Theory of Plasticity. Lecture Notes

Transcription

1 Theory of Plasticity Lecture Notes Spring 2012

2 Contents I Theory of Plasticity 1 1 Mechanical Theory of Plasticity Field Equations for A Mechanical Theory Strain-displacement Relations Equilibrium Equations Compatibility Conditions Constitutive Laws Uniaxial Inelastic Properties Yield Function and Surfaces General Framework for Plastic Constitutive Relations Uniqueness and Stability Postulates Properties of Yield Surfaces Initial Yield Surfaces von Mises Yield Criterion Tresca Yield Criterion ii

3 1.7.3 Properties of Initial Yield Surfaces Subsequent Yield Surfaces Isotropic Hardening Kinematic Hardening Isotropic Hardening vs. Kinematic Hardening Incremental Formulation for Isotropic Hardening Incremental Formulation for Kinematic Hardening Three Dimensional Plasticity Initial Yielding Isotropic Hardening Kinematic Hardening (3D) Combined Isotropic and Kinematic Hardening Implementation in the Finite Element Method Principle of Virtual Work Principle of Minimum Total Potential Energy Introduction to Finite Element Method An Example of Constant Stress Triangular Elements Plane Strain Elements Plane Stress Elements Incremental Plasticity Finite Element Numerical Implementation of the Elastoplastic Constitutive Relations iii

4 2.7 Iteration on the Global Equations Limit Analysis Limit Functions and Surfaces Generalized Stresses and Strains Limit Theorems Limit Analysis for Beams and Frames iv

5 List of Figures 1.1 Generalized deformation of a three-dimensional solid body Measurement of infinitesimal strains Uniaxial loading of a prismatic bar Rate dependent material Time dependent material The Baushinger effect Strain hardening A typical stress-strain curve for metals Rigid-perfectly plastic model Elastic-perfectly plastic model Elastic-linear hardening model Multiple stress strain states Yield function Elastic-perfectly plastic behavior Strain hardening behavior under uniaxial stress state v

6 1.16 Yield function in terms of the stress tensor and internal variable Unloading from a yield surface Neutral loading along a yield surface Normality of plastic strains to the yield surface Two representative material behaviors Monotonic loading Net work and net complementary work Two representative material behaviors both satisfying Postulate Moving of yield surfaces accounting for plastic strain increments Representation of the stress tensor as a vector in principal directions Stress traction acting on the surface of a general elastoplastic body with outward unit normal n Decomposition of the stress traction vector into normal and shear components A unit vector defining octahedral planes in the principal space von Mises yield surface represented in principal directions of stress tensor Projection of von Mises yield surface on the Π-plane Simultaneous projection of von Mises and Tresca yield surface on the Π-plane Yield surfaces of both von Mises and Tresca criterion plotted on a single biaxial stress coordinate The yield surface of a thin-walled tube under combined axial and torsional loading predicted by von Mises yield criterion vi

7 1.34 The yield surface of a thin-walled tube under combined axial and torsional loading predicted by von Mises and Tresca yield criteria Initial and subsequent yield surfaces of isotropic modeling from a uniaxial test Initial and subsequent yield surfaces of kinematic modeling from a uniaxial test The schematic of a thin-walled tube simultaneously subjected to axial and torsional loads Two possible loading paths of the tension-torsion test of a thin-walled tube Characterization of loading path 1 of the tension-torsion test of a thin-walled tube The schematic of a thick-walled tube simultaneously subjected to axial and torsional loads Elastic-linear hardening behavior of a thick-walled tube simultaneously subjected to axial and torsional loads A differential element of a thick-walled tube simultaneously subjected to axial and torsional loads Movement of the yield surface due to kinematic hardening An elastic and linear-hardening material Two possible loading paths of an incremental kinematic hardening model Tractions and displacements specified boundaries of a solid body A cantilever beam subjected to a few generalized loads A constant-stress triangular element A rectangular plate element vii

8 List of Tables viii

9 Part I Theory of Plasticity 1

10 Chapter 1 Mechanical Theory of Plasticity Notation: direct tensor notation as well as Cartesian index will be used throughout the manuscript. 1.1 Field Equations for A Mechanical Theory Here we review the governing equations for the infinitesimal deformations (small strain) of a continuous medium. We restrict ourselves to purely mechanical theory and hence neglect the first and second law of thermodynamics. Consider a body B as shown: As we deform the body particle Q, originally with position vector x measured in an arbitrary coordinate system, moves to a new point q. The distance vector traveled is denoted by the displacement vector ū( x). 2

11 Figure 1.1: Generalized deformation of a three-dimensional solid body Strain-displacement Relations ɛ = 1 [(ū ) + ( ū)], (1.1) 2 or ɛ ij = 1 2 [ ui + u ] j. (1.2) x j x i For example, ɛ 11 = u 1 / x 1 gives a measure of the change in length per unit length of a differential vector oriented in x 1 direction and ɛ 12 = 1 2 [ u 1/ x 2 + u 2 / x 1 ] of the angle change between x 1 and x 2 axes. Note that the engineering strain tensor ɛ is not valid for finite deformations. 3

12 Figure 1.2: Measurement of infinitesimal strains Equilibrium Equations t + ρ b = 0. (1.3) where t is the Cauchy stress tensor, b the body force intensity (usually ḡ), and ρ the density of the solid body under study. In terms of index notation, the above equation can be rewritten as t ij,i + ρb j = 0. (1.4) Compatibility Conditions The compatibility equations are restrictions placed on the strain field, ɛ( x), to ensure the existence of continuous and single valued displacements. Using index notation they take the from 2 ɛ ij x k x l + 2 ɛ kl x i x j 2 ɛ ik x j x l 2 ɛ jl x i x k = 0. (1.5) 4

13 Remark: The above expression represents 81 equations although only six of them are unique Constitutive Laws The constitutive law is unique from previous three sets of equations in that it is material dependent. For example, for a linear isotropic material we can write t = λtr( ɛ) 1 + 2µ ɛ, (1.6) or t ij = λɛ kk δ ij + 2µɛ ij, (1.7) where λ and µ are Lamé constants. The inverted form of the above is ɛ = 1 + ν E t ν E tr( t) 1. (1.8) where E is Young s modulus and ν is Poisson s ratio. More generally a constitutive law can be written symbolically as ɛ(t) C-law t(t). (1.9) Mathematically, we can write t(t) = 0 [ ɛ(t s)ds]. (1.10) Examples of (classes of) history dependent material include 5

14 1) Viscoelasticity 2) Viscoplasticity 3) Plasticity (subject of most of this part) 1.2 Uniaxial Inelastic Properties Consider the axial loading of a bar made from an isotropic material. The Cauchy stress tensor is given by t = t 11 = P A (1.11) where A is the initial cross sectional area. For small strains ɛ 11 = u 1 / x 1. When we first load the bar we have linear elastic behavior so that t 11 = Eɛ 11. Here we assume E is the same in tension and compression. This is true for most non-granular materials. The material remains elastic between two points t + 11 and t 11, where t + 11 may or may not be equal to t 11. In elastic region we have 1) rate independence 2) time independence 3) path independence 6

15 Collectively, these imply that there is a unique relationship between stress and strain. Again consider uniaxial loading of a bar Figure 1.3: Uniaxial loading of a prismatic bar. Definition: t + 11 and t 11 are referred to as the yield stress. Now consider increasing t 11 beyond the yield stress t + 11, we observe the stress-strain curve becomes nonlinear. Beyond the yield stress it is possible to have 1) rate dependence 2) time dependence 3) path dependence Both rate and time dependence are assumed to be negligible in classical plasticity theory. This excludes 1) Typical polymers 7

16 Figure 1.4: Rate dependent material. Figure 1.5: Time dependent material. 2) Metals at high temperature However, we do have path dependence! For instance, let s continue to load to t and then unload. We observe a new region of elastic behavior between t and t In this region, the constitutive law is governed by t 11 = E ɛ 11 where t 11 and ɛ 11 represent stress and strain increments. 8

17 Figure 1.6: The Baushinger effect. There are three major observations to be made from the nonlinear loading 1) The lower yield stress t 2 11 is typically smaller in magnitude than t 1 11, i.e. t 2 11 < t 1 11 (Baushinger Effect). 2) The material remains linear elastic up to its most recent yield stress. 3) Regardless of whether the material has yielded or not, the behavior in the elastic range is governed by E. A typical metal will behave as follows There are several idealizations to this behavior 9

18 Figure 1.7: Strain hardening. Figure 1.8: A typical stress-strain curve for metals. 1) Rigid-perfectly plastic. 2) Elastic-perfectly plastic. 3) Elastic-linear hardening. 10

19 Figure 1.9: Rigid-perfectly plastic model. Figure 1.10: Elastic-perfectly plastic model. 1.3 Yield Function and Surfaces Here we consider yielding of a material subject to multiaxial stress states. We have in general t ij 0 and we will indicate such stress states with 2-D sketches. 11

20 Figure 1.11: Elastic-linear hardening model. Figure 1.12: Multiple stress strain states. For multiaxial stress states we still assume: 1) Rate independence 2) Time independence 12

