an abstract of a dissertation hydrostatic stress...

AN ABSTRACT OF A DISSERTATION

HYDROSTATIC STRESS EFFECTS IN LOW CYCLE FATIGUE

Phillip A. Allen

Doctor of Philosophy in Engineering

Classical metal plasticity theory assumes that hydrostatic stress has negligible effect on the yield and postyield behavior of metals. Recent reexaminations of classical theory have revealed a significant effect of hydrostatic stress on the yield behavior of notched geometries. New experiments and nonlinear finite element analyses (FEA) of 2024-T851 and Inconel 100 (IN100) test specimens have revealed the effect of internal hydrostatic tensile stresses on yielding. Nonlinear FEA using the von Mises (yielding is independent of hydrostatic stress) and the Drucker-Prager (yielding is linearly dependent on hydrostatic stress) yield functions were performed. Mechanical tests were performed to characterize the material properties of two metals, IN100 and 2024-T851. In addition, monotonic and low cycle fatigue tests were performed on several notched round bar (NRB) geometries to use for comparison with the finite element results. To perform the cyclic finite element analyses, a pressure-dependent constitutive model was developed as an ABAQUS user subroutine (UMAT). This UMAT incorporates the Drucker-Prager yield theory with combined multilinear kinematic and isotropic hardening. Finite element models (FEM’s) of a variety of test specimens were created including: smooth tensile, smooth compression, NRB, and equal-arm bend geometries. For all cases, load-displacement or load-microstrain test data was compared to von Mises and Drucker-Prager finite element solutions. For the monotonic tensile loading, the Von Mises solutions overestimated experimental load-displacement curves, while the Drucker-Prager solutions essentially matched the test data. For the low cycle fatigue tests, using a yield function that is dependent on hydrostatic stress significantly altered the predicted hysteresis response of notched specimens, particularly for the first few cycles. Specifically, for the 2024-T851 and IN100 test specimens, the Drucker-Prager solutions more accurately predicted the specimen’s behavior for first few cycles compared to the von Mises solutions. However, once the stable material response was reached, neither the Drucker-Prager nor von Mises results were entirely satisfactory. Also, neither solution truly captured the shapes of the hysteresis loops.

HYDROSTATIC STRESS EFFECTS IN

LOW CYCLE FATIGUE

A Dissertation

Presented to

the Faculty of the Graduate School

Tennessee Technological University

by

Phillip A. Allen

In Partial Fulfillment

of the Requirements of the Degree

DOCTOR OF PHILOSPHY

Engineering

December 2002

ii

CERTIFICATE OF APPROVAL OF DISSERTATION

HYDROSTATIC STRESS EFFECTS IN

LOW CYCLE FATIGUE

by

Phillip A. Allen

Graduate Advisory Committee:

Chairperson date

Member date

Member date

Member date

Member date

Approved for the Faculty

Associate Vice President of Research and Graduate Studies

Date

iii

DEDICATION

This dissertation is dedicated to my wife Shannon.

This work could not have been accomplished without

her unselfish sacrifices, love, and support. She is

my constant source of encouragement and hope.

In addition to Shannon, this work is dedicated to our parents

Harold and Joyce Allen and Doug and Betty Hays.

They have given us tremendous love and guidance

and have provided us with many opportunities

which we never thought possible.

iv

ACKNOWLEDGEMENTS

I would like to thank my major professor, Dr. Chris Wilson, for his guidance, his

support, and his friendship. I would also like to express thanks to the other committee

members, Dr. George Buchanan, Dr. Brian O’Connor, Dr. Dale Wilson, and Dr. John

Zhu for their comments and assistance.

I am grateful to several people at Marshall Space Flight Center for their

assistance, guidance, and advice. These people include Dr. Greg Swanson, Jeff Rayburn,

Dr. Preston McGill, Doug Wells, Bill Malone, and Mike Watwood. I would like to thank

Bill Mitchell and Jerry Sheldon at Pratt and Whitney for providing Inconel 100

specimens and test data. In addition, I would like to express thanks to Bill Scherzinger

and Kevin Brown at Sandia National Laboratories for their assistance in developing the

constitutive models.

Funding for this research was provided by the National Aeronautics and Space

Administration Marshall Space Flight Center. Support was also provided by the Center

for Manufacturing Research and the Mechanical Engineering Department at Tennessee

Technological University.

v

TABLE OF CONTENTS

Page

LIST OF TABLES .............................................................................................................ix

LIST OF FIGURES.............................................................................................................x

LIST OF SYMBOLS, ACRONYMS, AND ABBREVIATIONS....................................xx

Chapter

1. INTRODUCTION..........................................................................................................1

2. TECHNICAL BACKGROUND ....................................................................................4

A Classical View of Metal Plasticity............................................................................4

Yield Functions .........................................................................................................5

Hardening Rules........................................................................................................9

Flow Rules...............................................................................................................13

Hydrostatic Stress – Deviations From Classical Theory............................................14

Hydrostatic Stress - Recent Developments ................................................................30

Low Cycle Fatigue......................................................................................................33

Strain-Life Methodology.........................................................................................34

Multiaxial Fatigue Research....................................................................................38

3. RESEARCH PLAN......................................................................................................42

Experimental Program................................................................................................42

Analytical Program.....................................................................................................45

vi

Chapter Page

4. MECHANICAL TESTING..........................................................................................47

Test Apparatus ............................................................................................................47

Alignment Testing ......................................................................................................48

2024-T851 Testing .....................................................................................................49

Elastic Constants Tests............................................................................................50

Smooth Uniaxial Tension Tests ..............................................................................52

Smooth Uniaxial Compression Tests ......................................................................54

Notched Round Bar Tension Tests..........................................................................58

NRB Low Cycle Fatigue Tests................................................................................65

Smooth Round Bar Low Cycle Fatigue Tests.........................................................70

Inconel 100 Testing ....................................................................................................71

Smooth Tensile Tests ..............................................................................................72

Smooth Compression Tests.....................................................................................75

Low Cycle Fatigue Tests.........................................................................................78

5. FINITE ELEMENT CONSTITUTIVE MODEL DEVLOPMENT ............................81

Pressure-Dependent Plasticity Model.........................................................................81

Elasticity..................................................................................................................81

Yield Function.........................................................................................................82

Flow Rule ................................................................................................................84

State Variable Equations .........................................................................................86

Numerical Integration .............................................................................................87

Hardening Models ......................................................................................................91

vii

Chapter Page

Basic Definitions .....................................................................................................92

Isotropic Hardening.................................................................................................93

Kinematic Hardening ..............................................................................................97

Combined Bilinear Hardening ................................................................................98

Combined Multilinear Hardening .........................................................................103

Pressure-Dependent Combined Hardening ...........................................................105

Constitutive Model Programming ............................................................................107

6. FINITE ELEMENT MODELING .............................................................................109

Material Property Inputs ...........................................................................................109

2024-T851 Material Property Inputs.....................................................................110

Inconel 100 Material Property Inputs ...................................................................113

Test Specimen Finite Element Models.....................................................................116

Smooth Tensile Bar Specimen ..............................................................................116

Smooth Compression Cylinder Specimen ............................................................125

Notched Round Bar Specimens.............................................................................129

Equal Arm Bend Specimen...................................................................................138

7. FINITE ELEMENT MODEL RESULTS ..................................................................144

UMAT Program Verification ...................................................................................144

2024-T851 Results....................................................................................................150

Smooth Tensile Bar Results ..................................................................................150

Smooth Compression Cylinder Results ................................................................152

Notched Tensile Bar Results .................................................................................154

viii

Chapter Page

Notched Round Bar Low Cycle Fatigue Results ..................................................161

Inconel 100 Results ..................................................................................................183

Smooth Tensile Bar Results ..................................................................................183

Smooth Compression Cylinder Results ................................................................185

Equal-Arm Bend Low Cycle Fatigue Results .......................................................187

8. CONCLUSIONS AND RECOMMENDATIONS.....................................................195

BIBLIOGRAPHY ...........................................................................................................200

APPENDICES

A. TENSILE AND COMPRESSION TEST DATA.................................................209

B. NRB LOW CYLCLE FATIGUE TEST PLOTS..................................................212

C. ABAQUS UMAT PROGRAMS ..........................................................................226

D. ABAQUS MATERIAL DATA TABLES ............................................................238

E. SCRIPT FILE........................................................................................................242

F. FINITE ELEMENT MODEL MESHES...............................................................244

G. NRB LOW CYCLE FATIGUE FEA PLOTS ......................................................268

VITA ...............................................................................................................................284

ix

LIST OF TABLES

Table Page

Table 2.1. Summary of Experimental Results for Constants in Equation (2.27) [26,27] 22 Table 3.1. Nominal Alloy Compositions of 2024-T851 and IN100 [58].........................43 Table 4.1. Summary of 2024-T851 Elastic Constants Test Results.................................51 Table 4.2. Summary of 2024-T851 Smooth Tensile Results ...........................................54 Table 4.3. Summary of 2024-T851 Smooth Compression Tests .....................................56 Table 4.4. Summary of NRB Low Cycle Fatigue Tests...................................................66 Table 4.5. Summary of IN100 Smooth Tensile Results...................................................74 Table 4.6. Summary of IN100 Smooth Compression Tests.............................................76

x

LIST OF FIGURES

Figure Page

Figure 2.1. von Mises Yield Surface in Principal Stress Space [15]..................................8 Figure 2.2. Isotropic Hardening for the von Mises Yield Function ...................................9 Figure 2.3. Illustration of the Bauschinger Effect ............................................................10 Figure 2.4. Kinematic Hardening for the von Mises Yield Function...............................11 Figure 2.5. True Stress-True Strain Curves in Tension and Compression for 4310 Steel

[17]...............................................................................................................12 Figure 2.6. True Stress-True Strain Curves in Tension and Compression for 4330 Steel

[17]...............................................................................................................12 Figure 2.7. Flow Stress (Effective Stress) as a Function of Strain for Tempered Pearlite

Tested at Various Pressures [19] .................................................................16 Figure 2.8. Flow Stress (Effective Stress) as a Function of Strain for Tempered

Martensite Tested at Various Pressures [19] ...............................................16 Figure 2.9. Plastic Stress-Strain Relations in Tension for Nittany No.2 Brass Under

Hydrostatic Pressure [23] ............................................................................18 Figure 2.10. Effect of Hydrostatic Pressure on the Stress-Strain Curves in Compression

for 4330 Steel [25] .......................................................................................19 Figure 2.11. Effect of Hydrostatic Pressure on the Stress-Strain Curves in Compression

for Aged Maraging Steel [25]......................................................................20 Figure 2.12. Dependence of Yielding on I1 in 4330 Steel [25] ........................................20 Figure 2.13. Dependence of Yielding on I1 for Aged Maraging Steel [25] .....................21 Figure 2.14. Plastic Volume Increase as a Function of True Plastic Strain for 4310 and

4330 Steels [25] ...........................................................................................23 Figure 2.15. Schematic of σeff versus I1 [15]....................................................................25

xi

Figure Page

Figure 2.16. Cohesive Force as a Function of the Separation Between Atoms (Adapted from [28]).....................................................................................................25

Figure 2.17. Comparison of Drucker-Prager and von Mises Yield Surfaces in Principal

Stress Space [15]..........................................................................................28 Figure 2.18. Hysteresis Loop of a Specimen Subjected to Cyclic Loading (Adapted from

[46]) .............................................................................................................35 Figure 2.19. Log-Log Plot Showing the Relationship Between Fatigue Life and Strain

Amplitude (Adapted from [46])...................................................................37 Figure 2.20. Correlation of Lefebvre’s Test Data and Fatigue Life Relation for Various

Strain Ratios [55].........................................................................................40 Figure 3.1. Schematic of Calculation of a from Uniaxial Tension and Uniaxial

Compression Data........................................................................................44 Figure 4.1. Engineering Drawing of the Rectangular Alignment Specimen (All

Dimensions are in Inches) ...........................................................................48 Figure 4.2. Engineering Drawing of the Cylindrical Alignment Specimen (All

Dimensions are in Inches) ...........................................................................49 Figure 4.3. Schematic of the 2024-T851 1 in. Thick Plate with Specimen Machining

Directions.....................................................................................................50 Figure 4.4. Engineering Drawing of the Poisson’s Ratio Specimen (All Dimensions are in

Inches)..........................................................................................................50 Figure 4.5. Engineering Drawing of the Young’s Modulus Specimen (All Dimensions

are in Inches)................................................................................................51 Figure 4.6. Engineering Drawing of the 2024-T851 Smooth Tension Specimen (All

Dimensions are in Inches) ...........................................................................52 Figure 4.7. Composite True Stress-True Strain Plot for 2024-T851 L Direction Smooth

Tensile Tests ................................................................................................53 Figure 4.8. Composite True Stress-True Strain Plot for 2024-T851 L-T Direction Smooth

Tensile Tests ................................................................................................53

xii

Figure Page

Figure 4.9. Engineering Drawing of the Smooth Compression Specimen (All Dimensions are in Inches) ...........................................................................54

Figure 4.10. Composite True Stress-True Strain Plot for 2024-T851 L Direction Smooth

Compression Tests.......................................................................................55 Figure 4.11. Composite True Stress-True Strain Plot for 2024-T851 L-T Direction

Smooth Compression Tests .........................................................................56 Figure 4.12. Comparison of 2024-T851 L Direction Tensile and Compressive True

Stress-True Strain Curves ............................................................................57 Figure 4.13. Comparison of 2024-T851 L-T Direction Tensile and Compressive True

Stress-True Strain Curves ............................................................................57 Figure 4.14. Engineering Drawing of the Notched Round Bar Specimen (All Dimensions

are in Inches)................................................................................................59 Figure 4.15. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB

Tests with ρ = 0.005 in. ...............................................................................60 Figure 4.16. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB





Tests with ρ = 0.120 in. ...............................................................................63 Figure 4.21. Composite Load-Gage Displacement Plot for All 2024-T851 L Direction

NRB Geometries..........................................................................................63 Figure 4.22. Comparison of Fracture Mode for the NRB and Smooth Tensile Specimens

.....................................................................................................................64

xiii

Figure Page

Figure 4.23. Load-Gage Displacement Plot of Selected Cycles for 2024-T851 L Direction NRB LCF Tests with ρ = 0.040 in. (Specimen 403) ...................67

Figure 4.24. Load-Gage Displacement Plot of Selected Cycles for 2024-T851 L

Direction NRB LCF Tests with ρ = 0.080 in. (Specimen 803) ...................68 Figure 4.25. Load-Gage Displacement Plot of Selected Cycles for 2024-T851 L

Direction NRB LCF Tests with ρ = 0.120 in. (Specimen 125) ...................69 Figure 4.26. Comparison of 2024-T851 L Direction Monotonic and Transitional Cyclic

True Stress-True Strain Curves ...................................................................71 Figure 4.27. Tension and Compression Specimen Layout in IN100 Disk (All Dimensions

are in Inches)................................................................................................72 Figure 4.28. Engineering Drawing of the IN100 Smooth Tensile Specimen (All

Dimensions are in Inches) ...........................................................................73 Figure 4.29. Composite True Stress-True Strain Plot for IN100 Smooth Tensile Tests..74 Figure 4.30. Composite True Stress-True Strain Plot for IN100 Smooth Compression

Tests.............................................................................................................76 Figure 4.31. Comparison of IN100 Tensile and Compressive True Stress-True Strain

Curves ..........................................................................................................77 Figure 4.32. Engineering Drawing of the Equal-Arm Bend Specimen (All Dimensions

are in Inches) [59] ........................................................................................78 Figure 4.33. Load-Microstrain Plot for the Equal-Arm Bend Three-Cycle Fatigue Test

[73]...............................................................................................................79 Figure 5.1. Linear Drucker-Prager Model: Yield Surface and Flow Direction in the p-t

Plane (Adapted from [5]).............................................................................83 Figure 5.2. Illustration of a Yield Surface in Deviatoric Stress Space (Adapted from [79])

.....................................................................................................................92 Figure 5.3. Illustration of the Relationship Between Yield Stress and Equivalent Plastic

Strain for the Bilinear Hardening Case........................................................95 Figure 5.4. Illustration of the Uniaxial True Stress versus True Strain Relationship for

the Bilinear Hardening Case........................................................................96

xiv

Figure Page

Figure 5.5. Geometric Interpretation of the Incremental Form of the Consistency Condition for Combined Hardening (Adapted from [79]) ........................102

Figure 5.6. Geometric Interpretation of the Incremental Form of the Radial Return

Correction (Adapted from [79]).................................................................102 Figure 5.7. Illustration of the Relationship Between Yield Stress and Equivalent Plastic

Strain for the Multilinear Hardening Case.................................................104 Figure 6.1. Comparison of Tensile Test Data and ABAQUS Input Data ......................112 Figure 6.2. Extended View of Comparison of Tensile Test Data and ABAQUS Input

Data............................................................................................................112 Figure 6.3. Comparison of Tensile Test Data and ABAQUS Input Data ......................115 Figure 6.4. Extended View of Comparison of Tensile Test Data and ABAQUS Input

Data............................................................................................................115 Figure 6.5. Schematic of Axisymmetric Model of a Smooth tensile Bar Specimen

Utilizing Two Planes of Symmetry ...........................................................117 Figure 6.6. Medium Mesh FEM of the 0.35 in. Diameter Smooth Tensile Bar.............118 Figure 6.7. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35

in. Diameter Smooth Tensile Bar FEM’s at Failure Load .........................119 Figure 6.8. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35 in.

Diameter Smooth Tensile Bar FEM’s at Failure Load...............................119 Figure 6.9. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35

in. Diameter Smooth Tensile Bar FEM’s at Failure Load .........................120 Figure 6.10. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine

Mesh 0.35 in. Diameter Smooth Tensile Bar FEM’s at Failure Load .......120 Figure 6.11. Medium Mesh FEM of the 0.25 in. Diameter Smooth Tensile Bar...........122 Figure 6.12. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh

0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load .................123 Figure 6.13. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.25

in. Diameter Smooth Tensile Bar FEM’s at Failure Load .........................123

xv

Figure Page

Figure 6.14. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load .........................124

Figure 6.15. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine

Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load .......124 Figure 6.16. Schematic of Axisymmetric Model of a Smooth Compression Cylinder

Specimen Utilizing Two Planes of Symmetry...........................................125 Figure 6.17. Medium Mesh FEM of the Smooth Compression Specimen ....................126 Figure 6.18. Effective Stress Across the Bottom Symmetry Plane of the Coarse,

Medium, and Fine Mesh Smooth Compression Cylinder FEM’s at Maximum Load .........................................................................................127

Figure 6.19. Mean Stress Across the Bottom Symmetry Plane of the Coarse, Medium,

and Fine Mesh Smooth Smooth Compression Cylinder FEM’s at Maximum Load ...........................................................................................................128

Figure 6.20. Radial Stress Across the Bottom Symmetry Plane of the Coarse, Medium,

and Fine Mesh Smooth Compression Cylinder FEM’s at Maximum Load...................................................................................................................128

Figure 6.21. Equivalent Plastic Strain Across the Bottom Symmetry Plane of the Coarse,

Medium, and Fine Mesh Smooth Compression Cylinder FEM’s at Maximum Load .........................................................................................129

Figure 6.22. Schematic of Axisymmetric Model of a Notched Round Bar Specimen

Utilizing Two Planes of Symmetry ...........................................................130 Figure 6.23. Medium Mesh FEM of NRB with ρ = 0.040 in.........................................131 Figure 6.24. Coarse Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in....132 Figure 6.25. Medium Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in. 132 Figure 6.26. Fine Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in........133 Figure 6.27. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh

NRB with ρ = 0.005 in. FEM’s at Failure Load ........................................134

xvi

Figure Page

Figure 6.28. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load..................................................134

Figure 6.29. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB

with ρ = 0.005 in. FEM’s at Failure Load..................................................135 Figure 6.30. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine

Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load...............................135 Figure 6.31. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh

NRB with ρ = 0.040 in. FEM’s at Failure Load ........................................136 Figure 6.32. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB

with ρ = 0.040 in. FEM’s at Failure Load..................................................136 Figure 6.33. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB

with ρ = 0.040 in. FEM’s at Failure Load..................................................137 Figure 6.34. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine

Mesh NRB with ρ = 0.040 in. FEM’s at Failure Load...............................137 Figure 6.35. Schematic of the Equal-Arm Bend Specimen Utilizing One Symmetry Plane

...................................................................................................................139 Figure 6.36. Medium Mesh Equal-Arm Bend FEM ......................................................140 Figure 6.37. Medium Mesh in the Fillet Region of the Equal-Arm Bend Specimen.....141 Figure 6.38. Effective Stress Across the Fillet Section of the Coarse, Medium, and Fine

Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load .................141 Figure 6.39. Mean Stress Across the Fillet Section of the Coarse, Medium, and Fine

Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load .................142 Figure 6.40. Stress in the x-Direction Across the Fillet Section of the Coarse, Medium,

and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load ..142 Figure 6.41. Equivalent Plastic Strain Across the Fillet Section of the Coarse, Medium,

and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load ..143

xvii

Figure Page

Figure 7.1. Comparison of the Built-In ABAQUS Models with Multilinear Isotropic Hardening and the Combined Multilinear Hardening UMAT for the Monotonic Loading of a NRB with ρ = 0.040 in. .....................................146

Figure 7.2. Comparison of the Built-In ABAQUS von Mises Model with Bilinear

Kinematic Hardening and the Combined Multilinear Hardening UMAT for the Cyclic Loading of a NRB with ρ = 0.040 in........................................147

Figure 7.3. Comparison of Drucker-Prager Combined Multilinear Hardening UMAT

Solutions with different β Values for the Smooth Compression Specimen...................................................................................................................148

Figure 7.4. Comparison of Drucker-Prager Combined Multilinear Hardening UMAT

Solutions with different β Values for the NRB with ρ = 0.040 in.............149 Figure 7.5. Load-Gage Displacement Results for 2024-T851 Smooth Tensile Bar ......151 Figure 7.6. Load-Gage Displacement Results for 2024-T851 Smooth Compression

Cylinder .....................................................................................................153 Figure 7.7. Load-Gage Displacement Results for NRB with ρ = 0.005 in. ...................155 Figure 7.8. Load-Gage Displacement Results for NRB with ρ = 0.010 in. ...................156 Figure 7.9. Load-Gage Displacement Results for NRB with ρ = 0.020 in. ...................157 Figure 7.10. Load-Gage Displacement Results for NRB with ρ = 0.040 in. .................158 Figure 7.11. Load-Gage Displacement Results for NRB with ρ = 0.080 in. .................159 Figure 7.12. Load-Gage Displacement Results for NRB with ρ = 0.120 in. .................160 Figure 7.13. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

...................................................................................................................163 Figure 7.14. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

...................................................................................................................164 Figure 7.15. Third Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

...................................................................................................................165

xviii

Figure Page

Figure 7.16. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in....................................................................................................................166

Figure 7.17. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

...................................................................................................................167 Figure 7.18. Ninth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

...................................................................................................................168 Figure 7.19. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.



...................................................................................................................172 Figure 7.22. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.

...................................................................................................................173 Figure 7.23. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.

...................................................................................................................174 Figure 7.24. Tenth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.

...................................................................................................................175 Figure 7.25. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.



...................................................................................................................179 Figure 7.28. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.

...................................................................................................................180 Figure 7.29. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.

...................................................................................................................181

xix

Figure Page

Figure 7.30. Tenth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in....................................................................................................................182

Figure 7.31. Load-Gage Displacement Results for IN100 Smooth Tensile Bar............184 Figure 7.32. Load-Gage Displacement Results for IN100 Smooth Compression Cylinder

...................................................................................................................186 Figure 7.33. First Cycle Load-Microstrain Small Strain Analysis Results for the Equal-

Arm Bend Specimen..................................................................................188 Figure 7.34. Second Cycle Load-Microstrain Small Strain Analysis Results for the

Equal-Arm Bend Specimen .......................................................................189 Figure 7.35. Third Cycle Load-Microstrain Small Strain Analysis Results for the Equal-

Arm Bend Specimen..................................................................................190 Figure 7.36. First Cycle Load-Microstrain Large Strain Analysis Results for the Equal-

Arm Bend Specimen..................................................................................192 Figure 7.37. Second Cycle Load-Microstrain Large Strain Analysis Results for the

Equal-Arm Bend Specimen .......................................................................193 Figure 7.38. Third Cycle Load-Microstrain Large Strain Analysis Results for the Equal-

Arm Bend Specimen..................................................................................194

xx

LIST OF SYMBOLS, ACRONYMS, AND ABBREVIATIONS

Symbol Description

a Slope of Effective Stress Versus the First Stress Invariant

ao Equilibrium Atomic Spacing

b Fatigue Strength Exponent

c Fatigue Ductility Exponent

d Modified Yield Strength

f Yield Function

g Plastic Potential Function

h Nonlinear Hardening Function

k Yield Strength in Pure Shear

nv

Flow Direction

p Hydrostatic Pressure

ppr Elastic Prediction of Hydrostatic Pressure

q1, q2 Adjustable Parameters in Tvergaard’s Modified Yield Function

r Ratio of Yield Stress in Triaxial Tension to Yield Stress in Triaxial

Compression

t (a) Pseudo-Effective Stress

(b) Time

x Separation Distance Between Atoms

A (a) Material Constant in Hu’s Yield Function

(b) Material Constant in Mowbray’s Equation

xxi

Symbol Description

B Material Constant in Hu’s Yield Function

C (a) Material Constant in Hu’s Yield Function

(b) Material Parameter in Brown and Miller’s Equation

D (a) Constant in Brown and Miller’s Equation

(b) Fourth Order Tensor of Elastic Coefficients

E Young’s Modulus

Et Tangent Modulus

F Void Volume Fraction

H Hardening Modulus

I1, 2, 3 Stress Invariants

J (a) Nonlinear Energy Release Rate

(b) Jacobian

J1, 2, 3 Deviatoric Stress Invariants

K Bulk Modulus

MF Multiaxiality Factor

Nf Number of Cycles to Failure

P Load

Q (a) Second Fracture Parameter in J-Q Theory

(b) Normal to the Yield Surface

R Magnitude of the Stress Difference

R Average of the Radial Displacements of an Ellipsoidal Void

Ro Initial Radius of Spherical Void

xxii

Symbol Description

S1, 2, 3 Deviatoric Principal Stresses

TF Triaxiality Factor

U1, 2, 3 Displacements in the x –Direction

V1, 2, 3 Displacements in the y –Direction

W1, 2, 3 Displacements in the z –Direction

Wf Fracture Work

Wpl Plastic Work

α Backstress Tensor

β Combined Hardening Parameter

∆ε Total Strain Range

ε True Strain

ε el Elastic Strain

εf’ True Strain Corresponding to Fracture in One Reversal

ε1, 2, 3 Principal Strains

ε pl Plastic Strain

pleqε Equivalent Plastic Strain

εp Volumetric Portion of Plastic Strain

εq Distortional Portion of Plastic Strain

φ Positive Constant in General Flow Rule

γ (a) Surface Energy

(b) Scalar Multiplier of Plastic Strain

xxiii

Symbol Description

λ Lattice Wavelength

λσ Hydrostatic Stress Ratio

µε Microstrain

µ Shear Modulus

ν Poisson’s Ratio

ρ Notch Root Radius

σ (a) True Stress

(b) Cohesive Stress

σ1, 2, 3 Principal Stresses

σc Theoretical Cohesive Strength

σ pr Elastic Stress Prediction

σeff Effective Stress

preffσ Elastic Prediction of Effective Stress

σf Failure Stress

σf’ True Stress Corresponding to Fracture in One Reversal

σm Mean Stress

σm,f Fatigue Mean Stress

σmax Maximum Applied Stress

σxx, yy, zz Normal Stresses

σult Ultimate Tensile Strength

xxiv

Symbol Description

σys Yield Strength

oysσ Initial Yield Strength

σysc Compressive Yield Strength

τxy, xz, yz Shear Stresses

θ Angle of the Slope of the Yield Surface in the p-t Stress Plane

ξ Stress Difference

ψ Dilation Angle

ζ Kinematic Hardening Material Parameter

iΓ State Variable

ijΙ Second-Order Identity Tensor

sed Unix Stream Editor

DENT Double-Edge Notch Tension

FEA Finite Element Analysis

FEM Finite Element Model

HCF High Cycle Fatigue

LCF Low Cycle Fatigue

MCPT Multiple Cycle Proof Testing

NASA National Aeronautics and Space Administration

NRB Notched Round Bar

S-D Strength-Differential

VLCF Very Low Cycle Fatigue

xxv

Symbol Description

2-D Two Dimensional

3-D Three Dimensional

1

CHAPTER 1

INTRODUCTION

Since the 1940’s, many have considered Bridgman’s [1] experiments on the

effects of hydrostatic pressure on metals the definitive study. Bridgman’s two

observations about metal behavior were that hydrostatic stress has a negligible effect on

yielding of metals and that metal is incompressible for plastic strain changes. These two

observations have become the standard tenets for studies in metal plasticity. Because of

the influence of Bridgman’s work, plasticity textbooks from the earliest (e.g. Hill [2]) to

the most modern (e.g. Lubliner [3]) infer that there is negligible hydrostatic stress effect

on the yielding of metals. Even modern finite element programs such as ANSYS [4] and

ABAQUS [5] direct the user to assume the same. Calculations are often made based on

the assumption that the effect of hydrostatic stress is negligible. In certain circumstances

though, the effects of hydrostatic stress can have a significant influence on material yield

behavior.

