[V6.04.221] SSNV221-Hydrostatic Testing with linear

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Titre : SSNV221 - Essai hydrostatique avec un comportement[...]
Responsable : BOTTONI Marina
Date : 08/08/2011 Page : 1/10
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SSNV221- Hydrostatic test with a behavior
DRUCK_PRAGER linear and parabolic
Summary:
The case test proposes a purely hydrostatic loading for the associated constitutive Drücker-Prager [R 7.01.16].
The formulation of this plastic model, often used for the soils, is made at the same time on the deviatoric and
hydrostatic part; nevertheless, surface criterion presents a singularity for a purely hydrostatic stress state. This
analytical benchmark is used to check correct hardening in this singularity.
The test is carried out on a material point with the command SIMU_POINT_MAT. One works with imposed
strains.
One makes a test with linear hardening (modelization A) and another with parabolic hardening (modelization
B).
Warning : The translation process used on this website is a "Machine Translation". It may be imprecise and inaccurate in whole or in part and is
provided as a convenience.
Licensed under the terms of the GNU FDL (http://www.gnu.org/copyleft/fdl.html)
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1
Reference problem
1.1
Material properties
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Elastic stripes :
E=3000 MPa Young modulus
Poisson ratio's
 =0,25
DRUCK_PRAGER linear (modelization A):
Coefficient of dependence in pressure
=0,20
p ultm=0,04
 Y =6 MPa
h=100 MPa
Ultimate cumulated plastic strain
Plastic stress
Modulus of hardening
DRUCK_PRAGER parabolic (modelization B):
=0,20
p ultm=0,04
 Y =6 MPa
 ult
Y =10 MPa
1.2
Coefficient of dependence in pressure
Ultimate cumulated plastic strain
Plastic stress
Ultimate plastic stress
Loadings and boundary conditions
A volumic strain is imposed
 v =tr  . The loading is not monotonous: one charges initially until the
volumic strain  v1 , by exceeding the threshold of plastification, then one discharge on a null level of
strain; then one still loads with the strain
 v 2 by thus exceeding the ultimate cumulated plastic strain
p ultm , beyond which one finds a perfect plasticity; one still discharge with null stress (strain equal to
p
the plastic strain  v 2 ) and one reloads while plasticizing later on until the strain  v 3 . The load time
(see Table 1.2-1) is fictitious because the plastic models are independent of time.
t
v
0
0
10
 v1 =0,018
14
0
26
 v 2=0,045
30
 vp2=0,03667
40
 v 3=0,06
Table 1.2-1: imposed volumic strain.
1.3
Initial conditions
All the components of the stresses and strains are null at the beginning of the loading.
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Reference solution
Modelization checks the behavior of the law with linear hardening.
2.1
Computation method
The equations which interest us for analytical computation are (
p
v
•
I 1=tr   : trace of the stress tensor,
 : volumic plastic strain):
plastic constitutive law on the volumic part:
I 1=3K  v −vp 
•
(eq 2.1-1)
surface criterion, by posing null the stress of Von Mises (  eq =0 ) :
F  , p= I 1−R  p
•
relation between the volumic plastic strain and the cumulated plastic strain (local variable of the plastic
model):
˙vp=3 ṗ
•
(eq 2.1-2)
thus by integrating:
 vp=3 p
(eq 2.1-3)
expression of hardening
•
linear:
R p= Y h p
R p= Y h pult = ult
Y
si p pult
si p p ult
(eq 2.1-4)
parabolic:
•
2
  
R p= Y 1− 1−
ult
Y

Y
p
pult
R p= ult
Y
It is observed that, as in the linear case,
si p p ult
(eq 2.1-5)
si p p ult
R p= Y if p=0 and there is perfect plasticity if
p pult .
2.1.1
Strain in extreme cases elastic initial
p
This strain is obtained for  v = p=0 .
If one poses F  , p=0 (plastic evolution) one a:
R p  Y
I el1 =
=


el
I
 elv = 1
3K
2.1.2
Ultimate strain
 ult
v that obtained for p= pult .
ult
pult
The trace of stresses easily is found I 1 and plastic strain  v corresponding:
ult
R p  Y
ult
I1 =
=


Ultimate strain is called
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 vpult =3  pult
I ult
1
 ult
=
 vpult
v
3K
2.1.3
Strain between the yield stress and the ultimate strain
One calculates initially the cumulated plastic strain.
•
By combining the equations (2.1-1), (2.1-2), (2.1-3) and (2.1-4) with
hardening one a:
p=
•
3 K A v1− Y
9 K 2h
By combining the equations (2.1-1), (2.1-2), (2.1-3) and (2.1-5) with
hardening one arrives at the equation of dismantled 2:
F  , p=0 for linear
(eq 2.1-6)
F  , p=0 for L«parabolic
A1 p 2B1 p C 1=0
2
A1= Y  1− 
B1=9 K  2 pult −2  Y  1− 
C 1= Y −3 K   v1
=


