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SSNL125 Traction of a fragile bar: damage with Date gradient
:
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E. Key LORENTZ
:
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Organization (S): EDF-R & D/AMA
Handbook 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 zone and the zone
healthy) and the other infinite length (it extends to the border from the part).
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SSNL125 Traction of a fragile bar: damage with Date gradient
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1
Problem of reference
1.1 Geometry
= 0
= E
0
X E X
E
X
X
L = 5 mm
X = 0
L
10 mm
0
-
1 =
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) to gradient
of damage (modeling * _GRAD_VARI).
ELAS ECRO_LINE
NON_LOCAL
E = 20.000 Mpa
SY = 2 Mpa
NAKED = 0
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 blocked: the problem
is purely 1D. En outre, as the Poisson's ratio is null, no constraint of fastening
develops in these directions.
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SSNL125 Traction of a fragile bar: damage with Date gradient
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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 zones. The differential equation of
behavior is as follows in the zones where the criterion is reached, i.e. where
the damage evolves/moves:
2
2
D
y
1+
W = K (D) - C
where
K (D) = W
éq
2.1-1
2
^x
1+ - D
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^ indicates the variable of space. One
henceforth standardize the variables of the problem while introducing:
W
y
2
E =
X = x^ (1+)
W
= 1+ - éq
2.1-2
W y (1+)
has
D
2
C
With the help of these changes of variables, the equation of behavior is written:
2
1
D has
E (X) =
+ 2
éq 2.1-3
2
2
has
dx
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 zones of the bar (subscripted by I):
da 2
1
- - E has = C
I
I
I
{,
0}
1
e0 = 0 e1 = E
éq
2.1-4
dx
has
To these two constants of integration Ci 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:
da
da
da
da
(L) =
(L) = 0
+
-
+
-
0
1
has (0) - has (0) = 0
(0) - (0) = 0 éq
2.1-5
dx
dx
dx
dx
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SSNL125 Traction of a fragile bar: damage with Date gradient
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One substitutes as of now for the two constants of integration C the two extreme values of
I
field of damage has = has (L) and has = has (L) by evaluating the integral first [éq 2.1-4] in
0
0
1
1
L and L:
0
1
1
1
C0 = -
1
C = -
- E 1
has
éq
2.1-6
a0
1
has
2.2
Resolution of the problem in the discharged zone
In the zone with null deformation (I = 0), the integral first [éq 2.1-4] seems an equation
differential with separable variable. Taking into account [éq 2.1-6] and definition of a0, its integration
conduit initially with the following implicit equation:
has (X)
1
1 - 1 2
X = L0 - - D éq
2.2-1
a0
0
has
In particular, in X = 0, one obtain:
has (0)
1
1 - 1 2
L0 = - D éq
2.2-2
a0
a0
As it is about a clean integral in a0, the second member has a finished value. Moreover,
the intégrande being positive and has ()
0 undervalued by, the ultimate value of the field has (which corresponds to one
total damage D = 1), one observes that 0
L is limited by:
1 1 - 12
0
L - D éq
2.2-3
a0
0
has
Consequently, if the length of the zone 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 length 0
L is higher on the terminal [éq 2.2-3]. The boundary layer
develop at a finite distance B completely included in this zone. It is about a news
0
unknown factor, but the extreme value of damage A is now known: indeed, by
0
continuity with the not damaged zone which is spread out of 0
L with 0
B, one a:
= 1 has
éq
2.2-4
0
+
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Finally, the equation [éq 2.2-1] must be corrected; it is written:
has (X)
1
1 - 1 2
X = 0
B - -
D
éq
2.2-5
1 +
+
1
By expressing this equation in X again = 0, one express 0
B according to A ()
0:
has (0)
1
1 - 1 2
0
B = -
D
éq
2.2-6
1 +
+
1
Finally, in substituent [éq 2.2-6] in [éq 2.2-5], one obtains the following implicit equation:
has (X)
has (
1
- 2
X
1
1
+ -
+ -
X = -
-
D =
(1+) 3
(1
)
)
1
2
- arccos
éq
2.2-7
1 +
1 +
1
+ has (0)
has (0)
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 zone charged.