21 3) Path dependence for inelastic strains Definition: The yield surface is a scaled valued function φ( t) of the 6-D stress space which separates the elastic region from the inelastic region. Figure 1.13: Yield function. Typically φ( t) < 0 = elastic. For an isotropic material in the elastic region, the constitutive law is governed by t = λtr( ɛ) 1 + 2µ ɛ. (1.12) Definition: Unloading is the process of inducing stress increments which cause only elastic strains when the stress state is on the yield surface. Definition: Loading is the process of inducing stress increments which cause plastic strains when the stress states is on the yield surface. 13

22 1.4 General Framework for Plastic Constitutive Relations In what follows, we restrict the development to materials which exhibit a hardening behavior, i.e. materials for which plastic strain increments can only occur with an increment in stress. This excludes the following behavior. Figure 1.14: Elastic-perfectly plastic behavior. Consider a uniaxial tension test where we load the material beyond yield and unload it. Figure 1.15: Strain hardening behavior under uniaxial stress state Hence, we can write the total strain as 11 = e11 + p11, (1.13) 14

23 where ɛ e 11 = ɛ 11/E. Now we extend this idea to 3-D theory and write ɛ = ɛ e + ɛ p, (1.14) Here, the elastic strains are governed by Hooke s law ɛ e = 2G t 1 ν ( ) tr t 1. (1.15) E Upon complete unloading, t = 0, and hence 1) ɛ e = 0. 2) ɛ = ɛ p. The plastic strains depend on the stress history. We need some means of accounting for this history. To do this, we introduce an internal state vector consisting of a collection of scalar variables, i.e. ξ = (ξ 1, ξ 2, ξ 3,, ξ n ). We assume these variables characterize the stress history in some sense. As an alternate viewpoint the variables may be assumed to characterize the current microstructural state of the material. Definition: The yield function is a scalar valued function of the stress, t, and the state vector, ξ, which separate regions of elastic and plastic behavior. φ( t, ξ) < 0: elastic. φ( t, ξ) = 0: right on the yield surface. 15

24 Figure 1.16: Yield function in terms of the stress tensor and internal variable φ( t, ξ) > 0: inaccessible states. We do not allow the yield function to be greater than zero. Rather, we require the yield function to move to acknowledge loading. This requirement is motivated by physical observation, i.e., unloading induces only elastic strains! We can summarize the behavior of the yield function as follows Elastic range: starts from φ( t, ξ) < 0 and ends at φ( t + d t, ξ = const) < 0. Unloading: starts from φ( t, ξ) = 0 and ends at φ( t + d t, ξ = const) < 0. Alternatively, unloading on a yield surface requires dφ ξ=const = φ t : d t < 0. Neutral loading: starts from φ( t, ξ) = 0 and ends at φ( t + d t, ξ) = 0. Alternatively, neutral loading along a yield surface requires dφ ξ=const = φ t : d t = 0. Loading: starts from φ( t, ξ) = 0 and requires φ t : d t > 0. This condition implies that dφ = φ t : d t + φ ξ d ξ = 0 or φ ξ d ξ = φ t : d t. Hence, the internal state vector must change to accommodate loading and the yield function ends at φ( t + d t, ξ + d ξ) = 0. 16

25 Figure 1.17: Unloading from a yield surface. Figure 1.18: Neutral loading along a yield surface. The fact that plastic strains only occur if the state vector changes allows us to write dɛ p ij = h k ijdξ k, (1.16) where h k ij is an array of second order tensors. 17

26 Furthermore, for a work hardening material, the state variables are allowed to change only if the stress changes. Consequently, the plastic strain may be expressed functionally as d ɛ p = d ɛ p ( t, ξ, d t). (1.17) To proceed further we need a math definition. Definition: A function F (x, y, z, u, v, w) is said to be homogeneous of degree n in variables u, v and w if for arbitrary n F (x, y, z, hu, hv, hw) = h n F (x, y, z, u, v, w). Now make the following observations: 1) Plastic strains cannot occur if t remains constant. Hence, the expression on the right hand side of equation (1.17) cannot contain terms that are independent of dt ij. 2) The constitutive law should be rate independent. Hence, we can write equation (1.17) in a rate form as ɛ p = ɛ p ( t, ξ, t). In order to satisfy rate independence we require ɛ p to be homogeneous of degree 1 in the stress rate ɛ p = ɛ p ( t, ξ, a t) = a ɛ p ( t, ξ, t). (1.18) These two physical observations collectively imply that equation (1.17) must be homogeneous of degree 1 in d t. d ɛ p = d ɛ p ( t, ξ, ad t) = ad ɛ p ( t, ξ, d t). (1.19) To proceed further we shall assume that for changes in the state vector ξ 18

27 1) The magnitude of d ξ is a function of ξ, t and d t; 2) The direction of d ξ is determined by ξ and t. These assumptions are based on experimental observation where the internal state vector was taken to be the plastic strains themselves, i.e. ξ = {ɛ P 11, ɛ P 22, ɛ P 33, ɛ P 12, ɛ P 23, ɛ P 13}. Hence, dɛ P ij is always normal to the yield function. Figure 1.19: Normality of plastic strains to the yield surface. They further allow us to write dξ k = bt k, (1.20) where b = b( t, ξ, d t) is a scalar valued function and T k = T k ( t, ξ). Now let us solve for b. Earlier we had for the case of loading dφ = φ t : d t + φ ξ d ξ = 0. (1.21) 19

28 Substituting for d ξ produces Hence, φ t : d t = φ ξ b T. (1.22) φ t b = : d t φ, (1.23) T ξ dξ k = φ t : d t φ ξ T T k. (1.24) Moreover, the plastic strains are given by where φ dɛ P ij = h k t ijdξ k = : d t φ ξ T hk ijt k. (1.25) Now let us simplify this expression by writing dɛ P ij = f ij ( t, ξ) φ t : d t, (1.26) f ij ( t = hk ijt k ξ T. (1.27) We further assume that f ij maybe represented as f ij ( t, ξ) = G( t, ξ) g( t, ξ) t ij, (1.28) where G( t, ξ) is called the scalar hardening function and g( t, ξ) the plastic potential. Remark: We will rigorously justify this later in the text. 20

29 Therefore, the constitutive law for plastic strain increments assumes the form dɛ P ij = G( t, ξ) g( t, ξ) t ij ( ) φ t : d t. (1.29) To complete the formulation we must find 1) the yield function φ; 2) the scalar hardening function G and 3) the plastic potential g. A brief summary to Section 1.4: 1) The total strain at a given stress state may be decomposed as ɛ = ɛ p + ɛ e ; 2) For the elastic component, the constitutive law may be written as ɛ e = 1 2µ t ν E tr( t) 1 or in an incremental form as d ɛ e = 1 2µ d t ν E tr(d t) 1; 3) The yield function φ( t, ξ) 0 represents all physically accessible states a. φ( t, ξ) < 0: elastic behavior b. c. d. φ t : d t < 0:unloading φ t : d t = 0: neutral loading φ : d t ( > 0: loading with plastic strain increments dɛ P g( t, ξ) t ij = G( t, ξ) φ : d t ) t ij. t 1.5 Uniqueness and Stability Postulates Before proceed further, we would like to restrict the materials in a specific manner. As shown in Figure 1.20, the type of material we will study is typically given by curve A. By not allowing type B, we will exclude material types 1) ice and 2) concrete to some extent. 21

30 Figure 1.20: Two representative material behaviors. Curve A has the representative characteristic dt ij dɛ ij 0, which curve B does not. This can be recast in an integrated form. Consider the following loading path t ij = t A ij + (t B ij t A ij)λ(t). (1.30) We restrict this to be a monotonically increasing stress by delimiting the increasing function of time λ(t) 0 λ(t) 1. (1.31) It now becomes obvious that, equation (1.30) represents a monotonic loading for t B ij t A ij and can also be referred to as a proportional loading. Its differential form is given by dt ij = (t B ij t A ij)dλ(t). (1.32) 22

31 Figure 1.21: Monotonic loading. Hence, we can write for curve A dt ij dɛ ij = (t B ij t A ij)dλ(t)dɛ ij 0. (1.33) On the other hand, it is obvious that (t B ij t A ij)dɛ ij (t ij t A ij)dɛ ij = dt ij dɛ ij 0. (1.34) Integrating equation (1.34) with respect to ɛ ij gives (t B ij t A ij)(ɛ B ij ɛ A ij) ɛ B ij ɛ A ij ( tij t A ij ) dɛij 0. (1.35) Definition: The term ɛb ij ɛ A ij ( tij t A ij by stresses in excess of the initial stress t A ij. ) dɛij is called Net Work, and it represents the work done Equation (1.35) shows that the net work is positive semi-definite. Remark: Using integration by parts, the following identity can be shown. 23

32 Figure 1.22: Net work and net complementary work. ɛ B ij ) t B ij dɛij + ( tij t A ij ( ɛij ɛ A ij ) dtij = ( ) ( ) t ij t A ij ɛij ɛ A ij. (1.36) ɛ A ij t A ij In view of the first inequality of equation (1.34) we can show t B ij ( ɛij ɛ A ij ) dtij 0. (1.37) t A ij Definition: The integral (1.37) is referred to as net complementary work. Therefore we have arrived the following postulate. Postulate 1: For an inelastic material subjected to a monotonically proportional load, t ij = t A ij + (t B ij t A ij)λ(t), the net stress work and the net complimentary stress work are both positive semi-definite for stress bounds t B ij t A ij and the increasing function of time 0 λ(t) 1. Now let us consider the following two representative material behaviors. As shown in Figure 1.23, both curves A and C satisfy Postulate 1. Although curve C represents a real material and 24

