Presentation is loading. Please wait.

Presentation is loading. Please wait.

An elastic material has a unique, natural, elastic reference state to which it will return when the deformation-causing forces are removed. The deformation.

Similar presentations


Presentation on theme: "An elastic material has a unique, natural, elastic reference state to which it will return when the deformation-causing forces are removed. The deformation."— Presentation transcript:

1

2

3 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 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.

5 We should specify: for which material? under which conditions? Note that we already assume that the loading/unloading is slow, and temperature is constant. These questions are not the right (complete) ones !

6 .

7 ultimate failure (maximum strain) ultimate strength (maximum stress) initial yield

8 Plastic strainTotal plastic strain Initial yield stress New yield stress During the plastic loadaing, by increasing total strain: 1) 1) The plastic strain increases. 2) 2) What about the elastic strain? Perfect plastic: it is constant. Hardening: it increases. Softening: it decreases. Elastic strain is proportional to stress.

9

10

11

12

13 .

14 . 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.

15 load path

16 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.

17 in terms of stress components in terms of principal stresses in terms of stress invariants

18 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:

19 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 q load

20 Yes / No This case will not be plastic at all because contains no shear at all.

21 p q step-by-step loading yield 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: It should be returned Stress is changing

22 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.

23 p q step-by-step loading yield Assuming elastic behavior: slope: 2sqrt(3)G 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.

24 Assuming elastic behavior: p q yield slope: 3 Trial stress: Good Trial stress: It should be returned Stress is changing What is going on? Something is wrong in this slide. What is the wrong point here? 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 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)

26 p q Initial yield stress slope: 2G/K  Initial yield stress    q Initial yield stress   p q Initial yield stress  Initial yield stress    q Initial yield stress   slope: 2G slope: Eslope: 3slope: E slope: H Constrained modulus: slope: K

27 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

28 Ideally plastic: Isotropic hardening: Kinematic hardening: Combined:

29 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)

30 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: Special case of perfect plasticity:

31 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):

32 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):

33 Assignment 1 pure math problem Plasticity equations from book chapter

34 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


Download ppt "An elastic material has a unique, natural, elastic reference state to which it will return when the deformation-causing forces are removed. The deformation."

Similar presentations


Ads by Google