2.3
Resolution of the problem in the zone charged
In the zone with null deformation (I = 1), 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 U = has - 1
has):
has (X) - a1
1 2
a1 (a1 + U)
X = L1 -
éq
2.3-1
E has u2
2
1
+ (ea1 -)
1 U
0
Again, it is about a clean integral, except in the case
2
E has
1
1 =, i.e.
the damage corresponding to the stress E in a homogeneous problem. That means that
the extreme damage 1
A is all the more close to the homogeneous solution the bar is
long: the boundary layer in the zone charged is not limited and asymptotically extends towards
homogeneous answer.
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By expressing the equation [éq 2.3-1] in X = 0, one obtain an equation making it possible to determine 1
has in
function of A ()
0:
has (0) - a1
1 2
a1 (a1 + U)
L1 =
éq
2.3-2
E has u2
2
1
+ (ea1 -)
1 U
0
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 2
E 1
has
= 1
-
1
= E has
éq
2.3-3
As for the profile of damage in the charged zone, it is him also completely parameterized by
has ()
0, since while combining [éq 2.3-1], [éq 2.3-2] and [éq 2.3-3] one obtains:
has (0)
has (0)
1 2
1
has
1
has
X =
D
= 2 1
has -
1
arg tanh has
éq
2.3-4
- 1
has
has (X)
has (X)
2.4
Determination of the damage to the interface
Let us reconsider the step of integration. Initially, we awaited four constants of integration:
C0, 1
C and two resulting from the integration of the integrals first [éq 2.1-4]. Then, to both
constant C and C were substituted the two extreme values of damage has and
0
1
0
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 a0 and 1
has, to see the equations [éq 2.2-1] and
[éq 2.3-1]. Finally, one substituted for the constants a0 and 1
with the value of the damage with
the interface has ()
0, equalize on the left and on the right since the jump of A is null there. This substitution is
operated by noticing that the damaged zone is of size finished in the discharged zone,
contrary to the zone 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 has ()
0 thanks to the last unutilised condition, the nullity of the jump of derived with the interface.
Thus, by evaluating the two integrals first in X = 0 and calculating their difference, one obtain:
1
1
E has ()
0 = E has +
-
éq
2.4-1
1
has
has
1
0
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:
has (= 2
)
0
-
1
éq
2.4-2
E
E (1+)
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2.5 Application
numerical
For elasticity, work hardening and the internal length, one adopts the following characteristics:
E = 2.104 MPa
y
= 2 MPa
B
L = 26 mm éq 2.5-1
= 0
T
E = 2
- .104 MPa
These choices lead to the following parameters in the differential equation [éq 2.1-1]:
W y = 10-4 MPa
= 1
C = 8.10-4 NR éq
2.5-2
As for standardization, it becomes:
X = x^
= 2 - D has
W
- 4
= 4.10
E
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 D = 1 in a homogeneous context. That is translated
for the imposed deformation:
1
- 4
- 4
E 1
10
10
.
2
éq
2.5-4
4
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2.6
Results of reference
The reference solution is obtained by taking a bar length L = 5
0
- and L
10
1 =
. One
examine the value of the field of damage D for three levels of loading and of them two
places, one in the discharged zone, the other in the zone charged.
E
D (X =)
1
-
D (X =)
1
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 - Résultats of reference
1
E = 0.50
E = 0.75
0,8
E = 1.00
0,6
0,4
Damage D
0,2
0-5
0
5
10
position X (mm)
Appear 2.6-a: Profil of damage for the reference solution
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3 Modeling
With
3.1
Characteristics of modeling and the grid
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 grid 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. Grid 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
V (
1 X = -)
1 1.414.104 2.5100000000000E-02 2.5300980013184E-02
0.801%
V (
1 X = -)
1 1.414.104 3.4370000000000E-01 3.4334039567418E-01
- 0.105%
V (
1 X = -)
1 1.732.104 1.1060000000000E-01 1.1070787251591E-01 0.098
%
V (
1 X = -)
1 1.732.104 6.0450000000000E-01 6.0422953501431E-01 - 0.045
%
V (
1 X = -)
1 2. 104 1.8770000000000E-01
1.8805237130425E-01
0.188
%
V (
1 X = -)
1 2. 104 7.8970000000000E-01
7.8950184194123E-01
- 0.025
%
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5
Summary of the results
One notes a very good agreement between modeling and the analytical solution.
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