#### Transcript ComputationalPlasticity

```• October 25: Concepts
• October 27: Formulations
• Introduction of the concepts in 1D.
• Extension of the concepts to 2D and 3D.
• Investigation of some special cases and important issues in 3D.
Assumptions:
• I am talking about the rate-independent plasticity. It means
• Temperature is almost constant.
• I am talking about the associative plasticity, in which it is
assumed that the flow direction (returning path to the yield
surface) is perpendicular to the yield surface.
Is the elastic model a validated model for this example?
(=Does the model represent the real world with enough accuracy?)
An elastic material has a unique,
Validation
Verification
natural,
elasticvs.
reference
state to
which it will return when the
Validation: Doesforces
our are
deformation-causing
(mathematical)
model
removed.
The deformation
between
represents
the real state
worldand
with
this
elastic reference
the
currents
enough
state
accuracy?
is reversible.
ThereVerification:
is a one-to-one
relationship
Does
our
between stresscode/software
and strain.
(computational)
represents the mathematical
The material
does not
have
model
with enough
accuracy?
memory.
Is the elastic-perfect plastic model a validated model?
(=Does the model represent the real world
with enough accuracy?)
There is a stress state, called yield
includes permanent (plastic)
deformation. A yielded material will
unload along a curve that is parallel
to the initial elastic curve. Perfectly
Plastic Hardening Law assumes the
stresses above yield are constant.
There is no one-to-one relationship
between stress and strain.
The material has memory.
Is the elastic model a validated model for this test?
Is the elastic-perfect plastic model a validated model for this test?
These questions are not the right (complete) ones !
We should specify:
for which material?
under which conditions?
and temperature is constant.
.
• Many metals exhibit nearly linear elastic behavior at low strain magnitudes.
•
Rubbers exhibit Hyper-elastic behavior, and they remain elastic up to large
strain values (often up to 100% strain and beyond).
• For metals, the yield stress usually occurs at .05% - .1% of the material’s
Elastic Modulus.
•
Based on my knowledge, there is almost no material showing the exact
elastic-perfect plastic behavior. Perfectly Plastic can be used as an
approximation which may be appropriate for some design processes.
ultimate strength
(maximum stress)
initial yield
ultimate failure
(maximum strain)
New yield stress
Initial yield stress
strain by increasing
Total plastic
During the plasticPlastic
totalstrain
strain:
1) The plastic strain increases.
2) What about the elastic strain?
Perfect plastic: it is constant.
Hardening: it increases.
Softening: it decreases.
Elastic strain is proportional to stress.
This is more common behavior in material plasticity, for example in metals. When the
material has already been yielded, it yields earlier in the opposite direction. This
effect is referred to as the Bauschinger effect.
.
• Isotropic hardening is commonly used to model drawing or other metal
forming operations.
• For many materials, the kinematic hardening model gives a better
represent either cyclic hardening or cyclic softening.
.
The initial hardening is assumed to be almost entirely isotropic,
but after some plastic straining, the elastic range attains an
essentially constant value (that is, pure kinematic hardening).
In this model, there is a variable proportion between the
isotropic and kinematic contributions that depends on the
extent of plastic deformation.
•
Combined Hardening is good for simulating the shift of the stress-strain
for a given stress state
r = (constant) x (equivalent shear)
r
Hydrostatic component:
z = (constant) x (pressure)
z
In the von Mises model, only equivalent shear is important in yielding.
This is a pressure-independent model.
in terms of stress components
in terms of principal stresses
in terms of stress invariants
y
In the uniaxial stress tension test, which is a common test to determine the yield
stress:
Stress:
 11

0

 0
0
0
0
 y
Stress at yield point:  0

 0
0

0

0 
0
0
0
0

0

0 
Equivalent shear at uniaxial tension test:
q 
( 11   22   33 )  3 ( 11 22   22  33   33 11   12   23   31 )   11
2
Equivalent shear at yield point: q  
y
2
y
2
2
The von Mises (J2) model is dependent only on equivalent stress
(=equivalent shear). Thus, we can think about that like a 1D model.
q
q
q
Yes / No
Yes / No
This case will not be plastic at all
because contains no shear at
Yesall.
/ No
Assuming elastic behavior:


ε  0

 0
0
0
0
0

0 ,

0 
 E (1   )

(1   )( 1  2 )

σ  
0


0


E
(1   )( 1  2 )

(1   )
3
 
E
3 (1  2 )
,
E 
(1   )( 1  2 )
0
2G: shear modulus
K : bulk modulus
p
0
q
E
(1   )




0


E 

(1   )( 1  2 ) 
0

slope: 2G
q
Trial stress: It should be returned
yield
slope: 2G/K
It is not a helpful diagram for our
question. Which diagram will be helpful?
Trial stress: Good
Stress is changing
Trial stress: Good
p
How can
When
willwe
thecalculate
stress bethe
constant
Total changes in strain during each step (load increment) contains two
parts: elastic and plastic.