33 Figure 1.23: Two representative material behaviors both satisfying Postulate 1. is typical of porous material such as snow and porous aluminum, we wish to eliminate it. To do this consider the complimentary work for a complete stress cycle, loading from t A ij to t B ij and unloading back to t A ij at a stress level within the strain hardening scope. The complimentary work is given by ɛ ij dt ij = = = t B ij t A ij t B ij t A ij (ɛ (C1) ij ɛ ij dt ij + ɛ ij dt ij t A ij t B ij t B ij t A ij ɛ (C2) ij ɛ ij dt ij ɛ ij dt ij ) dtij, (1.38) where ɛ (C1) ij and ɛ (C2) ij are strains on the loading curve C1 and the unloading curve C2, respectively. For such a loading-unloading stress cycle, ɛ (C1) ij ɛ (C2) ij and hence ɛ ij dt ij 0. (1.39) 25

34 If we perform the same analysis for curve C we would get the opposite sign. Postulate 2: For an inelastic material subjected to a stress cycle, the complimentary stress work is negative semi-definite. A mathematic statement of these two postulates are Postulate 1: ɛ B ij ( tij t A ij ) dɛij 0, t B ij ( ɛij ɛ A ij ) dtij 0, (1.40) ɛ A ij t A ij for a monotonically proportional load. Postulate 2: ɛ ij dt ij 0, (1.41) for a complete stress cycle. Let us consider the second postulate in more detail. For a stress path t A to t B, we can integrate the complementary work as t B ij t A ij ɛ B ij ɛ ij dt ij = t ij dɛ ij + t B ijɛ B ɛ A ij t A ijɛ A ij. (1.42) ij We now apply the above equation to a full stress cycle ɛ A ( ) ij ɛ ij dt ij = t ij dɛ ij + t A ɛ A ij ɛ A ij ɛ A ij ij = ɛ A ij ɛ A ij ( tij t A ij ) dɛij (1.43)

35 This requires ɛ A ij ( tij t A ij ) dɛij 0, (1.44) ɛ A ij which is true for any loading-unloading cycle at stress levels not exceeding the ultimate strength. As a result, the net stress work is positive semi-definite for any load cycle. 1.6 Properties of Yield Surfaces In what follows we use the stability postulates proposed in Section 1.5 to develop restrictions on 1) the shape of the yield surface; 2) the direction of the plastic strain. Consider a stress cycle where we 1) start inside the yield surface at t A ; 2) follow a load path to the yield surface at t; 3) deform the material plastically to t + d t; 4) return elastically to t A ; Let 1) ɛ pa and d ɛ p denote the plastic strains prior to the cycle and any additional plastic strain increments when we have arrived at t + d t, repectively; 27

36 Figure 1.24: Moving of yield surfaces accounting for plastic strain increments. 1) From t A to t, we have only elastic loading ɛ = ɛ e + ɛ pa = C : t + ɛ pa ; (1.45) 2) from t to t + d t, we have ɛ = ɛ e + ɛ pa + d ɛ p = C : t + ɛ pa + d ɛ p ; (1.46) 3) from t + d t to t A, we have ɛ = C : t + ɛ pa + d ɛ p. (1.47) Now consider the second postulate 0 ɛ ij dt ij 28

37 = t ij t A ij ( ɛ e ij + ɛ pa ) ij dtij + t ij +dt ij t ij ( ɛ e ij + ɛ pa ij + dɛ p ij) dtij = t A ij ( + ɛ e ij + ɛ pa ij + dɛ p ij) dtij t ij +dt ij t ɛ e ijdt ij + ɛ pa ij +dt ij t A ij dt ij + dɛ p ij ijdt ij + dɛ p ijdt ij. (1.48) t ij t ij +dt ij In the above equation 1) ɛ e ijdt ij = 0 since elastic complementary strain energy is recoverable; 2) ɛ pa ij dt ij = 0 because ɛ pa ij is a constant; 3) By the use of mean value theorem one can reason that t ij +dt ij t ij dɛ p ijdt ij O ( dɛ p ijdt ij ), (1.49) which is a second order differential quantity. 4) Finally, since dɛ p ij is independent of the unloading path from t ij + dt ij to t A ij t A ij t ij +dt ij dɛ p ijdt ij = ( t A ij t ij dt ij ) dɛ p ij = ( t A ij t ij ) dɛ p ij dt ij dɛ p ij. (1.50) Noting the above facts and neglecting the second order differential quantities in (1.49) and (1.50), equation (1.48) can be rewritten as ( t A ij t ij ) dɛ p ij 0, (1.51) 29

38 or ( t ij t A ij ) dɛ p ij 0. (1.52) Equation (1.51) or (1.52) has the following implications 1) The yield surface is convex. By convex, we mean that any straight line from t A contained inside the yield surface to another point t inside or on the yield surface must remain inside or on the yield surface. 2) The plastic strain increments must be perpendicular to the yield surface. The second implication allows us to write dɛ p ij = λ φ t ij, (1.53) which implies that the gradient of a level function φ is a vector perpendicular to the level surface defined by φ = 0. It can also be shown that λ 0. Summary: From the second postulate we have dɛ p ij = λ φ t ij. (1.54) Furthermore, we have previously developed the equation (1.29) dɛ P ij = G( t, ξ) g( t, ξ) t ij ( ) φ t : d t. (1.55) 30

39 1.7 Initial Yield Surfaces Up until now we have been working in a 6-D stress space. Now let us change to the 3-D principal space. Recall, t = t ij (ē i ē j ). From eigenvalue theory t n = λ n, ( t λ 1 ) n = 0. (1.56) (1.57) (1.58) It can be shown that the necessary and sufficient condition for nonzero eigenvectors of equation (1.56) or (1.57) is given by det ( t λ 1 ) = 0. (1.59) The above equation can always be solved to produce the following representation of the stress tensor t = t ij(ē i ē j), (1.60) where t ij = t I t II t III, (1.61) 31

40 Figure 1.25: Representation of the stress tensor as a vector in principal directions. and ē i are the principal directions of t. This principal representation of the stress tensor allows us to put the stress tensor in a vector form, i.e. t = t I ē 1 + t II ē 2 + t III ē 3, as shown in Figure To go further with plasticity theory we need to rely on physical observations. The most classical observation for materials is that the hydrostatic stress states have nothing to do with plastic behavior. Let v be a unit vector which makes equal angles with the three principal directions v = 1 3 (ē 1, ē 2, ē 3). (1.62) Planes with unit normal vector defined by v are referred to as the octahedral planes. It now becomes obvious that any stress vector with the form t = t v = t 3 (ē 1, ē 2, ē 3) represents a hydrostatic state of stress since each of the three components is equal t I = t II = t III = Here, the scalar t denotes the magnitude of this hydrostatic stress state. Transforming the stress represented by t v to any other coordinate systems will also produce a hydrostatic stress state. 32 t 3.

41 Now assume we are on the yield surface defined by φ = 0 and induce a stress change d t = dt v. This is a hydrostatic stress increment. From experimental observation we know the corresponding yield function increment dφ = 0 and ξ remains constant. Hence, we can write dφ = φ dt I + φ dt II + φ dt III. (1.63) t I t II t III We cannot have dφ > 0. However, dφ < 0 is also not possible since changing the sign of dt would make φ > 0. Note that d t = dt v = dt 3 (ē 1 + ē 2 + ē 3), (1.64) dt I = dt II = dt III = dt 3. (1.65) This allows us to write dφ = dt 3 ( φ t I + φ t II + φ t III ) = 0. (1.66) This is equivalent to require φ + φ + φ = 0. (1.67) t I t II t III Hence, stress increments which are hydrostatic must represent neutral loading. Our plasticity theory is independent of volumetric effects. Recall that v = 1 3 (ē 1 + ē 2 + ē 3). The initial yield surface is an open ended cylinder oriented along the vector v which is normal to octahedral planes. Clearly, shear stress must play an important role in yielding. 33

42 Figure 1.26: Stress traction acting on the surface of a general elastoplastic body with outward unit normal n. Consider the outer boundary of an elastoplastic body with outward unit normal n and stress traction t ( n) as shown in Figure 1.26 The stress traction is related to the stress vector by Cauchy s relation t ( n) = n t; t ( n) i = n j t ji. (1.68) (1.69) Relative to a principal set of axes, we express 34

43 t ij = t = t ij(ē i ē j), (1.70) t I t II 0. (1.71) 0 0 t III The components of the stress traction then become t ( n) 1 = n 1 t I, t ( n) 2 = n 2 t II, t ( n) 3 = n 3 t III. (1.72) To proceed further, let us denote the normal and shear component of the stress traction with N and τ, respectively, as illustrated in Figure The normal stress is related to the stress traction by N = t ( n) n = t I n t II n t III n 2 3. (1.73) Then, the square of the shear stress is given by τ 2 = t ( n) 2 N 2 = (n 1 t I ) 2 + (n 2 t II ) 2 + (n 3 t III ) 2 (t I n t II n t III n 2 3) 2. (1.74) 35

