Development of Yield Criteria for Describing the Behavior of Porous Metals with Tension-Compression Asymmetry

DEVELOPMENT OF YIELD CRITERIA FOR DESCRIBING THE BEHAVIOR OF POROUS METALS WITH TENSION-COMPRESSION ASYMMETRY

By JOEL B. STEWART

A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2009

1 c 2009 Joel B. Stewart

2 ACKNOWLEDGMENTS I would like to thank the members of my supervisory committee for their suggestions and support. I would like to especially thank Dr. Oana Cazacu for her support, her patience and her enthusiasm throughout this study. Her comments and direction have been invaluable to the completion of this research. I would also like to thank the Air Force Research Laboratory for providing the oppor- tunity to pursue a graduate education. Special recognition is extended to Dr. Lawrence Lijewski and Dr. Kirk Vanden for enthusiastically supporting my graduate research goals from the outset. I would also like to acknowledge the many helpful conversations and generous moral support received from Dr. Michael Nixon, Dr. Martin Schmidt and Dr. Brian Plunkett. Their steadfast support and honest critiques were always appreciated and a source of encouragement. I would like to extend special gratitude to Dr. Michael Nixon for reading through the dissertation and making a number of insightful comments and suggestions. Dr. Stefan Soare was of great assistance while he and I were studying for the qualifying exam; his helpful suggestions and engaging conversation were greatly appreciated. Finally, I would like to thank my friends and family who supported me throughout both this current research eﬀort and earlier academic endeavors. Special appreciation is extended to my wife, Kelly, for supporting me throughout my academic journey. The process would have been much more painful without her unfailing support and encouragement.

3 TABLE OF CONTENTS page ACKNOWLEDGMENTS...... 3 LIST OF TABLES...... 7 LIST OF FIGURES...... 8 ABSTRACT...... 11

CHAPTER 1 INTRODUCTION...... 13 2 HOMOGENIZATION APPROACH...... 27 2.1 Kinematic Homogenization Approach of Hill and Mandel...... 28 2.2 Yield Criterion for the Matrix Material...... 30 2.3 Plastic Multiplier Rate Derivation...... 33 2.3.1 Plastic multiplier rate when J3 ≤ 0...... 35 2.3.2 Plastic multiplier rate when J3 ≥ 0...... 38 2.3.3 General plastic multiplier rate expression...... 41 3 PLASTIC POTENTIAL FOR HCP METALS WITH SPHERICAL VOIDS... 46 3.1 Limit Solutions...... 48 3.1.1 Zero porosity and equal yield strengths limiting cases...... 48 3.1.2 Exact solution for a hydrostatically-loaded hollow sphere...... 49 3.1.2.1 Strain-displacement relations...... 49 3.1.2.2 Strain compatibility...... 50 3.1.2.3 Equations of motion...... 51 3.1.2.4 Elastic constitutive relation: Hooke’s law...... 52 3.1.2.5 Ultimate pressure...... 52 3.2 Choice of Trial Velocity Field...... 57 3.3 Calculation of the Local Plastic Dissipation...... 62 3.4 Development of the Macroscopic Plastic Dissipation Expressions...... 66 4 NUMERICAL IMPLEMENTATION OF THE MATRIX YIELD CRITERION. 78 4.1 Return Mapping Procedure...... 78 4.2 First Derivatives...... 85 4.2.1 Isotropic CPB06 ﬁrst derivatives: general loading...... 87 4.2.2 Isotropic CPB06 ﬁrst derivatives: biaxial loading...... 88 4.3 Second Derivatives...... 89 4.3.1 Von mises second derivatives...... 89 4.3.2 Isotropic CPB06 second derivatives: general loading...... 92 4.3.3 Isotropic CPB06 second derivatives: biaxial loading...... 96

4 5 ASSESSMENT OF THE PROPOSED SPHERICAL VOID MODEL BY FI- NITE ELEMENT CALCULATIONS...... 97 5.1 Modeling Procedure...... 97 5.2 Finite Element Results...... 102 5.3 Concluding Remarks...... 104 6 PLASTIC POTENTIALS FOR HCP METALS WITH CYLINDRICAL VOIDS 122 6.1 Limit Solutions...... 123 6.1.1 Zero porosity and von mises material limiting cases...... 123 6.1.2 Analysis of a hydrostatically-loaded hollow cylinder...... 124 6.1.2.1 Strain-displacement relations...... 125 6.1.2.2 Strain compatibility...... 126 6.1.2.3 Equations of motion...... 127 6.1.2.4 Elastic constitutive relation: Hooke’s law...... 128 σ 6.1.2.5 Relation between J3 and Σm ...... 128 6.2 Choice of Trial Velocity Field...... 133 6.3 Parametric Representation of the Porous Aggregate Yield Locus for Ax- isymmetric Loading...... 137 6.3.1 β ≥ 1: the matrix yield strength in tension is greater than in compression...... 138 Σ 6.3.1.1 Σm > 0 and J3 < 0...... 138 6.3.2 Discussion...... 144 6.4 Proposed Closed-Form Expression for a Plane Strain Yield Criterion.... 145 6.4.1 Calculation of the local plastic dissipation...... 145 6.4.2 Development of the macroscopic plastic dissipation expressions... 148 7 ASSESSMENT OF THE PROPOSED CYLINDRICAL VOID MODEL BY FI- NITE ELEMENT CALCULATIONS...... 162 7.1 Modeling Procedure...... 162 7.2 Finite Element Results...... 166 7.3 Concluding Remarks...... 168 8 ANISOTROPIC PLASTIC POTENTIAL FOR HCP METALS CONTAINING SPHERICAL VOIDS...... 180 8.1 Kinematic Homogenization Approach of Hill and Mandel...... 181 8.2 Yield Criterion for the Matrix Material...... 183 8.3 Choice of Trial Velocity Field...... 187 8.4 Calculation of the Local Plastic Dissipation...... 188 8.5 Development of the Macroscopic Plastic Dissipation Expression...... 189 8.6 Assessment of the Proposed Anisotropic Criterion through Comparison with Finite Element Calculations...... 192 8.7 Concluding Remarks...... 197

5 9 CONCLUSIONS...... 210

APPENDIX A PARAMETRIC REPRESENTATION DERIVATION OF THE AXISYMMET- RIC YIELD LOCUS...... 213 A.1 General Form of Equations...... 213 A.1.1 Plastic multiplier rate branches...... 213 A.1.2 Macroscopic plastic dissipation and derivatives...... 217 A.2 β ≥ 1: The Matrix Yield Strength in Tension is Greater than in Compres- sion...... 220 Σ A.2.1 J3 < 0...... 220 A.2.1.1 Σm > 0...... 220 A.2.1.2 Σm < 0...... 221 Σ A.2.2 J3 > 0...... 223 A.2.2.1 Σm > 0...... 223 A.2.2.2 Σm < 0...... 226 A.3 β ≤ 1: The Matrix Yield Strength in Tension is Less than in Compression. 229 Σ A.3.1 J3 < 0...... 230 A.3.1.1 Σm > 0...... 230 A.3.1.2 Σm < 0...... 233 Σ A.3.2 J3 > 0...... 236 A.3.2.1 Σm > 0...... 236 A.3.2.2 Σm < 0...... 237 B RELATIONSHIP BETWEEN HILL48 AND CPB06 COEFFICIENTS..... 240 B.1 Determine The Hill48 Coefficients Given The CPB06 Coefficients..... 245 B.2 Determine The CPB06 Coefficients Given The Hill48 Coefficients..... 246 REFERENCES...... 247 BIOGRAPHICAL SKETCH...... 251

6 LIST OF TABLES Table page 2-1 k-relations...... 44 2-2 z-parameters...... 44

5-1 k = 0, J3 > 0 and f0 = 0.01 spherical void computational test matrix...... 108

5-2 k = 0, J3 < 0 and f0 = 0.01 spherical void computational test matrix...... 108

5-3 k = −0.3098, J3 > 0 and f0 = 0.01 spherical void computational test matrix... 108

5-4 k = −0.3098, J3 < 0 and f0 = 0.01 spherical void computational test matrix... 109

5-5 k = 0.3098, J3 > 0 and f0 = 0.01 spherical void computational test matrix.... 109

5-6 k = 0.3098, J3 < 0 and f0 = 0.01 spherical void computational test matrix.... 109

5-7 k = 0, J3 > 0 and f0 = 0.04 spherical void computational test matrix...... 111

5-8 k = 0, J3 < 0 and f0 = 0.04 spherical void computational test matrix...... 111

5-9 k = −0.3098, J3 > 0 and f0 = 0.04 spherical void computational test matrix... 111

5-10 k = −0.3098, J3 < 0 and f0 = 0.04 spherical void computational test matrix... 112

5-11 k = 0.3098, J3 > 0 and f0 = 0.04 spherical void computational test matrix.... 112

5-12 k = 0.3098, J3 < 0 and f0 = 0.04 spherical void computational test matrix.... 112

5-13 k = 0, J3 > 0 and f0 = 0.14 spherical void computational test matrix...... 114

5-14 k = 0, J3 < 0 and f0 = 0.14 spherical void computational test matrix...... 114

5-15 k = −0.3098, J3 > 0 and f0 = 0.14 spherical void computational test matrix... 114

5-16 k = −0.3098, J3 < 0 and f0 = 0.14 spherical void computational test matrix... 115

5-17 k = 0.3098, J3 > 0 and f0 = 0.14 spherical void computational test matrix.... 115

5-18 k = 0.3098, J3 < 0 and f0 = 0.14 spherical void computational test matrix.... 115 7-1 k = 0 plane strain cylindrical void computational test matrix...... 172 7-2 k = −0.3098 plane strain cylindrical void computational test matrix...... 172 7-3 k = 0.3098 plane strain cylindrical void computational test matrix...... 173 8-1 Transversely isotropic CPB06 constants...... 198

7 LIST OF FIGURES Figure page 1-1 Particle cracking example...... 24 1-2 Particle decohesion example...... 25 1-3 Void in single crystal...... 26 2-1 CPB06 yield condition π-plane representation...... 43 2-2 CPB06 plastic multiplier rate π-plane representation...... 45 3-1 Ductile crack in aluminum plate...... 74 3-2 Representative volume element for a sphere containing a spherical void...... 75 3-3 Hydrostatically-loaded hollow sphere...... 75 3-4 Macroscopic ductile yield surfaces with tension-compression asymmetry...... 76 3-5 Evolution of macroscopic ductile yield surfaces with porosity...... 77 5-1 Axisymmetric unit cell for the spherical void...... 106

5-2 f0 = 0.01 axisymmetric ﬁnite element mesh for the unit cell...... 107

5-3 f0 = 0.04 axisymmetric ﬁnite element mesh for the unit cell...... 110

5-4 f0 = 0.14 axisymmetric ﬁnite element mesh for the unit cell...... 113 5-5 k = 0 with f = 0.01: FE versus analytical...... 116 5-6 k = 0 with f = 0.01, f = 0.04 and f = 0.14: FE versus analytical...... 116 5-7 k = −0.3098 with f = 0.01: FE versus analytical...... 117 5-8 k = 0.3098 with f = 0.01: FE versus analytical...... 117 5-9 k = −0.3098 with f = 0.04: FE versus analytical...... 118 5-10 k = 0.3098 with f = 0.04: FE versus analytical...... 118 5-11 k = −0.3098 with f = 0.14: FE versus analytical...... 119 5-12 k = 0.3098 with f = 0.14: FE versus analytical...... 119 5-13 k = 0 axisymmetric FE data versus analytical yield curves...... 120 5-14 k = −0.3098 axisymmetric FE data versus analytical yield curves...... 120 5-15 k = 0.3098 axisymmetric FE data versus analytical yield curves...... 121

8 6-1 Voids and shear band in titanium...... 157 6-2 Cylindrical RVE...... 158

6-3 σT /σC = 1.21 yield curve parametric representation for one quadrant...... 159

6-4 σT /σC = 1.21 yield curve parametric representation...... 159

6-5 σT /σC = 0.82 yield curve parametric representation...... 160 6-6 f = 0.01 cylindrical macroscopic yield curves...... 160 6-7 f = 0.04 cylindrical macroscopic yield curves...... 161 6-8 f = 0.14 cylindrical macroscopic yield curves...... 161 7-1 Plane strain geometry used in ﬁnite element calculations...... 169 7-2 Valid triaxiality angles for plane strain...... 170

7-3 f0 = 0.01 plane strain ﬁnite element mesh for the unit cell...... 170

7-4 f0 = 0.04 plane strain ﬁnite element mesh for the unit cell...... 171

7-5 f0 = 0.14 plane strain finite element mesh for the unit cell...... 171 7-6 Plain strain yield point determination...... 173 7-7 k = 0, no fitting parameters: FE versus analytical...... 174 7-8 k = −0.3098, no fitting parameters: FE versus analytical...... 175 7-9 k = 0.3098, no fitting parameters: FE versus analytical...... 176 7-10 k = 0, with fitting parameters: FE versus analytical...... 177 7-11 k = −0.3098, no fitting parameters: FE versus analytical...... 178 7-12 k = 0.3098, no fitting parameters: FE versus analytical...... 179 8-1 Transversely isotropic RVE...... 198 8-2 Anisotropic effective stress versus effective strain curves...... 199

8-3 Plane stress yield loci for materials A, B and C with σT = σC ...... 199

8-4 Plane stress yield loci for materials A, B and C with σT < σC ...... 200

8-5 Plane stress yield loci for materials A, B and C with σT > σC ...... 200 8-6 k = 0 Material A deviatoric plot...... 201 8-7 k < 0 Material A deviatoric plot...... 201

9 8-8 k > 0 Material A deviatoric plot...... 202 8-9 k = 0 Material B deviatoric plot...... 202 8-10 k < 0 Material B deviatoric plot...... 203 8-11 k > 0 Material B deviatoric plot...... 203 8-12 k = 0 Material C deviatoric plot...... 204 8-13 k < 0 Material C deviatoric plot...... 204 8-14 k > 0 Material C deviatoric plot...... 205 8-15 Material A theoretical yield curves versus FE data...... 206 8-16 Material B theoretical yield curves versus FE data...... 207 8-17 Material C theoretical yield curves versus FE data...... 208 8-18 Anisotropic ductile yield surfaces with tension-compression asymmetry...... 209

10 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy DEVELOPMENT OF YIELD CRITERIA FOR DESCRIBING THE BEHAVIOR OF POROUS METALS WITH TENSION-COMPRESSION ASYMMETRY By Joel B. Stewart August 2009 Chair: Oana Cazacu Major: Mechanical Engineering A significant difference between the behaviors in tension versus compression is obtained at the polycrystal level if either twinning or non-Schmid effects are included in the description of the plastic deformation at the single crystal level. Examples of materials that exhibit tension-compression asymmetry include hexagonal close packed (HCP) polycrystals and intermetallics (e.g., molybdenum compounds). Despite recent progress in modeling the yield behavior of such materials, the description of damage by void growth remains a challenge. This dissertation is devoted to the development of macroscopic plastic potentials for porous metallic aggregates in which the void-free, or matrix, material displays tension- compression asymmetry. Using a homogenization approach, new analytical plastic potentials for a random distribution of voids are obtained. Both spherical and cylindrical void geometries are considered for void-matrix aggregates containing an isotropic matrix, while spherical voids are considered for the case of an anisotropic matrix material. The matrix plastic behavior in all cases is described by a yield criterion that captures strength differential effects and can account for the anisotropy that may be exhibited in the void- free material. For the case when the matrix material is isotropic, the developed analytical potentials for the void-matrix aggregate are sensitive to the second and third invariants of the stress deviator and display tension-compression asymmetry. Furthermore, if the matrix material

11 has the same yield strength in tension and compression, the developed criteria reduce to the classical Gurson criteria for either spherical or cylindrical voids. It has also been demonstrated that the developed isotropic criterion for porous aggregates containing spherical voids captures the exact solution of a hollow sphere loaded in hydrostatic tension or compression. Finite element cell calculations with the matrix material obeying an isotropic yield criterion and displaying tension-compression asymmetry were performed and the comparison between finite element calculations and theoretical predictions demonstrate the versatility of the proposed formulations. A new anisotropic potential for the porous aggregate was also developed for the case when the matrix material is anisotropic and displays tension-compression asymmetry. If the matrix is isotropic, the proposed analytical anisotropic criterion reduces to the isotropic criterion developed in this dissertation for a void-matrix aggregate containing spherical voids. Comparison between finite element calculations and theoretical predictions show the predictive capabilities of the developed anisotropic formulation. The yield criteria developed in this dissertation are the only criteria available to capture the influence of damage by void growth in HCP metals and other materials that exhibit tension-compression asymmetry.

12 CHAPTER 1 INTRODUCTION The aim of this dissertation is to develop analytical plastic potentials for metals that exhibit tension-compression asymmetry (e.g., certain BCC materials such as molybdenum and HCP metals such as α-titanium) and that contain either cylindrical or spherical voids. Closed-form yield criteria for porous materials exhibiting tension-compression asymmetry (i.e., that have different yield strengths in tension versus compression) do not currently exist in the literature. The successful derivation of analytic yield criteria for these types of materials should give researchers an additional and more accurate tool for dealing with damage in these materials. Ductile failure in metals occurs due to the nucleation, growth and coalescence of voids (e.g., see McClintock, 1968; Rousselier, 1987). Additionally, some localization phenomena that commonly lead to failure in metallic structures (e.g., adiabatic shear band formation) are thought to be influenced by micro-voids in the matrix material (see, for example Batra and Lear, 2005; Tvergaard, 1980). The void distribution in a material can exist because of pre-existing voids (e.g., manufacturing defects) or because of nucleation at second- phase particles. For example, Figure 1-1 illustrates void nucleation due to cracking of the inclusion (as well as some decohesion at the inclusion-matrix interface). Figure 1-2 shows an example of void nucleation due to decohesion of the matrix material at the inclusions. Voids can also nucleate in single crystals that contain neither pre-existing voids nor inclusions (see, for example, Cuitiñoand Ortiz, 1996; Lubarda et al., 2004). Cuitiñoand Ortiz(1996) proposed that vacancy condensation can act as the nucleating mechanism in single crystals for low strain rates. Lubarda et al.(2004) found that vacancy condensation cannot account for void nucleation at the extreme conditions seen, for example, in laser- driven shock experiments (with stress pulse duration on the order of 10 ns) and proposed dislocation emission as an alternate nucleating mechanism at this higher strain rate

13 regime. Figure 1-3 shows an example from Lubarda et al.(2004) of a void that has formed in monocrystalline copper. There have been a great deal of experimental investigations to assess void evolution under loading and its influence on the load-carrying capacity. For example, Benzerga et al. (2004a) performed a number of experiments on medium carbon low alloy steel in order to investigate the role of plasticity and porous microstructure on this material’s failure. The steel being investigated exhibited anisotropic behavior (both due to anisotropic void evolution and plastic anisotropy) as well as mild tension-compression asymmetry. Both round bar specimens were used (to characterize the material’s anisotropy and investigate failure properties) and notched round bar specimens (to investigate the influence of stress triaxiality on failure). In these tests, the initial microstructure was characterized and interrupted tests were performed to track the evolution of porous properties including average values of void volume fraction, void aspect ratio and void spacing ratio. The experiments show void nucleation occurring at very low strain levels at MnS and oxide inclusions either by particle cracking followed by decohesion or debonding after cracking (depending on the loading direction). At a certain point, the micro voids were observed to coalesce due to micro-necking of the ligaments between neighboring voids. The specimens also exhibited anisotropic crack propagation (i.e., the propagation pattern depended on the loading direction). For all specimens, the void growth rate was found to be mainly confined to the center region of the specimen with the extensional void growth rate dominating at low triaxialities and the radial void growth rate becoming dominant with increasing stress triaxiality. The authors also compared their experimental data to theoretical results using finite element calculations (see Benzerga et al., 2004b, and the discussion of the model used therein later in this section). Because of the relationship between material porosity and ductile failure, the ability to accurately describe the evolution of voids in a ductile metal is crucial to being able to accurately predict the failure of the material. Unfortunately, computational constraints

14 make it prohibitively expensive to model each of the micro-voids in most engineering structures; therefore, the method of explicitly tracking the evolution of each micro-void is not practical at this time. An alternative to explicitly tracking the evolution of each void (i.e., tracking microscopic, or local, quantities) is to incorporate the effects of the micro-voids into the macroscopic, or average, properties (such as macroscopic stress, strain, yielding, etc.). Since the rate of dilatation of the porous solid is related to the void growth rate, plastic potentials for the porous solid must be developed in order to describe the void growth. The most widely used plastic potential for porous solids was proposed by Gurson(1977). For the case of a spherical void geometry, the unit cell or Representative Volume Element (RVE) considered was a spherical shell, while a cylindrical tube RVE was used for the cylindrical void analysis. To derive an analytic expression for the plastic potential (and, thus, for the yield criterion, assuming associated plasticity), Gurson performed a limit load analysis on the RVE. The work in this dissertation mainly follows the homogenization procedure outlined in Gurson(1977) where Gurson developed analytic yield criteria for ductile materials containing either spherical or cylindrical voids. In Gurson’s analysis, it was assumed that the virgin material (void-free) obeys the classical von Mises yield criterion. To obtain the plastic potential, minimization of the plastic energy was done for a specific velocity field compatible with uniform strain rate boundary conditions. Thus, the obtained criterion is an upper bound of the exact plastic potential (since the plastic energy was minimized for only one velocity field rather than for the complete set of kinematically admissible velocity fields). Gurson’s yield criterion is given for spherical voids as

2 S Σe 3Σm 2 ΦG = + 2f cosh − 1 − f = 0 (1–1) σY 2σY and for cylindrical voids as √ 2 ! C Σe 3Σγγ 2 ΦG = Ceqv + 2f cosh − 1 − f = 0 (1–2) σY 2σY

15 where   (1 + 3f + 24f 6)2 for plane strain Ceqv = (1–3)  1 for axisymmetry.

In the previous expressions, Σe is the macroscopic von Mises effective stress, σY is the yield strength of the virgin material, f is the void volume fraction (sometimes referred to as the porosity), Σm is the macroscopic mean stress and Σγγ is the sum of the in-plane stresses (e.g., Σγγ = Σ11 + Σ22 if the 3-direction is the out-of-plane direction). Note that Gurson’s criteria depends not only on the second invariant of the stress deviator, but also on the pressure (or the mean stress) and on the level of porosity in the material; therefore, the yield function incorporates the influence of voids on the plastic deformation in a material. The main premise behind incorporating this void effect is that a volume of virgin material will behave differently under an applied load than will a volume of material that has a reduced load-bearing area due to the presence of voids and their subsequent evolution (growth and coalescence). Gurson’s spherical void criterion is considered most often in the literature but the cylindrical void criterion is applicable to certain problems as well (e.g., the plane stress analysis of sheet metal). The void volume fraction, f, (the ratio of the void volume to the total volume) evolves both from the nucleation at second-phase particles and the growth of existing voids (see, for example, Chu and Needleman, 1980; Lemaitre and Desmorat, 2005) such that

˙ ˙ ˙ f = fgrowth + fnucleation (1–4) where the rate of change due to growth is determined from the assumption of plastic incompressibility as ˙ P fgrowth = (1 − f) dkk (1–5)

16 P with dij being the plastic part of the rate of deformation tensor. The void volume fraction’s rate of change due to nucleation is generally given as

B f˙ = A σ˙ + σ σ˙ (1–6) nucleation σ Y 3 kk

for stress-controlled nucleation, or as

˙ fnucleation = AN ¯˙ (1–7)

for plastic strain-controlled nucleation, where ¯ is obtained from the following relation:

P σijdij = (1 − f) σY .¯˙ (1–8)

In the case of plastic strain-controlled nucleation, the following statistical expression has been frequently used: " 2# fN 1 ¯− N AN = √ exp − (1–9) sN 2π 2 sN

where fN is the volume fraction of void-nucleating particles, N is the mean nucleation

strain and sN is the standard deviation. A modified version of Gurson’s spherical criterion was used in Spitzig et al.(1988) to compare with experimental tests on compacted iron specimens. The modified criterion can be written as Σ 2 3mΣ Φ = e + 2f m cosh m − 1 − f 2m = 0 (1–10) σ¯ 2¯σ whereσ ¯ is the flow stress and m is a material coefficient representing strain hardening (a power law matrix model was used in the paper to compare with the experimental data). A widely used modification of Gurson’s spherical yield criterion was suggested in Tvergaard(1981) and Tvergaard(1982) based on comparisons with finite element calculations of shear band instabilities (where the instability is determined by a loss of ellipticity of the governing equations). With finite element calculations, the minimization of the plastic energy is done over a larger set of kinematically admissible velocity fields than in Gurson’s analysis (in which a single kinematically admissible velocity field is

17 assumed); therefore, adjustments to Gurson’s yield criteria can be proposed based on these calculations. In Tvergaard(1982), the finite element calculations involved a cylindrical cell containing a spherical void which was compared with the modified Gurson spherical yield criterion (i.e., compared with the modified form proposed in Tvergaard, 1981). Specifically, these calculations were meant to represent a periodic array of spherical voids which were arranged such that hexagonal representative volume elements (RVEs) could be fit together to form the structure (the cylindrical RVE used in the calculations was an approximation of the hexagonal RVE). Based on these axisymmetric finite element calculations, the following modified form of Gurson’s yield criterion was suggested:

2 Σe 3Σm 2 Φ = + 2fq1 cosh q2 − 1 − q3f = 0 (1–11) σY 2σY

where the qi are the ﬁtting parameters introduced by Tvergaard (all equal to one in

Gurson’s original expression). Tvergaard recommended values of q1 = 1.5, q2 = 1 and

2 q3 = q1 based on the ﬁnite element results. The introduction of these ﬁtting parameters

(q1, q2 and q3) can be thought of as a necessary adjustment of the yield surface to account for the inﬂuence of neighboring voids. Leblond and Perrin(1990) used a self-consistent analysis of a void within a porous plastic sphere (thus explicitly accounting for two diﬀerent void geometries and their interaction) to recommend values for Tvergaard’s

2 fitting parameters as q1 = 4/e ≈ 1.47, q2 = 1 and q3 = q1. Tvergaard and Needleman(1984) further modified Gurson’s spherical yield criterion to account for the onset of void coalescence leading to final material fracture. The authors used this modified yield criterion in both numerical and finite element calculations to compare with experimental data of a copper rod fracturing under uniaxial tension (exhibiting cup-cone fracture). This final modification is generally referred to as the GTN criterion (after the three authors) in the literature and is what is typically found in the

18 ﬁnite element codes. The GTN criterion is given below:

2 Σe ∗ 3Σm ∗ 2 Φ = + 2f q1 cosh q2 − 1 − q3 (f ) = 0 (1–12) σY 2σY

where f ∗ is the effective void volume fraction which is a function of the actual void volume fraction, f, and represents the modification of the yield criterion to account for final material failure. The effective void volume fraction is given as   f for f ≤ fC ∗  f (f) = ∗U (1–13) f − fC  fC + (f − fC ) for f > fC fF − fC where fC is the critical void volume fraction of a material, fF is the void volume fraction

∗U ∗ at final failure and f = f (fF ) is the ultimate value of the effective void volume fraction (i.e., the effective void volume fraction at which the macroscopic stress carrying capacity

∗U 2 vanishes) such that f = 1/q1 if q3 = q1 is used as suggested by Tvergaard. Richelsen and Tvergaard(1994) performed three-dimensional unit cell ﬁnite element calculations with proportional loading (i.e., the applied macroscopic stresses on the boundaries of the unit cell were constant multiples of each other for the duration of the calculation) to compare with the axisymmetric model of Equation (1–12) (note that Gurson’s spherical model is an axisymmetric model since the geometry of the RVE and the assumed velocity ﬁeld were both axisymmetric in the limit load analysis). For equal applied transverse stresses, the

GTN model using q1 = 1.5 was found to agree well with the unit cell calculations. Unit cell calculations with unequal applied transverse stresses also showed good agreement with the GTN model; speciﬁcally, the unit cell results for largely unequal transverse stresses

were found to lie between the theoretical results of q1 = 1.5 and q1 = 1.0 in the GTN model. In many cases, the micro-voids that form within a loaded material are ellipsoidal rather than spherical due to the inﬂuences of asymmetrical loading and/or an anisotropic

19 microstructure. Gologanu et al.(1993) used a homogenization procedure on an axisymmetric, ellipsoidal RVE containing a confocal, prolate ellipsoidal void to arrive at an analytic expression for prolate, ellipsoidal voids rather than for spherical voids. The general expression is given as

2 Σe κA :Σ 2 Φ = + 2fq1 cosh − 1 − q3f = 0 (1–14) σY σY where 1 √ ln (e /e )−1 κ = √ + 3 − 2 1 2 (1–15) 3 ln f and the void anisotropy tensor, A is given as

A = α (e2)(~ex ⊗ ~ex + ~ey ⊗ ~ey) + [1 − 2α (e2)]~ez ⊗ ~ez. (1–16)

In the previous equations, e1 and e2 are the eccentricities of the ellipsoidal void and

RVE, respectively, and should not be confused with the Cartesian basis vectors ~ex, ~ey

and ~ez. The evolution of the eccentricities is governed by the evolution of the void shape

parameter, S = ln(a1/b1), where a1 and b1 are the major and minor semi-axes of the void, respectively. The function, α, is given as

2 1 1 − ei −1 α (ei) = 2 − 3 tanh (ei). (1–17) 2ei 2ei

Gologanu showed that the model reduced to Gurson’s spherical criterion in the case of

a spherical void (e1 = e2 = 0) and to Gurson’s cylindrical criterion in the case of a

cylindrical void (e1 = e2 = 1). The formulas given in this paragraph are for Gologanu’s prolate ellipsoidal void analysis; Gologanu developed similar expressions for oblate voids (see, for example, Gologanu et al., 2001). Garajeu et al.(2000) extended the work of Gologanu by investigating the inﬂuence of void distribution evolution in addition to the inﬂuence of void shape evolution. Liao et al.(1997) developed a Gurson-type criteria for transversely isotropic metal sheets under plane stress conditions using Hill’s 1948 yield criterion (see Hill, 1948, 1950)

20 for the matrix material. The proposed model is as follows: s ! Σ 2 1 + 2R 3Σ Φ = e + 2f cosh m − 1 − f 2 = 0 (1–18) σY 2 (1 + R) 2σY

where the anisotropy parameter, R, is deﬁned as “the ratio of the transverse plastic strain rate to the through-thickness plastic strain rate under in-plane uniaxial loading conditions.” Note that the previous model accounts for porosity in transversely isotropic materials and reduces to Gurson’s cylindrical criterion for isotropy (R = 1). Benzerga and Besson(2001) also developed a Gurson-type criteria, intended for orthotropic porous metals, by assuming a matrix material that could be characterized using Hill’s 1948 criterion. The model is given as

0 0 3Σ : H :Σ 3Σm 2 Φ = 2 + 2fq1 cosh q2 − 1 − q3f = 0 (1–19) 2σY hσY where h is called the anisotropy factor (h = 2 for isotropy) and is a function of the macroscopic anisotropy tensor, H. Equation (1–19) diﬀers from Equation (1–18) in that it assumes spherical rather than cylindrical voids and applies to the more general case of orthotropic materials versus transversely isotropic materials. Besson and Guillemer-Neel(2003) extended the GTN model of Equation (1–12) to include mixed isotropic and kinematic hardening within a thermodynamical framework. State variables related to isotropic and kinematic hardening are employed in the model and are evolved in a thermodynamically-consistent manner using a dissipation potential (see Lemaitre and Chaboche, 1990). The extended model can be written as

2 Σe ∗ 3Σm ∗ 2 Φ = + 2f q1 cosh q2 − 1 − q3 (f ) = 0 (1–20) σ∗ 2σ∗

where σ∗ is an eﬀective scalar stress and is a function of the state variables related to isotropic and kinematic hardening. Besson and Guillemer-Neel obtained good agreement between the previous model and ﬁnite element calculations of cyclically-loaded unit cells.

21 Benzerga et al.(2004b) used a yield criterion which combines many of the properties of Equations (1–14) and (1–19) in order to compare with the experimental results given in Benzerga et al.(2004a). The yield criterion, prior to void coalescence, used in the paper is as follows: 0 0 3Σ : H :Σ κA :Σ 2 Φ = 2 + 2fqw cosh − 1 − qwf = 0 (1–21) 2σ∗ hσ∗ where qw = 1 + (q1 − 1)/ cosh(S) was introduced by Gologanu (see, for example, Gologanu et al., 2001). In this yield criterion, σ∗ replaces σY in the denominators to account for hardening as in Equation (1–20). The authors use a diﬀerent criterion to account for yield after coalescence (this approach can be viewed as analogous to the f ∗ parameter in the GTN model; an accurate ductile yield criterion must consider void coalescence separately from initial void growth). This post-coalescence criterion is given as q 3 Σ0 : H :Σ0 2 1 |3Σm| 3 2 Φ = + − 1 − χ Ψf (χ, S) = 0 (1–22) σ∗ 2 σ∗ 2

where χ is the ligament size ratio (related to the void spacing) and Ψf is a function of both χ and the void shape parameter, S. The authors showed good agreement between the experimental results obtained in Benzerga et al.(2004a) and the yield criteria given in this paragraph. Researchers at Los Alamos National Laboratory (LANL) have extended Gurson’s model (see, for example, Addesio and Johnson, 1993; Bronkhorst et al., 2006; Maudlin et al., 1999) to solve computational problems involving highly dynamic material behavior (i.e., high strain rates and high temperatures). This modiﬁed model is referred to as TEPLA and uses a Mie-Gruneisen equation of state to specify the pressure along with a mechanical threshold strength (MTS) model to account for rate and temperature eﬀects. The above works and many others have done an enormous amount of research aimed at extending Gurson’s analysis both to more complex materials as well as to more complex void shapes and distributions. It is worth noting that all of the models mentioned above are applicable to metallic materials with cubic structure; i.e., the models were developed

22 for face-centered-cubic metals (e.g., aluminum) and body-centered-cubic metals (e.g., steel). The aim of this research is to develop models to describe the yielding and failure of hexagonal-closed-packed metals such as α-titanium, zirconium, and uranium (i.e., materials that have different yield strengths in tension versus compression). The outline of this dissertation is as follows. The kinematic homogenization approach of Hill-Mandel (Hill, 1967; Mandel, 1972) that is used to develop the analytical plastic potentials of the void-matrix aggregate is described in Chapter2. Chapter3 details the development of a closed-form plastic potential for an isotropic porous aggregate containing spherical voids when the matrix displays tension-compression asymmetry. The return mapping algorithms used in implementing the matrix, or void-free, plastic potential into a finite element code is outlined in Chapter4 and the necessary first and second derivatives are given for the chosen matrix plastic potential of Cazacu et al.(2006), which can account for tension-compression asymmetry in the void-free material. In Chapter 5, comparisons are made between the developed isotropic porous plastic potential for spherical voids and finite element unit cell calculations. The finite element calculations are designed such that a spherical void is explicitly meshed inside an axisymmetric cylinder whose response is governed by the Cazacu et al.(2006) plastic potential. Macroscopic plastic potentials for void-matrix aggregates containing cylindrical voids when the matrix displays tension-compression asymmetry are developed in Chapter6 and a proposed plane strain plastic potential is compared to finite element unit cell calculations in Chapter7. Finally, the isotropic plastic potential developed in Chapter3 for a void-matrix aggregate containing spherical voids and a matrix exhibiting tension-compression asymmetry is extended in Chapter8 to include the effects of matrix anisotropy on the yielding of the void-matrix aggregate. The developed anisotropic macroscopic plastic potential is compared to transversely isotropic unit cell finite element calculations.

23 Figure 1-1. Void nucleation at MnS inclusions in steel. (a) and (b) Longitudinal loading. (c) and (d) Transverse loading. (e) Decohesion at the poles of a tiny spherical MnS particle close to a grain boundary and 100 lm ahead of the crack tip (2% Nital solution etched). Field emission SEM imaging using back-scattered electrons in (a)(c) and (e) and secondary electrons with an in-lens detector in (d) to reveal the crack. [Reprinted with permission. Benzerga, Besson, and Pineau(2004a). Anisotropic ductile fracture. Part I: Experiments. Acta Materialia, 52 , 4623-4638.]

24 Figure 1-2. Reconstructed images of the same internal section at five different relevant deformation steps of a metal composite consisting of an aluminum matrix with embedded spherical ceramic particles (a) initial state; (b) = 0.065; (c) = 0.15; (d) = 0.35-0.48; (e) = 0.51-0.81. Detail (A) pre-existing holes induced by the extrusion process which start to grow during tension. Detail (B) particle/matrix decohesion during the tensile loading. Detail (C) coalescence. The tensile direction is vertical in the figure. The various true strain values on (d) and (e) convey the presence of necking in the sample. [Reprinted with permission. Babout, Maire, Buffière,and Fougères(2001). Characterization by X-ray computed tomography of decohesion, porosity growth and coalescence in model metal matrix composites. Acta Materialia, 49, 2055-2063.]

25 Figure 1-3. TEM micrograph of laser-shocked monocrystalline copper: dark ﬁeld image of an isolated void near the rear surface of the specimen and associated work-hardened layer (white rim). [Reprinted with permission. Lubarda, Schneider, Kalantar, Remington, and Meyers(2004). Void growth by dislocation emission. Acta Materialia, 52, 1397-1408.]

26 CHAPTER 2 HOMOGENIZATION APPROACH All engineering metals and alloys contain inclusions and second-phase particles at which micro-voids nucleate either by decohesion of the particle-matrix interface or by particle breaking. Subsequently, voids grow due to plastic deformation of the surrounding material until a localized internal necking of the intervoid matrix occurs that leads to the formation of some macroscopic crack. Investigations of the expansion of voids of cylindrical and spherical geometries in rigid ideal plastic matrices by McClintock(1968) and Rice and Tracey(1969) have established the effects of stress state (stress triaxiality) on the void growth rate. Gurson(1977) proposed approximate yield criteria and flow rules for ductile materials containing spherical or cylindrical cavities using an upper-bound approach. To better account for the interaction between voids, several modifications of Gurson’s criterion have been proposed based on results of two-dimensional finite-element studies (e.g., Koplik and Needleman, 1988; Tvergaard, 1981; Tvergaard and Needleman, 1984) or from rigorous estimates of the exact macroscopic potentials (for a review of those alternative approaches pioneered by Talbot and Willis 1985, see for example, Garajeu and Suquet 1997; Leblond et al. 1994). Recently, analytical criteria that account for the combined effects of void shape and matrix anisotropy on the macroscopic response of ductile porous solids were proposed (see for example Benzerga and Besson, 2001; Monchiet et al., 2008). In all the models mentioned it is assumed that the matrix has the same yield in tension and compression. However, in the absence of voids, some cubic materials such as high strength steels or molybdenum exhibit tension-compression asymmetry. This strength-differential (S-D) effect is a consequence of crystal slip that does not obey the well-known Schmid law (see for example Vitek et al., 2004). Also, twinning at the single crystal level may result in a strong tension-compression asymmetry at the aggregate level in some cubic metals (see Hosford and Allen, 1973). Hexagonal close packed (HCP)

27 metals exhibit tension-compression asymmetry as a result of twinning activation at the single crystal level. HCP metals can deform either by slip or twinning, with twinning becoming increasingly prominent with increasing strain rate. If a metal deforms by slip alone (a reversible shear mechanism), irreversible ﬂow depends only on the magnitude of the resolved shear stress such that the strengths in tension and compression are equal; however, if twinning also exists as a deformation mechanism then a diﬀerence between the strengths in tension and compression will exist. This chapter is organized as follows. Section 2.1 introduces the homogenization approach due to Hill and Mandel (Hill, 1967; Mandel, 1972) that is used in developing the macroscopic plastic potentials for the void-matrix aggregates considered in this dissertation. Section 2.2 presents the isotropic version of the Cazacu et al.(2006) plastic potential that is used to describe the matrix material in all analyses except for those discussed in Chapter8 (which focuses on anisotropic plastic potentials). Section 2.3 presents the derivation of the plastic multiplier rate associated to the isotropic version of the Cazacu et al.(2006) plastic potential. The derivation of this plastic multiplier rate is the main challenge in determining the local plastic dissipation in the matrix. 2.1 Kinematic Homogenization Approach of Hill and Mandel

The current work will consider both a spherical and a cylindrical representative volume element (RVE) containing a void of similar geometry (i.e., spherical and cylindrical, respectively). These RVEs were chosen both because the void geometries considered are typically seen in experiments and because the symmetry of the void and outer surface greatly simpliﬁes the homogenization analysis. If the matrix material in the RVE is assumed to be rigid plastic, then any volume change is due exclusively to void evolution such that the rate of void growth is easily obtained from the rate of dilatation (see Equation (1–5)). Consider a representative volume element V , composed of a homogeneous rigid- plastic matrix and a traction-free void. The matrix material is described by a convex yield

28 function ϕ(σ) in the stress space and an associated ﬂow rule

∂ϕ d = λ˙ , (2–1) ∂σ where σ is the Cauchy stress tensor, d = (1/2)(∇v +∇vT ) denotes the rate of deformation tensor with v being the velocity ﬁeld, and λ˙ ≥ 0 stands for the plastic multiplier rate. The yield surface is deﬁned as ϕ(σ) = 0. Let C denote the convex domain delimited by the yield surface such that C = {σ|ϕ(σ) ≤ 0} . (2–2)

The plastic dissipation potential of the matrix is deﬁned as

w(d) = sup (σ : d) (2–3) σ∈C where “:” denotes the tensor double contraction. Uniform rate of deformation boundary conditions are assumed on the boundary of the RVE, ∂V , such that

v = Dx for any x ∈ ∂V (2–4) with D, the macroscopic rate of deformation tensor, being constant. For the boundary conditions of Equation (2–4), the Hill-Mandel (Hill, 1967; Mandel, 1972) lemma applies; hence,

hσ : diV = Σ : D, (2–5) where h i denotes the average value over the representative volume V , and Σ = hσiV . Furthermore, there exists a macroscopic plastic dissipation potential W (D) such that

∂W (D) Σ = (2–6) ∂D with

W (D) = inf hw(d)iV , (2–7) d∈K(D)

29 where K(D) is the set of incompressible velocity fields satisfying Equation (2–4) (for more details see Gologanu et al., 1997; Leblond, 2003). The matrix material being considered obeys the isotropic version of the pressure-insensitive yield criterion that captures strength differential effects of Cazacu et al.(2006). 2.2 Yield Criterion for the Matrix Material

Twinning and martensitic shear are directional deformation mechanisms and, if they occur, yielding will depend on the sign of the stress (see Hosford, 1993). Early polycrystalline simulations results by Hosford and Allen(1973) who analyzed deformation by twinning in random FCC polycrystals, predicted a yield stress in uniaxial tension 22% lower than that in uniaxial compression. Based on these simulations and more recent results concerning the effects of non-Schmid type yield criteria at the single-crystal level on the polycrystalline response (e.g., see Vitek et al., 2004), it can be concluded that yield loci with a strong asymmetry between tension and compression should be expected in any isotropic pressure-insensitive material that deforms either by twinning or directional slip. To account for strength differential effects in pressure insensitive materials, Plunkett (2005) and Cazacu et al.(2006) proposed the following isotropic form for yielding:

2 2 2 F = (|s1| − ks1) + (|s2| − ks2) + (|s3| − ks3) (2–8)

where si are the principal values of the Cauchy stress deviator. In Equation (2–8), k takes into account the tension-compression asymmetry and is given by

1 − h (σ , σ ) k = T C (2–9) 1 + h (σT , σC )

where v u 2 u 2 − σT u σC h (σ , σ ) = u (2–10) T C t 2 2 σT − 1 σC in the isotropic case. The terms σT and σC in the previous expressions are the yield strengths in tension and compression, respectively. The current work will sometimes refer

30 to Equation (2–8) along with the deﬁnitions in Equations (2–9) and (2–10) as the “CPB06 isotropic criterion” for notational convenience.