It is well documented that large tensile hydrostatic stresses develop in sharply

notched or cracked geometries [6-9]. Wilson [10,11] and the author [12] have

demonstrated that for these cases, a yield criterion that is dependent on hydrostatic stress,

such as the Drucker-Prager yield criterion, produces results that better match monotonic

test data. Therefore, it was postulated that a pressure-dependent yield function would

2

also lead to more accurate strain prediction in low cycle fatigue (LCF) loadings of

notched components.

The strain-life approach is the most commonly used method to estimate the low

cycle fatigue life of a component. The strain-life method is a suitable approach if the

inelastic strains can be accurately evaluated, but the prediction or measurement of

inelastic strains at geometric discontinuities is not a trivial matter. Several different

methods have been used in an attempt to evaluate the inelastic notch strains including

experimental methods, robust methods, and elastic-plastic finite element analysis. The

problem of accurate inelastic strain prediction is further complicated by the fact that

many engineering components are subjected to multiaxial fatigue processes.

Many researchers have attempted to modify the basic strain-life equations by the

addition of additional hydrostatic dependent functions and empirical constants. Most of

these formulations, though, have limited application and are highly dependent on the

empirical constants. To the author’s knowledge, no current research has proposed using

a modified yield function that considers the effect of hydrostatic stress on the yield and

postyield behavior of the material. A yield function of this type should lead to more

accurate inelastic strain prediction and hence more accurate modeling of the transitional

LCF response from the first hysteresis loop to a stable hysteresis response. The author’s

[12] previous research demonstrated the hydrostatic dependent Drucker-Prager yield

function predicted notch strains much more accurately than the von Mises yield function

on the initial loading cycle. It is postulated that when the load is reversed the Drucker-

Prager solutions will more accurately match the hysteresis loops of subsequent cycles.

3

In the chapters that lie ahead, a classical view of metal plasticity is discussed.

Results of previous research that deviate from classical metal plasticity are presented.

Then, a discussion of low cycle fatigue including the strain-life method and multiaxial

fatigue research is given. Additionally, the experimental and analytical plans for this

research project are presented including, mechanical testing, constitutive model

development, and finite element modeling. The mechanical testing and finite element

results are presented and compared. Finally, conclusions and recommendations are

given.

It should be noted that this dissertation is written in a nontraditional style. All of

the results are not all presented at the end of the document as is traditionally done.

Instead certain results are presented as they become relevant to the reader. To avoid

confusion, all results not produced by the author have been explicitly documented.

4

CHAPTER 2

TECHNICAL BACKGROUND

In this chapter, a classical view of metal plasticity is presented including a

discussion of yield functions, hardening rules, and flow rules. Next, the results of several

researchers that studied the effects of hydrostatic stress on plastic material behavior are

presented. These results deviate from classical plasticity theory and illustrate the effects

of hydrostatic stress on yielding and ductile fracture. Next, a general introduction to low

cycle fatigue failure is presented. This discussion is followed with a section on current

research in the area of hydrostatic stress effects in LCF.

A Classical View of Metal Plasticity

Plastic material behavior is a more complex phenomenon than elastic material

behavior. In the elastic range, the strains are linearly related to the stresses by Hooke’s

law, and the strains are uniquely determined by the stresses. In general, plastic strains are

not uniquely determined by the stresses. Plastic strains depend on the whole loading

history or how the stress state was reached [13]. Therefore, to completely describe

material behavior in the plastic range, one must determine the appropriate yield function,

hardening rule, and flow rule.

5

Yield Functions

A yield function is a mathematical relationship that predicts the onset of yielding

in a material. Some background information defining various stress quantities must be

presented before a mathematical description of yield functions is given. The principal

stresses are given by σ1, σ2, and σ3. The cubic equation

0322

13 =−−− III σσσ (2.1)

is solved to give the principal stresses, where the three roots of σ are principal stresses

and I1, I2, and I3 are the stress invariants. The stress invariants are expressed in the

cartesian coordinate system as

zzyyxxI σσσ ++=1 , (2.2)

( )xxzzzzyyyyxxzxyzxyI σσσσσστττ ++−++= 2222 , (2.3)

and

( )2223 2 xyzzzxyyyzxxzxyzxyzzyyxxI τστστστττσσσ ++−+= , (2.4)

where σxx, σyy, and σzz are the normal stresses and τxy, τxz, and τ yz are the shear stresses in

the cartesian coordinate system [14]. The hydrostatic or mean stress is defined as

13

1Im =σ , (2.5)

and the hydrostatic pressure is

13

1Ip m −=−= σ . (2.6)

6

Bridgman [1] conducted experiments studying the effect of an externally applied

pressure on the yield and postyield behavior of metals. He found that there was no

significant effect on the yield point until high external pressures (450 ksi) were reached.

(Bridgman’s work is discussed in more detail in the “Hydrostatic Stress – Deviations

From Classical Theory” section.) Early developers of plasticity theory interpreted

Bridgman’s results to mean that hydrostatic stress, whether externally applied or

internally generated from constraint, has a negligible effect on the yield behavior of

metals. These early plasticity researchers, therefore, developed the first tenet of classical

metal plasticity—hydrostatic stress has no effect on yielding. This rationale led to the

development of a plasticity theory that subtracts the hydrostatic stress from the principal

stresses resulting in the deviatoric stresses S1, S2, and S3. The deviatoric stresses are

111 3

1IS −= σ , (2.7)

122 3

1IS −= σ , (2.8)

and

133 3

1IS −= σ . (2.9)

The deviatoric stresses are the roots of the cubic equation

0322

13 =−−− JSJSJS , (2.10)

where J1, J2, and J3 are the deviatoric stress invariants given by

01 =J , (2.11)

7

( ) ( ) ( )[ ]213

232

2212 6

1 σσσσσσ −+−+−=J , (2.12)

and

( )( )( )mmmJ σσσσσσ −−−= 3213 . (2.13)

In terms of the deviatoric stresses, the deviatoric stress invariants are

( )23

22

212 2

1SSSJ ++= (2.14)

and

3213 SSSJ = . (2.15)

In classical metal plasticity theory, a yield function, f , is a function of the

principal stresses written in the form

( )321 ,, σσσff = . (2.16)

By definition, when 0<f the material behaves elastically, and when 0=f yielding

occurs and the material behavior is plastic. Assuming that yield is independent of

hydrostatic stress leads to a yield function

( )32 , JJff = . (2.17)

The von Mises yield function is often used for classical metal plasticity

calculations. This function states that yield is independent of hydrostatic stress and is

only dependent on J2 in the form of

( ) 222 kJJf −= , (2.18)

where k is the yield strength in pure shear and is a function of plastic strain for hardening

materials. The von Mises or effective stress is defined as

8

( ) ( ) ( )[ ]213

232

2212 2

13 σσσσσσσ −+−+−== Jeff . (2.19)

Setting ( )2Jf equal to zero in Equation (2.18) leads to

22 kJ = , (2.20)

which can be interpreted as the von Mises yield surface in principal stress (Haigh-

Westergaard) space. The yield surface for the von Mises yield function is a circular

cylinder of radius, k, whose axis is defined in the direction of the hydrostatic pressure

(Figure 2.1). A yield locus can be made by intersecting the yield surface with a plane

perpendicular to the cylinder axis. For the von Mises yield function, a yield locus taken

anywhere along the hydrostatic pressure axis is a circle of radius k, thus demonstrating

the function’s hydrostatic independence. The hydrostatic stress is zero on the plane

passing through principal stress space origin. This plane is defined as the π plane and is

given by the equation σ1 + σ2 + σ3 = 0.

σ1

p

k

σ2

σ3

Figure 2.1. von Mises Yield Surface in Principal Stress Space [15]

9

Hardening Rules

If a material exhibits strain hardening, the yield surface may change shape or

location or both as the material deforms plastically. The shape-change effect can be

approximated for many materials by assuming isotropic hardening wherein the yield

surface expands equally in all directions. Considering the von Mises yield function, the

radius of the yield surface increases from k1 to k2 as the material hardens. A graphical

representation of isotropic hardening for the von Mises yield function is given in Figure

2.2. Points “a” and “b” represent arbitrary points on the yield surfaces, and the line from

“a” to “b” is an arbitrary path through principal stress space connecting the two points.

As implied by the name, a material that obeys isotropic hardening has the same

yield behavior in both tension and compression. This is approximately true for some

materials, but it is not an accurate description of material behavior in general. Many

materials exhibit a behavior referred to as the Bauschinger effect, which is illustrated

graphically in Figure 2.3. Upon initial loading of a specimen, stress and strain are

σ1 σ2

σ3

k1

k2 a

b

Figure 2.2. Isotropic Hardening for the von Mises Yield Function

10

ε

σmax

σ

σys

−σys

1

2

3

Figure 2.3. Illustration of the Bauschinger Effect

linearly related until the tensile yield strength, σys, is reached at point 1. The load is then

increased on the specimen causing plastic deformation and bringing the stress in the

specimen up to a maximum stress, σmax, at point 2. When the load is reversed, plastic

strains develop at point 3 before –σys is reached. This effect is very important when a

reversal of loading (locally or globally) is to be considered.

A kinematic hardening model attempts to describe the behavior of materials with

a significant Bauschinger effect. This is accomplished by shifting the axis of the yield

surface in principal stress space while maintaining the same radius as the initial yield

surface. Kinematic hardening for the von Mises yield function is graphically represented

11

in Figure 2.4. Because almost no material hardens in a pure isotropic or kinematic

fashion, a linear combination of both models is sometimes used to describe real materials.

A material behavior related to the Bauschinger effect is the strength-differential

(S-D) phenomenon. The term strength-differential refers to the difference between

tensile and compressive yield strengths. Many plasticity researchers including Drucker

[16] and Spitzig [17] have studied the causes of this phenomenon and the resultant effect

on material behavior. Spitzig conducted compression and tension tests on 4310 and 4330

steel to investigate the strength-differential effect in high strength steels, and the results

of his tests are given in Figure 2.5 and Figure 2.6. The yield strength in compression

forthe 4310 steel increases by approximately 4.5% from 151 ksi to 158 ksi, and the yield

strength in compression for the 4330 steel increases by approximately 4.3% from 210 ksi

to 219 ksi.

σ1 σ2

σ3

a

b

Figure 2.4. Kinematic Hardening for the von Mises Yield Function

12

Figure 2.5. True Stress-True Strain Curves in Tension and Compression for 4310 Steel [17]

Figure 2.6. True Stress-True Strain Curves in Tension and Compression for 4330 Steel [17]

13

Flow Rules

Flow rules for plastic behavior are analogous to Hooke’s law for elastic behavior.

Hooke’s law defines the relationship between stress and elastic strains, while flow rules

define the relationship between stresses and plastic strain increments. A general form of

a flow rule relating stresses to plastic strain increments is given by

φσ

ε dg

dij

plij ∂

∂= , (2.21)

where plijdε are the plastic strain increments, g is the plastic potential function, and dφ is

a positive constant [13]. The pair of indices i and j range from 1 to 3 or x to z.

Associated flow occurs when g = f, where f is the yield function.

Bridgman [1] made the observation that volume change during plastic

deformation is nearly elastic, and, therefore, he assumed metals were incompressible.

The influence of Bridgman’s observation lead to the second basic tenet of classical metal

plasticity—metal incompressibility. For an incompressible material, the sum of the

plastic strain increments (or plastic dilatation rate) must be zero. This can be written in

terms of the principal strain increments dε1, dε2, and dε3 as

0321 =++= plplplplii dddd εεεε , (2.22)

where pld 1ε is the plastic portion of dε1.

Equation (2.21) is written in associated form as

φσ

ε df

dij

plij ∂

∂= . (2.23)

14

Drucker and Prager [18] showed that pliidε can be summed from Equation (2.23) to

obtain

1

3I

fdd pl

ii ∂∂= φε . (2.24)

Because the hydrostatic stress is I1/3, pliidε must equal zero if the yield function does not

depend on hydrostatic stress.

Hydrostatic Stress – Deviations From Classical Theory

Since the 1940’s, many have considered Bridgman’s experiments on the effects of

hydrostatic pressure on metals the definitive study. In his study, Bridgman tested smooth

(unnotched) tensile bars made from a variety of common metals including aluminum,

copper, brass, bronze, and various steels. He conducted tensile tests under the conditions

of hydrostatic pressures up to 450 ksi and found that there was no significant effect on the

yield point until the higher pressures were reached. His studies revealed that the primary

effect of hydrostatic pressure was increased ductility. Bridgman also measured the

volume of the material in the gage section and found that this volume did not change,

even for very large changes in plastic strain. Because the volume in the gage section did

not change, he concluded that metals have incompressible plastic strains. His two

observations about metal behavior—no significant effect of hydrostatic pressure on

yielding and incompressibility for plastic strain changes—have become the standard

tenets for studies of metal plasticity [1].

15

Bridgman continued to study the effects of external hydrostatic pressure for many

years, and, in 1952, he wrote a comprehensive summary of his work in his book Studies

in Large Plastic Flow and Fracture with Special Emphasis on the Effects of Hydrostatic

Pressure [19]. In this book, he reexamined his earlier results and made observations that

many plasticity books failed to notice. On p. 64 of his book, Bridgman writes:

“The fact that a curve is obtained with haphazard pressures indicates that the effect of pressure as such on the strain hardening is unimportant, the role of pressure being merely to permit the large strains without fracture which determine the strain hardening. This is indeed the case to a first approximation. In nearly all the work tabulated above, no consistent correlation was apparent between pressure and the stress-strain points, in view of the sometimes large scatter arising from other factors. By the time the last series of measurements was being made under the arsenal contract, however, skill in making the measurements had so increased, and probably also the homogeneity of the material of the specimens had also increased because of care in preparation, that it was possible to establish a definite effect of pressure on the strain hardening curve [19].”

Representative results of Bridgman’s later tests are given in Figures 2.7 and 2.8.

These data clearly demonstrate a strong dependence of flow stress on hydrostatic pressure

for both tempered pearlite and tempered martensite. For example, the flow stress

(effective stress) for tempered pearlite at a strain of 2.75 increased from 255 ksi at

atmospheric pressure to 315 ksi when pressurized to approximately 360 ksi (Figure 2.7).

Therefore, Bridgman clearly demonstrated in his later work a definite external hydrostatic

pressure effect on yielding. These externally applied hydrostatic pressures were so large

that Bridgman concluded that they would not be seen in practical application.

Unfortunately, he failed to consider the effect of internally generated hydrostatic stresses.

16

Figure 2.7. Flow Stress (Effective Stress) as a Function of Strain for Tempered Pearlite Tested at Various Pressures [19]

Figure 2.8. Flow Stress (Effective Stress) as a Function of Strain for Tempered Martensite Tested at Various Pressures [19]

17

In the 1950’s and 60’s, Hu conducted several experiments to test the validity of a

hydrostatic independent yield condition. He postulated that the biaxial tension-tension

and biaxial tension-torsion experiments used by early plasticity researchers to check the

validity of Bridgman’s work were not sensitive enough to detect the effect of hydrostatic

stresses on the yielding of metals. Hu stated that the results of these experiments “have

led to the false conclusion that the effect of hydrostatic stresses on plastic behavior of

metals is insignificant as assumed, even though the influence of hydrostatic pressure on

simple tension and compression has long been known [20].”

In the 1950’s Hu conducted biaxial-stress tests on aluminum alloys to check the

validity of assumptions made in theories of plastic flow for metals. Tubular specimens

were stressed by applying an internal pressure and axial tension. His test results did not

agree with the stress-strain relations formulated using the von Mises yield criterion and

they did not support the theory of metal incompressibility by the classical flow theories

[21,22]. He later performed pressurized tension tests on Nittany No. 2 brass and found

the effect of hydrostatic pressure on plastic stress-strain relations to be quite significant as

shown in Figure 2.9 [23]. For example, Hu found the yield strength of the brass to be

approximately 45 ksi with no externally applied pressure (No. 1, Figure 2.9), but the yield

strength increased to about 55 ksi when the specimen was pressurized to 53.2 ksi (No. 7,

Figure 2.9). His tests also demonstrated a threefold increase in ductility with increased

external pressure. From his findings, Hu suggested that a yield criterion for metals

should include the influence of hydrostatic stress and could be written in simple form as

211

22 CIBIAJ ++= , (2.25)

18

Figure 2.9. Plastic Stress-Strain Relations in Tension for Nittany No.2 Brass Under Hydrostatic Pressure [23]

where A, B, and C are experimentally determined material constants. Hu [20] also

suggested that for many metals, C is negligible and therefore the yield function could be

written as

12

2 BIAJ += . (2.26)

In the 1970’s and early 1980’s Richmond, Spitzig, and Sober [17,24,25,26] also

conducted experiments that challenged the basic tenets of classic metal plasticity. They

studied the effects of hydrostatic pressure on yield strength for four steels (4310, 4330,

maraging steel, and HY80) and grade 1100 aluminum. They conducted compression and

tension tests in a Harwood hydrostatic-pressure unit at pressures up to 160 ksi (1100

MPa).

19

Richmond, et al. found that hydrostatic pressure had a significant effect on the

stress-strain response of the steels as shown in Figure 2.10 and Figure 2.11. For 4330

steel, the compressive yield strength increased from 1520 MPa to 1610 MPa as pressure

was increased to 1100 MPa, and for the aged maraging steel, the compressive yield

strength increased from 1810 MPa to 1890 MPa as pressure was increased to 1100 MPa.

Richmond also found that the yield strength was a linear function of hydrostatic

pressure. This is shown graphically in Figure 2.12 and Figure 2.13. Richmond proposed

that for high-strength steels the yielding process is described by the yield function

daIJJIf −+= 1221 3),( , (2.27)

where a and d are material constants proportional to those suggested by Drucker and

Figure 2.10. Effect of Hydrostatic Pressure on the Stress-Strain Curves in Compression for 4330 Steel [25]

20

Figure 2.11. Effect of Hydrostatic Pressure on the Stress-Strain Curves in Compression for Aged Maraging Steel [25]

Figure 2.12. Dependence of Yielding on I1 in 4330 Steel [25]

21

Figure 2.13. Dependence of Yielding on I1 for Aged Maraging Steel [25]

Prager [18] for soils. The experimental values of a and d obtained by Spitzig and

Richmond are listed in Table 2.1 along with the values for clay as reported by Chen [27]

for comparison.

22

Table 2.1. Summary of Experimental Results for Constants in Equation (2.27) [26,27]

Material Name a d, MPa a/d, MPa-1 HY80 Steel 0.008 606 13×10-6 Unaged Maraging Steel 0.017 1005 17×10-6 4310 Steel 0.025 1066 23×10-6 4330 Steel 0.025 1480 20×10-6 Aged Maraging Steel 0.037 1833 20×10-6 Polyethylene 0.022 13 17×10-4 Polycarbonate 0.011 36 31×10-5 Clay 0.118 6.7×10-2 1.76

Another interesting result of Richmond’s tests was a strong correlation between

the coefficients a and d. He found that the ratio of a/d was nearly constant for most of

the steels as listed in Table 2.1. Therefore, Richmond suggested that the ratio a/d is a

property of the bulk iron lattice similar to the elastic constants E and ν.

It was previously shown that the plastic dilatation rate, pliidε , (Equation (2.24))

will equal zero if the yield function is independent of hydrostatic stress. In contrast, if

Equation (2.27) is true, then pliidε will not be equal to zero. Instead, taking the partial

derivative of f with respect to I1 from Equation (2.27) results in

aI

f =∂∂

1

. (2.28)

Combining Equations (2.24) and (2.28) produces the expression

03 >= φε add plii . (2.29)

Therefore, the plastic volume must increase for increasing plastic strain. Richmond

measured the plastic volume increase for varying levels of plastic strain and compared

23

this data to the plastic volume increase predicted by Equation (2.29). His measured

values of plastic volume increase were only about one-fifteenth of that predicted by the

normality flow rule as illustrated in Figure 2.14. The true or equivalent plastic strain, pleqε ,

plotted in Figure 2.14 is defined as the sum of the plastic strain increments and can be

written as

( ) ( ) ( )[ ]∫ −+−+−=5.02

13

2

32

2

213

2 plplplplplplpleq dddddd εεεεεεε . (2.30)

Richmond also conducted pressurized compression and tension tests on two

polymers—crystalline polyethylene and amorphous polycarbonate. These tests were

performed to see if the plasticity theories developed for metals were compatible with

other materials. He found that hydrostatic pressure had a significant effect on the stress-

strain response of the polymers and that the effective stress was a linear function of

hydrostatic pressure. In other words, Richmond established that the plastic response of

Figure 2.14. Plastic Volume Increase as a Function of True Plastic Strain for 4310 and 4330 Steels [25]

24

the polymers could be described by the same plasticity theories that he developed for

metals.

The yield function developed by Richmond for the steels and polymers was

identical to a yield function formulated in the 1950’s by Drucker and Prager [18]. The

Drucker-Prager yield criterion is a modification of the von Mises criterion that includes a

linear dependence on hydrostatic pressure and was originally formulated to solve soil

mechanics problems. Therefore, the fact that soils, metals, and polymers are all affected

in a similar manner by hydrostatic pressure is a unifying concept. The Drucker-Prager

yield function is defined as

( ) ,3, 2121 dJaIJIf −+= (2.31)

where d is the modified yield strength in absence of mean stress and a is a material

constant related to the theoretical cohesive strength of the material, σc. The material

constant a is determined graphically as the slope of the graph of σeff versus I1, as

illustrated in Figure 2.15. The value of I1 for σeff = 0 is equal to the theoretical cohesive

strength of the material, and the value of I1 = 0 leads to σeff = d.

The theoretical cohesive strength can be defined as the stress required to

overcome the cohesive force between the neighboring atoms. The cohesive stress, σ,

between two atoms as a function of atomic separation is illustrated in Figure 2.16, where

25

0

σeff

σc I1

a

1d

Figure 2.15. Schematic of σeff versus I1 [15]

ao

σc

σ

x

λ/2

Figure 2.16. Cohesive Force as a Function of the Separation Between Atoms (Adapted from [28])

26

ao is the equilibrium spacing, λ is the lattice wavelength, and x is the separation between

atoms [28]. A good approximation of the cohesive stress is given by

λπσσ x

c

2sin= . (2.32)

For small displacements, sin(x) ≈ x, and therefore Equation (2.32) can be written as

λπσσ x

c

2= . (2.33)

Assuming elastic behavior leads to

oa

Ex=σ , (2.34)

where E is Young’s modulus. Combining Equations (2.33) and (2.34) and solving for σc

leads to

o

c a

E

πλσ

2= . (2.35)

Assuming ao ≈ λ/2 allows Equation (2.35) to be simplified to

π

σ Ec = . (2.36)

The value for cohesive strength given by Equation (2.36) generally overpredicts the

actual value of σc because it does not account for the dislocations and lattice

imperfections inherent in most engineering materials or that slip occurs plane by plane.

Dieter [28] lists a range of σc from E/15 to E/4 with a typical value for σc of E/5.5.

An expression for cohesive strength can also be derived by considering the

energetics of the fracture process. The fracture work done per unit area is given by

27

πλσ

λπσ

λ

ccf dxx

W == ∫ 2

0 2sin . (2.37)

If the expression for fracture work is equated to the energy required to form two new

fracture surfaces 2γ, one can solve for λ to give

cσ

πγλ 2= . (2.38)

Substituting Equation (2.38) into Equation (2.37) and solving for σc gives

o

c a

Eγσ = . (2.39)

The Drucker-Prager yield surface is a right-circular cone in principal stress space

as shown in Figure 2.17. The axis of the cone is the hydrostatic pressure axis, and the

apex of the cone is located at a hydrostatic stress equal to the cohesive strength. The

yield surface for an actual material probably does not come to a sharp apex as the linear

Drucker-Prager model predicts. The sharp point of the cone could cause numerical

difficulty for flow calculations, and, therefore, ABAQUS provides hyperbolic and

exponential Drucker-Prager constitutive models that blunt the end of the cone [5]. For

small amounts of hydrostatic stress, the cylinder of the von Mises yield criterion can

approximate the cone. As the hydrostatic stress increases, the deviation from the cylinder

can be considerable, and the Drucker-Prager yield surface is preferable (Figure 2.17).

Therefore, because of its hydrostatic dependency, the Drucker-Prager yield criterion

should result in more accurate modeling of geometries that have a large hydrostatic stress

influence, such as cracked or notched geometries.