44 Figure 1.27: Decomposition of the stress traction vector into normal and shear components. One can show that τ 2 is independent of the value of the hydrostatic stress given by t m = 1 (t 3 I + t II + t III ). This can be shown by incrementing all three components of the principal stress vector by t, i.e. t I t I + t, t II t II + t, t III t III + t. (1.75) that Substituting the above into equation (1.74), no changes will be found. Now consider the octahedral plane defined earlier by the unit normal v. This case implies n 1 = n 2 = n 3 = 1 3. (1.76) Definition: The shear stress on the octahedral plane is referred to as the octahedral shear stress τ 0. 36

45 Substituting the surface normal of octahedral planes (1.76) into the shear stress formulae (1.74), we obtain τ0 2 = 1 ( t 2 3 I + t 2 II + tiii) (t I + t II + t III ) 2 (1.77) = 1 ( (ti t II ) 2 + (t II t III ) 2 + (t III t I ) 2). 9 (1.78) Now let us explore some alternate forms for octahedral shear stress τ 0. Consider the deviatoric stress tensor defined by s = t t m 1, ( t m = 1 ) 3 tr( t). (1.79) In component form we have t 11 t m t 12 t 13 s ij = t 12 t 22 t m t 23 t 13 t 23 t 33 t m. (1.80) A few facts about the deviatoric stress tensor should be noted. 1) The trace of the deviatoric stress tensor is zero, i.e. tr s = tr ( t t m 1 ) = tr t t m tr 1 = 0. (1.81) 2) The deviatoric stress tensor has the same principal directions as t. Proof: let t n = t n, s n = s n. 37 (1.82) (1.83)

46 Now substituting the definition of deviatoric stress tensor (1.79) into its characteristic equation (1.83) ( t t m 1) n = s n, t n = (s + t m ) n. (1.84) Compare equation (1.82) and (1.84), we see that n = n, t = s + t m. (1.85) Hence, we can write t I = s I + t m, t II = s II + t m, t III = s III + t m. (1.86) Noting the above relations, the octahedral shear stress (1.77) can be rewritten in terms of the principal deviatoric stress as τ0 2 = 1 ( ) s 2 3 I + s 2 II + s 2 III. (1.87) 38

47 Definition: Three invariants of a given stress tensor t are I t = tr( t), II t = 1 [ (tr( t)) 2 tr( t 2 ) ], 2 III t = det( t). (1.88) For the deviatoric stress tensor, we can write I s = 0, II s = 1 2 tr( t 2 ), III s = det( s). (1.89) Notation: We define J 2 = II s = 1 2 tr( s2 ) = 1 2 s ijs ij. With respect to the principal directions s ij = s = s ij(ē i ē j); s I s II 0. (1.90) 0 0 s III J 2 can be simplified as J 2 = 1 2 ( s 2 I + s 2 II + s 2 III ). (1.91) This leads to τ 2 0 = 2 3 J 2 = 2 3 II s. (1.92) 39

48 1.7.1 von Mises Yield Criterion Figure 1.28: A unit vector defining octahedral planes in the principal space. The von Mises yield criterion states that yielding occurs when the octahedral shear stress reaches a critical value, denoted by k 0 for example. The yield surface in this case can be expressed as φ = τ 2 0 k 2 0 = 0, (1.93) where τ 0 denotes the octahedral sher stress and k 0 a constant to be determined experimentally. Noting previous relations for the octahedral shear stress we can rewrite the von Mises yield criterion as φ(t I, t II, t III, k 0 ) = 1 3 ( t 2 I + t 2 II + t 2 III) 1 9 (t I + t II + t III ) 2 k 2 0 = 0; (1.94) 40

49 Figure 1.29: von Mises yield surface represented in principal directions of stress tensor. φ(s I, s II, s III, k 0 ) = 1 ( ) s 2 3 I + s 2 II + s 2 III k 2 0 = 0; (1.95) φ(j 2, k 0 ) = 2 3 J 2 k0 2 = 0; (1.96) φ(i t, II t, k 0 ) = 2 9 I2 t 2 3 II t k0 2 = 0. (1.97) At this point we introduce a special plane for graphics purpose. Definition: The Π-plane is defined by the unit normal vector v = 1 3 (ē 1 + ē 2 + ē 3) and the mean stress t m = 1 3 tr t = 0. Thus, it is the octahedral plane located at the origin. From the above definition, we see that any stress vector t = t I ē 1 + t II ē 2 + t III ē 3 can be decomposed into a mean stress vector and a stress vector in the Π-plane. As demonstrated in Figure ) Clearly the vector component on the Π-plane is independent of the mean stress; 41

50 2) The Π-plane vector is precisely the octahedral shear stress. Looking down v on the π-plane we see Figure Figure 1.30: Projection of von Mises yield surface on the Π-plane. Recall that the von Mises yield criterion is given by equation (1.93) or τ 2 0 = k 2 0. Consequently, the von Mises yield criterion appears as a circle on the Π-plane as shown. It can be further shown that, the radius of this circle is R = 3k 0. The proof is left as an exercise. (Hint: consider any stress vectors belonging to the Π-plane, such as t = { 1 2, 1 2, 0}R or t = { 1 6, 1 6, 2 6 }R and substitute back to the yield criterion (1.94).) Tresca Yield Criterion The Tresca yield criterion states that plastic yielding initiates when the maximum shear stress (τ max ) reaches a critical value, denoted by k 1 for example. The yield surface in this case can be 42

51 expressed as φ = τ 2 max k 2 1 = 0. (1.98) In analogy to k 0 for the case of von Mises, k 1 is a constant to be determined experimentally. We can further generalize the above relation in terms of principal stress states as φ = (τ 2 1 k 2 1)(τ 2 2 k 2 1)(τ 2 3 k 2 1) = 0, (1.99) where τ 1 = 1 2 (t I t II ), τ 2 = 1 2 (t II t III ), τ 3 = 1 2 (t III t I ). (1.100) We can show that of τ 1, τ 2 and τ 3, one must be τ max by examining a 3-D Mohr s circle. The Tresca yield criterion plots as a hexagon on the Π-plane Properties of Initial Yield Surfaces Now let us examine some properties of the yield surface based on isotropic arguments. 1) The yield surface implies yielding is the same in tension and compression; 2) From isotropy the yield surface must be the same in each 60 0 sector; 3) Our stability postulates tell us the yield surface must be convex, this implies 43

52 a. The Tresca yield surface represents the inner bound of convexity. b. The outer bound on convexity is represented by the dashed line in Figure c. The maximum difference between the inner bound and the outer bound is 2 3 R 3 R R = 25%. (1.101) d. The maximum difference between the Tresca and von Mises occurs in pure shear and is R 3 2 R R 13.4%. (1.102) 4) The yield surface takes on different appearances depending on what stress space coordinates are used. For example, let us consider biaxial testing where t I, t II 0 and t III = 0. Under such a simplified condition, the von Mises yield criterion (1.94) becomes t 2 I + t 2 II t I t II 9 2 k2 0 = 0, (1.103) while three candidates of the maximum shear stress in Tresca criterion (1.99) now take the form τ 1 = 1 2 (t I t II ), τ 2 = 1 2 t II, τ 3 = 1 2 t I. (1.104) Figure 1.32 plots the yield surfaces of both von Mises and Tresca criterion on the same biaxial stress space. 44

53 5) To determine the values for k 0 and k 1, let us further suppose that the Tresca and von Mises yield criterion are chosen to agree in a uniaxial tension test, i.e. t I 0 and t II, t III = 0. This is conventional but absolutely not mandatory. For convenience, we will let σ 0 denote the initial yield stress. In this case, the von Mises yield criterion (1.103) can be further simplified to φ = σ k2 0 = 0 = k 0 = 2 3 σ 0. (1.105) For uniaxial testing, the maximum shear stress τ max = σ 0 /2. Hence, the yield criterion of Tresca (1.99) is φ = ( ) σ0 2 k 2 1 = 0 = k 1 = σ 0. (1.106) These results are of importance in the comparison of von Mises and Tresca yield criterion for especially biaxial stress states. As shown in Figure 1.32, the von Mises criterion is more economic while the Tresca criterion is more conservative based on a simple comparison on the area inside their yield contours. 1.8 Subsequent Yield Surfaces After initial yielding, the subsequent loading produces changes in the the yield surface. To further study this, let us consider a plane stress formulation consisting of a normal stress and a shear stress. A stress state of this nature is readily achieved using combined tension and torsion on a thin-walled tube. For this, let us use polar coordinates (r, θ, z). For this case, we have 45

54 Figure 1.31: Simultaneous projection of von Mises and Tresca yield surface on the Π-plane. Figure 1.32: Yield surfaces of both von Mises and Tresca criterion plotted on a single biaxial stress coordinate. t zz = P A = P 2πR, (1.107) t θz = T R J = T 2πR 2, (1.108) 46

55 where P, T, R, A, J, denote the axial load, twisting moment, median radius, cross-sectional area, polar moment of inertia, and tube wall thickness, respectively. The complete stress state can be written t ij = 0 0 t θz 0 t θz t zz. (1.109) Clearly, we can use the following equivalent stress state for such a case t 11 t 12 0 t ij = t (1.110) Let us find the corresponding principal stress components. t I,II = t avg ± R = t 11 2 ± (t11 2 ) 2 + t 2 12, t III = 0. (1.111) Note the order of the three principal stresses t I > t III > t II. Now consider the two yield criteria. 1) von Mises: φ = τ 2 0 k 2 0 = 1 ( t 2 3 I + t 2 II + tiii) (t I + t II + t III ) 2 k0 2 = 1 ( ) t t t (t 11) σ2 0 = 2 9 t t σ2 0 = (1.112)