Let (e1, e2, e3) be a Cartesian coordinate system associated with the principal directions of the stress tensor. Due to the tension-compression asymmetry, the projection of the yield surface in Equation (2–8) in the deviatoric plane (the plane with normal n =

√1 e + √1 e + √1 e ) has threefold symmetry. Let f be the projections of the eigenvectors 3 1 3 2 3 3 i ei, i = 1...3 on the deviatoric plane. As an example, Figure 2-1 illustrates deviatoric

π-plane representations of the yield curves given by Equation (2–8) for σT /σC = 0.82 and σT /σC = 1.21, along with the von Mises yield locus for comparison. Note a very drastic departure of the yield locus from the Von Mises circle. This departure is due to a strong inﬂuence of the third invariant of the stress deviator on yielding, which results from tension-compression asymmetry. In the π-plane representation, the radial coordinate is related to the second invariant of the stress deviator while the angular coordinate is related to the third invariant of the stress deviator. Therefore, the von Mises circle is independent of the third invariant, while the CPB06 criterion for non-zero k depends on both the radial and angular coordinates (i.e., it is not a circle) such that it is dependent on both the second and third invariants of the stress deviator. The yield function (a.k.a. the stress potential), ϕ, is given as follows:

ϕ = φ − Y where φ is a scalar eﬀective stress associated to Equation (2–8) and Y represents the material’s hardening condition. If Y is taken to be the yield strength in tension, σT , then the previous equation yields

ϕ (sα, k, σT ) = σe (sα, k) − σT (2–11)

where σe is now the eﬀective stress formulated such that it reduces to the yield strength in tension for uniaxial loading. The explicit expression for this eﬀective stress may be found

31 by writing Equation (2–8) for the uniaxial tension case and forcing σe equal to σT . The resulting expression is given as √ σe = m F v u 3 (2–12) uX 2 = mt (|si| − ksi) i=1 where s 9 1 m := . (2–13) 2 3k2 − 2k + 3 p Notice that for k = 0, m = 3/2 and σe simply becomes the von Mises eﬀective stress. Also, note that for the uniaxial compression case,

r3k2 + 2k + 3 σ = σ e C 3k2 − 2k + 3 σT = σC σC

= σT such that the yield surface of Equation (2–11) is indeed satisﬁed. Since the Cazacu et al.(2006) criterion is homogeneous of degree one, Equation (2–1) yields (assuming rigid plastic behavior in the matrix)

P P w d = σijdij ˙ ∂f = σijλ (2–14) ∂σij ˙ = λσT

such that the determination of the local plastic dissipation w(dP ) reduces to determining an expression for the plastic multiplier rate, λ˙ . In the next section, the plastic multiplier rate corresponding to the isotropic version of the Cazacu et al.(2006) criterion will be derived.

32 2.3 Plastic Multiplier Rate Derivation

A key step in obtaining the overall plastic potential, W, of the void-matrix aggregate is the calculation of the local plastic dissipation, w (see Equations (2–3) and (2–14)), associated with the Cazacu et al.(2006) criterion, the isotropic version of which is shown

in Equation (2–12). Since the matrix material is isotropic, the eigenvectors ei, i = 1...3 of the Cauchy stress tensor σ are also eigenvectors of the rate of deformation tensor d. Substituting in the ﬂow rule Equation (2–1) the expression in Equation (2–12) of the

plastic potential, ϕ(σ), it follows that the principal values of the strain rate tensor di are expressed as

P ˙ ∂σe dα = λ ∂σα ∂σ ∂s = λ˙ e i ∂si ∂σα 2 ∂σ 1 ∂σ 1 ∂σ = λ˙ e − e − e 3 ∂sα 3 ∂sβ 3 ∂sγ where α 6= β 6= γ. Making use of the relation

∂σ ∂σ ∂σ e = − e + e ∂sα ∂sβ ∂sγ yields P ˙ ∂σe dα = λ . (2–15) ∂sα The derivative terms in Equation (2–15) must now be determined. Noting that the three stress deviators are dependent functions of each other, the eﬀective stress is written as follows:

( 2 2s1 − s2 − s3 2s1 − s2 − s3 σe =m − k 3 3 2 2s2 − s1 − s3 2s2 − s1 − s3 + − k 3 3 1 2) 2 2s3 − s2 − s1 2s3 − s2 − s1 + − k 3 3

33 where the fact that s1 + s2 + s3 = 0 has been used. Now, taking derivatives of the above expression yields: ∂σe m 2 1 = √ (|s1| − ks1) (sgn(s1) − k) − (|s2| − ks2) (sgn(s2) − k) ∂s1 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 ∂σe m 2 1 = √ (|s2| − ks2) (sgn(s2) − k) − (|s1| − ks1) (sgn(s1) − k) ∂s2 F 3 3 (2–16) 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 ∂σe m 2 1 = √ (|s3| − ks3) (sgn(s3) − k) − (|s1| − ks1) (sgn(s1) − k) ∂s3 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 2 2 2

Note that ∂σ ∂σ ∂σ e + e + e = 0. ∂s1 ∂s2 ∂s3 Equations (2–15) and (2–16) can now be used in order to come up with expressions

P ˙ for the principal plastic rate of deformation components, dα , in terms of sα, λ, F and k as follows: ˙ P mλ 2 1 d = √ (|s1| − ks1) (sgn(s1) − k) − (|s2| − ks2) (sgn(s2) − k) 1 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 mλ˙ 2 1 dP = √ (|s | − ks ) (sgn(s ) − k) − (|s | − ks ) (sgn(s ) − k) 2 3 2 2 2 3 1 1 1 F (2–17) 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 ˙ P mλ 2 1 d = √ (|s3| − ks3) (sgn(s3) − k) − (|s1| − ks1) (sgn(s1) − k) 3 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 2 2 2 where the three components sum to zero since dP is a deviatoric tensor. Equation (2–17) contains the general expressions for the principal rate of deformation components in terms ˙ of sα, λ, F and k. In order to simplify the expressions further, some assumption regarding

34 the sign of the principal stress deviators, sα, must be made. There are two possible cases corresponding to the following stress states:

1. The lone principal stress deviator is negative such that s1 > 0; s2 ≥ 0; s3 < 0 ⇔ J3 ≤ 0.

2. The lone principal stress deviator is positive such that s1 > 0; s2 ≤ 0; s3 < 0 ⇔ J3 ≥ 0.

The term J3 above is the third invariant of the deviatoric stress tensor and can be expressed as follows:

J3 = s1s2s3.

Relationships between various functions of k, m and the ratio of the yield strengths in tension and compression can be derived using Equations (2–9), (2–10), and (2–13). Some useful relations are given in Table 2-1 and will be used throughout the remainder of this document.

2.3.1 Plastic multiplier rate when J3 ≤ 0

For the case when J3 ≤ 0, Equation (2–17) is used to obtain ˙ P mλ 1 2 2 d = √ (2s1 − s2) (1 − k) + (s1 + s2) (1 + k) 1 F 3 ˙ mλ 1 2 2 = √ (2s1 − s2) k − 2k + 1 + (s1 + s2) k + 2k + 1 (2–18) F 3 ˙ mλ 1 2 = √ s1 3k − 2k + 3 + 4ks2 F 3 and ˙ P mλ 1 2 2 d = √ (2s2 − s1) (1 − k) + (s1 + s2) (1 + k) 2 F 3 (2–19) ˙ mλ 1 2 = √ s2 3k − 2k + 3 + 4ks1 . F 3 Note that

P P P d3 = − d1 + d2 ˙ mλ 1 2 = √ s3 3k + 2k + 3 F 3

35 P such that d3 < 0 since s3 < 0. Also, ˙ P P mλ 2 d − d = √ (s1 − s2) (1 − k) 1 2 F

P P such that d1 > d2 since s1 > s2. Now, from Equation (2–18), √ " ! # 1 P 3 F s1 = d − 4ks2 3k2 − 2k + 3 1 mλ˙ and from Equation (2–19), √ " ! # 1 P 3 F s2 = d − 4ks1 . 3k2 − 2k + 3 2 mλ˙

Solving the above two expressions for s1 and s2 yields, √ ! F P P s1 = c1d + c2d mλ˙ 1 2 √ ! F P P s2 = c2d + c1d mλ˙ 1 2 where 3k2 − 2k + 3 c = 1 3k4 − 4k3 + 2k2 − 4k + 3 1 2 3k2 − 2k + 3 = 1 − k 3k2 + 2k + 3 1 2 σ 2 = C 1 − k σT and −4k c = 2 3k4 − 4k3 + 2k2 − 4k + 3 1 2 −4k = 1 − k 3k2 + 2k + 3 " # 1 2 σ 2 = C − 1 1 − k σT

36 using some of the relations given in Table 2-1. Now, using the above equations along with Equation (2–8) and the relation

s3 = − (s1 + s2) yields the following:

2 2 2 2 2 F = s1 + s2 (1 − k) + s3 (1 + k) F n h P P 2 P P 2i 2 = c1d + c2d + c2d + c1d (1 − k) m2λ˙ 2 1 2 1 2 2 P P 2 2 o + (c1 + c2) d1 + d2 (1 + k) F n P 2 P 2 2 2 2 2 2 = d + d c + c (1 − k) + (c1 + c2) (1 + k) m2λ˙ 2 1 2 1 2 P P 2 2 2 o + d1 d2 4c1c2 (1 − k) + 2 (c1 + c2) (1 + k) .

After some algebra, and taking into account the following relations,

1 2 c − c = 1 2 1 − k 3 c + c = 1 2 3k2 + 2k + 3 ˙ one can obtain the following expression for the plastic multiplier rate, λ1, corresponding to the case when J3 ≤ 0:

2 2 2 1 2 2 3k − 10k + 3 2 λ˙ = dP + dP + dP . 1 m (1 − k) 1 2 3k2 + 2k + 3 3

The above equation can be broken out into a contribution from the second invariant of the plastic rate of deformation tensor, and a contribution from the component associated with

P the lone negative principal component (i.e., d3 ). The expression is written below as

2 i ˙ h P 2 P 2 P 2 P 2 λ1 = A1 d1 + d2 + d3 + B1 d3 (2–20)

37 where

1 2 A = 1 m (1 − k)   (2–21) 2 1 =   3  2  2 − σT σC

and

1 2 −12k B = 1 m (1 − k) 3k2 + 2k + 3  2  σC (2–22) σ − 1 = 2  T  .  2  2 − σT σC

In the above expressions, m is deﬁned by Equation (2–13). Note that Equation (2–20)

reduces to von Mises if k = 0 (i.e., if σT = σC ). Also, note that the plastic multiplier rate ˙ for this case, λ1, is singular for k = 1.

2.3.2 Plastic multiplier rate when J3 ≥ 0 Equation (2–17) can be used again to determine the expressions for the ﬁrst two principal strain rates corresponding to this case as follows: ˙ P mλ 1 2 2 d = √ 2s1 (1 − k) + s1 (1 + k) 1 F 3 (2–23) ˙ mλ 1 2 = √ 3k − 2k + 3 s1 F 3 and ˙ P mλ 1 2 2 d = √ (3s2 + s1) (1 + k) − s1 (1 − k) 2 F 3 (2–24) ˙ mλ 1 2 = √ 3s2 (1 + k) + s1 (4k) . F 3

P Note that, similarly to the previous section, d1 > 0 since s1 > 0 and ˙ P P mλ 2 d − d = √ (s2 − s3) (1 + k) 2 3 F

P P such that d2 > d3 since s2 > s3.

38 Solving Equations (2–23) and (2–24) for the principal stress deviators similar to the procedure used in the previous section now yields √ ! 3 F 1 P s1 = d mλ˙ 3k2 − 2k + 3 1 and √ 2 " ! # 1 1 3 F P s2 = d − 4ks1 . 3 1 + k mλ˙ 2 Plugging the previous equations into Equation (2–8) yields the following:

2 2 2 2 2 F = s1 (1 − k) + s2 + s3 (1 + k) ( 2F 3 (k2 + 1) dP 2 −4k 2 1 2 = 1 + dP + dP m2λ˙ 2 3k2 − 2k + 3 2 3k2 − 2k + 3 1 1 + k ) 3 −4k + dP + dP dP 3k2 − 2k + 3 2 3k2 − 2k + 3 1 1

2 " 4 3 2 2F 1 2 9k + 6k + 10k + 6k + 9 = dP m2λ˙ 2 1 + k 1 (3k2 − 2k + 3)2 # P 2 P P + d2 + d1 d2

2 2 F 1 2 3k + 10k + 3 2 2 = dP + dP + dP . m2λ˙ 2 1 + k 1 3k2 − 2k + 3 2 3 Using the previous relation, one can now obtain the following expression for the plastic ˙ multiplier rate, λ2, corresponding to the case when J3 ≥ 0:

2 2 2 1 3k + 10k + 3 2 2 2 λ˙ = dP + dP + dP . 2 m (1 + k) 3k2 − 2k + 3 1 2 3

Writing the previous equation in the same form as in the previous section yields the following: 2 ˙ h P 2 P 2 P 2i P 2 λ2 = A2 d1 + d2 + d3 + B2 d1

39 where

1 2 A = 2 m (1 + k)   2 1 =   3  2  2 σT − 1 σC and

1 2 12k B = 2 m (1 + k) 3k2 − 2k + 3  2  σT σ − 1 = 2  C  .  2  2 σT − 1 σC

Note that the following relationship exists between the parameters A2 and B2 above with the parameters A1 and B1 of Equation (2–20):

A2 = C2A1 2 σT B2 = −C2 B1 σC where

1 − k 2 C = 2 1 + k  2  σT (2–25) 2 − σ =  C  .  2  2 σT − 1 σC

Using the relationship between A1, A2, B1 and B2 yields the following expression for the

plastic multiplier rate, corresponding to the case when J3 ≥ 0, in terms of A1 and B1:

2 n h i 2 o ˙ P 2 P 2 P 2 σT P 2 λ2 = C2 A1 d1 + d2 + d3 − B1 d1 . (2–26) σC

Note that, as was the case with Equation (2–20), Equation (2–26) reduces to von Mises if

k = 0 (i.e., if σT = σC ). Also, note that the above plastic multiplier rate also contains a singularity, this time at k = −1.

40 2.3.3 General plastic multiplier rate expression

It can be shown that both plastic multiplier rates (2–20) and (2–26) are equivalent for

the case when J3 = 0 and reduce to the following expression:

2 k + 1 2 λ˙ 2 = 4 dP 3k2 − 2k + 3 1 " 2# 2 σT P 2 = 1 + d1 3 σC

˙ 2 P 2 The previous expression further reduces to λ = (4/3) d1 for the von Mises case of k = 0 (i.e, if the material has the same strengths in tension and compression) as would

P P P be expected when J3 = 0 such that d2 = 0 and d3 = −d1 (refer to Equations (2–17) for k = 0). It is convenient to write a single, general expression for λ˙ that includes all possible stress states. Equations (2–20) and (2–26) can be combined to yield the following general plasticity multiplier:

2 ˙ P 2 h P 2i P 2 λ = z1 d1 + z2tr d + z3 d3 (2–27)

where  2  2 σT − 2 1 σC z := sgn2 (J ) + sgn (J )   (2–28a) 1 2 3 3  2  2 σT − 1 σC     2/3 1 2/3 2/3 z :=   + sgn2 (J ) + sgn (J )  −  (2–28b) 2  2  2 3 3  2 2  2 − σT 2 σT − 1 2 − σT σC σC σC  2   2  2 σC − 2 2 σC − 2 σT 1 σT z :=   − sgn2 (J ) + sgn (J )   (2–28c) 3  2  2 3 3  2  2 − σT 2 − σT σC σC with

sgn (J3) = 1 J3 > 0

sgn (J3) = 0 if J3 = 0

sgn (J3) = −1 J3 < 0.

41 Table 2-2 shows the values for the parameters z1, z2 and z3 as functions of the material yield strengths and sign of the third invariant of the stress deviator. Note that, by ˙ Equation (2–28), Equation (2–27) reduces to the expression for λ1, given by Equation ˙ ˙ (2–20), when J3 = 0. This equation is valid since both λ1 and λ2 are valid and, indeed,

equivalent for J3 = 0. An alternate way of writing Equation (2–27) is to express everything in terms of the

rate of deformation components. Noting that s2s3 ≥ 0 for J3 ≥ 0 and s1s2 ≥ 0 for J3 ≤ 0, the plastic multiplier rate can be expressed as follows:

 r s  2 (3β2 − 2) (dP )2 + (dP )2 + (dP )2 dP 1  1 2 3 if 1 ≥  2 q p  3 (2β − 1) dP dP 2 (β4 − β2 + 1) ˙  ij ij λ = r s  2 (dP )2 + (dP )2 + (3 − 2β2)(dP )2/β2 dP −β2  1 2 3 if 3 ≤  2 q p  3 (2 − β ) P P 2 (β4 − β2 + 1)  dijdij (2–29)

P P P where d1 ≥ d2 ≥ d3 are the ordered principal components of the plastic rate of

deformation tensor and β = σT /σC . Note that the ﬁrst branch in Equation (2–29) corresponds to J3 ≥ 0 and the second branch to J3 ≤ 0. Figure 2-2 shows the representation of the plastic multiplier rate in the deviatoric

− − plane for β = σT /σC = 0.82. Each of the half-lines L1 and L2 form with (−f3) an angle θ−, expressible in terms of β as √ 3 2 − β2 tan(θ−) = . (2–30) 3 β2

+ + The angle between L1 and (f3) or L2 and (f3) is given by √ 3 tan(θ+) = 2β2 − 1 . (2–31) 3

Note that if the matrix material displays tension-compression asymmetry (β 6= 1) then θ− 6= θ+. Using Equations (2–30) and (2–31) it can be easily shown that

π θ− + θ+ = . (2–32) 3

42 Equation (2–32) shows that the plastic multiplier rate (i.e. the exact dual of the plastic potential in Equation (2–12)) has three-fold symmetry, which is a consequence of the material isotropy. Note that when the yield strengths in tension and compression are equal (i.e., von Mises), the plastic multiplier rate has six-fold symmetry such that θ+ = θ− = π/6. Note also that in this case Equation (2–29) reduces to

r2 λ˙ = dP dP (2–33) 3 ij ij

which is the classic von Mises eﬀective plastic strain rate.

σT = σC 3 σT < σC σT > σC

1 2

Figure 2-1. Π-plane representation of the isotropic version of the CPB06 yield criterion for k = 0 (von Mises circle), k = −0.3098 (σT < σC ) and k = 0.3098 (σT > σC ).

43 Table 2-1. Various relationships between k-expressions and yield strength ratios. k-expressions strength-ratio expressions 2 2 3k + 2k + 3 σT 2 3k − 2k + 3 σC

2 2 3k + 10k + 3 σT 2 3 − 2 3k − 2k + 3 σC

2 2 3k − 10k + 3 σC 2 3 − 2 3k + 2k + 3 σT

2 −4k σC 2 − 1 3k + 2k + 3 σT

2 " 2 # k + 1 1 σT 2 + 1 3k − 2k + 3 6 σC

  1 2 2 1   m (1 − k) 3  2  2 − σT σC   1 2 2 1   m (1 + k) 3  2  2 σT − 1 σC

Table 2-2. z-parameters as functions of material yield strengths and sign of J3.

J3 > 0 J3 ≤ 0 2 σT 2 σ − 2 z = C z = 0 1 2 1 2 σT − 1 σC 2/3 2/3 z = z = 2 2 2 2 2 σT − 1 2 − σT σC σC 2 σC 2 σ − 2 z = 0 z = T 3 3 2 2 − σT σC

Figure 2-2. Π-planeFigure 2: representation Representation in the ofdeviat theoric plasticplane of the multiplier strain rate potential rate associated associated to with the the Cazacu et al (2006) plastic potential (k=-0.3098). isotropic version of the CPB06 criterion for k = −0.3098 (σT < σC ).

45 CHAPTER 3 PLASTIC POTENTIAL FOR HCP METALS WITH SPHERICAL VOIDS Spherical voids are commonly observed in isotropic metals prior to failure. As an example, Figure 3-1 shows the growth and coalescence of initially spherical voids in an aluminum flyer plate experiment. To describe the influence of voids on the plastic flow of a porous aggregate, Gurson(1977) proposed an analytical macroscopic yield criterion. Tvergaard(1981, 1982) modified Gurson’s (1977) criterion by introducing fitting

parameters q1, q2 and q3 to better agree with finite element unit cell calculations. Leblond and Perrin(1990) performed a self-consistent analysis of a void within a porous plastic sphere to provide a physical basis and a rationale for the numerical values of these fitting parameters. The recommended values for Tvergaard’s fitting parameters in Leblond and

2 Perrin(1990) were q1 = 4/e ≈ 1.47, q2 = 1 and q3 = q1. Tvergaard and Needleman(1984) replaced the parameter associated with the void volume fraction, f, in Gurson’s (1977) criterion with an effective void volume fraction, f ∗, as a way to account for the onset of void coalescence leading to final material fracture (see Chapter1). Ellipsoidal voids are generally observed in cold rolled sheets. Gologanu et al.(1993) used a homogenization procedure on an axisymmetric, ellipsoidal RVE containing a confocal, prolate ellipsoidal void to arrive at an analytic expression for the yield criterion of a porous aggregate containing prolate, ellipsoidal voids. Gologanu also considered the case of oblate voids and developed analytic expressions for such criteria (see, for example, Gologanu et al., 2001). Garajeu et al.(2000) extended the work of Gologanu and collaborators by investigating the influence of this void distribution evolution in addition to the influence of void shape evolution. Garajeu also captured the effects of matrix viscosity on ductile failure by considering a matrix described by a Norton-type power law. As mentioned previously, the models currently existing in the literature concern materials that have the same yield in tension and compression in the absence of voids. However, early results of Hosford and Allen(1973) have shown that tension-compression

46 asymmetry in yielding at the polycrystal level is observed if twinning is a contributor at the single crystal level to plastic deformation. Furthermore, metals with hexagonal close packed structure display a directional (orientation preferred) tension-compression asymmetry. The central goal of this dissertation is to provide yield criteria for such materials. This chapter focuses on developing a macroscopic yield criterion for a void- matrix aggregate where the voids have spherical geometry and the metal matrix is isotropic with tension-compression asymmetry. The outline of this chapter is as follows. Section 3.1 presents some limiting solutions which the new criterion should reduce to for some particular cases (zero porosity, equal yield strengths and hydrostatic loading, respectively). Secondly, the velocity ﬁeld used to calculate the local plastic dissipation potential will be presented in Section 3.2. Section 3.3 details the derivation of the corresponding local plastic dissipation and Section 3.4 develops the expressions for the macroscopic plastic dissipation of the void-matrix aggregate. Figure 3-2 illustrates the representative volume element (RVE) that will be used in this chapter. It is a sphere containing a concentrical spherical void. The outer radius of the spherical RVE is b while the inner radius (or the radius of the void) is denoted by a. Uniform rate of deformation boundary conditions are assumed for the RVE. The following assumptions are made with respect to the loading and matrix material response:

• The axial deformation is symmetric; thus, a principal system in Cartesian coordinates can be found such that D11 = D22 and Dij = 0 if i 6= j (with similar expressions for Σ).

P • The matrix material is fully plastic and incompressible such that dij = dij in Equation (2–1) (i.e., rigid plastic behavior).

47 Since axisymmetric loading is assumed, the macroscopic stress state, Σ, and associated deviator, Σ0, are given as   Σ11 0 0     Σ =  0 Σ 0  (3–1)  11    0 0 Σ33 and   1 (Σ11 − Σ33) 0 0  3  0   Σ =  0 1 (Σ − Σ ) 0  . (3–2)  3 11 33   2  0 0 − 3 (Σ11 − Σ33)

3.1 Limit Solutions

Before proceeding to the derivation of the macroscopic yield condition, it is prudent to determine some limiting solutions that are available for the problem of interest. These limiting solutions can be compared with the eventual yield criterion and can serve as guides in later sections for any assumptions that need to be made in developing the macroscopic yield expressions. Ideally, any analytic criterion that is developed will reduce to these limiting solutions. 3.1.1 Zero porosity and equal yield strengths limiting cases

The ﬁrst state where an analytic solution immediately comes to mind is when the material is void-free. When no void exists in the sphere, the homogenized material is exactly the matrix material and the yield criterion should reduce to the isotropic Cazacu et al.(2006) yield criterion (see Equation (2–11)). In other words,

˜ Σe = σe = σT . (3–3)

Secondly, if the matrix material has equal strengths in tension and compression, the yield condition of Equation (2–11) reduces to the von Mises yield condition such that the macroscopic yield condition from the homogenization performed in this section should reduce to Gurson’s (1977) macroscopic yield condition given in Chapter1 and rewritten

48 below as 2 S Σe 3Σm 2 ΦG = + 2f cosh − 1 − f = 0 (1–1) σY 2σY where σY is the yield strength of the von Mises matrix material. 3.1.2 Exact solution for a hydrostatically-loaded hollow sphere

Consider the hollow sphere shown in Figure 3-3 of inner radius a and outer radius b that is subjected to a hydrostatic pressure on its outer surface (either tensile or compressive). An exact solution for the problem of an elastic-plastic hollow sphere subject to hydrostatic loading exists in the case when the material is governed by the von Mises yield criterion (see, for example, Lubliner, 1990). The external pressure at which the hollow sphere becomes fully plastic (i.e., the ultimate, or limit, pressure) is given as

2 Σ = ± σ ln(f) (3–4) m 3 Y

3 3 where σY is the uniaxial yield strength and f = a /b is the void volume fraction. The choice of + or − in the equation depends on whether the applied load is compressive or tensile, respectively (since ln(f) < 0). The challenge here is to determine an exact solution for a hydrostatically-loaded hollow sphere when the matrix material exhibits strength differential effects. In the derivation below, the problem of an elastic-plastic material governed by the isotropic CPB06 yield condition of Equation (2–8) is considered. The rigid-plastic solution can then be deduced by letting the Young modulus tend to infinity. Strain-displacement relations, strain compatibility, Hooke’s law, the yield criterion and the equilibrium equation will all be used to obtain an expression for the applied external pressure as a function of the matrix yield stress and void geometry for a fully plastic sphere. 3.1.2.1 Strain-displacement relations

Assuming small strains, the microscopic strain tensor is deﬁned as

1 ←− −→ = (u∇ + ∇u), 2

49 where, in spherical coordinates, the gradient of the displacement vector u is deﬁned as (see Malvern, 1969)

 ∂u 1 ∂u u 1 ∂u u  r r − θ r − φ  ∂r r ∂θ r r sin θ ∂φ r  ←−  ∂u 1 ∂u u 1 ∂u u   θ θ r θ φ  [u∇] :=  + − cot φ   ∂r r ∂θ r r sin θ ∂φ r   ∂u 1 ∂u 1 ∂u u u  φ φ φ + θ cot θ + r ∂r r ∂θ r sin θ ∂φ r r with −→ ←− [u∇] = [u∇]T .

In the given problem, the external loading is hydrostatic such that, Σm = Σr. Due to the spherical symmetry of the problem, only the radial displacement component is diﬀerent

from zero. In other words, u = ur(r)~er. The only nonzero strain components are then

du = (3–5a) rr dr u = = . (3–5b) θθ φφ r

3.1.2.2 Strain compatibility

The St. Venant compatibility equations are required when the strains are considered known since, in the general case, the strain-displacement equations yield six equations for three displacement unknowns. The compatibility equations ensure that the assumed strains are physically possible. The St. Venant compatibility equations are expressed as (see Malvern, 1969) −→ ←− S = ∇ × E × ∇ = 0.

This expression represents six partial diﬀerential equations; however, it can be shown that only three of the six equations are independent. The six equations are rather tedious to write in spherical coordinates, so only the ﬁrst equation (the only one needed for this

50 hollow sphere problem) is expanded below as

1 ∂ ∂ ∂ S = φθ − cos θ − sin θ φφ + sin θ + cos θ 11 r2 sin θ ∂θ ∂φ φφ ∂θ rθ θθ ∂ 1 ∂ ∂ − θθ − cos θ − sin θ θφ − φθ cos θ − ∂φ sin θ ∂φ θφ ∂θ sin θ rφ ∂ ∂ ∂ + sin θ θr − (r ) + − sin θ + r sin θ φφ ∂θ ∂r θθ rr φφ ∂r ∂ − φr + cos θ + sin θ . ∂φ θr rr

The other ﬁve equations can similarly be determined using the procedure outlined in Malvern(1969) but are unnecessary for the present analysis. Thus, the compatibility condition yields the necessary relationship

d = (r ). (3–6) r dr θ

In the following, only one subscript will be used to denote the strain components since there are no shear components present such that the second subscript is extraneous. Note that for a radial displacement ﬁeld the compatibility equation given by Equation (3–6) is automatically satisﬁed. 3.1.2.3 Equations of motion

Neglecting body forces and assuming static equilibrium, the balance of linear momentum yields the following: ∇ · σ = 0.

The expanded form of the equilibrium equation in spherical coordinates is

∂σ 1 ∂σ 1 ∂σ 1 rr + θr + φr + (2σ − σ − σ ) = 0 ∂r r ∂θ r sin θ ∂φ r rr θθ φφ ∂σ 1 ∂σ 1 ∂σ 1 rθ + θθ + φθ + [3σ + (σ − σ ) cot θ] = 0 (3–7) ∂r r ∂θ r sin θ ∂φ r rθ θθ φφ ∂σ 1 ∂σ 1 ∂σ 1 rφ + θφ + φφ + (3σ + 2σ cot θ) = 0. ∂r r ∂θ r sin θ ∂φ r rφ θφ As with the displacement ﬁeld, the stress ﬁeld in the sphere must also be spherically symmetric due to the spherical geometry, the isotropic material and the applied boundary

51 conditions (on the boundary r = b, the state of stress is radial). The only nonzero

components of the stress tensor are then σrr, σθθ, and σφφ with σθθ = σφφ. Using these symmetries, the ﬁrst equation in Equations (3–7) now yields

dσ σ − σ r + 2 r θ = 0 (3–8) dr r

where only one subscript is now being used for the stress components as for the strain components. 3.1.2.4 Elastic constitutive relation: Hooke’s law

In the elastic regime, the stress-strain relations are

1 = [(1 + ν)σ − νσ δ ] (3–9) ij E ij kk ij

σθ = σφ, Equation (3–9) can be reduced to the following two relevant equations relating the two unknown strains to the two unknown stresses: 1 r = [σr − 2νσθ] E (3–10) 1 = [(1 − ν)σ − 2σ ]. θ E θ r

3.1.2.5 Ultimate pressure

Substituting Equation (3–10) into Equation (3–6) (strain compatibility), yields

d 1 + ν [(1 − ν)σ − νσ ] + (σ − σ ) = 0. dr θ r r θ r

Now, making use of the equilibrium equation, Equation (3–8), the above equation simpli- ﬁes further to yield d (σ + 2σ ) = 0. dr r θ

52 The previous equation along with Equation (3–8) represent a system of two ordinary diﬀerential equations in two unknown functions (σr and σθ). In order to solve the system, an operator method will be used (see Ross, 1984). Let the operator D = d/dr, such that the last equation along with the equilibrium equation become

Dσr + 2Dσθ = 0 and 2 2 D + σ − σ = 0. r r r θ

In order to determine σr the ﬁrst equation is multiplied by 1/r, the second equation by

D, and the resulting equations are added to obtain an expression in terms of σr alone.

Using a similar procedure yields an expression in terms of σθ alone. Replacing D with d/dr yields the resulting ordinary diﬀerential equations below as

d2σ dσ r2 r + 3r r = 0 dr2 dr d2σ dσ r2 θ + 3r θ = 0. dr2 dr The previous equations are ordinary linear differential equations with variable coefficients. Furthermore, the above equations correspond to a specific type of ordinary differential equation known as Cauchy-Euler equations. Cauchy-Euler equations are equations whose terms are constant multiples of an expression of the form xn(dny/dxn). The method of solution is to make the substitution x = et in order to reduce the equation to a linear differential equation in terms of t with constant coefficients. Making the substitution into the above two equations yields

d2σ dσ r + 3 r = 0 dt2 dt d2σ dσ θ + 3 θ = 0, dt2 dt

53 with general solutions

c σ = c + 2 r 1 r3 c σ = c + 4 . θ 3 r3

The constants c1–c4 are determined by substituting the above expressions back into the equilibrium equation (3–8). Doing so yields c1 = c3 and c2 = −2c4. Now, let c1 = c3 =: A and c4 = −(1/2)c2 =: B, such that the expressions for the stresses become 2B σ = A − r r3 B σ = A + . θ r3 The constants A and B are determined by imposing the boundary conditions for the problem. The boundary conditions for the given problem are as follows:   0 at r = a σr =  Σm = Σr at r = b

The constants A and B can now be solved for as Σ A = m 1 − f Σ a3 B = m 2(1 − f) where f := (a/b)3. Finally, the expressions for the stresses are given as

Σ a3 σ = m 1 − (3–11a) r 1 − f r3 Σ a3 σ = m 1 + . (3–11b) θ 1 − f 2r3

Substituting Equations (3–11a) and (3–11b) into the yield condition gives the level of applied external pressure Σm at which the material first yields and the location at which the plastic flow first develops. The deviator of the stress tensor whose non-zero

54 components are given by Equations (3–11a) and (3–11b) is as follows:   sr 0 0     s =  0 s 0   θ    0 0 sθ

such that   2 (σr − σθ) 0 0   1   s =  0 − (σ − σ ) 0  3  r θ    0 0 − (σr − σθ)   (3–12) − a3 0 0  r3  Σm  3  =  0 a 0  1 − f  2r3   a3  0 0 2r3 where 1 σ = (σ + 2σ ) m 3 r θ Σ = m . 1 − f

Note that Equation (3–12) dictates that the sign of the microscopic third invariant of

σ 2 the deviatoric stress tensor, J3 = srsθ, is opposite in sign to the external pressure (i.e.,

σ σ sgn (J3 ) = −sgn (Σm)). This is a very important point since the sign of J3 will be used to determine which branch of the plastic multiplier rate expression (which is related to the local plastic dissipation) given in Equation (2–27) must be used in computing the overall plastic dissipation. For a state of stress described by Equation eq:stressallelast, the isotropic CPB06 yield criterion of Equation (2–12) results in the following expressions:

σ (σr − σθ) = σT if J3 > 0

σ (σr − σθ) = −σC if J3 < 0.

55 Note that the above expressions are the same as what one would obtain with a von Mises material under hydrostatic loading if the yield strengths in tension and compression are

σ equal (i.e., if σT = σC ). Also, note that J3 < 0 implies that σr − σθ < 0 such that yielding is always governed by the absolute value of the diﬀerence between the two stress components. This quantity is determined using Equation (3–11) as

|Σ | 3a3 |σ − σ | = m r θ 1 − f 2r3

which has obviously a maximum on the inner boundary r = a. In conclusion, plastic ﬂow ﬁrst develops at the inner boundary r = a (the void boundary). If the assumption of rigid plastic behavior is used along with the assumption that the sphere is entirely plastic, Equation (3–8) can be integrated directly as follows:

Z Σm Z b dr dσr = 2ce 0 a r

which yields 2 Σ = c ln(f) (3–13) m 3 e

where ce is a constant (the “e” is to denote that it is dependent on the eﬀective stress) and is given as   −σC if Σm > 0 ce = .  σT if Σm < 0

σ In the above expressions, the relation sgn (J3 ) = −sgn (Σm) has been used to shift

σ the dependence from a microscopic quantity (i.e., J3 ) to a macroscopic quantity (the

macroscopic mean stress, Σm). Note that Equation (3–13) reduces to Equation (3–4) when

the yield strengths in tension and compression are equal (i.e., σT = σC = σY ). Equations (3–3), (1–1) and (3–13) provide limiting solutions that the macroscopic yield criterion developed in the next few sections can be compared against.

56 3.2 Choice of Trial Velocity Field

As mentioned in Section 2.1, the macroscopic plastic dissipation potential W (D) is deﬁned as

v = Dx for any x ∈ ∂V. (2–4)

Obtaining quality velocity fields which satisfy the uniform strain rate boundary conditions is challenging. Gurson(1977) proposed an incompressible velocity field compatible with a constant macroscopic rate of deformation tensor D. For the sake of clarity, the expression of this velocity field will be deduced in the following.

The macroscopic rate of deformation field, Dij, is related to the local rate of deformation field, dij as follows: 1 Z Dij = dijdV. V V The local, or microscopic, rate of deformation tensor is by definition (see Malvern, 1969) 1 ∂vi ∂vj dij = + 2 ∂xj ∂xi

where ~x is the position of a material point in Cartesian coordinates. The velocity ﬁeld, ~v, in the RVE is assumed to satisfy the compatibility conditions (displacement boundary conditions) as well as the incompressibility condition. These conditions are

~v|~x=b~er = D~x

= Dijxj (3–14)

div (~v) = 0.

57 The main assumption of Gurson(1977) is that a solution of Equation (3–14) can be obtained by using an additive decomposition of the velocity ﬁeld. Therefore,

~v = ~vV + ~vS

where ~vV is associated with volume changes and ~vS is associated with shape changes. The deviatoric portion of the macroscopic rate of deformation is

1 D0 = D − D = D − D . ij ij 3 kk ij m

The boundary condition in Equation (3–14) can be expressed in terms of the volumetric and shape-changing parts of the velocity ﬁeld as

S 0 ~vi |~x=b~er = Dijxj (3–15) V ~vi |~x=b~er = Dmxi.

The portion of the velocity ﬁeld responsible for the volume change, ~vV , should be a purely radial ﬁeld because of symmetry. In other words,

V V ~v = v (r)~er.