28

σ1

p

k

σ2

σ3

von Mises

Drucker-Prager

Figure 2.17. Comparison of Drucker-Prager and von Mises Yield Surfaces in Principal Stress Space [15]

Several approaches for dealing with hydrostatic pressure effects have been used in

the study of fracture mechanics. Void growth analysis during ductile failure is one area

of fracture mechanics research in which hydrostatic pressure has long been recognized as

a significant factor. In the 1960’s Rice and Tracey [8] developed a semi-empirical

relationship to approximate the growth of a single void that may form during ductile

failure. They assumed that the initial void was spherical but became ellipsoidal as it

deformed. This equation is dependent on mean stress (σm = I1/3) and can be written as

pleq

ys

dI

R

R pleq ε

σε

∫

=

0

1

0 2exp283.0ln , (2.40)

where R0 is the radius of the initial spherical void and R is the average of the radial

displacements of the ellipsoidal void. Expanding on this work, Gurson [9] developed a

29

failure criterion that assumes that the material behaves as a continuum and, therefore, the

effects of each void have an averaged effect on the global behavior. The Gurson yield

function is also a function of mean stress and is written as

( )2122 1

2cosh2

3F

IF

Jf

ysys

+−

+=

σσ, (2.41)

where F is the void volume fraction. When F = 0, Equation (2.41) reduces to the von

Mises yield failure theory. Tvergaard [29] modified the Gurson model by adding two

adjustable parameters q1 and q2 that are used to calibrate the equation with experimental

data. Tvergaard’s modified equation is

( )[ ]21

1212

2 12

cosh23

FqIq

FqJ

fysys

+−

+=

σσ. (2.42)

After calibrating Equation (2.42) with experimental data, Tvergaard found that failure

could be reasonably predicted when q1 = 2 and q2 = 1. These void growth models,

however, are physically reasonable for materials with little or no initial porosity.

Another method of dealing with hydrostatic stress effects in fracture mechanics is

through the use of two-parameter elastic-plastic solutions. Two-parameter solutions are

necessary when the plastic zone surrounding the crack tip is so large that the fracture

toughness is no longer independent of the size and geometry of the test specimen. One of

the most widely used two-parameter solution methods is J-Q theory, where J is the

nonlinear energy release rate and Q is the amplitude of the stress field shift in front of the

crack tip. In J-Q theory, fracture toughness is not viewed as a single value, rather, it is a

curve that defines a critical locus of J and Q values [30]. Henry and Luxmoore [31]

30

reported that Q is a linear function of the triaxiality factor (TF). The triaxiality factor is

the ratio of the hydrostatic stress to the von Mises stress and can be written as

( )

( ) ( ) ( )213

232

221

321

3

2

σσσσσσσσσ

σσ

−+−+−

++==eff

mTF , (2.43)

where σ1, σ2, and σ3 are the principal stresses. For example, for the case of uniaxial

tension TF = 1/3, and for the case of a thin-wall pressure vessel where σ2 = σ1/2 and σ3 =

0, 33=TF . Although useful in characterizing crack tip stress fields, J-Q theory is not

directly applicable to yield function calculations.

Hydrostatic Stress - Recent Developments

In recent years, the use of the Drucker-Prager yield function for non-geological

applications has become more prevalent. This is especially true in the analysis of

materials that are highly pressure sensitive. Several researchers have successfully used

the Drucker-Prager yield theory to account for material pressure sensitivity, and

summaries of their work are presented below.

Giannakopoulos and Larsson [32,33] studied the pyramid indentation of a variety

of hard metals and ceramics using the finite element method. In addition, they

investigated the influence of pressure sensitivity on the crack nucleation and growth

during pyramid indentation tests. For all cases, they reported that the pressure sensitivity

of the tests material was critical in the deduction of material properties from indentation

tests. Therefore, Giannakopoulos and Larsson successfully used the Drucker-Prager

31

constitutive model to account for the strength-differential (S-D) effects in their test

materials.

Lu [34] developed finite element models to study the stress and strain evolution in

cast refractory blocks during cooling. The blocks were made of aluminosilicate with

added zirconia for corrosion resistance. The compressive and tensile stress-strain

behaviors for this material are substantially different. Lu successfully used the Drucker-

Prager constitutive model to account for the significant S-D effect in his test material.

Pan, et al. [35-39] have also used the Drucker-Prager yield criterion to model

material pressure sensitivity. Some of the topics they investigated include:

1. plane strain stress and slip line fields near the tips of wedge-shaped notches in

perfectly plastic pressure-sensitive materials [35,36],

2. plane strain asymptotic crack-tip fields for perfectly plastic and power law

hardening pressure sensitive materials [37],

3. the effects of pressure sensitive yielding on J integral estimation for compact

tension specimens [38], and

4. finite element analysis of near-tip fields of notched specimens in non-porous

polymers [39].

In all of the analyses of Pan, et al., the Drucker-Prager yield theory proved useful in

modeling the pressure sensitivity of the materials.

Gill, Lissenden, and Lerch [40] performed combined tension-torsion tests on

Inconel 718 (IN718) tubes at elevated temperatures. These tests determined the initial

and subsequent yield loci of IN718. They found that the center of the initial yield loci

32

was eccentric due to a S-D effect, and that the eccentricity increased with temperature.

They proposed that the S-D effect could possibly be explained using a pressure

dependent yield function such as the Drucker-Prager yield criterion.

Wilson [10,11] conducted experiments and nonlinear finite element analyses

(FEA) of 2024-T351 notched round bars (NRB) to investigate the effect of hydrostatic

tensile stresses on yielding. He modeled the loading of a NRB to failure using the finite

element method and the von Mises yield function. The failure loads predicted by using

the von Mises yield function overestimated the experimental failure loads by

approximately 10%. The failure displacements predicted using the von Mises yield

function overestimated the experimental failure displacements by 20% to 65%,

depending on the notch root acuity. Wilson then conducted finite element analyses using

the Drucker-Prager yield function, and these FEA results essentially matched the

experimental test data.

The author [12] repeated Wilson’s [10,11] experiments and performed new

nonlinear finite element analyses of Inconel 100 (IN100) equal-arm bend and double-

edge notch tension (DENT) test specimens to study the effect of internal hydrostatic

tensile stresses on yielding. Nonlinear FEA using the von Mises and the Drucker-Prager

yield functions were performed. In all test cases, the von Mises constitutive model

overestimated the load for a given displacement or strain. Considering the failure

displacements or strains, the Drucker-Prager FEM’s predicted loads that were about 3%

lower than the von Mises values. For the failure loads, the Drucker Prager FEM’s

predicted strains that were 20% to 35% greater than the von Mises values. Therefore,

33

the Drucker-Prager yield function more accurately predicted the overall specimen

response of the IN100 geometries with significant internal hydrostatic stress influence.

The preceding sections discussed the work of various researchers that have

investigated various effects of hydrostatic stress on the monotonic loading process. Most

of the work focused on modifications to classical metal plasticity theory and elastic-

plastic fracture theory to include a hydrostatic pressure influence. Recently, some

researchers have used the Drucker-Prager yield criterion to model the pressure sensitivity

of a variety of non-geological materials. In a limited number of cases, some researchers

have also studied the effect of hydrostatic stress on the fatigue process. The following

section outlines some of the work done in the area hydrostatic stress effects in low cycle

fatigue.

Low Cycle Fatigue

The fatigue process is damage to or failure of a component from the application of

cyclic loading. The study of fatigue is usually divided into several areas depending on

the amount of plastic strain involved in the process. High cycle fatigue (HCF) processes

involve plastic strains at the microscopic level, which are small compared to the elastic

strains. In HCF the stress level normally remains below the yield stress, and the number

of cycles to failure, Nf, is usually larger than 105. Low cycle fatigue processes involve

plastic strains that are the same order of magnitude as the elastic strains. In LCF the

stress level is often greater than the yield stress, and Nf is approximately between 102 and

104. A transition from LCF to HCF typically occurs between 104 and 105 cycles for

34

metals. Very low cycle fatigue (VLCF) processes involve plastic strains that are much

larger that the elastic strains, and Nf is on the order of 10 [41].

A primary focus of this research is on finite element modeling of the transitional

low cycle fatigue response from the first hysteresis loop to a stable hysteresis response.

An understanding of LCF is needed in several different fields of industry, such as metal

forming [42], earthquake prevention in large structures [43], analysis of the residual

strength of structures after accidents [41], and multiple cycle proof testing (MCPT) of

rocket engine components [44]. In particular, Rocketdyne has been using MCPT since

1952 for the pressurized rocket engine components that they provide to NASA.

Rocketdyne recommends that MPCT be performed at a minimum proof pressure of 1.2

times the maximum operating pressure using three to five cycles [45].

Strain-Life Methodology

Service life estimates for components subjected to LCF and VLCF can be

calculated using strain-life methodology. The strain-life method assumes that the

response of the material in critical locations, such as notches, is strain dependent. Stress

concentrations often cause plastic strains to occur at the root of notches even when the

nominal stresses in a component are elastic. Deformation at the notch root plastic zone is

considered strain controlled because of the constraint of the surrounding elastically

stressed material.

A hysteresis loop represents the response of a specimen subjected to cyclic

inelastic loading (Figure 2.18). The total height of the loop is ∆σ, the total stress range,

35

σ

ε

E

∆σ

∆ε∆ε

∆εpl

el

Figure 2.18. Hysteresis Loop of a Specimen Subjected to Cyclic Loading (Adapted from [46])

and the total width of the loop is ∆ε, the total strain range. The total strain is the sum of

the elastic and plastic strain ranges,

plel εεε ∆+∆=∆ . (2.44)

Using Hooke’s law, ∆σ/E can be substituted for the elastic term in Equation (2.44) to

give

pl

Eεσε ∆+∆=∆ (2.45)

or can be written in terms of strain amplitude as

222

pl

E

εσε ∆+∆=∆. (2.46)

The area enclosed by the loop represents the energy per unit volume dissipated during a

cycle, and gives a measure of the plastic work done on the material.

36

The stress-strain behavior of many metals undergoes a transient period during

early fatigue life. During this period, the metal’s mechanical properties change causing

the material to cyclically harden or cyclically soften. As a general guideline, if σult/σys >

1.4 the metal will cyclically harden while if σult/σys < 1.2 the metal will cyclically soften,

where σult is the ultimate tensile strength of the metal. After approximately 20% to 40%

of the fatigue life, most metals achieve a cyclically stable condition characterized by a

stable material response [46].

Plastic strain-life and elastic strain-life data can be plotted as approximately

straight lines on a log-log scale (Figure 2.19). The equation of the plastic strain line in

Figure 2.19 is

( ) cff

pl

N22

εε ′=∆ , (2.47)

and the equation of the elastic strain line is

( )bf

fel

NE

22

σε ′=∆

, (2.48)

with the following terms defined.

• Fatigue ductility coefficient, εf’ is the true strain corresponding to fracture in one reversal.

• Fatigue strength coefficient, σf’ is the true stress corresponding to fracture in one reversal.

• Fatigue ductility exponent, c is the slope of the plastic strain line in Figure 2.19. • Fatigue strength exponent, b is the slope of the elastic strain line in Figure 2.19.

37

εf

2Nf

100 1072Nt

c

b

ElasticPlastic

Total = Elastic and Plastic

Log

Sca

le∆ε

/2

’

εf/E’

Figure 2.19. Log-Log Plot Showing the Relationship Between Fatigue Life and Strain Amplitude (Adapted from [46]) Equations (2.47) and (2.48) can be combined with Equation (2.46) to give

( ) ( )cff

bf

f NNE

222

εσε ′+

′=∆

, (2.49)

which is the Coffin-Manson relationship between fatigue life and total strain [47].

Care must be taken to differentiate between the hydrostatic or mean stress, σm and

the fatigue mean stress, σm,f. The fatigue mean stress is the average of the alternating

stresses and can have a significant effect on fatigue life. The effects of σm,f are seen

primarily in high cycle fatigue. In low cycle fatigue, stress relaxation due to the high

strain amplitudes occurs, and the fatigue mean stress often moves toward zero. Several

researchers including Morrow [48], Manson and Halford [49], and Smith, Watson and

Topper [50] modified Equation (2.49) in an attempt to account for σm,f effects. All of the

modifications are empirically based and are only valid for the ranges for which they were

developed [46].

38

Multiaxial Fatigue Research

The study of multiaxial fatigue failure of metals has demonstrated that the static

yield theories suggested by von Mises and Tresca are nonconservative. Bong-Ryul You

and Soon-Bok Lee suggested that Multiaxial fatigue life theories be grouped into five

categories [51]:

1. empirical formulas and modifications of the Coffin-Manson equation 2. use of stress or strain invariants 3. use of the critical plane 4. use of space averages of stress or strain 5. use of energy concepts.

In an attempt to better match test results, several researchers have developed low cycle

multiaxial fatigue failure theories that include a hydrostatic stress term. The research

from categories 1-3 is further discussed below. The reader is referred to Bong-Ryul You

and Soon-Bok Lee [51] for further discussion of categories 4 and 5.

A few researchers have used hydrostatic dependent functions and empirical

constants to modify the Coffin-Manson equation. Mowbray [52] modified Equation

(2.49) by considering the effect of hydrostatic stress under biaxial fatigue loading.

Mowbray’s modified equation is

( )( ) ( )( ) cff

bf

f NAgA

NfE

2,3

32,

2 σσ λενλσε ′

−+

′=∆

, (2.50)

where

( ) ( )( )21

1,

σσ

σσ

λλνλνλ

+−

−=f , (2.51)

39

( ) ( ) ( ) ( )( )2

2

16

1132,

σσ

σσσσσ λλ

λλλλλ

+−+−+−

−=A

Ag , (2.52)

A is an empirical constant, ν is Poisson’s ratio, and λσ is the hydrostatic stress ratio.

Considerable scatter exists in the comparison of Equation (2.50) with experimental

results.

Kalluri and Bonacuse [53] investigated the multiaxial fatigue behavior of Haynes

188 superalloy at high temperature. They formed the following equation with

multiaxiality factor (MF) terms to account for the hydrostatic stress effect:

( ) ( ) cf

fbf

cb

feq N

MFN

MF

E22

εσε

′+

′=∆ , (2.53)

where

( )

≥≤−

=1

121

TFforTF

TFforTFMF , (2.54)

TF is defined by Equation (2.43), and ∆εeq is the von Mises equivalent strain. Kalluri and

Bonacuse reported that Equation (2.53) accurately predicts the fatigue life at high

temperatures. This equation, however, is invalid for out-of-phase multiaxial loading

because TF and MF are not constant over the loading cycle.

Sines and Ohgi [54] proposed a relationship that includes the interaction between

the static stresses and the cyclic stresses in the form

11.

2 =+ staticcycl bIJa , (2.55)

where a and b are material constants. Lefebvre [55] wrote Equation (2.55) in a modified

form as

40

fstaticcycl bIJa σ=+ 1

.23 , (2.56)

where σf is the failure stress at a given fatigue life in the uniaxial case. He then solved

for a and b be in terms of TF to get

fstaticcycl I

TF

TFJ

TFσ=

++

+ 1.

2 23

2

2. (2.57)

Lefebvre conducted 44 fully-reversed, strain controlled, biaxial LCF tests on A-515

Grade 70 steel. The number of cycles to failure from his tests are compared to the fatigue

life curve predicted by Equation (2.57) in Figure 2.20. An excellent correlation was

obtained for fatigue lives between 103 and 105 cycles for all strain ratios, ρ.

The critical plane method of fatigue analysis defines a failure parameter based on

a combination of the maximum shear strain (or stress) and maximum normal strain (or

stress) acting on maximum shear strain (or stress) plane [51]. For the case of biaxial

stress, Brown and Miller [56] developed a critical plane failure parameter written as

Figure 2.20. Correlation of Lefebvre’s Test Data and Fatigue Life Relation for Various Strain Ratios [55]

41

DC n =+ εγ2max , (2.58)

where

22

21max εεγ −= , (2.59)

2

31 εεε +=n , (2.60)

ε1 and ε3 are principal strains, C is a material parameter, and D is a constant. Lohr and

Ellison [57] modified Equation (2.58) by replacing γmax with γ*, where γ* is the shear

strain that acts on a plane inclined at 45° to the surface. For the biaxial stress case, εn in

Equation (2.58) represents the mean or hydrostatic strain influence on fatigue failure.

In summary, this chapter began with a classical view of metal plasticity. Also, the

work of various researchers that challenged the tenets of classical metal plasticity was

presented. These researchers demonstrated that hydrostatic stress can have a significant

effect on the yield and postyield behavior of materials. Then, an overview of low cycle

fatigue theory was presented with an emphasis on multiaxial LCF. Finally, the work of

several researchers that include a hydrostatic stress effect in their LCF theories was

reviewed. The next chapter presents the author’s research plan to investigate the effects

of hydrostatic stress on the low cycle fatigue life of metals.

42

CHAPTER 3

RESEARCH PLAN

Previous finite element analyses (FEA) by Wilson [10,11] and the author [12]

have shown that a test specimen’s monotonic response is more closely modeled using a

hydrostatic dependent yield function as opposed to a hydrostatic independent yield

function. Therefore, it was theorized that for cyclic loading, the specimen’s hysteresis

response would be more accurately predicted using a pressure-dependent yield function.

To test this postulate, a combined experimental and analytical research program was

designed to study the effect of hydrostatic stress on low cycle fatigue. The following

sections describe this research program in detail.

Experimental Program

Two of the largest areas of uncertainty in the author’s previous research [12] were

the lack of accurate material properties to use in the finite element analyses and the

limited amount of mechanical test data for comparison with the FEA results. To rectify

this situation, the experimental program for this research consisted of two main tasks.

The first task was to completely characterize several material properties of two metals,

and the second task was to produce accurate low cycle fatigue load-displacement or load-

strain data for geometries with varying mean stress influence.

43

The two metals chosen for analysis were a high strength aluminum alloy, 2024-

T851 and a nickel-base superalloy, Inconel 100 (IN100). The 2024-T851 alloy was

chosen for its availability and for its widespread use in many engineering applications.

The IN100 was chosen because of NASA’s interest in this material and because of its

usefulness in high performance aerospace engineering applications. The nominal alloy

compositions of 2024-T851 and IN100 are given in Table 3.1 [58].

The following mechanical tests were performed in the course of this research:

1. alignment tests to qualify the test machines,

2. elastic constants tests to determine Young’s modulus and Poisson’s ratio,

3. smooth uniaxial tensile tests to determine the tensile stress-strain behavior,

4. smooth uniaxial compression tests to determine the compressive stress-strain

behavior,

5. notched round bar (NRB) tensile tests to determine monotonic notched load-

displacement records,

6. NRB low cycle fatigue tests to determine cyclic notched load-displacement

records,

7. smooth round bar low cycle fatigue tests to investigate the transitional cyclic

stress-strain behavior.

Table 3.1. Nominal Alloy Compositions of 2024-T851 and IN100 [58]

Nominal Composition % Alloy Designation Al Cr Co Cu Mg Mn Mo Ni Ti V 2024-T851 93.50 4.40 1.50 0.60 IN100 5.50 10.00 15.00 3.00 60.85 4.70 0.95

44

Only a limited amount of IN100 was available for testing, so only smooth uniaxial

tensile and smooth uniaxial compression tests were performed for this material. The low

cycle fatigue tests for IN100 were performed by Pratt and Whitney using the equal-arm

bend geometry [59]. Details of the mechanical test specimens and procedures are given

in Chapter 4.

The estimation of the Drucker-Prager material constant, a, deserves special

consideration. It was reported in Chapter 2 that the ratio of a/d in Richmond’s tests [25]

was nearly constant for the high strength steels, and, therefore, a/d is possibly a material

constant just as Young’s modulus and Poisson’s ratio are considered constants. It was

postulated that the value for a can be estimated for any material by conducting uniaxial

tension and uniaxial compression tests and plotting the yield results in σeff versus I1 space.

The slope a of the line connecting the compressive yield, σysc, and the tensile yield, σys, in

Figure 3.1 is

I10

a

1

σeff

σc

σysc

σysc σys

σys

Uniaxial Tensile Test

Uniaxial Compression Test

Figure 3.1. Schematic of Calculation of a from Uniaxial Tension and Uniaxial Compression Data

45

ysysc

ysyscaσσσσ

+−

= . (3.1)

Also, as mentioned in Chapter 2, the material cohesive strength, σc, can be estimated as

E/15 to E/4 with a typical value of E/5.5 [28]. Therefore, a can be estimated as initial

yield strength divided by an approximate cohesive strength. Another possible method to

determine the Drucker-Prager material constant is to relate a to fracture mechanics J-Q

theory. It was mentioned in Chapter 2 that Henry and Luxmoore [31] found Q to be a

linear function of the triaxiality factor. Because the triaxiality factor is the ratio of σm to

σeff, it is hypothesized that TF and hence Q can be used to determine the value for a. The

results of this research, though, demonstrate that none of these estimation procedures are

ideal, and other methods must be developed to calculate a.

Analytical Program

The analytical program consisted of three main tasks. The first task was to

develop a pressure-dependent constitutive model with combined multilinear kinematic

and multilinear isotropic hardening to be used in conjunction with the commercial finite

element code ABAQUS [5]. The second task was to develop finite element models of all

the test geometries. The final task was to compare the finite element results with the LCF

data produced in the experimental program.

ABAQUS has built-in plastic flow models that allow bilinear kinematic

hardening, multilinear isotropic hardening, or a nonlinear combination of kinematic and

46

isotropic hardening when using the von Mises yield function. The Drucker-Prager yield

function in ABAQUS, though, only allows isotropic hardening. Therefore, to

realistically model the low cycle fatigue process, it was necessary to write a pressure-

dependent constitutive model that incorporates a combination of kinematic and isotropic

hardening using the Drucker-Prager yield function. The pressure-dependent constitutive

model with combined hardening was implemented using the ABAQUS user subroutine

(UMAT) function. The development of the constitutive model and its implementation

into ABAQUS are discussed in detail in Chapter 5.

After completing the pressure-dependent constitutive model development, finite

element models (FEM’s) of all of the test specimen geometries were created. The

commercial finite element code Patran [60] and a Sandia National Laboratory program

FASTQ [61] were used for preprocessing of meshes and boundary conditions. ABAQUS

was used for the finite element analyses and postprocessing the results. The development

of the finite element models is discussed in detail in Chapter 6.

The final step in the analytical program was to compare the FEA results with the

physical test data. Load-displacement or load-microstrain data was taken from the test

records and compared to the finite element simulations. Details of the finite element

analysis results along with the comparisons with test data are presented in Chapter 7.

This chapter presented the research plan for investigating the effect of hydrostatic

stress on low cycle fatigue. A two-part plan was presented. First the experimental

program was discussed, and then a plan for analytical research was presented. The next

chapter gives details on the implementation of the experimental program along with the

experimental program results.

47

CHAPTER 4

MECHANICAL TESTING

In this chapter, the procedure used to conduct the mechanical testing portion of

this research project is presented. The chapter begins with a description of the test

apparatus. A description of test machine alignment testing is then presented. This

discussion is followed by descriptions of the mechanical property and low cycle fatigue

tests conducted on the 2024-T851 specimens along with summaries of the results.

Afterward, the test procedures used for the mechanical property and low cycle fatigue

tests on IN100 are presented along with summarized results.

Test Apparatus

All of the mechanical tests except for the uniaxial compression and the IN100 low

cycle fatigue tests were performed in the mechanical testing lab at Tennessee

Technological University (TTU). All of the tests at TTU were conducted on a MTS 810

servohydraulic test machine. This test platform has a 20 kip capacity and is equipped

with hydraulic grips with interchangeable grip faces. Flat-faced grip faces were used for

the Poisson’s ratio tests, and notched grip faces for holding cylindrical specimens were

used for the remainder of the tests. A MTS 458.20 MicroConsole was used to control the

test machine. A variety of extensometers or strain gages were used to measure gage

displacement or strain. All tests were performed at room temperature in air.

48

Alignment Testing

Alignment tests were performed to qualify the test machine before each

mechanical testing phase began and each time that the grip faces were changed. All

alignment tests were performed in accordance with ASTM E1012, “Standard Practice for

Verification of Specimen Alignment Under Tensile Loading” [62]. Two types of

alignment specimens were made, rectangular and cylindrical, and both geometries were

machined from 2024-T851 1 in. thick plate.

An engineering drawing of the rectangular alignment specimen is given in Figure

4.1. Four self-temperature compensating strain gages, Measurements Group Inc. type

EA-13-120LZ-120, were bonded to the rectangular alignment specimen. Strains were

read using a Measurements Group SB-10 switch and balance unit and a P-3500 strain

indicator. Throughout the testing process, the percent bending for the rectangular

alignment specimen was less than 6%.

An engineering drawing of the cylindrical alignment specimen is given in Figure

4.2. An array of four self-temperature compensating strain gages, Measurements Group

Inc. type EA-13-060LZ-120, were bonded every 90 degrees around the circumference of

Figure 4.1. Engineering Drawing of the Rectangular Alignment Specimen (All Dimensions are in Inches)

49

Figure 4.2. Engineering Drawing of the Cylindrical Alignment Specimen (All Dimensions are in Inches)

the cylindrical alignment specimen. Once again, strains were read using a Measurements

Group SB-10 switch and balance unit and a P-3500 strain indicator. Throughout the

testing process, the percent bending for the cylindrical alignment specimen was less than

2%.

2024-T851 Testing

All of the 2024-T851 specimens were machined from a single piece of 1 in. thick

plate. Specimens were machined in both the longitudinal (L) and the long transverse (L-

T) directions, depending on test requirements. The L direction is defined as the direction

parallel to the rolling direction, and the L-T direction is the direction perpendicular to the

rolling direction. A schematic of the 2024-T851 plate along with the machining

directions is shown in Figure 4.3. All specimens were assigned a code number that

designated machining direction, specimen type, and specimen number.

50

L Dire

ction

Rolling

Directi

on

L-T DirectionThickness

Figure 4.3. Schematic of the 2024-T851 1 in. Thick Plate with Specimen Machining Directions

Elastic Constants Tests

A Poisson’s ratio specimen was machined in the L direction as shown in Figure

4.4. Two 90 degree strain gage rosettes, Measurements Group, Inc. type EA-13-060RZ-

120, were bonded to the specimen. The strains were read using a Measurements Group

SB-10 switch and balance unit and a P-3500 strain indicator. The Poisson’s ratio tests

were performed in accordance with ASTM E132, “Standard Test Method for Poisson’s

Ratio at Room Temperature” [63], and a summary of the test results is given in Table 4.1.

Because Poisson’s ratio is a function of two orthogonal directions, it was not necessary to

test a L-T direction Poisson’s ratio specimen.

Figure 4.4. Engineering Drawing of the Poisson's Ratio Specimen (All Dimensions are in Inches)

51

Table 4.1. Summary of 2024-T851 Elastic Constants Test Results

Specimen Direction Statistical Measure Poisson’s Ratio

Young’s Modulus,

103 ksi (GPa) Average 0.323 10.6 (72.8)

L Standard Deviation 0.005 0.11 (0.84)

Average NA 10.7 (74.0) L-T

Standard Deviation NA 0.06 (0.40)

The geometry of the Young’s modulus test specimen is shown in Figure 4.5. Five

specimens were machined in the L direction and three in the L-T direction. A MTS

632.25B-20 extensometer with a 2 in. gage length was used to measure the gage

displacement. The 2 in. gage length was used to capture the average elastic response

over as much of the length of the test specimen as possible. All Young’s modulus tests

were performed in accordance with ASTM E111, “Standard Test Method for Young’s

Modulus, Tangent Modulus, and Chord Modulus” [64]. A summary of the Young’s

modulus test results is given in Table 4.1. Young’s modulus is almost identical for the

two directions with the L-T direction’s being slightly higher.

