Constitutive models
Part 2
Elastoplastic
Elastoplastic material models
• Elastoplastic materials are assumed to
behave elastically up to a certain stress limit
after which combined elastic and plastic
behaviour occurs.
• Plasticity is path dependent – the changes in
the material structure are irreversible
Stress-strain curve of a hypothetical material
Idealized results of one-dimensional tension test
  force/ initial area
Johnson’s limit … 50% of Young modulus value
Engineering stress
Yield point
Yield stress
l
 0
l
Engineering strain
Real life 1D tensile test, cyclic loading
Conventional yield point
Where is the yield point?
Lin. elast. limit
Hysteresis loops move
to the right - racheting
Mild carbon steel
before and after heat treatment
Conventional yield point … 0.2%
The plasticity theory covers the
following fundamental points
• Yield criteria to define specific stress
combinations that will initiate the non-elastic
response – to define initial yield surface
• Flow rule to relate the plastic strain increments to
the current stress level and stress increments
• Hardening rule to define the evolution of the
yield surface. This depends on stress, strain and
other parameters
Yield surface, function
F ( ij , ijP , K...)  0
• Yield surface, defined in stress space separates stress states
that give rise to elastic and plastic (irrecoverable) states
• For initially isotropic materials yield function depends on
the yield stress limit and on invariant combinations of
stress components
• As a simple example Von Mises … F   effective   yield  0
• Yield function, say F, is designed in such a way that
F  0 stressstatewithin thesurface
F  0 on thesurface
F  0 outside,inadmissible for analyticalplasticity
Three kinematic conditions are to be
distinguished
• Small displacements, small strains
– material nonlinearity only (MNO)
• Large displacements and rotations, small strains
– TL formulation, MNO analysis
– 2PK stress and GL strain substituted for engineering
stress and strain
• Large displacements and rotations, large strains
– TL or UL formulation
– Complicated constitutive models
Rheology models for plasticity
Ideal or perfect plasticity, no hardening
Isotropic hardening
stress
+ new yield stress 1
reloading
loading
unloading
- new yield stress 1
initial yield stress
new yield stress 2
Loading, unloading, reloading and cyclic loading in 1D
strain
Isotropic hardening in principal stress space
 - plane
arccos (2/sqrt(3))
von Mises expressedby principalstressesand1D yieldstressin tension
F  [( 1   2 )2  ( 2   3 )2  ( 3   1 )2 ]  2 Y2  0
T rescaexpressedby principalstressesand1D yieldstressin tension
F  ( 1   3 )   Y  0,  1   2   3
Kinematic hardening
reloading
loading
unloading
+ new yield stress 1
stress
initial yield stress
Loading, unloading, reloading and cyclic loading in 1D
strain
Kinematic hardening in principal stress space
insteadof F ( ij )  0 (as in case of isotropichardening)
we takeF ( ij  ij )  0, where ij  c ijP , c... constant
Von Mises yield condition, four hardening models
1. Perfect plasticity – no hardening
3. Kinematic hardening
2. Isotropic hardening
4. Isotropic-kinematic
Different types of yield functions
F  F ( ij )  perfect plast icit y
meansno hardening,mat erialst art st o flow and is inclided t o do so forever.
It pract iceit is st abilized by t he' healt hy'mat erialst ruct urewhich exist s
around t heplast icit yregion.
P last ic mat erialflow is caused by mot ionof dislocat ions.
Definit ionof dislocat ions ...
Generally,t hehardeningdepends on blocking t hemot ionof dislocat ions (free flow)
which depends on t hepermanentplast icst rain ijP .
F  F ( ij   ij )  kinemat ichardening
where  ij  c  ijP and c is a const ant .
F  F ( ij ,  ijP )  non - isot ropichardening
hardeningdepends on everycomponentof  ij in a different way
F  F ( ij , K )  isot ropichardening
where K  K ( ijP ) is a scalar funct ionof  ijP , usually an invariant .
Generally,which is not general at all, we could have
F  F ( ij ,  ijP , K )
Plasticity models – physical relevance
• Von Mises
- no need to analyze the state of stress
- a smooth yield sufrace
- good agreement with experiments
• Tresca
- simple relations for decisions (advantage for hand calculations)
- yield surface is not smooth (disadvantage for programming,
the normal to yield surface at corners is not uniquely defined)
• Drucker Prager
a more general model
1D example, bilinear characteristics
d T  d E  d P
stress
d d d