If the incompressibility constraint is imposed on the volumetric portion of the velocity ﬁeld, the following expression results:

∂ r2vV = 0. ∂r r

A solution to the equation above is c vV = 1 r r2

where c1 is a constant. Enforcing the boundary condition of Equation (3–15), yields an

expression for c1 as follows:

3 c1 = Dmb

58 Using the result above for c1 yields the expression for the volumetric part of the radial velocity component as b3 vV = D . r m r2 The shape-changing part of the velocity ﬁeld is found by assuming that it is of the following form: ~vS = B~x

where B is a constant, symmetric and deviatoric (i.e., tr(B) = 0) tensor. Equation (3–15) shows that the tensor B is identically equal to the deviatoric part of the macroscopic rate

0 0 of deformation tensor, D . Note that due to axisymmetry, Dij = 0 if i 6= j in Cartesian coordinates. In summary, a velocity ﬁeld compatible with the uniform rate of deformation boundary conditions is ~v = ~vV + ~vS where

b3 ~vV = D ~e and ~vS = D0~x (3–16) m r2 r

Now, the deviatoric portion of the local rate of deformation tensor is simply dS = D0. The volumetric part of the rate of deformation ﬁeld can be found by using the deﬁnition of the rate of deformation in spherical coordinates (see Malvern, 1969) along with the assumption of axisymmetric deformation as follows:

∂vV dV = r rr ∂r b3 = −2D m r

and vV dV = dV = r θθ φφ r b3 = D m r

59 Note that ~vV is expressed in spherical coordinates, while ~vS is in Cartesian coordinates such that a change of reference frame is necessary to write ~v in a single coordinate system. In order to transform the expression for the shape-changing part of the rate of deformation to the spherical coordinate system, the following transformation equation must be used: ¯S T S d(r,θ,φ) = A d(1,2,3)A where A is now the spherical-to-Cartesian transformation matrix. Since

x1 = r sin θ cos φ

x2 = r sin θ sin φ

x3 = r cos θ

and ∂~x ~e = = sin θ cos φ~i + sin θ sin φ~i + cos θ~i r ∂r 1 2 3 ∂~x ~e = = cos θ cos φ~i + cos θ sin φ~i − sin θ~i θ ∂θ 1 2 3 ∂~x ~e = = − sin φ~i + cos φ~i φ ∂φ 1 2 the transformation matrix is then

r A = [ac] h i = ~ir · ~ec   sin θ cos φ cos θ cos φ − sin φ     =  sin θ sin φ cos θ sin φ cos φ  .     cos θ − sin θ 0

60 The following expressions for the nonzero components of the shape-changing part of the microscopic rate of deformation tensor in spherical coordinates are now obtained:

1 dS = D0 = D0 sin2 θ − 2 cos2 θ = − D0 (1 + 3 cos 2θ) rr rr 11 2 11 1 dS = D0 = D0 cos2 θ − 2 sin2 θ = − D0 (1 − 3 cos 2θ) θθ θθ 11 2 11 S 0 0 dφφ = Dφφ = D11 3 dS = D0 = 3D0 sin θ cos θ = D0 sin 2θ. rθ rθ 11 2 11 Therefore, the nonzero components in spherical coordinates of the microscopic rate of deformation tensor become

b3 d = −2D + D0 rr m r rr b3 d = D + D0 θθ m r θθ b3 d = D + D0 φφ m r φφ

0 drθ = Drθ

The principal values for the rates of deformation are needed in the expression for the plastic multiplier (Equation (2–27)). The unordered principal values are, s d + d d − d 2 d˜ = rr θθ ± rr θθ + (d )2 2,3 2 2 rθ b3 d˜ = d = D + D0 . 1 φφ m r φφ

The above expressions result in s D b3 D0 3 b6 2 b3 d˜ = − m − φφ ± D2 − D (D0 − D0 ) + D0 2 2,3 2 r 2 2 m r 3 m r rr θθ φφ (3–17) b3 d˜ = D + D0 . 1 m r φφ

61 3.3 Calculation of the Local Plastic Dissipation

As in the case of Gurson’s analysis, for arbitrary loading the macroscopic plastic dissipation W (D) cannot be determined analytically without making some simplifying assumptions. Due to the tension-compression asymmetry of the matrix, fresh diﬃculties are encountered when estimating the local plastic dissipation, w(d). This is because the plastic multiplier rate λ˙ has multiple branches (see Equation (2–29)) and depends on all the principal values of the local rate of deformation tensor, d. However, for the special cases of purely hydrostatic or purely deviatoric loading, the macroscopic plastic dissipation can be solved for explicitly. The corresponding macroscopic stresses at yielding are given in Section 3.4. The expressions for the microscopic rates of deformation given by (3–17) are in principal components as needed for Equation (2–27); however, identiﬁcation of the lone positive or negative principal component is also needed. In order to determine which is the only positive or negative principal component (hereafter referred to as the “lone” ˜ ˜ component), the product d2d3 will be investigated such that if the product is positive then ˜ ˜ ˜ dlone = d1, whereas if the product is negative then either d2 or d3 is the lone component. Accordingly, ˜ ˜ 2 2 2 0 0 2 d2d3 = −2Dmu − (4 − 9 sin θ)D11Dmu − 2 (D11) (3–18)

3 3 where u = b /r . Taking the discriminant with respect to Dm of this quadratic polynomial yields the following:

2 ∆Dm = b − 4ac

2 2 0 2 2 2 0 2 = (4 − 9 sin θ) (D11) u + 4(2u )(−2 (D11) )

2 0 2 2 2 = 9u (D11) sin θ(9 sin θ − 8).

62 Now, there are three cases corresponding to the discriminant being positive, negative or zero. These three cases can be represented as

˜ ˜ 2 sgn(d2d3) = −sgn(−2u ) = 1 if ∆Dm > 0 ˜ ˜ 2 sgn(d2d3) = sgn(−2u ) = −1 if ∆Dm < 0 ˜ ˜ sgn(d2d3) = −1 if ∆Dm = 0

0 where the last relation results from either Dm = 0 or D11 = 0 (i.e., a zero discriminant corresponds to either the purely hydrostatic or purely deviatoric case). Note that in the ˜ ˜ ˜ ﬁrst case dlone = d1 whereas in the ﬁnal two cases the lone component is either d2 or d3. Note also that the sign of the discriminant is simply a function of the angular coordinate, θ ∈ (0, π), such that

θ ∈ (θ1, θ2) if ∆Dm > 0

θ ∈ (0, θ1) ∪ (θ2, π) if ∆Dm < 0

θ ∈ (0, π) if ∆Dm = 0 √ √ where θ1 = arcsin( 8/3) and θ2 = π − arcsin( 8/3). Note that these results imply that one expression for the microscopic plastic multiplier rate is valid over the entire domain of θ for the purely hydrostatic and purely deviatoric solutions, whereas there exist two expressions for the microscopic plastic mul-

0 tiplier rate for the more general case of non-zero Dm and D11, depending on the value of θ. This becomes an issue in the next section when the macroscopic plastic dissipation is derived (which involves integrating the microscopic plastic multiplier rate over θ) since there are two separate expressions with no apparent method of combining them. However, since the purely deviatoric and purely hydrostatic cases provide the intercepts of the yield

criterion (in the ΣmΣe-plane), the microscopic plastic multiplier rate corresponding to these special cases will be developed. The reasoning here is that 1) it is easier to ﬁnd analytical solutions for these special cases, and 2) the yield criterion, at the very least, should satisfy these special cases.

63 ˜ ˜ Keeping the previous argument in mind, dlone corresponds to either d2 or d3 (since their product is negative when the coupled term in Equation 3–18 is zero). Recognizing ˜ that dlone will ﬂip in sign whenever d1 ﬂips in sign, the lone principal deviator can be written as s D D0 D 3 D 2 2 D d = − m − φφ − sgn m + D0 m − m (D0 − D0 ) + D0 2 lone 2ρ 2 ρ φφ 2 ρ 3 ρ rr θθ φφ (3–19) where ρ = r3/b3. Equation (2–27) requires squared values of the principal rates of deformation. First

0 0 0 note that for the case of axially symmetric deformation where D33 = −2D11 = −2D22 (in

Cartesian coordinates), the term De can be deﬁned as a measure of the second invariant of the deviatoric part of the macroscopic rate of deformation tensor as follows:

r2 D = D0 D0 e 3 ij ij 0 (3–20) = 2 |D11|

0 = 2 Dφφ .

Note that De as deﬁned corresponds to the equivalent rate of deformation for a von Mises

material. In this section, De is simply used for notational convenience and as a measure for the second invariant of the deviatoric rate of deformation tensor.

2 0 Using Equation (3–20) to relate De in terms of Dφφ, Equation (3–17) yields s 3 D D 2 2 D D 2 d˜2 = ∓ m + D0 m − m (D0 − D0 ) + e 2,3 2 ρ φφ ρ 3 ρ rr θθ 2 " # 1 D 2 9 D 2 2 D D 2 + m + D0 + m − m (D0 − D0 ) + e 4 ρ φφ 4 ρ 3 ρ rr θθ 2 D 2 d˜2 = m + D0 1 ρ φφ such that the second invariant of the microscopic rate of deformation tensor becomes

3 D 2 D d d = D2 + 6 m − 6 m D0 ij ij 2 e ρ ρ rr

64 which can be shown as being equivalent to the expression given in Gurson(1977) for

dijdij. Also, the square of the lone positive or negative microscopic rate of deformation principal component can be expressed as follows: s 3 D D D 2 2 D (d )2 = m + D0 sgn m + D0 m − m (D0 − D0 ) + D0 2 lone 2 ρ φφ ρ φφ ρ 3 ρ rr θθ φφ " # 1 D 2 9 D 2 2 D + m + D0 + m − m (D0 − D0 ) + D0 2 . 4 ρ φφ 4 ρ 3 ρ rr θθ φφ

(3–21)

Now, the plastic multiplier rate from Equation (2–27) can be expressed as

q ˙ 2 λ = z2dijdij + (z1 + z3)(dlone) (3–22)

where z6 = 4z1 + 6z2 + 4z3 and z1–z3 were deﬁned in Table 2-2 of Chapter2. Note that the second term in the previous expression is due to the tension-compression asymmetry of Equation (2–8) and does not appear in the Gurson analysis using a von Mises material. The complexity of Equations (3–19) or (3–21) makes the practice of obtaining an analytic solution from the volume integral in the next section an exceedingly complicated task. However, noting that in the special cases of purely hydrostatic and purely deviatoric loading (see the rationale presented earlier in this section for focusing on these particular

cases), dlone can be expressed simply as

D d = −2 m + D0 (3–23) lone ρ φφ

such that D 2 (d )2 = 4 m + 4 D0 2 (3–24) lone ρ φφ since the crossed term is zero for both purely hydrostatic and purely deviatoric loading. Now, v u " 2 2# u Dm De Dm λ˙ ≈ tz + − 6z D0 . (3–25) 6 ρ 2 2 ρ rr

65 Note that z6 can be expressed in terms of the tensile and compressive yield strengths as follows:   σ  4 if J3 > 0 z6 = 2 (3–26)  4 σC if J σ ≤ 0.  σT 3

3.4 Development of the Macroscopic Plastic Dissipation Expressions

Let W +(D) denote the macroscopic plastic dissipation corresponding to the particular velocity ﬁeld v given by Equation (3–16). In other words,

1 Z W +(D) = w (d) dV V V

where V = (4/3)πb3 is the volume of the spherical RVE considered. Note that W + is not the true plastic dissipation since a speciﬁc velocity ﬁeld is being assumed. The microscopic plastic dissipation is determined as follows:

w = σijdij ˙ ∂ϕ = σijλ ∂σij ˙ = σT λ

˙ where σT is the yield strength in uniaxial tension (assumed constant) and λ is the plastic multiplier rate associated to the Cazacu et al.(2006) yield criterion (see Equation (2–27) or, for the speciﬁc case presented in this section, Equation (3–25)). Note that the expression for λ˙ in Equation (3–25) contains a non-invariant term (i.e.,

0 the Drr term) under a square root which does not lend the expression to being easily integrated. The approach of Leblond(2003) will be followed, involving the application of the Cauchy-Schwartz inequality, to obtain an upper-bound estimate of the yield locus.

66 This procedure will result in a new upper bound, W ++, for W + as follows:

1 Z W + = w (d) dV V V 1 Z b Z 2π Z π = wr2 sin θdθdφdr V a 0 0 Z b Z 2π Z π 4π 2 1 2 = r 2 wr sin θdθdφ dr V a 4πr 0 0 4π Z b 1 Z = r2 wdS dr V a S S Z b 4π 2 = hwiS(r)r dr V a where the surface area of the sphere at a location r is S = 4πr2 with a diﬀerential element of dS = r2 sin θdθdφ. Now, using the Cauchy-Schwartz inequality of

Z Z 1/2 Z 1/2 (fg) dV ≤ fdV gdV V V V an upper bound for the average of w on a surface of radius r can be determined as

1 Z hwiS(r) = wdS S S 1 Z 1/2 Z 1/2 ≤ 1dS w2dS S S S s 1 Z = w2dS S S

2 1/2 = hw iS(r) .

Since, Z 2 1 2 hw iS(r) = w dS S S σ2 Z = T λ˙ 2dS S S

67 the upper bound macroscopic plastic dissipation becomes

4π Z b W ++ = r2 h(w+)2i 1/2 dr V S(r) √ a 4πσ Z b Z 2π Z π 1/2 = T r2 λ˙ 2 sin θdθdφ dr V a 0 0 √ 4πσ Z 1 Z 2π Z π 1/2 b3 = T r2 λ˙ 2 sin θdθdφ dρ 4 3 2 3 πb f 0 0 3r σ Z 1 Z 2π Z π 1/2 = √ T λ˙ 2 sin θdθdφ dρ 4π f 0 0 where f = a3/b3 is the porosity in the RVE and ρ = r3/b3 was deﬁned in the previous subsection.

0 0 2 2 Note that the integral of Drr = D11 sin θ − 2 cos θ over the surface of the sphere is equal to zero. Also, notice that in Gurson’s analysis, the upper bound assumption (resulting in λ˙ 2 in the integrand instead of λ˙ ) eﬀectively removes the dependence of the

Σ macroscopic third invariant of the stress deviator, J3 , on porous yielding. In the present

Σ analysis, a J3 -dependence on yielding still exists due to the macroscopic eﬀective stress

σ being deﬁned in terms of the isotropic CPB06 criterion of Equation (2–8) and a J3

σ dependence still exists due to z6 (although the dependence on J3 will need to be shifted to a dependence on macroscopic quantities in the later sections). Using the previous assumption, the integral expression for the upper bound macroscopic plastic dissipation becomes

σ Z 1 Z 2π Z π 1/2 W ++ = √ T λ˙ 2 sin θdθdφ dρ 4π f 0 0 1/2 Z 1 " 2 2# √ Dm De = σT z6 + dρ f ρ 2 r 2 2 1 De 1/f u + √ Z 4 Dm = z6 |Dm| σT 2 du 1 u where the substitution u = ρ−1 has been used.

68 The above integral is of the form √ Z u2 + A2 I = du u2

which has the solution (see, for example, Zwillinger, 2003) √ u2 + A2 √ I = − + ln u + u2 + A2 √ u u2 + A2 u = − + sinh−1 + constant u A The expression for the upper bound macroscopic plastic dissipation is now obtained by

replacing A in the expression above with De/(2 |Dm|) and is given as

√ " r # 1/f z 2 |D | D2 ++ 6 −1 m e 2 W = σT 2 |Dm| sinh u − 2 + 4Dm . (3–27) 2 De u 1 The above equation for the upper-bound macroscopic plastic dissipation differs from that √ obtained using a von Mises matrix material by z6/2 (which is equal to one for k = 0). The isotropic CPB06 effective stress is given by Equations (2–12), and (2–13). For axisymmetric loading, the form of the effective stress depends on whether Σ11 or Σ33 is the larger stress component; alternatively, it could be stated that the form of the effective

Σ stress depends on the sign of the macroscopic third invariant of the stress deviator, J3 .

Σ The following expression gives the eﬀective stress depending on the sign of J3 :

˜ Σe = ze |Σ33 − Σ11| r3 = z Σ0 Σ0 e 2 ij ij where σT 1 Σ σT ze = + 1 + sgn J3 1 − . (3–28) σC 2 σC ˜ Expressions for Σe and Σm in terms of the upper-bound macroscopic plastic dissipation of Equation (3–27) next need to be developed. This is done by using the following

69 relation: ∂W Σij = ∂Dij ∂W ++ ≈ ∂Dij ∂W ++ ∂D ∂W ++ ∂D = m + e ∂Dm ∂Dij ∂De ∂Dij ++ 0 ++ 1 ∂W 2 Dij ∂W = δij + . 3 ∂Dm 3 De ∂De

0 Note that Σij = Σij + Σmδij; comparing this with the last line of the equation above yields (dropping the ≈),

1 ∂W ++ Σm = 3 ∂Dm 0 ++ 0 2 Dij ∂W Σij = . 3 De ∂De Also note that

++ 2 0 0 2 ∂W ΣijΣij = 3 ∂De ++ ˜ ∂W ⇒ Σe = ze . ∂De ˜ The two derivative terms that appear in the previous equations for Σe and Σm are obtained from Equation (3–27) as

  1/f

++ √ 2 ∂W z6  2 |Dm| 2 |Dm| u 1 2De/u  = σT  − −  ∂D 2 r 2 D2 2 q D2 e  2D e e 2  m u + 1 u2 + 4Dm De 1 1/f √ " 2 2 # z6 −γ u − 1 = σT 2 upγ2u2 + 1 1 √ " # 1/f z pγ2u2 + 1 = 6 σ − 2 T u 1

70 and   1/f

++ ∂W √  −1 2 |Dm| 2Dmu/De 2Dm  = z6σT sgn(Dm) sinh u + −  ∂D D r 2 q D2 m  e 2D e 2  m u + 1 u2 + 4Dm De 1 1/f √ −1 = sgn(Dm) z6σT sinh (γu) 1 where the substitution γ = 2 |Dm| /De has been used. The stress invariants can now be expressed as follows:

++ ˜ ∂W Σe = ze ∂De √ 1/f z z pγ2u2 + 1 e 6 = σT − 2 u √ 1 z z e 6 p 2 2 p 2 = σT − γ 1 + f + γ + 1 √2 ze z6 p p = σ γ2 + 1 − γ2 + f 2 2 T and 1 ∂W ++ Σ = m 3 ∂D √ m z 1/f 6 −1 = sgn(Dm)σT sinh (γu) √3 1 z γ = 6 sgn(D )σ sinh−1 − sinh−1 (γ) . 3 m T f

˜ Eliminating γ from the equations for Σe and Σm given at the end of the previous subsection can be done by rearranging the expression for the mean stress and taking the hyperbolic cosine of both sides of the equation as follows: 3Σm −1 γ −1 sgn(Dm)√ = sinh − sinh (γ) z6σT f 3Σ γ ⇒ cosh √ m = cosh sinh−1 − sinh−1 (γ) z6σT f 1 p p = γ2 + f 2 γ2 + 1 − γ2 f

71 where the relations cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y) and cosh(sinh−1(x)) = √ ± 1 + x2 have been used. Now, squaring the expression for the eﬀective stress yields

!2 ˜ 2 2 Σe (ze) z6 p p = γ2 + 1 − γ2 + f 2 σT 4 2 (ze) z6 p p = 1 + f 2 − 2 γ2 + 1 γ2 + f 2 + 2γ2 . 4 The last two equations can now be combined to obtain

!2 Σ˜ 4 3Σ e 2 m (3–29) 2 = 1 + f − 2f cosh √ σT (ze) z6 z6σT

Σ σ Note that the parameters ze and z6 are functions of sgn J3 and sgn (J3 ), respectively, as well as both being functions of the ratio of the yield strengths. Any macroscopic yield criterion should only be a function of macroscopic quantities since the reason for using a macroscopic yield criterion becomes unclear if one has access to the microscopic quantities.

σ Therefore, the dependence of z6 on the microscopic quantity J3 needs to be shifted to a dependence on macroscopic quantities. The following assumptions will be used to shift the

σ Σ dependence of sgn (J3 ) to the macroscopic quantities Σm and J3 :

σ 1. The sign of the microscopic quantity J3 on the right hand side of Equation (3–29) is σ equal to the negative sign of the macroscopic mean stress, Σm, such that sgn (J3 ) = −sgn (Σm).

σ 2. The sign of the microscopic quantity J3 on the left hand side of Equation (3–29) is Σ equal to the sign of the macroscopic third invariant of the stress deviator, J3 , such σ Σ that sgn (J3 ) = sgn J3 . Note that the ﬁrst assumption ensures that the hydrostatic solution given by Equation (3–13) is recovered while the second assumption ensures that the criterion matches the limiting solution of Equation (3–3). Equation (3–29) describes the macroscopic yielding and can now be written in the form of a yield function as follows:

˜ !2 Σe 3Σm 2 Φ = + 2f cosh zs − 1 + f = 0 (3–30) σT 2σT

72 where the parameter zs is deﬁned as 1 2 σT zs = 1 + sgn (Σm) − sgn (Σm) − 1 (3–31) 2 σC

and the isotropic CPB06 eﬀective stress is given by Equations (2–12) and (2–13) and is rewritten below for the general loading case as s 9 (|S | − kS )2 + (|S | − kS )2 + (|S | − kS )2 Σ˜ = 1 1 2 2 3 3 e 2 3k2 − 2k + 3

where Si are the macroscopic principal stress deviators.

Equation (3–30) reduces to Gurson’s result as shown in Equation (1–1) for σT = σC

since zs = 1 for that case. The yield function can be broken out into the following

expressions which explicitly show the inﬂuence of σT and σC on yielding:    2  Σ˜ e 3Σm 2  + 2f cosh − 1 − f = 0 if Σm > 0  σT 2σT Φ =   ˜ 2  Σe 3Σm 2  + 2f cosh − 1 − f = 0 if Σm < 0  σT 2σC and ˜ Σe = (1 − f) σT if Σm = 0.

Note that Equation (3–30) reduces to the two exact solutions derived earlier and given in ˜ Equations (3–3) and (3–13) for the cases when f = 0 and Σe = 0, respectively. Figure 3-4 shows a comparison between the yield curves represented by Equation (3–30) for two different materials (k = ±0.3098) assuming a porosity of f = 0.01. The exact hydrostatic solutions for both tensile mean stress and compressive mean stress are captured by the yield curves for the two different materials and are labeled in Figures 3-4A and 3-4B. Notice in both figures that the new yield curves are not symmetric about the vertical axis. This is to be expected when considering the exact hydrostatic solution shown in the figures and represented by Equation (3–13) since for asymmetric yield strengths, a

73 material that behaves according to the isotropic CPB06 criterion of Equation (2–8) has hydrostatic solutions with different magnitudes. Figure 3-5 shows the yield curves for a material with σT < σC (k = −0.3098) but for three different void volume fractions (f = 0.01, f = 0.04 and f = 0.14). Note that as the void volume fraction increases, the material softens due to the decrease in load-bearing area. In Chapter5, finite element cell calculations will be conducted to validate the developed criterion for the porous aggregate.

A B

Figure 3-1. Ductile crack in aluminum plate due to ﬂyer plate impact. A) Ductile crack by void coalescence. B) Tip of ductile crack shown in (A) at higher magniﬁcation. Source: Antoun, Seaman, and Curran 1998, pg. 35. Dynamic failure of materials: Volume 1 - experiments and analyses. Tech. Rep. DSWA-TR-96-77-V1, Defense Special Weapons Agency.

74 b

Figure 3-2. Representative volume element for a sphere containing a spherical void.

σr σθ Σm

σr σθ r

b a

Figure 3-3. Hydrostatically-loaded hollow sphere.

75 1.2

f = 01.0 Σe 1

σ T 0.8

0.6

σ σ CT = 21.1 0.4

Proposed Criterion (J3>0) 0.2 Σ 2 Proposed Criterion (J3<0) Σ 2σ m = ln()f m −= C ln()f σ T 3 σ T 3σ T 0 -4 -3 -2 -1 0 1 2 3 4

Σm

σ T A

1.2 f = 01.0

Σe 1

σ T 0.8

0.6

σ σ CT = 82.0 0.4 Proposed Criterion (J3>0)

0.2 Proposed Criterion (J3<0) Σ 2 Σ 2σ m = ln()f m −= C ln()f σ T 3 σ T 3σ T 0 -4 -3 -2 -1 0 1 2 3 4

Σm

σ T B

Figure 3-4. Macroscopic yield surfaces deﬁned by Equation (3–30) for a matrix material containing a spherical void of porosity f = 0.01. A) k = 0.3098 (σT > σC ). B) k = −0.3098 (σT < σC ).

76 1.2

Σe 1

σ T 0.8

0.6

0.4

σ σ CT = 82.0 0.2

f = 14.0 04.0 01.0 0 -4 -3 -2 -1 0 1 2 3 4

Σm

σ T A

1.2

Σe 1

σ T 0.8

0.6

0.4

σ σ CT = 82.0 0.2

f = 14.0 04.0 01.0 0 -4 -3 -2 -1 0 1 2 3 4

Σm

σ T B

Figure 3-5. Macroscopic yield surfaces deﬁned by Equation (3–30) for a matrix material with k = −0.3098 (σT < σC ) and containing spherical voids: normalized von Mises macroscopic eﬀective stress versus normalized macroscopic mean stress Σ Σ for various porosities. A) J3 > 0. B) J3 < 0.

77 CHAPTER 4 NUMERICAL IMPLEMENTATION OF THE MATRIX YIELD CRITERION The analytical expression for the stress potential derived in Section 3.4 was obtained using the trial velocity ﬁeld due to Gurson(1977) and given by Equation (3–16). However, the true macroscopic plastic dissipation is deﬁned in Equation (2–7) and rewritten below for convenience:

W (D) = inf hw(d)iV (2–7) d∈K(D) where K(D) is the set of incompressible velocity fields satisfying the uniform rate of deformation boundary conditions on the boundary of the RVE, ∂V . Thus, the proposed criterion is an upper-bound of the exact stress potential since one particular velocity field was considered instead of the set of kinematically admissible fields. In order to assess the validity of the yield criteria proposed in this dissertation, calculations of W (D) for arbitrary incompressible velocity fields need to be performed. These calculations cannot be done analytically; hence, finite element calculations of unit cells will be performed instead. In these calculations, the matrix material is modeled using the isotropic CPB06 criterion and the void boundary is explicitly meshed into the geometry. The average yield behavior of the unit cell under various loading conditions can then be compared to the macroscopic yield functions developed in this dissertation. A key step in performing these finite element calculations is the implementation of the specific yield criterion for the matrix. This chapter details the stress update algorithms and the derivatives of the isotropic CPB06 criterion necessary to implement the yield criterion into a finite element code. The computational testing procedure used in the unit cell finite element calculations is given for axisymmetric calculations in Chapter5 and for plane strain calculations in Chapter7. 4.1 Return Mapping Procedure

Before ﬁnite element calculations can be performed, the yield criterion of Equation (2–8) must be implemented into a ﬁnite element framework (e.g., into the commercial

78 finite element code Abaqus, 2008). In order to implement a criterion into a finite element framework, a stress update algorithm must be used to correctly update the stresses at each time increment based on the amount of plastic flow that has occurred in the increment. Return mapping schemes are widely-used for numerically implementing these stress updates. Return mapping schemes first predict the trial stresses based on the current total strain increment (an elastic predictor step) and then return the stresses back to the yield surface using an iterative procedure (a plastic corrective step). The next few paragraphs outline the return mapping procedure used to implement the isotropic CPB06 criterion of Equation (2–8) into a finite element framework (into ABAQUS in this case, although the return mapping scheme is simply a numerical algorithm for integrating the constitutive equations and as such is independent of the finite element code being used). The scheme is presented below as a small strain algorithm; however, since commercial finite element codes generally update the stress tensor to account for large deformations and rotations before calling the stress update subroutines, the scheme presented below is valid also for large displacements (with the Cauchy stress replaced by some objective stress measure and the small strain increments replaced by the rate of deformation increments). See Belytschko et al.(2000) or Simo and Hughes(1998) for more details. The yield condition can be stated below as

ϕ = σe − σT = 0 (4–1)

where σe is the effective stress of the matrix material and σT is the matrix material’s yield strength in tension. For associated flow (i.e., the plastic flow potential coincides with the yield function, ϕ) and constant yield strength,

∂ϕ dP = λ˙ ∂σ ∂σ = λ˙ e ∂σ

79 where dP is the plastic part of the strain rate tensor, λ˙ the plastic multiplier rate and σ is the Cauchy stress tensor. The loading-unloading conditions (also known as the Kuhn-Tucker conditions) are given as follows:

λ˙ ≥ 0, ϕ ≤ 0, λϕ˙ = 0,

where the ﬁrst expression states that the plastic multiplier rate is always positive or zero, the second expression states that the stress is always on or within the yield surface, and the last expression states that the stresses remain on the yield surface during plastic loading (i.e., when λ˙ > 0). This last condition can also be stated as

ϕ˙ = 0 (4–2)

since ϕ = 0 during plastic loading. Equation (4–2) is known as the consistency condition. For the rate-independent macroscopic yield function given by Equation (3–30), all of the unit cell calculations will be performed as quasi-static. In quasi-static analyses, the rate of deformation is not physically meaningful since there is no time-dependence (although one could still define a pseudo-rate by using the step size). Because of the rate-independence of this problem, the plastic multiplier, λ, will be employed for the rest of this section as opposed to the plastic multiplier rate, λ˙ . The plastic multiplier is defined for associated flow as ∂ϕ P = λ ∂σ ∂σ = λ e ∂σ where P is the plastic part of the strain tensor. Even though small strains are implied by using instead of E, this is not an assumption needed for the return mapping scheme; if the finite element code computes finite strains and objective stresses (i.e., a stress measure that is frame-invariant such that large rotations and stretchings are accounted

80 for correctly) and passes these to the stress update subroutine, then the return mapping scheme presented below will also be valid for large deformations. The return mapping scheme iteratively updates the stresses such that the yield condition of Equation (4–1) is satisﬁed. Therefore, the ﬁrst step in deriving the return mapping scheme is to expand Equation (4–1) using a Taylor series expansion and set the result equal to zero. Doing so, while dropping second-order and higher terms, yields

dϕ (j) ϕ(j+1) = ϕ(j) + ∆σ(j) dσ (4–3) (j) (j) = ϕ + ϕσ∆σ = 0 where the subscript notation has been used to denote a derivative with respect to the subscript and j = 0 corresponds to the trial values obtained from the elastic predictor step. The previous equation contains the unknown plastic corrector to the stress tensor, ∆σ(j). The following set of equations can be used in order to determine this unknown:

P E n+1 = n+1 + n+1 (4–4) P P n+1 = n + ∆λn+1r where the n indicate the increment number of the converged quantities and should not be confused with the iterative counter j. Also, the plastic flow direction, r, is defined from the relation P = λr and is distinguished from ϕσ since the two are only equal for associated flow. The converged stress tensor can now be expressed as follows:

E σn+1 = Cn+1

P = C n+1 − n+1

P P = C n + ∆n+1 − n − ∆n+1

P = σn + C∆n+1 − C∆n+1

81 such that

trial P σn+1 = σn+1 − C∆n+1

trial = σn+1 − ∆λn+1Cr

trial = σn+1 + ∆σn+1 where C is the elasticity tensor. Note that nothing has yet been said regarding whether the plastic ﬂow direction, r,

is taken as the previously converged direction, rn, or as the current increment’s direction,

rn+1. It turns out that either value can be used, although the simplicity and accuracy of the routine will depend on this choice. If the previously converged value for the plastic ﬂow direction is used to update the stresses then the return mapping scheme presented below is referred to as a semi-implicit backward Euler scheme since the algorithm is

implicit in the plastic multiplier (λn+1) but explicit in the plastic ﬂow direction (rn).

Alternatively, if the current increment’s value, rn+1, is used for the plastic flow direction in the plastic corrector phase, then the return mapping algorithm presented here is referred to as a fully implicit backward Euler scheme since it is implicit in both the plastic multiplier and the plastic flow direction. For associated plasticity, a geometric interpretation can be made regarding the difference between the two schemes. The semi-implicit routine projects the predicted stresses onto the yield surface along the normal to the previous yield surface while the fully implicit routine projects the predicted stresses onto the yield surface along the normal to the current yield surface (closest point projection). Obviously, as the step or increment size decreases, the two schemes converge to the same result since the two normals will become approximately the same. The semi- implicit scheme is commonly used in explicit finite element methods since the time step required is very small compared to implicit methods due to the Courant-Friedrichs-Lewy condition (for example, the CFL condition generally limits the time step such that a wave cannot pass through the smallest element in the grid in less than one time increment).

82 A semi-implicit scheme can also be used in implicit ﬁnite element methods although care must be taken to ensure that the step size used is small enough to achieve acceptable results. Obviously, a semi-implicit scheme will require smaller step sizes for a given problem than will a fully implicit scheme since the plastic ﬂow direction is explicit in the semi-implicit scheme.

An expression for the plastic corrector to the stress tensor, ∆σn+1 = −∆λn+1r still needs to be determined for the iterative plastic corrector phase. In order to do so, the second expression in Equation (4–4) will be rewritten as follows:

P an+1 := −n+1 + n + ∆λn+1r (4–5) −1 = C ∆σn+1 + ∆λn+1r = 0.

In order to obtain an expression for the increment in the plastic multiplier during the plastic correction (i.e., δλ(j)), a Taylor series expansion of the vector equation a will be performed as follows (again dropping second-order terms and higher):

da (j) da (j) a(j+1) = a(j) + δλ(j) + ∆σ(j) d∆λ dσ

The previous expression represents a vector equation and results in two separate expressions depending on whether the plastic loading direction is explicit or implicit. The two expressions are as follows:

(j+1) (j) (j) (j) −1 (j) (j) (j) (j) aFI = a + r δλ + C ∆σ + ∆λ rσ ∆σ = 0

(j+1) (j) (j) −1 (j) aSI = a + rnδλ + C ∆σ = 0

where the subscripts “FI” and “SI” refer to the fully implicit and semi-implicit solution, respectively. The previous equations can be solved for the plastic corrector to the stress

83 tensor, ∆σ(j). Doing so yields,

(j) −1 (j) (j)−1 j (j) (j) ∆σFI = − C + ∆λ rσ a + r δλ

=: −C¯(j) aj + r(j)δλ(j) (4–6)

(j) (j) ∆σSI = −Crnδλ

Finally, the previous results can be substituted back into the Taylor series expansion of the yield condition, Equation (4–3), and solved for the plastic corrector iteration’s increment in the plastic multiplier, δλ(j). Doing so yields

(j) (j) ¯(j) (j) (j) ϕ − ϕσ C a δλFI = ϕ(j)C¯(j)r(j) σ (4–7) (j) (j) ϕ δλSI = (j) ϕσ Crn where ¯ −1 (j) (j)−1 C = C + ∆λ rσ and a(j) is given by Equation (4–5). Once the iteration’s increment in the plastic multiplier, δλ(j), is determined from Equation (4–7), the iteration’s increment in the plastic corrector to the stress tensor, ∆σ(j) can be determined from Equation (4–6). The iterative loop is stopped once the updated stresses satisfy the yield condition of Equation (4–1) and once Equation (4–5) is satisﬁed within a given tolerance (note that Equation (4–5) is always satisﬁed at every iteration for the semi-implicit scheme). Note from Equation (4–7) that the fully implicit routine requires the second derivative of the yield function

(j) ϕ with respect to the stress tensor (in the form of rσ embedded in a , and assuming associative plasticity) while the semi-implicit routine only requires the ﬁrst derivative. The second-order tensor r = ϕσ (assuming associative plasticity here) and the fourth-order tensor rσ will be derived in the following sections (these two tensors will become a vector and a matrix, respectively, during the ﬁnite element implementation since in this phase the symmetric stress and strain tensors will be vectorized to reduce computational cost).

84 Implicit ﬁnite element codes such as the implicit version of ABAQUS require an algorithmic tangent modulus to be computed every increment. The algorithmic tangent modulus that will be used for both the semi-implicit and fully implicit schemes is as follows: ¯ ¯ ¯ Cr ⊗ ϕσC Calg = C − ¯ ϕσCr

where the plastic ﬂow direction, r, is taken to be rn for the semi-implicit scheme and rn+1 for the fully implicit scheme while C¯ = C for the semi-implicit scheme. Additionally, for the semi-implicit scheme, the value of σ is taken to be the previous increment’s value such that ϕσ = rn and the algorithmic tangent modulus is symmetric. Note that Belytschko et al.(2000) gives an alternate unsymmetric algorithmic tangent modulus for the semi- implicit scheme. 4.2 First Derivatives

Both the semi-implicit and fully implicit return mapping schemes use the first derivative ϕσ = r in order to perform the plastic correction to the elastic predictor (see Equation (4–7)). This section details the derivation of the first derivative of the yield criterion. Taking the first derivative of Equation (4–1) gives

dϕ ϕ = σ dσ dσ = e dσ

where the yield strength, σT , has been taken as a constant. Applying the chain rule to the last term in the previous expression yields

∂σ ∂σ ∂s e = e α ∂σ ∂s ∂σ ij α ij (4–8) ∂σ ∂s ∂J ∂s ∂J = e α 2 + α 3 ∂sα ∂J2 ∂σij ∂J3 ∂σij

with α = 1, 2 and 3 and with J2 and J3 being the second and third invariants of the

deviatoric stress tensor, respectively. The expressions for J2 and J3 in terms of the

85 principal deviatoric stresses are written below as

1 J = s2 + s2 + s2 2 2 1 2 3 1 J = s3 + s3 + s3 . 3 3 1 2 3 The ﬁrst derivative term on the right hand side of Equation (4–8) was derived in Section 2.3 and is repeated below for convenience as ∂σe m 2 1 = √ (|s1| − ks1) (sgn(s1) − k) − (|s2| − ks2) (sgn(s2) − k) ∂s1 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 ∂σe m 2 1 = √ (|s2| − ks2) (sgn(s2) − k) − (|s1| − ks1) (sgn(s1) − k) ∂s2 F 3 3 (2–16) 1 − (|s | − ks ) (sgn(s ) − k) 3 3 3 3 ∂σe m 2 1 = √ (|s3| − ks3) (sgn(s3) − k) − (|s1| − ks1) (sgn(s1) − k) ∂s3 F 3 3 1 − (|s | − ks ) (sgn(s ) − k) . 3 2 2 2

Expressions for ∂sα/∂J2 and ∂sα/∂J3 can now be found by diﬀerentiating the characteristic equation

3 η − J2η − J3 = 0 where the principal stress deviators (s1, s2 and s3) are the roots of the previous third- order algebraic expression. Diﬀerentiating the characteristic equation with respect to η yields the following:

2 3η dη − J2dη − dJ2η − dJ3 = 0 ∂η η ⇒ = 2 ∂J2 3η − J2 ∂η 1 = 2 ∂J3 3η − J2

86 where η can be any of the principal stress deviators, sα, with α = 1, 2, 3. Thus, ∂s s α = α ∂J 3s2 − J 2 α 2 (4–9) ∂sα 1 = 2 ∂J3 3sα − J2 p Note that the derivatives in the previous expression have singularities at sα = ± J2/3.

This condition holds when any two of the sα’s are equal (biaxial loading) or when they

are all equal (i.e., sα = 0 as in pure hydrostatic loading). Only the ﬁrst case is of interest here since plasticity does not typically occur in non-porous metals during pure hydrostatic p loading. For the two singular cases corresponding to biaxial loading (s1 = s2 = J2/3 p and s2 = s3 = − J2/3), ∂σe/∂σij must be calculated directly. The expressions for the

∂σe/∂σij corresponding to the two relevant loading states (general and biaxial) will now be presented. 4.2.1 Isotropic CPB06 ﬁrst derivatives: general loading

For the general loading situation when s1 6= s2 6= s3, Equation(4–8) can be used directly along with Equation(4–9) to obtain the following: ∂σe ∂σe sα ∂J2 1 ∂J3 = 2 + 2 ∂σij ∂sα 3sα − J2 ∂σij 3sα − J2 ∂σij with ∂J 2 = σ0 ∂σ ij ij (4–10) ∂J3 0 0 2 = σipσpj − J2δij ∂σij 3

0 where σij is the deviatoric stress tensor (not to be confused with the principal deviatoric

stress components, sα). Therefore, the previous expression can be written as follows: ∂sα 1 0 0 0 2 = 2 σijsα + σipσpj − J2δij (4–11) ∂σij 3sα − J2 3

such that ∂σe ∂σe 1 0 0 0 2 = 2 σijsα + σipσpj − J2δij (4–12) ∂σij ∂sα 3sα − J2 3

87 where the derivative terms on the right hand side in the previous expression are given by Equation (2–16). 4.2.2 Isotropic CPB06 ﬁrst derivatives: biaxial loading

In the biaxial loading case, lets ˜ denote the lone principal stress deviator such that r J s˜ = ±2 2 3

with the other two principal stress deviators being identical and equal to minus one half of

s˜. In this loading state, the loading direction, ∂σe/∂σij, can be found directly as follows:

∂σ ∂σ ∂F e = e ∂σ ∂F ∂σ ij ij (4–13) ∂σ ∂F ∂s˜ ∂J = e 2 ∂F ∂s˜ ∂J2 ∂σij where F was deﬁned in Equation (2–8) and can be written for the particular case of biaxial loading as s˜2 F = 3k2 − 2sgn(˜s)k + 3 . 2 Also, note that √ σe = m F such that ∂σ m e = √ . ∂F 2 F The previous equations can be used to evaluate Equation (4–13) and yield the following result:

∂σe m 2 0 = √ 3k − 2sgn(˜s)k + 3 σij. (4–14) ∂σij 3 F p Since k = 0 yields m = 3/2 and F = 2J2, Equation (4–14) reduces to √ ∂σe 3 0 = √ σij ∂σij 2 J2

88 for k = 0, which is the known result for a von Mises material (see Equation (4–15)). Note that the previous result for a von Mises material holds for all loading conditions, not just the special case of biaxial loading. The first derivatives derived in this section yield all the necessary ingredients for implementing a semi-implicit return mapping algorithm into a finite element code (see Section 4.1). The next section will detail the second derivatives which will be needed along with the first derivatives derived in this section to implement a fully implicit return mapping algorithm into a finite element code. 4.3 Second Derivatives

The fully implicit scheme requires the second derivative of the yield condition in Equation (4–1) with respect to the stress tensor in addition to the first derivative derived in the previous section. As was the case for the first derivatives in Section 4.2, the second derivatives will also need to be handled differently for the special case of biaxial loading. This section first derives the second derivatives for a von Mises material, then derives the second derivatives for the general loading case when s1 6= s2 6= s3, and lastly derives the derivatives for the special case of biaxial loading. 4.3.1 Von mises second derivatives

Before deriving the isotropic CPB06 second derivatives, it is instructive to derive the second derivatives for a von Mises material since isotropic CPB06 reduces to von Mises for k = 0 and since, at least numerically, these von Mises second derivatives can validate the longer and more complex isotropic CPB06 second derivatives. Therefore, the von Mises relations can be written as,

r3 σVM = σ0 σ0 e 2 ij ij p = 3J2

89 such that ∂σVM ∂σVM ∂J e = e 2 ∂σij ∂J2 ∂σij √ (4–15) 3σ0 = √ ij 2 J2 and √ ! ∂ ∂σVM ∂ 3σ0 e = √ ij ∂σmn ∂σij ∂σmn 2 J2 √ 0 0 3 1 σijσmn (4–16) = √ δimδjn − δijδmn − 2 J2 3 2J2 3 ˆ = VM I 2σe where Iˆ = Idev − nˆ ⊗ nˆ

with r(0) nˆ = kr(0)k r2 = r(0) 3 σ0 = √ 2J2 and 1 Idev = I − I ⊗ I 3 where I is the fourth-order unit tensor given by

Iijmn = δimδjn.

