Weibull PIA
Weibull KTA mod
Weibull CEA mod
ASME Failure Assessment Validation Exercise
 VP-01 Bending to Tensile
x
x
x
 VP-02 Rectangular Notched Tensile Specimen
x
x
x
x
x
Yves Lejeail
 VP-06 L-shaped Specimen
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Hypotheses (1)
 1  s m 
Ps  exp  
  s 0  dv 
 V0 Vtot


 Weibull uniaxial
 Weibull PIA:
replace sm above by
s éqm  s Im  s IIm  s IIIm
 Weibull KTA mod :
replace s above by s
if si>0 s  s
S
s s
if si<0
S
eqMSE
i
i
i
i
 s 12  s 22  s 32  2 (s 1s 2  s 2s 3  s 1s 3 )
t
c
 Weibull CEA mod
mS
 1


 s 
 s

 * exp   1


Ps( structure)  exp 

dv
V 
S



 Vos Vtot  s 0 S 

 Vov Vtot  s 0V



v + s =
1
07-09, February 2007
Lyon, France
mV


 dv 



ASME Failure Assessment Validation Exercise
Hypotheses (2)
 No Volume effect taken into account
V0=Vtot (1 type of defect)
V0s=V0v=Vtot (2 types of defect)
Why ?
- No data
- Time constraint
- No large effects in experiments (theory seems
to overestimate this effect)
But…
Interest in case of several type of defects (more
control on the analyses)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Hypotheses (3)
 Further with the Weibull CEA mod
• Surface defects and volume defects are in two
separate regions

 s éq
1

Ps  exp  
 VS  Vv V  s 0 S
S


 s éq
1
 dv 


V

V
S
v VV  s 0V

mS
mV


 dv 



• These defects come from fabrication including
machining (heat, forces,…)
• Parametric studies to look at the influence of the
parameters (no experimental data)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
VP-01 Bending to tensile
Classical formulas from Weibull uniaxial (assuming
s=0 if s<0)
Maximum bending stress must be less than:
Sb = (2(m+1))1/mSm
Sb3 pt  (m  1)
Sb 4 pt


2 1/ m
Sb  Sm 2(m  1)
1/ m
2 1/ m
 2(m  1)  
 2(m  1) 

 Sb  S m 

 m2 
 m2 
1/ m
Equal survival prob. between tensile and bending
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
flexion 3 points et 4 points
Bending
material limits from Weibull
4,0
Sb/Sm
Sf/St
flexion
3pts 3 pts
flexion
4pts 4 pts
3,0
pure pure
flexion
2,0
1,0
0
5
10
m
15
20
25
• Note that it is possible to apply a more general rule
following Weibull initial equation and the Loading factor
Lmax 
1
Vtot

Vto t
 s éq

s
 éq max
07-09, February 2007
m

 dV


Lyon, France
ASME Failure Assessment Validation Exercise
Two possible approaches for design, with the same Weibull initial equation
1,20
1,00
E1/ Peq is a result of a projection on
the characterization curve and can be
compared to Sm
Characterization
Specimen
Structure
1/ m
PR
0,80
1

Peq    s eqm dV 
 V0 V

0,60
s T  s eq max
V 
 
 V0 
1/ m
Lmax 1 / m < Sm
E2/ The maximum local Stress
is compared to a modified Sm
0,40
0,20
1/ m
Sm
Allowable Stress
of the Material s
eq max
0,00
0,0000 0,2000 0,4000 0,6000 0,8000 1,0000 1,2000 1,4000 1,6000
s/s 0
07-09, February 2007
1/ m
V 
V 
Sm  0 
sT  0 
V 
V  

