and introduction of hardening laws October 25 Concepts October 27 Formulations Lecturer Alireza Sadeghirad Contents Introduction of the concepts in 1D Extension of the concepts to 2D and 3D ID: 303043
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Slide1
Validation of the plasticity modelsandintroduction of hardening laws
October 25: Concepts
October 27: Formulations
Lecturer:
Alireza
SadeghiradSlide2
Contents: 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 loading/unloading is slow.
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.Slide3
Uniaxial stress – strain diagram:
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, natural, elastic reference state to which it will return when the deformation-causing forces are removed. The deformation between this elastic reference state and the currents state is reversible.
There
is a one-to-one relationship between stress and
strain.
The material does not have
memory
.
Validation vs. Verification
Validation
: Does our (mathematical) model represents the real world with enough accuracy?
Verification
: Does our (computational) code/software represents the mathematical model with enough accuracy?Slide4
Uniaxial stress – strain diagram:
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 stress, which loading beyond that 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
.
Loading/unloading behaviorSlide5
Uniaxial stress – strain diagram:
Is the elastic-perfect plastic model a
validated
model for this test?
We should specify: for which
material
?
under which
conditions
?
Note that we already assume that the loading/unloading is slow, and temperature is constant.
Is the elastic model a
validated
model for this test?
These questions are not the right (complete) ones !Slide6
.
Validation of elastic and elastic-perfect plastic models:
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.Slide7
Typical uniaxial stress–strain diagram for an
elasto
-plastic material
ultimate failure
(maximum strain)
ultimate strength
(maximum stress)
initial yieldSlide8
Plastic strain
Total plastic strain
Initial yield stress
New yield stress
Typical
uniaxial
stress–strain diagram for an
elasto
-plastic material
Loading/unloading behavior
During the plastic
loadaing
, by increasing total strain:
The plastic strain increases.
What about the elastic strain?
Perfect plastic: it is constant.
Hardening: it increases.
Softening: it decreases.
Elastic strain is proportional to stress.Slide9
Three types of plastic behaviors are considered here: perfect plastic
isotropic hardening
kinematic hardeningSlide10
Ideally plastic:
Loading/Unloading behaviorSlide11
Isotropic hardening:
Loading/Unloading behaviorSlide12
Kinematic hardening:
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.
Loading/Unloading behaviorSlide13
.
Validation of isotropic and kinematic hardening:
Isotropic hardening is commonly used to model drawing or other metal forming operations.
For many materials, the kinematic hardening model gives a better representation of loading/unloading behavior than the isotropic hardening model. For cyclic loading, however, the kinematic hardening model cannot represent either cyclic hardening or cyclic softening. Slide14
.
Combined hardening: isotropic + kinematic
Combined Hardening is good for simulating the shift of the stress-strain curve apparent in a cyclical loading (hysteresis).
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.
Validation of combined hardeningSlide15
Multi-axial hardening behavior (2D):
Isotropic hardening:
size of the yield surface changes; location of the yield surface
does not
change
Kinematic hardening:
size of the yield surface
does not
change; location of the yield surface changes
Combined hardening:
size of the yield surface changes; location of the yield surface changes
Is this 1D stress-strain diagram related to isotropic or kinematic hardening?
What is the similar 2D to this diagram?
load pathSlide16
Multi-axial hardening behavior (3D) – von Mises (or J2) model:
for a given stress state
r
z
Radial component:
r = (constant) x (equivalent shear)
Hydrostatic component:
z = (constant) x (pressure)
In the von
Mises
model, only equivalent shear is important in yielding.
This is a pressure-independent model.Slide17
in terms of stress componentsin terms of principal stressesin terms of stress invariants
What is the relation between in above equation and axial yield stress in
uniaxial
tension test?Slide18
In the uniaxial stress tension test, which is a common test to determine the yield stress:
Stress:
Stress at yield point:
Equivalent shear at
uniaxial
tension test:
Equivalent shear at yield point:
in von
Mises
(J2) model and axial yield stress in
uniaxial
tension test are the same.Slide19
The von Mises (J2) model is dependent only on equivalent stress (=equivalent shear). Thus, we can think about that like a 1D model.
Ideally plastic:
Isotropic hardening:
Kinematic hardening
q
q
q
load
load
loadSlide20
Is the stress constant during the plastic loading in the perfect plastic in 3D (assume associative von Mises (J2) plasticity and small deformations)?