(eq 2.1-7)
ult
Y
Y
p
p=
pult
p
One uses the equations then (2.1-3) (2.1-1) to find the plastic strain  v and plots it stresses I 1 .
If one makes discharge elastic material of way until null stress, one finds a residual strain equal to the
plastic strain; it is on the other hand necessary to charge material in compression to obtain a null total
strain. This second branch is also elastic, because the material of Drücker-Prager cannot plasticize in
a hydrostatic state of compression. In this last case, the trace of the stresses, negative, is:
I c1=−3 K  vp
2.1.4
(eq 2.1-8)
Strain higher than the ultimate strain
All the quantities of interest are easily found, because the trace of stresses is known a priori and equal
ult
to I 1 .
I ult
 =v − 1
3K
p

p= v
3
p
v
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2.2
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Quantities and reference results
The modulus of compressibility K is:
K=
2.2.1
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E
=2000 MPa
31−2 
Strain in extreme cases elastic
For two modelizations one finds easily:
el
I 1 =30 MPa
el
 v =0,005
2.2.2
Ultimate strain
For two modelizations one finds:
ult
I 1 =50 MPa
pult
 v =0,024
ult
 v ≈0,03233
2.2.3
•
Strain equal to 0.018 and discharge with null strain
This value of strain  v1=0,018 is higher than the yield stress  el
and lower than  ult
. One
v
v
calculates initially the cumulated plastic strain with the equations (2.1-7) and (2.1-8), then plastic strain
and the trace of the stresses:
linear hardening:
3 K A v1− Y
≈0,019
9 K  2h
p
 v1
=3  p1=0,0114
1
p
I 1=3 K  v1 − v1
≈39,51 MPa
p 1=
•
parabolic hardening:
p 1≈0.0192
p
 v1
=3  p1≈0.0115
1
p
I 1=3 K  v1 − v1
≈38.956 MPa
•
The trace of the stresses with null strain is:
linear hardening:
•
parabolic hardening:
p
I 1c
1 =−3 K  v1≈−68,49 MPa
p
I 1c
1 =−3 K  v1≈−69,044 MPa
Indeed, the difference between the parabolic and linear case is very weak.
2.2.4
Loading until strain EGA to 0.045 and 0.06
ult
One reloads material up to the values of strain  v 2=0,045 and  v3 =0,06 , higher than  v .
The results are the same for two modelizations.
For
 v 2=0,045 :
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 vp2= v 2−
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I ult
1
≈0,03667
3K
vp2
p 2= ≈0,0611
3
p
Following the elastic discharge (until null stress), one finds  v = v 2 , p= p 2 .
For
 v 3=0,06 :
I ult
 = v 3− 1 ≈0,051667
3K
p

p 3= v 3 ≈0,0861
3
p
v3
2.2.5
Stress-strain curves
In Figures (2.2.5-a) and (2.2.5-b) one represents the curve (  v , I 1 ) for linear and parabolic hardening.
In red are the points tested by the benchmark.
Figure 2.2.5-a: stress-strain curves for linear hardening.
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Figure 2.2.5-b: stress-strain curves for parabolic hardening.
2.3
Uncertainties on the solution
The solution is analytical.
2.4
Bibliographical references
[1] Document [R 3.01.16], Intégration of the elastoplastic mechanical behavior of Drücker-Prager
DRUCK_PRAGER and postprocessing. Manual of Code_Aster reference.
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Modelization A
3.1
Characteristics of modelization
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The test is carried out on a material point with the command SIMU_POINT_MAT . One works with
imposed strains.
Hardening is linear.
3.2
Not on Figure
2.2.5-a
Checked quantity
Value of reference
Type of reference
Tolerance (relative)
1
Trace of the stresses
I 11=39,51 MPa
ANALYTIQUE
10−6 %
2
Trace of the stresses
I 1c
1 =−68,49 MPa
ANALYTIQUE
10−6 %
3 or 4
Spherical part of the
plastic strain
 vp2=0,03667
ANALYTIQUE
10−6 %
3 or 5
Trace of stresses
I ult
1 =50 MPa
ANALYTIQUE
10−6 %
5
plastic strain
 vp3=0,051667
ANALYTIQUE
10−6 %
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Modelization B
4.1
Characteristics of modelization
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The test is carried out on a material point with the command SIMU_POINT_MAT . One works with
imposed strains.
Hardening is parabolic.
4.2
Not on Figure
2.2.5-b
Checked quantity
Value of reference
Type of reference
Tolerance (relative)
1 or 2
plastic strain
 vp2=0,03667
ANALYTIQUE
10−6 %
1 or 3
Trace of stresses
I ult
1 =50 MPa
ANALYTIQUE
10−6 %
3
plastic strain
 vp3=0,051667
ANALYTIQUE
10
−6
%
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Summary of results
The results of the benchmark are satisfactory, Code_Aster reproduced the analytical results with a high
accuracy.

[V6.04.221] SSNV221-Hydrostatic Testing with linear

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