Equation (4–16) is given in tensorial form as a fourth-order tensor. Commercial ﬁnite element codes typically have the symmetric stresses (and strains) in vector form for computational eﬃciency. Equation (4–16) can be transformed using Voight notation to a

90 form that can be readily implemented into a ﬁnite element code as follows:

√   2 VM ∂ σ 3 V11 V12 e = √   . (4–17) ∂σij∂σmn 2 J2  T  V12 V22

Note that in the ﬁrst quadrant where V11 resides, i = j and m = n with i = j = m = n on the diagonal. Therefore,

 0 2 σ0 σ0 0 0  2 − (σx) − 1 − x y − 1 − σxσz 3 2J2 3 2J2 3 2J2  2  0 0 0 0 0  1 σxσy 2 (σy) 1 σyσz  V11 =  − − − − −  . (4–18)  3 2J2 3 2J2 3 2J2   0 0 σ0 σ0 0 2  − 1 − σxσz − 1 − y z 2 − (σz) 3 2J2 3 2J2 3 2J2

Similarly, in the second and third quadrants where V12 resides, i = j and m 6= n such that

 0 0 0  − σxσxy − σxσxz − σxσyz  2J2 2J2 2J2  0 0 0  σyσxy σyσxz σyσyz  V12 =  − − −  . (4–19)  2J2 2J2 2J2   0 0 0  − σzσxy − σzσxz − σzσyz 2J2 2J2 2J2

Last of all, i 6= j and m 6= n in the fourth quadrant where V22 resides. Going through the necessary manipulation yields

 2  1 − (σxy) − σxyσxz − σxyσyz  2J2 2J2 2J2  2  σxzσxy (σxz) σxzσyz  V22 =  − 1 − −  . (4–20)  2J2 2J2 2J2   2  − σyzσxy − σyzσxz 1 − (σyz) 2J2 2J2 2J2

The previous expressions detail the second derivatives corresponding to a von Mises material. These second derivatives are not of much practical use since the von Mises yield surface is a circle in deviatoric stress space and, thus, the stresses can be updated by using the radial return algorithm which does not explicitly use the second derivatives. However, the second derivatives of a von Mises material can be used to check for errors in the second derivatives derived in the next few paragraphs for an isotropic CPB06 material since the two should be equivalent when the yield in tension is equal to the yield in compression (i.e., when k = 0).

91 4.3.2 Isotropic CPB06 second derivatives: general loading

The next few paragraphs will focus on deriving the expressions for the second derivatives of an isotropic CPB06 material given by Equation (4–1) for the general loading situation when s1 6= s2 6= s3. Therefore, the following derivatives are needed: ∂ ∂σe ϕσ = ∂σmn ∂σij ∂ ∂σ ∂s = e α ∂σmn ∂sα ∂σij ∂2s ∂σ ∂2σ ∂s ∂s = α e + e β α ∂σmn∂σij ∂sα ∂sβ∂sα ∂σmn ∂σij

= T1 + T2 where the tensors T1 and T2 are deﬁned as

2 ∂ sα ∂σe T1 = ∂σmn∂σij ∂sα 2 (4–21) ∂ σe ∂sβ ∂sα T2 = . ∂sβ∂sα ∂σmn ∂σij

Note that the ﬁrst derivative term in T1 is given by Equation (2–16) and the ﬁrst derivatives in T2 are given in Section 4.2; therefore, the only unknowns in Equation (4–21) are the second derivatives.

The second derivative term appearing in the expression for T1 in Equation (4–21) is found as follows: ∂2s ∂ ∂s ∂J ∂s ∂J α = α 2 + α 3 ∂σmn∂σij ∂σmn ∂J2 ∂σij ∂J3 ∂σij ∂s ∂2J ∂s ∂2J = α 2 + α 3 ∂J2 ∂σij∂σmn ∂J3 ∂σij∂σmn ∂J ∂2s ∂J ∂2s + 2 α + 3 α ∂σij ∂σmn∂J2 ∂σij ∂σmn∂J3

= TA + TB + TC + TD

92 where the tensors TA, TB, TC and TD are deﬁned as

2 ∂sα ∂ J2 TA = ∂J2 ∂σij∂σmn 2 ∂sα ∂ J3 TB = ∂J3 ∂σij∂σmn 2 (4–22) ∂J2 ∂ sα TC = ∂σij ∂σmn∂J2 2 ∂J3 ∂ sα TD = . ∂σij ∂σmn∂J3

The ﬁrst derivative terms in tensors TA–TD were already derived in Section 4.2. Only

the second derivatives are currently unknown. The second derivative term in tensor TA is found by taking the derivative of the ﬁrst expression in Equation (4–10). Doing so yields

2 ∂ J2 ∂ σkk = σij − δij ∂σij∂σmn ∂σmn 3 1 = δ δ − δ δ . im jn 3 mn ij The previous expression is in tensorial form and must, as before, be transformed to matrix form using Voight’s notation as follows:   2 ∂ J S1 0 2   =   (4–23) ∂σij∂σmn 0 I

where   2 −1 −1   1   S1 =  −1 2 −1  (4–24) 3     −1 −1 2 and I is the identity matrix.

Similarly, the second derivative term of tensor TB can be found by using the second expression in Equation (4–10) which yields,

2 ∂ J3 ∂ 0 0 2 = σipσpj − J2δij ∂σij∂σmn ∂σmn 3 2 = δ σ0 + δ σ0 − δ σ0 + δ σ0 . im nj jn im 3 mn ij ij mn

93 Using Voight notation, the previous expression can be presented as,   2 ∂ J S11 S12 3 =   (4–25) ∂σij∂σmn  T  S12 S22 where   0 0 0 0 0 σx − σx + σy − (σx + σz) 2    0 0 0 0 0  S11 =  − σ + σ σ − σ + σ  , (4–26) 3  x y y y z    0 0 0 0 0 − (σx + σz) − σy + σz σz   σxy σxz −2σyz   1   S12 =  σ −2σ σ  (4–27) 3  xy xz yz    −2σxy σxz σyz and   0 0 σ + σ σyz σxz  x y   0 0  S22 =  σ σ + σ σ  . (4–28)  yz x z xy   0 0  σxz σxy σy + σz

The second derivative terms of tensors TC and TD for the general loading state of

s1 6= s2 6= s3 can be determined using the chain rule. Doing so for the J2-term yields

2 ∂ sα ∂ sα = 2 ∂σmn∂J2 ∂σmn 3sα − J2 (4–29) 1 ∂sα sα ∂sα ∂J2 = − 6sα − . 2 2 2 3sα − J2 ∂σmn (3sα − J2) ∂σmn ∂σmn

Similarly, for the derivative with respect to J3,

2 ∂ sα ∂ 1 = 2 ∂σmn∂J3 ∂σmn 3sα − J2 (4–30) −1 ∂sα ∂J2 = 6sα − . 2 2 (3sα − J2) ∂σmn ∂σmn The equations derived up to this point along with the ﬁrst derivatives derived in

Section 4.2 determine the tensor T1 in Equation (4–21). The tensor T2 in Equation (4–21) must now be determined. The ﬁrst derivative term was derived in Section 4.2; therefore,

94 the only unknown is the second derivative term and it is found using Equation (2–16) which is rewritten as follows: ∂σe m 2 1 = √ (|sα| − ksα) (sgn(sα) − k) − (|sη| − ksη) (sgn(sη) − k) ∂sα F 3 3 1 − (|s | − ks ) (sgn(s ) − k) 3 γ γ γ with α 6= η 6= γ in the previous equation and α, η, γ = 1, 2, or 3. Using the previous equation yields an expression for the second derivative term in the tensor T2 of Equation (4–21) as 2 ∂ σe Hαβ := (4–31) ∂sα∂sβ such that

m n 2 2 1 2 1 2 1 ∂σ 2 o √ e Hαβ = (sgn (sα) − k) + (sgn (sη) − k) + (sgn (sγ) − k) − 2 F 3 3 3 m ∂sα (4–32) for α = β, and

m h 2 2 1 1 ∂σ ∂σ i √ 2 2 2 e e Hαβ = − (sgn (sα) − k) − (sgn (sβ) − k) + (sgn (sγ) − k) − 2 F 9 9 9 m ∂sα ∂sβ (4–33) for α 6= β.

The previous equations for Hαβ complete the derivation of the second derivatives of

the yield function given by Equation (4–1) for the case of general loading when s1 6= s2 6= s3. The derivatives presented in this section can be easily implemented into a subroutine using the fully implicit scheme presented in Section 4.1 for the purpose of running ﬁnite element calculations.

95 4.3.3 Isotropic CPB06 second derivatives: biaxial loading

In the biaxial loading case, the second derivatives can be calculated directly using Equation (4–14) as follows: ∂ ∂σe ∂ m 2 0 = √ 3k − sgn(˜s)k + 3 σij ∂σmn ∂σij ∂σmn 3 F " # m 1 ∂σ0 σ0 ∂F 2 √ ij ij = 3k − sgn(˜s)k + 3 − 3/2 3 F ∂σmn 2 (F ) ∂σmn

which leads to the result ∂ ∂σ m e = √ 3k2 − sgn(˜s)k + 3 ∂σmn ∂σij 3 F (4–34) 1 1 σ0 σ0 δ δ − δ δ − 3k2 − sgn(˜s)k + 3 ij mn im jn 3 ij mn 3 F

for biaxial loading conditions. Similar to what was seen with the biaxial first derivatives, Equation (4–34) reduces to the von Mises result of Equation (4–16) when k = 0. Equation (4–34) along with the second derivatives derived previously for the general loading case and the first derivatives of Section 4.2 yield all that is necessary to implement a fully implicit return mapping algorithm into a commercial finite element code (see Section 4.1).

96 CHAPTER 5 ASSESSMENT OF THE PROPOSED SPHERICAL VOID MODEL BY FINITE ELEMENT CALCULATIONS The analytic expression for the macroscopic yield function of a porous aggregate containing spherical voids (see Equation (3–30)) was obtained by assuming a specific RVE geometry, specific loading conditions, and a number of approximations. These assumptions were necessary to obtain an analytic solution; however, it is now desirable to determine how well these analytic yield curves compare to more general loading conditions and RVEs. One way that this validation can be conducted is by comparing the analytic yield curves to finite element unit cell calculations. In the finite element unit cell calculations, the void boundary is explicitly meshed and the matrix material is modeled as an elastic- plastic material with the plastic response governed by the isotropic CPB06 criterion of Equation (2–8). Since in finite element calculations, the minimization of the macroscopic plastic dissipation W D) is performed over a large set of kinematically admissible velocity fields, the accuracy of the yield surface, developed for a single assumed velocity field, can be assessed. The axisymmetric unit cell calculations detailed in this section are under constant strain triaxiality and similar to those performed by Ristinmaa(1997) to assess the predictive capabilities of a cyclic model for a porous aggregate. 5.1 Modeling Procedure

The aim of this section is to compare the analytical yield locus given by Equation (3–30) with numerical yield points obtained by performing axisymmetric ﬁnite element cell calculations. The continuum is considered to consist of a periodic assemblage of hexagonal cylindrical unit cells which are approximated by right circular cylinders as shown in Figure 5-1A. Due to symmetry, only one half of the unit cell is meshed (see Figure 5-1B). Every cell of initial length L0 and radius R0 contains a spherical void of radius a and is subject to homogeneous radial and axial displacements. The boundary conditions for this unit cell

97 are as follows:

u1 = ur2 = ur3 = 0 for x1 = 0

u3 = ur1 = ur2 = 0 for x3 = 0

u1 = U1 for x1 = A

u3 = U3 for x3 = B where ui are the displacements in the xi directions and uri are the rotations about the xi axes. The void is considered to be traction free. For this axisymmetric unit cell, the initial porosity f0 is deﬁned as

Vvoid f0 = Vtotal 3 2a0 = 2 . 3A0B0 A constant strain triaxiality is imposed on the surface of the unit cell. The procedure that is used was proposed by Ristinmaa(1997) and is detailed in the following. The main idea is that a constant strain triaxiality is equivalent to maintaining a constant stress triaxiality during the elastic portion of the calculation. The macroscopic principal strains are as follows: A A0 + U1 E1 = ln = ln A0 A0 B B0 + U3 E3 = ln = ln B0 B0 where A0 and B0 are the initial side lengths of the unit cell as shown in Figure 5-1B while U1 and U3 are the prescribed displacements on those sides. These displacements will be prescribed such that a constant strain triaxiality is maintained throughout the calculations. Note that for axisymmetric loading, E1 = E2. The strain triaxiality, TE, is

98 deﬁned as

Ekk TE = 3Ee 2E + E = 1 3 2 |E1 − E3| where Ekk is the trace of the macroscopic strain tensor and Ee is the macroscopic eﬀective strain deﬁned as r2 E = E0 E0 (5–1) e 3 ij ij

0 with Eij being the macroscopic deviatoric strain tensor. The equation above shows that the strains E1 and E3 must be linearly related in order for the strain triaxiality TE to be constant. Indeed,

2cE + 1 TE = sgn(E3) (5–2) 2 |cE − 1| where cE = E1/E3. Alternatively,

2 2TE + 1 ± 3TE cE = 2 (5–3) 2TE − 2

Note that if the strain triaxiality TE is considered constant for a given calculation, then cE may be calculated using the above equation; however, the sign of the strain components

(e.g., E3) are now ﬁxed such that the sign of the strain triaxiality given by Equation (5–2)

(using the calculated value for cE) must agree with the sign of the strain triaxiality that was chosen.

The prescribed boundary displacements, U1 and U3 can now be related using the above equations. Doing so yields

cE B0 + U3 U1 = A0 − 1 . B0

The macroscopic stress, Σ, is also axisymmetric such that

F Σ = Σ = 1 1 2 2πAB F Σ = 3 3 πA2

99 where F1 is the total radial force at X1 = A and F3 is the total axial force at X3 = B. In the elastic regime, the macroscopic strains are related to the macroscopic stresses via Hooke’s law such that

EE1 = (1 − ν)Σ1 − νΣ3

EE3 = Σ3 − 2νΣ1

where E is Young’s modulus and ν is Poisson’s ratio. The above expression can be solved

for Σ1 and Σ3 using the fact that E1 = cEE3 to yield the following: c + ν Σ = E EE 1 1 − ν − 2ν2 3 2c ν − ν + 1 Σ = E EE 3 1 − ν − 2ν2 3

such that

Σ1 = cΣΣ3

with

cE + ν cΣ = . 2cEν − ν + 1 Now, notice that

Σ sgn(J3 ) = sgn(Σ3 − Σ1)

= sgn(1 − cE)sgn(E3)

E = sgn(J3 ) such that

Σ E sgn(J3 ) = sgn(J3 )

= sgn(TE)sgn(1 − cE)sgn(2cE + 1)

which results in the third invariant of the macroscopic stress being controlled by the

Σ ± sign in Equation (5–3) (at least until macroscopic yield). In other words, J3 in the elastic portion of the calculation is positive or negative if the negative or positive sign, respectively, is chosen in Equation (5–3). This is an important point that will be used to

100 control the sign of the third invariant in the ﬁnite element calculations presented in the next section. The macroscopic stress triaxiality is deﬁned analogous to the macroscopic strain triaxiality as

Σkk TΣ = 3Σe 2Σ + Σ = 1 3 3 |Σ3 − Σ1| 2cΣ + 1 = sgn (Σ3) 3 |1 − cΣ|

where Σe is the von Mises eﬀective stress deﬁned as

r3 Σ = Σ0 Σ0 (5–4) e 2 ij ij

˜ The isotropic CPB06 eﬀective stress, Σe, is related to the von Mises eﬀective stress, for ˜ s s the axisymmetric case, by Σe = zeΣe where ze is given by Equation (3–28). Note that the macroscopic stress triaxiality as given by the above equation is constant for a given stress ratio, cΣ, since the signs of the macroscopic stresses are also constant for a displacement controlled calculation in the elastic region. Also note that the stress equations in this section are valid only for the elastic region since Hooke’s law was used to relate the macroscopic stresses to the macroscopic strains. The stress triaxiality will typically not be constant once macroscopic yielding occurs, and this macroscopic yield behavior is what will be compared to the analytic yield curve given by Equation (3–30). A relationship which is useful for obtaining the desired spacing of data points is the slope of a given calculation in the elastic region in the 0.5Σm-Σe plane:

6 |1 − cΣ| slope = sgn(Σ3) 2cΣ + 1

such that θ = tan−1 (slope)

101 where θ is the angle that the elastic curve makes with respect to the horizontal (i.e.,

0.5Σm) axis. This is used, along with the other equations presented in this section, to obtain the computational test matrix presented in the next section. 5.2 Finite Element Results

Finite element results for three different materials under axisymmetric loading will be presented in this section. All calculations were performed using a CPB06 fully implicit user material subroutine that was implemented into the ABAQUS finite element code (Abaqus, 2008). The three materials have the same yield strength in tension; however, one material has an equal yield strength in compression (k = 0), one has a yield in compression that is less than the yield in tension corresponding to a bcc material (k = 0.3098) and one has a yield in compression that is greater than the yield in tension corresponding to a fcc material (k = −0.3098). These specific values for k were taken from polycrystal calculations done by Cazacu et al.(2006) for randomly-oriented polycrystals (either fcc or bcc) in which twinning was the only deformation mode at the single crystal level. A Poisson’s ratio of ν = 0.32 was used for each material along with a tensile yield

strength to Young’s modulus ratio of σT /E = 0.00124. All stress and strain quantities presented in this section are macroscopic or average quantities unless otherwise noted. The computational test matrix used to obtain the data presented in this section is shown in Tables 5-1—5-18. Figures 5-2, 5-3 and 5-4 show the ﬁnite element meshes

used for the calculations and correspond to initial porosities of f0 = 0.01, f0 = 0.04

and f0 = 0.14, respectively. The unit cells have dimensions of A0 = B0 = 6 inches (the units are of limited practical importance but are provided to add meaning to the

prescribed displacements U1 and U3 given in Tables 5-1—5-18). The meshes are comprised of axisymmetric quadrilateral elements with reduced integration and use ABAQUS’ enhanced hourglass control. A mesh reﬁnement study was performed on a sample of the calculations to ensure mesh convergence and the selected meshes were found to provide satisfactory results (note that since all quantities being investigated are average

102 quantities, even fairly coarse meshes can provide reasonably close agreement in terms of the average response; the meshes shown were also found to provide adequate local detail with reasonable computational cost). The macroscopic yield surface deﬁned in Equation (3–30) can be modiﬁed to include the parameters introduced in Tvergaard(1981) to account for void interaction as follows:

2 Σe 3Σm 2 Φ = + 2fq1 cosh zsq2 − 1 + q3f = 0 (5–5) σT 2σT where zs was deﬁned by Equation (3–31) as 1 2 σT zs = 1 + sgn (Σm) + sgn (Σm) − 1 . (3–31) 2 σC

The parameters qi are taken to be the values suggested in Leblond and Perrin(1990);

2 thus, q1 = 4/e ≈ 1.47, q2 = 1 and q3 = q1. Figures 5-5 and 5-6 show the ﬁnite element results of a von Mises material (k = 0) versus the GTN criterion of Equation (1–11) for various void volume fractions using

2 q1 = 4/e, q2 = 1 and q3 = q1. Since the macroscopic stress triaxiality in the finite element calculations remains constant only in the elastic region, the macroscopic yield points were identified when the stress triaxiality would change from its initial constant value. Note the good agreement between the theoretical yield curves and the finite element results. The positive mean stress half of the figures is what has been typically reported in the literature, although no indication regarding the sign of the third invariant is typically given. Notice that even for the von Mises calculations (k = 0) shown in Figures 5-5 and 5-6 where the matrix material’s yield strengths are equal, the first half of the curve alone does not contain the complete information. For example, notice that the GTN yield curve seems to match reasonably well with the finite element data corresponding to

Σ sgn(J3 ) = −sgn(Σm). This conclusion based on the ﬁnite element data cannot be drawn by investigating solely the region with positive mean stress.

103 Figures 5-7 to 5-12 compare the ﬁnite element results for materials exhibiting tension- compression asymmetry (using CPB06 values of k = −0.3098 and k = 0.3098, respectively)

when the initial porosity is f0 = 0.01, f0 = 0.04 and f0 = 0.14 with the proposed model

2 of Equation (5–5) (q1 = 4/e, q2 = 1 and q3 = q1). The results seem to indicate distinct trends; namely, that the yield surface depends strongly on the sign of the third invariant

Σ of the stress deviator, J3 . Also, note that the ﬁgures illustrate material softening as the void volume fraction increases; this behavior is to be expected since the load-bearing area decreases with increasing void volume fraction. Figures 5-13, 5-14 and 5-15 show comparisons of Equation (5–5) with the unit cell axisymmetric ﬁnite element calculations when k = 0, k = −0.3098 and k = 0.3098,

respectively, for initial porosities of f0 = 0.01, f0 = 0.04 and f0 = 0.14; the figures show the yield points along with the complete curves obtained from the finite element calculations. Notice that these last figures use the normalized CPB06 effective stress as the vertical axis such that the third invariant is implicit and the yield curves for a

Σ particular porosity which depend on the sign of J3 (for non-zero k) collapse to a single curve (this choice of vertical axis was made simply to provide all of the information for a given material on one ﬁgure with minimal confusion). Notice that an implicit assumption in all of these ﬁgures is that the initial porosity,

f0, is approximately equal to the ﬁnal porosity, f, at yield for all calculations. This assumption should not be too bad since only the behavior up to initial yield is of interest here and, thus, the loading on all of the unit cells is fairly small. 5.3 Concluding Remarks

Yielding of materials in which the matrix displays tension-compression asymmetry and contains spherical voids has been studied. An analytical yield criterion which is an upper-bound estimate was developed in Chapter3 by extending Gurson’s (1977) analysis of the hollow sphere to the case when the matrix plastic behavior is described by Cazacu et al.’s (2006) isotropic yield criterion. Due to the tension-compression asymmetry of the

104 matrix response, fresh difficulties were encountered when estimating the local plastic dissipation, w(d). This is because the plastic multiplier rate associated to the Cazacu et al. (2006) yield criterion has multiple branches (see Equation (2–29)) and depends on each of the principal values of the local rate of deformation tensor, d. Certain approximations were introduced in order to obtain the analytical, closed-form expression in Equation (3–30) of the overall plastic potential. Yet, the approximate yield criterion in Equation (3–30) has the property that it reproduces the exact solution for all kinematically- admissible velocity fields in the case of purely hydrostatic loading and the exact solution for the assumed Gurson(1977) velocity field in the case of deviatoric loading. Comparison between the theoretical predictions using the criterion of Equation (3–30) and results of finite element cell calculations show an overall good agreement. The derived criterion of Equation (3–30) is sensitive to the third invariant of the stress deviator and exhibits tension-compression asymmetry (i.e., it is no longer symmetric with respect to the vertical axis, Σm = 0). Although the expression of the proposed criterion for the void-matrix aggregate is similar to that of Gurson’s (1977) criterion, there are distinct differences: q ˜ P3 0 0 2 • First, the von Mises equivalent stress is replaced by Σe = m i=1 (|Σi| − kΣi) , which depends on all principal values of the stress deviator Σ0 and on the ratio between the uniaxial yield in tension and the uniaxial yield in compression of the matrix through the parameter m (see Equation (2–13) for the definition of the constant m).

• Secondly, Equation (3–30) involves a new coefficient zs defined in Equation (3–31) such that for tensile hydrostatic loading, yielding of the void-matrix aggregate occurs when Σm = −(2/3)σC ln(f), while for compressive hydrostatic loading yielding occurs when Σm = (2/3)σT ln(f) (both correspond to the exact solutions obtained in Section 3.1.2). Thus, for arbitrary loadings the effective yield locus is no longer symmetric with respect to the vertical axis, Σm = 0, as it is in the case of Gurson’s (1977) yield locus. If there is no difference in response between the yield in tension and compression, the ˜ coefficient zs = 1 and Σe = Σe (since Cazacu et al.’s (2006) isotropic criterion reduces

105 to the von Mises criterion); hence, the proposed criterion of Equation (3–30) reduces to the classical analytic criterion of Gurson(1977). In the absence of voids, the proposed criterion reduces to Cazacu et al.’s (2006) yield criterion. The accuracy of the analytical criterion was assessed through comparison with finite-element cell calculations. To improve the agreement, the proposed analytic yield criterion of Equation (3–30) was modified to include additional parameters, qi, as was done by Tvergaard(1981) and Tvergaard and Needleman(1984) for the Gurson’s (1977) yield criterion (see Equation (5–5)). In this manner, for k = 0 (Von Mises matrix), the criterion reduces to the GTN model commonly found in the commercial finite element codes. The agreement between the theoretical predictions using this criterion in Equation (5–5) and results of finite element cell calculations are quite good.

A0 X3

X1 2B0

X1 A0

A B

Figure 5-1. Axisymmetric unit cell for the spherical void. A) Unit cell geometry. B) Slice needed for ﬁnite element calculations.

106 Figure 5-2. f0 = 0.01 axisymmetric ﬁnite element mesh for the unit cell.

107 Table 5-1. k = 0, J3 > 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S1 0.2500 -0.00300 0.01502 73.0 S2 0.7500 0.00214 0.01502 47.5 S3 1.1000 0.00429 0.01502 36.6 S4 1.5000 0.00600 0.01502 28.6 S5 2.7500 0.00901 0.01502 16.6 S6 -0.2500 -0.01124 0.01126 -73.0 S7 -0.7500 -0.01777 0.00356 -47.5 S8 -1.1000 -0.01799 -0.00113 -36.6 S9 -1.5000 -0.01895 -0.00474 -28.6 S10 -2.7500 -0.01565 -0.00843 -16.6

Table 5-2. k = 0, J3 < 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S11 0.2500 0.01130 -0.01120 73.0 S12 0.7500 0.01780 -0.00360 47.5 S13 1.1000 0.01800 0.00110 36.6 S14 1.5000 0.01900 0.00470 28.6 S15 2.7500 0.01570 0.00840 16.6 S16 -0.2500 0.00300 -0.01500 -73.0 S17 -0.7500 -0.00210 -0.01500 -47.5 S18 -1.1000 -0.00430 -0.01500 -36.6 S19 -1.5000 -0.00600 -0.01500 -28.6 S20 -2.7500 -0.00900 -0.01500 -16.6

Table 5-3. k = −0.3098, J3 > 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S21 0.2500 -0.00300 0.01502 73.0 S22 0.7500 0.00214 0.01502 47.5 S23 1.1000 0.00429 0.01502 36.6 S24 1.5000 0.00600 0.01502 28.6 S25 2.7500 0.00901 0.01502 16.6 S26 -0.2500 -0.01124 0.01126 -73.0 S27 -0.7500 -0.01777 0.00356 -47.5 S28 -1.1000 -0.01799 -0.00113 -36.6 S29 -1.5000 -0.01895 -0.00474 -28.6 S30 -2.7500 -0.01565 -0.00843 -16.6

108 Table 5-4. k = −0.3098, J3 < 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S31 0.2070 0.01004 -0.01124 73.0 S32 0.6200 0.01868 -0.00632 47.5 S33 0.9090 0.01746 -0.00113 36.6 S34 1.2400 0.01939 0.00267 28.6 S35 2.2720 0.01381 0.00633 16.6 S36 -0.2070 0.00364 -0.01498 -73.0 S37 -0.6200 -0.00111 -0.01498 -47.5 S38 -0.9090 -0.00321 -0.01498 -36.6 S39 -1.2400 -0.00495 -0.01498 -28.6 S40 -2.2720 -0.00812 -0.01498 -16.6

Table 5-5. k = 0.3098, J3 > 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S41 0.2500 -0.00300 0.01502 73.0 S42 0.7500 0.00214 0.01502 47.5 S43 1.1000 0.00429 0.01502 36.6 S44 1.5000 0.00600 0.01502 28.6 S45 2.7500 0.00901 0.01502 16.6 S46 -0.2500 -0.01124 0.01126 -73.0 S47 -0.7500 -0.01777 0.00356 -47.5 S48 -1.1000 -0.01799 -0.00113 -36.6 S49 -1.5000 -0.01895 -0.00474 -28.6 S50 -2.7500 -0.01565 -0.00843 -16.6

Table 5-6. k = 0.3098, J3 < 0 and f0 = 0.01 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S51 0.3030 0.01297 -0.01124 73.0 S52 0.9070 0.01706 -0.00113 47.5 S53 1.3310 0.01479 0.00267 36.6 S54 1.8150 0.01800 0.00633 28.6 S55 3.3280 0.01389 0.00844 16.6 S56 -0.3030 0.00227 -0.01498 -73.0 S57 -0.9070 -0.00320 -0.01498 -47.5 S58 -1.3310 -0.00535 -0.01498 -36.6 S59 -1.8150 -0.00700 -0.01498 -28.6 S60 -3.3280 -0.00979 -0.01498 -16.6

109 Figure 5-3. f0 = 0.04 axisymmetric ﬁnite element mesh for the unit cell.

110 Table 5-7. k = 0, J3 > 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S61 0.2500 -0.00365 0.01830 73.0 S62 0.7500 0.00261 0.01830 47.5 S63 1.1000 0.00522 0.01830 36.6 S64 1.5000 0.00731 0.01830 28.6 S65 2.7500 0.00987 0.01646 16.6 S66 -0.2500 -0.01330 0.01333 -73.0 S67 -0.7500 -0.01878 0.00376 -47.5 S68 -1.1000 -0.01886 -0.00118 -36.6 S69 -1.5000 -0.01855 -0.00464 -28.6 S70 -2.7500 -0.01621 -0.00873 -16.6

Table 5-8. k = 0, J3 < 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S71 0.2500 0.01333 -0.01330 73.0 S72 0.7500 0.01884 -0.00376 47.5 S73 1.1000 0.01892 0.00118 36.6 S74 1.5000 0.01860 0.00465 28.6 S75 2.7500 0.01625 0.00874 16.6 S76 -0.2500 0.00365 -0.01824 -73.0 S77 -0.7500 -0.00261 -0.01824 -47.5 S78 -1.1000 -0.00522 -0.01824 -36.6 S79 -1.5000 -0.00730 -0.01824 -28.6 S80 -2.7500 -0.00986 -0.01642 -16.6

Table 5-9. k = −0.3098, J3 > 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S81 0.2500 -0.00365 0.01830 73.0 S82 0.7500 0.00261 0.01830 47.5 S83 1.1000 0.00522 0.01830 36.6 S84 1.5000 0.00731 0.01830 28.6 S85 2.7500 0.00987 0.01646 16.6 S86 -0.2500 -0.01330 0.01333 -73.0 S87 -0.7500 -0.01878 0.00376 -47.5 S88 -1.1000 -0.01886 -0.00118 -36.6 S89 -1.5000 -0.01855 -0.00464 -28.6 S90 -2.7500 -0.01621 -0.00873 -16.6

111 Table 5-10. k = −0.3098, J3 < 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S91 0.2070 0.01321 -0.01478 73.0 S92 0.6200 0.01880 -0.00637 47.5 S93 0.9090 0.01830 -0.00118 36.6 S94 1.2400 0.01792 0.00247 28.6 S95 2.2720 0.01716 0.00787 16.6 S96 -0.2070 0.00444 -0.01824 -73.0 S97 -0.6200 -0.00135 -0.01824 -47.5 S98 -0.9090 -0.00391 -0.01824 -36.6 S99 -1.2400 -0.00603 -0.01824 -28.6 S100 -2.2720 -0.00890 -0.01642 -16.6

Table 5-11. k = 0.3098, J3 > 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S101 0.2500 -0.00365 0.01830 73.0 S102 0.7500 0.00261 0.01830 47.5 S103 1.1000 0.00522 0.01830 36.6 S104 1.5000 0.00731 0.01830 28.6 S105 2.7500 0.00987 0.01646 16.6 S106 -0.2500 -0.01330 0.01333 -73.0 S107 -0.7500 -0.01878 0.00376 -47.5 S108 -1.1000 -0.01886 -0.00118 -36.6 S109 -1.5000 -0.01855 -0.00464 -28.6 S110 -2.7500 -0.01621 -0.00873 -16.6

Table 5-12. k = 0.3098, J3 < 0 and f0 = 0.04 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S111 0.3030 0.01382 -0.01197 73.0 S112 0.9070 0.01987 -0.00131 47.5 S113 1.3310 0.01875 0.00339 36.6 S114 1.8150 0.01812 0.00637 28.6 S115 3.3280 0.01599 0.00972 16.6 S116 -0.3030 0.00276 -0.01824 -73.0 S117 -0.9070 -0.00390 -0.01824 -47.5 S118 -1.3310 -0.00651 -0.01824 -36.6 S119 -1.8150 -0.00768 -0.01642 -28.6 S120 -3.3280 -0.01073 -0.01642 -16.6

112 Figure 5-4. f0 = 0.14 axisymmetric ﬁnite element mesh for the unit cell.

113 Table 5-13. k = 0, J3 > 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S121 0.2500 -0.00503 0.02523 73.0 S122 0.7500 0.00360 0.02523 47.5 S123 1.1000 0.00720 0.02523 36.6 S124 1.5000 0.00907 0.02270 28.6 S125 2.7500 0.01361 0.02270 16.6 S126 -0.2500 -0.01833 0.01838 -73.0 S127 -0.7500 -0.02586 0.00519 -47.5 S128 -1.1000 -0.02597 -0.00163 -36.6 S129 -1.5000 -0.02554 -0.00640 -28.6 S130 -2.7500 -0.02232 -0.01203 -16.6

Table 5-14. k = 0, J3 < 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S131 0.2500 0.01838 -0.01833 73.0 S132 0.7500 0.02597 -0.00518 47.5 S133 1.1000 0.02608 0.00163 36.6 S134 1.5000 0.02565 0.00640 28.6 S135 2.7500 0.02240 0.01205 16.6 S136 -0.2500 0.00504 -0.02512 -73.0 S137 -0.7500 -0.00360 -0.02512 -47.5 S138 -1.1000 -0.00719 -0.02512 -36.6 S139 -1.5000 -0.00906 -0.02262 -28.6 S140 -2.7500 -0.01358 -0.02262 -16.6

Table 5-15. k = −0.3098, J3 > 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S141 0.2500 -0.00453 0.02270 73.0 S142 0.7500 0.00324 0.02270 47.5 S143 1.1000 0.00648 0.02270 36.6 S144 1.5000 0.00907 0.02270 28.6 S145 2.7500 0.01225 0.02043 16.6 S146 -0.2500 -0.01650 0.01654 -73.0 S147 -0.7500 -0.02328 0.00467 -47.5 S148 -1.1000 -0.02338 -0.00146 -36.6 S149 -1.5000 -0.02299 -0.00576 -28.6 S150 -2.7500 -0.02009 -0.01083 -16.6

114 Table 5-16. k = −0.3098, J3 < 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S151 0.2070 0.01639 -0.01833 73.0 S152 0.6200 0.02333 -0.00790 47.5 S153 0.9090 0.02271 -0.00146 36.6 S154 1.2400 0.02471 0.00340 28.6 S155 2.2720 0.02129 0.00976 16.6 S156 -0.2070 0.00550 -0.02262 -73.0 S157 -0.6200 -0.00168 -0.02262 -47.5 S158 -0.9090 -0.00485 -0.02262 -36.6 S159 -1.2400 -0.00748 -0.02262 -28.6 S160 -2.2720 -0.01103 -0.02036 -16.6

Table 5-17. k = 0.3098, J3 > 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S161 0.2500 -0.00453 0.02270 73.0 S162 0.7500 0.00324 0.02270 47.5 S163 1.1000 0.00648 0.02270 36.6 S164 1.5000 0.00907 0.02270 28.6 S165 2.7500 0.01225 0.02043 16.6 S166 -0.2500 -0.01650 0.01654 -73.0 S167 -0.7500 -0.02328 0.00467 -47.5 S168 -1.1000 -0.02338 -0.00146 -36.6 S169 -1.5000 -0.02299 -0.00576 -28.6 S170 -2.7500 -0.02009 -0.01083 -16.6

Table 5-18. k = 0.3098, J3 < 0 and f0 = 0.14 spherical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] S171 0.3030 0.01715 -0.01485 73.0 S172 0.9070 0.02466 -0.00163 47.5 S173 1.3310 0.02327 0.00420 36.6 S174 1.8150 0.02248 0.00791 28.6 S175 3.3280 0.01983 0.01205 16.6 S176 -0.3030 0.00343 -0.02262 -73.0 S177 -0.9070 -0.00483 -0.02262 -47.5 S178 -1.3310 -0.00807 -0.02262 -36.6 S179 -1.8150 -0.01058 -0.02262 -28.6 S180 -3.3280 -0.01331 -0.02036 -16.6

115 1.2

f = 01.0 Σe 1

σ T 0.8

0.6

σ = σ CT 0.4 FE Yield Points (J3>0)

0.2 FE Yield Points (J3<0)

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-5. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for k = 0 (i.e., von Mises) and an initial porosity of f0 = 0.01.

1.2 σ σ =1 GTN Criterion CT FE Data (J3>0) Σ 1 e FE Data (J3<0) σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-6. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for k = 0 (i.e., von Mises) with initial porosities of f0 = 0.01, f = 0.04 and f = 0.14.

116 1.2 f = 01.0

Σe 1

σ T 0.8

0.6

σ σ CT = 82.0 0.4 FE Yield Points (J3>0)

0.2 FE Yield Points (J3<0)

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-7. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for an FCC material with k = −0.3098 (σT < σC ) and an initial porosity of f0 = 0.01.

1.2

f = 01.0 Σe 1

σ T 0.8

0.6

σ σ CT = 21.1 0.4 FE Yield Points (J3>0)

0.2 FE Yield Points (J3<0)

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-8. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for a BCC material with k = 0.3098 (σT > σC ) and an initial porosity of f0 = 0.01.

117 1.2 f = 04.0

Σe 1

σ T 0.8

0.6

σ σ CT = 82.0 0.4 FE Yield Points (J3>0)

0.2 FE Yield Points (J3<0)

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-9. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for an FCC material with k = −0.3098 (σT < σC ) and an initial porosity of f0 = 0.04.

1.2 f = 04.0

Σe 1 σ σ CT = 21.1 σ T 0.8

0.6

0.4 FE Yield Points (J3>0)

0.2 FE Yield Points (J3<0)

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-10. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for a BCC material with k = 0.3098 (σT > σC ) and an initial porosity of f0 = 0.04.

118 1.2 f = 14.0 FE Yield Points (J3>0) FE Yield Points (J3<0) Σe 1 σ T σ σ CT = 82.0 0.8

0.6

0.4

0.2

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-11. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for an FCC material with k = −0.3098 (σT < σC ) and an initial porosity of f0 = 0.14.

1.2 f = 14.0 FE Yield Points (J3>0) FE Yield Points (J3<0) Σe 1 σ σ CT = 21.1 σ T 0.8

0.6

0.4

0.2

0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

Σm

σ T

Figure 5-12. Axisymmetric ﬁnite element results versus analytic yield curve using q1 = 4/e for a BCC material with k = 0.3098 (σT > σC ) and an initial porosity of f0 = 0.14.

119 1 Σ J3 > 0 Σ J3 < 0 0.8

0.6 e T ˜ Σ σ 0.4

0.2

f=0.14 f=0.04 f=0.01 0 −1.5 −1 −0.5 0 0.5 1 1.5 Σm 2σT

Figure 5-13. Axisymmetric ﬁnite element yield points versus analytical yield curve for k = 0 (i.e., von Mises) and various porosities (the vertical axis is the von Mises eﬀective stress).

1 Σ J3 > 0 Σ J3 < 0 0.8

0.6 e T ˜ Σ σ 0.4

0.2

f=0.14 f=0.04 f=0.01 0 −1.5 −1 −0.5 0 0.5 1 1.5 Σm 2σT

Figure 5-14. Axisymmetric ﬁnite element results versus analytical yield curve for an FCC material with k = −0.3098 (σT < σC ) and various porosities (the vertical axis is the isotropic CPB06 eﬀective stress, containing an implicit dependence on Σ the third invariant, J3 ).

120 1 Σ J3 > 0 Σ J3 < 0 0.8

0.6 e T ˜ Σ σ 0.4

0.2

f=0.14 f=0.04 f=0.01 0 −1.5 −1 −0.5 0 0.5 1 1.5 Σm 2σT

Figure 5-15. Axisymmetric ﬁnite element results versus analytical yield curve for a BCC material with k = 0.3098 (σT > σC ) and various porosities (the vertical axis is the isotropic CPB06 eﬀective stress, containing an implicit dependence on the Σ third invariant, J3 ).

121 CHAPTER 6 PLASTIC POTENTIALS FOR HCP METALS WITH CYLINDRICAL VOIDS This chapter focuses on developing a macroscopic yield criterion for a void-matrix aggregate where the voids have cylindrical geometry and the metal matrix is isotropic with tension-compression asymmetry. Nominally cylindrical voids are often found in experimental specimens due, for example, to the cracking of cylindrical inclusions (as in Figure 1-1), the elongation of voids in necking regions or to the linking up of elliptical voids in some preferred direction (see, for example, Figure 6-1 where spherical voids inside a shear band have coalesced to form a long, cylindrical void). Gurson(1977) developed a macroscopic yield criterion for a void-matrix aggregate containing cylindrical voids and with the matrix material obeying a von Mises yield criterion (see Chapter1). Leblond et al.(1994) extended both Gurson’s (1977) cylindrical and spherical criteria to the case when the matrix material obeys a Norton ﬂow rule. Liao et al.(1997) extended Gurson’s (1977) cylindrical criterion in the case of transversely isotropic metal sheets under plane stress loading conditions when the matrix material obeys Hill’s 1948 yield criterion (see Hill, 1950). The outline of the chapter is as follows. Section 6.1 discusses some limit solutions which the developed criterion should reduce to under the respective scenarios. Section 6.2 presents the choice of velocity ﬁeld. Section 6.3 details the derivation of parametric representations of the yield locus for the void-matrix aggregate when the voids are cylindrical, the matrix exhibits tension-compression asymmetry and the loading is axisymmetric. Section 6.4.1 develops the local plastic dissipation for either plane stress or plane strain loading and Section 6.4.2 develops the corresponding approximate analytical and macroscopic yield criterion. In this chapter, the representative volume element (RVE) is axisymmetric and contains a through-thickness, circular, cylindrical void as shown in Figure 6-2. The following assumptions are made concerning the loading and matrix material response:

122 • The axial deformation is uniform; i.e., d33 = D33 = constant.