56 We normalize φ as φ = t t 2 12 σ 2 0 = 0, (1.113) or t 2 11 σ t2 12 ( σ0 3 ) 2 = 1, (1.114) as graphically represented by an ellipse in Figure Figure 1.33: The yield surface of a thin-walled tube under combined axial and torsional loading predicted by von Mises yield criterion. 2) Tresca: φ = τ 2 max k 2 1 = 1 4 (t I t II ) σ2 0 for t I t III t II = 1 4( t t 2 12 ) 1 4 σ2 0 (1.115) = 0 48 (1.116)

57 Normalization gives φ = t t 2 12 σ 2 0 = 0, (1.117) or φ = t2 11 σ t = 1. (1.118) (σ 0 /2) A comparison on the graphical representations of von Mises and Tresca yield criteria are shown in Figure Figure 1.34: The yield surface of a thin-walled tube under combined axial and torsional loading predicted by von Mises and Tresca yield criteria. In what follows we will write the initial yield surface as φ = t a t 2 12 σ 2 0 = 0, (1.119) where a = 3 for von Mises and a = 4 for Tresca criterion, respectively. For subsequent yield surface, two popular models have evolved 49

58 1) isotropic hardening 2) kinematic hardening Isotropic Hardening This model, while simple, appears to work well for most metals subject to monotonic loading and generally small plastic strains. This model assumes that 1) the shape of the yield surface does not change, 2) the yield surface remains centered at the origin, i.e. it does not translate, 3) the yield surface merely expands, i.e. from φ = t a t 2 12 σ 2 0 = 0, (1.120) to φ = t a t 2 12 σ 2 = 0, (1.121) where σ is the largest value reached by (t at 2 12) Kinematic Hardening To accommodate a yield surface which translates we can write φ = (t 11 ˆσ) 2 + a (t 12 ˆτ) 2 σ 2 0 = 0, (1.122) 50

59 where ˆσ and ˆτ are state variables reflecting the history of the inelastic loading, i.e. ξ = {ˆσ, ˆτ}. From a geometric point of view, ˆσ and ˆτ represent the coordinates of the center of the ellipse. The yield function may be expressed as φ = φ(t 11, t 12, ˆσ, ˆτ), (1.123) Recall that during loading, φ = 0 and dφ = 0, hence dφ = φ t 11 dt 11 + φ dt 12 + φ t 12 ˆσ Substituting for φ from equation (1.122) results in φ dˆσ + dˆτ = 0. (1.124) ˆτ (t 11 ˆσ)dt 11 + a (t 12 ˆτ)dt 12 = (t 11 ˆσ)dˆσ + a (t 12 ˆτ)dˆτ. (1.125) This gives us one equation to determine increments in ˆσ and ˆτ, i.e. (dˆσ, dˆτ). To find another relation, recall from Section that we assumed d ξ = b T( t, ξ), (1.126) i.e. we assumed t and ξ determined the direction of d ξ. The simplest form that has been observed to work well is φ( t, ξ) dξ i = α. ξ i (1.127) The above implies the vector d ξ is parallel to the gradient of yield function φ with respect to ξ. This is an experimental result. For our 2-D theory, it can be further simplified to dˆσ = α φ ˆσ, dˆτ = α φ ˆτ, 51 (1.128)

60 or, diving these two state vectors gives dˆσ dˆτ = φ/ ˆσ φ/ ˆτ. (1.129) Rearrangement of the above yields dˆσ = φ/ ˆσ φ/ ˆτ dˆτ. (1.130) With the aid of the translated yield function (1.122), the above equation reads dˆσ = (t 11 ˆσ) dˆτ. (1.131) a (t 12 ˆτ) The above equation represents another relation necessary for solving dˆσ and dˆτ. Joining of equations (1.131) and (1.125) produces [ ] (t11 ˆσ)dt 11 + a (t 12 ˆτ)dt 12 dˆσ = (t 11 ˆσ) (t 11 ˆσ) 2 + a 2 (t 12 ˆτ) 2 [ ] (t11 ˆσ)dt 11 + a (t 12 ˆτ)dt 12 dˆτ = a (t 12 ˆτ). (1.132) (t 11 ˆσ) 2 + a 2 (t 12 ˆτ) 2 Note that the above is only valid under the condition of loading, i.e. φ = 0, φ dt 11 + φ dt 12 > 0. (1.133) t 11 t 12 For both unloading and neutral loading dˆσ = dˆτ = 0. 52

61 1.8.3 Isotropic Hardening vs. Kinematic Hardening Let us now explore the difference between isotropic and kinematic hardening under the general framework of uniaxial testing for stress component t 11, while maintaining the shear component as a constant in our example of combined tension and torsion on thin-walled tubes. 1) Isotropic Hardening: As shown in Figure 1.35, the yield surface is a monotonically increasing function for isotropic hardening model. To account for this mathematically, we can relate hardening to the increments in plastic work, i.e. dw p = t ij dɛ p ij. (1.134) Hence, we can write φ = φ( t, W p ). Since W p is also a monotonically increasing function we often refer to isotropic hardening as work hardening. Figure 1.35: Initial and subsequent yield surfaces of isotropic modeling from a uniaxial test. 53

62 2) Kinematic hardening: As shown in Figure 1.36, after one complete cycle ɛ p 11 = 0 and moreover the yield surface is the same as the original yield surface. This suggests that the yield surface for kinematic hardening can be expressed as a function of the plastic strains φ = φ( t, ɛ) = 0. Later we will prove there is an one-to-one relationship between (ˆσ, ˆτ) and (ɛ p 11, ɛ p 12) in our 2-D theory. For this reason, we refer to kinematic hardening as strain hardening. Figure 1.36: Initial and subsequent yield surfaces of kinematic modeling from a uniaxial test. 1.9 Incremental Formulation for Isotropic Hardening Engineering shear strain: Recall the constitutive law for plastic deformation dɛ p ij = G ( ) φ φ t : d t (1.135) t ij 54

63 Implicit in the above equation is symmetry with respect to shear components of stress and strain. For instance, for our 2-D theory we ve written: φ = t at 2 12 σ 2 = 0. (1.136) In order to use equation (1.135) we must write the yield function as φ = t a 2 (t t 2 21) σ 2 = 0. (1.137) For instance: φ = t at 12 t 21 σ 2 = 0 dɛ p 11 = G( t, ξ) ( φ t 11 dt 11 + φ t 12 dt 12 + φ t 21 dt 21 ) φ t 11 dɛ p 12 = G( t, ξ) ( φ t 11 dt 11 + φ t 12 dt 12 + φ t 21 dt 21 ) φ t 12 dɛ p 21 = G( t, ξ) ( φ t 11 dt 11 + φ t 12 dt 12 + φ t 21 dt 21 ) φ t 21 (1.138) However, we can use equation (1.136) by recognizing the shear strain as the engineering shear strain, γ 12. Hence, we can write: i) φ = t at 2 12 σ 2 = 0 ii) dɛ p 11 = G ( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 11 (1.139) iii) dγ p 12 = G ( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 12 Exercise: Expand the derivatives in equations (1.138) and (1.139) and verify dγ p 12 = 2dɛ p 12 (1.140) Let s now develop a constitutive law for our 2-D plasticity theory, where t 11, t 12 0 and all other t ij = 0. 55

64 Elastic strains: dɛ e ij = 1 + ν E dt ij ν E dt kkδ ij (1.141) dt kk = dt 11 + dt 22 + dt 33 = dt 11 (1.142) d ɛ e = 1 + ν E d t ν E (tr d t) 1 (1.143) dɛ e 11 = dt 11 E, dɛe 12 = 1 + ν E dt 12 (1.144) or dγ e 12 = 2(1+ν) E dt 12 = dt 12 µ. (1.145) Plastic strains: dɛ p 11 = C G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 11 (1.146) dγ p 12 = C G( φ dt 11 + φ dt 12 ) φ (1.147) t 11 t 12 t 12 1, loading where C = 0, unloading Remarks: 1) Remember, during loading we have both elastic and plastic strain increments. 2) Notice the coupling between normal and shear in stress and strain for the plastic constitutive law. For our 2D theory the yield function is: φ = t at 2 12 σ 2 (1.148) 56

65 3 for Von Mise a = 4 for Treaca To completely specify the constitutive law, a functional for the scalar hardening coefficient must be determined. We have from earlier work: G = G(t 11, t 12, σ 2 ) From isotropy arguments one can prove: G = G( σ 2 ) only! (1.149) We can determine a form for G using experimental data. Let s consider a uniaxial tension test, dt 11 0, dt 12 = 0. Then φ = t 2 11 σ 2 The normal plastic strain increment is: dɛ p 11 = G ( φ t 11 dt 11 ) φ t 11 We have φ t 11 = 2t 11 Therefore: dɛ p 11 = 4Gt 2 11dt 11. Further during loading t 2 11 = σ 2. Hence, dɛ 11 p = 4G σ 2 dt