ET
E
EP
d

tan   ET
Y
elastic
plastic

total
Strain hardening parameter
ET E
ET
 EP 

 H
E  ET 1  ET / E
tan   E
… elastic modulus
… tangent modulus
strain
d  d T  d EP
… means total or elastoplastic
Strain hardening parameter again
Upon unloading and reloading the effective stress must exceed
Initial yield
Elastic strains removed
Geometrical meaning of the strain hardening parameter is
the slope of the stress vs. plastic strain plot
How to remove elastic part
ET E
EP 
E  ET
1D example, bar (rod) element
elastic and tangent stiffness
L
F
F

A
  Y
Elastic stiffness
kE 
F

Tangent stiffness
kT 

EA
L
  Y
dF d A
EP d P A


d d T L d E  d P  L
EP A
d /E P
EA 
EP 
1 

kT 

L d / E  d / EP
L  E  EP 
Results of 1D experiments must
be correlated to theories capable
to describe full 3D behaviour of
materials
• Incremental theories relate stress increments to strain increments
• Deformation theories relate total stress to total strain
Relations for incremental theories
isotropic hardening example 1/9
d
Relation between increments and rates : lim
 
t 0 dt
Parameter only
Let theyield surface is F ( ij ,  ijP )  0
incrementof deformat ion depends on F and  eff
if
F  0 elastic
F  0 and  eff  0 elastic
F  0 and  eff  0 elastoplastic
F  0 and  eff  0 neutral- it meansthatijP  0
F  0 go back t oyield surface
Relations for incremental theories
isotropic hardening example 2/9
Flow rule is assumed in theform(Drucker,1947)
F
ijP  
 q Eq. (i) … increment of plastic deformation has a direction
 ij
normal to F while its magnitude (length of vector) is not yet known
where  is so far unknownscalar and
F
q {
 11
F T

}
 31
defines outer normal to F
in six dimensional stress space
F can be expressedas a totaldifferential
F
F
F
F P
P
dF 
d ij  P d ij 
 ij  P ij
 ij
 ij
 ij
 ij
which must be zero during plasticdeformations, so dF  0
Relations for incremental theories
isotropic hardening example 3/9
F
Denotingp  { P
11
F T

}
P
 31
theconditiondF  0 can be expressedin theform
q d T  pT dε P  q  T  pT ε P  0
stress increment sare
σ  E ε E  E (ε  ε P )
eq. (ii)
eq. (iii)
elastic total plastic deformations
matrix of elastic moduli
Relations for incremental theories
isotropic hardening example 4/9
Combiningtherelationsfor flow rule (i),
dF  0 (ii) and for stressincrements(iii) we get
Row vector
T

q
E

 T
T
p q  q Eq
Still to be determined
Column vector
Dot product and quadratic form … scalar
Lambda is the scalar quantity determining the magnitude
of plastic strain increment in the flow rule
Relations for incremental theories
isotropic hardening example 5/9
Now, for thestressincrementwe can write
σ  Eε E  E(ε  ε P ) with ε P   q
Substituting for  we get thestress increment
as a functionof totalstrain incrementin theform
σ  EEPε
diadic product
with
T
Eq
(
Eq
)
EEP  E  T
p q  q T Eq
where p still has to be determined
equal to zero for perfect plasticity
Relations for incremental theories
isotropic hardening example 6/9
Determination of p  {
F
11P

F T
}
P
 31
At time t
Assume von Mises yield conditionF  J D2  13 t Y2  0
where J D2  12 sij sij is theseconddeviatoricinvariant
to evaluate
F
P
we
need


f
(

)
Y
ij
P
 ij
Experiments suggest that
 Y  f (W P ), W P    ij d ijP  work done by plasticincrements
t
A new constant defined
Chain rule
t
F
F  t Y W P

Y
t
2
 t
  3 Y
 ij   A  ij
P
P
P
P
 ij   Y W  ij
W
using
F
 t Y
  23 t Y
W P
and
  ij
P
 ij
Relations for incremental theories
isotropic hardening example 7/9
Y
t
ET
EP
W
Y
0
E
WP


