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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
1/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
Organization (S):
EDF-R & D/AMA
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
Document: V6.02.125
SSNL125 - Traction of a fragile bar:
damage with gradient
Summary:
This test allows the validation of the fragile law of damage gradient in a unidimensional situation
nonhomogeneous. From its character 1D, this problem admits an analytical solution which exhibe two modes
boundary layers: one finite length (existence of a free border enters the damaged area and the area
healthy) and the other infinite length (it extends to the border from the part).
Code_Aster
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
:
06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
2/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
1
Problem of reference
1.1 Geometry
The studied structure is a 15 mm length bar. The problem being purely 1D, its section is
without influence.
1.2
Properties of material
The material obeys a law of fragile elastic behavior (
ENDO_FRAGILE
) with gradient
of damage (modeling
* _GRAD_VARI
).
ELAS ECRO_LINE
NON_LOCAL
E
= 20.000 Mpa
NAKED
= 0
SY
= 2 Mpa
D_SIGM_EPSI
= - 20.000 MPa
LONG_CARA
= 5.099 mm
1.3
Conditions of loading
One forces the left part of the bar (5 mm length) to remain rigid (blocking of the degrees of
freedom of displacement). As for the right part of the bar, it is subjected to an axial deformation
uniform
0
, i.e. with an imposed displacement whose spatial distribution is linear. Only one
parameter thus controls the intensity of the loading: the level of imposed deformation
0
.
In the directions perpendicular to the axis of the bar, displacements are locked: the problem
is purely 1D. Moreover, as the Poisson's ratio is null, no stress of fastening
develops in these directions.
X
X
E
E
=
0
0
=
X
mm
10
1
=
L
mm
5
0
-
=
L
0
=
X
X
E
Code_Aster
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
3/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
2
Reference solution
In the case of the law of behavior to gradients of damage [R5.03.18], two equations with
derivative partial must be solved: the equilibrium equation and the equation of behavior. To obtain
an analytical solution proves generally delicate, even for unidimensional structures.
To validate nevertheless this model, one sticks to a simpler problem for which the equation
of balance need does not have to be solved, i.e. the field of displacement is fixed everywhere.
The equation of behavior is then controlled by the elastic deformation energy
W
known in all
not space.
2.1
Characterization of the solution
More precisely, one considers a bar of which a part is obligation to remain without deformation
while the other is subjected to a homogeneous deformation. One studies the boundary layer then
of damage which develops with the interface of these two areas. The differential equation of
behavior is as follows in the areas where the criterion is reached, i.e. where
the damage evolves/moves:
2
2
2
1
1
)
K (
^
)
K (
-
+
+
=
-
=
D
W
D
X
D
C
D
W
y
where
éq
2.1-1
where
y
W
,
and
C
are parameters of material, to see again [R5.03.18] for the correspondence
with the sizes provided in
DEFI_MATERIAU
, while
x^
indicate the variable of space. One
henceforth standardize the variables of the problem while introducing:
(
)
(
)
D
has
C
W
X
X
W
W
E
y
y
-
+
=
+
=
+
=
1
2
1
^
1
2
éq
2.1-2
With the help of these changes of variables, the equation of behavior is written:
2
2
2
2
1
)
(
dx
has
D
has
X
E
+
=
éq 2.1-3
There one recognizes an equation of motion in a gravitational field under an imposed force.
It admits an integral first in each of the two areas of the bar (subscripted by
I
):
{}
E
E
E
I
C
has
E
has
dx
da
I
I
=
=
=
-
-
1
0
2
0
1
,
0
1
éq
2.1-4
With these two constants of integration
I
C
come to be added two other constants resulting from
the integration of [éq 2.1-4]. These four constants are fixed by the two conditions of edge and them
conditions of jump to the interface:
() ()
() ()
0
0
0
0
0
0
0
)
(
)
(
1
0
=
-
=
-
=
=
-
+
-
+
dx
da
dx
da
has
has
L
dx
da
L
dx
da
éq
2.1-5
Code_Aster
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
4/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
One substitutes as of now for the two constants of integration
I
C
two extreme values of
field of damage
)
(
0
0
L
has
has
=
and
)
(
1
1
L
has
has
=
by evaluating the integral first [éq 2.1-4] in
0
L
and
1
L
:
1
1
1
0
0
1
1
has
E
has
C
has
C
-
-
=
-
=
éq
2.1-6
2.2
Resolution of the problem in the discharged area
In the area with null deformation (
0
=
I
), the integral first [éq 2.1-4] seems an equation
differential with separable variable. Taking into account [éq 2.1-6] and definition of
0
has
, its integration
conduit initially with the following implicit equation:
-
-
=
-
)
(
2
1
0
0
0
1
1
X
has
has
D
has
L
X
éq
2.2-1
In particular, in
0
=
X
, one obtains:
-
=
-
)
0
(
2
1
0
0
0
1
1
has
has
D
has
L
éq
2.2-2
As it is about a clean integral in
0
has
, the second member has a finished value. Moreover,
the intégrande being positive and
)
0
(
has
undervalued by
, the ultimate value of the field
has
(which corresponds to one
total damage
1
=
D
), it is observed that
0
L
is limited by:
-
-
0
2
1
0
0
1
1
has
D
has
L
éq
2.2-3
Consequently, if the length of the area noncharged is larger than this terminal, the solution
[éq 2.2-1] is not valid any more. That comes owing to the fact that it was supposed that the criterion was reached everywhere for
to write the equation [éq 2.1-1].