Figure 4.5. Engineering Drawing of the Young’s Modulus Specimen (All Dimensions are in Inches)

52

Smooth Uniaxial Tension Tests

Ten smooth round bar tension specimens, five from the L direction and five from

the L-T direction, were used in uniaxial tension tests. An engineering drawing of the

smooth tension specimen is shown in Figure 4.6. Gage displacement was measured using

a MTS 634-31E-24 adjustable gage length extensometer, set to a 1 in. gage length. All

smooth uniaxial tension tests were performed in accordance with ASTM E8, “Standard

Test Methods for Tension Testing of Metallic Materials” [65].

Composite true stress-true strain plots for the L and L-T directions are shown in

Figures 4.7 and 4.8. All of the curves are very consistent up to the ultimate strength.

After the ultimate strength, a small amount of scatter exists due to the differences in

necking behavior and failure mode of individual specimens. The true stress-true strain

data after the ultimate strength was not corrected for the effects of triaxial stress. A

summary of the tensile data is given in Table 4.2. The tensile test Young’s modulus

values are identical to those obtained from the E111 tests. The yield strength, ultimate

strength, and true fracture strain are all slightly higher for the L direction. For both

directions, σult/σys §��LQGLFDWLQJ�WKDW�WKH�PDWHULDO�ZLOO�OLNHO\�F\FOLFDOO\�VRIWHQ��$�

summary of tensile test data for individual test specimens is given in Appendix A.

Figure 4.6. Engineering Drawing of the 2024-T851 Smooth Tension Specimen (All Dimensions are in Inches)

53

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.02 0.04 0.06 0.08 0.1

AT03

AT04

AT05

AT06

AT07

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

1 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

Figure 4.7. Composite True Stress-True Strain Plot for 2024-T851 L Direction Smooth Tensile Tests

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.02 0.04 0.06 0.08 0.1

BT01

BT02

BT03

BT04

BT05

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

1 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

Figure 4.8. Composite True Stress-True Strain Plot for 2024-T851 L-T Direction Smooth Tensile Tests

54

Table 4.2. Summary of 2024-T851 Smooth Tensile Results

Specimen Direction Statistical Measure

Young’s Modulus,

103 ksi (GPa)

0.2% Offset Yield Strength,

ksi (MPa)

Ultimate Tensile Strength, ksi (MPa)

True Fracture Strain

Average 10.6 (72.4) 67.8 (467) 76.6 (528) 0.086 L

Standard Deviation 0.05 (0.55) 0.84 (5.81) 0.55 (3.83) 0.003 Average 10.7 (73.6) 66.8 (460) 73.6 (507) 0.079

L-T Standard Deviation 0.08 (0.55) 0.45 (3.13) 2.19 (15.34) 0.002

Smooth Uniaxial Compression Tests

All of the uniaxial compression tests were performed at NASA’s Marshall Space

Flight Center (MSFC). Ten smooth compression cylinders (Figure 4.9), five from the L

direction and five from the L-T direction, were tested. The tests were performed with a

100 kip capacity servohydraulic test frame controlled by a MTS 458.20 MicroConsole.

Gage displacement was measured using a MTS 632.26E-21 extensometer with a 0.3 in.

gage length. All smooth uniaxial compression tests were performed in accordance with

ASTM E9, “Standard Test Methods of Compression Testing of Metallic Materials at

Room Temperature” [66]. The test apparatus was qualified per E9 without using

lubrication between the specimens and the compression platens, and therefore all of the

specimens were tested without lubricated ends. All of the tests were interrupted at 0.025

to 0.030 true strain to prevent damage to the extensometer.

Figure 4.9. Engineering Drawing of the Smooth Compression Specimen (All Dimensions are in Inches)

55

Composite true stress-true strain plots for the L and L-T direction compression

tests are shown in Figures 4.10 and 4.11. The compression curves for each direction are

practically coincident. A summary of the compression data is given in Table 4.3.

Young’s modulus is slightly higher for the L-T direction, but the 0.2% offset yield

strength is identical for the two directions. A summary of the test data for individual

compression specimens is given in Appendix A.

Comparisons of representative compression and tensile true stress-true strain

curves for both directions are given in Figures 4.12 and 4.13. In both cases the

compressive and tensile behaviors are almost identical. Young’s modulus is

approximately 5% higher for the compression tests, but the 0.2% offset yield strength is

essentially the same.

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

AC01

AC02

AC03

AC04

AC05

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

0.3 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

Figure 4.10. Composite True Stress-True Strain Plot for 2024-T851 L Direction Smooth Compression Tests

56

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

BC01

BC02

BC03

BC04

BC05

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

0.3 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

Figure 4.11. Composite True Stress-True Strain Plot for 2024-T851 L-T Direction Smooth Compression Tests

Table 4.3. Summary of 2024-T851 Smooth Compression Tests

Specimen Direction Statistical Measure

Young’s Modulus,

103 ksi (GPa)


ksi (MPa) Average 11.1 (76.2) 67.8 (467)

L Standard Deviation 0.12 (0.84) 0.45 (2.68)

Average 11.3 (77.6) 67.8 (467) L-T

Standard Deviation 0.20 (1.52) 0.45 (2.68)

57

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

Compression, Specimen AC03Compression, Specimen AC04Tension, Specimen AT04Tension, Specimen AT06

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

Compresssion Tests - 0.3 in. Gage LengthTensile Tests - 1 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Figure 4.12. Comparison of 2024-T851 L Direction Tensile and Compressive True Stress-True Strain Curves

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

Compression, Specimen BC01

Compression, Specimen BC02

Tension, Specimen BT02

Tension, Specimen BT03

Tru

e S

tre

ss,

σ (

ksi)

True Strain, ε

Compression Tests - 0.3 in. Gage LengthTensile Tests - 1 in. Gage Length

Tru

e S

tre

ss,

σ (

MP

a)

Figure 4.13. Comparison of 2024-T851 L-T Direction Tensile and Compressive True Stress-True Strain Curves

58

It was originally postulated that the Drucker-Prager material constant, a could be

calculated using Equation (3.1) by comparing uniaxial tension and compression results.

For the 2024-T851 tests, though, the compressive and tensile yield strengths were

practically identical. This implies a Drucker-Prager constant of approximately zero.

Several researchers [67,68,69] have demonstrated that lubricating the ends of the test

specimens can shift the load displacement record for a compression test significantly

upward. For example, Chait and Curll [67] found that the Teflon lubricated compressive

σ – ε curve was from 2% to 15% higher than the unlubricated σ – ε curve for 4340 steel.

Equation (3.1) can be solved for σysc to give

ysysc a

a σσ−+=

1

1. (4.1)

Substituting the values of a for 2024-T851 used in this research (a = 0.029 – 0.041; See

Chapter 6 for more detail on the selection of a.) into Equation (4.1) results in a σysc that is

6% to 8% higher that σys. Therefore, if the same geometry was tested with proper

lubrication on the ends, it is probable that the compressive load-displacement test record

would shift upward. This would produce a differential between the compressive and

tensile yield strengths allowing for a more accurate experimental prediction for a.

Notched Round Bar Tension Tests

Notched round bar specimens were machined in the L direction for use in the

NRB tension tests. An engineering drawing of the NRB specimen is shown in Figure

4.14. The NRB geometry was chosen because of the large internal hydrostatic stress

59

ρ

Figure 4.14. Engineering Drawing of the Notched Round Bar Specimen (All Dimensions are in Inches) produced by the constraint of the circumferential notch. Six notch radii, ρ were used:

0.005, 0.010, 0.020, 0.040, 0.080, and 0.120 inch. These notch radii were chosen to

ensure a wide range of hydrostatic stress influence. Due to a limited number of

specimens, only 3 NRB’s were tested for ρ equal to 0.005, 0.010, and 0.020 in. and only

2 for ρ equal to 0.040, 0.080, and 0.120 inch. Gage displacement was measured using a

MTS 634-31E-24 adjustable gage length extensometer set to a 0.5 in. gage length. All

notched round bar tension tests were performed in accordance with ASTM E602,

“Standard Test Method for Sharp-Notch Tension Testing with Cylindrical Specimens”

[70].

Load- gage displacement plots for all 6 notch radii are shown in Figures 4.15

through 4.20. Overall the test data is very consistent with slightly more scatter in the

curves for the 3 smaller notch radii. As shown in Figure 4.15, the NRB’s with ρ = 0.005

60

in. exhibit very little nonlinear specimen response before fracture. Conversely, the

NRB’s with ρ = 0.120 in. demonstrate a pronounced nonlinear specimen response before

fracture similar to a smooth tensile test (Figure 4.20). The difference in tensile load-gage

displacement response for all the notch radii is shown in Figure 4.21.

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

501502503

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)

Gage Displacement, v (in)

Load

, P (

N)

Gage Displacement, v (mm)

0.5 in. Gage Length

Specimen No.

Figure 4.15. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.005 in.

61

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

101102103

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Specimen No.


0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

201205206

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Specimen No.


62

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

401

406

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Specimen No.


0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

801

806

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Specimen No.


63

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

121

126

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Specimen No.


0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 501, ρ = 0.005 in.Specimen 101, ρ = 0.010 in.Specimen 201, ρ = 0.020 in.Specimen 401, ρ = 0.040 in.Specimen 801, ρ = 0.080 in.Specimen 121, ρ = 0.120 in.

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Figure 4.21. Composite Load-Gage Displacement Plot for All 2024-T851 L Direction NRB Geometries

64

The fracture surfaces of broken NRB’s specimens with ρ = 0.005, 0.040, and

0.120 in. and a smooth tensile specimen are shown in Figure 4.22. The NRB with ρ =

0.005 in. has a large amount of constraint due to the sharp notch and therefore has an

almost completely flat fracture surface. As ρ is increased, the constraint decreases, and

the fracture surfaces begin to exhibit shear lips around the edge. This trend is most

evident in the NRB with ρ = 0.120 in., which has a shear lip around the entire

circumference of the fracture plane. A smooth tensile specimen is shown for comparison

and demonstrates a specimen geometry with very little radial constraint. The tensile

specimen has a flat angled fracture indicating failure along a plane of maximum shear

stress.

Figure 4.22. Comparison of Fracture Mode for the NRB and Smooth Tensile Specimens

NRB ρ = 0.005 in.

Smooth Tensile

NRB ρ = 0.040 in.

NRB ρ = 0.120 in.

65

NRB Low Cycle Fatigue Tests

Notched round bar specimens, identical to those used in the NRB tension tests

(Figure 4.14), were machined in the L direction for use in the NRB low cycle fatigue

tests. Initially all six notch radii (0.005, 0.010, 0.020, 0.040, 0.080, and 0.120 in.) were

tried for LCF testing, but the NRB’s with ρ equal to 0.005 through 0.020 in. did not

produce hysteresis loops of useful size and were therefore abandoned. Once again, the

tests were performed on a MTS 810 servohydraulic test machine, and gage displacement

was measured using a MTS 634-31E-24 adjustable gage length extensometer set to a 0.5

in. gage length. The frequency for all the tests was 0.1 Hz using a triangular waveform.

All of the NRB low cycle fatigue tests were performed in accordance with ASTM E606,

“Standard Practice for Strain-Controlled Fatigue Testing” [71].

Some special comments on test control must be made. An MTS 458.20

MicroConsole along with MTS 759.20 TestWare® [72] was used to control the LCF

tests. This controller software allows one to choose the maximum and minimum gage

displacement desired for the LCF test. Unfortunately, the software does not ramp

completely to the maximum and minimum desired displacement levels on the first cycle.

Instead, the controller starts under the desired amount and slowly increases the level until

the desired value is reached. The ramping process usually takes between four and six

cycles to achieve the desired maximum and minimum gage displacement. For tests

where the expected number of cycles is in the hundreds or thousands, this ramping

process will have no appreciable effect on the results. However, for very low cycle

66

fatigue tests, the ramping process may consume a large percentage of the life of the

specimen.

For all of these tests the maximum and minimum gage displacement were chosen

to be 0.004 in. and –0.004 in. respectively. These limits were chosen to provide as large

of a hysteresis loop as possible while still allowing a reasonable number of cycles before

specimen failure. Due to the limitations of the controller though, the first cycle of the

tests ran to approximately 0.0035 in. and -0.0035 in. and slowly ramped to the limiting

values somewhere between cycle 4 and 6.

A summary of the low cycle fatigue tests is given in Table 4.4. The number of

cycles to failure, Nf range from a minimum of 10 cycles for the ρ = 0.040 in. NRB’s to a

maximum of 53 cycles for a ρ = 0.120 in. NRB. Therefore, all of the LCF tests

performed can easily be classified as very low cycle fatigue (VLCF) tests.

Representative load-gage displacement plots for selected cycles of the LCF tests are

Table 4.4. Summary of NRB Low Cycle Fatigue Tests

Specimen Number

Notch Root Radius, in. (mm)

Gage Length, in.

(mm)

Max. Gage Displacement,

in. (mm)

Min. Gage Displacement,

in. (mm) Frequency

(Hz)

Number of Cycles to Failure

402 0.04 (1.016) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 10 403 0.04 (1.016) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 10 404 0.04 (1.016) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 10 405 0.04 (1.016) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 12 802 0.08 (2.032) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 23 803 0.08 (2.032) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 27 804 0.08 (2.032) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 29 805 0.08 (2.032) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 29 122 0.12 (3.048) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 35 123 0.12 (3.048) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 53 124 0.12 (3.048) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 49 125 0.12 (3.048) 0.5 (12.7) 0.004 (0.1016) -0.004 (-0.1016) 0.1 40

67

plotted in Figures 4.23 through 4.25. The ramping of the gage displacement is evident in

the difference between the maximum and minimum gage displacement values of the first

through the fifth cycles. For all cases, the specimen response became stable within

approximately seven cycles, and afterward the subsequent hysteresis loops essentially

traced previous ones. Composite load-displacement plots of selected cycles for each

NRB notch root radii are included in Appendix B.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Cycle 1

Cycle 5

Cycle 9

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Load

, P (

lbs)

Gage Displacement (in.)

Load

, P (

N)

Gage Displacement (mm)

Figure 4.23. Load-Gage Displacement Plot of Selected Cycles for 2024-T851 L Direction NRB LCF Tests with ρ = 0.040 in. (Specimen 403)

68

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Cycle 1

Cycle 5

Cycle 10

Cycle 20

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



69

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Cycle 1

Cycle 5

Cycle 10

Cycle 20

Cycle 30

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



70

Smooth Round Bar Low Cycle Fatigue Tests

A limited number of smooth round bar LCF tests were performed to investigate

unnotched cyclic stress strain behavior. Three smooth round bar tension specimens

(Figure 4.6) machined from the L direction were used in the smooth round bar LCF tests.

Gage displacement was measured using a MTS 634-31E-24 adjustable gage length

extensometer, set to a 1 in. gage length. A MTS 458.20 MicroConsole along with MTS

759.20 TestWare® [72] was used to control the LCF tests. The tests were performed in

strain control to ± 0.015 gage strain. This strain range was chosen to ensure that Nf for

both the NRB’s and the smooth bars were approximately equal. Two specimens were

cycled to failure, and fracture occurred after 36 and 39 cycles, respectively.

The LCF test of the third specimen, AD01, was stopped after 10 cycles. This

specimen was then loaded in tension to failure. The resulting stress-strain curve is shown

in Figure 4.26 along with a representative monotonic stress-strain curve. The cyclic

stress-strain curve does not necessarily represent the stable fatigue stress-strain response,

but instead is representative of a transitional cyclic stress-strain curve. In this way, some

understanding of the cyclic stress-strain properties can be reached without resorting to

more complex methods for producing a true stable stress-strain curve, such as companion

sample tests or incremental step tests [46]. Young’s modulus and the ultimate tensile

strength for the monotonic and the transitional cyclic tests are essentially identical. The

cyclic response does not have a distinct yield point though, and the difference in area

under the two curves is representative of the strain energy lost due to damage

accumulation.

71

0

10

20

30

40

50

60

70

80

0

100

200

300

400

500

0 0.02 0.04 0.06 0.08 0.1

Monotonic, Specimen AT06

Transitional Cyclic, Specimen AD01

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

1 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Figure 4.26. Comparison of 2024-T851 L Direction Monotonic and Transitional Cyclic True Stress-True Strain Curves

Inconel 100 Testing

A very limited amount of Inconel 100 (IN100) was available for testing. Pratt and

Whitney generously donated two IN100 ring forgings labeled 4A and 7A. These two

rings were machined to obtain 8 smooth tensile and 24 smooth compression specimens.

The dimensions of the rings and the specimen layouts are shown in Figure 4.27. All

specimens were assigned a code number that designated specimen type and number.

Because no material was available for machining into LCF test specimens, Pratt and

Whitney test LCF test data [73] was used for the IN100 LCF analysis. Also, no specific

tests were performed to determine the elastic constants E and ν. Instead, E was estimated

72

Tensile Specimen,two through thickness

CompressionSpecimen

1.1252.450t = 1.2

Figure 4.27. Tension and Compression Specimen Layout in IN100 Disk (All Dimensions are in Inches)

from the smooth tensile test data, and a value for ν of 0.298 was obtained from the

Aerospace Structural Metals Handbook [74]. The details of the IN100 tests are presented

below.

Smooth Tensile Tests

The geometry of the IN100 smooth tensile specimen is given in Figure 4.28. Four

tensile specimens were machined from each of the two rings. Gage displacement was

measured using a MTS 634-31E-24 adjustable gage length extensometer, set to a 0.5 in.

gage length. All smooth uniaxial tension tests were performed in accordance with ASTM

E8 [65].

73

Figure 4.28. Engineering Drawing of the IN100 Smooth Tensile Specimen (All Dimensions are in Inches)

A composite true stress-true strain plot for the IN100 tensile tests is shown in

Figure 4.29. All of the curves are very consistent up to final fracture. This material

exhibits a unique stress-strain response that includes an upper and lower yield strength .

The ratio of σult/σys §��ZKLFK�LQGLFDWHV�WKDW�WKH�PDWHULDO�ZLOO�OLNHO\�F\FOLFDOO\�KDUGHQ��

A summary of the tensile data is given in Table 4.5, and tensile test data for individual

specimens is provided in Appendix A.

74

0

50

100

150

200

250

300

0 0.05 0.1 0.15 0.2 0.25

4A-TS-14A-TS-24A-TS-34A-TS-47A-TS-17A-TS-27A-TS-37A-TS-4

0

500

1000

1500

2000

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

0.5 in. Gage Length

Figure 4.29. Composite True Stress-True Strain Plot for IN100 Smooth Tensile Tests

Table 4.5. Summary of IN100 Smooth Tensile Results

Statistical Measure

Young’s Modulus,

103 ksi (GPa)

Upper Yield Strength, ksi (MPa)


ksi (MPa)



Average 31.8 (220) 172 (1184) 167 (1150) 235 (1618) 0.230 Standard Deviation 0.32 (1.99) 1.49 (10.42) 2.96 (20.26) 0.82 (5.72) 0.010

75

Smooth Compression Tests

All of the IN100 uniaxial compression tests were performed at NASA MSFC at

the same time as the 2024-T851 compression tests. Three smooth compression cylinders

(Figure 4.9) from each ring were tested. The tests were performed with a 100 kip

capacity servohydraulic test frame controlled by a MTS 458.20 MicroConsole. Gage

displacement was measured using a MTS 632.26E-21 extensometer with a 0.3 in. gage

length. All smooth uniaxial compression tests were performed in accordance with ASTM

E9 [66]. As before, all of the specimens were tested without end lubrication, and all of

the tests were interrupted at 0.025 to 0.030 true strain to prevent damage to the

extensometer.

A composite true stress-true strain plot for the IN100 compression tests is given in

Figure 4.30. The compression curves for each test are fairly consistent, but demonstrate

some scatter with increasing plastic strain. The deformed specimens showed evidence of

side-slip buckling, which usually is a result of misalignment of the loading train or loose

end tolerance on specimen dimensions [69]. Side-slip buckling tends to lower the true

stress-true strain curve and may have contributed to the test data scatter. A summary of

the compression data is given in Table 4.6, and compression test data for individual test

specimens is given in Appendix A.

A comparison of representative compression and tensile true stress-true strain

curves is given in Figure 4.31. The compressive and tensile behaviors are very similar.

Young’s modulus is approximately 2.5% higher for the compression tests, but the upper

76

0

50

100

150

200

0

200

400

600

800

1000

1200

0 0.01 0.02 0.03 0.04 0.05

4A-CP-4

4A-CP-5

4A-CP-6

7A-CP-3

7A-CP-4

7A-CP-5

Tru

e S

tres

s, σ

(ks

i)

True Strain, ε

0.3 in. Gage Length

Tru

e S

tres

s, σ

(M

Pa)

Specimen No.

Figure 4.30. Composite True Stress-True Strain Plot for IN100 Smooth Compression Tests

Table 4.6. Summary of IN100 Smooth Compression Tests

Statistical Measure

Young’s Modulus,

103 ksi (GPa)

Upper Yield Strength, ksi (MPa)


ksi (MPa) Average 32.6 (225) 172 (1185) 167 (1153)

Standard Deviation 0.29 (2.17) 0.41 (2.86) 0.52 (3.61)

77

0

50

100

150

200

0

200

400

600

800

1000

1200

0 0.004 0.008 0.012 0.016 0.02 0.024 0.028

Compression, Specimen 4A-CP-5

Compression, Specimen 7A-CP-5

Tension, Specimen 4A-TS-3

Tension, Specimen 7A-TS-2

Tru

e S

tre

ss,

σ (

ksi)

True Strain, ε

Compresssion Tests - 0.3 in. Gage LengthTensile Tests - 0.5 in. Gage Length

Tru

e S

tre

ss,

σ (

MP

a)

Figure 4.31. Comparison of IN100 Tensile and Compressive True Stress-True Strain Curves

and 0.2% offset yield strengths are approximately the same. Therefore, as for the 2024-

T851, the Drucker-Prager constant for IN100 could not be calculated using Equation

(3.1). Substituting the value of a for IN100 used in this research (a = 0.022; See Chapter

6 for more detail on the selection of a.) into Equation (4.1) results in a σysc that is 4%

higher that σys. Therefore, if the IN100 compression cylinders were tested with proper

lubrication on the ends, it is possible that the compressive load-displacement test record

would shift upward. This would produce a differential between the compressive and

tensile yield strengths and allow a to be calculated using Equation (3.1).

78

Low Cycle Fatigue Tests

Due to a limited amount of IN100 available for testing, the author was not able to

perform LCF tests on IN100 specimens. Instead, Pratt and Whitney provided IN100 LCF

data [73] for a unique test specimen, the equal-arm bend specimen [59]. An engineering

drawing of the equal-arm bend specimen is given in Figure 4.32. This specimen was

designed to simulate the geometry and loading condition of a highly stressed area in the

Space Shuttle main rocket engine fuel turbopump housing.

A three-cycle proof test was performed on the equal-arm bend specimen. The

specimen was pin loaded. A strain gage was bonded in the fillet, and the test was run in

load control to achieve approximate strain levels. On the first cycle, the specimen was

loaded to 3% strain and then unloaded

Figure 4.32. Engineering Drawing of the Equal-Arm Bend Specimen (All Dimensions are in Inches) [59]

79

to 1.3% strain. The second cycle reloaded the specimen back to 3% strain and then

unloaded to 1.3% strain. The third cycle was a repeat of the second cycle. The proof test

was performed at 80º F in air.

A plot of load-microstrain for the equal-arm bend three-cycle proof test is shown

in Figure 4.33. A large amount of plastic deformation occurs on the initial loading cycle.

Also, because the test was run in load control, there is some variance in the maximum

and minimum strain values achieved for each cycle.

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Cycle 1

Cycle 2

Cycle 3 -500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

Figure 4.33. Load-Microstrain Plot for the Equal-Arm Bend Three-Cycle Fatigue Test [73]

80

In summary, this chapter presented the mechanical testing portion of the research.

Test methods, specimens, and procedures for testing both 2024-T851 and IN100 were

discussed. The results of the monotonic mechanical property tests and the low cycle

fatigue tests were summarized and presented. The next chapter begins the analytical

portion of the research by presenting the development of a pressure-dependent

constitutive model with combined kinematic and isotropic hardening.

81

CHAPTER 5

FINITE ELEMENT CONSTITUTIVE MODEL DEVLOPMENT

This chapter presents the development of a pressure-dependent constitutive model

with combined kinematic and isotropic hardening. First the pressure-dependent plasticity

model is derived. Next the equations for isotropic and kinematic hardening are

developed. Following this, the combined bilinear and combined multilinear hardening

equations are developed for von Mises plasticity theory. The hardening rule equations

are then modified to include pressure dependency. Finally, method for implementing the

new constitutive model into ABAQUS is presented.

Pressure-Dependent Plasticity Model

This section details the development of the governing equations for a pressure-

dependent plasticity model including numerical methods for the integration of the

constitutive equations. The equations are derived using the method developed by Aravas

[75,76]. The tensor components are given with respect to a cartesian coordinate system

with indices ranging from 1 to 3.

Elasticity

To begin, the strain tensor is assumed composed of an elastic and plastic part

plij

elijij εεε += , (5.1)

82

where elijε and pl

ijε are the elastic and plastic portions of the strain tensor, respectively.

Assuming isotropic linear elasticity, one can write

ijelkk

elijij K Ι+= εµεσ 2 , (5.2)

where σij is the stress tensor, µ is the shear modulus, K is the bulk modulus, and ijΙ is the

second-order identity tensor.

Yield Function

Next the yield function equations are developed. In this case, the chosen

pressure-dependent yield function is the Drucker-Prager yield function. The Drucker-

Prager yield function is written in ABAQUS [5] notation as

0tan =−−= dptf θ , (5.3)

where t is a pseudo-effective stress written as

−−+=

3

2

32

3

11

113

2

1

J

J

rrJt , (5.4)

θ is the slope of the linear yield surface in the p-t stress plane, p is the hydrostatic

pressure, d is the effective cohesion of the material, and r is the ratio of the yield stress in

triaxial tension to the yield stress in triaxial compression. In metal plasticity, d is

equivalent to the current yield stress, σys. In general, σys is a function of the equivalent

plastic strain, pleqε , which is given by the equation

plij

plij

pleq εεε

3

2= . (5.5)

83

The variables used by the linear Drucker-Prager yield function are shown graphically in

Figure 5.1.

The flow potential, g, for the linear Drucker-Prager model is defined as

ψtanptg −= , (5.6)

where ψ is the dilation angle in the p-t plane (Figure 5.1). The dilation angle controls the

movement of an arbitrary point on the yield surface during the hardening process.

Setting ψ = θ results in associated flow. Therefore, the original Drucker-Prager model is

available by setting ψ = θ and r = 1.

0

t

p

θ

d

θ

ψdε

hardening

pl

Figure 5.1. Linear Drucker-Prager Model: Yield Surface and Flow Direction in the p-t Plane (Adapted from [5])

84

To conveniently compare ABAQUS Drucker-Prager material property variables

with those used by Richmond, et al. in their material testing, one must correlate the

variables in Equations (2.31) and (5.3). Recalling that 131 Ip −= and 23Jt = for r =

1, Equation (5.3) can be written as

dIJf −+= θtan3

13 12 . (5.7)

Comparing Equations (2.31) and (5.7) leads to

θtan3

111 IaI = , (5.8)

and solving for θ yields

( )a3tan 1−=θ . (5.9)

Finally, Equation (5.7) can be written as

( ) 03 =−−= pleqyseff apf εσσ , (5.10)

thus relating the material constants of the original Drucker-Prager theory with the

material constants used in the ABAQUS linear Drucker-Prager constitutive model.

