Presentation on theme: "Validation of the plasticity models introduction of hardening laws"— Presentation transcript:
1Validation of the plasticity models introduction of hardening laws andintroduction of hardening lawsOctober 25: ConceptsOctober 27: FormulationsLecturer: Alireza Sadeghirad
2Contents: 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.
3Validation 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. VerificationValidation: Does our (mathematical) model represents the real world with enough accuracy?Verification: Does our (computational) code/software represents the mathematical model with enough accuracy?
4The material has memory. Uniaxial stress – strain diagram:Is the elastic-perfect plastic model a validated model?(=Does the model represent the real worldwith 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
5Uniaxial 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.
6Validation 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.
7Typical uniaxial stress–strain diagram for an elasto-plastic material ultimate strength(maximum stress)ultimate failure(maximum strain)initial yield
8Typical uniaxial stress–strain diagram for an elasto-plastic material Loading/unloading behaviorNew yield stressInitial yield stressPerfect 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 strainTotal plastic strain
9Three types of plastic behaviors are considered here: perfect plasticisotropic hardeningkinematic hardening
12Kinematic hardening:Loading/Unloading behaviorThis 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.
13Validation 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.
14Combined 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 hardeningCombined Hardening is good for simulating the shift of the stress-strain curve apparent in a cyclical loading (hysteresis).
15Multi-axial hardening behavior (2D): load pathIs 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 changeKinematic hardening: size of the yield surface does not change; location of the yield surface changesCombined hardening: size of the yield surface changes; location of the yield surface changes
16This is a pressure-independent model. Multi-axial hardening behavior (3D) – von Mises (or J2) model:for a given stress stateRadial component:r = (constant) x (equivalent shear)rzHydrostatic component:z = (constant) x (pressure)In the von Mises model, only equivalent shear is important in yielding.This is a pressure-independent model.
17in terms of stress components in terms of principal stressesin terms of stress invariantsWhat is the relation between in above equation and axial yield stress in uniaxial tension test?
18Equivalent 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.
19The 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:qqloadloadqloadKinematic hardening
20This 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 StrainPure ShearYes / NoYes / NoHydrostatic Tension / CompressionThis case will not be plastic at all because contains no shear at all.Yes / No
21Uniaxial Strain q p Assuming elastic behavior: step-by-step loading K : bulk modulus2G: shear moduluspqstep-by-step loadingyieldslope: 2GTrial stress: It should be returnedslope: 2G/KTrial stress: GoodStress is changingIt is not a helpful diagram for our question. Which diagram will be helpful?Trial stress: Good
22How 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.pqyield
23Pure Shear q p Assuming elastic behavior: step-by-step loading yieldslope: 2sqrt(3)GTrial stress: It should be returnedTrial stress: GoodStress is constantTrial stress: GoodThe whole changes in strain during the plastic loading is plastic. There is no elastic strain.
24Is the stress constant during the uniaxial stress loading after yielding? Assuming elastic behavior:pqyieldTrial stress: It should be returnedNOslope: 3Trial stress: GoodWhat is going on? Something is wrong in this slide. What is the wrong point here?Stress is changingTrial stress: GoodYESIn 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.
25qIf 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.yieldslope: 3For 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 Strain11Initial yield stressseqInitial yield stresse11pqInitial yield stressslope: Kslope: 2Gslope: 2G/Kslope: HUiniaxial Stress11Initial yield stressseqInitial yield stresse11pqInitial yield stressslope: Eslope: 3slope: E
27Microscopic 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.stressstrain
28Simply 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.
30Simple 1D isotropic hardening example: Yield function:Initial yield stressPlastic modulusHardening law:We also know the following elasticity relation:We want to obtain the following relation during the plastic loading:Special case of perfect plasticity:
31Simple 3D isotropic hardening in associative J2 plasticity example: Yield function:: plastic strain-increment normFlow rule:: unit tensor normal to the yield surfacePerfect 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.
32Simple 3D isotropic hardening in associative J2 plasticity example: Yield function:: plastic strain-increment normFlow rule:: unit tensor normal to the yield surfaceHardening: We always can see the effects of hardening as quantity H in the consistency conditionConsistency 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
33Assignment 1 pure math problem Plasticity equations from book chapter
34References:A. Anandarajah, Computational Methods in Elasticity and Plasticity, Springer, 2010units.civil.uwa.edu.au/teaching/CIVIL8140?f=284007