66 The total strain increment is given by: dɛ 11 = dɛ e 11 + dɛ p 11 (1.150) dɛ 11 = 1 E dt G σ 2 dt 11 (1.151) dɛ 11 dt 11 = 1 E + 4G σ2 (1.152) Definition: The tangent modulus, E T, is the slope of the stress strain curve in the nonlinear region. E T = dt 11 dɛ 11 = E T ( σ) (1.153) We have 1 E T ( σ) = 1 E + 4G σ2 G = 1 [ 1 4 σ 2 ET ( σ) 1 ] E 0 (1.154) Note: during loading σ 2 = t at 2 12 Remarks: 1) We assume the above form of G is valid for all 2-D stress states. This assumption is essential to the success (or failure) of plasticity theory. 2) We could just as well have used a torsion test to define G( σ). 58

67 Figure 1.37: The schematic of a thin-walled tube simultaneously subjected to axial and torsional loads. G = 1 4a σ 2 [ 1 µ T 1 µ ] Example: Path dependence of plastic strains. Consider a tension-torsion test of a thin-walled tube as shown in Figure We will use the Von Mises yield criterion (a = 3). From equilibrium t 11 = P 2πR, t 12 = T 2πR 2 (1.155) We will consider two different paths in stress space: 1) The first path is 59

68 t 11 = t 12, 0 2σ 0 P = 2πR T T = P R 2πR 2 Figure 1.38: Two possible loading paths of the tension-torsion test of a thin-walled tube. 2) We have two parts to the second path as shown in Figure a. t 11 = 0 2σ 0 ; t 12 = 0 b. t 11 = 2σ 0 ; t 12 = 0 2σ 0 For simplicity, we will assume G( σ) = α =constant, e.g. 1 4 σ 2 ( 1 E T E T = E 4α σ 2 E+1 1 E ) = α This type of behavior is characterized by Figure

69 Figure 1.39: Characterization of loading path 1 of the tension-torsion test of a thin-walled tube. Path (a): φ = t t 2 12 σ 2. Since t 11 = t 12 we can write: 1) dt 11 = dt 12 2) The plastic strains are: dɛ p 11 = G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 11 = 16Gt 2 11dt 11 (1.156) dγ p 12 = G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 12 = 48Gt 2 11dt 12 (1.157) We require the initial yield point be the lower limit of integration. φ = t t 2 12 σ 2 0 = 0 (1.158) t 11 = t 12 = 1 2 σ 0 (1.159) We can now integrate equation (1.156) and (1.157) 61

70 ɛ p 11 = 2σ 0 σ Gt 2 11dt 11 = 42Gσ 3 0 γ p 12 = 2σ 0 σ Gt 2 12dt 12 = 126Gσ 3 0 Path (b): 1) t 11 : 0 2σ 0, t 12 = 0 γ p 12 = 0 dɛ p 11 = G( φ t 11 dt 11 ) φ t 11 = 4Gt 2 11dt 11 ɛ p 11 = 2σ 0 σ 0 4Gt 2 11dt 11 = 28 3 Gσ3 0 2) t 11 = 2σ 0, t 12 : 0 2σ 0 dɛ p 11 = G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 11 = G(2t 11 dt t 12 dt 12 )2t 11 = 24Gσ 0 t 12 dt 12 dγ p 12 = G(6t 12 dt 12 )6t 12 = 36Gt 2 12dt 12 ɛ p 11 = 28 3 Gσ σ Gσ 0 t 12 dt 12 = ( )Gσ3 0 γ p 12 = 2σ Gt 2 12dt 12 = 96Gσ 3 0 Path dependence! Example: Tension-torsion of a thick-walled cylinder with inner radius a and outer radius b, as shown in Figure An elastic-linear hardening behavior is assumed, as illustrated in Figure Let s look at a differential element in cylindrical coordinate as illustrated in Figure From St. Venant s torsion theory: t zz 0, t zθ 0, all other t ij = 0. we can use our 2-D theory. 62

71 Figure 1.40: The schematic of a thick-walled tube simultaneously subjected to axial and torsional loads. Elastic strains: dɛ e zz = 1 E dt zz dγzθ e = 2(1+ν) dt E zθ 63 (1.160)

72 Figure 1.41: Elastic-linear hardening behavior of a thick-walled tube simultaneously subjected to axial and torsional loads. Figure 1.42: A differential element of a thick-walled tube simultaneously subjected to axial and torsional loads. 64

73 Plastic strains: dɛ p zz = C G( φ t zz dt zz + φ t zθ dt zθ ) φ t zz dγ p zθ = C G( φ t zz dt zz + φ t zθ dt zθ ) φ t zθ (1.161) Assuming a Von Mises yield criterion, we can write φ = t 2 zz + 3t 2 zθ σ 2 and σ = σ 0 for initial yield. The scalar hardening coefficient is G = 1 4 σ 2 ( 1 E T 1 E ) = 1 4 σ 2 E ( 1 β Carrying out the differentiation of φ 1) (1.162) dɛ p zz = dγ p zθ = The total strain increments: C ( 1 E σ 2 β 1)(t2 zzdt zz + 3t zz t zθ dt zθ ) C ( 1 1)(3t E σ 2 β zθt zz dt zz + 9t 2 zθdt zθ ) (1.163) dɛ zz = dɛ e zz + dɛ p zz (1.164) dγ zθ = dγ e zθ + dγ p zθ Hence, collecting terms: dɛ zz = [ 1 + C ( 1 E E σ 2 β 1)t2 zz]dt zz + [ C ( 1 1)3t E σ 2 β zzt zθ ]dt zθ = α 1 dt zz + α 2 dt zθ (1.165) dγ zθ = [ C ( 1 1)3t E σ 2 β zθt zz ]dt zz + [ 2(1+ν) + C ( 1 E E σ 2 β 1)9t2 zθ]dt zθ = β 1 dt zz + β 2 dt zθ In matrix form: dɛ zz α 1 α 2 dt zz = dγ zθ β 1 β 2 dt zθ 65 (1.166)

74 Note: We can not solve these equations in their present form. The reason is, we do not know the shear stress distribution. Therefore let s invert the equations. dt zz dt zθ = 1 α 1 β 2 α 2 β 1 β 2 α 2 β 1 α 1 dɛ zz dγ zθ (1.167) Now letting δ and Θ denote the axial elongation and angle of twist per unit length, respectively. Then ɛ zz = δ, γ L zθ = r Θ, (Ref. St. Venant theory of torsion) L Rewriting the above in an incremental form: ɛ zz = δ L, γ zθ = r L Θ t zz t zθ = 1 α 1 β 2 α 2 β 1 β 2 α 2 β 1 α 1 ɛ zz γ zθ (1.168) The loads required to produce this deformation are p = t zz A and T = A t zθr da = b 2π a 0 t zθ r rdθdr ********************************* 1.10 Incremental Formulation for Kinematic Hardening Recall the yield function for kinematic hardening is φ = (t 11 ˆσ) 2 + a(t 12 ˆτ) 2 σ 2 0 (1.169) 66

75 In the above equation, (ˆσ, ˆτ) denotes the coordinates of the center of the yield function (an ellipse). Assume initially that (ˆσ, ˆτ) = (0, 0). The evolution equations for (ˆσ, ˆτ) are dˆσ = (t 11 ˆσ) (t 11 ˆσ)dt 11 +a(t 12 ˆτ)dt 12 (t 11 ˆσ) 2 +a 2 (t 12 ˆτ) 2 dˆτ = a(t 12 ˆτ) (t 11 ˆσ)dt 11 +a(t 12 ˆτ)dt 12 (t 11 ˆσ) 2 +a 2 (t 12 ˆτ) 2 (1.170) The plastic strains are dɛ p 11 = G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 11 dγ p 12 = G( φ t 11 dt 11 + φ t 12 dt 12 ) φ t 12 (1.171) Figure 1.43: Movement of the yield surface due to kinematic hardening. Differentiating the yield function φ gives: φ t 11 = 2(t 11 ˆσ), φ t 12 = 2a(t 12 ˆτ). The plastic strain, hence, can be rewritten as dɛ p 11 = 4G(t 11 ˆσ)[(t 11 ˆσ)dt 11 + a(t 12 ˆτ)dt 12 ] dγ p 12 = 4Ga(t 12 ˆτ)[(t 11 ˆσ)dt 11 + a(t 12 ˆτ)dt 12 ] (1.172) If we compare equations (1.170) and (1.172) we find dˆσ = dɛ p 11 4G[(t 11 ˆσ) 2 + a 2 (t 12 ˆτ) 2 ] 67 (1.173)

76 dˆτ = dγ p 12 4G[(t 11 ˆσ) 2 + a 2 (t 12 ˆτ) 2 ] (1.174) These equations can be integrated to provide one to one relation for ˆσ ɛ p 11, ˆτ γ p 12. (1.175) Hence, we can express the yield function in either of the following two forms φ = φ(t 11, t 12, ˆσ, ˆτ) = φ(t 11, t 12, ɛ p 11, γ p 12). (1.176) In the above equations the scalar hardening coefficient, G, is still arbitrary. Therefore, we can write G = G(t 11, t 12, ˆσ, ˆτ) = where G 0 is the principle G 0 (t 11 ˆσ) 2 + a 2 (t 12 ˆτ) 2, (1.177) G 0 = G 0 (t 11, t 12, ˆσ, ˆτ). (1.178) The simplest possible form for G 0 is to assume it is a constant. This implies, however, that the stress-strain curve is bilinear as will be shown. Returning to equations (1.170) and (1.172) we find dˆσ = dɛp 11 4G 0, dˆτ = dγp 12 4G 0. (1.179) Now let us determine a value for G 0 using a tension test. We will approximate the curve as being bilinear, as shown in Figure For this case, dɛ 11 = 1 E dt G(t 11 ˆσ) 2 dt 11 (1.180) 68

