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SSLP311 - Biblio_65. Fissure central oblique in a plate
Date:
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Author (S):
S. GRANET, I. CORMEAU, E. LECLERE
Key: V3.02.311-A Page: 1/8

Organization (S): EDF-R & D/AMA, CS IF
Handbook of Validation
V3.02 booklet: Linear statics of the plane systems
V3.02.311 document

SSLP311 - Biblio_65. Fissure central oblique in
a finished rectangular plate, with two materials,
subjected to uniform traction

Summary:

This test results from the validation independent of version 3 in breaking process.

It is about a two-dimensional test in statics with Bi-material in the presence of an internal fissure of interface
oblique.

The behavior of the structure (Bi-material) is elastic linear isotropic.

The case test includes/understands four modelings in plane constraints in which the influence of the slope of
the fissure is studied (4 cases).

Handbook of Validation
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SSLP311 - Biblio_65. Fissure central oblique in a plate
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Key: V3.02.311-A Page: 2/8


1
Problem of reference

1.1 Geometry


One considers 4 values of the angle = 15°, 30°, 45° and 60°.
Other dimensions are selected such as H = 2W = 4a.
The value of A is worth 1.E-3 Mr.

1.2
Properties of materials

Material n° 1

Rubber band, linear, isotropic, Young modulus E1 = 2e+12 Pa and Poisson's ratio 1 = 0,3.

Material n° 2

Rubber band, linear, isotropic, Young modulus E2 = 2e+11 Pa and Poisson's ratio 2 = 0,3.

1.3
Boundary conditions and loading

· The loading being autoéquilibré, one is satisfied to block the 3 rigid modes:

UX = UY = 0 with the left lower corner of the complete model.
UY = 0 with the corner lower right of the complete model.

· On the lower edge, we impose UY = 0

· Loading: uniform tension yy = 0 on the higher edge:

The value of 0 is worth 100MPa.
Handbook of Validation
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SSLP311 - Biblio_65. Fissure central oblique in a plate
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Key: V3.02.311-A Page: 3/8


2
Reference solution

2.1
Method of calculation used for the reference solution

Method of the elements of border, with quadratic elements [bib1].

The calculation of KI and KII is carried out by an integral of contour (integral M [bib2]) in which
the constraints and displacements calculated in the part intervene, as well as the constraints and
displacements deduced from analytically definite asymptotic solutions, in which KI and KII
are alternatively null. The calculation of K is also carried out by the method of virtual extension,
as comparison.

2.2
Results of reference

Method


On the left-hand side


On the right-hand side



= 15°
= 30°
= 45°
= 60°
= 15°
= 30°
= 45°
= 60°
integral FI
1,0115 0,7868 0,5211 0,2770 1,1266 0,9910 0,7646 0,4919
MR. FII - 0,4434
- 0,6244
- 0,6723
- 0,5804
0,0862 0,2961 0,4056 0,4057
extension FI
1,0110 0,7864 0,5210 0,2769 1,1260 0,9904 0,7643 0,4919
virtual FII - 0,4429
- 0,6240
- 0,6720
- 0,5801
0,0865 0,2960 0,4055 0,4056

K J
In this table one a: F =
J = I, II
J
0 A

The relation between the total rate of restitution of energy G and Kj is written as follows [bib3]:

3 -

I
=
I = 1 2
,
I
1 + I
µ
I.E.(internal excitation)
=
I
(21+i)
1

1
1 2
1
=
ln

+

+



2
µ1 µ2 µ2 µ1


1+
1
1
+

+
2
µ1
µ
=
2

CH
16
2 ()
G = (K2 + K2
I
II)

2.3
Uncertainty on the solution

Estimated at less than 0,1%.

2.4 References
bibliographical

[1]
Stress intensity Factor analysis off interface ace using boundary element method. Application
off contour-integral method. NR. MIYAZAKI, T. IKEDA, T.SODA and T. MUNAKATA.
Engng.Fract.Mechs., 45, n°5, 599-610, 1993.
[2]
Year analysis off interface aces between dissimilar isotropic materials using conservation
integrals in elasticity. J.F. YAU and T.C. CHANG. Engng.Fract.Mechs., 20, 423-432, 1984.
[3]
The strength off adhesive joints using the theory off aces. B. Mr. MALYSHEV and
R.L. SALGANIK. Int.J.Fract.Mech., 1, 114-128, 1965.
Handbook of Validation
V3.02 booklet: Linear statics of the plane systems
HT-66/02/001/A

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SSLP311 - Biblio_65. Fissure central oblique in a plate
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Key: V3.02.311-A Page: 4/8


3
Modelings A, B, C, D

3.1
Characteristics of modeling

Various modelings are identical to share the slope of the fissure.

Complete grid for an angle of 60 °

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Key: V3.02.311-A Page: 5/8

radius
center

Zoom on the point of fissure

The radius is worth 7.5E-5 Mr.

