Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
1/10
Organization (S): EDF/IMA/MN
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
V6.04.121 document
SSNV121 - Rotation and traction hyper-rubber band
of a bar
Summary:
This test of quasi-static mechanics consists in making turn of 90° a parallelepipedic bar, with
to subject to an important traction for finally letting it return in a discharged state. One validates thus
kinematics of the great deformations hyper-rubber bands (command STAT_NON_LINE [U4.32.01], key word
COMP_ELAS), and thus in particular great rotations, for a relation of elastic behavior linear.
The bar is modelled by a voluminal element (HEXA8, modeling A) or plan (QUAD4, assumption
plane deformations, modeling B).
The results obtained by Code_Aster do not differ from the theoretical solution.
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
2/10
1
Problem of reference
1.1 Geometry
y
1.000 (mm)
1
2
3
4
X
1.000 (mm)
1.2
Material properties
Behavior hyper-rubber band of St Venant - Kirchhoff:
E
E
E = 200.000.MPa
S
= (
tr
+
1 +) (1 -
2)
(E) 1
E
1 +
= 0 3
.
1.3
Boundary conditions and loadings
The loading is applied in two times: first of all, an overall rotation of the structure,
followed by a traction in the new configuration:
Overall rotation (0 < T < 1 S)
Traction (1 S < T < 2 S)
T
2 '
4 '
1
2
2 '
4 '
3
4
1 '
1 '
3
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
3/10
2
Reference solution
2.1
Method of calculation used for the reference solution
It is about a plane problem. One can seek the solution in the form of a rigid rotation and one
lengthening of a factor in direction Y.
- X - Y
U (X, Y, Z) =
(1+) X - Y
0
The gradient of the transformation and the deformation of Green-Lagrange are then:
0
- 1
0
E 0
0
(+ 2)
F
=
1 +
0
0 E
= 0 0
0
with E =
2
0
0
1
0 0 0
The relation of behavior leads then to a tensor of Lagrangian constraints diagonal:
(1) E
S
=
xx
(1+) (1 - 2) E
E
S
= S
=
yy
zz
(1+) (1 - 2) E
The boundary condition of the equilibrium equation then enables us to determine the value of
lengthening:
(1)
+1 +
= (
2
FS)
= (1+)
E
(
) (
)
T
S
(
=
1 +) (1 - 2)
T
yx
xx
2
The constraint of Cauchy is given by:
S yy
1
xx
=
zz
=
=
F S F T
1 +
Det F
yy
= (1+)
S xx
Lastly, the force exerted on the faces:
· [2,4]: F
=
=
y
yy
[
S 2,4]
yy So [2,4]
· [4,3]: F
=
= 1+
X
xx
[
S 4,] 3
xx (
) So [4,] 3
· [1,2,3,4]: F
=
= 1+
Z
zz
[
S 1,2,3,4]
zz (
) So [1,2,3,4]
where S [
O]
initial surfaces of the faces represent.
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
4/10
2.2
Results of reference
One adopts like results of reference displacements, the constraint of Cauchy and the force
exerted on the faces [2,4] and [4,3].
At time T = 2 S:
One seeks T such as lengthening
= 01.
==> T = 31 096.154 MPa.
The constraint of Cauchy is then:
=
=
xx
zz
11 013.986 MPa
=
yy
31 096.154 MPa
The exerted forces are:
F
= 12.115.385
.
×
X
So [4,] NR
3
F
= 31 096.154 ×
y
So [2,4] NR
F
= 12 115.385 ×
Z
So [, 12, 34] NR
At time T = 3 S:
The bar returned in its initial state:
= 0
= 0
F =
0
2.3
Uncertainty on the solution
Analytical solution.
2.4 References
bibliographical
[1]
Eric LORENTZ “Une nonlinear relation of behavior hyperelastic” internal Note
EDF/DER HI-74/95/011/0
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
5/10
Intentionally white left page.