Flow Rule

Next the flow rule is developed. The previously developed flow rule given by

Equation (2.21) is written in rate form as

ij

plij

g

σφε

∂∂= && , (5.11)

85

where the dot notation indicates a derivative with respect to time.

For the case of associated flow, let g = f which gives

ij

plij

f

σφε

∂∂= && . (5.12)

The flow rule is written in a more general form as

Ι+= pijqpl

ij n εεε &v

&&

3

1, (5.13)

where

eff

ijij

Sn

σ2

3=v, (5.14)

eff

q

f

σφε

∂∂= && , (5.15)

p

fp ∂

∂−= φε && , (5.16)

and qε& and pε& are the distortional and volumetric portions of the plastic strain rate,

respectively. Eliminating φ& from Equations (5.15) and (5.16) gives

0=∂

∂+∂∂

effpq

f

p

f

σεε && . (5.17)

The partial derivatives of f (see Equation (5.10)) for Equation (5.17) are

ap

f3−=

∂∂

(5.18)

and

86

1=∂

∂

eff

f

σ. (5.19)

Therefore, Equation (5.17) is rewritten as

( ) 03 =+− pq a εε && . (5.20)

State Variable Equations

Now the state variable equations are developed. For the case of the linear

Drucker-Prager model, only one state variable, 1Γ& is required and is defined as

pleqε&& ≡Γ1 . (5.21)

The plastic work equation is used to derive a usable form of the state variable equation.

The rate of plastic work, plW& is defined in terms of the stress and plastic strain rate

tensors as

plijij

plW εσ && = (5.22)

or in terms of the effective stress and equivalent plastic strain rate as

( ) pleq

pleqys

plW εεσ && = . (5.23)

Combining Equations (5.22) and (5.23) and solving for pleqε& gives

( )pleqys

plijijpl

eq εσεσ

ε&

& = . (5.24)

Expanding the numerator of Equation (5.24) gives

qijijpij

ijpl

eqij n εσεσεσ &v

&& +Ι

=3

, (5.25)

87

which can be written as

qeffppl

eqij p εσεεσ &&& +−= . (5.26)

Finally, substituting the expression for pleqijεσ & in Equation (5.26) into Equation (5.24)

gives

( )pleqys

qeffppleq

p

εσεσε

ε&&

&+−

= . (5.27)

Numerical Integration

Next a proper numerical integration method must be determined and then applied

to the elastoplastic differential equations. Several integration procedures can be used

including the forward Euler method, backward Euler method, and the midpoint method.

The forward Euler integration methods are usually simple, but the integration method is

explicit and therefore has a stability limit [77]. The backward Euler method is an implicit

scheme and is usually more complicated than the forward Euler method. Ortiz and

Popov [78] found that the backward Euler method is very accurate for strain increments

several times the size of the yield surface in strain space and that the method is

unconditionally stable. For these reasons, the backward Euler method was chosen for the

integration of the elastoplastic equations.

Using the backward Euler method to integrate equations (5.2), (5.13), (5.20) and

(5.27) leads to the following incremental forms

ijpnijqpr

ijnij Kn Ι∆−∆−=++

εεµσσ11

2v

, (5.28)

88

ijpnijqpl

ij n Ι∆+∆=∆+

εεε3

11

v, (5.29)

( ) 03 =∆+−∆ pq a εε , (5.30)

and

( )pleqys

qeffppleq

p

εσεσε

ε∆+∆−

=∆ . (5.31)

The “n+1” subscript denotes values at the end of the increment at time, tttnn

∆+=+1

,

and the “pr” superscript indicates a predicted value based on the purely elastic solution.

Next, it is shown that 1+nijn

v can be found from the elastic stress predictor, pr

ijσ .

Equation (5.14) can be rewritten as

1

1

1 23

+

++

=neff

nij

nij

Sn

σv

, (5.32)

and the deviatoric part of Equation (5.32) can be expressed as

1

1 31

+

+ ∆+

=

neff

q

prij

nij

SS

σεµ , (5.33)

thus demonstrating that 1+nijS and pr

ijS are collinear. Finally, substituting Equation (5.33)

into Equation (5.32), one finds

prijpr

eff

prij

nij nS

nvv ==

+ σ23

1, (5.34)

which is a known quantity.

89

With 1+nijn

v known, the final step in the integration of the elastoplastic equations

is the determination of ∆εp and ∆εq. By projecting Equation (5.28) onto 1+nijn

v and ijΙ and

taking into account that 23=ijijnnvv

and effijijn σσ =vv, one finds

qpreffneff εµσσ ∆−=

+3

1 (5.35)

and

ppr

nKpp ε∆+=

+1. (5.36)

Summarizing the primary equations (Equations (5.10), (5.30), (5.35),(5.36), and

(5.31)) and dropping the “n+1” subscript for brevity, one can write

( ) 03 =−−= pleqyseff apf εσσ , (5.37)

( ) 03 =∆+−∆ pq a εε , (5.38)

qpreffeff εµσσ ∆−= 3)( , (5.39)

ppr Kpp ε∆+= , (5.40)

and

( )pleqys

qeffppleq

p

εσεσε

ε∆+∆−

=∆ . (5.41)

This forms a set of five nonlinear equations that is solved for p, σeff, ∆εp, ∆εq, and pleqε∆

[76]. Solving Equations (5.37) through (5.41) for the five unknowns gives

( )

µεσσµ

+−+

=23Ka

KaKapp

pleqys

preff

pr

, (5.42)

90

( )

µεµσσµ

σ+

++=

2

2

3

33

Ka

aKap pleqys

preff

pr

eff , (5.43)

( )[ ]

µεσσ

ε+

−−=∆

23

3

Ka

apa prpleqys

preff

p , (5.44)

( )

µεσσ

ε39

32 +

−−=∆

Ka

ap prpleqys

preff

q , (5.45)

and

( )

µεσσ

ε39

32 +

−−=∆

Ka

ap prpleqys

preffpl

eq . (5.46)

In general, Newton iteration is used to solve Equation (5.46) for pleqε∆ and ( )pl

eqys εσ , and

then Equations (5.42) through (5.45) are solved by direct substitution. Strain increments

and stresses are then determined by

ijqijppl

ij nvεεε ∆+Ι∆=∆

3

1 (5.47)

and

ijeffijij npvσσ

3

2+Ι−= (5.48)

thereby completing the solution of the elastoplastic equations.

In an implicit finite element code, such as ABAQUS, the equilibrium equations

are written at the end of the increment resulting in a set of nonlinear equations in terms of

the nodal unknowns. If a Newton scheme is used to solve the global nonlinear equations,

the Jacobian (tangent stiffness) must be calculated [76]. The Jacobian, J is defined as

91

1

1

+

+

∂∂

=n

nJεσ

. (5.49)

The accuracy of Jacobian effects convergence, and, therefore, errors in the Jacobian

formulation may result in analyses that require more iterations or, in some cases, diverge

[77].

The Jacobian for von Mises plasticity was employed in this research. This

Jacobian proved sufficient for all of the finite element analyses except for the smooth

tensile bar. Once the smooth tensile bar FEM began “necking” the finite element

solution diverged. Therefore, the built-in Drucker-Prager model with a pressure

dependent Jacobian was used for modeling the monotonic loading of the smooth tensile

bar.

Hardening Models

Next, the development of the isotropic, kinematic, and combined kinematic and

isotropic hardening models is presented. For simplicity, the hardening equations for the

classical von Mises theory along with a bilinear hardening theory are developed first.

The derivation of the bilinear hardening equations closely parallels the work of Taylor

and Flanagan [79]. These equations are then modified to form the more complex

nonlinear (multilinear) hardening equations. Finally, the bilinear and multilinear

equations are modified to include the Drucker-Prager pressure dependency. To begin,

several terms that are common to each of the hardening models are defined.

92

Basic Definitions

To begin a discussion of plastic hardening, several basic terms are developed.

The center of the yield surface in deviatoric stress space (the backstress) is defined by the

tensor αij, which is illustrated in Figure 5.2. Recalling from Chapter 2, the deviatoric

stress tensor, Sij is defined as

13

1IS ijij −= σ . (5.50)

The stress difference, ξij is defined by

ijijij S αξ −= . (5.51)

The magnitude of the stress difference, R, is then written as

ijijijR ξξξ == . (5.52)

The geometric relationship between ξij, αij, and Sij leads to taking the magnitude of the

αij

S ij

ξij

Qij

Figure 5.2. Illustration of a Yield Surface in Deviatoric Stress Space (Adapted from [79])

93

stress difference as a radius, and therefore R is used. The von Mises yield surface in

terms of the stress difference is

2

2

1kf ijij == ξξ , (5.53)

where k is the yield strength in pure shear, and the effective stress is

ijijeff ξξσ2

3= . (5.54)

Combining Equations (5.52) and (5.54), one defines R as

effR σ3

2= . (5.55)

The normal to the yield surface, Qij is then determined from Equation (5.53) as

Rf

fQ ij

ij

ijij

ξσσ

=∂∂

∂∂= . (5.56)

Assuming a normality condition, the plastic part of the strain is

ijpl

ij Qγε = , (5.57)

where γ is a scalar multiplier that must be determined [79].

Isotropic Hardening

In the isotropic hardening case, the yield surface changes size but does not change

location in deviatoric stress space. Therefore, αij is zero and ξij = Sij for the isotropic

case. A consistency equation is written to ensure that the state of stress remains on the

94

yield surface at all times. For the isotropic case, the consistency condition is formed by

taking the time derivative of Equation (5.53)

kkf && 2= . (5.58)

Using the chain rule and Equation (5.56), the consistency condition is

ijijij

ijij

Qff

f σσ

σσ

&&&

∂∂=

∂∂= , (5.59)

where from Equations (5.52) and (5.53)

RSf

ijij

==∂∂σ

. (5.60)

Combining Equations (5.58) through (5.60) gives

RSR ijij

&& =σ1. (5.61)

Noting that because Sij is deviatoric,

ijijijij SSS && =σ (5.62)

and

( )

ijeffeff

ijijijij dt

dSS

dt

dSS σσ

σ&&

3

2

32

12

=

=

= . (5.63)

Equation (5.61) can then be written as

effR σ&&

3

2= . (5.64)

For bilinear hardening effσ& is

pleqeff Hεσ =& , (5.65)

95

where H is the slope of the equivalent stress versus equivalent plastic strain as shown in

Figure 5.3. Uniaxial σ-ε data, as shown in Figure 5.4, can be used to determine H by

utilizing the equation

t

t

EE

EEH

−= . (5.66)

Combining Equations (5.61), (5.64), and (5.65) gives

ijijpl

eq QH σε &=3

2. (5.67)

The elastic predictor stress rate, prijσ& is

klijklpr

ij D εσ = , (5.68)

where Dijkl is the fourth-order tensor of elastic coefficients defined by Equation (5.2).

σys

εeqpl

0σys

H

1

σys n

σys n+1

plεeq n ∆εeqpl

pl

∆σys

εeq n+1

Figure 5.3. Illustration of the Relationship Between Yield Stress and Equivalent Plastic Strain for the Bilinear Hardening Case

96

σ

ε

E

Et

1

1σys0

Figure 5.4. Illustration of the Uniaxial True Stress versus True Strain Relationship for the Bilinear Hardening Case

Combining the equations (5.67) and (5.68) with the additive strain decomposition given

in Equation (5.1) gives

pleqijklij

prijij

plq DQQH εσε −= &

3

2. (5.69)

Noting that because Qij is deviatoric, ijklijkl QQD µ2= and µ2=klijklij QDQ . Then using

the normality condition and the definition of equivalent plastic strain, Equation (5.69)

becomes

µγσγ 232 −= pr

ijijQH & . (5.70)

Also, because Qij is deviatoric, ijijpr

ijij QQ εµσ 2= , and therefore γ can be determined as

ijijQH

ε

µ

γ

+=

31

1, (5.71)

which completes the derivation of the bilinear isotropic hardening equations.

97

Kinematic Hardening

In the kinematic hardening case, the yield surface does not change size but does

change location in deviatoric stress space. As for the isotropic case, a consistency

equation is written to ensure that the state of stress remains on the yield surface at all

times. For the kinematic case, the consistency condition is formed by taking the rate of

Equation (5.53)

0=f& . (5.72)

Using the chain rule on Equation (5.72) gives

0=∂∂

ijij

f ξξ

& , (5.73)

and

ijijijij

RQQff =

∂∂=

∂∂

ξξ. (5.74)

Combining Equations (5.73) and (5.74) gives

( ) .0=− ijijij SQ α&&& (5.75)

The definition of ijα& must now be determined. Recall that for the isotropic case

γεσ HHQ pleqijij 3

232 ==& . (5.76)

Assuming that ijα& is a function of plastic strain leads to the equation

ijpl

ijij Qζγζεα == , (5.77)

98

where ζ is a material parameter to be determined. Equation (5.77) combined with (5.75)

gives a result identical to the isotropic hardening case, Equation (5.76), if 32H=ζ .

Therefore, 32H=ζ is used to ensure consistency between the isotropic and kinematic

cases.

Combining the strain rate decomposition, the elastic strain rate, and the

consistency condition for kinematic hardening gives

plklijkl

prijij DQH εσγ −= &

3

2. (5.78)

Then taking the tensor inner product of both sides of the previous equation with Qij gives

( )ijpr

ijijijij QQQHQ µγσγ 232 −= & . (5.79)

Therefore, γ can be determined as

ijijQH

ε

µ

γ

+=

31

1, (5.80)

which is the same as Equation (5.71) for the isotropic case, thus completing the

derivation of the bilinear isotropic hardening equations.

Combined Bilinear Hardening

Next, the equations for the combined isotropic and kinematic bilinear hardening

case are derived. Assuming a linear combination of the two hardening types, a scalar

99

parameter, β, can be defined which determines the amount of each type of hardening. It

is a requirement that

10 ≤≤ β . (5.81)

A value of β = 1 indicates only isotropic hardening, and a value of β = 0 indicates only

kinematic hardening.

New equations are developed for R and αij by taking the results already derived

for the independent hardening cases and multiplying by the appropriate hardening

fraction. Equations (5.64) and (5.77) are rewritten as

βε pleqHR

3

2=& (5.82)

and

( )βγα −= 132

ijij QH& . (5.83)

Again, the consistency equation is formed as

RQ ijij&& =ξ (5.84)

or

( ) βεα pleqijijij HSQ

32=− && . (5.85)

Using the additive strain rate decomposition and the elastic stress rate with Equation

(5.85) and taking the tensor product with the normal, Qij gives

( ) ijijijijklijklijpr

ijij QHQQHQQDQQ

=

−−− βγβγγσ

32

32

132

& . (5.86)

100

Solving for γ gives

ijijQH

ε

µ

γ

+=

31

1, (5.87)

which is the same equation for γ obtained for the two independent cases.

The governing equations for the combined hardening theory are summarized as:

( ) prij

plijijijklij D σεεσ && =−= , (5.88)

γββε HHR pleq 3

2

3

2 ==& , (5.89)

( )βεα −= 13

2 plijij H& , (5.90)

( )( )

≥<

= 2

2

;plastic,

;elastic,0

kfQ

kf

ij

plij ξλ

ξε , (5.91)

ijijQH

ε

µ

γ

+=

31

1, (5.92)

and

Rf

fQ ij

ij

ijij

ξξξ

=∂∂∂∂

= . (5.93)

Equations (5.88) through (5.93) must now be written in incremental form for

implementation into a finite element algorithm. The incremental forms of Equations

(5.88) through (5.90) are

ijn

prijnij Qµγσσ 2

11∆−=

++, (5.94)

101

γβ∆+=+

HRRnn 3

21

, (5.95)

and

( )βγαα −∆+=+

132

1 ijnijnij QH (5.96)

where, as before, the subscripts “n” and “n+1” refer to the beginning and end of a time

step, respectively.

An incremental form of the consistency condition is also required and is written

as

111 +++

=+nijijnnij SQRα . (5.97)

A graphical interpretation of Equation (5.97) is shown in Figure 5.5. Substituting

Equations (5.94) through (5.96) into the consistency condition of (5.97) gives

( ) ijn

prijijnijnij QSQHRQH µγγββγα 2

32

132

1∆−=

∆++

−∆+

+. (5.98)

Taking the tensor product of both sides of the previous equation with Qij and solving for

∆γ gives

( )nn

prij R

H−

+=∆

+1

312

1 ξ

µµ

γ . (5.99)

Equation (5.99) demonstrates that the plastic strain increment is proportional to the

magnitude of the distance of the elastic predictor stress past the yield surface (see Figure

5.6). Having now determined ∆γ from Equation (5.99) Equations (5.94) through (5.96)

can be solved. In addition, one also computes

102

αij n+1

Qij n+1

S ij n+1

R n+1 Qij n+1

Figure 5.5. Geometric Interpretation of the Incremental Form of the Consistency Condition for Combined Hardening (Adapted from [79])

αij n+1

ξ n+1

pr

ξ n+1 -R npr

Figure 5.6. Geometric Interpretation of the Incremental Form of the Radial Return Correction (Adapted from [79])

103

γε ∆=∆ ijpl

il Q (5.100)

and

γε ∆=∆32pl

eq , (5.101)

which completes the bilinear combined hardening incremental solution [79].

Combined Multilinear Hardening

Next, the equations for the combined isotropic and kinematic multilinear

hardening case are derived. The equations for the multilinear hardening case are similar

to those for the bilinear case, but an added level of complexity is required because the

slope of the equivalent stress versus equivalent plastic strain, H is no longer constant.

This complication is clearly seen by examining Figure 5.7, where for a given plastic

strain increment, H may take on several values, Hi and comparing it with Figure 5.3.

Recall that for bilinear hardening effσ∆ was written as

γεσ ∆=∆=∆3

2HH pl

eqeff . (5.102)

Similarly, for the multilinear hardening case effσ∆ can be written as

( ) ( ) γεεεσ ∆=∆=∆3

2pleq

pleq

pleqeff hh , (5.103)

where ( )pleqh ε is a hardening function. Substituting ( )pl

eqh ε for H in Equations (5.89) and

(5.90) gives

104

σys

εeqpl

0σys

σys n

σys n+1

pl εeq n ∆εeqpl

εeq n+1

pl

∆σys

1

1

1

11

H1

H2

H3

H4H5

Figure 5.7. Illustration of the Relationship Between Yield Stress and Equivalent Plastic Strain for the Multilinear Hardening Case

( ) βεε pleq

pleqhR

3

2=& , (5.104)

and

( )βγεα −= 1)(32

ijpl

eqij Qh& . (5.105)

Equations (5.104) and (5.105) are integrated exactly and written in incremental form as

( ) ( )[ ]n

pleq

pleqn

pleq hhR εεεβ −∆+=∆

32

, (5.106)

and

( ) ( ) ( )[ ]13

21

+−∆+−=∆

nijn

pleq

pleqn

pleqij Qhh εεεβα . (5.107)

105

Rewriting Equation (5.97) in terms of Equations (5.106) and (5.107) (where

RRRnn

∆+=+1

, etc.) gives

( ) ( )[ ]13

223

2+

−∆+++∆=−nijn

pleq

pleqn

pleqn

pleqnij

prij QhhRS εεεεµα . (5.108)

Multiplying both sides of the previous equation by 1+nijQ leads to

( ) ( )[ ] ( )[ ]nnij

prijn

pleq

pleqn

pleq

pleq RShh −−=−∆++∆ αεεεεµ

32

32

2 , (5.109)

which is a nonlinear equation that can be solved for pleqε∆ . Once pl

eqε∆ is known, the

remaining incremental elastoplastic equations can be solved following the method

demonstrated for the bilinear hardening case.

Pressure-Dependent Combined Hardening

To conveniently compare the pressure-independent combined hardening

equations with the previously derived Drucker-Prager elastoplastic equations, the

following substitutions are made:

nysn

R σ32= , (5.110)

preffn

pr

n

prij RS σξ

32

1==−

+, (5.111)

pleqεγ ∆=∆

2

3, (5.112)

and for the multilinear case

106

( ) ( )1+

=∆+n

pleqeff

pleqn

pleqh εσεε , (5.113)

and

( )neffn

pleqh σε = . (5.114)

Making the previous substitutions into Equation (5.99) gives

H

nyspreffpl

eq +

−=∆

µσσ

ε3

. (5.115)

Recall the equation for pressure-dependent equivalent plastic strain, Equation (5.46)

( )

µεσσ

ε39

32

1

+

−−=∆ +

Ka

ap pr

n

pleqys

preffpl

eq . (5.116)

Substituting a = 0 for pressure-independent plasticity and

( ) pleqneffn

pleqeff H εσεσ ∆+=

+1 (5.117)

for bilinear hardening results in

H

nyspreffpl

eq +

−=∆

µσσ

ε3

, (5.118)

which is identical to Equation (5.115).

In a similar manner for the multilinear hardening case, substituting Equations

(5.110) through (5.114) into Equation (5.109) and solving for pleqε∆ gives

( )

µσσεσσ

ε3

1 nysneffn

pleqeff

preffpl

eq

−+−=∆ + . (5.119)

For the case of pure isotropic hardening nysneff σσ = , and Equation (5.119) can be

written as

107

( )

µεσσ

ε3

1+−

=∆ n

pleqeff

preffpl

eq , (5.120)

which is identical to Equation (5.116) for pressure-independent (a = 0) plasticity.

Therefore, adding the terms forneffσ and

nysσ to the numerator of Equation (5.116) results

in the proper form of the equation for pleqε∆ for pressure-dependent combined multilinear

hardening, which is written in its final form as

( )

µσσεσσ

ε39

32

1

+

−−+−=∆ +

Ka

ap pr

nysneffn

pleqys

preffpl

eq . (5.121)

Therefore, Equation (5.121) completes the derivation of the pressure-dependent

plasticity model with combined multilinear isotropic and kinematic hardening.

Constitutive Model Programming

The next step in the analytical research program was to implement the combined

hardening Drucker-Prager constitutive model equations into the commercial finite

element code ABAQUS. ABAQUS provides a useful user subroutine interface called

UMAT that allows one to define complex or novel constitutive models that are not

available with the built-in ABAQUS material models. UMAT’s are written in

FORTRAN code, and these FORTRAN subroutines are linked, compiled, and used by

ABAQUS in the finite element analysis.

Two UMAT’s were developed for this research. The first UMAT embodied a

Drucker-Prager constitutive model with combined bilinear hardening. This subroutine

was developed as a test case to confirm the validity of the pressure-dependent and

108

combined hardening equations. The second UMAT embodied a Drucker-Prager

constitutive model with combined multilinear hardening. This subroutine was developed

for use in the low cycle fatigue finite element analyses for this research. Both

subroutines were written in FORTRAN and are included in Appendix C.

This chapter presented the development of a pressure-dependent constitutive

model with combined kinematic and isotropic hardening. First the pressure-dependent

plasticity model was derived. Next the combined bilinear and combined multilinear

hardening equations were developed for von Mises plasticity theory. The hardening rule

equations were then modified to include pressure dependency. Finally, the method for

implementing the new constitutive model into ABAQUS was presented. The next

chapter presents the development of the finite element models.

109

CHAPTER 6

FINITE ELEMENT MODELING

This chapter presents the second portion of the analytical research program, the

development of the finite element models (FEM’s). The chapter begins with a

description of the required material property inputs for the FEM’s followed by a

discussion of the specific material property inputs for 2024-T851 and IN100. Next, a

general discussion of the finite element modeling procedure is given. Finally, several test

specimen geometries are presented, and specific details concerning the creation of finite

element models for each geometry are given.

Material Property Inputs

Several material property inputs are required for the bilinear and multilinear

hardening UMAT’s in ABAQUS. Both UMAT’s require Young’s modulus, E, Poisson’s

ratio, ν, the Drucker-Prager material constant, a, and the combined hardening parameter,

β. The two UMAT’s differ in how they relate changes in yield stress with changes in

plastic strain. The bilinear model requires values for initial yield strength, osyσ and the

hardening modulus, H. The multilinear hardening model requires a table of effective

stress, σeff versus equivalent plastic strain, pleqε , which is equivalent to true stress, σ versus

plastic strain, ε pl data for a uniaxial test. The bilinear hardening model was used for

developmental purposes only and, therefore, will not be discussed further.

110

2024-T851 Material Property Inputs

The 2024-T851 material properties were obtained from the mechanical property

test program discussed in Chapter 4. Only the L direction material properties are

considered here, because all of the low cycle fatigue specimens were machined from this

direction. The values for Young’s modulus and Poisson’s ratio used were 10.6×106 psi

and 0.323, respectively. The hardening parameter, β, was adjusted as required to best

match the first cycle hysteresis loop of the individual LCF tests.

The Drucker-Prager constant, a was 0 for von Mises plasticity, but the estimation

of a for Drucker-Prager plasticity was more difficult. First, no pressurized tensile and

compressive data was available for high strength aluminums. Therefore, it was originally

postulated that Equation (3.1) could be used to calculate a. As previously discussed in

Chapter 4, the tensile and compressive yield strengths were so close that an estimation for

a using Equation (3.1) was impractical. In the author’s previous research [12], a

Drucker-Prager constant of 0.029 was estimated for 2024-T351 by using a ratio of

estimated theoretical cohesive strength over the initial tensile yield strength. This proved

to be a reasonable estimate of a for 2024-T351. Therefore, for this research, a for 2024-

T851 was originally assumed to be 0.029. As will be shown in the following chapter, this

proved to be a reasonable choice for a, but perhaps not the best choice. For all the

geometries tested, a was found to have a range of approximately 0.029 to 0.041, which

corresponds to a θ (Figure 5.1) of 5º to 7º.

111

Considerable care was taken in developing the true stress versus plastic strain data

table. The test record of specimen AT06 was chosen as representative of the L direction

tensile test data. Data points from the actual test record up to the maximum load were

used for the initial portion of the table values. After the maximum load, the test record

no longer represents a uniaxial stress-strain response due to necking of the test specimen,

therefore, a different method was employed to generate the remainder of the data table.

First, the dimensions of the broken specimen were used to estimate a true fracture stress

corresponding to the known fracture strain. Next, a power law was fit to the three points

leading up to the maximum load point and the estimated fracture point. This power law

equation was then used to complete the table out to a plastic strain of 1.0. The ABAQUS

input data table values are plotted alongside the test record of specimen AT06 in Figure

6.1, and an extended view out to a plastic strain of 1.0 is shown in Figure 6.2. The

complete data table for 2024-T851 used in this research is given in Appendix D.