77 Figure 1.44: An elastic and linear-hardening material. Solving for G gives G = 1 4(t 11 ˆσ) ( E T E ) (1.181) Furthermore, for this stress state the form assumed for G is given by G = G 0 (t 11 ˆσ) 2. (1.182) This implies that G 0 = 1 4 ( 1 1 ) constant. (1.183) E T E Example: (path dependence) consider an elastic and linear-hardening material as illustrated in Figure E = 30MPa, ν = 0.3, β = 0.2, σ 0 = 20MPa. We ll examine two load paths 1) t 11 = t 12, 0 2σ 0 2) t 12 = 0, t 11 = 0 2σ 0 t 11 = 2σ 0, t 12 = 0 2σ 0 69

78 Figure 1.45: Two possible loading paths of an incremental kinematic hardening model. Path (a): t 11 = t 12, initial yield surface: φ = t t 2 12 σ 2 0 = 0 dˆσ = (t 11 ˆσ) (t 11 ˆσ)dt (t 12 ˆτ)dt 12 (t 11 ˆσ) 2 + 9(t 12 ˆτ) 2 dˆτ = 3(t 12 ˆτ) (t 11 ˆσ)dt (t 12 ˆτ)dt 12 (t 11 ˆσ) 2 + 9(t 12 ˆτ) 2 (1.184) The above equations must be integrated numerically. Path (b): 1) t 12 = 0, t 11 = 0 2σ 0 dˆτ = 0, dˆσ = dt 11 ˆσ = 2σ 0 σ 0 dt 11 = σ 0 ˆσ = σ 0, ˆτ = 0 2) t 11 = 2σ 0, t 12 = 0 2σ 0 70

79 dˆσ = dˆτ = (1.185) Numerical integration is subsequently performed... Results: Path (a): ˆσ = 23 KPa, ˆτ = 33.8 KPa; ɛ p 11 = 3.03(10 3 ), γ p 12 = 4.53(10 3 ) Path (b): ˆσ = 35.2 KPa, ˆτ = 29.6 KPa; ɛ p 11 = 4.39(10 3 ), γ p 12 = 3.93(10 3 ) 1.11 Three Dimensional Plasticity In 3D, the elastic strain increments are d ɛ e = 1 + ν E d t ν E (tr t) 1 (1.186) For the inelastic strains we have d ɛ p = G( φ t : d t) φ t (1.187) where φ = φ( t, ξ) is the yield function, and G is the scalar hardening coefficient (G > 0). Recall that during loading φ t : d t = φ t kl dt kl > 0 (1.188) Note: the yield function φ is a scalar function of a second order tensor t. This implies that φ may be expressed as a function of the invariants of t, i.e. φ = φ(i t, II t, III t, ξ), or the principal 71

80 values of t, i.e. φ = φ(t I, t II, t III, ξ). Now note, we have assumed that plastic deformation is independent of hydrostatic effects. This implies tr ɛ p = ɛ p kk = 0 = V V (1.189) Recall that the constitutive law is given by equation (1.187). Taking the trace of d ɛ p gives tr(dɛ p ) = G( φ t : d t)tr( φ t ) = 0 (1.190) During loading G > 0, φ t : d t > 0 tr( φ t ) = 0 φ t kk = 0 The above results may be used to show φ is not a function of I T, i.e. φ = φ(ii t, III t ξ). The proof is left as an exercise. Since the volumetric effect does not play a role in φ, it is more appropriate to express the yield function in terms of deviatoric stress. Recall that s = t 1tr( 1)Ĩ) and 3 I s = 0 = J 1, II s = 1s 2 ijs ij = J 2, III s = det( s) = J 3. Therefore φ = φ(j 2, J 3, ξ) Initial Yielding 1) Von Mises yield criterion is given by φ = τ 2 0 k 2 0, (1.191) where τ 0 is the octahedral shear stress τ 2 0 = 1 3 s ijs ij = 2 3 J 2. (1.192) Hence, we have φ = 2 3 J 2 k 2 0. (1.193) 72

81 2) Tresca yield criterion is given by φ = 4J J k 4 1J 2 64k 6 1. (1.194) This function is generally not used due to its complexity. However, the Tresca criterion assumes a much simpler form when expressed in terms of Principal stresses. Hence, it may be used when the principal stress are known (e.g. Thick walled pressure vessels) Isotropic Hardening For initial yielding: φ = 2J 3 2 k0 2 = 0, ( 2J 3 2 = k0 2 = 2 9 σ2 0 J 2 = 1 3 σ2 0). For subsequent yielding: φ = 2(J 3 2 J 2 ), where J 2 is the highest recorded value of J 2 for all time. Note: J σ2 0 as this is the value of J 2 at initial yield in a uniaxial tension test. Hence, our internal state vector is ξ = J 2. (1.195) Let us now look at the constitutive law dɛ p ij = G( φ t : d t) φ t ij (1.196) Definition: let ẽ denote the deviatoric strain given by ẽ = ɛ 1 3 (tr ɛ) 1. (1.197) Since ɛ p kk = 0, we have ẽ p = ɛ p 1 3 ɛp kk 1 ẽ p = ɛ p. (1.198) 73

82 Also, since φ = φ(j 2, ξ) = φ( s, ξ), we can write equation (1.196) as de p ij = G ( ) φ φ s : d s. (1.199) s ij Its proof is left as an exercise. (Hint: φ : d t = φ t s : d s and φ t = φ s ) To complete the problem, we need a form for the scalar hardening function G G = G( s, ξ). (1.200) From isotropy: G = G(J 2, J 3, ξ). Recall that ξ = J 2. The form we adopt subsequently is G = G( J 2 ). This is consistent with our 2-D theory where G = G( σ). Now let us expand the derivatives in equation (1.199) φ = 2 3 J J 2, (1.201) J 2 = 1 2 s ijs ij, (1.202) φ = φ J 2 = 2 s ij J 2 s ij 3 s ij, (1.203) de p ij = G ( 2 3 s klds kl ) 2 3 s ij, (1.204) de p ij = 4 9 G (s klds kl ) s ij. (1.205) Now consider a tension test t ij = t s ij = 2 t t t 11 (1.206) 74

83 de p 11 = 4 9 G(s 11ds 11 + s 22 ds 22 + S 33 ds 33 )S 11 = 4 ( ) [ ( ) ( ) 1 2 ( ) G 3 t 11 t 11dt 11 + t 11dt 11 + t 11dt 11] = ( 2 3 ) 4 Gt 2 11dt 11 (1.207) The total strain increment is dɛ 11 = dɛ e 11 + dɛ p 11 = 1 ( ) 2 4 E dt 11 + Gt dt 11 (1.208) Now let E T = dt 11 /dɛ 11 and hence 1 = 1 ( ) 2 4 E T E + Gt ( 1 1 G = ( 2 1 ) 3 )4 t 2 11 E T E ( 1 1 = ( 2 3 )3 2 J 1 ) 2 E T E = 27 ( 1 16 J 1 ) = G( 2 E T E J 2 ), (1.209) where E T = E T ( J 2 ). Note: During loading J 2 = J 2 = 1 2 s ijs ij = 1 2 ( 4 9 t t t2 11 ) = 1 3 t2 11 t 2 11 = 3 J 2. (1.210) Recall that de p ij = 4 9 G (s klds kl ) s ij, (1.211) 75

84 Hence, de p ij = 3 ( 1 4 J 1 ) (s kl ds kl )s ij. (1.212) 2 E T E Kinematic Hardening (3D) Assuming a Von Mises yield criterion, the initial yield surface is φ = τ 2 0 k 2 0 ( τ 2 0 = 1 3 s ijs ij ), (1.213) where s is the deviatonic stress tensor. Therefore, φ = φ( s) for initial yielding. For subsequent yielding we assume the yield function φ = φ( s, ẽ p ). (1.214) This is similar to our 2D theory, where we assumed φ = φ(t 11, t 12, ˆσ, ˆτ), (1.215) or φ = φ(t 11, t 12, ɛ p 11, γ p 12). (1.216) since there is an one-to-one correspondence between (ˆσ, ˆτ) and (ɛ p 11, γ p 12). Recall the plastic strain increments are given by de p ij = G( φ s kl ds kl ) φ s ij (1.217) We also have a requirement during loading that dφ = φ s kl ds kl + φ ξ k dξ k = 0 (1.218) 76

85 In this case: ξ = ( ξ 1, ξ 2, ξ 3,..., ξ 9 ) = (e p 11, e p 12,..., e p 33). (1.219) Noting equation (1.218) we can write φ ds kl = φ s kl e p de p kl (1.220) kl Substitute equation (1.220) into (1.217) gives or de p ij = G de p ij = G ( ) φ φ e p de p kl, (1.221) kl s ij ( φ e p kl s ij ) φ de p kl. (1.222) The above implies G ( φ e p kl s ij ) φ = δ ik δ jl. (1.223) From the above we can readily show G = 1 φ e p kl φ. s kl (1.224) Note that the scalar hardening function G is completely determined by the yield function φ. Prager s Hardening rule A commonly assumed form of the yield function is φ( s, ẽ p ) = ( s hẽ p ) : ( s hẽ p ) 3k 2 0, (1.225) 77