Which conditions are required?
Changes in stress = (Elasticity Tensor) X (Elastic part of changes in strain)
We do not have any changes in stress when there is no elastic part in changes
Stressinisstrain
constant
if each
step I’ll talk about the formulations later.
during
plastic
only in equivalent shear not in
q
pressure. In this case, stress
yield
path during returning to yield
surface coincides the stress path
during the initial elastic stress
increment.
p
Assuming elastic behavior:
0

ε 

 0

0
0
0

0 ,

0 
 0

σ  2G 

 0
2G 
0
0
0

0 ,

0 
p  0 , q  2 3G 
slope: 2sqrt(3)G
q
Trial stress: It should be returned
yield
Trial stress: Good
Stress is constant
Trial stress: Good
The whole changes in strain during the plastic loading is plastic. There is no elastic strain.
p
Assuming elastic behavior:


ε  0

 0
0
 
0
E

σ  0

 0
0 

0 ,

  
0
0
0
0

0 ,

0 
p
E
, q  E
3
q
Trial stress: It should be returned
yield
slope: 3
What is going on? Something
is wrong in this slide. What
is the wrong point here?
Trial stress: Good
Stress is changing
Trial stress: Good
In uniaxial stress case, because of boundary conditions, the stress is always of
the above form even during the plastic loading. It means that q   11 , and
after yielding  11  q   y i.e. stress in constant.
p
q
IfHow
we assume
can we that
knowthe
thiswhole
is thechanges
right path
in strain
after yielding?
is
elastic,
Actually
the
westress
do not
path
know.
should be like this.
yield
slope: 3
For calculate trial stress:
Stress increment = (Elasticity Tensor)x(Total strain Increment)
p
Constrained modulus: H 
q
11
q
(1   ) E
(1   )( 1  2 )
slope: K
Initial yield
stress
Initial yield
stress
Initial yield
stress
slope: 2G
slope: H
slope: 2G/K
e 11
e 11
p
11
q
q
Initial yield
stress
Initial yield
stress
Initial yield
stress
slope: 3
slope: E
slope: E
e 11
p
e 11
• The motion of dislocations (or other imperfections like porosity in
geomaterials) allows plastic deformation to occur.
• Hardening is due to obstacles to this motion; obstacles can be particles,
precipitations, grain boundaries.
stress
strain
• Ideally plastic:
F ( ij )  
y
0
• Isotropic hardening:
F ( ij )   y ( e )  0
• Kinematic hardening:
F ( ij   ij ( e ))  
• Combined:
F ( ij   ij ( e ))   y ( e )  0
p
p
p
y
0
p
ε  ε  ε
e
σ  E : ε
p
e
What we need from a plasticity model to be introduced to the host code, which solves the
equations of motion (EOMs)? What should be the contribution from a plasticity model in the
host code?
The answer is simple: A relationship between stress increment and strain increment.
The goal of solving plasticity equations, is to obtain this relationship.
σ  E
E
ep
ep
: ε
= Elastoplastic modulus (tensor)
In the next slides, the plasticity equations are solved in some special 1D and 3D cases.
Yield function: F ( ,  y )     y
Initial yield stress
Plastic modulus
p
0
p
Hardening law:  y ( e )   y  E e
p
e
We also know the following elasticity relation:  σ  E  ε
We want to obtain the following relation during the plastic loading:  σ  E ep  ε
E
ep
Special case of perfect plasticity: E p  0

EE
p
E  E
p
E
ep
0
Yield function: F (σ ,  y )  q   y
Flow rule:  ε p  (   ) N
  ε
p
 F 
N 


σ


Perfect plasticity:  y ( e p )   y0
: plastic strain-increment norm
F
σ
: unit tensor normal to the
yield surface
N : σ  0
We also know the following elasticity relation:  σ  E :  ε e
We want to obtain the following relation during the plastic loading:  σ  E ep :  ε
E
ep
ijkl
 E ijkl 
E ijab N ab N cd E cdkl
N pq E pqrs N rs
Yield function: F (σ ,  y )  q   y
  ε
Flow rule:  ε p  (   ) N
p
 F 
N 


σ


: plastic strain-increment norm
F
σ
: unit tensor normal to the
yield surface
Hardening: We always can see the effects of hardening as quantity H in the consistency
condition
Consistency condition (during plastic loading): N :  σ  H (   )
We also know the following elasticity relation:  σ  E :  ε e
We want to obtain the following relation during the plastic loading:  σ  E ep :  ε
E
ep
ijkl
 E ijkl 
E ijab N ab N cd E cdkl
N pq E pqrs N rs  H
Hardening: H>0, and Softening: H<0
Assignment 1 pure math problem
Plasticity equations from book chapter
A. Anandarajah, Computational Methods in Elasticity and Plasticity, Springer, 2010
units.civil.uwa.edu.au/teaching/CIVIL8140?f=284007