112

55

60

65

70

75

80

85

400

450

500

550

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

Specimen AT06

ABAQUS Input Data

Tru

e S

tres

s, σ

(ks

i)

Plastic Strain, εpl

Tru

e S

tres

s, σ

(M

Pa)

Figure 6.1. Comparison of Tensile Test Data and ABAQUS Input Data

55

60

65

70

75

80

85

400

450

500

550

0 0.2 0.4 0.6 0.8 1

Specimen AT06

ABAQUS Input Data

Tru

e S

tres

s, σ

(ks

i)


Tru

e S

tres

s, σ

(M

Pa)

Figure 6.2. Extended View of Comparison of Tensile Test Data and ABAQUS Input Data

113

Inconel 100 Material Property Inputs

The tensile IN100 material properties were obtained from the mechanical property

test program discussed in Chapter 4 and from the Aerospace Structural Metals Handbook

[74]. The value for Poisson’s ratio was 0.30, and the Drucker-Prager constant, a, was 0

for von Mises plasticity. Because Young’s modulus for IN100 and steel is similar,

Richmond’s pressurized tension and compression tests results (Table 2.1) for high

strength steels was used to estimate a value for a of 0.022 (θ of 3.8º) for Drucker-Prager

plasticity. This value of a for IN100 proved reasonable in the author’s previous thesis

research [12]. The hardening parameter, β, was adjusted as required to best match the

first cycle hysteresis loop of the equal-arm bend test.

The values for Young’s modulus used were 26.0×106 psi for the equal-arm bend

FEM and 31.8×106 psi for the rest of the FEM’s. The lower value of E was used for the

equal-arm bend specimen to match the test data in the linear range. The equal-arm test

specimen was taken from a different lot of material than the material tested in this

research. Also, Pratt and Whitney [59] tensile test data for IN100 has widely scattered

values for E ranging from 25.0×106 to 35.0×106 psi. Apparently the values for E reported

by Pratt and Whitney were not actually measured separately but were merely estimated

from tensile test data. In addition, some change in Young’s modulus can be attributed to

the variability in product form, such as castings versus forgings. The reason for the

change in elastic modulus is a matter of debate, but it is also possible that the measured

value of E for IN100 changes with loading type [80]. This is supported by the fact that in

114

addition to normal stresses, the equal-arm bend specimen was under the influence of

large bending stresses.

Once again, considerable care was taken in developing the true stress versus

plastic strain table. The test record of specimen 4A-TS-3 was chosen as representative of

the tensile test data. Discrete data points from the actual test record were used to

generate the table up to the point that the σ - ε pl curve starts rolling over (ε pl §��

The upper yield point was neglected in picking points for this portion of the table. After

the σ - ε pl curve rolls over, the test record no longer represents a uniaxial stress strain

response due to necking of the test specimen. Therefore to generate the rest of the table,

a straight line was fit between the last two points before the σ - ε pl curve starts rolling

over, and this line was projected out to a plastic strain of 1.0 to complete the data table.

The ABAQUS input data table values are plotted alongside the test record of specimen

4A-TS-3 in Figure 6.3, and an extended view out to a plastic strain of 1.0 is shown in

Figure 6.4. The complete data table for IN100 used in this research is given in Appendix

D.

115

100

150

200

250

300

800

1000

1200

1400

1600

1800

2000

0 0.05 0.1 0.15 0.2 0.25

Specimen 4A-TS-3ABAQUS Input Data

Tru

e S

tres

s, σ

(ks

i)


Tru

e S

tres

s, σ

(M

Pa)

Figure 6.3. Comparison of Tensile Test Data and ABAQUS Input Data

200

400

600

800

1000

1000

2000

3000

4000

5000

6000

0 0.2 0.4 0.6 0.8 1

Specimen 4A-TS-3ABAQUS Input Data

Tru

e S

tres

s, σ

(ks

i)


Tru

e S

tres

s, σ

(M

Pa)

Figure 6.4. Extended View of Comparison of Tensile Test Data and ABAQUS Input Data

116

Test Specimen Finite Element Models

Several finite element models were created for this research. The Sandia National

Laboratory program FASTQ [61] was used for preprocessing of meshes and boundary

conditions for all of the FEM’s except for the equal-arm bend model. The commercial

finite element code Patran [60] was used for the generation of the meshes and boundary

conditions for the equal-arm bend geometry, due to its more complex geometry.

ABAQUS [5] was used for the finite element analyses and postprocessing the results.

Large strain analysis and full integration were used, and all of the FEM’s were loaded in

displacement control. The load-gage displacement data was gleaned from the ABAQUS

database files using a Unix stream-editor (sed) script, which is given in Appendix E.

Smooth Tensile Bar Specimen

Axisymmetric FEM’s of the 0.35 in diameter 2024-T851 smooth tensile

specimens and of the 0.25 in. diameter IN100 smooth tensile specimens were created

using Q4 axisymmetric elements (type CAX4 in ABAQUS). As illustrated in Figure 6.5,

only one-quarter of the tensile bar gage section was modeled by using two symmetry

planes, and a uniform displacement was applied to the top nodes of the FEM. The neck

diameter of both FEM’s was reduced approximately 0.005 in. to ensure necking in the

gage region of the model.

Three levels of mesh refinement were used in developing the 0.35 in. diameter

smooth tensile FEM. The coarse, medium, and fine meshes consisted of 212, 780, and

117

Uniform Displacement

Figure 6.5. Schematic of Axisymmetric Model of a Smooth tensile Bar Specimen Utilizing Two Planes of Symmetry

1440 elements, respectively. An illustration of the 0.35 in. diameter medium mesh FEM

is shown in Figure 6.6. The complete set of 0.35 in. diameter smooth tensile FEM’s is

given in Appendix F.

A convergence study was performed to determine the variation of effective stress,

σeff, mean stress, σm, radial stress, σrr, and equivalent plastic strain, pleqε across the neck

region at failure load for the three mesh densities. The results of the convergence study

are plotted in Figures 6.7 through 6.10. All plotted values are averaged at the nodes and

were generated using von Mises plasticity theory. For all cases, the variation between the

three mesh densities is very small until the outer surface of the neck region is reached.

The medium mesh provided essentially the same results as the fine mesh and

computationally took approximately the same time as the coarse mesh to run. Therefore,

118

Figure 6.6. Medium Mesh FEM of the 0.35 in. Diameter Smooth Tensile Bar

119

80150

80200

80250

80300

80350

80400

80450

80500

0 0.035 0.07 0.105 0.14 0.175

Coarse Mesh

Medium Mesh

Fine Mesh

552.8

553.2

553.6

554

554.4

554.8

0 0.5 1 1.5 2 2.5 3 3.5 4

Effe

ctiv

e S

tres

s, σ

eff (

psi)

Distance Across Neck (in.)

Distance Across Neck (mm)

Effe

ctiv

e S

tres

s, σ

eff (

Mpa

)

Figure 6.7. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

25000

30000

35000

40000

45000

0 0.035 0.07 0.105 0.14 0.175

Coarse Mesh

Medium Mesh

Fine Mesh

180

200

220

240

260

280

300

0 0.5 1 1.5 2 2.5 3 3.5 4

Mea

n S

tre

ss, σ

m (

psi)



Mea

n S

tres

s, σ

m (

Mpa

)

Figure 6.8. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

120

0

5000

10000

15000

20000

0 0.035 0.07 0.105 0.14 0.175

Coarse Mesh

Medium Mesh

Fine Mesh

0

20

40

60

80

100

120

0 0.5 1 1.5 2 2.5 3 3.5 4

Rad

ial S

tres

s, σ

rr (

psi)



Rad

ial S

tres

s, σ

rr (

Mpa

)

Figure 6.9. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.35 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

0.195

0.200

0.205

0.210

0.215

0.220

0.225

0.230

0 0.035 0.07 0.105 0.14 0.175

Coarse Mesh

Medium Mesh

Fine Mesh

0 0.5 1 1.5 2 2.5 3 3.5 4

Equ

ival

ent P

last

ic S

trai

n, ε

pl

eq



Figure 6.10. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine Mesh 0.35 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

121

after completion of the convergence study, the medium mesh FEM was chosen for use in

this research .

Three levels of mesh refinement were also used in developing the 0.25 in.

diameter smooth tensile FEM. The coarse, medium, and fine meshes consisted of 158,

552, and 1070 elements respectively. An illustration of the 0.25 in. diameter medium

mesh FEM is given in Figure 6.11. All of the 0.25 in. diameter smooth tensile FEM’s are

illustrated in Appendix F.

A convergence study was performed to determine the variation σeff, σm, σrr, and

pleqε across the neck region at failure load for the three mesh densities. The results of the

convergence study are plotted in Figures 6.12 through 6.15. All of the values plotted are

averaged at the nodes and were generated using von Mises plasticity theory. Again, the

variation between the three mesh densities is very small except close to the outer surface

of the neck region. After completion of the convergence study, the medium mesh FEM

was chosen as a compromise between computation speed and accuracy.

122

Figure 6.11. Medium Mesh FEM of the 0.25 in. Diameter Smooth Tensile Bar

123

285000

290000

295000

300000

305000

310000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

1980

2000

2020

2040

2060

2080

2100

2120

0 0.5 1 1.5 2 2.5 3

Effe

ctiv

e S

tres

s, σ

eff (

psi)



Effe

ctiv

e S

tre

ss, σ

eff (

Mpa

)

Figure 6.12. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

96000

98000

100000

102000

104000

106000

108000

110000

112000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

680

700

720

740

760

0 0.5 1 1.5 2 2.5 3

Mea

n S

tre

ss, σ

m (

psi)



Mea

n S

tres

s, σ

m (

Mpa

)

Figure 6.13. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

124

-1000

0

1000

2000

3000

4000

5000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

-5

0

5

10

15

20

25

30

0 0.5 1 1.5 2 2.5 3

Rad

ial S

tres

s, σ

rr (

psi)



Rad

ial S

tres

s, σ

rr (

Mpa

)

Figure 6.14. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

0.170

0.175

0.180

0.185

0.190

0.195

0.200

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

0 0.5 1 1.5 2 2.5 3

Equ

ival

ent P

last

ic S

trai

n, ε

pl

eq



Figure 6.15. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine Mesh 0.25 in. Diameter Smooth Tensile Bar FEM’s at Failure Load

125

Smooth Compression Cylinder Specimen

Axisymmetric finite element models of a smooth compression cylinder were

created to simulate the uniaxial compression tests. One quarter of the compression

specimen geometry was modeled by using two symmetry planes (Figure 6.16) and Q4

axisymmetric elements (type CAX4 in ABAQUS). A uniform downward displacement

was applied to the top nodes of the FEM for the loading boundary condition.

Three levels of mesh refinement were used in developing the compression

specimen FEM. The coarse, medium, and fine meshes consisted of 150, 534, and 1008

elements respectively, and an illustration of the medium mesh FEM is given in Figure

6.17. The complete set of compression specimen FEM’s is given in Appendix F.


Figure 6.16. Schematic of Axisymmetric Model of a Smooth Compression Cylinder Specimen Utilizing Two Planes of Symmetry

126

Figure 6.17. Medium Mesh FEM of the Smooth Compression Specimen

127

A convergence study was performed to determine the variation σeff, σm, σrr, and

pleqε across the bottom symmetry plane at maximum test load for the three mesh densities.

The results of the convergence study are given in Figures 6.18 through 6.21. All of the

values plotted are averaged at the nodes and were generated using von Mises plasticity

theory. Similar to the tensile bar FEM’s, the variation between the three mesh densities

is very small except in the region close to the outer surface cylinder. After completion of

the convergence study, the medium mesh FEM was chosen as a compromise between

computation speed and accuracy.

75260

75280

75300

75320

75340

75360

75380

75400

75420

0 0.04 0.08 0.12 0.16 0.2

Coarse Mesh

Medium Mesh

Fine Mesh519

519.2

519.4

519.6

519.8

5200 1 2 3 4 5

Effe

ctiv

e S

tres

s, σ

eff (

psi)

Distance Across Bottom Symmetry Plane (in.)

Distance Across Bottom Symmetry Plane (mm)

Effe

ctiv

e S

tres

s, σ

eff (

Mpa

)

Figure 6.18. Effective Stress Across the Bottom Symmetry Plane of the Coarse, Medium, and Fine Mesh Smooth Compression Cylinder FEM's at Maximum Load

128

-25200

-25100

-25000

-24900

-24800

-24700

-24600

-24500

0 0.04 0.08 0.12 0.16 0.2

Coarse Mesh

Medium Mesh

Fine Mesh

-173.6

-172.8

-172

-171.2

-170.4

-169.6

0 1 2 3 4 5

Mea

n S

tres

s, σ

m (

psi)



Me

an S

tres

s, σ

m (

Mpa

)

Figure 6.19. Mean Stress Across the Bottom Symmetry Plane of the Coarse, Medium, and Fine Mesh Smooth Smooth Compression Cylinder FEM’s at Maximum Load

0

50

100

150

200

0 0.04 0.08 0.12 0.16 0.2

Coarse Mesh

Medium Mesh

Fine Mesh

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

Rad

ial S

tres

s, σ

rr (

psi)



Rad

ial S

tres

s, σ

rr (

Mpa

)

Figure 6.20. Radial Stress Across the Bottom Symmetry Plane of the Coarse, Medium, and Fine Mesh Smooth Compression Cylinder FEM’s at Maximum Load

129

0.0237

0.0238

0.0239

0.0240

0.0241

0.0242

0.0243

0.0244

0.0245

0 0.04 0.08 0.12 0.16 0.2

Coarse Mesh

Medium Mesh

Fine Mesh

0 1 2 3 4 5

Equ

iva

lent

Pla

stic

Str

ain,

εp

l eq



Figure 6.21. Equivalent Plastic Strain Across the Bottom Symmetry Plane of the Coarse, Medium, and Fine Mesh Smooth Compression Cylinder FEM’s at Maximum Load

Notched Round Bar Specimens

Axisymmetric finite element models of the notched round bar geometries were

created to simulate the NRB tensile and low cycle fatigue tests. The NRB finite element

models were composed of Q4 axisymmetric elements (type CAX4 in ABAQUS). Two

planes of symmetry were utilized as illustrated in Figure 6.22. A uniform displacement

was applied to the top nodes of the FEM’s for the loading boundary condition. For cyclic

loading, the displacement boundary condition was iteratively applied to ensure that the

maximum and minimum gage displacement for each cycle matched with the test

specimen maximum and minimum displacements. Six variations of the NRB finite

130


Figure 6.22. Schematic of Axisymmetric Model of a Notched Round Bar Specimen Utilizing Two Planes of Symmetry

element models were created by making the notch root radius, ρ, equal to 0.005, 0.010,

0.020, 0.040, 0.080, and 0.120 in., respectively.

Each variation of the NRB model was meshed with three levels of mesh

refinement. The coarse, medium, and fine meshes for each variation typically had 250,

500, and 1000 elements in the notch region, respectively. An illustration of the medium

mesh NRB FEM with ρ = 0.040 in. is given in Figure 6.23, and representative meshes in

the notch region for each level of mesh refinement are illustrated in Figures 6.24 through

6.26. The complete set of NRB finite element models is given in Appendix F.

Convergence studies were performed on two of the NRB geometries. The two

geometries chosen represent a case of severe constraint, ρ = 0.005 in., and a median

131

Figure 6.23. Medium Mesh FEM of NRB with ρ = 0.040 in.

132

Figure 6.24. Coarse Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in.

Figure 6.25. Medium Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in.

133

Figure 6.26. Fine Mesh FEM in the Notch Region of the NRB with ρ = 0.040 in.

constraint case, ρ = 0.040 in. Once again, the variation σeff, σm, σrr, and pleqε across the

notch region at failure load was plotted as a nodal average for the three mesh densities

using von Mises plasticity. The results of the convergence study for the NRB with ρ =

0.005 in. are given in Figures 6.27 through 6.30, and the results for the NRB with ρ =

0.040 in. are given in Figures 6.31 through 6.34. In all cases, the variation between the

three mesh densities is negligible until the outer surface of the notch is reached. After

completion of the convergence study, the medium mesh FEM’s were chosen for all notch

round bar geometries as a compromise between computation speed and accuracy.

134

20000

30000

40000

50000

60000

70000

80000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

150

200

250

300

350

400

450

500

550

0 0.5 1 1.5 2 2.5 3

Effe

ctiv

e S

tres

s, σ

eff (

psi)



Effe

ctiv

e S

tre

ss, σ

eff (

Mpa

)

Figure 6.27. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load

40000

50000

60000

70000

80000

90000

100000

110000

120000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

300

400

500

600

700

800

0 0.5 1 1.5 2 2.5 3

Mea

n S

tre

ss, σ

m (

psi)



Mea

n S

tres

s, σ

m (

Mpa

)

Figure 6.28. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load

135

0

10000

20000

30000

40000

50000

60000

70000

80000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

0

100

200

300

400

500

0 0.5 1 1.5 2 2.5 3

Rad

ial S

tres

s, σ

rr (

psi)



Rad

ial S

tres

s, σ

rr (

Mpa

)

Figure 6.29. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

0 0.5 1 1.5 2 2.5 3

Equ

ival

ent P

last

ic S

trai

n, ε

pl

eq



Figure 6.30. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.005 in. FEM’s at Failure Load

136

66000

68000

70000

72000

74000

76000

78000

80000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

460

480

500

520

540

0 0.5 1 1.5 2 2.5 3

Effe

ctiv

e S

tres

s, σ

eff (

psi)



Effe

ctiv

e S

tre

ss, σ

eff (

Mpa

)

Figure 6.31. Effective Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.040 in. FEM’s at Failure Load

30000

40000

50000

60000

70000

80000

90000

100000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh 240

320

400

480

560

640

0 0.5 1 1.5 2 2.5 3

Mea

n S

tre

ss, σ

m (

psi)



Mea

n S

tres

s, σ

m (

Mpa

)

Figure 6.32. Mean Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.040 in. FEM’s at Failure Load

137

0

10000

20000

30000

40000

50000

60000

70000

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

0

80

160

240

320

400

480

0 0.5 1 1.5 2 2.5 3

Rad

ial S

tres

s, σ

rr (

psi)



Rad

ial S

tres

s, σ

rr (

Mpa

)

Figure 6.33. Radial Stress Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.040 in. FEM’s at Failure Load

0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.025 0.05 0.075 0.1 0.125

Coarse Mesh

Medium Mesh

Fine Mesh

0 0.5 1 1.5 2 2.5 3

Equ

ival

ent P

last

ic S

trai

n, ε

pl

eq



Figure 6.34. Equivalent Plastic Strain Across the Neck of the Coarse, Medium, and Fine Mesh NRB with ρ = 0.040 in. FEM’s at Failure Load

138

Equal Arm Bend Specimen

The equal-arm bend finite element models were created using Q4 plane strain

elements (type CPE4 in ABAQUS) and one symmetry plane. The symmetry plane and

boundary conditions are illustrated in Figure 6.35. Plane strain elements were used

because the thickness to width ratio in the fillet region was approximately 5 to 1. The

load boundary condition was applied to the FEM by filling the hole for the loading-pin

with elements and applying a displacement to the node in the center of the loading pin

hole.

Three levels of mesh refinement were used in developing the equal-arm bend

FEM. The coarse, medium, and fine meshes consisted of 1079, 1388, and 2469 elements,

respectively. An illustration of the medium mesh FEM is given in Figure 6.36, and a

closer view of the medium mesh in the fillet region is shown in Figure 6.37. The

complete set of equal-arm bend FEM’s is given in Appendix F.

A convergence study was performed to determine the variation σeff, σm, σxx (the

stress in the x-direction), and pleqε across the fillet region at the first cycle maximum load

for the three mesh densities. The results of the convergence study are given in Figures

6.38 through 6.41. All of the values plotted are averaged at the nodes and were generated

using von Mises plasticity theory. The mean stress and equivalent plastic strain variation

is very small for all three mesh densities. The coarse mesh shows some divergence from

the other two meshes for σeff and σxx though. After completion of the convergence study,

the medium mesh FEM was chosen as a compromise between computation speed and

accuracy.

139

Nodal Displacement

Figure 6.35. Schematic of the Equal-Arm Bend Specimen Utilizing One Symmetry Plane

140

Figure 6.36. Medium Mesh Equal-Arm Bend FEM

141

Figure 6.37. Medium Mesh in the Fillet Region of the Equal-Arm Bend Specimen

0

50000

100000

150000

200000

0 0.03 0.06 0.09 0.12 0.15

Coarse Mesh

Medium Mesh

Fine Mesh

0

200

400

600

800

1000

1200

0 0.5 1 1.5 2 2.5 3 3.5

Effe

ctiv

e S

tres

s, σ

eff (

psi)

Distance Across Fillet Section (in.)

Distance Across Fillet Section (mm)

Effe

ctiv

e S

tres

s, σ

eff (

Mpa

)

Figure 6.38. Effective Stress Across the Fillet Section of the Coarse, Medium, and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load

142

-100000

-50000

0

50000

100000

150000

0 0.03 0.06 0.09 0.12 0.15

Coarse Mesh

Medium Mesh

Fine Mesh

-600

-400

-200

0

200

400

600

800

1000

0 0.5 1 1.5 2 2.5 3 3.5

Mea

n S

tre

ss, σ

m (

psi)



Mea

n S

tres

s, σ

m (

Mpa

)

Figure 6.39. Mean Stress Across the Fillet Section of the Coarse, Medium, and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load

-10000

0

10000

20000

30000

40000

50000

60000

70000

0 0.03 0.06 0.09 0.12 0.15

Coarse Mesh

Medium Mesh

Fine Mesh

0

100

200

300

400

0 0.5 1 1.5 2 2.5 3 3.5

Str

ess

in th

e x-

Dire

ctio

n, σ

xx (

psi

)



Str

ess

in th

e x-

Dire

ctio

n, σ

xx (

Mpa

)

Figure 6.40. Stress in the x-Direction Across the Fillet Section of the Coarse, Medium, and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load

143

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0 0.03 0.06 0.09 0.12 0.15

Coarse Mesh

Medium Mesh

Fine Mesh

0 0.5 1 1.5 2 2.5 3 3.5

Equ

iva

lent

Pla

stic

Str

ain,

εp

l eq



Figure 6.41. Equivalent Plastic Strain Across the Fillet Section of the Coarse, Medium, and Fine Mesh Equal-Arm Bend FEM’s at First Cycle Maximum Load

This chapter presented the development of the finite element models. First, a

description of the required material property inputs for the FEM’s was given, followed by

a discussion of the specific material property inputs for 2024-T851 and IN100. Next, a

general discussion of the finite element modeling procedure was discussed. Finally,

several test specimen geometries were presented, and specific details concerning the

creation of finite element models for each geometry were given. The next chapter

presents the finite element model results.

144

CHAPTER 7

FINITE ELEMENT MODEL RESULTS

This chapter presents the finite element model results, which is the final portion of

the analytical research program. First, results are presented to verify the accuracy of the

user developed constitutive model. Next, finite element results are compared to 2024-

T851 load-gage displacement test data for both monotonic and fatigue loadings. Finally,

FEA solutions are compared to load-gage displacement or load-microstrain test records

for the IN100 test program.

UMAT Program Verification

The user defined constitutive model (UMAT) was compared to several of

ABAQUS’s built-in material models to test the accuracy of the elastoplastic equations

and the corresponding FORTRAN code. The 2024-T851 material properties were used

in all the verification FEM’s. Two geometries were used in the verification tests, the

NRB with ρ = 0.040 in. and a smooth compression cylinder. The NRB specimen was

chosen to represent a specimen with a high hydrostatic stress influence. The smooth

compression cylinder was chosen as an example of a geometry with an approximately

uniaxial stress state.

The first test case was the monotonic loading of the NRB FEM using the

multilinear isotropic von Mises and multilinear isotropic Drucker-Prager built-in

145

constitutive models in ABAQUS. For this case, β = 1 (pure isotropic hardening) for the

combined multilinear hardening UMAT. The results of this test are shown in Figure 7.1,

and clearly show that the global responses of the UMAT and the built-in ABAQUS

models are identical.

The next test case was the cyclic loading of the NRB FEM using the bilinear

kinematic von Mises model in ABAQUS. As previously mentioned, ABAQUS does not

have a built-in model for the Drucker-Prager model with kinematic hardening or for the

von Mises model with multilinear kinematic hardening. For this case, only the first and

last values in the σ – εpl data table were used with the combined multilinear hardening

UMAT to simulate bilinear hardening, and β = 0 (pure kinematic hardening). The results

of this test are shown in Figure 7.2, which once again show that the global responses of

the UMAT and the built-in ABAQUS model are identical. The Drucker-Prager UMAT

curve is included in Figure 7.2 for comparison and follows the expected trend.

The final test case was to examine the response of the combined multilinear

hardening UMAT for varying β values. This was first done by conducting finite element

analyses of the uniaxial compression cylinder with β = 0 and β = 1. The results of this

test are given in Figure 7.3, and as expected both FEA’s give identical load–gage

displacement responses on the initial loading cycle. Next, cyclic finite element analyses

with the NRB with ρ = 0.040 in. were conducted with β = 0, 0.5, and 1. The results from

the first cycle are shown in Figure 7.4. The three FEM’s have practically identical load-

gage displacement responses until the negative loads are reached on the first unloading

146

0

1000

2000

3000

4000

5000

6000

7000

0 0.002 0.004 0.006 0.008 0.01

ABAQUS von Mises UMAT von MisesABAQUS Drucker-PragerUMAT Drucker-Prager

0

5000

10000

15000

20000

25000

30000

0 0.05 0.1 0.15 0.2 0.25

Load

, P (

lbs)

Gage Displacement, v (in.)

Load

, P (

N)


0.5 in. Gage Length

Figure 7.1. Comparison of the Built-In ABAQUS Models with Multilinear Isotropic Hardening and the Combined Multilinear Hardening UMAT for the Monotonic Loading of a NRB with ρ = 0.040 in.

147

-6000

-4000

-2000

0

2000

4000

-0.006 -0.004 -0.002 0 0.002 0.004 0.006

ABAQUS von Mises

UMAT von Mises

UMAT Drucker-Prager -24000

-16000

-8000

0

8000

16000

-0.15 -0.1 -0.05 0 0.05 0.1 0.15Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length

Figure 7.2. Comparison of the Built-In ABAQUS von Mises Model with Bilinear Kinematic Hardening and the Combined Multilinear Hardening UMAT for the Cyclic Loading of a NRB with ρ = 0.040 in.

148

0

2000

4000

6000

8000

10000

0 0.002 0.004 0.006 0.008 0.01

Drucker-Prager, β = 1

Drucker-Prager, β = 0

0

5000

10000

15000

20000

25000

30000

35000

40000

0 0.05 0.1 0.15 0.2 0.25Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length

Figure 7.3. Comparison of Drucker-Prager Combined Multilinear Hardening UMAT Solutions with different β Values for the Smooth Compression Specimen

149

-6000

-4000

-2000

0

2000

4000

6000

-0.006 -0.004 -0.002 0 0.002 0.004 0.006

Drucker-Prager UMAT, β = 1.0



-24000

-16000

-8000

0

8000

16000

24000

-0.15 -0.1 -0.05 0 0.05 0.1 0.15Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length

Figure 7.4. Comparison of Drucker-Prager Combined Multilinear Hardening UMAT Solutions with different β Values for the NRB with ρ = 0.040 in.

150

cycle. From this point on, the three solutions diverge in the expected manner with the β

= 0.5 solution falling between the pure isotropic and pure kinematic curves.

2024-T851 Results

The finite element analysis results for the 2024-T851 test geometries are

presented in this section. The finite element solutions are compared to load-gage

displacement results from Chapter 4. Both monotonic and low cycle fatigue loadings are

examined.