Lmax 1 / m
Lmax 1 / m
Lyon, France
ASME Failure Assessment Validation Exercise
VP-01 Weibull CEA mod
Ls
H
• ms, mv, s0s, s0v imposed  build the Weibull curve
of the specimen
• determine the effect of Ls/H , and compare
between tension and (pure) bending
• definition of apparent Weibull parameters
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Matrix of the study
• ms=50 , mv=5, sos = 35 MPa
sov = 10 MPa
sov = 25 MPa
sov = 33 MPa
• ms=5, mv=50, sos = 35 MPa
sov = 10 MPa
sov = 25 MPa
sov = 33 MPa
• ms=10, mv=20, sos = 33 MPa, sov = 35 MPa
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
ms=50 , mv=5, sos = 35 MPa, sov = 10 MPa
Pure
bending
Flexion
pure
mv = 5
Sos = 35 MPa
1,2
1
Ls = 0 mm
0,8
Ls = 0,5 mm
Ls = 1 mm
0,6
Ls = 1,5 mm
Ls = 2 mm
0,4
Ls = 2.46 mm
Lv = 0.02 micron
0,2
Ls = 2,5 mm
0
0
10
20
-0,2
30
40
the structure
probability
Survivalprobabilité
de survieof
de l'éprouvette
structure
probability
Survival
Probabilité
de survieof
de the
l'éprouvette
Sos = 35 Mpa
tension
Traction
simple
Sov = 10 Mpa ms = 50
contrainte (MPa)
Stress
(MPa)
Sov = 10 MPa
ms = 50
1
0,9
0,8
0,7
Ls = 0 mm
Ls = 0,5 mm
0,6
Ls = 1 mm
0,5
Ls = 1,5 mm
Ls = 1,8 mm
0,4
Ls = 2 mm
Ls = 2,5 mm
0,3
0,2
0,1
0
5
10
15
20
25
30
contrainte (MPa)
Stress (MPa)
07-09, February 2007
mv = 5
Lyon, France
35
40
45
ASME Failure Assessment Validation Exercise
ms=50 , mv=5, sos = 35 MPa, sov = 10 MPa
60
50
40
ms
mv
m apparent
30
20
10
modulus
Weibull
module de Weibull
Weibull modulus
modules de Weibull
50
40
ms
30
mv
m app
20
10
0
0
0
0,2
0,4
0,6
0,8
1
0
1,2
0,2
0,4
0,8
1
1,2
45
35
30
Sos
Sov
So apparent
25
20
15
cointrainte de Weibull
40
Weibull stress (MPa)
45
stress
Weibull de
contraintes
Weibull (MPa)
0,6
Ls/H
2Ls/H
Ls/H
2Ls/H
40
35
30
Sos
Sov
25
So app
20
10
15
5
10
0
0
0,2
0,4
0,6
0,8
1
1,2
0
Ls/H
2Ls/H
0,2
0,4
0,6
0,8
Ls/H
2Ls/H
07-09, February 2007
Lyon, France
1
1,2
ASME Failure Assessment Validation Exercise
ms=5 , mv=50, sos = 35 MPa, sov = 25 MPa
Pure Flexion
bending
pure
Sos = 35 MPa
1
0,9
0,8
Ls = 0 mm
Ls = 0,06 mm
Ls = 0,5 mm
Ls = 1 mm
Ls = 1,5 mm
Ls = 2 mm
Ls = 2,4 mm
Ls = 2.4998
Ls = 2,5 mm
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
5
15
25
35
45
55
probability of the structure
Survival
probabilité de survie de l'éprouvette
probability of the structure
Survival
Probabilité de survie de l'éprouvette
tension
Traction
Simple
Sos = 35 MPa Sov = 25 MPa ms = 5 mv = 50
contrainte
(MPa)
Stress
(MPa)
Sov = 25 MPa
1
0,9
0,8
0,7
Ls = 0 mm
Ls = 0.3 mm
0,6
Ls = 0,5 mm
0,5
Ls = 1 mm
Ls = 1,5 mm
0,4
Ls = 2 mm
Ls = 2,5 mm
0,3
0,2
0,1
0
20
30
40
50
60
Stress (MPa)
contrainte (MPa)
07-09, February 2007
ms = 5 mv = 50
Lyon, France
70
80
ASME Failure Assessment Validation Exercise
ms=5 , mv=50, sos = 35 MPa, sov = 25 MPa
60
55
45
40
ms
30
mv
mapparent
20
10
modulus
Weibull
module de Weibull
Weibull modulus
modules de Weibull
50
0
0,2
0,4
2Ls/H
Ls/H
0,6
0,8
mv
m app
15
-5 0
1
36
0,2
0,4
0,6
2Ls/H
Ls/H
0,8
1
1,2
34
32
Sos
30
Sov
So apparent
28
26
Weibull stress (MPa)
