#### Transcript ComputationalPlasticity

• October 25: Concepts • October 27: Formulations Lecturer: Alireza Sadeghirad • 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. 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 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 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. . • 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) Loading/unloading behavior New yield stress Initial yield stress strain by increasing Total plastic During the plasticPlastic loadaing, 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. Loading/Unloading behavior Loading/Unloading behavior 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. . • 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. . 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 curve apparent in a cyclical loading (hysteresis). load path for a given stress state Radial component: 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 load load q load 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 step-by-step loading 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 changes in stress during the plastic loading? When willwe thecalculate stress bethe constant during the plastic loading? 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 load step I’ll talk about the formulations later. during plastic loading. (increment) leads to changes 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 step-by-step loading 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 During plastic loading: 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 Consistency condition (during plastic loading): dF 0 : 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 Even without hardening, stress may change during the plastic loading. 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 www.cadfamily.com/download/CAE/Marc.../mar120_lecture_09.ppt