• The axial deformation is symmetric; thus, a principal system in Cartesian coordinates can be found such that D11 = D22 and Dij = 0 if i 6= j (with similar expressions for Σ).

P • The matrix material is fully plastic and incompressible such that dij = dij in Equation (2–1) (i.e., rigid plastic behavior). In the above assumptions, d represents the microscopic rate of deformation tensor and D represents the macroscopic rate of deformation tensor. The outer radius of the RVE is b while the inner radius (or the radius of the void) is denoted by a. The macroscopic stress state, Σ, and associated deviator, Σ0, corresponding to these assumptions are   Σ11 0 0     Σ =  0 Σ 0  (6–1)  11    0 0 Σ33 and   1 (Σ11 − Σ33) 0 0  3  0   Σ =  0 1 (Σ − Σ ) 0  . (6–2)  3 11 33   2  0 0 − 3 (Σ11 − Σ33)

6.1 Limit Solutions

Before proceeding to the derivation of the macroscopic yield criterion for the porous aggregate, various limiting solutions will be investigated in order to provide direction for certain assumptions made in the later sections in order to arrive at a closed-form expression for the criterion. The macroscopic analytic criterion to be developed in the later sections should agree with the limiting solutions laid out in the following. 6.1.1 Zero porosity and von mises material limiting cases

The ﬁrst state where an analytic solution is immediately clear is when the porosity tends toward zero. In this case, the homogenized material and the matrix material become

123 the same and the yield criterion reduces to Equation (2–11). In other words,

Σe = σe = σT . (6–3)

Another limiting case of interest is the macroscopic yield criterion obtained by Gurson (1977) in which the void-matrix aggregate is comprised of cylindrical voids and a von Mises matrix material. Gurson’s macroscopic yield criterion for cylindrical voids was given in Chapter1 (see Equation (1–2)) and is rewritten below for the reader’s convenience: √ 2 ! C Σe 3Σγγ 2 ΦG = Ceqv + 2f cosh − 1 − f = 0 (1–2) σY 2σY where σY is the yield strength of the von Mises material and   (1 + 3f + 24f 6)2 for plane strain Ceqv = (1–3)  1 for axisymmetry.

Since the isotropic CPB06 yield criterion reduces to the von Mises yield criterion for k = 0, the macroscopic yield criterion for the porous aggregate containing cylindrical voids and with the matrix described by the isotropic CPB06 should also, for k = 0, reduce to Gurson’s macroscopic yield criterion. 6.1.2 Analysis of a hydrostatically-loaded hollow cylinder

The problem of a cylinder containing a cylindrical void and under axisymmetric loading has been investigated over the years in some detail when the matrix material is governed by a von Mises criterion (see, for example, Kachanov, 2004; Lubliner, 1990). Rather than obtain the full solution to the elastic-plastic problem of an isotropic CPB06 thick-walled cylinder, the next few paragraphs will be devoted to the assessment of the signs of the principal deviator stresses that are the solution to the problem. This information is crucial in determining the local plastic dissipation—speciﬁcally, the ordering of the principal values of the microscopic rate of deformation tensor (see Equation (2–27)).

124 Consider a thick-walled tube of inner radius a and outer radius b that is subjected to a hydrostatic pressure on its outer surface. As already mentioned, the more general problem of an elastic-plastic material obeying the isotropic CPB06 yield condition of Equation (2–8) will be assumed even though in the derivation of the macroscopic yield criterion for porous aggregates, the limiting solution will correspond to the rigid-plastic case. This is done because the elastic-plastic sphere is a more general problem with the rigid plastic state resulting as the Young’s modulus tends toward inﬁnity. The derivation carried out below will follow much of the same procedure used in Lubliner(1990) with the various governing equations obtained from Malvern(1969). Strain-displacement relations, strain compatibility, Hooke’s law, and the equilibrium equation will all be used to obtain a relationship between the third invariant of the local deviatoric stress tensor and the global applied pressure or mean stress. 6.1.2.1 Strain-displacement relations

Assuming small strains, the microscopic strain tensor is deﬁned as

1 ←− −→ = (u∇ + ∇u), 2

where, in cylindrical coordinates, the gradient of a vector u is deﬁned as

 ∂u 1 ∂u u ∂u  r r − θ r  ∂r r ∂θ r ∂z  ←−  ∂uθ 1 ∂uθ ur ∂uθ  [u∇] :=  +   ∂r r ∂θ r ∂z   ∂uz 1 ∂uz ∂uz  ∂r r ∂θ ∂z with −→ ←− [u∇] = [u∇]T .

In the given problem, the only external loading in the r-θ plane is the in-plane mean stress, Σγγ = Σr, which acts normal to the cylinder’s outer surface. Therefore, the displacement ﬁeld must be axially symmetric and the only nonzero displacement in the r-θ plane is the radial displacement which varies with r alone due to the axial symmetry of

125 the problem (i.e., ur = ur(r)). Also, note that if the cylinder is suﬃciently long, the axial stresses and strains away from the ends may be regarded as independent of z. With the conditions of axial symmetry and a “long” cylinder, the result is that all the shear strains and stresses vanish and the strain-displacement relation above yields the only nonzero components of the strain tensor as follows:

du = r (6–4a) r dr u = r (6–4b) θ r du = z (6–4c) z dz

6.1.2.2 Strain compatibility

The St. Venant compatibility equations are required when the strains are considered known since then the strain-displacement equations yield six equations for three displacement unknowns (in the general case). The compatibility equations ensure that the assumed strains are physically possible. (If the displacements are included as unknowns in the problem, as in Navier’s displacement equations of motion, then the compatibility equations are not needed.) The St. Venant compatibility equations are expressed below as

−→ ←− S = ∇ × E × ∇ = 0.

The above expression represents six partial diﬀerential equations. However, it can be shown that only three of these six equations are independent. The six equations are rather tedious to write in cylindrical coordinates, so only the equation needed for the present problem is expanded below as

2 ∂2 ∂2 1 ∂2 1 ∂ S = θr − θθ − rr + rr zz r ∂θ∂r ∂r2 r2 ∂θ2 r ∂r 2 ∂ 2 ∂ + rθ − θθ = 0. r2 ∂θ r ∂r

126 The other ﬁve compatibility equations in cylindrical coordinates can be found in Malvern (1969). The compatibility condition above yields the relationship

d = (r ) (6–5) r dr θ

where only one subscript is now being used for the strain components since there are no shears present and the second subscript is extraneous. Equation (6–5) is obviously satisﬁed by Equations (6–4). 6.1.2.3 Equations of motion

Neglecting body forces and assuming static equilibrium in Cauchy’s equations of motion (also known as Cauchy’s ﬁrst law of motion or simply as the balance of linear momentum), yields the following equations:

∇ · σ = 0.

The above result is commonly referred to as the equilibrium equations and represents in

3 R three ﬁrst-order, linear partial diﬀerential equations with solution σij = σij(x1, x2, x3). The expanded form of the equilibrium equation in cylindrical coordinates is listed below as

∂σ 1 ∂σ ∂σ 1 rr + θr + zr + (σ − σ ) = 0 ∂r r ∂θ ∂z r rr θθ ∂σ 1 ∂σ ∂σ 2 rθ + θθ + zθ + σ = 0 (6–6) ∂r r ∂θ ∂z r rθ ∂σ 1 ∂σ ∂σ 1 rz + θz + zz + σ = 0. ∂r r ∂θ ∂z r rz As with the displacement ﬁeld, the microscopic stress ﬁeld must also be axially symmetric since, in the r-θ plane, the problem contains only boundary stresses acting normal to the outer surface of an isotropic cylinder and since the cylinder is considered long enough such that the strains and stresses in the interior (away from the edges) are independent of z.

The only nonzero components of the stress tensor are then σr, σθ, and σz. Using these symmetries, the ﬁrst equation in Equations (6–6) is the only non-trivial equation and

127 yields: dσ σ − σ r + r θ = 0 (6–7) dr r where only one subscript is now being used for the stress components as with the strain components. 6.1.2.4 Elastic constitutive relation: Hooke’s law

The elastic response of the material is assumed to be governed by Hooke’s law. Therefore, the stress-strain relation becomes,

1 = [(1 + ν)σ − νσ δ ] (6–8) ij E ij kk ij

where E is the Young’s modulus, ν is Poisson’s ratio, and δij is the Kronecker delta. Note that for rigid plastic behavior, the Young’s modulus, E, tends toward inﬁnity such that the elastic stresses produce no strains (and, hence, no displacements). Equation (6–8) can be reduced to the following relevant equations relating the unknown strains to the unknown stresses: 1 = [σ − ν(σ + σ )] r E r θ z 1 = [σ − ν(σ + σ )] θ E θ r z 1 = [σ − ν(σ + σ )]. z E z r θ

The above expressions can be simpliﬁed by solving the last equation for σz and substituting the resultant expression into the other two. Doing so yields,

1 + ν r = [(1 − ν)σr − −νσθ] − νz E (6–9) 1 + ν = [(1 − ν)σ − −νσ ] − ν . θ E θ r z

σ 6.1.2.5 Relation between J3 and Σm In order to derive the equations for the elastic solution, the cylinder will be regarded as purely elastic. The in-plane problem contains two unknown stresses and two unknown strains which will be solved for using two equations from Hooke’s law, one equation from

128 compatibility and one equation from equilibrium (i.e., four equations for four unknowns). The axial strain and axial stress will be determined by assuming either plane stress

(σz = 0) or plane strain (z = 0) loading conditions. Substituting the equations from Hooke’s law, Equation (6–9), into the equation from strain compatibility, Equation (6–5), yields

d σ − σ [(1 − ν)σ − νσ ] − r θ = 0. dr θ r r

Now, making use of the equilibrium equation, Equation (6–7), the above equation simpli- ﬁes further to yield d (σ + σ ) = 0. dr r θ The above equation along with Equation (6–7) represent a system of two ordinary

diﬀerential equations in two unknown functions (σr and σθ). In order to solve the system, an operator method will be used (see Ross, 1984). Let the operator D = d/dr, such that the last equation along with the equilibrium equation become,

Dσr + Dσθ = 0

and 1 1 D + σ − σ = 0. r r r θ

In order to determine σr the ﬁrst equation is multiplied by 1/r, the second equation by

D, and the resulting equations are added to obtain an expression in terms of σr alone.

Using a similar procedure yields an expression in terms of σθ alone. Replacing D with d/dr yields the resulting ordinary diﬀerential equations below as,

d2σ dσ r2 r + 2r r = 0 dr2 dr d2σ dσ r2 θ + 2r θ = 0. dr2 dr The above equations are ordinary linear differential equations with variable coefficients. Furthermore, the above equations correspond to a specific type of ordinary

129 differential equation known as Cauchy-Euler equations. Cauchy-Euler equations are equations whose terms are constant multiples of an expression of the form xn(dny/dxn). The method of solution for a Cauchy-Euler equation is to make the substitution x = et in order to reduce the equation to a linear differential equation in terms of t with constant coefficients. Making the substitution into the above two equations yields,

d2σ dσ r + 2 r = 0 dt2 dt d2σ dσ θ + 2 θ = 0, dt2 dt with general solutions

c σ = c + 2 r 1 r2 c σ = c + 4 . θ 3 r2

The constants c1 to c4 are determined by substituting the above expressions back into the equilibrium equation (6–7). Doing so yields c1 = c3 and c2 = −c4. Now, let c1 = c3 =: A and c4 = −c2 =: B, such that the expressions for the stresses become, B σ = A − r r2 B σ = A + . θ r2 The constants A and B are determined by imposing the boundary conditions for the problem. The boundary conditions for the given problem, again assuming that the cylinder is totally elastic, are as follows:   0 at r = a σr =  Σr = Σγγ/2 at r = b

The constants A and B can now be solved for and are given as,

Σ A = γγ 1 − f Σ a2 B = γγ 1 − f 2

130 where f := (a/b)2. Finally, the expressions for the stresses are given as follows:

Σ /2 a2 σ = γγ 1 − (6–10a) r 1 − f r2 Σ /2 a2 σ = γγ 1 + . (6–10b) θ 1 − f r2

Now, let plane stress loading conditions be assumed such that σz = 0. If this is the case, then the microscopic mean stress, σm, is given as

2 Σ /2 σPST = γγ m 3 1 − f and the microscopic stress deviator for the plane stress case, sPST , becomes   sr 0 0   PST   s =  0 s 0   θ    0 0 sz   (6–11) 1 − a2 0 0  3 r2  Σγγ/2  2  =  0 1 + a 0  . 1 − f  3 r2   2  0 0 − 3

Note that Equation (6–11) dictates the following: √ PST sgn (sr) = −sgn (Σγγ) for r < 3a =: c

sgn (sθ) = sgn (Σγγ) for r ∈ (a, b)

sgn (sz) = −sgn (Σγγ) for r ∈ (a, b) which implies that   sgn (Σ ) if r < cPST σ  γγ sgn (J3 ) = PST  −sgn (Σγγ) if r > c where √ cPST = 3a.

131 If plane strain loading conditions are assumed instead such that σz = ν(σr + σθ), then

the microscopic mean stress, σm, will become

2 Σ /2 σPSN = (1 + ν) γγ m 3 1 − f

and the microscopic stress deviator for the plane strain case, sPSN , becomes   sr 0 0   PSN   s =  0 s 0   θ    0 0 sz   (6–12) 1−2ν − a2 0 0  3 r2  Σγγ/2  2  =  0 1−2ν + a 0  . 1 − f  3 r2   1−2ν  0 0 −2 3

Note that the term 1 − 2ν is always positive or zero since ν ≤ 0.5. Therefore, Equation (6–12) implies the following:

q 3 PSN sgn (sr) = −sgn (Σγγ) for r < 1−2ν a =: c

sgn (sθ) = sgn (Σγγ) for r ∈ (a, b)

sgn (sz) = −sgn (Σγγ) for r ∈ (a, b)

which implies that   sgn (Σ ) if r < cPSN σ  γγ sgn (J3 ) = PSN  −sgn (Σγγ) if r > c where r 3 cPSN = a. 1 − 2ν Note that in both the plane stress and plane strain case, the following is observed:

sgn (sr) = −sgn (Σγγ) for r < c

sgn (sθ) = sgn (Σγγ) for r ∈ (a, b) (6–13)

sgn (sz) = −sgn (Σγγ) for r ∈ (a, b)

132 which implies that   sgn (Σ ) if r < c σ  γγ sgn (J3 ) = (6–14)  −sgn (Σγγ) if r > c where  √  cPST = 3a for plane stress conditions c = q (6–15)  PSN 3  c = 1−2ν a for plane strain conditions. Equations (6–3), (1–2) and (6–14) provide limiting solutions that can be used to guide in the ordering of the principal values of the local rate of deformation. As already mentioned, the crucial challenge in obtaining a closed-form expression for the macroscopic criterion of the porous aggregate is the estimate of the expression of the plastic dissipation in a matrix exhibiting strength diﬀerential eﬀects. 6.2 Choice of Trial Velocity Field

The first step in determining the macroscopic yield function is to determine a particular form for the microscopic velocity field in the RVE and determine the corresponding microscopic rate of deformation. The macroscopic rate of deformation field, Dij, is defined by the following equation (see Gurson, 1977):

1 Z 1 Dij = (vinj + vjni) dS V S 2

where V is the volume of a unit cube RVE, S is the outer surface of the RVE, ~n is the unit outward normal to the surface S, and ~v is the microscopic velocity ﬁeld. The above equation can be rewritten using Gauss’s theorem and the deﬁnition for the rate of deformation as 1 Z Dij = dijdV. V V

In the above expression, dij is the microscopic rate of deformation tensor, which is by deﬁnition (see Malvern, 1969) 1 ∂vi ∂vj dij = + 2 ∂xj ∂xi where ~x is the position of a material point in Cartesian coordinates.

133 The velocity ﬁeld, ~v, in the RVE is required to be kinematically admissible. In other words, the velocity ﬁeld should satisfy the compatibility conditions (displacement boundary conditions) as well as the incompressibility condition. These conditions are written as follows:

~v|~x=b~er = D~x

= Dijxj (6–16)

div (~v) = 0.

The first requirement is a statement of the displacement boundary condition. The second requirement states that the matrix material is incompressible, which is generally the assumption made for a fully plastic state. To obtain the upper bound estimate of the overall plastic potential, the classical local velocity field v proposed by Gurson(1977); Rice and Tracey(1969) will be used. This velocity field contains a part that controls the volume change, vV , and a part that controls the change in shape, vS: v = vV + vS.

The in-plane deviatoric portion of the macroscopic rate of deformation is now deﬁned as follows: 1 D0 = D − D . αβ αβ 2 γγ In the above equation and throughout the remainder of this section, Greek subscripts range from 1 to 2. Note that in the special case of axisymmetric deformation that was

0 assumed, Dαβ = 0 since D11 = D22. The term (1/2)Dγγ in the previous expression represents the mean in-plane rate of deformation. This nomenclature is consistent with that used in Gurson(1977) and Liao et al.(1997). The boundary condition in Equation (6–16) can now be expressed in terms of the volumetric and shape-changing parts of the

134 velocity ﬁeld as follows:

S 0 vα |~x=b~er = Dαβxβ (6–17) 1 vV | = D x . α ~x=b~er 2 γγ α The incompressibility constraint can now be enforced on the volumetric portion of the velocity ﬁeld (noting that the velocity ﬁeld in the z-direction is purely volumetric) to obtain 1 ∂ 1 ∂vV ∂v rvV + θ + z = 0 r ∂r r r ∂θ ∂z ∂vV r + vV r + D = 0 ∂r r 33 or ∂vV vV r = − r + D . ∂r r 33 A solution to the equation above is

c D vV = 1 − 33 r r r 2

where c1 is a constant. Enforcing the boundary condition of Equation (6–17), yields an

expression for c1 as follows: b2 c = (D + D + D ) . 1 2 11 22 33

Using the previous result for c1, the volumetric part of the radial velocity component can be expressed as

1 b2 r vV = (D + D + D ) − D r 2 11 22 33 r 33 2 b2 r = D¯ − D r 33 2

where the new quantity D¯ is deﬁned as follows:

1 D¯ = (D + D + D ) . 2 11 22 33

135 The radial and angular portions of the volumetric part of the rate of deformation ﬁeld can now be written as (see Malvern, 1969)

∂vV dV = r rr ∂r b2 D = −D¯ − 33 r 2

and vV dV = r θθ r b2 D = D¯ − 33 . r 2

The shape-changing part of the velocity ﬁeld is found by assuming that it is of the following form: ~vS = B~x

where B is a constant, symmetric and deviatoric (i.e., tr(B) = 0) tensor. Using Equa- tion (6–17), the tensor B is shown to be identically equal to the deviatoric part of the macroscopic rate of deformation tensor, D0, which is zero due to the assumption of axially symmetric and uniform deformation (pure in-plane volume change). Therefore, the velocity ﬁeld can be expressed as

b2 r vV = D¯ − D e and vS = D0x (6–18) r 33 2 r

¯ 0 where, once again, D = (1/2)(D11 + D22 + D33) and D = D − (1/2)(D11 + D22). The principal values (unordered) of the local strain rate ﬁeld d corresponding to the velocity ﬁeld v given by Equation (6–18) are as follows:

b2 D d = −D¯ − 33 r r 2 b2 D d = D¯ − 33 (6–19) θ r 2

dz = d33 = D33.

136 Since principal values of a second-order tensor are invariant (see Malvern, 1969), the above components yield the principal microscopic rate of deformation components with respect to any arbitrary choice of initial coordinate system. The unordered principal components ˜ of the microscopic rate of deformation tensor, dα, can then be expressed as follows:

˜ d1 = dr ˜ d2 = dθ (6–20) ˜ d3 = dz = D33.

6.3 Parametric Representation of the Porous Aggregate Yield Locus for Axisymmetric Loading

This section details the development of an exact (for the assumed velocity ﬁeld given by Equation (6–18)), parametric representation for the yield locus of a void-matrix aggregate when the matrix is governed by the Cazacu et al.(2006) criterion. Here the void geometry is cylindrical and the loading is axisymmetric. The following constants are deﬁned for use in the analysis: σ β = T σC and 3 p = β2 + 1 4 m = 3 1 − β2

q = 3β2 − 1 s 3 (2 − β2) g = . 4β2

Also, β2 − 2 s = 3 2β2 and

s¯3 = −s3.

137 The necessary derivatives of the macroscopic plastic dissipation, W +, are obtained using ¯ the chain rule on the two variables D and B = D33 such that ∂W + ∂W + ∂D¯ 1 ∂W + Σ11 = = = ∂D11 ∂D¯ ∂D11 2 ∂D¯ ∂W + Σ = Σ + Σ = 2Σ = γγ 11 22 11 ∂D¯ and ∂W + ∂W + ∂D¯ ∂W + ∂B 1 ∂W + ∂W + ∂W + Σ33 = = + = + = Σ11 + ∂D33 ∂D¯ ∂D33 ∂B ∂D33 2 ∂D¯ ∂B ∂B + ∂W Σe = |Σ33 − Σ11| = . ∂B Lastly, let the parameter u be deﬁned such that

D¯ u = . (6–21) |B|

This ratio and the previously-deﬁned constants will be used in the subsequent analysis. 6.3.1 β ≥ 1: the matrix yield strength in tension is greater than in compression

A derivation of the parametric representations is given in the following for a material

Σ where β = σT /σC ≥ 1 with Σm > 0 and J3 < 0. The derivations for the other cases (and for β ≤ 1) are given in AppendixA. The parametric representations are then plotted for this material (β ≥ 1) for multiple values of void volume fraction.

Σ 6.3.1.1 Σm > 0 and J3 < 0 Σ ¯ If Σm > 0 and J3 < 0 then D > 0 and B = D33 < 0 such that

D¯ u = − B

by Equation (6–21). The relevant branches of the plastic multiplier rate (see Equation (2–29)) must now be determined. The expression for the plastic multiplier rate λ˙ is

138 rewritten below for the reader’s convenience as  s r 2 2 2 2  2 (3β − 2) d + d + d d1 1  1 2 3 if ≥  2 p p 4 2  3 (2β − 1) dijdij 2 (β − β + 1) λ˙ = s (6–22) r 2 2 2 2 2 2  2 d + d + (3 − 2β ) d /β d3 −β  1 2 3 if ≤ .  2 p p 4 2  3 (2 − β ) dijdij 2 (β − β + 1)

Let x denote the quantity x = Db¯ 2/r2 such that

B d = − x + r 2 B d = x − θ 2

dz = B.

¯ Evidently, dθ ≥ dr ≥ dz for this case (D > 0 and B = D33 < 0). The result is that the relevant branch from Equation (6–22) either corresponds to dθ = d1 being the principal ˙ ˙ value that is alone in sign (with notation λ = λθ) or dz = d3 being the principal value that ˙ ˙ is alone in sign (with notation λ = λz). ˙ Let λz be the relevant branch. The range where this branch is valid is determined by solving the relevant inequality from Equation (6–22) for any roots, xi. Thus, for

d −β2 z ≤ , p p 4 2 dijdij 2 (β − β + 1) ¯ the valid root (noting that x ≥ 0, β > 1, D > 0 and D33 < 0) is obtained as

x¯3 =s ¯3B.

The plastic multiplier rate for this range is

2 ˙ p 2 2 2 λz = x + g B p3 (2 − β2)

˙ ˙ Similarly, for λ = λθ, the inequality

d 1 θ ≥ p p 4 2 dijdij 2 (β − β + 1)

139 yields the following valid root:

x¯3 =s ¯3B.

The plastic multiplier rate for this branch can be written as s 2 p λ˙ = pB2 + mBx + qx2. θ 3 (2β2 − 1)

˙ ˙ In summary, it can be seen that the valid range for λ = λz is 0 ≤ u ≤ s¯3 and the valid ˙ ˙ range for λ = λθ iss ¯3 ≤ u ≤ ∞. In general, the macroscopic plastic dissipation can be written as

1 Z W +(D) = w(d)dV V V σ Z = T λdV.˙ V V In order to determine an expression for the macroscopic plastic dissipation, W +(D), there are three ranges that must be looked at separately that correspond to the relevant limits of integration.

First range: 0 ≤ u ≤ s¯3f. In this range, the macroscopic plastic dissipation is given as

¯ Z D/f¯ 2 σT D p dx W + = √ x2 + g2B2 p 2 2 3 2 − β D¯ x ¯ D/f¯ 2 σT D h i = √ p h4(x) 3 2 − β2 D¯ where

Z px2 + g2B2 h (x) = dx 4 x2 p x2 + g2B2 p = − + ln x + x2 + g2B2 . x

140 The derivatives can now be written as

¯ ∂W + W + 2 σ D¯ ∂h ∂x D/f = + √ T 4 ¯ ¯ p 2 ¯ ∂D D 3 2 − β ∂x ∂D D¯ ¯ ∂W + 2 σ D¯ ∂h ∂x ∂h D/f = √ T 4 + 4 p 2 ∂B 3 2 − β ∂x ∂B ∂B D¯ with ∂h px2 + g2B2 4 = ∂x x2 and ∂h Bg2 4 = − ∂B x2 + xpx2 + g2B2 such that, at yielding,

+ p 2 2 2 ! ∂W 2 σT u + u + g f 1 Σγγ = = √ ln ∂D¯ 3 p2 − β2 u + pu2 + g2 f

and + ∂W 2 σT p 2 2 2 p 2 2 Σe = = √ u + g f − u + g . ∂B 3 p2 − β2

Second range:s ¯3f ≤ u ≤ s¯3. In this next range, the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x3 x ¯ Z x3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x r ¯ D/f¯ 2 σT D mB = qh1(x) + h2(x) + h3(x) 3 p 2 2 2β − 1 x3 ¯ x 2 σT D h i 3 + √ p h4(x) 3 2 − β2 D¯

where x3 = s3B. In the previous equation, the integral solution of √ Z X mB dx = qh (x) + h (x) + h (x) x2 1 2 2 3

141 has been used with X = pB2 + mBx + qx2 and

Z dx 1 p h1(x) = √ = √ ln 2 qX + 2qx + mB X q ! Z dx −1 2ppB2X + mBx + 2pB2 h2(x) = √ = ln x X ppB2 x √ − X h (x) = . 3 x The relevant derivatives can now be written as

+ + r ¯ ( D/f¯ D/f¯ ∂W W 2 σT D ∂h1 ∂x mB ∂h2 ∂x = + q + ∂D¯ D¯ 3 p 2 ∂x ∂D¯ 2 ∂x ∂D¯ 2β − 1 x3 x3 ¯ s x3  D/f 2 ∂h3 ∂x 2 (2β − 1) ∂h4 ∂x  + + ∂x ∂D¯ 2 − β2 ∂x ∂D¯ x3 D¯  ( ¯ ¯ ∂W + r2 σ D¯ ∂h ∂x ∂h D/f mB ∂h ∂x ∂h D/f = T q 1 + 1 + 2 + 2 ∂B 3 p 2 ∂x ∂B ∂B 2 ∂x ∂B ∂B 2β − 1 x3 x3 ¯ s ) ∂h ∂x ∂h D/f 2 (2β2 − 1) ∂h ∂x ∂h x3 + 3 + 3 + 4 + 4 ∂x ∂B ∂B 2 − β2 ∂x ∂B ∂B x3 D¯ with ∂h 1 1 = √ ∂x X ∂h 1 2 = √ ∂x x X ∂h − (mBx/2 + qx2 − X) 3 = √ ∂x x2 X and √ √ ! ∂h1 1 q (mx + 2pB) + m X = √ √ √ ∂B q 2 qX + (2qx + mB) X √ ! ∂ (Bh ) −sgn(B) pB (4pB2 + 3mBx + 2qx2) + (mx + 4pB) ppB2 X 2 = √ √ ∂B p 2pB2X + (mBx + 2pB2) X ∂h pB + mx/2 3 = − √ . ∂B x X

142 This yields the following stress invariants at yielding:

r "p 2 p 2 2 ! 2 σT 2 (2β − 1) s3 − s3 + g Σγγ = ln − 3 p2β2 − 1 p2 − β2 u + pu2 + g2 √ p ! √ 2 q pf 2 − muf + qu2 + 2qu − mf 1 + q ln √ · p 2 f 2 q p + ms3 + qs3 − 2qs3 − m √ p 2 2 !# m 2 p pf − muf + qu − mu + 2pf s3 + √ ln − √ · 2 p p 2 u 2 p p + ms3 + qs3 + ms3 + 2p

and " # r2 ppf 2 − muf + qu2 p2 (u2 + g2)

Σe = σT p − p . 3 2β2 − 1 2 − β2

Third range:s ¯3 ≤ u ≤ ∞. In this last range, the macroscopic plastic dissipation is as follows:

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 D¯ x r ¯ D/f¯ 2 σT D mB = qh1(x) + h2(x) + h3(x) p 2 3 2β − 1 2 D¯ with derivatives ( ¯ ¯ ∂W + W + r2 σ D¯ ∂h ∂x D/f mB ∂h ∂x D/f = + T q 1 + 2 ¯ ¯ p 2 ¯ ¯ ∂D D 3 2β − 1 ∂x ∂D D¯ 2 ∂x ∂D D¯ D/f¯ ) ∂h3 ∂x + ¯ ∂x ∂D D¯ ( ¯ ¯ ∂W + r2 σ D¯ ∂h ∂x ∂h D/f mB ∂h ∂x ∂h D/f = T q 1 + 1 + 2 + 2 p 2 ∂B 3 2β − 1 ∂x ∂B ∂B D¯ 2 ∂x ∂B ∂B D¯ ¯ ) ∂h ∂x ∂h D/f + 3 + 3 ∂x ∂B ∂B D¯ such that, at yielding,

r " √ p 2 2 ! 2 σT √ 2 q pf − muf + qu + 2qu − mf 1 Σγγ = q ln √ 3 p2β2 − 1 2 qpp − mu + qu2 + 2qu − m f √ p !# m 2 p pf 2 − muf + qu2 − mu + 2pf + √ ln √ 2 p 2 ppp − mu + qu2 − mu + 2p

143 and r 2 σT p 2 2 p 2 Σe = pf − muf + qu − p − mu + qu . 3 p2β2 − 1 Σ Figure 6-3 illustrates the three branches derived here for the case of β ≤ 1, J3 < 0

and Σm > 0. The horizontal axis in the ﬁgure is the in-plane mean stress 0.5Σγγ =

0.5(Σ11 + Σ22) normalized by the tensile yield strength. The curves shown are for a material with a yield-strength ratio of β = 1.21. Curves corresponding to three diﬀerent void volume fractions are given in Figure 6-3 and show a decreasing yield locus with increasing porosity. This trend is to be expected since a void-matrix aggregate has less load-bearing area as the void volume fraction increases. 6.3.2 Discussion

Section 6.3.1 detailed the development of parametric representations for a material whose yield strength in tension was greater than the yield strength in compression and for

Σ the particular case of J3 < 0 and Σm > 0. For this type of material (β ≥ 1) there are three additional cases that need to be considered corresponding to permutations on the

Σ sign of J3 and Σm. The derivation of and expressions for these parametric representations are given in AppendixA. Figure 6-4 plots the complete parametric representation of the yield curve for the particular strength ratio of β = 1.21 and f = 0.01. As can be seen, the curves are symmetric about the vertical axis and depend on the sign of the third invariant of the

Σ macroscopic stress deviator J3 . Parametric representations were also derived for a material whose tensile yield strength is less than the compressive yield strength (i.e., β < 1 such that k < 0). The relevant equations are presented in AppendixA. Figure 6-5 plots the complete parametric representation of the yield curve corresponding to a particular strength ratio of β = 0.82 and void volume fraction f = 0.01. Again, the curves are symmetric about the vertical

Σ axis but are ﬂipped with respect to Figure 6-4. In other words, the J3 > 0 curve is the

144 Σ one that reaches the highest point on the vertical axis in Figure 6-4 whereas the J3 < 0 curve is the one that reaches the highest point on the vertical axis in Figure 6-5. Lastly, the CPB06 isotropic criterion reduces to the von Mises yield criterion when no tension-compression asymmetry exists in the matrix (i.e., for β = 1 such that k = 0). This results in the exact parametric representations presented here reducing to the cylindrical criterion developed by Gurson(1977) using the same velocity ﬁeld given by Equation (6–18). 6.4 Proposed Closed-Form Expression for a Plane Strain Yield Criterion

In the previous section, a parametric representation of the yield surface was developed. It is to be noted that the integral representing the macroscopic plastic dissipation for the given velocity field was calculated exactly. The main drawback with the parametric representations derived in the previous section is that multiple branches appear and ¯ u = D/|D33| cannot be easily eliminated from any of them. For ease in applications and finite element implementation, a closed form of the surface rather than a parametric form is more desirable. With the goal of deriving such a closed-form expression, certain terms will be neglected in the expression of the local plastic dissipation in this section. The following will focus on deriving analytical expressions for the yield locus when the void-matrix aggregate contains cylindrical voids, the matrix exhibits tension-compression asymmetry and the loading condition corresponds to plane strain (i.e., u = ∞ using the terminology of the previous section). It is demonstrated that this approximation results in a much simpler form for Σe and Σγγ. The validity of the simplified analytical form of the criterion for cylindrical voids will be further assessed by comparisons with finite element calculations. 6.4.1 Calculation of the local plastic dissipation

Equation (2–27) contains the squared values of the rate of deformation components; therefore, the expressions for these squared values are developed below, using the

145 substitution ρ = r2/b2 and Equations (6–19) and (6–20):

D¯ 2 D 2 D¯ D d˜2 = + 33 + 2 33 1 ρ 2 ρ 2 D¯ 2 D 2 D¯ D d˜2 = + 33 − 2 33 2 ρ 2 ρ 2 ˜2 2 d3 = D33 such that the second invariant of the rate of deformation tensor can be expressed as

D¯ 2 D 2 d d = 2 + 6 33 . ij ij ρ 2

Now the plastic multiplier rate from Equation (2–27) can be written as

˙ 2 2 λ = z2 (dijdij) + (z1 + z3) dlone

˙ where z1–z3 were deﬁned in Table 2-2 of Chapter2. The previous expression for λ requires that the principal rate of deformation component corresponding to the lone positive or

lone negative principal stress deviator be identiﬁed (i.e., dlone). Notice from Equation

(6–13) that the lone principal rate of deformation components are dz and dθ for r > c and r < c, respectively, for the case of an elastic and hydrostatically-loaded hollow cylinder.

For simplicity, the same dlone will be assumed in this section such that two diﬀerent expressions result for the plastic multiplier rate depending on whether the lone component is dz or dθ (i.e., depending on whether r > c or r < c, respectively). The ﬁrst of the plastic multiplier rates corresponding to the lone component of dθ is written below as,

2 ˙ 2 λθ = z2 (dijdij) + (z1 + z3) dθ " # D¯ 2 D 2 = z 2 + 6 33 2 ρ 2 " # D¯ 2 D 2 D¯ D + (z + z ) + 33 − 2 33 . 1 3 ρ 2 ρ 2

The coupled term in the last bracketed expression makes the integration more cumbersome and the prospect of ﬁnding an analytic expression for the macroscopic yield criterion

146 more uncertain. In order to simplify the subsequent calculations, the coupled term will be neglected; the validity of this approximation can be assessed, for example, using ﬁnite element calculations. Neglecting the last term in the previous expression yields, s ¯ 2 2 ˙ D D33 λθ ≈ (z1 + 2z2 + z3) + (z1 + 6z2 + z3) ρ 2 (6–23)

θp 2 = z1 x + gθ where

θ √ z1 := z1 + 2z2 + z3 2 z1 + 6z2 + z3 D33 gθ := z1 + 2z2 + z3 2 D 2 = zθ 33 2 2 and D¯ x := . ρ Likewise,

2 ˙ 2 λz = z2 (dijdij) + (z1 + z3) dz " # D¯ 2 D 2 = z 2 + 6 33 2 ρ 2

2 + (z1 + z3) D33.

Therefore, s ¯ 2 2 ˙ D D33 λz = 2z2 + (4z1 + 6z2 + 4z3) ρ 2 (6–24)

zp 2 = z1 x + gz

147 where

z √ z1 := 2z2 2 4z1 + 6z2 + 4z3 D33 gz := 2z2 2 D 2 = zz 33 2 2

and D¯ x = ρ

as before. Note that these deﬁnitions of x and gθ,z reduce to the ones used in Gurson

(1977) when k = 0 (since then z1 = z3 = 0 and z2 = 2/3). Equations (6–23) and (6–24) provide tractable forms for the plastic multiplier rate, λ˙ , which can be used in deriving the macroscopic plastic dissipation in the RVE. 6.4.2 Development of the macroscopic plastic dissipation expressions

The upper-bound, macroscopic plastic dissipation, W +, is deﬁned in terms of the microscopic plastic dissipation, w, as

1 Z W + = w (d) dV. V V

The microscopic plastic dissipation is determined as follows:

w = σijdij ˙ ∂f = σijλ ∂σij ˙ = σT λ

˙ where σT is the yield strength in uniaxial tension (assumed constant) and λ is given by Equation (2–27) or, speciﬁcally for the case presented in this subsection, by Equations (6–23) and (6–24).