Henceforth, it is supposed that the length
0
L
is higher on the terminal [éq 2.2-3]. The boundary layer
develop at a finite distance
0
B
completely included in this area. It is about a news
unknown factor, but the extreme value of the damage
0
has
is now known: indeed, by
continuity with the not damaged area which is spread out
0
L
with
0
B
, one a:
+
=
1
0
has
éq
2.2-4
Code_Aster
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
5/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
Finally, the equation [éq 2.2-1] must be corrected; it is written:
+
-
-
=
+
-
)
(
1
2
1
0
1
1
1
X
has
D
B
X
éq
2.2-5
By expressing this equation again in
0
=
X
, one expresses
0
B
according to
)
0
(
has
:
+
-
=
+
-
)
0
(
1
2
1
0
1
1
1
has
D
B
éq
2.2-6
Finally, in substituent [éq 2.2-6] in [éq 2.2-5], one obtains the following implicit equation:
(
)
(
)
)
(
)
0
(
2
3
)
(
)
0
(
2
1
1
1
arccos
1
1
1
1
1
1
X
has
has
X
has
has
D
X
+
-
+
-
+
-
+
+
=
+
-
-
=
-
éq
2.2-7
Thus, it appears that the profile of damage is completely controlled by its value into 0, i.e.,
taking into account the condition of continuity [éq 2.1-5], by what occurs in the area charged.
2.3
Resolution of the problem in the area charged
In the area with null deformation (
1
=
I
), the integral first [éq 2.1-4] also seems
a differential equation with separable variable. Taking into account [éq 2.1-6] and definition of
1
has
,
its integration leads initially to the following implicit equation (with the help of
change of variable
1
has
has
U
-
=
):
(
)
(
)
U
has
E
U
has
E
U
has
has
L
X
has
X
has
-
+
+
-
=
-
1
)
(
0
2
1
2
1
2
1
1
1
1
1
éq
2.3-1
Again, it is about a clean integral, except in the case
1
2
1
=
has
E
, i.e.
the damage corresponding to the stress
E
in a homogeneous problem. That means that
the extreme damage
1
has
is all the more close to the homogeneous solution the bar is
long: the boundary layer in the area charged is not limited and asymptotically extends towards
homogeneous answer.
Code_Aster
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
:
06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
6/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
By expressing the equation [éq 2.3-1] in
0
=
X
, one obtains an equation allowing to determine
1
has
in
function of
)
0
(
has
:
(
)
(
)
U
has
E
U
has
E
U
has
has
L
has
has
-
+
+
=
-
1
)
0
(
0
2
1
2
1
2
1
1
1
1
1
éq
2.3-2
However, to simplify the analytical resolution of this problem, it is supposed henceforth that the bar
is sufficiently long, so that an approximate solution of [éq 2.3-2] is given by:
2
1
1
2
1
1
-
=
=
E
has
has
E
éq
2.3-3
As for the profile of damage in the charged area, it is him also completely parameterized by
)
0
(
has
, since while combining [éq 2.3-1], [éq 2.3-2] and [éq 2.3-3] one obtains:
)
0
(
)
(
1
1
1
)
0
(
)
(
1
2
1
1
tanh
arg
2
has
X
has
has
X
has
has
has
has
D
has
has
X
-
=
-
=
éq
2.3-4
2.4
Determination of the damage to the interface
Let us reconsider the step of integration. Initially, we awaited four constants of integration:
0
C
,
1
C
and the two resulting one from the integration of the integrals first [éq 2.1-4]. Then, to both
constants
0
C
and
1
C
were substituted the two extreme values of damage
0
has
and
1
has
, they
also unknown. By exploiting the boundary conditions of Neumann in [éq 2.1-6], one has
implicitly given two complementary constants of integration not to express the profiles
of damage that according to the only values
0
has
and
1
has
, to see the equations [éq 2.2-1] and
[éq 2.3-1]. Finally, one substituted for the constants
0
has
and
1
has
the value of the damage with
the interface
)
0
(
has
, equalizes on the left and on the right since the jump of
has
y
is null. This substitution is
operated by noticing that the damaged area is of size finished in the discharged area,
contrary to the area charged where we privileged an approached solution, simpler on
analytical plan.