60
contrainte de Weibull (MPa)
Weibull stress (MPa)
ms
25
5
0
contraintes de Weibull
35
55
50
45
Sos
40
Sov
So app
35
30
25
20
24
0
0,2
0,4
0,6
2Ls/H
0,8
1
0
Ls/H
07-09, February 2007
0,2
0,4
0,6
Ls/H
2Ls/H
0,8
Lyon, France
1
1,2
ASME Failure Assessment Validation Exercise
VP-01 Weibull CEA mod (conclusions)
Ls
H
• the apparent Weibull parameters may vary with
the surface thickness layer
• as expected, this variation is favoured in case of
bending
• if thickness of layer depends on machining
Parameters  size effect again even if volume
effect is neglected
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
VP-02
Rectangular Notched Tensile Specimen
Some preliminary remarks on Schmidt’s document
• No material indicated for this case, from where comes the
Weibull law used to interpret the experiments?
To validate a design approach it is necessary to take the
results of characterization to interpret component’s tests
• How many tests performed to determine the mean load to
rupture?
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Description of the model
P
Other details
r
• 2D plane stress hypothesis
• Total load = 11kN  5.5 kN
here
• E = 6500 MPa
  = 0.15
• post-treatment for Weibull
analysis  build the whole
curve(s)
• Ls=0.3 mm and Ls=5 mm
View of a typical
mesh (9000 elements
for Ls=5mm)
Loading conditions
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Mean experimental Values of Fracture Loads (N)
Survival
probability
radius
Probabilités
de survie for
pourdifferent
les différents
rayons d'entaille
m=12, s0=17.7MPa
100,00%
90,00%
80,00%
Ps
r=20mm
Probabilité
70,00%
R = 20 mm
R = 50 mm
R = 75 mm
R = 100 mm
R = 120 mm
R = 150 mm
Traction Simple
60,00%
50,00%
40,00%
30,00%
20,00%
10,00%
Von Mises stress
0,00%
7000
9000
11000
13000
Charge (N)
Load(N)
07-09, February 2007
Lyon, France
15000
ASME Failure Assessment Validation Exercise
m=12, s0=17.7MPa
Survival Probability R = 20 mm
100,00%
Surface
Volume
Total
90,00%
80,00%
70,00%
50,00%
40,00%
30,00%
20,00%
10,00%
0,00%
0
2000
4000
6000
8000
10000
12000
14000
Load (N)
Ls=5mm
90,00%
80,00%
70,00%
60,00%
50,00%
40,00%
30,00%
20,00%
10,00%
0,00%
10,00
15,00
80,00%
70,00%
60,00%
50,00%
40,00%
30,00%
10,00%
0,00%
6000
Surface
Volume
Total
7000
8000
9000
10000
11000
12000
25,00
30,00
R = 150 mm
90,00%
20,00%
20,00
Stress (MPa)
Contrainte
(MPa)
Survival Probability R = 150 mm
100,00%
Proba
100,00%
13000
14000
Load (N)
07-09, February 2007
the structure
Survival probability
Probabilitéof
de survie
Proba
60,00%
Probabilité of
de survie
the structure
Survival probability
R = 20 mm
35,00
40,00
Ls=0.3mm
100,0%
90,0%
80,0%
70,0%
60,0%
50,0%
40,0%
30,0%
20,0%
10,0%
0,0%
10,00
15,00
20,00
25,00
Stress (MPa)
Contrainte (MPa)
Lyon, France
30,00
35,00
40,00
ASME Failure Assessment Validation Exercise
As we have obtained the same results with PIA and MSE
for this study (as in others), we will apply MSE in all the
others calculations
As a reminder, for an homogeneous stress state :