There are four crowns defined by command CALC_THETA:

crown 1: Rinf = 0.
Rsup = 1.875E-5m
crown 2: Rinf = 1.875E-5m
Rsup = 3.750E-5m
crown 3: Rinf = 3.750E-5m
Rsup = 5.625E-5m
crown 4: Rinf = 5.625E-5m
Rsup = 7.500E-5m

The direction of propagation is defined by: cos, sin

3.2
Characteristics of the grid

The grid consists of 10676 nodes and 4584 elements, including 1392 elements QUA8 and 3168
elements TRI6.

3.3 Functionalities
tested


Commands




AFFE_MODELE
MECANIQUE
C_PLAN
TOUT

MECA_STATIQUE

AFFE_CHAR_MECA
FORCE_CONTOUR

CALC_THETA
THETA_2D

CALC_G_THETA
OPTION
CALC_G


The calculation of KI and KII is not valid for a bimatériau.
Handbook of Validation
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HT-66/02/001/A

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Key: V3.02.311-A Page: 6/8


4
Results of modeling A

4.1 Values
tested

Identification Reference
Aster %
difference
Left end, = 15°



G, crown 1
9,67362E+1
9,2428E+1
- 4,45
G, crown 2
9,67362E+1
9,6392E+1
- 0,356
G, crown 3
9,67362E+1
9,6417E+1
- 0,330
G, crown 4
9,67362E+1
9,6421E+1
- 0,326
KI 5,6694E+6
-
-
KII
- 2,4852E+6
- -
Right end, = 15°



G, crown 1
1,0125E+2
9,6763E+1
- 4,33
G, crown 2
1,0125E+2
1,0093E+2
- 0,315
G, crown 3
1,0125E+2
1,0095E+2
- 0,295
G, crown 4
1,0125E+2
1,0095E+2
- 0,291
KI 6,3145E+6
-
-
KII 4,8309E+5
-
-

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Key: V3.02.311-A Page: 7/8


5
Results of modeling B

5.1 Values
tested

Identification Reference
Aster %
difference
Left end, = 30°



G, crown 1
8,0017E+1
7,6431E+1
- 4,48
G, crown 2
8,0017E+1
7,9707E+1
- 0,387
G, crown 3
8,0017E+1
7,9730E+1
- 0,358
G, crown 4
8,0017E+1
7,9734E+1
- 0,353
KI 4,4100E+6
-
-
KII
- 3,499E+6
- -
Right end, = 30°



G, crown 1
8,48417E+1
8,1080E+1
- 4,433
G, crown 2
8,48417E+1
8,4583E+1
- 0,305
G, crown 3
8,48417E+1
8,4602E+1
- 0,282
G, crown 4
8,48417E+1
8,4602E+1
- 0,282
KI 5,5545E+6
-
-
KII 1,6596E+6
-
-

6
Results of modeling C

6.1 Values
tested

Identification Reference
Aster %
difference
Left end, = 45°



G, crown 1
5,73826E+1
5,48161E+1
- 4,473
G, crown 2
5,73826E+1
5,71687E+1
- 0,373
G, crown 3
5,73826E+1
5,71865E+1
- 0,342
G, crown 4
5,73826E+1
5,7189E+1
- 0,337
KI 2,92076E+6
-
-
KII
- 3,7682E+6
- -
Right end, = 45°



G, crown 1
5,94122E+1
5,7039E+1
- 3,994
G, crown 2
5,94122E+1
5,9505E+1
0,157
G, crown 3
5,94122E+1
5,9516E+1
0,175
G, crown 4
5,94122E+1
5,9518E+1
0,179
KI 4,28557E+6
-
-
KII 2,27338E+6
-
-

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Date:
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Key: V3.02.311-A Page: 8/8


7
Results of modeling D

7.1 Values
tested

Identification Reference
Aster %
difference
Left end, = 60°



G, crown 1
3,28015E+1
3,10680E+1
- 5,285
G, crown 2
3,28015E+1
3,24037E+1
- 1,213
G, crown 3
3,28015E+1
3,24140E+1
- 1,181
G, crown 4
3,28015E+1
3,24156E+1
- 1,177
KI 1,55258E+6
-
-
KII
- 3,2531E+6
- -
Right end, = 60°



G, crown 1
3,22436E+1
3,11825E+1
- 3,291
G, crown 2
3,22436E+1
3,25321E+1
0,895
G, crown 3
3,22436E+1
3,25383E+1
0,914
G, crown 4
3,22436E+1
3,25398E+1
0,919
KI 2,75709E+6
-
-
KII 2,27394E+6
-
-

7.2 Remarks

To obtain G on the bottom of fissure, one calculates the rate of refund of energy using the relation
between G and Kj [bib3]:

=
= 2 076923
,
1
2
µ = 7 6923
,
E + 11
1
µ = 7 6923
,
E + 10
2
= - 9 37742
,
E - 2
= 2 524488
,
E - 12
G = (2
2
K + K
I
II)

8
Summary of the results

The calculation of G is not precise on the first crown in all the cases of slope of the fissure.
With regard to the other crowns, the variations are about 0,4%. In the case of slope
= 60° the variation exceeds 1%. As a whole the results are satisfactory for G.

The calculation of KI and KII is not available for a fissure located at the interface of a bimatériau.

Handbook of Validation
V3.02 booklet: Linear statics of the plane systems
HT-66/02/001/A

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