Consequently, there remains nothing any more but one constant of integration to be determined, the damage with
the interface
)
0
(
has
thanks to the last unutilised condition, the nullity of the jump of derived with the interface.
Thus, by evaluating the two integrals first in
0
=
X
and by calculating their difference, one obtains:
0
1
1
1
1
)
0
(
has
has
has
E
has
E
-
+
=
éq
2.4-1
By taking account of the expressions [éq 2.2-4] and [éq 2.3-3], one deduces the expression from it from
the damage with the interface:
()
+
-
=
1
1
2
)
0
(
E
E
has
éq
2.4-2
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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V6.02.125-A
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7/10
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V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
2.5 Application
numerical
For elasticity, work hardening and the internal length, one adopts the following characteristics:
mm
26
MPa
10
.
2
MPa
2
0
MPa
10
.
2
4
4
=
-
=
=
=
=
B
T
y
L
E
E
éq
2.5-1
These choices lead to the following parameters in the differential equation [éq 2.1-1]:
NR
10
.
8
1
MPa
10
4
4
-
-
=
=
=
C
W
y
éq
2.5-2
As for standardization, it becomes:
E
W
D
has
X
X
4
10
.
4
2
^
-
=
-
=
=
éq
2.5-3
The load evolves/moves between the value of initiation of the damage and that for which
the damage would reach its maximum value
1
=
D
in a homogeneous context. That is translated
for the imposed deformation:
4
4
10
.
2
10
1
4
1
-
-
E
éq
2.5-4
Code_Aster
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Version
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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E. LORENTZ
Key
:
V6.02.125-A
Page:
8/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
2.6
Results of reference
The reference solution is obtained by taking a bar length
5
0
-
=
L
and
10
1
=
L
. One
examine the value of the field of damage
D
for three levels of loading and in two
places, one in the discharged area, the other in the area charged.
E
(
)
1
-
=
X
D
(
)
1
=
X
D
1.414 10
- 4
0.50 0.0251 0.3437
1.732 10
- 4
0.75 0.1106 0.6045
2.000 10
- 4
1.00 0.1877 0.7897
Count 2.6-1 - Results of reference
- 5
0
5
10
position X (mm)
0
0,2
0,4
0,6
0,8
1
Damage D
E = 0.50
E = 0.75
E = 1.00
Appear 2.6-a: Profile of damage for the reference solution
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Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
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V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
3 Modeling
With
3.1
Characteristics of modeling and the mesh
It is about an axisymmetric modeling (
AXIS_GRAD_VARI
). Geometry
corresponding is a rectangle, i.e. the bar is laid out of manner
vertical and its section (without influence) are circular.
The mesh consists of only one element according to the radius. According to the axis, smallest
elements have a size of 0.1 mm along the interface and grow in progression
geometrical of reason 1.05 while moving away from the interface. Mesh thus generated
finally consists of 59 quadrangular elements with 8 nodes.
4
Results of modeling A
4.1
Sizes tested and results
One validates the modeling and the algorithm of integration of nonlocal laws by examining the level
of damage (variable V1 intern) on the various levels of loading and the various places
geometrical listed in [Table 2.6-1]. The results are joined together in the extract of the file of result
below.
Identification Moment
Reference
Aster
Difference
(
)
1
1
-
=
X
V
1.414.10
4
2.5100000000000E-02
2.5300980013184E-02
0.801%
(
)
1
1
-
=
X
V
1.414.10
4
3.4370000000000E-01
3.4334039567418E-01
- 0.105%
(
)
1
1
-
=
X
V
1.732.10
4
1.1060000000000E-01
1.1070787251591E-01
0.098
%
(
)
1
1
-
=
X
V
1.732.10
4
6.0450000000000E-01
6.0422953501431E-01 - 0.045
%
(
)
1
1
-
=
X
V
2. 10
4
1.8770000000000E-01
1.8805237130425E-01
0.188
%
(
)
1
1
-
=
X
V
2. 10
4
7.8970000000000E-01
7.8950184194123E-01
- 0.025
%
Code_Aster
®
Version
6.4
Titrate:
SSNL125 Traction of a fragile bar: damage with gradient
Date
:
06/10/03
Author (S):
E. LORENTZ
Key
:
V6.02.125-A
Page:
10/10
Manual of Validation
V6.02 booklet: Nonlinear statics of the linear structures
HT-66/03/008/A
5
Summary of the results
One notes a very good agreement between modeling and the analytical solution.