(m=15)
Tube with a
temperature
gradient in
the thickness
Tube with
internal
pressure
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Application of the Weibull CEA mod
Calculation Cases for r=20 mm and r=150 mm, Ls = 0.3mm
Sos
Sov
17,7
35
35
35
35
33
33
33
33
25
25
25
25
10
10
10
10
ms
17,7
33
33
33
33
35
35
35
35
35
35
35
35
35
35
35
35
mv
12
5
10
20
50
5
10
20
50
5
10
20
50
5
10
20
50
12
50
20
10
5
50
20
10
5
50
20
10
5
50
20
10
5
This is a sensitivity study, nothing more
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Survival probability R = 20 mm, Ls = 0,3 mm
Sos = 25 MPa Sov = 35 MPa ms = 20 mv = 10
Survival probability R = 20 mm, Ls = 0,3 mm
Sos = 35 MPa Sov = 33 MPa ms = 50 mv = 5
100,00%
100,00%
90,00%
90,00%
80,00%
80,00%
70,00%
70,00%
60,00%
Proba
Proba
60,00%
50,00%
40,00%
50,00%
40,00%
30,00%
30,00%
Surface
Volume
20,00%
Total
Tensile
10,00%
Volume
Total
10,00%
Tensile
Bending
Bending
0,00%
0,00
10,00
20,00
30,00
40,00
0,00%
0,00
50,00
Surface
20,00%
5,00
10,00
15,00
20,00
Stress (MPa)
Survival probability R = 20 mm, Ls = 0,3 mm
Sos = 35 MPa Sov = 33 MPa ms = 20 mv = 10
100,00%
90,00%
80,00%
80,00%
70,00%
70,00%
40,00%
45,00
50,00
50,00%
40,00%
Surface
30,00%
Volume
Total
20,00%
Tensile
10,00%
Bending
0,00%
0,00
40,00
60,00%
50,00%
Proba
Proba
60,00%
10,00%
35,00
Survival probability R = 20 mm, Ls = 0,3 mm
90,00%
20,00%
30,00
Sos = 10 MPa Sov = 35 MPa ms = 10 mv = 20
100,00%
30,00%
25,00
Stress (MPa)
10,00
20,00 30,00 40,00 50,00
60,00 70,00 80,00
Stress (MPa)
07-09, February 2007
0,00%
0,00
Surface
Volume
Total
Tensile
Bending
5,00
10,00
15,00
20,00
Stress (MPa)
Lyon, France
25,00
30,00
35,00
40,00
ASME Failure Assessment Validation Exercise
An example of a curve for r=150 mm, Ls = 0.3mm
Comparison of the apparent curves
Survival Probability
Sos = 33 MPa
100%
90%
80%
70%
60%
50%
40%
30%
20%
10%
0%
0,00
Sov = 35 Mpa
ms = 50
mv = 5, Ls=0.3 mm
R = 20 mm
R = 150 mm
Tensile
Pure Bending
10,00
20,00
30,00
40,00
Stress (MPa)
50,00
The influence of the surface defects is less important
but the r=150 mm results are significantly different from
pure tensile specimens
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
VP-02 Weibull and Weibull CEA mod (conclusions)
• See the preliminary remarks about the Weibull
parameters and what is necessary to validate a
design rule
• Again, it may be possible to explain the different
apparent Weibull parameters for different
structures with (at least) two populations of
defects
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
VP-06 Weibull and Weibull CEA mod
L
L
L
L
L
L
•Study of the 4 points bending
•Study of the L shaped specimens, with Ls=0.2 mm
and Ls=2 mm
• No PIA
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
Study of the 4 points bending
•Loadings
•A special methodology to represent the support
400N
A
B1
B2
M1
SsH1
Ai
SsH2
Ci
M2
Sv1
Fi
Sv2
M3
SsB1
F
C
Di
SsB2
M4
07-09, February 2007
E
D
Lyon, France
ASME Failure Assessment Validation Exercise
mesh