148 Therefore, the upper bound macroscopic plastic dissipation becomes

σ Z W + = T λdV˙ V V Z L Z 2π Z b σT ˙ = 2 λ (r) rdrdθdz πb L 0 0 a "Z fˆ Z 1 # ˙ ˙ = σT λθ (ρ) dρ + λz (ρ) dρ f fˆ "Z D/f¯ Z D/¯ fˆ # ¯ ˙ dx ˙ dx = σT D λθ (x) 2 + λz (x) 2 D/¯ fˆ x D¯ x

Z D/f¯ p 2 Z D/¯ fˆ p 2 ! ¯ θ x + gθ z x + gz = σT D z1 2 dx + z1 2 dx D/¯ fˆ x D¯ x where f = a2/b2 is the void volume fraction in the RVE with ρ = r2/b2 and x = D/ρ¯ being defined in the previous section. The parameter fˆ is the location where the third invariant of the stress deviator flips sign in the hydrostatic analysis of Section 6.1.2 and is defined as

 c2  if c < b fˆ := b2  1 if c ≥ b where c is deﬁned in Equation (6–15) and depends on whether plane stress or plane strain loading conditions are assumed. The solution to the integral expression for W + can be easily found in a table of integrals (see, for example, Zwillinger(2003)) and is given below as p p Z x2 + g x2 + g p dx = − + ln x + x2 + g . x2 x Using the previous equation, the integral in the expression for the upper-bound macroscopic plastic dissipation, W +, can be solved to yield the following equation:

 D/f¯ " p 2 #  x + gθ p W + = σ D¯ zθ − + ln x + x2 + g T 1 x θ  D/¯ fˆ (6–25) D/¯ fˆ " p 2 # x + gz p  + zz − + ln x + x2 + g . 1 x z D¯ 

149 The equation above will now be used in order to derive expressions for the macroscopic

effective stress, Σe, and the in-plane mean stress, (1/2)Σγγ. The CPB06 effective stress is given by Equations (2–12) and (2–13). For the particular problem illustrated by Equation (6–2), the form of the effective stress depends on whether Σ11 or Σ33 is the larger stress component; alternatively, it could be stated that the form of the effective stress depends on the sign of the macroscopic third invariant of the

Σ stress deviator, J3 . The following expression gives the eﬀective stress depending on the

Σ sign of J3 :  σT Σ  − (Σ33 − Σ11) if J3 < 0  σC  Σ˜ = Σ . e (Σ33 − Σ11) if J3 > 0   Σ  0 if J3 = 0 These expressions can be combined into a single expression using the sign function as follows: ˜ (Σ33 − Σ11) σT 2 Σ Σ σT Σe = − 2 − sgn J3 + sgn J3 + 1 2 σC σC (6–26) c = ze (Σ33 − Σ11)

where c σT 1 2 Σ Σ σT ze = − + sgn J3 + sgn J3 + 1 . (6–27) σC 2 σC

An expression for the in-plane mean stress, (1/2)Σγγ, is now needed. This is done simply by noting that Σ11 = Σ22 by Equation (6–1); therefore, (1/2)Σγγ = Σ11. Next, expressions for Σ11 and Σ33 in terms of the upper-bound macroscopic plastic dissipation of Equation (6–25) need to be developed. This is done by using the following relation:

∂W Σ = ij ∂D ij (6–28) ∂W + ≈ . ∂Dij

150 Using the chain rule in Equation (6–28) yields the following:

∂W + ∂W + ∂D¯ ∂W + ∂x ∂D¯ = + ∂D11 ∂D¯ ∂D11 ∂x ∂D¯ ∂D11 ∂W + ∂W + ∂D¯ ∂W + ∂x ∂D¯ ∂W + ∂g ∂W + ∂g = + + θ + z ∂D33 ∂D¯ ∂D33 ∂x ∂D¯ ∂D33 ∂gθ ∂D33 ∂gz ∂D33 where the individual derivatives are ∂D¯ ∂D¯ 1 = = ∂D11 ∂D33 2 ∂x 1 = ∂D¯ ρ

∂gθ,z θ,z D33 = z2 ∂D33 2 ∂W + W + = ∂D¯ D¯ with

 " # D/f¯ +  p 2 ∂W ¯ θ gθ x + gθ + x = σT D z + ∂x 1 x2px2 + g xpx2 + g + x2 + g  θ θ θ D/¯ fˆ " # D/¯ fˆ g px2 + g + x  + zz z + z 1 x2px2 + g xpx2 + g + x2 + g z z z D¯   " # D/f¯ " # D/¯ fˆ  px2 + g px2 + g  = σ D¯ zθ θ + zz z T 1 x2 1 x2  D/¯ fˆ D¯  and " # ∂W + σ 1 1 = T Dz¯ θ,z − + 1 p 2 p 2 2 ∂gθ,z 2 x x + gθ,z x x + gθ,z + x + gθ,z

" p 2 # σT ¯ θ,z 1 x + gθ,z = Dz1 − . 2 gθ,z gθ,zx

151 The expressions for Σ11 and Σ33 can now be written as follows:

+ h i D/f¯ h i D/¯ fˆ ∂W σT θ p 2 z p 2 Σ11 ≈ = z1 ln x + x + gθ + z1 ln x + x + gz ∂D11 2 D/¯ fˆ D¯ ∂W + ∂W + Σ33 ≈ = ∂D33 ∂D11  " # D/f¯ " # D/¯ fˆ  p 2 p 2  σT ¯ θ θ D33 x + gθ z z D33 x + gz + D z1z2 − + z1 z2 − . 2 2 gθx 2 gzx  D/¯ fˆ D¯ 

Therefore (dropping the ≈), the in-plane mean stress and the eﬀective stress are determined as follows:     ¯ p ¯ 2 2 ˆ! Σγγ σT  θ D + D + gθf f = z1 ln  q   2 2 ¯ ¯ 2 ˆ2 f  D + D + gθf (6–29)  q   D¯ + D¯ 2 + g fˆ2 z z 1  + z1 ln    ¯ p ¯ 2 ˆ D + D + gz f 

and " σ zθpzθ q ˜ c T 1 2 p ¯ 2 2 ¯ 2 ˆ2 Σe = ze √ D + gθf − D + gθf 2 gθ √ (6–30) zz zz q 1 2 ¯ 2 ˆ2 p ¯ 2 + √ D + gzf − D + gz . gz

The macroscopic yield function gives a relationship between the macroscopic eﬀective ˜ stress from Cazacu et al.(2006), Σe, and the in-plane mean stress, Σγγ/2. First, Equations (6–29) and (6–30) can be written as follows:

θ z Σγγ = Σγγ + Σγγ (6–31) ˜ ˜ θ ˜ z Σe = Σe + Σe

˜ θ θ where the terms due to r < c in Equations (6–29) and (6–30) are deﬁned as Σe and Σγγ ˜ z z while the terms due to r > c are deﬁned as Σe and Σγγ. Breaking up Equations (6–29) and (6–30) in this way allows the invariant terms in each of the regions (r < c and r > c) to be grouped together. For example, consider the region where r < c and the following

152 substitutions: q θ ˆ2 ¯ 2 ¯ Y1 := gθf + D + D

θ p 2 ¯ 2 ¯ Y2 := gθf + D + D √ ! Σ˜ θ 2 g T θ := e θ II zcσ θp θ e T z1 z2 θ ˆ θ ! θ Σγγ fY2 TI := θ = ln θ z1σT fY1

f suchthatT θ = Y θ − Y θ exp T θ II 1 fˆ 1 I T θ Y θ = II 1 1 − f exp T θ fˆ I f θ θ ˆTII exp TI Y θ = f . 2 1 − f exp T θ fˆ I θ θ The expressions for Y1 and Y2 can now be combined as follows:

θ2 ˆ2 ¯ θ Y1 − gθf = 2DY1

θ2 2 ¯ θ Y2 − gθf = 2DY2

which yields

θ2 ˆ2 θ2 2 Y1 − gθf Y2 − gθf θ = θ . Y1 Y2

The previous relations can be combined to obtain

T θ 2 II 2 ˆ2 ˆ 1 θ θ = f + f − 2ff exp TI + exp −TI gθ 2 2 ˆ2 ˆ θ = f + f − 2ff cosh TI

such that !2 !2 ! Σ˜ θ 2 Σθ e = f 2 + fˆ2 − 2ffˆcosh γγ (6–32) zcσ θp θ zθσ e T z1 z2 1 T

153 and, using a similar procedure in the region where r > c,

!2 Σ˜ z 2 2 Σz e √ ˆ2 ˆ γγ c z z = 1 + f − 2f cosh z . (6–33) zeσT z1 z2 z1 σT

While the two relations in Equations (6–32) and (6–33) have a very similar form to the cylindrical yield criterion in Gurson(1977), the criterion must be expressed in terms ˜ of the macroscopic invariants Σe and Σγγ. Having the yield criterion broken up by region (i.e., r < c and r > c) is not practically useful since, in general, only the macroscopic

quantities on the boundary of the RVE are known (e.g., the value for Σγγ calculated

θ on the boundary of the RVE is typically known rather than the values of either Σγγ or

z Σγγ). Therefore, Equations (6–32) and (6–33) will be treated as particular limiting cases and these two expressions will be combined to have a general criterion. The validity of this general criterion can then be assessed using ﬁnite element unit cell calculations, for example. The ﬁrst case to be considered is when fˆ → f; in other words, c ≈ a. Based on the elastic analysis of Section 6.1.2, this is only an approximate condition which should hold for very small void volume fraction. Under this scenario, the macroscopic yield criterion is given by Equation (6–33) and can be written as follows:

2 p Σ ! 3J2 2 Σγγ σ = 1 + f − 2f cosh z for J3 > 0 σT σT z1 2 (6–34) p Σ ! 3J2 2 2 Σγγ σ β = 1 + f − 2f cosh z for J3 < 0 σT σT z1

where β = σT /σC .

σ There are two issues with Equations (6–34). First, J3 is a microscopic quantity and must be expressed in terms of average quantities to be useful in ﬁnite element calculations where only average quantities on the boundary and not the microscopic ﬁelds in the

σ RVE are known. Secondly, neither of Equations (6–34) contain the solution for J3 = 0.

σ This is because under the assumption of axisymmetric loading, J3 = 0 is equivalent to

154 Σγγ = 0 which is the trivial case of no loading. Even though this criterion is derived under axisymmetric loading assumptions, the analytical form makes it an attractive criterion to use for more general loading cases such as the plane strain non-axisymmetric loading analyzed in Chapter7. The previous two issues will be addressed as follows. First, note

σ Σ that for f = 0, J3 = J3 ; this ensures that Equation (6–34) reduces to the Cazacu et al. (2006) criterion of Equation (2–12) for f = 0. Secondly, for f 6= 0 the two expressions given in Equations (6–34) will be combined in order to have a single equation for general

use (since these two equations should agree when Σγγ = 0). The combined form is obtained by taking the average of the constant on the left-hand-side of each expression in Equations (6–34) to yield

2 p Σ ! z 3J2 2 Σγγ Φ = cz − 1 − f + 2f cosh z = 0 (6–35) σT σT z1

where     (β2 + 1) 1 for f 6= 0  2 cz = (6–36)    sgn2(J Σ) + sgn(J Σ) (1 − β2) 1 + β2 for f = 0.  3 3 2

The second case to be considered occurs when fˆ = 1 such that c = b. This scenario is governed by Equation (6–32) and yields

2 p Σ ! 2 3J2 4β − 2 2 Σγγ σ 2 = 1 + f − 2f cosh θ for J3 > 0 σT β + 1 σT z1 2 (6–37) p Σ ! 2 4 3J2 4β − 2β 2 Σγγ σ 2 = 1 + f − 2f cosh θ for J3 < 0. σT β + 1 σT z1

Following a similar rationale as before (except that f = 0 is no longer a concern since fˆ = 1), a general expression for the yield condition can be obtained as

2 p Σ ! θ 3J2 2 Σγγ Φ = cθ − 1 − f + 2f cosh θ = 0 (6–38) σT σT z1

155 where −β4 + 4β2 − 1 c = . (6–39) θ β2 + 1 Now, for the general case when a ≤ c ≤ b, Equations (6–35) and (6–38) can be combined as follows:

Φ = wθΦθ + wzΦz (6–40) where wθ and wz are weighting constants given as

fˆ− f w = θ 1 − f (6–41) 1 − fˆ w = z 1 − f ˆ ˆ such that Φ = Φθ if f = 1 and Φ ≈ Φz if f ≈ f. The general yield condition can now be written in a more complete expression as:

!2 p3J Σ Σ Φ = 2 (c w + c w ) − 1 + f 2 + 2f w cosh γγ σ θ θ z z θ zθσ T 1 T (6–42) Σγγ +wz cosh z = 0 z1 σT

z θ where z1 and z1 were defined in the previous section, cz is defined in Equation (6–36), cθ is defined in Equation (6–39) while wz and wθ are given in Equation (6–41).

Note that both cθ and cz reduce to unity when σT = σC such that Equation (6–42) reduces to the cylindrical yield criterion obtained in Gurson(1977) when there is no tension-compression asymmetry in the matrix. It should be noted that the expression obtained in Equation (6–42) was derived under the assumption of axisymmetric loading. In the next section the analytical result will be compared to ﬁnite element calculations of a unit cell under plane strain loading conditions. The aim of this comparison is to explore the validity of the proposed criterion of Equation (6–42) for more general (non-axisymmetric) loading conditions. Figures 6-6–6-8 show a comparison between the yield curves represented by Equation (6–42) for three diﬀerent materials corresponding to k = −0.3098, k = 0 and k = 0.3098

156 using the plane strain version of the transition boundary c in Section 6.1.2 and ν = 0.32.

p Σ The vertical axis in the figures is the von Mises effective stress (Σe = 3J2 ). The same tensile yield strength was used for all materials (i.e., only the compressive yield strengths were varied). As expected, the criterion predicts that increasing porosity results in softening of the material. Notice that the Gurson(1977) yield curves (corresponding to k = 0) are symmetric about the vertical axis while the new yield curves are not. This difference is to be expected since the Cazacu et al.(2006) yield criterion accounts for asymmetric yield strengths while the von Mises (and, by extension, Gurson) yield criterion does not. Also, note that the yield curves of Figure 6-6 and Figure 6-8 are qualitatively dissimilar for k 6= 0. This is because for ν = 0.32, the transition boundary c equals the outer boundary b when the porosity reaches a level of f = 0.12 (see Equation (6–15) in

Section 6.1.2) such that the yield criterion in Figure 6-6 has non-zero weights on both Φθ and Φz since a < c < b (see Equations (6–40)) whereas the yield criterion plotted in Figure

6-8 (corresponding to f = 0.14) is simply Φ = Φz since the transition boundary is greater than the outer boundary of the RVE (i.e., c > b).

A B

Figure 6-1. Voids inside a shear band coalescing into a crack in titanium due to tension. A) Voids. B) Crack. [Reprinted with permission of John Wiley & Sons, Inc. Meyers(1994). Dynamic behavior of materials. (Page 457, Figure 15.9). New York: John Wiley & Sons, Inc. Copyright 1994 by John Wiley & Sons, Inc.]

157 X3

a b

X1 L

Figure 6-2. Axisymmetric cylindrical representative volume element containing a through-thickness cylindrical void.

158 1 1st Branch σ σ = 21.1 CT 2nd Branch

Σe 0.8 3rd Branch

σ T

0.6

0.4

0.2 f = 14.0 f = 04.0 f = 01.0

0 0 0.5 1 1.5 2 2.5 3

Σ + Σ2211

2σ T

Figure 6-3. Parametric representation of the void-matrix aggregate for σT /σC = 1.21 Σ (k = 0.3098), J3 < 0 and Σm > 0. Curves presented are for void volume fractions of f = 0.01, f = 0.04 and f = 0.14.

1.4

Σ σ σ CT = 21.1 D33>0J3 > 0 1.2 Σe Σ D33<0J3 < 0 σ T 1

0.8

0.6

0.4

0.2

0 -4-3-2-101234

Σ + Σ2211

2σ T

Figure 6-4. Parametric representation of the void-matrix aggregate for σT /σC = 1.21 (k = 0.3098) and f = 0.01.

159 1.4 B>0J Σ > 0 σ σ CT = 82.0 3 1.2 Σ Σ e B<0J3 < 0 σ T 1

0.8

0.6

0.4

0.2

0 -4-3-2-101234

Σ + Σ2211

2σ T

Figure 6-5. Parametric representation of the void-matrix aggregate for σT /σC = 0.82 (k = −0.3098) and f = 0.01.

1.2

1 Σe

σ T 0.8

0.6

0.4

k=0 (σt=σc) k=0 (σT=σC) 0.2 k<0k<0 ( σ(σt0k>0 ( σ(σt>Tσ>c)σC) 0 -5 -3 -1 1 3 5

Σ + Σ2211

2σ T

Figure 6-6. Macroscopic yield surfaces from Equation (6–42) corresponding to a void volume fraction of f = 0.01 for a cylindrical RVE containing a long cylindrical void: materials shown correspond to a ratio between the yield strength in tension and compression of σT /σC = 0.82, σT /σC = 1 (von Mises) and σT /σC = 1.21 (i.e., k = −0.3098, 0 and 0.3098, respectively).

160 1.2

k=0k=0 ( (σσt=T=σσc)C) k<0 (σ <σ ) Σ 1 k<0 (σt0 (σ >σ ) k>0 (σt>T σc)C σ T 0.8

0.6

0.4

0.2

0 -5 -3 -1 1 3 5

Σ + Σ2211

2σ T

Figure 6-7. Macroscopic yield surfaces from Equation (6–42) corresponding to a void volume fraction of f = 0.04 for a cylindrical RVE containing a long cylindrical void: materials shown correspond to a ratio between the yield strength in tension and compression of σT /σC = 0.82, σT /σC = 1 (von Mises) and σT /σC = 1.21 (i.e., k = −0.3098, 0 and 0.3098, respectively).

1.2

k=0k=0 ( (σσt==σσc) ) Σ 1 T C e k<0k<0 ( (σσt0 (σ >σ ) σ T k>0 (σt>T σc)C 0.8

0.6

0.4

0.2

0 -5 -3 -1 1 3 5

Σ + Σ2211

2σ T

Figure 6-8. Macroscopic yield surfaces from Equation (6–42) corresponding to a void volume fraction of f = 0.14 for a cylindrical RVE containing a long cylindrical void: materials shown correspond to a ratio between the yield strength in tension and compression of σT /σC = 0.82, σT /σC = 1 (von Mises) and σT /σC = 1.21 (i.e., k = −0.3098, 0 and 0.3098, respectively).

161 CHAPTER 7 ASSESSMENT OF THE PROPOSED CYLINDRICAL VOID MODEL BY FINITE ELEMENT CALCULATIONS The analytical expression of the macroscopic yield function for a porous aggregate with randomly-oriented, cylindrical voids (see Equation (6–42) in Chapter6) was obtained by assuming a specific geometry for the representative volume element (RVE), specific loading conditions and a number of approximations. These assumptions were necessary to obtain a closed-form expression for the yield surface. One way to validate the developed criterion is to compare the theoretical yield surface to finite element unit cell calculations. In the finite element unit cell calculations, the void boundary is explicitly meshed and the matrix material is modeled as an elastic-plastic material with the plastic response governed by the isotropic CPB06 criterion of Equation (2–8). This same matrix yield criterion was used in the homogenization procedure to derive the macroscopic yield criterion in Chapter6. The purpose of the unit cell calculations is to perform a minimization of the plastic dissipation for a large set of velocity fields compatible with the uniform strain rate boundary conditions and thus test whether the developed yield surface adequately captures the yield behavior for velocity fields other than the one assumed in Section 6.2. 7.1 Modeling Procedure

The unit cell calculations detailed in this chapter are analogous to those done by Richelsen and Tvergaard(1994) for a cylindrical void under plane strain loading conditions and Ristinmaa(1997) for a spherical void under axially symmetric loading conditions. This section details the computational testing procedure used in setting up and running the finite element unit cell calculations described in the introduction of Chapter4. The procedures used for the plane strain cylindrical void calculations and the axisymmetric spherical void calculations are very similar. The basic idea in both cases is to keep the strain triaxiality constant throughout a particular calculation and to define the yield point for that particular calculation as the increment in which the global maximum occurs on the effective stress-strain curve. The first issue then is obviously the need to

162 ensure that the strain triaxiality is constant in the elastic regime. It will be described in the following how this requirement of a constant strain triaxiality is ensured for plane strain loading. The geometry of the unit cell for the cylindrical void geometry is shown in Figure 7-1.

Note that for this unit cell, the initial porosity f0 is deﬁned as

Vvoid f0 = Vtotal πa2 = 0 4A0B0 A constant stress triaxiality for this problem is maintained in a similar manner to that proposed by Ristinmaa(1997) for axisymmetric loading conditions and a spherical void. In the elastic regime, maintaining a constant stress triaxiality is equivalent to maintaining a constant strain triaxiality in the calculation (this can be seen in the equations presented below). First, let the macroscopic strains be deﬁned as follows: A A0 + U1 E1 = ln = ln A0 A0 B B0 + U2 E2 = ln = ln B0 B0

where A0 and B0 are the initial side lengths of the unit cell as shown in Figure 7-1 while

U1 and U2 are the prescribed displacements to those sides. Note that for plane strain

loading, E3 = 0. The strain triaxiality, TE, is deﬁned as

Ekk TE = 3Ee E + E = 1 2 q 2 2 (E1 + E2) − 3E1E2

where Ekk is the trace of the macroscopic strain tensor and Ee is the macroscopic eﬀective

q 0 0 0 strain deﬁned as Ee = (2/3)EijEij with Eij being the macroscopic deviatoric strain tensor. The equation above shows (after a little algebra) that the strains E1 and E2 must be linearly related in order for the strain triaxiality to be constant. This relation is stated

163 as follows:

E1 = cEE2 with √ 2 p 2 2TE + 1 ± 2 3TE 1 − TE cE = 2 . 4TE − 1

Note that the above equation implies the constraint TE ≤ 1 for cE to be a real. The ± give two values for cE which are the inverse of each other; thus, the two cE values give the same result since plane strain calculations are symmetric with respect to the x1 and x2 directions. An alternative to the previous expression with TE now in terms of cE is as follows: 1 + c T = E sgn(E ). E p 2 2 2 1 − cE + cE

The prescribed boundary displacements, U1 and U2 can now be related using the above equations. Doing so yields

cE B0 + U2 U1 = A0 − 1 . B0

In the elastic regime, the macroscopic strains are related to the macroscopic stresses via Hooke’s law such that

2 2 EE1 = 1 − ν Σ1 − ν + ν Σ2

2 2 EE2 = 1 − ν Σ2 − ν + ν Σ1 where E is Young’s modulus and ν is Poisson’s ratio. The above expression can be solved for Σ1 and Σ2 using the fact that E1 = cEE2 and thus obtain

Σ1 = cΣΣ2 with 2 2 cE (1 − ν ) + ν + ν cΣ = 2 2 . cE (ν + ν) + 1 − ν

164 The macroscopic stress triaxiality is deﬁned analogous to the macroscopic strain triaxiality as

Σkk TΣ = 3Σe (1 + cΣ)(1 + ν) = sgn(Σ2) g1(k, cΣ, ν)

= g2(k, cΣ, ν)

where gi(k, cΣ, ν) are simply constant functions of k, cΣ and ν and are, therefore, constant

in the elastic region of a given calculation if the strain triaxiality (and, thus, cE and cΣ) are constant. The stress triaxiality will typically not be constant once macroscopic yielding occurs; hence, the deviation from a constant triaxiality in the ﬁnite element unit cell calculations correspond to the onset of yielding. A relationship which is useful for obtaining a desired spacing of the FE data points in

the 0.5Σγγ-Σe plane is the slope, for k = 0, of a given calculation in the elastic region in the respective plane:

p 2 2 2 6(1 + cΣ) + (4ν − 4ν − 2)(1 + cΣ) slope = sgn(Σ2). (7–1) (1 + cΣ)

Equation (7–1) can be converted to an angle θ = atan(slope), and is plotted versus the

strain ratio E1/E2 in Figure 7-2. Figure 7-2 illustrates that these plane strain calculations are constrained regarding what numerical points can be obtained (i.e., the slopes cannot go below the minimum obtained when cE = 1). Note that while the above equation was derived for a von Mises material (k = 0), the calculations where k = 0 will be used as baseline calculations such that the spacing of the data points is set for the baseline case and all the cE values used in the computational test matrix for this baseline case will then be used for the other values of k. Also, the previous equation does provide a quick and easy estimate for the slope when there are non-zero values of k.

165 Lastly, the macroscopic stresses obtained from the ﬁnite element calculations are deﬁned as follows: F Σ = 1 1 B F Σ = 1 2 A

Σ3 = avg(σ3) where F1 and F2 are the total forces on the cell boundaries and σ3 are the individual out-of-plane stresses in the ﬁnite elements. All of these quantities are readily available from any standard ﬁnite element code. 7.2 Finite Element Results

Finite element results for three different materials under plane strain loading will be presented in this section. For this purpose a user material routine (UMAT) was developed (see also Chapter4) to implement the Cazacu et al.(2006) yield criterion. All calculations were performed in ABAQUS (Abaqus, 2008) using a fully implicit return mapping scheme. A mesh refinement study was performed to ensure mesh convergence. Figures 7-3 to 7-5 show the finite element meshes used with initial void volume fractions of f = 0.01, f = 0.04 and f = 0.14, respectively. All three materials considered have the same yield strength in tension; however, one material has an equal yield strength in compression (k = 0), one has a yield in compression that is less than the yield in tension corresponding to a randomly oriented BCC polycrystal (k = 0.3098) and one has a yield in compression that is greater than the yield in tension corresponding to a randomly oriented FCC polycrystal (k = −0.3098). These specific values for k were reported in Cazacu et al.(2006). A Poisson’s ratio of ν = 0.32 was used for each material along with a tensile yield strength to Young’s modulus ratio of σT /E = 0.00124 (i.e., the elastic properties were arbitrarily chosen to correspond to a steel). Tables 7-1 to 7-3 tabulate the strain triaxialities and resulting displacements prescribed in the unit cell calculations for k = 0, k = −0.3098 and k = 0.3098, respectively.

166 Figures 7-7 to 7-9 show a comparison between the finite element data and the analytical yield surfaces obtained using Equation (6–42) for three different initial porosities (f = 0.14, f = 0.04 and f = 0.01) and the three different materials (k = 0, k = −0.3098 and k = 0.3098). As can be seen in Figures 7-7 to 7-9, the agreement between the finite element data and Equation (6–42) is less than ideal. In order to provide better agreement, the proposed analytical yield criterion of Equation (6–42) can be modified in the following way:

!2 p3J Σ Σ Φ = C 2 (c w + c w ) − 1 + q2f 2 + 2qf w cosh γγ eqv σ θ θ z z θ zθσ T 1 T (7–2) Σγγ +wz cosh z = 0 z1 σT with

q fˆ− f w = f θ 1 − f (7–3) 1 − q fˆ w = f . z 1 − f and 3 fˆ = f (7–4) 1 − 2ν for plane strain loading (see Equation (6–15) in Section 6.1.2).

Three modiﬁcations were introduced in Equation (7–2). First, the term Ceqv is included as introduced by Gurson(1977) for cylindrical voids. The form of Ceqv being used in this section is

3 62 Ceqv = 1 + 3f − 81f + 24f (7–5) which differs from the plane strain term introduced in Gurson(1977) by the addition of a f 3 term. This additional term was introduced based on the k = 0 (i.e., von Mises) finite element calculations to obtain better agreement between the finite element results and analytical yield surfaces for increasing porosity. Secondly, the parameter q was introduced by Tvergaard(1981) and is also used in Equation (7–2) with q = 4/e (this is the same

167 value as that used in Chapter5 for q1). Lastly, the fitting parameter qf = 0.55 was ˆ ˆ introduced as a multiplier on f. The rationale for the inclusion of qf is that f was derived for a cylindrical RVE rather than the cubic RVE being employed here. The parameter qf allows for better agreement as the porosity, f, increases. Figures 7-10 to 7-12 now show a comparison between the finite element data and the analytical yield curves obtained using Equation (7–2) for three different initial porosities (f = 0.14, f = 0.04 and f = 0.01) and three different materials (k = 0, k = −0.3098 and k = 0.3098). As can be seen, the agreement with the finite element data using the modified criterion of Equation (7–2) is fairly good. 7.3 Concluding Remarks

An analytical yield criterion which is an upper-bound estimate has been developed by extending Gurson’s (1977) analysis of the hollow cylinder to the case when the matrix plastic behavior is described by Cazacu et al.’s (2006) isotropic yield criterion. Due to the tension-compression asymmetry of the matrix response, fresh difficulties were encountered when estimating the local plastic dissipation, w(d). This is because the plastic multiplier rate associated to the Cazacu et al.(2006) yield criterion has multiple branches (see Equation (2–29)) and depends on each of the principal values of the local rate of deformation tensor, d. If there is no difference in response between the yield in tension and compression the proposed criterion in Equation (6–42) reduces to the classical analytical cylindrical criterion proposed in Gurson(1977) (since the isotropic criterion of Cazacu et al.(2006) reduces to von Mises for k = 0). In the absence of voids, the proposed criterion reduces to Cazacu et al.’s (2006) yield criterion. The accuracy of the analytical criterion was assessed through comparison with finite-element cell calculations. To improve the agreement, the proposed analytical yield criterion of Equation (6–42) was modified to include additional parameters, q, qf and Ceqv (see Equation (7–2)) based partly on suggestions made in Gurson(1977); Richelsen and Tvergaard(1994); Tvergaard(1981).

168 The agreement between the theoretical predictions using the criterion given in Equation (7–2) and results of ﬁnite element cell calculations is fairly good.

X1 A0

Figure 7-1. Plane strain geometry used in ﬁnite element calculations.

169 100 sgn(Σ2) > 0 sgn(Σ2) < 0

0 [deg] θ

−50

−100 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 E1 E2

Figure 7-2. Valid triaxiality angles available for k = 0 using the procedure outlined in Section 7.1 (see Equation (7–1)); angles are with respect to the 0.5Σγγ − Σe plane and measured from the 0.5Σγγ axis with +θ referring to slopes in the positive Σγγ domain and −θ referring to slopes in the negative Σγγ domain.

Figure 7-3. f0 = 0.01 plane strain ﬁnite element mesh for the unit cell.

170 Figure 7-4. f0 = 0.04 plane strain ﬁnite element mesh for the unit cell.

Figure 7-5. f0 = 0.14 plane strain ﬁnite element mesh for the unit cell.

171 Table 7-1. k = 0 plane strain cylindrical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] C1 -1.0000 -0.0135 -0.0135 -19.7989 C2 -0.7100 -0.0192 -0.0052 -26.8869 C3 -0.5300 -0.0198 -0.0008 -34.1861 C4 -0.3200 -0.0180 0.0047 -48.3665 C5 -0.1900 -0.0178 0.0089 -62.1759 C6 -0.0900 -0.0150 0.0109 -75.9638 C7 0.0900 -0.0109 0.0150 75.9638 C8 0.1900 -0.0075 0.0150 62.1759 C9 0.3200 -0.0039 0.0150 48.3665 C10 0.5300 0.0006 0.0150 34.1861 C11 0.7100 0.0041 0.0150 26.8869 C12 1.0000 0.0135 0.0135 19.7989

Table 7-2. k = −0.3098 plane strain cylindrical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] C13 -1.0000 -0.0135 -0.0135 -19.7989 C14 -0.5900 -0.0192 -0.0023 -26.8805 C15 -0.4400 -0.0180 0.0015 -34.1198 C16 -0.2800 -0.0177 0.0058 -47.4640 C17 -0.1700 -0.0164 0.0089 -61.5341 C18 -0.0800 -0.0144 0.0109 -76.0444 C19 0.0800 -0.0113 0.0150 76.6962 C20 0.1900 -0.0075 0.0150 61.3407 C21 0.3200 -0.0039 0.0150 48.0132 C22 0.5300 0.0006 0.0150 34.1758 C23 0.7000 0.0039 0.0150 26.7508 C24 0.9000 0.0085 0.0150 19.7240

172 Table 7-3. k = 0.3098 plane strain cylindrical void computational test matrix.

Test # TE U1 [in] U3 [in] Angle [deg] C25 -1.0000 -0.0135 -0.0135 -19.7989 C26 -0.8100 -0.0175 -0.0072 -26.9583 C27 -0.6400 -0.0189 -0.0034 -33.9966 C28 -0.3900 -0.0181 0.0028 -48.0278 C29 -0.2200 -0.0182 0.0080 -62.5359 C30 -0.1000 -0.0155 0.0109 -76.3462 C31 0.1000 -0.0105 0.0150 75.5189 C32 0.2000 -0.0072 0.0150 62.0157 C33 0.3300 -0.0037 0.0150 47.9415 C34 0.5300 0.0006 0.0150 34.2013 C35 0.6700 0.0033 0.0150 28.7276 C36 1.0000 0.0135 0.0135 23.5401

0.8

0.7

0.6

0.5 ¦ ¤ £ ¥ 0.4

0.3

0.2

0.1

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 ¢¡ x 10−3

Figure 7-6. Plane strain eﬀective stress criterion used to pick the yield point (data shown from Test# C1).

173 1.2 Proposed Criterion σ σ CT =1 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-7. Yield surface of the void-matrix aggregate for σT /σC = 1 (von Mises) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (6–42).

174 1.2 Proposed Criterion σ σ CT = 82.0 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-8. Yield surface of the void-matrix aggregate for σT /σC = 0.82 (k = −0.3098) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (6–42).

175 1.2 Proposed Criterion σ σ CT = 21.1 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-9. Yield surface of the void-matrix aggregate for σT /σC = 1.21 (k = 0.3098) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (6–42).

176 1.2 Proposed Criterion σ σ CT =1 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-10. Yield surface of the void-matrix aggregate for σT /σC = 1 (von Mises) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (7–2).

177 1.2 Proposed Criterion σ σ CT = 82.0 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-11. Yield surface of the void-matrix aggregate for σT /σC = 0.82 (k = −0.3098) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (7–2).

178 1.2 Proposed Criterion σ σ CT = 21.1 FE Data Σe 1

σ T 0.8

0.6

0.4

0.2

f = 14.0 04.0 01.0 0 -3 -2 -1 0 1 2 3

Σ + Σ2211

2σ T

Figure 7-12. Yield surface of the void-matrix aggregate for σT /σC = 1.21 (k = 0.3098) with f = 0.01, f = 0.04 and f = 0.14. Comparison between the ﬁnite element results and the proposed criterion of Equation (7–2).

179 CHAPTER 8 ANISOTROPIC PLASTIC POTENTIAL FOR HCP METALS CONTAINING SPHERICAL VOIDS This chapter focuses on extending the macroscopic yield criterion developed in Chapter3 for a void-matrix aggregate where the voids have spherical geometry and the metal matrix is isotropic with tension-compression asymmetry to the case where the metal matrix still exhibits tension-compression asymmetry but is now anisotropic. We begin with a brief review of the main contributions in modeling yielding of porous aggregates containing an anisotropic matrix material. As previously mentioned, Gurson(1977) developed widely used macroscopic yield criteria for void-matrix aggregates containing either spherical or cylindrical voids and with the matrix obeying a von Mises isotropic yield condition. Liao et al.(1997) extended Gurson’s (1977) cylindrical criterion to account for transverse isotropy using Hill’s 1948 yield criterion (see Hill, 1948, 1950) for the matrix material; the proposed criterion is 2D (plane stress conditions) and was applied to model the behavior of thin metal sheets (steels and aluminum). Benzerga and Besson(2001) extended Gurson’s (1977) spherical criterion for orthotropic metals by assuming a matrix material that could be characterized using Hill’s 1948 criteria. Using a Hill-Mandel homogenization procedure on an axisymmetric, ellipsoidal RVE containing a confocal, prolate ellipsoidal void, Gologanu et al.(1993) arrived at an analytic expression for prolate, ellipsoidal voids. Benzerga et al. (2004b) used a yield criterion which combined properties from both Gologanu et al.’s (1993) and Benzerga and Besson’s (2001) criteria to account for both void shape and orthotropy. This criterion compares well with the experimental results given in Benzerga et al.(2004a). All the previous works cited involve cubic metals. Modeling the yielding behavior of hexagonal close packed (HCP) metals poses tremendous challenges due to their unusual deformation characteristics—namely, strong asymmetry between tensile and compressive behavior and highly pronounced anisotropy. Rigorous mathematical methods to introduce

180 anisotropy in any given isotropic yield criterion exist in the literature (e.g., the linear transformation approach and the generalized invariants approach). A major difficulty encountered in formulating analytical expressions for the yield behavior of HCP metals is related to the description of the tension versus compression asymmetry which is due to the combined effects of crystal structure and twinning. In the following sections, the isotropic yield criterion developed in Chapter3 which already accounts for the presence of voids and tension-compression asymmetry in the matrix will be extended to account for anisotropy in the matrix. The structure of this chapter is as follows. In Section 8.1, the homogenization approach due to Hill and Mandel (Hill, 1967; Mandel, 1972) that is used in developing the macroscopic plastic potential for the void-matrix aggregate is briefly recalled. Next, the anisotropic version of the Cazacu et al.(2006) yield criterion that describes the matrix material behavior in the void-matrix aggregate is presented in Section 8.2. Thirdly, the choice of velocity field used to calculate the local plastic dissipation potential is introduced in Section 8.3. The corresponding local plastic dissipation is then derived in Section 8.4 and the expressions for the macroscopic plastic dissipation of the void-matrix aggregate are developed in Section 8.5. Section 8.6 shows comparisons between the macroscopic yield criterion developed in this chapter for a void-matrix aggregate with axisymmetric (transversely isotropic) finite element unit cell calculations. 8.1 Kinematic Homogenization Approach of Hill and Mandel

Consider a representative volume element V , composed of a homogeneous rigid- plastic matrix and a traction-free void. The matrix material is described by a convex yield function ϕ(σ) in the stress space and an associated ﬂow rule

∂ϕ d = λ˙ , (8–1) ∂σ where σ is the Cauchy stress tensor, d = (1/2)(∇v +∇vT ) denotes the rate of deformation tensor with v being the velocity ﬁeld, and λ˙ ≥ 0 stands for the plastic multiplier rate. The

181 yield surface is deﬁned as ϕ(σ) = 0. Let C denote the convex domain delimited by the yield surface such that C = {σ|ϕ(σ) ≤ 0} . (8–2)

The plastic dissipation potential of the matrix is deﬁned as

w(d) = sup (σ : d) (8–3) σ∈C

where “:” denotes the tensor double contraction. Uniform rate of deformation boundary conditions are assumed on the boundary of the RVE, ∂V , such that

v = Dx for any x ∈ ∂V (8–4)

with D, the macroscopic rate of deformation tensor, being constant. For the boundary conditions of Equation (2–4), the Hill-Mandel (Hill, 1967; Mandel, 1972) lemma applies; hence,

hσ : diV = Σ : D, (8–5)

where h i denotes the average value over the representative volume V , and Σ = hσiV . Furthermore, there exists a macroscopic plastic dissipation potential W (D) such that

∂W (D) Σ = (8–6) ∂D

with

W (D) = inf hw(d)iV , (8–7) d∈K(D) where K(D) is the set of incompressible velocity fields satisfying Equation (8–4) (for more details see Gologanu et al., 1997; Leblond, 2003). The matrix material being considered obeys the anisotropic version of the pressure-insensitive yield criterion that captures strength differential effects of Cazacu et al.(2006).

182 8.2 Yield Criterion for the Matrix Material

The Cazacu et al.(2006) criterion (using a = 2) can be written as

whereσ Î ,σ ÎI andσ ÎII are the principal components ofσ ˆ. The anisotropic version of the Cazacu et al.(2006) yield criterion involves a linear transformation, L, of the stress deviator, σ0. In this work an additional constraint that the transformed stress

σˆ = Lσ0 (8–9) be deviatoric will be imposed. AppendixB shows that the CPB06 criterion can reduce to the Hill48 criterion if the constraint that the transformed stress be deviatoric is imposed and if k = 0 such that there is no tension-compression asymmetry (since, in this case, both criterion have the same numbers of degrees of freedom). Note that this condition of a deviatoric transformed stress is more restrictive than the original anisotropic criterion presented in Cazacu et al.(2006). With respect to the coordinate system associated with orthotropy, the linear transformation, L, can be written in Voight notation as follows:   L11 L12 L13 0 0 0      L L L 0 0 0   12 22 23       L13 L23 L33 0 0 0  L =   .    0 0 0 L44 0 0       0 0 0 0 L 0   55    0 0 0 0 0 L66

183 Note that, in order forσ ˆ to be traceless, the following restrictions on the components of L must hold:

L11 + L12 + L13 = 1

L12 + L22 + L23 = 1

L13 + L23 + L33 = 1.

The constant on the right hand side of the previous equations has been chosen to be unity such that the linear transformation reduces to the identity tensor for the case of isotropy. The CPB06 anisotropic yield criterion can be written as

T ϕ = σe − σ1 = 0 (8–10) where √ σe =m ˆ F v u 3 (8–11) uX 2 =m ˆ t (|σˆi| − kσˆi) i=1

T and σ1 is the uniaxial yield strength along an axis of orthotropy of the material, say the rolling direction, which is denoted the 1-direction. The constant,m ˆ , is the anisotropic version of the eﬀective stress constant and is given by s 1 mˆ = 2 2 2 (8–12) (|Φ1| − kΦ1) + (|Φ2| − kΦ2) + (|Φ3| − kΦ3)

where 2 1 1 Φ = L − L − L 1 3 11 3 12 3 13 2 1 1 Φ = L − L − L (8–13) 2 3 12 3 22 3 23 2 1 1 Φ = L − L − L . 3 3 13 3 23 3 33 Recall from Section 8.1 that determining the macroscopic plastic dissipation W (D) requires the determination of the local plastic dissipation expression, w(d). Since the

184 anisotropic criterion is a homogeneous function of ﬁrst order in stresses, the local plastic

˙ T ˙ dissipation reduces to w = λσ1 where λ is the anisotropic plastic multiplier rate (or the

T dual of the Cazacu et al., 2006, anisotropic yield criterion) and σ1 is the tensile yield strength in the rolling direction. The dual of the Cazacu et al.(2006) anisotropic yield criterion can be determined as follows. Note that

0 σˆmn = Lmnklσkl = LmnklJklijσij (8–14)

where, in Voight notation,   2 −1 −1 0 0 0      −1 2 −1 0 0 0        1  −1 −1 2 0 0 0  J =   . 3    0 0 0 3 0 0       0 0 0 0 3 0      0 0 0 0 0 3

The plastic rate of deformation tensor dP can now be written as follows:

P ˙ ∂φ dij = λ ∂σij ∂φ ∂σˆ = λ˙ mn ∂σˆ ∂σ mn ij (8–15) ˙ ∂φ = λ LmnrsJrsij ∂σˆmn ˙ ∂φ = λ Lmnij ∂σˆmn where LJ = L if L is a deviatoric tensor. Let b denote the following ﬁrst order transformed rate of deformation tensor:

P −1 brs = dijLijrs. (8–16)

Therefore, ˙ ∂φ brs = λ . (8–17) ∂σˆrs

185 Thus, the expression for λ˙ in the anisotropic case can now be found in the same manner as in the isotropic case (see Section 2.3) such that d in the isotropic expression for λ˙ (see Equation (2–29)) is replaced with b to obtain the anisotropic expression given as

˙ 2 2 λ = αbijbij + βblone (8–18)

where blone is the principal component of b associated with the lone positive or negative

principal transformed stress deviators (i.e.,σ Î ,σ ÎI orσ ÎII ). The parameters α and β are functions of the tension-compression asymmetry parameter k and can be expressed as follows:      2 2  1 bIII − (3k + 2k + 3)  if J σˆ ≤ 0 ⇔ ≤  mˆ (1 − k) 3 pb2 + b2 + b2 p6 (k2 + 3) (3k2 + 1)  I II III α = (8–19)     2 2  1 bI 3k − 2k + 3  if J σˆ ≥ 0 ⇔ ≥  mˆ (1 + k) 3 pb2 + b2 + b2 p6 (k2 + 3) (3k2 + 1)  I II III

and      2  −12k bIII − (3k + 2k + 3)  α if J σˆ ≤ 0 ⇔ ≤  3k2 + 2k + 3 3 pb2 + b2 + b2 p6 (k2 + 3) (3k2 + 1)  I II III β =     2  12k bI 3k − 2k + 3  α if J σˆ ≥ 0 ⇔ ≥  3k2 − 2k + 3 3 pb2 + b2 + b2 p6 (k2 + 3) (3k2 + 1)  I II III (8–20)

where bI , bII and bIII are the ordered principal components of bij and associated with the

ordered transformed stress deviators,σ Î ≥ σÎI ≥ σÎII .

