1. Validation of the plasticity models introduction of hardening laws
and
introduction of hardening laws
October 25: Concepts
October 27: Formulations
Lecturer: Alireza Sadeghirad

2. 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.

3. Validation vs. Verification
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?

4. The material has memory.
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 behavior

5. Uniaxial stress – strain diagram:
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?
Note that we already assume that the loading/unloading is slow, and temperature is constant.

6. 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.

7. Typical uniaxial stress–strain diagram for an elasto-plastic material
ultimate strength
(maximum stress)
ultimate failure
(maximum strain)
initial yield

8. Typical uniaxial stress–strain diagram for an elasto-plastic material
Loading/unloading behavior
New yield stress
Initial yield stress
Perfect plastic: it is constant.
Hardening: it increases.
Softening: it decreases.
Elastic strain is proportional to stress.
During the plastic loadaing, by increasing total strain:
The plastic strain increases.
What about the elastic strain?
Plastic strain
Total plastic strain

9. Three types of plastic behaviors are considered here:
perfect plastic
isotropic hardening
kinematic hardening

10. Ideally plastic:
Loading/Unloading behavior

11. Isotropic hardening:
Loading/Unloading behavior

12. Kinematic hardening:
Loading/Unloading behavior
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.

13. 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.

14. Combined hardening: isotropic + kinematic.
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 hardening
Combined Hardening is good for simulating the shift of the stress-strain curve apparent in a cyclical loading (hysteresis).

15. Multi-axial hardening behavior (2D):
load path
Is this 1D stress-strain diagram related to isotropic or kinematic hardening?
What is the similar 2D to this diagram?
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

16. This is a pressure-independent model.
Multi-axial hardening behavior (3D) – von Mises (or J2) model:
for a given stress state
Radial component:
r = (constant) x (equivalent shear) r z
Hydrostatic component:
z = (constant) x (pressure)
In the von Mises model, only equivalent shear is important in yielding.
This is a pressure-independent model.

17. in terms of stress components
in terms of principal stresses
in terms of stress invariants
What is the relation between in above equation and axial yield stress in uniaxial tension test?

18. Equivalent shear at uniaxial tension test:
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.

19. 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: q q
load
load q
load
Kinematic hardening

20. This case will not be plastic at all because contains no shear at all.
Consider the following prescribed deformation (strain-control) cases:
Is the stress constant during the plastic loading in the perfect plastic in 3D (assume associative von Mises (J2) plasticity and small deformations)?
Uniaxial Strain
Pure Shear
Yes / No
Yes / No
Hydrostatic Tension / Compression
This case will not be plastic at all because contains no shear at all.
Yes / No

21. Uniaxial Strain q p Assuming elastic behavior: step-by-step loading
K : bulk modulus
2G: shear modulus p q
step-by-step loading
yield
slope: 2G
Trial stress: It should be returned
slope: 2G/K
Trial stress: Good
Stress is changing
It is not a helpful diagram for our question. Which diagram will be helpful?
Trial stress: Good

22. 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.
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. p q
yield

23. Pure Shear q p Assuming elastic behavior: step-by-step loading
yield
slope: 2sqrt(3)G
Trial stress: It should be returned
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.

24. Is the stress constant during the uniaxial stress loading after yielding?
Assuming elastic behavior: p q
yield
Trial stress: It should be returned NO
slope: 3
Trial stress: Good
What is going on? Something is wrong in this slide. What is the wrong point here?
Stress is changing
Trial stress: Good
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.

25. q
If we assume that the whole changes in strain is elastic, the stress path should be like this.
How can we know this is the right path after yielding? Actually we do not know.
yield
slope: 3
For calculate trial stress:
Stress increment = (Elasticity Tensor)x(Total strain Increment) p

26. (You should remember them very well to not be confused)
Common diagrams in uniaxial stress and uniaxial strain examples
(You should remember them very well to not be confused)
Constrained modulus:
Uiniaxial Strain 11
Initial yield stress s e q
Initial yield stress e 11 p q
Initial yield stress
slope: K
slope: 2G
slope: 2G/K
slope: H
Uiniaxial Stress 11
Initial yield stress s e q
Initial yield stress e 11 p q
Initial yield stress
slope: E
slope: 3
slope: E

27. Microscopic interpretation of plasticity and hardening:
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

28. Simply showing the effects of hardening in the yield function:
Ideally plastic:
Isotropic hardening:
Kinematic hardening:
Combined:
I will present the more general forms in the next slides.

29. = Elastoplastic modulus (tensor)
Solving plasticity governing equations:
During plastic loading:
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.
= Elastoplastic modulus (tensor)
In the next slides, the plasticity equations are solved in some special 1D and 3D cases.

30. Simple 1D isotropic hardening example:
Yield function:
Initial yield stress
Plastic modulus
Hardening law:
We also know the following elasticity relation:
We want to obtain the following relation during the plastic loading:
Special case of perfect plasticity:

31. Simple 3D isotropic hardening in associative J2 plasticity example:
Yield function:
: plastic strain-increment norm
Flow rule:
: unit tensor normal to the yield surface
Perfect plasticity:
Consistency condition (during plastic loading):
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.

32. Simple 3D isotropic hardening in associative J2 plasticity example:
Yield function:
: plastic strain-increment norm
Flow rule:
: 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):
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

33. Assignment 1 pure math problem
Plasticity equations from book chapter

34. References:
A. Anandarajah, Computational Methods in Elasticity and Plasticity, Springer, 2010
units.civil.uwa.edu.au/teaching/CIVIL8140?f=284007