Von Mises Equivalent Stresses
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
For Ls=0.2 mm:
ms=15 , mv=8, sos = 23.7 MPa, sov = 23 MPa
ms=8 , mv=15, sos = 19.8 MPa, sov = 24.6 MPa
For Ls=2 mm:
ms=15 , mv=8, sos = 25.3MPa, sov = 16.6 MPa
ms=8 , mv=15, sos = 23.6 MPa, sov = 18 MPa
Comparison of calculations with experiments (4points
loadings)
1
0,9
Survival probability
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
400
Psurvie
expérimentale
Weibull2
(ms=8 ;
mv=15)
Weibull2
(ms=15 ;
mv=8)
Weibull1
(m=12)
Weibull1
(m=15)
600
800
1000
Load (N)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
A
Bi
L specimens :
B
D
Di
C
E
Ei
Cote 5
O
Figure 6 : modèle utilisé
07-09, February 2007
Von Mises Stresses
Lyon, France
ASME Failure Assessment Validation Exercise
R = 2 mm and Ls = 2 mm
R = 2 mm and Ls = 0,2 mm
Some results of Von Mises Stresses
(a mesh size study has been performed)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
L specimens : some curves compared to experiments for Ls = 0.2mm
ms=8 ; sos=19.8MPa ; mv=15 ; sov=24.6MPa
R = 4 mm
R = 2 mm
100,00%
100,00%
90,00%
90,00%
80,00%
80,00%
70,00%
Probabilité
Psurvie Surface
60,00%
Psurvie Volume
50,00%
Psurvie Eprouvette
40,00%
expérience
Psurvie Surface
60,00%
Psurvie Volume
50,00%
Psurvie Eprouvette
40,00%
30,00%
30,00%
20,00%
20,00%
10,00%
10,00%
0,00%
Psurvie
0,00%
250
350
450
550
650
750
850
250
350
450
Charge (N)
550
650
750
850
Charge (N)
R = 1 mm
100,00%
90,00%
80,00%
70,00%
Probabilité
Probabilité
70,00%
Psurvie Surface
60,00%
Psurvie Volume
50,00%
Psurvie Eprouvette
40,00%
expérience
30,00%
20,00%
10,00%
0,00%
250
350
450
550
650
750
850
Charge (N)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
VP-06 Weibull and Weibull CEA mod (conclusions)
Some material variations found in the experiments
difficult to account for
Some Weibull parameters applied to 4 points
bending then to L specimens (1 or 2 kind of defects)
 seems better with 2 type of defects
 but the parameters were not optimized
 possibility to take in account the volume
effect of the two population of defects
 is it interesting to optimize without
experimental validations ? (i.e. microscopic
examinations of the cracks leading to the rupture)
07-09, February 2007
Lyon, France
ASME Failure Assessment Validation Exercise
CEA Cadarache general conclusions about the benchmark
• Difficult to find all the consistent experimental datas
(some are missing)
• Microscopic examinations are missing  materials are
unknown
• For the design it must be shown that a stress to initiation
in a structure is related to a stress to initiation in another
• A special care should be taken for manufacturing and
machining to develop a design method
• Concerning the Weibull laws (2 type of populations),
it should be possible to better adjust several type of tests
but what about the physical meaning?
if a Weibull law fails to represent different experiments,
is there something simple and better?
07-09, February 2007
Lyon, France