an elastic material has a unique, natural, elastic reference state to which it will return when the...

Validation of the plasticity models

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.

• 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.

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?


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


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 !

.

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.

Typical uniaxial stress–strain diagram for an elasto-plastic material

ultimate failure(maximum strain)

ultimate strength(maximum stress)

initial yield

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: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.

Three types of plastic behaviors are considered here:

perfect plastic isotropic hardening kinematic hardening

Ideally plastic:

Loading/Unloading behavior

Isotropic hardening:Loading/Unloading behavior

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 behavior

.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.

.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 hardening

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 path

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.

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?

y

In the uniaxial stress tension test, which is a common test to determine the yield stress:

000

000

0011

000

000

00yStress: Stress at yield point:

Equivalent shear at uniaxial tension test:

11231

223

212113333222211

2332211 )(3)( q

Equivalent shear at yield point: yq

in von Mises (J2) model and axial yield stress in uniaxial tension test are the same.y

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

load

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 StrainPure Shear

Hydrostatic Tension / Compression

Yes / NoYes / No

Yes / No

This case will not be plastic at all because contains no shear at all.

,)21(33

)1(

)21)(1(

EEp

p

q

step-by-step loading

yield

Uniaxial Strain

Assuming elastic behavior: ,

000

000

00

ε

)21)(1(00

0)21)(1(

0

00)21)(1(

)1(

E

E

E

σ

)1(

E

q

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 changing

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.

,0p

p

q

step-by-step loading

yield

Pure Shear


000

00

00

ε ,

000

002

020

G

G

σ

slope: 2sqrt(3)G

Trial stress: Good

Trial stress: Good


Stress is constant

Gq 32

The whole changes in strain during the plastic loading is plastic. There is no elastic strain.

Is the stress constant during the uniaxial stress loading after yielding?


000

000

00

Eσ,

00

00

00

ε ,

3Ep Eq

p

q

yield

slope: 3

Trial stress: Good

Trial stress: Good


Stress is changing

NO

What is going on? Something is wrong in this slide. What

is the wrong point here?

YESIn 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.

11q

yq 11

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)

Common diagrams in uniaxial stress and uniaxial strain examples

(You should remember them very well to not be confused)

p

q


slope: 2G/K

Uiniaxial Strain

Uiniaxial Stress

11


s

e11

q


e11

p

q


11


s

e11

q


e11

slope: 2G

slope: E slope: 3 slope: E

slope: H

)21)(1(

)1(

EHConstrained modulus:

slope: K

• 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:

• Ideally plastic:

• Isotropic hardening:

• Kinematic hardening:

• Combined:

0)()( pyijF

0)( yijF

0)())(( py

pijijF

Simply showing the effects of hardening in the yield function:

I will present the more general forms in the next slides.

0))(( yp

ijijF

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.

pe εεε eεEσ :

εEσ :ep

epE = Elastoplastic modulus (tensor)

Simple 1D isotropic hardening example:

yyF ),(

ppy

py E 0)(

Yield function:

Hardening law:

Plastic modulusInitial yield stress

We also know the following elasticity relation:

We want to obtain the following relation during the plastic loading:

eεEσ

εEσ ep

p

pep

EE

EEE

Special case of perfect plasticity: 0pE 0epE

Simple 3D isotropic hardening in associative J2 plasticity example:

yy qF ),(σYield function:

Flow rule:



eεEσ :

εEσ :ep

rspqrspq

cdklcdabijabijkl

epijkl NEN

ENNEEE

Even without hardening, stress may change during the plastic loading.

0)( yp

y Perfect plasticity:

Nε )( p

pε

σσN

FF

: plastic strain-increment norm

: unit tensor normal to the yield surface

Consistency condition (during plastic loading): 0dF 0: σN

Simple 3D isotropic hardening in associative J2 plasticity example:

yy qF ),(σYield function:

Flow rule:



eεEσ :

εEσ :ep

HNEN

ENNEEE

rspqrspq

cdklcdabijabijkl

epijkl

Hardening: H>0, and Softening: H<0

Hardening: We always can see the effects of hardening as quantity H in the consistency condition

Nε )( p

pε

σσN

FF

: plastic strain-increment norm

: unit tensor normal to the yield surface

Consistency condition (during plastic loading): )(: HσN

Assignment 1 pure math problem Plasticity equations from book chapter

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

an elastic material has a unique, natural, elastic reference state to which it will return when the...

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