Smooth Tensile Bar Results

Load-gage displacement curves from the 0.35 in. diameter tensile bar FEM’s are

plotted along with a representative 2024-T851 tensile test record in Figure 7.5. because

the smooth tensile bar does not have a large hydrostatic pressure influence until necking,

it was expected that the von Mises and Drucker-Prager constitutive models would give

similar results. The von Mises curve overestimates the load for a given value of

displacement in the plastic region. The Drucker-Prager curve with a = 0.029 is lower

than the von Mises curve slightly overestimates the test data up to the maximum load.

The Drucker-Prager solution with a = 0.041 essentially matches the test data up to the

maximum load.

151

0

2000

4000

6000

8000

0 0.02 0.04 0.06 0.08 0.1

Specimen AT06von Mises FEADrucker-Prager FEA, a = 0.029Drucker-Prager FEA, a = 0.041

0

5000

10000

15000

20000

25000

30000

35000

0 0.5 1 1.5 2 2.5

Load

, P (

lbs)


Load

, P (

N)


1.0 in. Gage Length

Figure 7.5. Load-Gage Displacement Results for 2024-T851 Smooth Tensile Bar

152

It is not surprising that all of the FEM’s solutions overestimate the load-

displacement curve after the maximum load. As reported in Chapter 6, the σ – ε pl table

used by the FEM’s only matches the test record up to the maximum load and then is a

power law estimation of the material behavior. In addition, many of the smooth tensile

specimens failed with an angled fracture surface along a line of maximum shear.

Axisymmetric finite element models cannot accurately model failure along slip planes.

Instead, the FEM’s assume that the neck region continues to decrease in diameter

uniformly until final fracture.

Smooth Compression Cylinder Results

Load-gage displacement curves from the smooth compression cylinder FEM’s are

plotted along with a representative 2024-T851 compression test record in Figure 7.6. The

von Mises curve is a close match to the test record. Both Drucker-Prager curves

overestimate the load for a given value of displacement in the plastic region. In

compression the Drucker-Prager curves are higher than the von Mises curve because the

radius of the Drucker-Prager yield surface is larger than the radius of the von Mises yield

surface for negative mean stress values.

153

0

2000

4000

6000

8000

10000

0 0.002 0.004 0.006 0.008 0.01

Specimen AC03von Mises FEADrucker-Prager FEA, a = 0.029Drucker-Prager FEA, a = 0.041

0

5000

10000

15000

20000

25000

30000

35000

40000

0 0.05 0.1 0.15 0.2 0.25Lo

ad, P

(lb

s)


Load

, P (

N)


0.3 in. Gage Length

Figure 7.6. Load-Gage Displacement Results for 2024-T851 Smooth Compression Cylinder

154

Notched Tensile Bar Results

Representative load-gage displacement data from the NRB tensile tests are plotted

along with the NRB finite element solutions in Figures 7.7 through 7.12. The von Mises

FEA is a close match to the test data for the NRB with ρ = 0.005 in., but for the rest of

the NRB geometries, the von Mises solution overestimates the load for a given value of

gage displacement in the plastic region. Considering the gage displacements at specimen

failure, the von Mises FEM’s overestimate the failure loads approximately 2% (ρ = 0.005

in.) to 12% (ρ = 0.040 in.). The Drucker-Prager finite element solutions with a = 0.029

slightly underestimate the test data for the 3 NRB’s with smaller notch root radii (ρ =

0.005, 0.010, 0.020 in.). For the remaining three NRB geometries, the a = 0.029

solutions slightly overestimate the load for a given value of displacement in the plastic

region. Considering the gage displacements at specimen failure, the Drucker-Prager

FEM’s (a = 0.029) underestimate the failure loads by less than 1% (ρ = 0.005 in.) and

overestimate the failure loads a maximum of approximately 5% (ρ = 0.040 in.). The

Drucker-Prager finite element solutions with a = 0.041 underestimate the test data for the

3 NRB’s with smaller notch root radii. For the remaining three NRB geometries, the a =

0.041 solutions skim the top of the test data. Considering the gage displacements at

specimen failure, the Drucker-Prager FEM’s (a = 0.041) underestimate the failure loads

by a maximum of approximately 2% (ρ = 0.005 in.) and overestimate the failure loads a

maximum of approximately 2% (ρ = 0.040 in.).

155

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 502

von Mises FEA

Drucker-Prager FEA, a = 0.029


0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2

Load

, P (

lbs)


Load

, P (

N)


0.5 in. Gage Length

Figure 7.7. Load-Gage Displacement Results for NRB with ρ = 0.005 in.

156

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 102von Mises FEADrucker-Prager FEA, a =0.029Drucker-Prager FEA, a =0.041

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length


157

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 206

von Mises FEA



0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length


158

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 406von Mises FEADrucker-Prager FEA, a = 0.029Drucker-Prager FEA, a = 0.041

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length


159

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 806

von Mises FEA



0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length


160

0

1000

2000

3000

4000

5000

6000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

Specimen 126von Mises FEADrucker-Prager FEA, a = 0.029Drucker-Prager FEA, a = 0.041

0

5000

10000

15000

20000

25000

0 0.05 0.1 0.15 0.2Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length


161

At this point in the analyses, some observations can be made concerning the most

appropriate value for the Drucker-Prager constant, a. An a of 0.029 for 2024-T351 was

used in the author’s previous research [12], and this value produced results that

essentially matched the load-gage displacement response of a smooth tensile bar and

NRB’s with ρ = 0.005, 0.010, and 0.020 in. For this research, an a of 0.029 for 2024-

T851 also closely approximates the response of the NRB’s with ρ = 0.005, 0.010, and

0.020 in., but overestimates the load-gage displacement record for the smooth tensile bar

and the larger notch geometries. Using a of 0.041 produces results that essentially match

the test data for the smooth tensile bar and the NRB’s with ρ = 0.040, 0.080, and 0.120

in. Therefore, a range for a from 0.029 to 0.041 for 2024-T851 can be considered

reasonable.

Notched Round Bar Low Cycle Fatigue Results

In this section, the low cycle fatigue results from three different NRB geometries

(ρ = 0.040, 0.080, and 0.120 in.) are compared with finite element solutions. For all

cases, the load-gage displacement data for the first five cycles are given to illustrate the

development of the test specimen and FEA solution hysteresis loops. Also, data from a

later cycle is given to illustrate an approximately stable hysteresis response. For all the

FEM’s, the combined hardening parameter, β was iteratively chosen to best match the

first cycle hysteresis loop. As mentioned in the previous section, an a of 0.041 best

162

matched the tensile data for NRB’s with ρ = 0.040, 0.080, and 0.120 in. Therefore, an a

of 0.041 was used for all the Drucker-Prager low cycle fatigue FEA’s.

Representative load-gage displacement data from selected cycles of the LCF tests

of the NRB with ρ = 0.040 are plotted along with NRB low cycle fatigue finite element

solutions in Figures 7.13 through 7.18. A β of 0.57 was used for the von Mises FEA and

a β of 0.27 was used for the Drucker-Prager FEA to provide a best fit to the first cycle

unloading test data. For the first cycle, the von Mises solution overestimates the

maximum load by 6% but otherwise matches the test data well. The Drucker-Prager

solution closely matches the first cycle test data. For the second through fifth cycles, the

von Mises solution slightly overestimates the minimum load, but consistently

overpredicts the maximum load by 10% to 12%. Conversely, for the second through fifth

cycles, the Drucker-Prager solution closely estimates the maximum load, but consistently

overpredicts the minimum load by 4% to 10%. The results from the cycle before

specimen failure (ninth cycle) are shown in Figure 7.18. For this cycle, the von Mises

FEM overpredicts the maximum load by 16% and the minimum load by 7%. The

Drucker-Prager FEM almost matches the maximum test load, but overestimates the

minimum load by 17%. A complete set of the finite element results for the first nine

cycles of the NRB with ρ = 0.040 is given in Appendix G.

163

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)

1st Cycle, ρ = 0.040 in.


β = 0.27

β = 0.57

Figure 7.13. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

164

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)

2nd Cycle, ρ = 0.040 in.


β = 0.27

β = 0.57

Figure 7.14. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

165

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)

3rd Cycle, ρ = 0.040 in.


β = 0.27

β = 0.57

Figure 7.15. Third Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

166

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)

4th Cycle, ρ = 0.040 in.


β = 0.27

β = 0.57

Figure 7.16. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

167

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.57


168

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-30000

-20000

-10000

0

10000

20000

30000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.57

Figure 7.18. Ninth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.

169








von Mises solution closely matches minimum load, but consistently overpredicts the

maximum load by 10% to 12%. Conversely, for the second through fifth cycles, the

Drucker-Prager solution closely estimates the maximum load, but slightly overpredicts

the minimum load by up to 5%. The results from the tenth cycle are shown in Figure

7.24. For this cycle, the von Mises FEM overpredicts the maximum load by 17% and

approximately matches the minimum load. The Drucker-Prager FEM overpredicts the

maximum load by 8%, and overestimates the minimum load by 8%. After ten cycles, the

test data and FEA hysteresis loops are essentially stable, and therefore, no results are

given for cycles greater than ten. A complete set of finite element results for the first ten

cycles of the NRB with ρ = 0.080 is given in Appendix G.

170

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.52


171

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.52


172

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.52


173

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.52


174

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.27

β = 0.52


175

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.52

Figure 7.24. Tenth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.

176








von Mises solution closely matches minimum load, but consistently overpredicts the

maximum load by 9% to 11%. Conversely, for the second through fifth cycles, the

Drucker-Prager solution closely estimates the maximum and minimum loads. The

results from the tenth cycle are shown in Figure 7.30. For this cycle, the von Mises FEM

overpredicts the maximum load by 14% and approximately matches the minimum load.

The Drucker-Prager FEM overpredicts the maximum load by 6%, and overestimates the

minimum load by 3%. After ten cycles, the test data and FEA hysteresis loops are

essentially stable, and therefore, no results are given for cycles greater than ten. A

complete set of finite element results for the first ten cycles of the NRB with ρ = 0.120 is

given in Appendix G.

177

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50


178

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50


179

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50


180

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50


181

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50


182

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1Lo

ad, P

(lb

s)


Load

, P (

N)



β = 0.23

β = 0.50

Figure 7.30. Tenth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.

183

Inconel 100 Results

The finite element results for the IN100 test geometries are presented in this

section. The finite element solutions are compared to load-gage displacement or load-

microstrain results from Chapter 4. Both monotonic and low cycle fatigue loadings are

examined.

Smooth Tensile Bar Results

Load-gage displacement curves from the 0.25 in. diameter tensile bar FEM’s are

plotted along with a representative IN100 tensile test record in Figure 7.31. The von

Mises curve slightly overestimates the load for a given value of displacement in the

plastic region. The Drucker-Prager curve essentially matches the test data up to

approximately 0.075 in. gage displacement. It is not surprising that both FEM’s diverge

from the test data after approximately 0.075 in. gage displacement because the σ – ε pl

table used by the FEM’s also diverges from the test data at this point. In addition, both

FEM’s truncate the upper yield point due to the influence of the σ – ε pl table data.

184

0

2000

4000

6000

8000

10000

12000

0 0.02 0.04 0.06 0.08 0.1

Specimen 4A-TS-3

Von Mises FEA

Drucker-Prager FEA

0

10000

20000

30000

40000

50000

0 0.5 1 1.5 2 2.5Lo

ad, P

(lb

s)


Load

, P (

N)


0.5 in. Gage Length

Figure 7.31. Load-Gage Displacement Results for IN100 Smooth Tensile Bar

185

Smooth Compression Cylinder Results

Load-gage displacement curves from the smooth compression cylinder FEM’s are

plotted along with a representative IN100 compression test record in Figure 7.32. The

von Mises solution slightly underestimates the load for a given displacement in the

plastic region. Conversely, the Drucker-Prager solution slightly overpredicts the load for

a given displacement in the plastic region. After approximately 0.009 in. gage

displacement, both finite element solutions diverge from the test data.

186

0

4000

8000

12000

16000

20000

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014

Specimen 4A-CP-5

von Mises FEA

Drucker-Prager FEA

0

20000

40000

60000

80000

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35Lo

ad, P

(lb

s)


Load

, P (

N)


0.3 in. Gage Length

Figure 7.32. Load-Gage Displacement Results for IN100 Smooth Compression Cylinder

187

Equal-Arm Bend Low Cycle Fatigue Results

In this section, the low cycle fatigue data from Pratt and Whitney [73] for the

equal-arm bend test specimen is compared with finite element solutions. Data from all

three cycles of the proof test is presented. A β of zero was chosen for both FEM’s to best

match the first cycle hysteresis loop. Both large and small strain analysis was used for

the equal arm bend analyses due to the large bending component in the fillet.

Load-microstrain test data for the first cycle is compared with the first cycle finite

element small strain analysis solutions in Figure 7.33. The von Mises and Drucker-

Prager FEM’s produce very similar results due to the combination of tensile and

compressive bending stresses applied to the fillet region, which essentially negates the

hydrostatic pressure influence in the fillet region. For the first cycle, both finite element

solutions approximately match the test data until the negative loads are reached. Both

models overestimate the first cycle minimum load by 34%. The second and third cycle

equal arm bend small strain analysis results are given in Figures 7.34 and 7.35. The

results for the second and third cycles are very similar. For both cycles, the FEM’s

underpredict the load for a given strain value on the loading cycles, but approximately

match the maximum load. Also, both FEM’s closely match the unloading data for cycles

two and three.

188

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

1st Cycle

Figure 7.33. First Cycle Load-Microstrain Small Strain Analysis Results for the Equal-Arm Bend Specimen

189

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

2nd Cycle

Figure 7.34. Second Cycle Load-Microstrain Small Strain Analysis Results for the Equal-Arm Bend Specimen

190

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

3rd Cycle

Figure 7.35. Third Cycle Load-Microstrain Small Strain Analysis Results for the Equal-Arm Bend Specimen

191

Load-microstrain test data is compared with the finite element large strain

analysis solutions in Figures 7.36 through 7.38. The von Mises and Drucker-Prager

FEM’s results are very similar to the small strain analysis finite element results. The von

Mises and Drucker-Prager FEM’s both overpredict the maximum first cycle load by

approximately 4% and the minimum first cycle load by 34%. The second and third cycle

results follow the same trends as the second and third cycle small strain results. For both

cycles, the FEM’s underpredict the load for a given strain value on the loading cycles, but

slightly overpredict the maximum load. Also, both FEM’s closely match the unloading

data for cycles two and three.

192

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

1st Cycle

Figure 7.36. First Cycle Load-Microstrain Large Strain Analysis Results for the Equal-Arm Bend Specimen

193

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

2nd Cycle

Figure 7.37. Second Cycle Load-Microstrain Large Strain Analysis Results for the Equal-Arm Bend Specimen

194

-300

-200

-100

0

100

200

300

400

500

600

700

0 5000 10000 15000 20000 25000 30000 35000

Specimen Test Data

von Mises FEA

Drucker-Prager FEA

-1000

-500

0

500

1000

1500

2000

2500

3000

Load

, P (

lbs)

Microstrain, µε

Load

, P (

N)

3rd Cycle

Figure 7.38. Third Cycle Load-Microstrain Large Strain Analysis Results for the Equal-Arm Bend Specimen This chapter presented the finite element model results. First, results were

presented to verify the accuracy of the user developed constitutive model. Next, finite

element results were compared to 2024-T851 load-gage displacement test data for both

monotonic and fatigue loadings. Finally, FEA solutions were compared to load-gage

displacement or load-microstrain test records for the IN100 test program. The next

chapter presents the conclusions and recommendations for this research program.

195

CHAPTER 8

CONCLUSIONS AND RECOMMENDATIONS

The results presented in the previous chapters demonstrate that hydrostatic stress

plays an important role in the low cycle fatigue process. Most traditional analysis

methods ignore the transition from the first cycle hysteresis loop to a stable hysteresis

response. The overall conclusion of this research is that using a yield function that is

dependent on hydrostatic stress can significantly alter the predicted hysteresis response

of notched specimens, particularly for the first few cycles. Specifically, compared to the

von Mises solutions, the Drucker-Prager solutions more accurately predicted the behavior

of the 2024-T851 and IN100 test specimens for first few cycles. The following six

supporting conclusions lend credence to the overall conclusion.

Accurate material property records for 2024-T851 and IN100 were developed for

both tensile and compressive loadings. This information was then carefully translated

into finite element material property input data, which gave confidence in the validity of

the finite element analyses. In addition all of the 2024-T851 LCF tests were consistent

and proved useful for testing the LCF response of specimens with a high hydrostatic

stress influence. Therefore, the first supporting conclusion is that careful experimental

testing to accurately quantify material properties is a necessity when attempting to model

a specimen’s LCF behavior.

196

The development of the Drucker-Prager constitutive model with combined

multilinear hardening was essential to this research. The constitutive model was written

in FORTRAN as an ABAQUS user material subroutine (UMAT). By simply changing

the value of two input variables, a and β, numerous comparative analyses can be

performed. By varying the constant a, the analysis can be changed from classical von

Mises to highly pressure-dependent plasticity. Similarly, by varying the constant β, the

hardening behavior can be changed from pure kinematic to pure isotropic or any linear

combination thereof. Therefore, the second supporting conclusion is that the UMAT

developed for this research should prove useful as a building block for future studies in

LCF behavior.

For the tensile monotonic test cases, the von Mises finite element model results

overpredicted the loads up to 12% for a given value of plastic strain. Conversely, the

Drucker-Prager FEM results closely matched the monotonic tensile test data for judicious

choices of the Drucker-Prager constant. Therefore, the third supporting conclusion is that

the Drucker-Prager constitutive model is superior to the von Mises model for simulating

tensile monotonic test behavior.

For the monotonic compression test cases, it is difficult to say whether the von

Mises or Drucker-Prager solution more accurately simulates the compression specimen

behavior. The top of the FEM’s were loaded with an applied displacement in the

longitudinal direction but had no constraint in the radial direction. In an actual

compression test, frictional loads between the specimen and the platens reduce the radial

expansion of the ends of the specimen. As mentioned in Chapter 4, several researchers

197

[67,68,69] have demonstrated that lubricating the ends of the test specimens can shift the

load displacement record for a compression test significantly upward. Therefore, if the

same geometries were tested with proper lubrication on the ends, it is possible that the

compressive load-displacement test records would shift upward toward the Drucker-

Prager FEA solutions. Therefore, the fourth supporting conclusion is that it is unclear

whether the von Mises or Drucker-Prager model is superior for simulating monotonic

compressive behavior.

The NRB low cycle fatigue FEA’s produced mixed results. For the first 5 cycles,

the von Mises solutions consistently overpredicted the maximum loads from 6% to 12%

but closely predicted the minimum loads in general. Conversely, for the first 5 cycles,

the Drucker-Prager solutions approximately predicted the maximum loads but

overpredicted the minimum loads by up to 10%. After a stable hysteresis response is

achieved, neither FEM is clearly superior. For cycle 9 or 10, the von Mises FEM

predicted maximum loads were 14% to 17% greater than the test data, but reasonably

predicted minimum loads. For cycle 9 or 10, the Drucker-Prager FEM predicted

maximum loads up to 8% greater and minimum loads from 3% to 17% greater than the

test data. Overall, the Drucker-Prager FEM’s closely simulated the first five cycles.

Conversely, after a stable hysteresis response was reached, so much damage had been

done to the test specimens that the stress-strain response had changed. Therefore, it is

unreasonable to expect an FEM with constant material properties to accurately model this

damage accumulation process.

198

In general, the hysteresis loops produced by the NRB low cycle fatigue FEM’s

were wider than the test data hysteresis loops. This behavior may be partially explained

by comparing the monotonic and transitional cyclic stress-strain curves shown in Figure

4.26. The finite element solutions use the monotonic stress-strain data to estimate the

relationship between the stresses and strains. Conversely, the test specimens accumulate

damage over time, and therefore develop a different stress-strain response as represented

by the transitional cyclic curve in Figure 4.26. The FEM’s solutions must follow a larger

path dictated by the monotonic stress-strain response. Therefore, the fifth supporting

conclusion is that the Drucker-Prager model is superior to the von Mises model for

simulating the first few cycles of the NRB low cycle fatigue process, but afterwards

neither FEM is clearly better than the other.

Both the von Mises and the Drucker-Prager FEM’s closely simulated the equal-

arm bend three-cycle fatigue test. Both finite element models closely matched the first

cycle data until the negative loads were reached. The FEM’s also accurately predicted

the shape of the hysteresis loops and the maximum and minimum loads for the second

and third cycles. Therefore, the sixth supporting conclusion is that both the Drucker-

Prager model and the von Mises model performed equally well in simulating the equal-

arm bend three-cycle fatigue test.

The following recommendations for future research in the area of hydrostatic

stress effects in low cycle fatigue are offered.

199

1. A straightforward and economical method for determining the Drucker-Prager

constant, a needs to be developed. One possible method to obtain a would be

uniaxial compression testing with properly lubricated specimens.

2. The pressure-dependent Jacobian matrix for the existing UMAT needs to be

developed. Although the currently used Jacobian leads to convergence for most

FEM geometries, using a pressure-dependent Jacobian should increase the rate of

convergence and ensure stability for all FEM geometries.

3. The existing pressure-dependent UMAT needs to be modified to improve the

modeling capability after the first few cycles. Some possible modifications might

include using a hyperbolic Drucker-Prager model similar to the one in ABAQUS

for isotropic hardening or some other higher order theory to describe the pressure

dependence. Other modifications could simulate damage effects by modeling

void growth, or by making material properties such as E, a, and β functions of

damage accumulation or plastic strain.

4. Additional LCF data from notched specimens needs to be generated. Low cycle

fatigue tests and finite element analyses of IN100 NRB specimens need to be

performed. The test program could mirror the 2024-T851 program in this

research. NRB low cycle fatigue tests and finite element analyses of materials

with a larger differential between σys and σult, such as overaged 2024-T851 or

low-carbon steel, also need to be performed. These materials should allow LCF

tests to be performed on NRB’s with smaller notch root radii.

200

BIBLIOGRAPHY

201

1. Bridgman, P.W., “The Effect of Hydrostatic Pressure on the Fracture of Brittle Substances,” Journal of Applied Physics, Vol. 18, 1947, p. 246.

2. Hill, R., “The Mathematical Theory of Plasticity,” Clarendon Press, Oxford,

1950. 3. Lubliner, J., “Plasticity Theory,” Macmillan, New York, 1990. 4. Kohnke, Peter, ed., “ANSYS User’s Manual for Revision 5.0,” Volume IV:

Theory, Feb. 1994. 5. ABAQUS Theory Manual, Version 5.5, Hibbit, Karlsson, and Sorensen, Inc.,

1995. 6. Neuber, Heinz, “Theory of Notch Stresses: Principles for Exact Calculation of

Strength with Reference to Structural Form and Material,” U.S. Atomic Energy Commission, ACE-tr-4547, Translated from publication of Springer-Verlag, Berlin, Gottingen, Heidelberg, 1958.

7. Hertzberg, Richard W., “Deformation and Fracture Mechanics of Engineering

Materials,” John Wiley & Sons, Inc., New York, 1996. 8. Rice, J.R., and D.M. Tracey, “On the Ductile Enlargement of Voids in Triaxial

Stress Fields,” Journal of the Mechanics and Physics of Solids, Volume 17, 1969, pp. 201-217.

9. Gurson, A.L., “Continuum Theory of Ductile Rupture by Void Nucleation and

Growth: Part 1 – Yield Criteria and Flow Rules for Porous Ductile Media,” Journal of Engineering Materials and Technology, Volume 99, 1977, pp 2-15.

10. Wilson, Christopher D., “Fracture Toughness Testing with Notched Round Bars,”

A Dissertation Presented for the Doctor of Philosophy Degree, The University of Tennessee, Knoxville, August 1997.

11. Christopher D. Wilson, “A Critical Reexamination of Classical Metal Plasticity,”

Journal of Applied Mechanics, Vol. 69, January 2002, pp. 63-68. 12. Allen, Phillip A., “Hydrostatic Stress Effects in Metal Plasticity,” A Thesis

Presented for the Master of Science Degree in Mechanical Engineering, Tennessee Technological University, August 2000.

13. Mendelson, Alexander, “Plasticity: Theory and Application,” Krieger Publishing,

Florida, 1983, originally published by Macmillan, 1968.

202

14. Boresi, Arthur, R. Schmidt, and O. Sidebottom, “Advanced Mechanics of

Materials,” Fifth Edition, John Wiley & Sons, New York, 1993. 15. Wilson, Christopher D., “Hydrostatic Stress Effects on Yielding and Fatigue

Life,” NASA Summer Faculty Fellowship Reports, 1999. 16. Drucker, D.C., “Plasticity Theory, Strength-Differential Phenomenon, and

Volume Expansion in Metals and Plastics,” Metallurgical Transactions, Volume 4, 1972, pp. 667-673.

17. Spitzig, W.A., R.J. Sober, and O. Richmond, “Pressure Dependence of Yielding

and Associated Volume Expansion in Tempered Martensite,” ACTA Metallurgica, Volume 23, July 1975, pp. 885-893.

18. Drucker, D.C., and W. Prager, “Soil Mechanics and Plastic Analysis for Limit

Design,” Quarterly of Applied Mathematics, Volume 10, 1952, pp. 157-165. 19. Bridgman, P.W., “Studies in Large Plastic Flow and Fracture with Special

Emphasis on the Effects of Hydrostatic Pressure,” McGraw-Hill, New York, 1952.

20. Hu, L.W., and K.D. Pae, “Inclusion of the Hydrostatic Stress Component in

Formulation of the Yield Condition,” Journal of the Franklin Institute, Volume 275, 1963, pp. 491-502.

21. Marin, J., and L.W. Hu, “On the Validity of Assumptions Made in Theories of

Plastic Flow for Metals,” Transactions of the ASME, Aug. 1953, pp. 1181-1190. 22. Hu, L.W., and J.F. Bratt, “Effect of Tensile Plastic Deformation on Yield

Condition,” Journal of Applied Mechanics, Sept. 1958, p. 411. 23. Hu, L.W., “Plastic Stress-Strain Relations and Hydrostatic Stress,” Proceedings of

the Second Symposium on Naval Structural Mechanics: Plasticity, April 1960, pp.194-201.

24. Spitzig, W.A., R.J. Sober, and O. Richmond, “The Effect of Hydrostatic Pressure

on the Deformation Behavior of Maraging and HY-80 Steels and its Implications for Plasticity Theory,” Metallurgical Transactions A, Volume 7A, Nov. 1976, pp. 377-386.

203

25. Richmond, O., and W.A. Spitzig, “Pressure Dependence and Dilatancy of Plastic Flow,” International Union of Theoretical and Applied Mechanics Conference Proceedings, 1980, pp. 377-386.