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
6/10
3 Modeling
With
3.1
Characteristics of modeling
Voluminal modeling:
1 mesh HEXA 8
1 mesh QUAD4
Z
5
6
y
7
8
1
2
1.000 (mm)
3
4
X
Boundary conditions:
R T
()
(3,7): DX = 0
DY = 0
1
(1,5): DX = - 1.000 R (T)
DY = - 1.000 R (T)
(2,6): DX = - 2.000 R (T)
(4,8): DX = - 1.000 R (T)
0
1
2
3
T
Loading: Traction on the face [2,4,8,6]
net [2,4,8,6] (QUAD4): FY = 31 096.154 F (T) MPa
F T
()
1
0
1
2
3
T
3.2
Characteristics of the grid
A number of nodes: 8
A number of meshs: 2
1 HEXA8
1 QUAD4
3.3 Functionalities
tested
STAT_NON_LINE
COMP_ELAS
DEFORMATION: “GREEN”
CALC_NO
OPTION: “FORC_NODA”
GEOMETRIE: “DEFORMEE”
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
7/10
4
Results of modeling A
4.1 Values
tested
Identification
Reference
Aster
% difference
T = 2 Déplacement DX (NO2)
100
100
0
T = 2 Déplacement DY (NO4)
1100
1100
0
T = 2 Contraintes SIGXX (PG1)
11013.986
11013.986
0
T = 2 Contraintes SIGYY (PG1)
31096.154
31096.154
0
T = 2 Contraintes SIGZZ (PG1)
11013.986
11013.986
0
T = 2 Contraintes SIGXY (PG1)
0
109
/
T = 2 Contraintes SIGXZ (PG1)
0
1010
/
T = 2 Contraintes SIGYZ (PG1)
0
1010
/
T = 3 Déplacement DX10 (NO2)
0
1011
/
T = 3 Déplacement DY (NO4)
0
1012
/
T = 3 Contraintes SIGXX (PG1)
0
109
/
T = 3 Contraintes SIGYY (PG1)
0
1010
/
T = 3 Contraintes SIGZZ (PG1)
0
109
/
T = 3 Contraintes SIGXY (PG1)
0
1011
/
T = 3 Contraintes SIGXZ (PG1)
0
1010
/
T = 3 Contraintes SIGYZ (PG1)
0
1011
/
T = 2 Force nodal DX (NO8)
3.0289 109
3.0288 109
0.002%
T = 2 Force nodal DY (NO8)
7.774 109
7.774 109
0
T = 2 Force nodal DZ (NO8)
3.0289 109
3.0288 109
0.002%
4.2 Remarks
Calculation of the nodal force:
The force applied F to a face described by a linear mesh is distributed by:
1/4
1/4
F
= 1
node
F
4
1/4
1/4
4.3 Parameters
of execution
Version: 3.03.30
Machine: CRAY C90
Obstruction memory:
8 MW
Time CPU To use:
62.3 seconds
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
8/10
5 Modeling
B
5.1
Characteristics of modeling
Modeling 2D plane deformations
y
1
2
3
4
X
Boundary conditions:
R T
()
3:
DX = 0
DY = 0
1
1:
DX = - 1.000 R (T)
DY = - 1.000 R (T)
2:
DX = - 2.000 R (T)
4:
DX = - 1.000 R (T)
0
1
2
3
T
Loading:
Traction on the face [2,4]
net [2,4]: FY = 31 096.154 F (T) MPa
F T
()
1
0
1
2
3
T
5.2
Characteristics of the grid
A number of nodes: 4
A number of meshs: 2
1 QUAD4
1 SEG2
5.3 Functionalities
tested
STAT_NON_LINE
COMP_ELAS
DEFORMATION: “GREEN”
CALC_NO
OPTION: “FORC_NODA”
GEOMETRIE: “DEFORMEE”
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
9/10
6
Results of modeling B
6.1 Values
tested
Identification
Reference
Aster
% difference
T = 2 Déplacement DX (NO2)
100
100
0
T = 2 Déplacement DY (NO4)
1100
1100
0
T = 2 Contraintes SIGXX (PG1)
11013.986
11013.986
0
T = 2 Contraintes SIGYY (PG1)
31096.154
31096.154
0
T = 2 Contraintes SIGZZ (PG1)
11013.986
11013.986
0
T = 2 Contraintes SIGXY (PG1)
0
10-10
/
T = 3 Déplacement DX (NO2)
0
1012
/
T = 3 Déplacement DY (NO4)
0
1012
/
T = 3 Contraintes SIGXX (PG1)
0
1010
/
T = 3 Contraintes SIGYY (PG1)
0
1010
/
T = 3 Contraintes SIGZZ (PG1)
0
1010
/
T = 3 Contraintes SIGXY (PG1)
0
1010
/
T = 2 Force nodal DX (NO4)
6.0577 106
6.0577 106
0
T = 2 Force nodal DY (NO4)
15.5481 106
15.5481 106
0
6.2 Remarks
Calculation of the nodal force:
The force applied F to a face described by a linear mesh is distributed by:
1/2
1/2
F
= 1
node
F
2
6.3 Parameters
of execution
Version: NEW 3.03.30
Machine: CRAY C90
Obstruction memory:
8 MW
Time CPU To use:
48.3 seconds
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A
Code_Aster ®
Version
3
Titrate:
SSNV121 Rotation and traction hyper-rubber band of a bar
Date:
23/07/99
Author (S):
E. LORENTZ
Key:
V6.04.121-A Page:
10/10
7
Summary of the results
It appears at the end of this test that the numerical solution coincides remarkably with the solution
analytical. It will be noticed however that the strong not linearity due to great rotations requires
a relatively fine discretization in time, without being penalizing on the precision since,
contrary to an incremental law of behavior, the errors do not cumulate a step of
time on the other.
Handbook of Validation
V6.04 booklet: Nonlinear statics of the voluminal structures
HI-75/96/044 - Ind A