186 Alternatively, λ˙ can be expressed in the same manner as was done in Section 2.3:  s 2 2  1 3k + 10k + 3 bI 3k − 2k + 3  b2 + b2 + b2 if ≥  mˆ (1 + k) 3k2 − 2k + 3 I II III pb b p6 (k2 + 3) (3k2 + 1) λ˙ = ij ij s 2 2  1 3k − 10k + 3 bIII − (3k + 2k + 3)  b2 + b2 + b2 if ≤  I II 2 III p p 2 2  mˆ (1 − k) 3k + 2k + 3 bijbij 6 (k + 3) (3k + 1) (8–21) where bI , bII and bIII are the ordered principal components of bij. 8.3 Choice of Trial Velocity Field

The velocity field, ~v, in the RVE is taken to be the same incompressible velocity field proposed by Gurson(1977) (i.e., the same velocity field assumed in Chapter3). The velocity field is written as follows: ~v = ~vV + ~vS where b3 ~vV = D ~e and ~vS = D0~x (8–22) m r2 r such that

b3 d = −2D + D0 rr m r rr 3 b 0 dθθ = Dm + Dθθ r (8–23) b3 d = D + D0 φφ m r φφ

0 drθ = Drθ with principal values s D b3 D0 3 b6 2 b3 d˜ = − m − φφ ± D2 − D (D0 − D0 ) + D0 2 2,3 2 r 2 2 m r 3 m r rr θθ φφ (8–24) b3 d˜ = D + D0 . 1 m r φφ

187 8.4 Calculation of the Local Plastic Dissipation

˙ T The calculation of the local plastic dissipation w = λσ1 even in the isotropic case is non-trivial (see Section 3.3) due to the multiple branches in the expression for the plastic multiplier rate, λ˙ (see Equation (2–29) or (8–21)). Recall from Section 3.3 that a form for an approximate macroscopic plastic dissipation W ++(D) which captured the hydrostatic and deviatoric solutions was obtained and that it could be written as

Z 1 1/2 ++ σT h ˙ 2 i W (D) = √ hλ iS(r) dρ 4π f (8–25) Z 1 2 σT 1/2 = α + β √ hdijdijiS(r) dρ 3 4π f where 1 Z 2π Z π hxiS(r) = x sin θdθdφ. (8–26) 4π 0 0 This expression for W ++(D) results from the fact that

h i1/2 2 hλ˙ 2i = α + β hd d i 1/2 S(r) 3 ij ij S(r)

0 since the surface integral of Drr is zero (note that the parameter z6 deﬁned in Section 3.3 is related to the parameters α and β as z6 = 6α + 4β). In order to obtain an anisotropic extension of the yield criterion developed in Chapter3, the result from Section 8.2 that the anisotropic plastic multiplier rate λ˙ is the same form as the isotropic version but with the rate of deformation tensor d replaced with the transformed rate of deformation tensor b = L−1d will be used. Similarly, in this section, d will be replaced with b in the expression for the surface integral of the plastic multiplier rate to obtain

2 hλ˙ 2i = α + β hb b i . (8–27) S(r) 3 ij ij S(r)

This expression will be used in the next section to determine an approximate macroscopic plastic dissipation W ++(D) that will incorporate the eﬀects of tension-compression asymmetry (through k) and anisotropy (through L) while reducing to the isotropic expression obtained in Chapter3 for L = I.

188 8.5 Development of the Macroscopic Plastic Dissipation Expression

This section focuses on developing an expression for the macroscopic plastic dissipation for the anisotropic case. Separating the rate of deformation tensor into a volumetric part and a deviatoric part, the volumetric portion can be written as follows in the spherical reference frame (see Equation (8–23)):     ¯V  drr   −2           d¯V   1   θθ         d¯V   1  V  φφ    d = Dmu = Dmu  0   0               0   0           0   0  where u = b3/r3 with r being the radial coordinate and b being the outer radius of the spherical RVE. Now, note that the scalar bijbij can be rewritten as follows:

P −1 P −1 bijbij = dklLklijdpqLpqij (8–28) P ˆ P = dklLklpqdpq

ˆ where Lklpq is a diagonal matrix when written in Voight notation such that   ˆ l1 0 0 0 0 0      0 ˆl 0 0 0 0   2     ˆ   0 0 l3 0 0 0  Lˆ =   . (8–29)  ˆ   0 0 0 l4 0 0       0 0 0 0 ˆl 0   5   ˆ  0 0 0 0 0 l6

189 Thus, the macroscopic plastic dissipation expression can be written as

T Z b 1/2 ++ σ1 2 h ˙ 2 i W = 4πr hλ iS(r) dr V a T Z b 1/2 σ1 2 2 = 4πr α + β hb : biS(r) dr (8–30) V a 3 T Z b 1/2 σ1 2 2 V ˆ V S ˆ S V ˆ S = 4πr α + β hd : L : d + d : L : d + 2d : L : d iS(r) dr. V a 3 Noting that V ˆ S V ˆ S hd : L : d iS(r) = hd iS(r) : L : d

and that

V hd iS(r) = h[−2~er ⊗ ~er + ~eθ ⊗ ~eθ + ~eφ ⊗ ~eφ]iS(r)

= 0

the following expression is obtained for the estimated macroscopic plastic dissipation:

σT Z b 2 1/2 W ++ = 1 4πr2 α + β hdV : Lˆ : dV + dS : Lˆ : dSi dr 4 3 S(r) 3 πb a 3 s s 2 3σT Z b b6 D2 1 2 2 ˆ e (8–31) = α + β 3 r Dm 6 hχ(li, θ, φ)iS(r) + 2 dr 3 b a r α + 3 β Z 1/f √ T 2 2 du = σ1 Dmh u + A 2 1 u where u = b3/r3, 2 D2 = α + β hdS : Lˆ : dSi , (8–32) e 3 S(r) D 1 A = e (8–33) Dm h and s 2 h = α + β hχ(ˆl , θ, φ)i . (8–34) 3 i S(r) ˆ The function χ is dependent on θ and φ along with the anisotropic parameters, li, and is deﬁned, for the particular velocity ﬁeld being used, as

ˆ χ = [−2~er ⊗ ~er + ~eθ ⊗ ~eθ + ~eφ ⊗ ~eφ]: L :[−2~er ⊗ ~er + ~eθ ⊗ ~eθ + ~eφ ⊗ ~eφ] . (8–35)

190 The anisotropy factor, h, which is applied to the hydrostatic part of the macroscopic yield criterion can now be determined in a similar manner as in Benzerga and Besson(2001). The second-order tensor Lˆ must be transformed from the material axes (where it is a diagonal matrix in Voight notation) to the spherical coordinate system (where it is a full ˆ ˆ ˆ matrix). The new notation will be LMN = LMN li, θ, φ . Now, h can be M,N=I,...,V I determined as follows: 2 h2 = α + β hχi 3 S(r) 2 2 2 2 = α + β hLˆ d¯V + Lˆ d¯V + Lˆ d¯V 3 I,I rr II,II θθ III,III φφ ˆ ¯V ¯V ˆ ¯V ¯V ˆ ¯V ¯V (8–36) + 2LI,II drrdθθ + 2LI,III drrdφφ + 2LII,III dθθdφφiS(r) 2 = α + β h4Lˆ + Lˆ + Lˆ 3 I,I II,II III,III ˆ ˆ ˆ − 4LI,II − 4LI,III + 2LII,III iS(r) such that 2 4 6 h2 = α + β ˆl + ˆl + ˆl + ˆl + ˆl + ˆl (8–37) 3 5 1 2 3 5 4 5 6 where Matlab was used to perform the coordinate transformation of Lˆ needed to obtain Equation (8–37). When k = 0, α = 2/3 with β = 0 and Lˆ = Hˆ such that Equation (8–37) reduces to

8 4 h2 = ˆl + ˆl + ˆl + ˆl + ˆl + ˆl (8–38) 15 1 2 3 5 4 5 6 which matches the result obtained in Benzerga and Besson(2001). Also, for isotropy (i.e., Lˆ = Iˆ), Equation (8–37) yields

2 h = 6α + 4β = z6 (8–39) which matches the result obtained in Chapter3 (see Equation (3–29)). The resulting macroscopic yield criterion incorporating anisotropy is as follows:

˜ !2 Σe 3Σm 2 Φ = T + 2f cosh T − 1 − f = 0. (8–40) σ1 hσ1

191 8.6 Assessment of the Proposed Anisotropic Criterion through Comparison with Finite Element Calculations

Transversely isotropic behavior is often seen, for example, in rolled sheets where the in-plane properties are isotropic with the through-thickness yield behavior being either weaker or stronger. In this section the proposed criterion developed in the previous section will be assessed by comparing the theoretical predictions with finite element unit cell calculations. This in-plane symmetry allows for the assessment of material behavior by conducting axisymmetric cell calculations. Table 8-1 tabulates the anisotropic parameters used in all materials being considered in this section. The specific plasticity parameters correspond to an isotropic material (material A), a material whose through-thickness yield strength is greater than its in-plane yield strength (material B) and a material whose through-thickness yield strength is less than its in-plane yield strength (material C). The last two materials (materials B and C) correspond for k = 0 to two of the transversely isotropic materials considered in Benzerga and Besson(2001). Figure 8-1 shows the axisymmetric section used in the finite element calculations for the transversely isotropic materials considered in this section. Note that under these axisymmetric loading conditions, the third invariant of the macroscopic stress

Σ deviator, J3 , is negative when Σ3 < Σ1 and positive otherwise. Figure 8-2 illustrates the effective stress-strain curves for the materials being considered in this section when k = 0 (a strain triaxiality of 1.5 was used in the calculations illustrated by Figure 8-2). Note that material B tends to yield last due to its larger through-thickness yield strength while material C yields first due to its lower through-thickness yield strength (the in-plane yield strengths were equal among the three calculations). Figures 8-3 to 8-5 show the plane stress yield loci for the nine different materials looked at in this study (3 different sets of L-coefficients combined with 3 different k- values). Note that all materials have roughly the same in-plane yield strengths for fixed values of k (and the same in-plane tensile yield strength throughout). All materials have

192 either equal yield strengths in tension and compression (k = 0), smaller yield strengths in tension versus compression (both in the plane and in the through-thickness direction) or larger yield strengths in tension versus compression (again, both in the plane as well as through the thickness). Figures 8-6 to 8-14 show the π-plane representations of the ductile criterion given by Equation (8–40) for the 9 materials considered in this study. Note that the yield surface shrinks in the deviatoric plane as the pressure increases toward the tensile hydrostatic limit pressure, Pmax. Figures 8-6 to 8-14 imply that the anisotropy of the matrix, together with any strength differential, influences the shape of the yield locus in the deviatoric plane while the presence of voids in the aggregate leads to a decrease in the size of the yield locus with increasing pressure (positive or negative). The plots only show curves for a single void volume fraction (f = 0.01); however, the yield locus simply decreases in size with increasing void volume fraction (for a fixed pressure). In summary, the shape of the yield locus (but not the size) is independent of the void volume fraction and independent of the applied pressure; the shape is governed by the matrix anisotropy and tension-compression asymmetry. Figures 8-15 to 8-17 show the finite element results where the 9 different materials are used as the matrix material in a unit cell containing a spherical void (1% void volume fraction). The analytical yield curves are obtained using the following equation:

˜ !2 Σe Σkk 2 Φ = T + 2qf cosh T − 1 − (qf) (8–41) σ1 hσ1

where q = 4/e (as was used in Chapter5). Also, the ﬁrst term contains the anisotropic CPB06 eﬀective stress which is given as

√ ˜ Σe =m ˆ F.

For the special case of transverse isotropy (with the 3-direction being the through- thickness direction), the diagonal components of the matrix Lˆ given by Equation (8–29)

193 are

ˆ −1 −12 −1 −1 2 l1 = L11 − L12 = 2L11 + L13 − 1 ˆ ˆ l2 = l1 (8–42) ˆ −12 −12 −1 −1 −1 −1 −1 −1 −1 −1 l3 = 3 L13 + L33 + 2L11 L12 − 2L11 L13 − 2L12 L13 − 2L13 L33

−12 −1 −1 −1 −1 −12 = 13 L13 − 8L13 − 2L11 L13 + 2L11 − 2 L11 + 1

and

ˆ ˆ −12 l4 = l5 = L44 (8–43) ˆ ˆ l6 = l1

where

2 −1 L11 − 2L11L13 − L13 L11 = 2 4L13 + 2L11 − 6L11L13 − 3L13 − 1 2 −1 L13 − 2L11L13 − L13 L13 = 2 4L13 + 2L11 − 6L11L13 − 3L13 − 1 −1 1 L44 = . L44 Again, for transverse isotropy, the hydrostatic parameter, h, is given as

r 2 h = (3α + 2β) 7ˆl + 2ˆl + 6ˆl 15 1 3 4 r 2 h i = (3α + 2β) 24 L−12 + 33 L−12 + 24L−1L−1 − 24L−1 − 30L−1 + 6 L−12 + 9 . 15 11 13 11 13 11 13 44 (8–44)

The α and β parameters are given by Equations (8–19) and (8–20) and each contain two branches depending on the sign of J3. However, in this section, the branching dependence

Σ will be transferred to the sign of the macroscopic mean stress (sgn(J3 ) = −sgn(Σm)) since h appears in the hydrostatic term of the yield condition (this is similar to what was done when deriving the isotropic yield criterion and ensures that the anisotropic criterion presented here reduces to the isotropic criterion presented in Chapter3).

194 Figures 8-15 to 8-17 show the normalized macroscopic von Mises eﬀective stress

1 1 Σe/σT versus the normalized macroscopic mean stress Σm/σT using CPB06 L-coeﬃcients corresponding to materials A, B and C of Table 8-1 (with k varied as shown in the ﬁgures). Note that the data presented in Figure 8-15 for Material A is the same data presented in Chapter3 since Material A is isotropic (the data is replotted in this section to compare with the transversely isotropic behavior of Materials B and C). The hydrostatic solutions in the negative plane (the left side) are seen to be independent of k in Figure 8-15. This is because the isotropic hydrostatic solution in the negative plane is simply (2/3) ln(f) as derived in Section 3.1.2 of Chapter3. The transversely isotropic materials illustrated in Figures 8-16 and 8-17 show very weak dependence on k for the hydrostatic solution in the negative plane (similar to the isotropic case where there is no k-dependence on the negative hydrostatic solution). The hydrostatic solutions for these transversely isotropic cases can be determined from Equation (8–41) as

Σm h T = ± ln(qf) σ1 3 such that  v  u ˆ ˆ ˆ !  u 2 7l1 + 2l3 + 6l4  −t ln(qf) for Σ > 0  15m ˆ 2 3k2 + 2k + 3 m Σm  T = (8–45) σ1  v  u ˆ ˆ ˆ !  u 2 7l1 + 2l3 + 6l4  t ln(qf) for Σm < 0  15m ˆ 2 3k2 − 2k + 3 where the two expressions diﬀer by the ±2k in the denominator. Figures 8-18A and 8-18B show the proposed criterion with the hydrostatic solutions of Equation (8–45) explicitly labeled for material B with the tensile yield strengths greater than the compressive yield strengths and vice versa, respectively. Notice that the curves are not symmetric with respect to the vertical axis. This results from Equation (8–45) where the two solutions are only equivalent if there is no tension-compression asymmetry in the matrix.

195 The macroscopic von Mises effective stress Σe can be related to the macroscopic ˜ CPB06 effective stress Σe as follows. If the 3-direction is taken to be the out-of-plane, or through-thickness, direction then the stress deviators for the axisymmetric loading case can be written as 1 Σ0 = Σ0 = (Σ − Σ ) 11 22 3 11 33 2 Σ0 = − (Σ − Σ ) 33 3 11 33 such that the von Mises effective stress becomes

q Σ Σe = 3J2 = |Σ11 − Σ33|. (8–46)

The CPB06 eﬀective stress can be determined as

√ q ˜ 2 2 Σe =m ˆ F =m ˆ 2 (|σˆ1| − kσˆ1) + (|σˆ3| − kσˆ3) . (8–47)

Note that for the case of axisymmetric loading under consideration

1 σˆ =σ ˆ = (L + L − 2L ) (Σ − Σ ) 1 2 3 11 12 13 11 33 2 σˆ = (L − L ) (Σ − Σ ) 3 3 13 33 11 33 such that for all materials under consideration in this section (i.e., Materials A, B and C given in Table 8-1)

sgn (ˆσ1) = −sgn (Σ11 − Σ33)

sgn (ˆσ3) = sgn (Σ11 − Σ33)

and

Σ σˆ sgn J3 = sgn J3 .

The von Mises eﬀective stress given in Equation (8–46) can now be related to the CPB06 eﬀective stress given in Equation (8–47) as follows:

˜ Σe = |Σ11 − Σ33| = gΣe

196 where  3  q for Σ11 > Σ33  2 2 2 2  mˆ 2 (L11 + L12 − 2L13) (1 − k) + 4 (L13 − L33) (1 + k) g = 3  q for Σ11 < Σ33.  2 2 2 2  mˆ 2 (L11 + L12 − 2L13) (1 + k) + 4 (L13 − L33) (1 − k)

8.7 Concluding Remarks

An analytical yield criterion has been developed by extending the spherical void analysis of Chapter3 to account for matrix anisotropy. The proposed criterion reduces to that given in Benzerga and Besson(2001) if there is no difference in response between the yield in tension and compression. In the absence of voids, the proposed criterion reduces to the anisotropic criterion of Cazacu et al.(2006) if there is tension-compression asymmetry in the matrix and to Hill’s (1948) yield criterion when no strength differential exists. The validity of the proposed anisotropic criterion was assessed through comparisons with transversely isotropic finite element calculations. The theoretical predictions provided by the anisotropic yield criterion for the void-matrix aggregate compare well with the predictions obtained by the finite element unit cell calculations.

197 Table 8-1. Transversely isotropic CPB06 constants used in the axisymmetric ﬁnite element calculations: Material A has the same in-plane and through-thickness yield strengths (isotropic), material B has a through-thickness yield strength greater T T than the in-plane yield strength (σ1 < σ3 ) and material C has a through-thickness yield strength lower than the in-plane yield strength T T (σ1 > σ3 ).

CPB06 Parameters Material A Material B Material C

L11 = L22 1.000 1.054 0.963

L33 1.000 0.850 1.064

L13 = L23 0.000 0.075 -0.032

L12 0.000 -0.129 0.069

a L44 = L55 1.000 0.775 1.817

a L66 1.000 1.183 0.894 a For axisymmetric calculations ABAQUS does not use either L55 or L66.

a0 X1 A0

Figure 8-1. Axisymmetric section of the RVE for the transversely isotropic material used in the ﬁnite element calculations. Axis X3 is the through-thickness direction for the transversely isotropic material with the X1-X2 plane being the plane of symmetry.

198 0.8

0.7

Σ−Σ 31 0.6 T σ1 0.5

0.4

0.3

Matl A 0.2 Matl B Matl C 0.1

0 0.00E+00 5.00E-04 1.00E-03 1.50E-03 2.00E-03

− EE 31

Figure 8-2. Typical effective stress versus effective strain curves illustrating the influence of the different through-thickness yield strengths (the in-plane yield strengths are held constant) for materials A, B and C of Table 8-1 (curves shown are for k = 0 such that there is no tension-compression asymmetry). Material A has the same in-plane and through-thickness yield strengths (isotropic), material B has a through-thickness yield strength greater than the in-plane yield strength T T (σ1 < σ3 ) and material C has a through-thickness yield strength lower than T T the in-plane yield strength (σ1 > σ3 ).

Material A k = 0 Material B σ z Material C T σ1

σ x T σ1

Figure 8-3. Plane stress yield loci for void-free materials A (isotropic), B and C (transversely isotropic) according to the CPB06 yield criterion. x is an in-plane direction with z being the through-thickness direction. It is assumed here that all these materials do not have strength diﬀerential eﬀects (k = 0).

199 k −= 3098.0 σ z T σ1

Material A Material B Material C

σ x T σ1

Figure 8-4. Plane stress yield loci for void-free materials A (isotropic), B and C (transversely isotropic) according to the CPB06 yield criterion. x is an in-plane direction with z being the through-thickness direction. It is assumed here that all these materials display tension-compression asymmetry with the yield strengths in tension less than the yield strengths in compression.

Material A k = 3098.0 Material B σ z Material C T σ1

σ x T σ1

Figure 8-5. Plane stress yield loci for void-free materials A (isotropic), B and C (transversely isotropic) according to the CPB06 yield criterion. x is an in-plane direction with z being the through-thickness direction. It is assumed here that all these materials display tension-compression asymmetry with the yield strengths in tension greater than the yield strengths in compression.

200 3 f=0 k = 0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-6. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for an isotropic material (material A in Table 8-1) with the yield strengths in tension and compression equal. Pmax is the tensile hydrostatic yield pressure.

3 f=0 k = − 3098.0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f-0.01

1 2

Figure 8-7. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for an isotropic material (material A in Table 8-1) with the yield strengths in tension less than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

201 3 f=0 k = 3098.0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-8. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for an isotropic material (material A in Table 8-1) with the yield strengths in tension greater than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

f=0 3 P=0, f=0.01 k = 0 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-9. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength greater than the in-plane yield strength (material B in Table 8-1) and with the yield strengths in tension and compression equal. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

202 f=0 3 P=0, f=0.01 k = − 3098.0 0.75 Pmax, f=0.01 0.90 Pmax, f-0.01

1 2

Figure 8-10. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength greater than the in-plane yield strength (material B in Table 8-1) and with the yield strengths in tension less than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

3 f=0 k = 3098.0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-11. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength greater than the in-plane yield strength (material B in Table 8-1) and with the yield strengths in tension greater than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

203 3 f=0 k = 0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-12. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength less than the in-plane yield strength (material C in Table 8-1) and with the yield strengths in tension and compression equal. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

3 f=0 k = − 3098.0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f-0.01

1 2

Figure 8-13. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength less than the in-plane yield strength (material C in Table 8-1) and with the yield strengths in tension less than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

204 3 f=0 k = 3098.0 P=0, f=0.01 0.75 Pmax, f=0.01 0.90 Pmax, f=0.01

1 2

Figure 8-14. Representation in the deviatoric plane of the ductile yield criterion given by Equation (8–40) for a material with the through-thickness yield strength less than the in-plane yield strength (material C in Table 8-1) and with the yield strengths in tension greater than the yield strengths in compression. Pmax is the tensile hydrostatic yield pressure and f is the void volume fraction.

205 1.2

Σ−Σ 31 T σ1 0.8

0.6

0.4 k=0 k=-0.3098 0.2 k=0.3098 FE Data (J3>0) 0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

2Σ + Σ31 T 3σ1 A

1.2

Σ−Σ 31 T σ1 0.8

0.6

0.4 k=0 k=-0.3098 0.2 k=0.3098 FE Data (J3<0) 0 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5

2Σ + Σ31 T 3σ1 B

Figure 8-15. Anisotropic, axisymmetric ﬁnite element yield points versus analytic curves when initial void volume fraction is f0 = 0.01 and the yield strength in the in-plane and through-thickness directions are equal (i.e., Material A in Table Σ Σ 8-1). A) J3 > 0 (Σ3 > Σ1). B) J3 < 0 (Σ3 < Σ1).

206 1.8

1.6

Σ−Σ 31 1.4 T σ1 1.2

0.8

0.6 k=0 0.4 k=-0.3098 0.2 k=0.3098 FE Data (J3>0) 0 -4.5 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 4.5

2Σ + Σ31 T 3σ1 A

1.8

1.6

Σ−Σ 31 1.4 T σ1 1.2

0.8

0.6 k=0 0.4 k=-0.3098 k=0.3098 0.2 FE Data (J3<0) 0 -4.5 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 4.5

2Σ + Σ31 T 3σ1 B

Figure 8-16. Anisotropic, axisymmetric ﬁnite element yield points versus analytic curves when initial void volume fraction is f0 = 0.01 and the yield strength in the through-thickness direction is greater than the yield strength in the in-plane Σ Σ direction (i.e., Material B in Table 8-1). A) J3 > 0 (Σ3 > Σ1). B) J3 < 0 (Σ3 < Σ1).

207 1.2

Σ−Σ 31 T σ1 0.8

0.6

0.4 k=0 k=-0.3098 0.2 k=0.3098 FE Data (J3>0) 0 -3 -2 -1 0 1 2 3

2Σ + Σ31 T 3σ1 A

1.2

Σ−Σ 31 T σ1 0.8

0.6

0.4 k=0 k=-0.3098 0.2 k=0.3098 FE Data (J3<0) 0 -3 -2 -1 0 1 2 3

2Σ + Σ31 T 3σ1 B

Figure 8-17. Anisotropic, axisymmetric ﬁnite element yield points versus analytic curves when initial void volume fraction is f0 = 0.01 and the yield strength in the through-thickness direction is less than the yield strength in the in-plane Σ Σ direction (i.e., Material C in Table 8-1). A) J3 > 0 (Σ3 > Σ1). B) J3 < 0 (Σ3 < Σ1).

208 1.8

1.6 f = 01.0 k = 3098.0

Σ−Σ 31 1.4 T σ1 1.2

0.8

0.6 Proposed Criterion (J3>0)

0.4 Proposed Criterion (J3<0)

0.2 Σm h Σm h T = ln()f T −= ln()f σ1 3 σ1 3 0 -5 -4 -3 -2 -1 0 1 2 3 4 5

2Σ + Σ31 T 3σ1 A

1.8

1.6 f = 01.0 k −= 3098.0

Σ−Σ 31 1.4 T σ1 1.2

0.8

0.6 Proposed Criterion (J3>0)

0.4 Proposed Criterion (J3<0)

0.2 Σm h Σm h T = ln()f T −= ln()f σ1 3 σ1 3 0 -5 -4 -3 -2 -1 0 1 2 3 4 5

2Σ + Σ31 T 3σ1 B

Figure 8-18. Anisotropic yield surfaces deﬁned by Equation (8–41) with q = 1 for a matrix material containing a spherical void of porosity f = 0.01 and with the through-thickness yield strength greater than the in-plane yield strength. The hydrostatic parameter h is deﬁned by Equation (8–44) for transverse isotropy. A) Yield strengths in tension greater than the yield strengths in compression. B) Yield strengths in tension less than the yield strengths in compression.

209 CHAPTER 9 CONCLUSIONS This dissertation was devoted to the development of macroscopic plastic potentials for void-matrix aggregates with the matrix displaying plastic anisotropy and different yield strengths in tension versus compression. A micromechanically-based approach was adopted for modeling the mechanical response of the porous ductile metals. Using a Hill-Mandel homogenization approach with homogeneous rate of deformation boundary conditions (see Chapter2), an upper bound of the yield surface was derived for a hollow sphere, and a hollow cylinder, made of a perfectly plastic matrix obeying the Cazacu et al.(2006) yield criterion. A void-matrix aggregate consisting of spherical voids in an isotropic matrix exhibiting tension-compression asymmetry was first considered (see Chapter3). An analytical criterion that depends on the second and third invariants of the stress deviator and that accounts for the tension-compression asymmetry of the plastic flow of the matrix was developed. The developed criterion captures the derived exact hydrostatic solution. If the matrix has the same yield in tension and compression, the new criterion reduces to Gurson’s (1977) criterion for a ductile metal with the matrix obeying the von Mises yield criterion and a random distribution of spherical voids. A void-matrix aggregate consisting of cylindrical voids in an isotropic matrix was also considered (see Chapter6). For axisymmetric loading, a parametric representation of the yield surface was obtained. An expression of the yield surface applicable to plane strain conditions was also developed, which reduces to Gurson’s (1977) cylindrical criterion when the yield in tension is equal to the yield in compression. Lastly, the combined effects of matrix plastic anisotropy and tension-compression asymmetry on the plastic flow of the porous aggregate was investigated in Chapter 8. The micromechanical analysis was done for arbitrary loading conditions but the principal axes of loading were assumed to align with the symmetry axes of the matrix.

210 The proposed anisotropic criterion reduces to that of Benzerga and Besson(2001) if there is no tension-compression asymmetry in the void-free material. This developed model provides information about how the plastic anisotropy influences the strength of the porous material. Namely, the anisotropy of the matrix, together with any strength differential, influences the shape of the yield locus in the deviatoric plane while the presence of voids in the aggregate leads to a decrease in the size of the yield locus with increasing pressure (positive or negative). The evolution of the voids result in the yield locus decreasing in size with increasing void volume fractions (the shape of the yield locus, by contrast, is independent of the void volume fraction and governed by the matrix anisotropy and tension-compression asymmetry). The analytical expressions developed for the macroscopic yield functions were obtained by assuming a specific geometry for the representative volume element, specific loading conditions and other simplifying assumptions. These assumptions and simplifica- tions were necessary to obtain analytical solutions but needed to be validated against more general scenarios. The validity of the proposed yield criteria was assessed by comparing the analytical yield curves to finite element unit cell calculations. In the finite element unit cell calculations, the void boundary was explicitly meshed and the matrix material was modeled as an elastic-plastic material with the plastic response governed by the void-free yield criterion. The theoretical yield criterion developed in Chapter3 for a void-matrix aggregate containing spherical voids was compared to axisymmetric unit cell finite element calculations in Chapter5. The agreement between the finite element results and the proposed criterion was found to be quite good. Plane strain unit cell calculations were also performed and compared in Chapter7 with the plane strain criterion developed in Chapter 6. Again, the agreement was found to be satisfactory. The proposed analytical yield curve developed in Chapter8 compares well with transversely isotropic finite element calculations (i.e., the void-free material was considered to be transversely isotropic).

211 All of the yield criteria proposed in this dissertation are significant in that they are the only models available that account for both void growth and tension-compression asymmetry in the void-free material (commonly seen in hexagonal close packed metals). All of the proposed criteria depend on the third invariant of the stress deviator and on both the tensile and compressive yield strengths. They all reduce to well-accepted models when no strength differential effect is present in the void-free material. The material models that resulted from this effort should provide the engineering community with a more accurate tool for analyzing the failure of porous hexagonal close packed metals. Future research could extend the models developed herein in a similar manner to how Gurson’s (1977) criteria have been extended (see, for example, the literature review done in Chapter1). Specifically, the yield criteria developed in Chapters3,6 and8 could be extended to capture the behavior of materials during large deformations and high strain rates by incorporating temperature sensitivity, strain hardening and strain-rate dependence. Also, while the void-growth stage of ductile failure has been the primary focus here, both a nucleation and a coalescence criterion may be considered in conjunction with the yield criteria developed in this dissertation in order to provide complete models for ductile damage. Either existing criteria for void nucleation and coalescence could be employed (see, for example, the criteria presented in Chapter1) or additional criteria, if required, could be developed. Still, the criteria developed in this dissertation fill a gap in the literature; these criteria are the only ones that capture the particularities of damage by void growth in textured, hexagonal close packed materials.

212 APPENDIX A PARAMETRIC REPRESENTATION DERIVATION OF THE AXISYMMETRIC YIELD LOCUS This appendix details the development of an exact (for the assumed velocity ﬁeld given by Equation (6–18)), parametric representation for the yield locus of a void-matrix aggregate when the matrix is governed by the Cazacu et al.(2006) criterion, the void geometry is cylindrical and the loading is axisymmetric. A.1 General Form of Equations

The general forms of the relevant equations used throughout the parametric representation derivation for the yield surface of the porous aggregate (cylindrical voids and axisymmetric loading) are given in the following. First, the diﬀerent branches of the plastic multiplier rate, λ˙ , will be given followed by the general form of the upper-bound macroscopic plastic dissipation, W +, (it is an upper bound since only one velocity ﬁeld is considered) and associated derivatives. A.1.1 Plastic multiplier rate branches

In order to determine the macroscopic plastic dissipation and associated derivatives, the expression of the plastic multiplier rate must ﬁrst be determined. The plastic multiplier associated with the Cazacu et al.(2006) yield criterion has multiple branches depending on the sign and ordering of the principal values of the rate of deformation ten-

¯ 2 2 ¯ sor (see Equation (2–29)). If the parameter x is deﬁned as x := Db /r where D := Dkk/2, then the non-zero microscopic rate of deformation components can be expressed, using the assumed velocity ﬁeld of Equation (6–18), as follows:

B d = − x + r 2 B d = x − θ 2

dz = B

213 where B := D33. Now, the plastic multiplier rate associated with the Cazacu et al.(2006)

P isotropic yield criterion can be written as follows (where di = di for rigid plastic ﬂow):  s r 2 2 2 2  2 (3β − 2) d + d + d d1 1  1 2 3 if ≥  2 p p 4 2  3 (2β − 1) dijdij 2 (β − β + 1) λ˙ = s (A–1) r 2 2 2 2 2 2  2 d + d + (3 − 2β ) d /β d3 −β  1 2 3 if ≤  2 p p 4 2  3 (2 − β ) dijdij 2 (β − β + 1) where d1 ≥ d2 ≥ d3 are the ordered principal components of the microscopic rate of deformation tensor and β = σT /σC . Note that the ﬁrst branch in Equation (A–1)

σ σ corresponds to J3 ≥ 0 and the second branch to J3 ≤ 0. Now, the following constants are deﬁned for use in the later sections:

3 p = β2 + 1 4 m = 3 1 − β2

q = 3β2 − 1 s 3 (2 − β2) g = 4β2 with

m¯ = −m

2 q2 = 3 − β r 3 (2β2 − 1) g = 2 4 and 1 β2 + 1 s = 1 2 β2 − 1 1 s = 2β2 − 1 2 2 β2 − 2 s = 3 2β2

214 with

s¯1 = −s1

s¯2 = −s2

s¯3 = −s3.

˙ σ If λz is the relevant branch, then there are two cases (depending on the sign of J3 ) corresponding to dz = d1, the maximum principal value, and dz = d3, the minimum principal value. The valid range is then determined by solving the relevant inequality from Equation (A–1) for the roots, x (note that one or more roots may be invalid for speciﬁc loading scenarios depending on the value of β and the known sign of x). Thus, for

d −β2 z ≤ p p 4 2 dijdij 2 (β − β + 1) the plastic multiplier rate becomes

2 ˙ p 2 2 2 λz = x + g B p3 (2 − β2) with roots

x3 = s3B

x¯3 =s ¯3B.

Likewise, for d 1 z ≥ p p 4 2 dijdij 2 (β − β + 1) the plastic multiplier rate is written as

2 q ˙ 2 2 2 λz = x + g B p3 (2β2 − 1) 2 with roots

x2 = s2B

x¯2 =s ¯2B.

215 ˙ If the loading condition is such that λθ is the relevant branch, then the analysis proceeds in a similar manner as that used previously. In other words, there are again two cases depending on whether dθ = d1 or dθ = d3 with a corresponding range of validity. When d −β2 θ ≤ p p 4 2 dijdij 2 (β − β + 1) the plastic multiplier rate is s 2 p λ˙ = pB2 +mBx ¯ + q x2 θ 3β2 (2 − β2) 2 with roots

x¯1 =s ¯1B

x¯2 =s ¯2B.

Similarly, if d 1 θ ≥ p p 4 2 dijdij 2 (β − β + 1) then s 2 p λ˙ = pB2 + mBx + qx2 θ 3 (2β2 − 1) with roots

x3 = s3B

x1 = s1B.

˙ σ If λr is the relevant branch then J3 ≤ 0 with dr = d3 is equivalent to

d −β2 r ≤ p p 4 2 dijdij 2 (β − β + 1)

such that s 2 p λ˙ = pB2 + mBx + q x2 r 3β2 (2 − β2) 2

216 with roots

x1 = s1B

x2 = s2B.

σ Similarly, when the sign of J3 is positive and dr = d1, this is equivalent to

d 1 r ≥ p p 4 2 dijdij 2 (β − β + 1)

such that s 2 p λ˙ = pB2 +mBx ¯ + qx2 r 3 (2β2 − 1) with roots

x¯3 =s ¯3B

x¯1 =s ¯1B.

A.1.2 Macroscopic plastic dissipation and derivatives

The necessary derivatives of the macroscopic plastic dissipation, W +, are obtained ¯ using the chain rule on the two variables D and B = D33 such that

∂W + ∂W + ∂D¯ 1 ∂W + Σ11 = = = ∂D11 ∂D¯ ∂D11 2 ∂D¯ ∂W + Σ = Σ + Σ = 2Σ = γγ 11 22 11 ∂D¯ and ∂W + ∂W + ∂D¯ ∂W + ∂B 1 ∂W + ∂W + ∂W + Σ33 = = + = + = Σ11 + ∂D33 ∂D¯ ∂D33 ∂B ∂D33 2 ∂D¯ ∂B ∂B + ∂W Σe = |Σ33 − Σ11| = . ∂B The general form of the integrals associated with the macroscopic plastic dissipation remains the same regardless of which case is under consideration. Let the parameter u be deﬁned such that D¯ u = . |B|

217 For demonstration purposes, the integral solutions using the integration limits for one speciﬁc case will be given in the following equations (the example shown corresponds to ¯ β = σT /σC > 1, D > 0 and B < 0 wheres ¯3f ≤ u ≤ s¯3). The macroscopic plastic dissipation is written as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x3 x ¯ Z x3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x r ¯ D/f¯ 2 σT D mB = qh1(x) + h2(x) + h3(x) 3 p 2 2 2β − 1 x3 ¯ x 2 σT D h i 3 + √ p h4(x) 3 2 − β2 D¯ where x3 = s3B and where the integral solution of √ Z X mB dx = qh (x) + h (x) + h (x) x2 1 2 2 3 has been used with X = pB2 + mBx + qx2 and

Z dx 1 p h1(x) = √ = √ ln 2 qX + 2qx + mB X q ! Z dx −1 2ppB2X + mBx + 2pB2 h2(x) = √ = ln x X ppB2 x √ − X h (x) = . 3 x Also,

Z px2 + g2B2 h (x) = dx 4 x2 p x2 + g2B2 p = − + ln x + x2 + g2B2 . x

218 The relevant derivatives can now be written as

+ + r ¯ ( D/f¯ D/f¯ ∂W W 2 σT D ∂h1 ∂x mB ∂h2 ∂x = + q + ∂D¯ D¯ 3 p 2 ∂x ∂D¯ 2 ∂x ∂D¯ 2β − 1 x3 x3 ¯ s x3  D/f 2 ∂h3 ∂x 2 (2β − 1) ∂h4 ∂x  + + ∂x ∂D¯ 2 − β2 ∂x ∂D¯ x3 D¯  ( ¯ ¯ ∂W + r2 σ D¯ ∂h ∂x ∂h D/f mB ∂h ∂x ∂h D/f = T q 1 + 1 + 2 + 2 ∂B 3 p 2 ∂x ∂B ∂B 2 ∂x ∂B ∂B 2β − 1 x3 x3 ¯ s ) ∂h ∂x ∂h D/f 2 (2β2 − 1) ∂h ∂x ∂h x3 + 3 + 3 + 4 + 4 ∂x ∂B ∂B 2 − β2 ∂x ∂B ∂B x3 D¯ with ∂h 1 1 = √ ∂x X ∂h 1 2 = √ ∂x x X ∂h − (mBx/2 + qx2 − X) 3 = √ ∂x x2 X ∂h px2 + g2B2 4 = ∂x x2 and √ √ ! ∂h1 1 q (mx + 2pB) + m X = √ √ √ ∂B q 2 qX + (2qx + mB) X √ ! ∂ (Bh ) −sgn(B) pB (4pB2 + 3mBx + 2qx2) + (mx + 4pB) ppB2 X 2 = √ √ ∂B p 2pB2X + (mBx + 2pB2) X ∂h pB + mx/2 3 = − √ ∂B x X ∂h Bg2 4 = − . ∂B x2 + xpx2 + g2B2

The resultant solutions using the previous derivatives are given for the various branches of β, D¯ and B in the following sections.

219 A.2 β ≥ 1: The Matrix Yield Strength in Tension is Greater than in Compression.

When dealing with the yield criterion of Cazacu et al.(2006), there are multiple regions and branches within regions to consider when deriving the parametric representation of the porous yield surface. This section will focus on those materials for which the matrix

has a yield in tension greater than in compression (i.e., β > 1 such that σT > σC ).

Σ A.2.1 J3 < 0

Σ If J3 < 0 then there are two cases to consider: 1) Σm > 0 and 2) Σm < 0. Also, when Σ ¯ J3 < 0, note that B = D33 < 0 such that u = −D/B.

A.2.1.1 Σm > 0 ¯ When Σm > 0, note that D > 0. There are three diﬀerent ranges of u to consider in this case which dictate the limits of integration used to evaluate W +(D) and, thus, determine the ﬁnal form of the stress invariants.

First Branch: 0 ≤ u ≤ s¯3f. In this ﬁrst branch the macroscopic plastic dissipation is given as

¯ Z D/f¯ 2 σT D p dx W + = √ x2 + g2B2 p 2 2 3 2 − β D¯ x

such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln ∂D¯ 3 p2 − β2 u + pu2 + g2 f and + ∂W 2 σT hp p i = √ u2 + g2f 2 − u2 + g2 . ∂B 3 p2 − β2

Second Branch:s ¯3f ≤ u ≤ s¯3. In this next branch the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x3 x ¯ Z x3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x

220 where x3 = s3B. This yields the following derivatives: " ! ∂W + r2 σ p2 (2β2 − 1) s − ps2 + g2 = T ln − 3 3 ∂D¯ 3 p2β2 − 1 p2 − β2 u + pu2 + g2 √ p ! √ 2 q pf 2 − muf + qu2 + 2qu − mf 1 + q ln √ · p 2 f 2 q p + ms3 + qs3 − 2qs3 − m √ p 2 2 !# m 2 p pf − muf + qu − mu + 2pf s3 + √ ln − √ · 2 p p 2 u 2 p p + ms3 + qs3 + ms3 + 2p and " # ∂W + r2 ppf 2 − muf + qu2 p2 (u2 + g2) = σT − . ∂B 3 p2β2 − 1 p2 − β2

Third Branch:s ¯3 ≤ u ≤ ∞.

Lastly, whens ¯3 ≤ u ≤ ∞ the macroscopic plastic dissipation is as follows:

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 D¯ x

such that

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf − muf + qu + 2qu − mf 1 = q ln √ ∂D¯ 3 p2β2 − 1 2 qpp − mu + qu2 + 2qu − m f √ p !# m 2 p pf 2 − muf + qu2 − mu + 2pf + √ ln √ 2 p 2 ppp − mu + qu2 − mu + 2p and + r ∂W 2 σT hp p i = pf 2 − muf + qu2 − p − mu + qu2 . ∂B 3 p2β2 − 1

A.2.1.2 Σm < 0 ¯ When Σm < 0 this results in D < 0. In this case, there are again three ranges of u for which distinct expressions of the macroscopic plastic dissipation exist.