86 where h is a scalar. Note that when ẽ p = 0, we can set φ = s ij s ij 3k 2 0 = s ij s ij 2 3 σ2 0. (1.226) The scalar hardening coefficient becomes G = 1 φ ẽ p ij φ. s ij (1.227) Noting equation (1.225), φ e p ij = 2h(s ij he p ij), (1.228) and we obtain φ s ij = 2(s ij he p ij), (1.229) G = 1 4h(s ij he p ij)(s ij he p ij) = 1. (1.230) 12hk0 2 The scalar h may be determined from a tension test and this is left as an exercise Combined Isotropic and Kinematic Hardening We can develop a combined isotropic/kinematic theory by assuming φ = φ(s ij, e p ij, α), (1.231) where α is the plastic work given by α = s ij de p ij, (1.232) 78

87 integrated over the entire stress history. Hence our internal state vector becomes ξ = (e p 11, e p 12,..., e p 33, α). (1.233) Our constitutive law is still given by de p ij = G( φ s kl ds kl ) φ s ij (1.234) Now consider the case of loading dφ = φ ds kl + φ s kl e p de p kl + φ dα = 0 (1.235) kl α φ [ φ ds kl = s kl e p de p kl + φ ] kl α dα Furthermore, dα = s kl de p kl. Noting the above one can readily show G = φ e p kl (1.236) 1 φ s kl + φ s α kl φ (1.237) s kl A combined isotropic/kinematic hardening model can be constructed by assuming φ = (s ij he p ij)(s ij he p ij) 3k 0 (k 0 gα). (1.238) where g is a constant. 79

88 Chapter 2 Implementation in the Finite Element Method Here we present an introduction to the finite element method as it applies to plasticity in structural mechanics. To do this we must first establish a principle of virtual work. 2.1 Principle of Virtual Work This principle is the fundamental variational principle of solid mechanics. Consider a body B in static equilibrium under the action of specified body forces and surface tractions. Let S denote the bounding surface S = S t + S u, 80 (2.1)

89 where S t and S u denote surface portions with tractions t ( n) or displacements ū( x) specified, respectively, as illustrated in Fig Figure 2.1: Tractions and displacements specified boundaries of a solid body. Recall that the equilibrium equations are t + f = 0, (2.2) where f = ρ b and b is the body force density (usually ḡ if generated due to gravity). A statically admissible stress field is one which satisfies the equilibrium equations over the interior of B and all stress boundary conditions on S t ; 81

90 A kinematically admissible displacement field is one satisfying all displacement boundary conditions on S u and possessing continuous first-order derivative in the interior of B. Now consider a body with statically admissible stress distributions and subjected to kinematically admissible virtual displacements. The virtual work due to external loads is δw E = V or in indicia form f δūdv + S t ( n) δūds = V f δūdv + ( n t ( n) ) δūds, (2.3) S δw E = V f i δu i dv + S n j t ( n) ji δu i ds. (2.4) Applying the divergence theorem, i.e. S n rds = V rdv, to the above equation we have δw E = = V f i δu i dv + ( δu i tji t ji + + f i x j x j V (t ji δu i )dv x j ) δu i dv. (2.5) Noting that t ji x j + f i = 0; x j δu i = δ u i x j = δ [ 1 2 ( u i + u j ) + 1 x j x i 2 ( u i u ] j ) = δɛ ij + δw ij ; x j x i t ij δw ij = 0 ( t = t T, w = w T ), (2.6) we can conclude V f i δu i dv + S t ( n) i δu i ds = V t ij δɛ ij dv. (2.7) The above relation is an alternative statement of equilibrium. Sometimes we write this as δw E = δw I. 82 (2.8)

91 Namely, the virtual work done by statically admissible external loads against kinematically admissible virtual displacements is equal to the virtual work done by internal stresses against virtual strain field. Remarks: 1) In equation (2.7) the body forces, stresses, and the surface tractions are constants; 2) The stresses are independent of the virtual deformation, i.e. there is no relationship between the statically admissible stress field and the kinematically admissible displacement field; 3) This is not an energy principle! It is valid when energy is not conserved (plasticity); 4) This principle is independent of any constitutive law. Example: Consider an isotropic, linear elastic cantilever beam. Deriving the governing equations for the deflection of the beam along with the boundary conditions. Figure 2.2: A cantilever beam subjected to a few generalized loads. 83

92 Solution: we begin by writing the statement of the virtual work for the system V ( a t ij δɛ ij dv = qδvdx p 0 δv(b) + M 0 δ dv ). (2.9) 0 dx x=l For small deflections we have ɛ = y d2 v dx 2 l t ij δɛ ij dv = yσ d2 δv V 0 A dx dadx 2 l d 2 δv l d 2 δv = yσdadx = 0 dx 2 A 0 dx Mdx 2 [ ] ( l M d2 δv 0 dx + qh(a x)δv + p 0ˆδ(x b)δv dx M 2 0 δ dv ) dx x=l = 0. (2.10) where H(a x) is a unit step (heavy side) function and ˆδ(x b) is the Dirac delta function. Integrating the first term by parts twice yields: + [ d 2 ] ( M dx + qh(a x) + p 0ˆδ(x b) δvdx M 2 0 δ dv ) dx x=l ( ) dv Mδ dm ] l dx dx δv = 0 (2.11) l 0 [ 0 or l 0 [ d 2 ] ( M dx + qh(a x) + p 0ˆδ(x b) δvdx + 2 ) ( Mδ dv ) + dx ( dm dx δv x=l x=0 (M M 0 )δ dv dx ) x=l ( ) dm dx δv = 0 (2.12) x=0 Using the standard variational arguments there follows 1) d2 M dx 2 = qh(a x) p 0ˆδ(x b) (The Differential Equation); 84

93 2) at x = 0; v = 0, dv dx = 0 at x = l; M = M 0, dm dx = 0 (Boundary Conditions) Now assuming linearly elastic behavior we have M = EI d2 v dx 2 d 2 ( ) EI d2 v = qh(a x) p 0ˆδ(x b) dx 2 dx 2 v(0) = 0, dv dx (0) = 0 at x = l; EI d2 v dx 2 = EIM 0, ( ) d EI d2 v = 0. (2.13) dx dx Principle of Minimum Total Potential Energy In what follows we develop a special case of the principle of virtual work which is restricted to elastic materials. Recall, for elastic materials there exists a strain energy density, U 0 (ɛ ij ), such that t ij = U 0 / ɛ ij. For instance, for a linearly elastic material U 0 = 1 2 t ijɛ ij. (2.14) Noting the above, let us invoke the principle of virtual work V t ij δɛ ij dv = = V V U 0 δɛ ij dv = ɛ ij f i δu i dv + S V δu 0 dv = δ U 0 dv = δu V t ( n) i δu i ds (2.15) Definition: we define the potential energy of the applied loads as V = f i u i dv t ( n) i u i ds. (2.16) V S 85

94 For f i and t ( n) i prescribed there follows δv = f i δu i dv t ( n) i δu i ds. (2.17) V S Hence, in view of equation (2.15) the principle of virtual work becomes δ(u + V ) = δπ = 0. (2.18) The above relation is often referred to as the principle of minimum total potential energy. Remarks: 1) Π is the total potential energy (PE) of the system; 2) The total potential energy is actually a minimum at equilibrium; 3) The principle of minimum total potential energy (2.18) is restricted to elastic materials. 2.3 Introduction to Finite Element Method Here we present a brief review of the finite element method as it applies to linearly elastic structural mechanics. The field equations to be solved are Equilibrium: t + ρ b = 0; Strain-displacement relationship: ɛ = 1 2 [ū + ū ] ; Constitutive law: ɛ = (1+ν) t ν tr( t) 1 + ɛ 0, where ɛ 0 is the non-elastic strain, i.e. thermal E E strain. The inverse relation is t = λ [tr( ɛ) tr( ɛ 0 )] 1 + 2µ( ɛ ɛ 0 ). 86

95 The finite element method consists of dividing the body into a number of sub-domains called elements, each of which contains a few grid points (nodes). Assuming each element has n nodes and furthermore, each node is allowed to have m generalized degrees of freedom (dof), e.g. typically displacements and/or slopes. T Notation: column vectors: {.}; row vectors: = {.} ; matrices: [...]. Examples of 2D elements: Constant stress triangle: as shown in Fig The generalized nodal displacements for this case can be chosen as u i u i ū e = {ū i } {ū j } {ū k } = v i u j v j u k = v i u j v j u k. (2.19) v k v k Rectangular plate element: as shown in Fig The generalized nodal displacements for 87

96 Figure 2.3: A constant-stress triangular element. this case can be chosen as w i w i w i x w i x w i y w i y {ū i } w j w j x w j w j x ū e = {ū j } {ū k } {ū l } = w j y w k w k x = w j y w k w k x. (2.20) w k y w k y w l w l w l x w l x w l y 88 w l y

97 Figure 2.4: A rectangular plate element. We can also have elements containing mid-side and/or interior nodes. Curved surfaces can be readily handled with the aid of more complex elements, e.g. an 8-node iso-parametric 2D element. Consider an element with n-nodal points. We select one interpolation function for each node point (N 1, N 2,, N n ) which relate the field variables of an arbitrary point, i.e. displacements ū( x), to the generalized nodal displacement ū e 89