Consider the following prescribed deformation (strain-control) cases:
Uniaxial
Strain
Pure Shear
Hydrostatic Tension / Compression
Yes / No
Yes / No
Yes / No
This case will not be plastic at all because contains no shear at all.Slide21
p
q
step-by-step loading
yield
Uniaxial
Strain
Assuming elastic
behavior
:
K : bulk modulus
2G: shear modulus
It is not a helpful diagram for our question. Which diagram will be helpful?
slope: 2G
slope: 2G/K
Trial stress: Good
Trial stress: Good
Trial stress: It should be returned
Stress is changingSlide22
p
q
yield
When will the stress be constant during the plastic loading?
Which conditions are required?
Stress is constant if each load step (increment) leads to changes
only
in equivalent shear not in pressure. In this case, stress path during returning to yield surface coincides the stress path during the initial elastic stress increment.
How can we calculate the changes in stress during the plastic loading?
Total changes in strain during each step (load increment) contains two parts: elastic and plastic.
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 in strain during plastic loading. I’ll talk about the formulations later.Slide23
p
q
step-by-step loading
yield
Pure Shear
Assuming elastic
behavior
:
slope: 2sqrt(3)G
Trial stress: Good
Trial stress: Good
Trial stress: It should be returned
Stress is constant
The whole changes in strain during the plastic loading is plastic. There is no elastic strain.Slide24
Is the stress constant during the uniaxial stress loading after yielding?
Assuming elastic
behavior
:
p
q
yield
slope: 3
Trial stress: Good
Trial stress: Good
Trial stress: It should be returned
Stress is changing
NO
What is going on? Something is wrong in this slide. What is the wrong point here?
YES
In
uniaxial
stress case, because of boundary conditions, the stress is always of the above form even during the plastic loading. It means that , and after yielding i.e.
stress in constant
.Slide25
p
q
yield
slope: 3
How can we know this is the right path after yielding? Actually we do not know.
If we assume that the whole changes in strain is elastic, the stress path should be like this.
For calculate trial stress:
Stress increment = (Elasticity Tensor)x(Total strain Increment)Slide26
Common diagrams in uniaxial stress and
uniaxial
strain examples
(
You should remember them very well to not be confused
)
p
q
Initial yield stress
slope: 2G/K
Uiniaxial
Strain
Uiniaxial
Stress
11
Initial yield stress
s
e
11
q
Initial yield stress
e
11
p
q
Initial yield stress
11
Initial yield stress
s
e
11
q
Initial yield stress
e
11
slope: 2G
slope: E
slope: 3
slope: E
slope: H
Constrained modulus:
slope: KSlide27
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.
strain
stress
Microscopic interpretation of plasticity and hardening:Slide28
Ideally plastic:
Isotropic hardening:
Kinematic hardening:
Combined:
Simply showing the effects of hardening in the yield function:
I will present the more general forms in the next slides.Slide29
Solving plasticity governing equations:During plastic loading:
The answer is simple:
A relationship between stress increment and strain increment
.The goal of solving plasticity equations, is to obtain this relationship.
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?
In the next slides, the plasticity equations are solved in some special 1D and 3D cases.
=
Elastoplastic
modulus (tensor)Slide30
Simple 1D isotropic hardening example:
Yield function:
Hardening law:
Plastic modulus
Initial yield stress
We also know the following elasticity relation:
We want to obtain the following relation during the plastic loading:
Special case of perfect plasticity:Slide31
Simple 3D isotropic hardening in associative J2 plasticity example:Yield function:
Flow rule:
We also know the following elasticity relation:
We want to obtain the following relation during the plastic loading:
Even without hardening, stress may change during the plastic loading.
Perfect plasticity:
: plastic strain-increment norm
: unit tensor normal to the yield surface
Consistency condition (during plastic loading):Slide32
Simple 3D isotropic hardening in associative J2 plasticity example:Yield function:
Flow rule:
We also know the following elasticity relation:
We want to obtain the following relation during the plastic loading:
Hardening: H>0, and Softening: H<0
Hardening: We always can see the effects of hardening as quantity H in the consistency condition
: plastic strain-increment norm
: unit tensor normal to the yield surface
Consistency condition (during plastic loading):Slide33
Assignment 1 pure math problemPlasticity equations from book chapterSlide34
References:A. Anandarajah
, Computational Methods in Elasticity and Plasticity, Springer, 2010
units.civil.uwa.edu.au/teaching/CIVIL8140?f=284007
www.cadfamily.com/download/CAE/Marc.../mar120_lecture_09.ppt