First Branch: 0 ≥ u ≥ s3f. In this ﬁrst branch the macroscopic plastic dissipation is given as

¯ Z D/f¯ 2 σT D p dx W + = √ x2 + g2B2 p 2 2 3 2 − β D¯ x

221 such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln ∂D¯ 3 p2 − β2 u + pu2 + g2 f and + ∂W 2 σT hp p i = √ u2 + g2f 2 − u2 + g2 . ∂B 3 p2 − β2

Second Branch: s3f ≥ u ≥ s3.

For the second branch s3f ≥ u ≥ s3 such that the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 x¯3 x ¯ Z x¯3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x

wherex ¯3 =s ¯3B. This yields the following derivatives: " ! ∂W + r2 σ p2 (2β2 − 1) s¯ − ps¯2 + g2 = T ln − 3 3 ∂D¯ 3 p2β2 − 1 p2 − β2 u + pu2 + g2 √ p ! √ 2 q pf 2 − muf¯ + qu2 + 2qu − mf¯ 1 + q ln √ · p 2 f 2 q p +m ¯ s¯3 + qs¯3 − 2qs¯3 − m¯ √ p 2 2 !# m¯ 2 p pf − muf¯ + qu − mu¯ + 2pf s¯3 + √ ln − √ · 2 p p 2 u 2 p p +m ¯ s¯3 + qs¯3 +m ¯ s¯3 + 2p and " # ∂W + r2 ppf 2 − muf¯ + qu2 p2 (u2 + g2) = σT − . ∂B 3 p2β2 − 1 p2 − β2

Third Branch: s3 ≥ u ≥ −∞. The last branch has the following macroscopic plastic dissipation:

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 D¯ x

222 such that

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf − muf¯ + qu + 2qu − mf¯ 1 = q ln √ ∂D¯ 3 p2β2 − 1 2 qpp − mu¯ + qu2 + 2qu − m¯ f √ p !# m¯ 2 p pf 2 − muf¯ + qu2 − mu¯ + 2pf + √ ln √ 2 p 2 ppp − mu¯ + qu2 − mu¯ + 2p and + r ∂W 2 σT hp p i = pf 2 − muf¯ + qu2 − p − mu¯ + qu2 . ∂B 3 p2β2 − 1

Σ A.2.2 J3 > 0 Σ ¯ When J3 > 0 then B = D33 > 0 such that u = D/B. As before, there are two cases that must be treated individually: the case of Σm > 0 and the case of Σm < 0.

A.2.2.1 Σm > 0 ¯ For the case of Σm > 0, D > 0. There are ﬁve diﬀerent ranges of u that must be accounted for and that result in separate expressions for the macroscopic plastic dissipation.

First Branch: 0 ≤ u ≤ s2f. The macroscopic plastic dissipation for this range is given as

2 σ D¯ Z D/f¯ q dx W + = √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln 2 ¯ p 2 p 2 2 f ∂D 3 2β − 1 u + u + g2 and ∂W + 2 σ q q = √ T u2 + g2 − u2 + g2f 2 . ∂B 3 p2β2 − 1 2 2

Second Branch: s2f ≤ u ≤ s1f.

223 In the second branch, the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB p 2 2 2 3 β (2 − β ) x2 x 2 σ D¯ Z x2 q dx + √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x where x2 = s2B. This yields the following derivative expressions: " ! ∂W + r2 σ p2β2 (2 − β2) s + ps2 + g2 = T ln 2 2 2 ¯ 3 p 2 2 p 2 p 2 2 ∂D β (2 − β ) 2β − 1 u + u + g2 √ p 2 2 ! √ 2 q2 pf + muf + q2u + 2q2u + mf 1 + q2 ln √ · p 2 f 2 q2 p + ms2 + q2s2 + 2q2s2 + m √ p 2 2 !# m 2 p pf + muf + q2u + mu + 2pf s2 − √ ln √ · 2 p p 2 u 2 p p + ms2 + q2s2 + ms2 + 2p

and + r "p 2 2 p 2 2 # ∂W 2 2 (u + g2) pf + muf + q2u = σT − . ∂B 3 p2β2 − 1 pβ2 (2 − β2)

Third Branch: s1f ≤ u ≤ s2.

The third branch contains the range s1f ≤ u ≤ s2 such that the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x1 x r ¯ Z x1 2 σT D p 2 2 dx + q2x + mBx + pB p 2 2 2 3 β (2 − β ) x2 x r 2 σ D¯ √ Z x2 q dx + T 2 x2 + g2B2 . p 2 2 2 3 2β − 1 D¯ x

224 This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf + muf + qu + 2qu + mf 1 = q ln √ · ¯ 3 p 2 p 2 f ∂D 2β − 1 2 q p + ms1 + qs1 + 2qs1 + m √ p 2 2 ! m 2 p pf + muf + qu + mu + 2pf s1 − √ ln √ · 2 p p 2 u 2 p p + ms1 + qs1 + ms1 + 2p ! # √ s + ps2 + g2 √ ps2 + g2 pp + ms + qs2 + 2 ln 2 2 2 − 2 2 2 + 1 1 p 2 2 s s u + u + g2 2 1 r " √ p 2 ! 2 σT √ 2 q2 p + ms1 + q2s1 + 2q2s1 + m + q2 ln √ 3 p 2 2 p 2 β (2 − β ) 2 q2 p + ms2 + q2s2 + 2q2s2 + m √ p 2 ! m 2 p p + ms1 + q2s1 + ms1 + 2p s2 − √ ln √ · 2 p p 2 s 2 p p + ms2 + q2s2 + ms2 + 2p 1 # pp + ms + q s2 pp + ms + q s2 − 1 2 1 + 2 2 2 s1 s2 and + r ∂W 2 σT q p = 2 (u2 + g2) − pf 2 + muf + qu2 . ∂B 3 p2β2 − 1 2

Fourth Branch: s2 ≤ u ≤ s1. In the fourth branch the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x1 x r ¯ Z x1 2 σT D p 2 2 dx + q2x + mBx + pB . p 2 2 2 3 β (2 − β ) D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf + muf + qu + 2qu + mf 1 = q ln √ · ¯ 3 p 2 p 2 f ∂D 2β − 1 2 q p + ms1 + qs1 + 2qs1 + m √ p 2 2 !# m 2 p pf + muf + qu + mu + 2pf s1 − √ ln √ · 2 p p 2 u 2 p p + ms1 + qs1 + ms1 + 2p r " √ p 2 ! 2 σT √ 2 q2 p + ms1 + q2s1 + 2q2s1 + m + q2 ln p 2 2 √ p 2 3 β (2 − β ) 2 q2 p + mu + q2u + 2q2u + m √ p 2 !# m 2 p p + ms1 + q2s + ms1 + 2p u − √ ln 1 · √ p 2 2 p 2 p p + mu + q2u + mu + 2p s1

225 and

+ r "p 2 p 2 2 ∂W 2 p + ms1 + qs1 − pf + muf + qu = σT ∂B 3 p2β2 − 1 # pp + mu + q u2 − pp + ms + q s2 + 2 1 2 1 . pβ2 (2 − β2)

Fifth Branch: s1 ≤ u ≤ ∞.

The ﬁfth and ﬁnal branch contains the range s1 ≤ u ≤ ∞ such that the expression for the macroscopic plastic dissipation becomes

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 . p 2 2 3 2β − 1 D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf + muf + qu + 2qu + mf 1 = q ln √ · ∂D¯ 3 p2β2 − 1 2 qpp + mu + qu2 + 2qu + m f √ p !# m 2 p pf 2 + muf + qu2 + mu + 2pf − √ ln √ 2 p 2 ppp + mu + qu2 + mu + 2p and + r ∂W 2 σT hp p i = p + mu + qu2 − pf 2 + muf + qu2 . ∂B 3 p2β2 − 1

A.2.2.2 Σm < 0 ¯ If Σm < 0 then D < 0 resulting in ﬁve relevant branches depending on the range of u being considered.

First Branch: 0 ≥ u ≥ s¯2f. In this ﬁrst branch the macroscopic plastic dissipation is given as 2 σ D¯ Z D/f¯ q dx W + = √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln 2 ¯ p 2 p 2 2 f ∂D 3 2β − 1 u + u + g2

226 and ∂W + 2 σ q q = √ T u2 + g2 − u2 + g2f 2 . ∂B 3 p2β2 − 1 2 2

Second Branch:s ¯2f ≥ u ≥ s¯1f.

In the ranges ¯2f ≥ u ≥ s¯1f, the expression for the macroscopic plastic dissipation becomes

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) x¯2 x 2 σ D¯ Z x¯2 q dx + √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x wherex ¯2 =s ¯2B. This yields the following derivative expressions: " ! ∂W + r2 σ p2β2 (2 − β2) s¯ + ps¯2 + g2 = T ln 2 2 2 ¯ 3 p 2 2 p 2 p 2 2 ∂D β (2 − β ) 2β − 1 u + u + g2 √ p 2 2 ! √ 2 q2 pf +muf ¯ + q2u + 2q2u +mf ¯ 1 + q2 ln √ · p 2 f 2 q2 p +m ¯ s¯2 + q2s¯2 + 2q2s¯2 +m ¯ √ p 2 2 !# m¯ 2 p pf +muf ¯ + q2u +mu ¯ + 2pf s¯2 − √ ln √ · 2 p p 2 u 2 p p +m ¯ s¯2 + q2s¯2 +m ¯ s¯2 + 2p

and + r "p 2 2 p 2 2 # ∂W 2 2 (u + g2) pf +muf ¯ + q2u = σT − . ∂B 3 p2β2 − 1 pβ2 (2 − β2)

Third Branch:s ¯1f ≥ u ≥ s¯2. The third branch results in a macroscopic plastic dissipation given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 x¯1 x r ¯ Z x¯1 2 σT D p 2 2 dx + q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) x¯2 x r 2 σ D¯ √ Z x¯2 q dx + T 2 x2 + g2B2 . p 2 2 2 3 2β − 1 D¯ x

227 This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf +muf ¯ + qu + 2qu +mf ¯ 1 = q ln √ · ¯ 3 p 2 p 2 f ∂D 2β − 1 2 q p +m ¯ s¯1 + qs¯1 + 2qs¯1 +m ¯ √ p 2 2 ! m¯ 2 p pf +muf ¯ + qu +mu ¯ + 2pf s¯1 − √ ln √ · 2 p p 2 u 2 p p +m ¯ s¯1 + qs¯1 +m ¯ s¯1 + 2p ! # √ s¯ + ps¯2 + g2 √ ps¯2 + g2 pp +m ¯ s¯ + qs¯2 + 2 ln 2 2 2 − 2 2 2 + 1 1 p 2 2 s¯ s¯ u + u + g2 2 1 r " √ p 2 ! 2 σT √ 2 q2 p +m ¯ s¯1 + q2s¯1 + 2q2s¯1 +m ¯ + q2 ln √ 3 p 2 2 p 2 β (2 − β ) 2 q2 p +m ¯ s¯2 + q2s¯2 + 2q2s¯2 +m ¯ √ p 2 ! m¯ 2 p p +m ¯ s¯1 + q2s¯1 +m ¯ s¯1 + 2p s¯2 − √ ln √ · 2 p p 2 s¯ 2 p p +m ¯ s¯2 + q2s¯2 +m ¯ s¯2 + 2p 1 # pp +m ¯ s¯ + q s¯2 pp +m ¯ s¯ + q s¯2 − 1 2 1 + 2 2 2 s¯1 s¯2 and + r ∂W 2 σT q p = 2 (u2 + g2) − pf 2 +muf ¯ + qu2 . ∂B 3 p2β2 − 1 2

Fourth Branch:s ¯2 ≥ u ≥ s¯1. The fourth branch yields the following expression for the macroscopic plastic dissipation:

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 x¯1 x r ¯ Z x¯1 2 σT D p 2 2 dx + q2x +mBx ¯ + pB . p 2 2 2 3 β (2 − β ) D¯ x

228 This, in turn, yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf +muf ¯ + qu + 2qu +mf ¯ 1 = q ln √ · ¯ 3 p 2 p 2 f ∂D 2β − 1 2 q p +m ¯ s¯1 + qs¯1 + 2qs¯1 +m ¯ √ p 2 2 !# m¯ 2 p pf +muf ¯ + qu +mu ¯ + 2pf s¯1 − √ ln √ · 2 p p 2 u 2 p p +m ¯ s¯1 + qs¯1 +m ¯ s¯1 + 2p r " √ p 2 ! 2 σT √ 2 q2 p +m ¯ s¯1 + q2s¯1 + 2q2s¯1 +m ¯ + q2 ln p 2 2 √ p 2 3 β (2 − β ) 2 q2 p +mu ¯ + q2u + 2q2u +m ¯ √ p 2 !# m¯ 2 p p +m ¯ s¯1 + q2s¯ +m ¯ s¯1 + 2p u − √ ln 1 · √ p 2 2 p 2 p p +mu ¯ + q2u +mu ¯ + 2p s¯1 and

+ r "p 2 p 2 2 ∂W 2 p +m ¯ s¯1 + qs¯1 − pf +muf ¯ + qu = σT ∂B 3 p2β2 − 1 # pp +mu ¯ + q u2 − pp +m ¯ s¯ + q s¯2 + 2 1 2 1 . pβ2 (2 − β2)

Fifth Branch:s ¯1 ≥ u ≥ −∞.

The ﬁnal branch contains the ranges ¯1 ≥ u ≥ −∞ such that the macroscopic plastic dissipation becomes

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 . p 2 2 3 2β − 1 D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q pf +muf ¯ + qu + 2qu +mf ¯ 1 = q ln √ · ∂D¯ 3 p2β2 − 1 2 qpp +mu ¯ + qu2 + 2qu +m ¯ f √ p !# m¯ 2 p pf 2 +muf ¯ + qu2 +mu ¯ + 2pf − √ ln √ 2 p 2 ppp +mu ¯ + qu2 +mu ¯ + 2p and + r ∂W 2 σT hp p i = p +mu ¯ + qu2 − pf 2 +muf ¯ + qu2 . ∂B 3 p2β2 − 1

A.3 β ≤ 1: The Matrix Yield Strength in Tension is Less than in Compression.

This section focuses on those materials where β < 1 such that σT < σC .

229 Σ A.3.1 J3 < 0 Σ ¯ If J3 < 0 then B < 0 such that u = −D/B. The two cases that must be taken into consideration here are Σm > 0 and Σm < 0.

A.3.1.1 Σm > 0 ¯ When Σm > 0 then D > 0 and there are ﬁve domains of u that must be treated individually.

First Branch: 0 ≤ u ≤ s¯3f. In the ﬁrst branch the macroscopic plastic dissipation is given as

¯ Z D/f¯ 2 σT D p dx W + = √ x2 + g2B2 p 2 2 3 2 − β D¯ x

such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln ∂D¯ 3 p2 − β2 u + pu2 + g2 f and + ∂W 2 σT hp p i = √ u2 + g2f 2 − u2 + g2 . ∂B 3 p2 − β2

Second Branch:s ¯3f ≤ u ≤ s¯1f.

The second branch contains the ranges ¯3f ≤ u ≤ s¯1f such that the expression for the macroscopic plastic dissipation in this region is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 + mBx + pB2 p 2 2 3 2β − 1 x3 x ¯ Z x3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x

where x3 = s3B. This yields the following derivative expressions: " ! ∂W + r2 σ p2 (2β2 − 1) −s + ps2 + g2 = T ln 3 3 ∂D¯ 3 p2β2 − 1 p2 − β2 u + pu2 + g2 √ p ! √ −2 q pf 2 − muf + qu2 − 2qu + mf 1 + q ln √ · p 2 f −2 q p + ms3 + qs3 + 2qs3 + m √ p 2 2 !# m 2 p pf − muf + qu − mu + 2pf s3 + √ ln − √ · 2 p p 2 u 2 p p + ms3 + qs3 + ms3 + 2p

230 and " # ∂W + r2 p2 (u2 + g2) ppf 2 − muf + qu2 = σT − − . ∂B 3 p2 − β2 p2β2 − 1

Third Branch:s ¯1f ≤ u ≤ s¯3.

Whens ¯1f ≤ u ≤ s¯3, the macroscopic plastic dissipation becomes

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB p 2 2 2 3 β (2 − β ) x1 x r ¯ Z x1 2 σT D p dx + qx2 + mBx + pB2 p 2 2 3 2β − 1 x3 x r ¯ Z x3 2 σT D √ p p dx + 2 β2 x2 + g2B2 . p 2 2 2 3 β (2 − β ) D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf + q2u − 2q2u + mf 1 = q2 ln √ · ¯ 3 p 2 2 p 2 f ∂D β (2 − β ) −2 q2 p + ms1 + q2s1 + 2q2s1 + m √ p 2 2 ! m 2 p pf − muf + q2u − mu + 2pf s1 + √ ln − √ · 2 p p 2 u 2 p p + ms1 + q2s1 + ms1 + 2p p 2 2 ! p 2 2 p 2 # p 2 −s3 + s3 + g p 2 s3 + g p + ms1 + q2s1 + 2β ln p + 2β − u + u2 + g2 s3 s1

r " √ p 2 ! 2 σT √ −2 q p + ms1 + qs1 + 2qs1 + m + q ln √ 3 p 2 p 2 2β − 1 −2 q p + ms3 + qs3 + 2qs3 + m √ p 2 ! m 2 p p + ms1 + qs1 + ms1 + 2p s3 + √ ln √ · 2 p p 2 s 2 p p + ms3 + qs3 + ms3 + 2p 1 # pp + ms + qs2 pp + ms + qs2 + 1 1 − 3 3 s1 s3

and + r h i ∂W 2 σT p 2 2 2 p 2 2 = − 2β (u + g ) + pf − muf + q2u . ∂B 3 pβ2 (2 − β2)

Fourth Branch:s ¯3 ≤ u ≤ s¯1.

231 In this branch the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB p 2 2 2 3 β (2 − β ) x1 x r ¯ Z x1 2 σT D p dx + qx2 + mBx + pB2 . p 2 2 3 2β − 1 D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf + q2u − 2q2u + mf 1 = q2 ln √ · ¯ 3 p 2 2 p 2 f ∂D β (2 − β ) −2 q2 p + ms1 + q2s1 + 2q2s1 + m √ p 2 2 !# m 2 p pf − muf + q2u − mu + 2pf s1 + √ ln − √ · 2 p p 2 u 2 p p + ms1 + q2s1 + ms1 + 2p r " √ p 2 ! 2 σT √ −2 q p + ms1 + qs1 + 2qs1 + m + q ln √ 3 pβ2 (2β2 − 1) −2 qpp − mu + qu2 − 2qu + m

√ p 2 !# m 2 p p + ms1 + qs1 + ms1 + 2p u + √ ln − √ p · 2 p 2 p p − mu + qu2 − mu + 2p s1

and

+ r "p 2 2 p 2 ∂W 2 pf − muf + q2u − p + ms1 + q2s1 = σT ∂B 3 pβ2 (2 − β2) # pp + ms + qs2 − pp − mu + qu2 + 1 1 . p2β2 − 1

Fifth Branch:s ¯1 ≤ u ≤ ∞. In this last branch, the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB . p 2 2 2 3 β (2 − β ) D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf + q2u − 2q2u + mf 1 = q2 ln · ¯ p 2 2 √ p 2 ∂D 3 β (2 − β ) −2 q2 p − mu + q2u − 2q2u + m f √ p 2 2 !# m 2 p pf − muf + q2u − mu + 2pf + √ ln √ p 2 2 p 2 p p − mu + q2u − mu + 2p

232 and + r h i ∂W 2 σT p 2 2 p 2 = pf − muf + q2u − p − mu + q2u . ∂B 3 pβ2 (2 − β2)

A.3.1.2 Σm < 0

There are another ﬁve branches that must be taken into account when Σm < 0 such that D¯ < 0.

First Branch: 0 ≥ u ≥ s3f.

First, attention will be given to the range of 0 ≥ u ≥ s3f. In this branch, the macroscopic plastic dissipation is given as

¯ Z D/f¯ 2 σT D p dx W + = √ x2 + g2B2 p 2 2 3 2 − β D¯ x

such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln ∂D¯ 3 p2 − β2 u + pu2 + g2 f and + ∂W 2 σT hp p i = √ u2 + g2f 2 − u2 + g2 . ∂B 3 p2 − β2

Second Branch: s3f ≥ u ≥ s1f. In the second branch the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ 2 σT D p dx W + = qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 x¯3 x ¯ Z x¯3 2 σT D p dx + √ x2 + g2B2 p 2 2 3 2 − β D¯ x

wherex ¯3 =s ¯3B. This yields the following derivative expressions: " ! ∂W + r2 σ p2 (2β2 − 1) −s¯ + ps¯2 + g2 = T ln 3 3 ∂D¯ 3 p2β2 − 1 p2 − β2 u + pu2 + g2 √ p ! √ −2 q pf 2 − muf¯ + qu2 − 2qu +mf ¯ 1 + q ln √ · p 2 f −2 q p +m ¯ s¯3 + qs¯3 + 2qs¯3 +m ¯ √ p 2 2 !# m¯ 2 p pf − muf¯ + qu − mu¯ + 2pf s¯3 + √ ln − √ · 2 p p 2 u 2 p p +m ¯ s¯3 + qs¯3 +m ¯ s¯3 + 2p

233 and " # ∂W + r2 p2 (u2 + g2) ppf 2 − muf¯ + qu2 = σT − − . ∂B 3 p2 − β2 p2β2 − 1

Third Branch: s1f ≥ u ≥ s3.

When s1f ≥ u ≥ s3 the macroscopic plastic dissipation is as follows:

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) x¯1 x r ¯ Z x¯1 2 σT D p dx + qx2 +mBx ¯ + pB2 p 2 2 3 2β − 1 x¯3 x r ¯ Z x¯3 2 σT D √ p p dx + 2 β2 x2 + g2B2 . p 2 2 2 3 β (2 − β ) D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf¯ + q2u − 2q2u +mf ¯ 1 = q2 ln √ · ¯ 3 p 2 2 p 2 f ∂D β (2 − β ) −2 q2 p +m ¯ s¯1 + q2s¯1 + 2q2s¯1 +m ¯ √ p 2 2 ! m¯ 2 p pf − muf¯ + q2u − mu¯ + 2pf s¯1 + √ ln − √ · 2 p p 2 u 2 p p +m ¯ s¯1 + q2s¯1 +m ¯ s¯1 + 2p p 2 2 ! p 2 2 p 2 # p 2 −s¯3 + s¯3 + g p 2 s¯3 + g p +m ¯ s¯1 + q2s¯1 + 2β ln p + 2β − u + u2 + g2 s¯3 s¯1

r " √ p 2 ! 2 σT √ −2 q p +m ¯ s¯1 + qs¯1 + 2qs¯1 +m ¯ + q ln √ 3 p 2 p 2 2β − 1 −2 q p +m ¯ s¯3 + qs¯3 + 2qs¯3 +m ¯ √ p 2 ! m¯ 2 p p +m ¯ s¯1 + qs¯1 +m ¯ s¯1 + 2p s¯3 + √ ln √ · 2 p p 2 s¯ 2 p p +m ¯ s¯3 + qs¯3 +m ¯ s¯3 + 2p 1 # pp +m ¯ s¯ + qs¯2 pp +m ¯ s¯ + qs¯2 + 1 1 − 3 3 s¯1 s¯3

and + r h i ∂W 2 σT p 2 2 2 p 2 2 = − 2β (u + g ) + pf − muf¯ + q2u . ∂B 3 pβ2 (2 − β2)

Fourth Branch: s3 ≥ u ≥ s1.

234 The fourth branch yields an expression for the macroscopic plastic dissipation as

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) x¯1 x r ¯ Z x¯1 2 σT D p dx + qx2 +mBx ¯ + pB2 . p 2 2 3 2β − 1 D¯ x

This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf¯ + q2u − 2q2u +mf ¯ 1 = q2 ln √ · ¯ 3 p 2 2 p 2 f ∂D β (2 − β ) −2 q2 p +m ¯ s¯1 + q2s¯1 + 2q2s¯1 +m ¯ √ p 2 2 !# m¯ 2 p pf − muf¯ + q2u − mu¯ + 2pf s¯1 + √ ln − √ · 2 p p 2 u 2 p p +m ¯ s¯1 + q2s¯1 +m ¯ s¯1 + 2p r " √ p 2 ! 2 σT √ −2 q p +m ¯ s¯1 + qs¯1 + 2qs¯1 +m ¯ + q ln √ 3 pβ2 (2β2 − 1) −2 qpp − mu¯ + qu2 − 2qu +m ¯

√ p 2 !# m¯ 2 p p +m ¯ s¯1 + qs¯1 +m ¯ s¯1 + 2p u + √ ln − √ p · 2 p 2 p p − mu¯ + qu2 − mu¯ + 2p s¯1

and

+ r "p 2 2 p 2 ∂W 2 pf − muf¯ + q2u − p +m ¯ s¯1 + q2s¯1 = σT ∂B 3 pβ2 (2 − β2) # pp +m ¯ s¯ + qs¯2 − pp − mu¯ + qu2 + 1 1 . p2β2 − 1

Fifth Branch: s1 ≥ u ≥ −∞. In the ﬁfth and ﬁnal branch the expression for the macroscopic plastic dissipation becomes r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB . p 2 2 2 3 β (2 − β ) D¯ x This yields the following derivative expressions:

+ r " √ p 2 2 ! ∂W 2 σT √ −2 q2 pf − muf¯ + q2u − 2q2u +mf ¯ 1 = q2 ln · ¯ p 2 2 √ p 2 ∂D 3 β (2 − β ) −2 q2 p − mu¯ + q2u − 2q2u +m ¯ f √ p 2 2 !# m¯ 2 p pf − muf¯ + q2u − mu¯ + 2pf + √ ln √ p 2 2 p 2 p p − mu¯ + q2u − mu¯ + 2p

235 and + r h i ∂W 2 σT p 2 2 p 2 = pf − muf¯ + q2u − p − mu¯ + q2u . ∂B 3 pβ2 (2 − β2)

Σ A.3.2 J3 > 0

Σ When J3 > 0 then the two cases to be considered separately are Σm > 0 and Σ < 0. Σ ¯ Likewise for J3 > 0, B > 0 such that u = D/B.

A.3.2.1 Σm > 0 ¯ If Σm > 0 then D > 0 and there are three branches containing distinct expressions for the macroscopic plastic dissipation W +(D).

First Branch: 0 ≤ u ≤ s2f. In the ﬁrst branch the macroscopic plastic dissipation is given as

Second Branch: s2f ≤ u ≤ s2. In the second branch the expression for the macroscopic plastic dissipation is given as

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB p 2 2 2 3 β (2 − β ) x2 x 2 σ D¯ Z x2 q dx + √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x

236 where x2 = s2B. This yields the following derivatives: " ! ∂W + r2 σ p2β2 (2 − β2) s + ps2 + g2 = T ln 2 2 2 ¯ 3 p 2 2 p 2 p 2 2 ∂D β (2 − β ) 2β − 1 u + u + g2 √ p 2 2 ! √ 2 q2 pf + muf + q2u + 2q2u + mf 1 + q2 ln √ · p 2 f 2 q2 p + ms2 + q2s2 + 2q2s2 + m √ p 2 2 !# m 2 p pf + muf + q2u + mu + 2pf s2 − √ ln √ · 2 p p 2 u 2 p p + ms2 + q2s2 + ms2 + 2p and + r "p 2 2 p 2 2 # ∂W 2 2 (u + g2) pf + muf + q2u = σT − . ∂B 3 p2β2 − 1 pβ2 (2 − β2)

Third Branch: s2 ≤ u ≤ ∞.

Finally, when s2 ≤ u ≤ ∞ the macroscopic plastic dissipation is as follows:

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x + mBx + pB p 2 2 2 3 β (2 − β ) D¯ x such that

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q2 pf + muf + q2u + 2q2u + mf 1 = q2 ln · ¯ p 2 2 √ p 2 ∂D 3 β (2 − β ) 2 q2 p + mu + q2u + 2q2u + m f √ p 2 2 !# m 2 p pf + muf + q2u + mu + 2pf − √ ln √ p 2 2 p 2 p p + mu + q2u + mu + 2p and + r h i ∂W 2 σT p 2 p 2 2 = p + mu + q2u − pf + muf + q2u . ∂B 3 pβ2 (2 − β2)

A.3.2.2 Σm < 0 ¯ If Σm < 0 then D < 0 and, again, there are three diﬀerent branches that must be described.

First Branch: 0 ≥ u ≥ s¯2f.

When 0 ≥ u ≥ s¯2f the macroscopic plastic dissipation is given as

2 σ D¯ Z D/f¯ q dx W + = √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x

237 such that ! ∂W + 2 σ u + pu2 + g2f 2 1 = √ T ln 2 · ¯ p 2 p 2 2 f ∂D 3 2β − 1 u + u + g2 and ∂W + 2 σ q q = √ T u2 + g2 − u2 + g2f 2 . ∂B 3 p2β2 − 1 2 2

Second Branch:s ¯2f ≥ u ≥ s¯2. In the second branch the macroscopic plastic dissipation becomes

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) x¯2 x 2 σ D¯ Z x¯2 q dx + √ T x2 + g2B2 p 2 2 2 3 2β − 1 D¯ x wherex ¯2 =s ¯2B. This yields the following derivatives: " ! ∂W + r2 σ p2β2 (2 − β2) s¯ + ps¯2 + g2 = T ln 2 2 2 ¯ 3 p 2 2 p 2 p 2 2 ∂D β (2 − β ) 2β − 1 u + u + g2 √ p 2 2 ! √ 2 q2 pf +muf ¯ + q2u + 2q2u +mf ¯ 1 + q2 ln √ · p 2 f 2 q2 p +m ¯ s¯2 + q2s¯2 + 2q2s¯2 +m ¯ √ p 2 2 !# m¯ 2 p pf +muf ¯ + q2u +mu ¯ + 2pf s¯2 − √ ln √ · 2 p p 2 u 2 p p +m ¯ s¯2 + q2s¯2 +m ¯ s¯2 + 2p

and + r "p 2 2 p 2 2 # ∂W 2 2 (u + g2) pf +muf ¯ + q2u = σT − . ∂B 3 p2β2 − 1 pβ2 (2 − β2)

Third Branch:s ¯2 ≥ u ≥ −∞.

The third branch contains the ranges ¯2 ≥ u ≥ −∞ and yields an expression for the macroscopic plastic dissipation as follows:

r ¯ Z D/f¯ + 2 σT D p 2 2 dx W = q2x +mBx ¯ + pB p 2 2 2 3 β (2 − β ) D¯ x

238 such that

+ r " √ p 2 2 ! ∂W 2 σT √ 2 q2 pf +muf ¯ + q2u + 2q2u +mf ¯ 1 = q2 ln · ¯ p 2 2 √ p 2 ∂D 3 β (2 − β ) 2 q2 p +mu ¯ + q2u + 2q2u +m ¯ f √ p 2 2 !# m 2 p pf +muf ¯ + q2u +mu ¯ + 2pf − √ ln √ p 2 2 p 2 p p +mu ¯ + q2u +mu ¯ + 2p and + r h i ∂W 2 σT p 2 p 2 2 = p +mu ¯ + q2u − pf +muf ¯ + q2u . ∂B 3 pβ2 (2 − β2)

239 APPENDIX B RELATIONSHIP BETWEEN HILL48 AND CPB06 COEFFICIENTS The following demonstrates that the CPB06 (see Cazacu et al., 2006) criterion can reduce to Hill’s 1948 anisotropic criterion (see Hill, 1948, 1950) for the case where the yield strengths in tension are equal to the yield strengths in compression (i.e., k = 0). Speciﬁcally, the case of transverse isotropy (with the 1-2 plane being the plane of symmetry and the 3-direction being the out-of-plane direction) will be considered in the following analysis since analytical expressions become more tractable; however, the procedure is the same for the more general case of orthotropy. In order to show this reduction to Hill48, letσ ˆ = LΣ0 with the constraint that the transformed stress,σ ˆ, be deviatoric (note that this condition is more restrictive than the original CPB06 criterion presented in Cazacu et al., 2006). The linear transformation, L, can be written in Voight notation as follows:   L11 L12 L13 0 0 0      L L L 0 0 0   12 22 23       L13 L23 L33 0 0 0  L =      0 0 0 L44 0 0       0 0 0 0 L 0   55    0 0 0 0 0 L66 which, for transverse isotropy, becomes   L11 L12 L13 0 0 0      L L L 0 0 0   12 11 13       L13 L13 L33 0 0 0  L =   .    0 0 0 L44 0 0       0 0 0 0 L 0   44    0 0 0 0 0 L11 − L12

240 Note that, in order for the previous transformation to yield a deviatoric tensor when applied to the stress deviator, the following equations must hold:

L11 + L12 + L13 = 1

L12 + L22 + L23 = 1

L13 + L23 + L33 = 1 where the constant on the right hand side of the previous equations has been chosen to be 1 such that the linear transformation reduces to the identity tensor in the case of isotropy. In general, the CPB06 criterion (using a = 2) can be written as

2 2 2 F =σ Î +σ ÎI +σ ÎII

σˆ = 2J2

σˆ whereσ Î ,σ ÎI andσ ÎII are the principal components ofσ ˆ and J2 is the second invariant of the transformed stress tensor,σ ˆ (note that the principal components are deviators given the imposed constraint that the result of applying the linear transformation, L, to a deviatoric tensor is itself a deviatoric tensor). Now, the CPB06 yield criterion can be written as

2 T 2 ϕ =m ˆ F − σ1 = 0 (B–1) such that, for k = 0,

2 σˆ T 2 mˆ 2J2 = σ1 (B–2)

T where σ1 is the uniaxial yield strength in the 1-direction (tension-compression symmetry

T is assumed here but the notation of σY = σ1 will be retained throughout the following paragraphs to maintain notational consistency with the case when k 6= 0). The constant,

241 mˆ , is the anisotropic version of the eﬀective stress constant and is given by s 1 mˆ = 2 2 2 (|Φ1| − kΦ1) + (|Φ2| − kΦ2) + (|Φ3| − kΦ3)

where 2 1 1 Φ = L − L − L 1 3 11 3 12 3 13 2 1 1 Φ = L − L − L 2 3 12 3 22 3 23 2 1 1 Φ = L − L − L 3 3 13 3 23 3 33 such that, assuming k = 0, s 9 mˆ = 2 2 2 (3L11 − 1) + (3L12 − 1) + (3L13 − 1) which can be further reduced to the following for transverse isotropy (using the constraint that the L-tensor transforms a deviatoric tensor to another deviatoric tensor): s 3 mˆ = 2 . (B–3) 6(L13 − 1)(L11 + L13) + 6L11 + 2

Hill’s 1948 criterion can be written as

0 0 2 0 0 2 0 0 2 2 2 2 2 F (Σ22 − Σ33) + G(Σ33 − Σ11) + H(Σ11 − Σ22) + 2LΣ23 + 2MΣ13 + 2NΣ12 = σY (B–4) where the parameters, F , G, H, L, M and N are deﬁned in terms of the uniaxial yield strengths in the 1, 2 and 3 directions as

2 2 σY 1 T = 1 = G + H = σ1 R11 2 2 σY 1 T = H + F = σ2 R22 2 2 σY 1 T = F + G = σ3 R33

242 and

σ 2 3 2 Y = M = σ¯13 2R13 σ 2 3 2 Y = L = σ¯23 2R23 σ 2 3 2 Y = N = σ¯12 2R12

T where σY = σ1 is used and where theσ ¯ij are the yield stresses in shear with respect to the respective principal axes of the material. The parameters, Rij, are introduced here since ABAQUS uses this terminology in its internal material library for Hill48. For references see Hill(1948), Hill(1950) and Abaqus(2008). Note that the scalar Hill48 Parameter, L, is an unfortunate coincidental notation and should not be confused with the fourth-order linear transformation tensor, L, or its Voight-notation counterpart, L. Also, the Hill48 parameter, F , should not be confused with the notation used to describe the CPB06 yield function in Equation (B–2) (in the remaining paragraphs, ‘F ’ will solely refer to the Hill48 parameter). Equating Equation (B–2) with Equation (B–4) yields

2 σˆ 0 0 2 0 0 2 0 0 2 2 2 2 2m ˆ J2 = F (Σ22 − Σ33) + G(Σ33 − Σ11) + H(Σ11 − Σ22) + 2LΣ23 + 2MΣ13 + 2NΣ12

0 2 0 2 0 2 = (2H + 2G − F ) (Σ11) + (2H + 2F − G) (Σ22) + (2F + 2G − H) (Σ33)

2 2 2 + 2L (Σ23) + 2M (Σ13) + 2N (Σ12)

0 2 0 2 0 2 = (2H + G) (Σ11) + (2H + G) (Σ22) + (4G − H) (Σ33)

2 2 2 + 2M (Σ23) + 2M (Σ13) + 2 (2H + G) (Σ12) (B–5)

where the last of the previous equations uses the fact that F = G, L = M and N = G+2H for the case of transverse isotropy under consideration. The main task now is to evaluate

0 the left-hand-side of Equation (B–5) and equate the terms of Σij with those obtained in the right-hand-side of the last expression in Equation (B–5). With that in mind, the

243 transformed stress can be expressed as   0 0 0  L11Σ11 + L12Σ22 + L13Σ33       L Σ0 + L Σ0 + L Σ0   12 11 11 22 13 33     0 0 0   L13Σ11 + L13Σ22 + L33Σ33  σˆ = (B–6)  L Σ   44 23       L44Σ13       (L11 − L12)Σ12 

σˆ where transverse isotropy has been assumed. Now, the second invariant, J2 , of the transformed stress deviator can be expressed as follows:

σˆ 2J2 =σ ˆ :σ ˆ =σ ˆijσˆij

2 2 2 0 2 = L11 + L12 + L13 + 2L12L13 + 2L11L13 + 2L13L33 (Σ11)

2 2 2 0 2 + L11 + L12 + L13 + 2L12L13 + 2L11L13 + 2L13L33 (Σ22)

2 2 0 2 2 2 2 2 + 4L13 + L33 + 4L11L12 (Σ33) + 2L44 (Σ23) + 2L44 (Σ13)

2 2 + 2 (L11 − L12) (Σ12) .

0 Equating the Σij terms between the previous equation and Equation (B–5) yields

2 2 2 2 2H + G =m ˆ L11 + L12 + L13 + 2L12L13 + 2L11L13 + 2L13L33

2 2 2 4G − H =m ˆ 4L13 + L33 + 4L11L12 (B–7)

2 2 M =m ˆ L44 which, along with

L11 + L12 + L13 = 1 (B–8)

2L13 + L33 = 1 from the deviatoric constraint on L yields 5 equations for the 5 unknowns L11, L33, L12,

L13 and L44 (assuming that the Hill48 coeﬃcients, H, G and M are known). The previous

244 5 equations can be used to determine the Hill48 parameters given the CPB06 coefficients or to determine the CPB06 coefficients given the Hill48 coefficients. B.1 Determine The Hill48 Coefficients Given The CPB06 Coefficients

Given a set of CPB06 coeﬃcients, the Hill48 parameters are readily determined from Equations (B–7) and (B–8). Neglecting the algebra, the following relations are ultimately obtained:

2 2 2 (L13 − 1) (L13 − L11) + 2L13 − 2L11 + 1 G = 2 6(L13 − 1)(L11 + L13) + 6L11 + 2 2 2 −4 (L13 − 1) (L13 − L11) − 4L13 + 4L11 + 1 H = 2 (B–9) 6(L13 − 1)(L11 + L13) + 6L11 + 2 2 3L44 M = 2 . 6(L13 − 1)(L11 + L13) + 6L11 + 2 Note that for transverse isotropy, only two anisotropic coeﬃcients are independent; either H or G can be calculated given the other using G + H = 1. For transverse isotropy, the material anisotropy is often reported in terms of the plastic strain ratios R and Rh. The Hill48 coeﬃcients are readily obtained from these plastic strain ratios as follows: d H R = RL = 22 = d33 G and 2 σY T = 1 = G + H σ1 from before, such that, 1 G = F = R + 1 and R H = . R + 1 Also, 2M − (H + G) R = RTS = . h 2(H + G)

245 B.2 Determine The CPB06 Coeﬃcients Given The Hill48 Coeﬃcients

If a set of Hill48 coeﬃcients are given, both Equations (B–7) and (B–8) must be used to determine the CPB06 coeﬃcients. Neglecting the tedious algebra once again, the following relations are obtained:

1 1r8 − 4G 1 G + 1 − p3G (G + 1) L = + − 11 2 4 G + 1 2 3 (G + 1) G + 1 − p3G (G + 1) L = (B–10) 13 3 (G + 1) r M L = . 44 G + 1 The last of the previous equations yields (see Equation (B–7))

mˆ 2 = G + 1 (B–11) for transverse isotropy.

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250 BIOGRAPHICAL SKETCH In May of 2002, Joel Benjamin Stewart graduated from the University of Florida with a Bachelor of Science in aerospace engineering. He began working in October of 2002 for the Air Force Research Laboratory at Eglin Air Force Base in Florida. In December of 2005 while working at Eglin, he completed a Master of Engineering degree in aerospace engineering at the Research and Engineering Education Facility in Shalimar, Florida. Joel Stewart was admitted in April of 2008, while working as a full-time graduate student on an Air Force fellowship, as a University of Florida doctoral candidate in the Department of Mechanical and Aerospace Engineering.

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