, HI-74/November 96,/013 1996 (handbook of reference
course).
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Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
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Key: V6.02.112-A Page: 11/14
3 Modeling
With
3.1
Characteristics of modeling
The model is made up D `an element of bar (BARRE).
Law of behavior: elastoplasticity with linear isotropic work hardening - Critère of Von Mises
3.2
Characteristics of the grid
2 nodes.
1 mesh SEG2
3.3
Functionalities tested
Commands
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 105
1.0000 105 0
normal effort NR
2
1.0250 105
1.0250 105 0
normal effort NR
3
9.7500 104
9.7500 104 0
normal effort NR
4
1.0395 105
1.0395 105 0
normal effort NR
5
9.6050 104
9.6050 104 0
normal effort NR
6
1.0587 105
1.0587 105 0
normal effort NR
7
4.4129 104
4.4129 104 0
Handbook of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 12/14
5 Modeling
B
5.1
Characteristics of modeling
The model is made up D `an element of bar (BARRE).
Law of behavior: elastoplasticity with linear kinematic work hardening - Critère of Von Mises
5.2
Characteristics of the grid
2 nodes.
1 mesh SEG2
5.3
Functionalities tested
Commands
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 105
1.0000 105 0
normal effort NR
2
1.0250 105
1.0250 105 0
normal effort NR
3
9.7500 104
9.7500 104 0
normal effort NR
4
9.9000 104
9.9000 104 0
normal effort NR
5
1.0100 105
1.0100 105 0
normal effort NR
6
1.0300 105
1.0300 105 0
normal effort NR
7
4.7000 104
4.7000 104 0
Handbook of Validation
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Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 13/14
7 Modeling
C
7.1
Characteristics of modeling
The model is made up D `an element of bar (BARRE).
Law of behavior: model of Pinto-Menegotto without buckling (value of ELAN lower than 5).
7.2
Characteristics of the grid
2 nodes.
1 mesh SEG2
7.3
Functionalities tested
Commands
DEFI_MATERIAU PINTO_MENEGOTTO
ELAN
:
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 105
1.0000 105 0
normal effort NR
2
1.0470 105
1.0470 105 0
normal effort NR
3
6.0777 104
6.0777 104 0
normal effort NR
4
9.1430 104
9.1430 104 0
normal effort NR
5
7.6082 104
7.6082 104 0
normal effort NR
6
1.0125 105
1.0125 105 0
normal effort NR
7
3.7965 104
3.7965 104 0
Handbook of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/02/001/A
Code_Aster ®
Version
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Titrate:
SSNL112 - Subjected Barre has a cyclic thermal loading
Date:
09/04/02
Author (S):
C. CHAVANT
Key: V6.02.112-A Page: 14/14
9 Modeling
D
9.1
Characteristics of modeling
The model is made up D `1 element of bar (BARRE).
Law of behavior: model of Pinto-Menegotto with buckling (value of ELAN higher than 5).
9.2
Characteristics of the grid
2 nodes.
1 mesh SEG2
9.3
Functionalities tested
Commands
DEFI_MATERIAU PINTO_MENEGOTTO
ELAN
:
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 105
1.0000 105 0
normal effort NR
2
1.0470 105
1.0470 105 0
normal effort NR
3
6.0556 104
6.0556 104 0
normal effort NR
4
8.9078 105
8.9078 105 0
normal effort NR
5
7.6905 105
7.6905 105 0
normal effort NR
6
1.0125 105
1.0125 105 0
normal effort NR
7
3.8119 104
3.8119 104 0
11 Summary of the results
The results calculated by Code_Aster are in excellent agreement with the analytical solutions.
Handbook of Validation
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