0 P
P
t
P
in 1D theelastic work done W P  12 ( t  Y  0 Y ) t P
1D bilinear characteristics t Y ( 0 Y  E P t P )
1 t 2 0 2
( Y  Y)
P
2E
W P t Y
2 t
E P 2 P 2 EE T
 P  A  Y t
 E 
t
 Y
E
3
Y 3
3 E  ET
W P 
so finaly p  A{ 11  22   31}T
Relations for incremental theories
isotropic hardening example 8/9
Summary.For given  ij and  Y and ε ij we can comput eσ as follows
 m  13 ( 11   22   33 )
s  {s11 s22 s33 s12 s23 s31}T  { 11   m  22   m  33   m  12  23  31}T
q  {s11 s22 s33 2 s12 2 s23 2 s31}T
2 EE T
A
3 E  ET
p  A { 11  22  33  12  23  31}T
a  p T q, b  Eq , c  q T Eq  q T b
T
bb
EEP  E 
ac
σ  EEPε
J2 theory, perfect plasticity 1/6
alternative notation … example of numerical treatment
{ }  [ E ]{ }...Hooke's law
{ }  { xx  yy  zz  xy  yz  zx }
T
{ }  { xx  yy  zz  xy  yz  zx }T
 m  13 ( xx   yy   zz ) mean stress
stress deviator
{s}  { xx   m  yy   m  zz   m  xy  yz  zx }T
second invariantof stress deviator
2
2
J D2  J 2  12 ( s xx
 s 2yy  szz2  2 s xy
 2 s 2yz  2 szx2 )
or J 2  12 {s}T [ M ]{s}, with [ M ]  diag(1,1,1,2,2,2)
J2 theory, numerical treatment …2/6
one can provethat
{s} [ M ]{ }  {s} [ M ]{s}, since sxx  s yy  szz  0
T
T
and also [ E ][M ]{s}  2G{s}, with G  E /(1   )
von Mises effectivestress eff  3J 2  3{s} [ M ]{s} / 2
T
yield criterionfor perfectlyplasticbehaviour eff   Y
J2 theory, numerical treatment …3/6
Flow rule according to P randtl- Reuss hypothesis
F
{}  
 [ M ]{s} ...  is so far unknownparameter
T
{ }
{ }  [ E ]{ E }  [ E ]{   P } ... Hooke's law
{ }  [ E ]{}  [ E ]{ P }  [ E ]{}  2G{s} ... its timederivative, increment
no plasticdeformation in elastic region can be expressedby
if  eff   Y then   0,
else   0
endif
Six nonlinear differential equations + one algebraic constraint (inequality)
There is exact analytical solution to this. In practice we proceed numerically
J2 theory, numerical treatment …4/6
Differentiat ingplasticitycondition eff   Y
 eff
 eff
3sT Ms

s 
0
T
s
2 eff
 sT Ms  0 and also sT Mσ  0
Subst ituting for σ
sT MEε  2GsT Ms
and realizing that
2
2GsT Ms  4GJ 2  4G eff
/ 3  4G Y2 / 3
we get
3sT MEε 3sT ε


2
4G Y
2 Y2
System of six nonlinear
differential equations
to be integrated
finally
σ  EEPε with EEP  E 
3G
 Y2
ssT
J2 theory, numerical treatment …5/6
predictor-corrector method, first part: predictor
1. known stress σt
3b. plastic part of increment (1  r ) sT
2. test stress (elastic shot)
σT  σt  σT  σt  E ε
4. sc  st  r sT
5.   3(1  r) sTc ε /(2 Y2 )
6. σ't t  σT  2G  sc
3a. elastic part of increment r sT
J2 theory, numerical treatment …6/6
predictor-corrector method, second part: corrector
Correction
For
st  t   s't  t
find  in such a way that
 eff (st  t )   Y
 eff (  s't  t )   Y
  eff (s't  t )   Y
Y
 
 eff (s't  t )
s't  t  st  t  (1   ) s't  t
and since thesphericalpart of thestress tensor
does not enterintoplasticityconsiderations we have
σ t  t  σ 't  t  (1   ) s't  t
Secant stiffness method and the method of radial return