26. Spitzig, W.A. and O. Richmond, “The Effect of Pressure on the Flow Stress of

Metals,” ACTA Metallurgica, Vol. 32, No. 3, 1984, pp. 457-463. 27. Chen, W.F., and X.L. Liu, 1990, “Limit Analysis in Soil Mechanics,” Elsevier,

New York. 28. Dieter, G.E., “Mechanical Metallurgy,” Third Edition, McGraw-Hill, New York,

1976. 29. Tvergaard, V., “On Localization in Ductile Materials Containing Spherical

Voids,” International Journal of Fracture, Volume 18, 1982, pp. 237-252. 30. Anderson, T.L., “Fracture Mechanics Fundamentals and Applications,” Second

Edition, CRC Press, Boca Raton, 1995. 31. Henry, B.S. and A.R. Luxmoore, “The Stress Triaxiality Constraint and the Q-

Value as a Ductile Fracture Parameter,” Engineering Fracture Mechanics, Vol. 57, No. 4, 1997, pp. 375-390.

32. Giannakopoulos, A.E. and P.L. Larsson, “Analysis of Pyramid Indentation of

Pressure-Sensitive Hard Metals and Ceramics,” Mechanics of Materials, Vol. 25, 1997, pp. 1-35.

33. Larsson, Per-Lennart, and A.E. Giannakopoulos, “Tensile Stresses and Their

Implication to Cracking at Pyramid Indentation of Pressure-Sensitive Hard Metals and Ceramics,” Materials Science and Engineering A254, 1998, 268-281.

34. Lu, Tian Jian, “Stress and Strain Evolution in Cast Refractory Blocks During

Cooling,” Journal of the American Ceramic Society, Vol. 81, No. 4, 1998, pp. 917-925.

35. Al-Abduljabbar, A. and J. Pan, “Plane-Strain Near-Tip Fields for Wedge-Shaped

Notches in Pressure-Sensitive Drucker-Prager Materials,” Proceedings of the ASME Aerospace Division, AD-Vol. 52, 1996, pp. 175-188.

36. Al-Abduljabbar, A. and J. Pan, “Effects of Pressure Sensitivity and Notch

Geometry on Notch-Tip Fields,” Polymer Engineering and Science, Vol. 38, No. 7, July 1998, pp. 1031-1038.

204

37. Chang, W.J. and J. Pan, “Effects of Yield Surface Shape and Round-Off Vertex on Crack-Tip Fields for Pressure-Sensitive Materials,” International Journal of Solids and Structures, Vol. 34, No. 25, September 1997, pp. 3291-3320.

38. Al-Abduljabbar, and J. Pan, “Effects of Pressure Sensitivity on the η Factor and

the J Integral Estimation for Compact Tension Specimens,” Journal of Materials Science, Vol. 34, 1999, pp. 4321-4328.

39. Al-Abduljabbar, A. and J. Pan, “Numerical Analysis of Notch-Tip Fields in

Rubber-Modified Epoxies,” Polymer Engineering and Science, Vol. 39, No. 4, April 1999, pp. 663-675.

40. Gil, Christopher M., C.J. Lissenden, and B.A. Lerch, “Yield of Inconel 718 by

Axial-Torsional Loading at Temperatures up to 649°C,” Journal of Testing and Evaluation, Vol. 27, No. 5, Sept. 1999, pp. 327-336.

41. Dufailly, J. and J. Lemaitre, “Modeling Very Low Cycle Fatigue,” International

Journal of Damage Mechanics, Vol. 4, April 1995. 42. Morris, J.W., S.J. Hardy, A.W. Lees, and J.T. Thomas, “Cyclic Behavior

Concerning the Response of Material Subjected to Tension Leveling,” International Journal of Fatigue, Vol. 22, No. 2, pp. 93-100, 2000.

43. Kuroda, Masatoshi, S. Yamanaka, J. Komotori, and M. Shimizu, “Consideration

on Dominant Factors for Extremely Low Cycle Fatigue Properties,” Technology Reports of the Osaka University, Vol. 49, No. 2355, pp. 147-154, Oct. 1999.

44. Chell, G.C., R.C. McClung, and C.J. Kuhlman, “Guidelines for Proof Test

Analysis,” NASA/CR–1999-209427, July, 1999. 45. McClung, R.C., G.C. Chell, and H.R. Milwater, “A Comparison of Single-Cycle

Versus Multiple-Cycle Proof Testing Strategies,” NASA/CR–1999-209426, July, 1999.

46. Bannatine, Julie A., J.J. Comer, and J.L Handrock, “Fundamentals of Metal

Fatigue Analysis,” Prentice Hall, New Jersey, 1990. 47. Shigley, Joseph, E., and C.R. Mischke, “Mechanical Engineering Design,” Fifth

Edition, McGraw-Hill, New York, 1989. 48. Morrow, J., “Fatigue Design Handbook”, Advances in Engineering, Vol. 4,

American Society for Testing and Materials, Philadelphia, pp. 3-36, 1970.

205

49. Manson, S.S. and G.R. Halford, “Practical Implementation of the Double Linear Damage Rule and Damage Curve Approach for Treating Cumulative Fatigue Damage,” International Journal of Fracture, Vol. 17, No. 2, pp.169-172, 1981.

50. Smith, K.N., P. Watson, and T.H. Topper, “A Stress-Strain Function for the

Fatigue of Metals,” J. Mater., Vol. 5, No. 4, pp. 767-778, 1970. 51. You, Bong-Ryul and Soon-Bok Lee, “A Critical Review on Multiaxial Fatigue

Assessments of Metals,” International Journal of Fatigue, Vol. 18, No. 4, 1996, pp.235-244.

52. Mowbray, D. F., “A Hydrostatic Stress-Sensitive Relationship for Fatigue Under

Biaxial Stress Conditions,” Journal of Testing and Evaluation, Vol. 8, No. 1, 1980, pp. 3-8.

53. Kalluri, S. and P.J. Bonacuse, “Advances in Multiaxial Fatigue,” ASTM STP

1191, American Society for Testing and Materials, PA, 1993, pp. 133. 54. Sines, George and G. Ohgi, “Fatigue Criteria Under Combined Stresses or

Strains,” Journal of Engineering Materials and Technology, Vol. 103, 1981, pp. 82-90.

55. Lefebvre, D., “Hydrostatic Pressure Effect on Life Prediction in Biaxial Low-

Cycle Fatigue,” Biaxial and Multiaxial Fatigue, MEP, London,1989. 56. Brown, M.W. and K.J. Miller, “A Theory for Fatigue Failure Under Multiaxial

Stress-Strain Conditions,” Proceedings of the Institution of Mechanical Engineers, Vol. 187, No. 65, 1973, pp. 745-755.

57. Lohr, R.D. and E.G. Ellison, “A Simple Theory for Low Cycle Multiaxial

Fatigue,” Fatigue of Engineering Materials and Structures, Vol. 3, 1980, pp. 1-17.

58. Avallone, Eugene, Ed., “Marks’ Standard Handbook for Mechanical Engineers,”

Tenth Edition, McGraw-Hill, Boston, 1996. 59. Pratt and Whitney, “Alternate Turbopump Development (ATD) Materials

Manual,” December, 1991. 60. MSC Software Corporation, "Patran Users Manual", Version 9.0, 2000. 61. Blacker, T.D., “FASTQ Users Manual Version 1.2,” Sandia Report SAND88-

1326, UC-705, Sandia National Laboratories, June 1988.

206

62. American Society of Testing and Materials, E1012 “Standard Practice for

Verification of Specimen Alignment Under Tensile Loading,” 1999 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1999.

63. American Society of Testing and Materials, E132 “Standard Test Method for

Poisson’s Ratio at Room Temperature,” 1997 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1997.

64. American Society of Testing and Materials, E111 “Standard Test for Young’s

Modulus, Tangent Modulus, and Chord Modulus,” 1997 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1997.

65. American Society of Testing and Materials, E8 “Standard Test Methods for

Tension Testing of Metallic Materials,” 1998 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1998.

66. American Society of Testing and Materials, E9 “Standard Test Methods of

Compression Testing of Metallic Materials at Room Temperature,” 1989 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1989.

67. Chait, R. and C.H. Curll, “Evaluating engineering Alloys in Compression,”

Recent Developments in Mechanical Testing, ASTM STP 608, American Society for Testing and Materials, 1976, pp. 3-19.

68. Hsü, T.C., “A Study of the Compression Test for Ductile Materials,” Materials

Research and Standards, December, 1969. 69. Kuhn, Howard A., “Uniaxial Compression Testing,” ASM Handbook Vol. 8,

Mechanical Testing and Evaluation, 2000. 70. American Society of Testing and Materials, E602 “Standard Test Method for

Sharp-Notch Tension Testing with Cylindrical Specimens,” 1991 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1991.

71. American Society of Testing and Materials, E606 “Standard Practice for Strain

Controlled Fatigue Testing,” 1998 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1998.

72. MTS Systems Corporation, “MTS TestWare® 759 Low Cycle Fatigue Test

Operators Guide,” MTS Document # 115721-03A, Minneapolis, MN, 1990. 73. Pratt and Whitney, Report Number MME 42030, Nov. 1999.

207

74. Manson, S.S., “Aerospace Structural Metals Handbook,” Syracuse University

Research Institute, Mechanical Properties Data Center, Mich. 1988. 75. Aravas, N., “On the Numerical Integration of a Class of Pressure-Dependent

Plasticity Models,” International Journal for Numerical Methods in Engineering, Vol. 24, 1987, pp. 1395-1416.

76. Aravas, N., “Use of Pressure-Dependent Plasticity Models in ABAQUS,”

ABAQUS Users’ Conference, Newport, Rhode Island, May 1996. 77. ABAQUS Training Course Notes, “Writing User Subroutines with ABAQUS,”

Hibbitt, Karlsson, & Sorensen, Inc., 2001. 78. Ortiz, M. and E.P. Popov, “Accuracy and Stability of Integration Algorithms for

Elastoplastic Constitutive Relations,” International Journal for Numerical Methods in Engineering, Vol. 21, 1985, pp. 1561-1576.

79. Taylor, L.M. and D.P. Flanagan, “PRONTO 2D: A Two-Dimensional Transient

Solid Dynamics Program,” Sandia Report, SAND86-0594, UC-32, April 1998. 80. American Society of Testing and Materials, E855 “Standard Test Methods for

Bend Testing of Metallic Flat Materials for Spring Applications Involving Static Loading,” 1990 Annual Book of ASTM Standards, Vol. 03.01, Philadelphia, 1990.

208

APPENDICES

209

APPENDIX A – TENSILE AND COMPRESSION TEST DATA

210

Table A.1. 2024-T851 L Direction Smooth Tensile Test Data

Specimen Number

Gage Diameter, in. (mm)

Gage Length, in. (mm)

Young’s Modulus,

103 ksi (GPa)


ksi (MPa)



AT03 0.3440 (8.738) 1.0 (25.4) 10.6 (73) 67 (462) 76 (524) 0.0837 AT04 0.3440 (8.738) 1.0 (25.4) 10.6 (73) 68 (468) 77 (531) 0.0902 AT05 0.3440 (8.738) 1.0 (25.4) 10.5 (72) 69 (476) 77 (531) 0.0877 AT06 0.3430 (8.712) 1.0 (25.4) 10.5 (72) 68 (468) 77 (531) 0.0850 AT07 0.3440 (8.738) 1.0 (25.4) 10.5 (72) 67 (462) 76 (524) 0.0818

Table A.2. 2024-T851 L-T Direction Smooth Tensile Test Data

Specimen Number



Young’s Modulus,

103 ksi (GPa)


ksi (MPa)



BT01 0.3440 (8.738) 1.0 (25.4) 10.8 (74) 67 (462) 72 (496) 0.0772 BT02 0.3440 (8.738) 1.0 (25.4) 10.6 (73) 67 (462) 72 (496) 0.0779 BT03 0.3430 (8.712) 1.0 (25.4) 10.7 (74) 66 (455) 72 (496) 0.0799 BT04 0.3440 (8.738) 1.0 (25.4) 10.7 (74) 67 (462) 76 (524) 0.0815 BT05 0.3430 (8.712) 1.0 (25.4) 10.6 (73) 67 (462) 76 (524) 0.0783

Table A.3. 2024-T851 L Direction Smooth Compression Test Data

Specimen Number



Young’s Modulus,

103 ksi (GPa)


ksi (MPa) AC01 0.3740 (9.500) 0.3 (7.62) 11.2 (77) 67 (462) AC02 0.3740 (9.500) 0.3 (7.62) 10.9 (75) 68 (468) AC03 0.3750 (9.525) 0.3 (7.62) 11.1 (76) 68 (468) AC04 0.3750 (9.525) 0.3 (7.62) 11.1 (76) 68 (468) AC05 0.3750 (9.525) 0.3 (7.62) 11.2 (77) 68 (468)

Table A.4. 2024-T851 L-T Direction Smooth Compression Test Data

Specimen Number



Young’s Modulus,

103 ksi (GPa)


ksi (MPa) BC01 0.3750 (9.525) 0.3 (7.62) 11.3 (78) 68 (468) BC02 0.3750 (9.525) 0.3 (7.62) 11.5 (79) 68 (468) BC03 0.3750 (9.525) 0.3 (7.62) 11.5 (79) 67 (462) BC04 0.3750 (9.525) 0.3 (7.62) 11.1 (76) 68 (468) BC05 0.3750 (9.525) 0.3 (7.62) 11.1 (76) 68 (468)

211

Table A.5. IN100 Smooth Tensile Test Data

Specimen Number



Young’s Modulus,

103 ksi (GPa)

Upper Yield Strength, ksi

(MPa)


ksi (MPa)



4A-TS-1 0.2503 (6.357) 1.0 (25.4)* 31.6 (218) 170 (1172) 163 (1124) 250 (1724)** *** 4A-TS-2 0.2474 (6.284) 1.0 (25.4) 31.7 (219) 172 (1186) 165 (1138) 234 (1613) *** 4A-TS-3 0.2470 (6.274) 0.5 (12.7) 31.9 (220) 173 (1193) 168 (1158) 234 (1613) 0.2326 4A-TS-4 0.2473 (6.281) 0.5 (12.7) 32.0 (221) 173 (1193) 169 (1165) 235 (1620) 0.2262 7A-TS-1 0.2500 (6.350) 1.0 (25.4)* 31.4 (217) 169 (1165) 162 (1117) 290 (1999)** *** 7A-TS-2 0.2475 (6.286) 0.5 (12.7) 31.6 (218) 172 (1186) 169 (1165) 235 (1620) 0.2185 7A-TS-3 0.2473 (6.281) 0.5 (12.7) 32.4 (223) 173 (1193) 169 (1165) 234 (1613) 0.2461 7A-TS-4 0.2475 (6.286) 0.5 (12.7) 32.0 (221) 172 (1186) 169 (1165) 236 (1627) 0.2263

Notes: * 1 in. gage length used on initial loading. 0.5 in. gage length used on reloading. ** Reported ultimate strength is value obtained from specimen reloading. These values were

neglected in the calculation of the average ultimate strength. *** Values not reported due to saturation of 1 in. gage length extensometer before fracture. Table A.6. IN100 Smooth Compression Test Data

Specimen Number



Young’s Modulus,

103 ksi (GPa)

Upper Yield Strength, ksi

(MPa)


ksi (MPa) 4A-CP-4 0.3750 (9.525) 0.3 (7.62) 32.4 (223) 172 (1186) 168 (1158) 4A-CP-5 0.3750 (9.525) 0.3 (7.62) 32.1 (221) 172 (1186) 167 (1151) 4A-CP-6 0.3750 (9.525) 0.3 (7.62) 32.8 (226) 172 (1186) 167 (1151) 7A-CP-3 0.3750 (9.525) 0.3 (7.62) 32.6 (225) 172 (1186) 168 (1158) 7A-CP-4 0.3750 (9.525) 0.3 (7.62) 32.9 (227) 171 (1179) 167 (1151) 7A-CP-5 0.3750 (9.525) 0.3 (7.62) 32.7 (225) 172 (1186) 167 (1151)

212

APPENDIX B – NRB LOW CYLCLE FATIGUE TEST PLOTS

213

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.1. First Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.2. Second Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

214

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.3. Third Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.4. Fourth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

215

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.5. Fifth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Lo

ad

, P

(lb

s)


Lo

ad

, P

(N

)



Figure B.6. Sixth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

216

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.7. Seventh Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.8. Eighth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

217

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 402

Specimen 403

Specimen 404

Specimen 405-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.9. Ninth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Lo

ad

, P

(lb

s)


Lo

ad

, P

(N

)




218

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




219

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




220

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.15. Tenth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Lo

ad

, P

(lb

s)


Lo

ad

, P

(N

)



Figure B.16. Fifteenth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.

221

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 802

Specimen 803

Specimen 804

Specimen 805-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.17. Twentieth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




222

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




223

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)




-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Lo

ad

, P

(lb

s)


Lo

ad

, P

(N

)




224

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.23. Tenth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.24. Twentieth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.

225

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004

Specimen 122

Specimen 123

Specimen 124

Specimen 125-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



Figure B.25. Thirtieth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.

226

APPENDIX C – ABAQUS UMAT PROGRAMS

227

ABAQUS UMAT for Pressure-Dependent Plasticity with Combined Bilinear Isotropic and Bilinear Kinematic Hardening SUBROUTINE UMAT(STRESS,STATEV,DDSDDE,SSE,SPD,SCD, 1 RPL,DDSDDT,DRPLDE,DRPLDT,STRAN,DSTRAN, 2 TIME,DTIME,TEMP,DTEMP,PREDEF,DPRED,CMNAME,NDI,NSHR,NTENS, 3 NSTATV,PROPS,NPROPS,COORDS,DROT,PNEWDT,CELENT, 4 DFGRD0,DFGRD1,NOEL,NPT,KSLAY,KSPT,KSTEP,KINC) C INCLUDE ’ABA_PARAM.INC’ C CHARACTER*80 CMNAME DIMENSION STRESS(NTENS),STATEV(NSTATV), 1 DDSDDE(NTENS,NTENS),DDSDDT(NTENS),DRPLDE(NTENS), 2 STRAN(NTENS),DSTRAN(NTENS),TIME(2),PREDEF(1),DPRED(1), 3 PROPS(NPROPS),COORDS(3),DROT(3,3), 4 DFGRD0(3,3),DFGRD1(3,3) C C LOCAL ARRAYS C --------------------------------------------------------- C EELAS - ELASTIC STRAINS C EPLAS - PLASTIC STRAINS C AKPHA - SHIFT TENSOR C FLOW - DIRECTION OF PLASTIC FLOW C OLDS - STRESS AT START OF INCREMENT C OLDPL - PLASTIC STRAINS AT START OF INCREMENT C --------------------------------------------------------- C DIMENSION EELAS(6),EPLAS(6),ALPHA(6),FLOW(6),OLDS(6),OLDPL(6) C DOUBLE PRECISION EMOD,ENU,A,BETA,I1,SYIEL0,HARD C PARAMETER (ZERO=0.D0,ONE=1.D0,TWO=2.D0,THREE=3.D0,SIX=6.D0, 1 ENUMAX=.4999D0,TOLER=1.0D-6) C C ----------------------------------------------------------- C UMAT FOR ISOTROPIC ELASTICITY AND DRUCKER-PRAGER PLASTICITY C WITH BILINEAR ISOTROPIC AND KINEMATIC HARDENING C CANNOT BE USED FOR PLANE STRESS C ----------------------------------------------------------- C PROPS(1) - E C PROPS(2) - NU C PROPS(3) - A, DRUCKER-PRAGER CONSTANT, LET A=ZERO FOR C CLASSICAL VON MISES PLASTICITY C PROPS(4) - SYIELD C PROPS(5) - HARD C PROPS(6) - BETA, SCALAR PARAMETER TO DETERMINE C PERCENTAGE OF EACH TYPE OF HARDENING C 0.0 <= BETA <= 1.0 C -----------------------------------------------------------

228

C IF (NDI.NE.3) THEN WRITE(6,1) 1 FORMAT(//,30X,’***ERROR - THIS UMAT MAY ONLY BE USED FOR ’, 1 ’ELEMENTS WITH THREE DIRECT STRESS COMPONENTS’) ENDIF C C ELASTIC PROPERTIES C EMOD=PROPS(1) ENU=MIN(PROPS(2),ENUMAX) EBULK3=EMOD/(ONE-TWO*ENU) EG2=EMOD/(ONE+ENU) EG=EG2/TWO EG3=THREE*EG ELAM=(EBULK3-EG2)/THREE EBULK=EBULK3/THREE A=PROPS(3) SYIEL0=PROPS(4) HARD=PROPS(5) BETA=PROPS(6) C C ELASTIC STIFFNESS C DO K1=1,NDI DO K2=1,NDI DDSDDE(K2,K1)=ELAM END DO DDSDDE(K1,K1)=EG2+ELAM END DO DO K1=NDI+1,NTENS DDSDDE(K1,K1)=EG END DO C RECOVER ELASTIC AND PLASTIC STRAINS AND AND SHIFT TENSOR AND ROTATE C NOTE: USE CODE 1 FOR (TENSOR) STRESS, CODE 2 FOR (ENGINEERING) STRAIN C CALL ROTSIG(STATEV( 1),DROT,EELAS,2,NDI,NSHR) CALL ROTSIG(STATEV( NTENS+1),DROT,EPLAS,2,NDI,NSHR) CALL ROTSIG(STATEV(2*NTENS+1),DROT,ALPHA,1,NDI,NSHR) EQPLAS=STATEV(3*NTENS+1) C C SAVE STRESS AND PLASTIC STRAINS AND C CALCULATE PREDICTOR STRESS AND ELASTIC STRAIN C DO K1=1,NTENS OLDS(K1)=STRESS(K1) OLDPL(K1)=EPLAS(K1) EELAS(K1)=EELAS(K1)+DSTRAN(K1) DO K2=1,NTENS STRESS(K2)=STRESS(K2)+DDSDDE(K2,K1)*DSTRAN(K1) END DO END DO

229

C C CALCULATE EQUIVALENT VON MISES STRESS C SMISES=(STRESS(1)-ALPHA(1)-STRESS(2)+ALPHA(2))**2 1 +(STRESS(2)-ALPHA(2)-STRESS(3)+ALPHA(3))**2 1 +(STRESS(3)-ALPHA(3)-STRESS(1)+ALPHA(1))**2 DO K1=NDI+1,NTENS SMISES=SMISES+SIX*(STRESS(K1)-ALPHA(K1))**2 END DO SMISES=SQRT(SMISES/TWO) SHYDRO=(STRESS(1)+STRESS(2)+STRESS(3))/THREE I1=SHYDRO*THREE C CALCULATE EQUIVALENT DRUCKER-PRAGER STRESS C EQUATION (5.10) DPYIEL=SMISES+A*I1 C C CALCULATE PREVIOUS YIELD VALUE SYIELD=SYIEL0+HARD*EQPLAS*BETA C C DETERMINE IF ACTIVELY YIELDING C IF (DPYIEL.GT.(ONE+TOLER)*SYIELD) THEN C C IF ACTIVELY YIELDING C SEPARATE THE HYDROSTATIC FROM THE DEVIATORIC STRESS C CALCULATE THE FLOW DIRECTION C EQUATION (5.14) C DO K1=1,NDI FLOW(K1)=(STRESS(K1)-ALPHA(K1)-SHYDRO)/SMISES END DO DO K1=NDI+1,NTENS FLOW(K1)=(STRESS(K1)-ALPHA(K1))/SMISES END DO C C SOLVE FOR EQUIVALENT PLASTIC STRAIN INCREMENT C EQUATION (5.46) C DEQPL=(SMISES-SYIELD+A*I1)/(THREE*EBULK3*A*A+EG3+HARD) C SOLVE FOR UPDATED YIELD STRESS SYIELD=SYIEL0+HARD*(EQPLAS+DEQPL)*BETA C CALCULATE NEEDED CONSTANTS C EQUATIONS (5.42) THROUGH (5.45) P=(-EBULK*A*SYIELD-SHYDRO*EG+EBULK*A*SMISES)/(EBULK3*A*A+EG) Q=(EBULK3*A*A*SMISES+SYIELD*EG-A*I1*EG)/(EBULK3*A*A+EG) DEP=(A*(SMISES-SYIELD+A*I1))/(EBULK3*A*A+EG) DEQ=(SMISES-SYIELD+A*I1)/(THREE*EBULK3*A*A+EG3) C C UPDATE STRESS, ELASTIC AND PLASTIC STRAINS AND SHIFT TENSOR C DO K1=1,NDI C EQUATION (5.90)

230

ALPHA(K1)=ALPHA(K1)+HARD*FLOW(K1)*DEQPL*(ONE-BETA) C CHANGE IN PLASTIC STRAIN FROM EQUATION (5.47) EPLAS(K1)=EPLAS(K1)+DEP/THREE+THREE*DEQ*FLOW(K1)/TWO EELAS(K1)=EELAS(K1)-(DEP/THREE+THREE*DEQ*FLOW(K1)/TWO) C EQUATION (5.48) STRESS(K1)=ALPHA(K1)-P+Q*FLOW(K1) END DO DO K1=NDI+1,NTENS C EQUATION (5.90) ALPHA(K1)=ALPHA(K1)+HARD*FLOW(K1)*DEQPL*(ONE-BETA) C CHANGE IN PLASTIC STRAIN FROM EQUATION (5.47) EPLAS(K1)=EPLAS(K1)+THREE*DEQ*FLOW(K1) EELAS(K1)=EELAS(K1)-THREE*DEQ*FLOW(K1) C EQUATION (5.48) STRESS(K1)=ALPHA(K1)+Q*FLOW(K1) END DO C STORE UPDATED VALUE OF EQUIVALENT PLASTIC STRAIN EQPLAS=EQPLAS+DEQPL C C CALCULATE THE PLASTIC DISSIPATION C SPD=ZERO DO K1=1,NTENS SPD=SPD+(STRESS(K1)+OLDS(K1))*(EPLAS(K1)-OLDPL(K1))/TWO END DO C C FORMULATE THE JACOBIAN (MATERIAL TANGENT) C FIRST CALCULATE THE EFFECTIVE MODULI C EFFG=EG*(SYIELD+HARD*DEQPL*(ONE-BETA))/SMISES EFFG2=TWO*EFFG EFFG3=THREE*EFFG EFFLAM=(EBULK3-EFFG2)/THREE EFFHRD=EG3*HARD/(EG3+HARD)-EFFG3 DO K1=1,NDI DO K2=1,NDI DDSDDE(K2,K1)=EFFLAM END DO DDSDDE(K1,K1)=EFFG2+EFFLAM END DO DO K1=NDI+1,NTENS DDSDDE(K1,K1)=EFFG END DO DO K1=1,NTENS DO K2=1,NTENS DDSDDE(K2,K1)=DDSDDE(K2,K1)+EFFHRD*FLOW(K2)*FLOW(K1) END DO END DO ENDIF C C STORE ELASTIC STRAINS, PLASTIC STRAINS AND SHIFT TENSOR C IN STATE VARIABLE ARRAY

231

C DO K1=1,NTENS STATEV(K1)=EELAS(K1) STATEV(K1+NTENS)=EPLAS(K1) STATEV(K1+2*NTENS)=ALPHA(K1) END DO STATEV(3*NTENS+1)=EQPLAS C RETURN END

232

ABAQUS UMAT for Pressure-Dependent Plasticity with Combined Multilinear Isotropic and Multilinear Kinematic Hardening SUBROUTINE UMAT(STRESS,STATEV,DDSDDE,SSE,SPD,SCD, 1 RPL,DDSDDT,DRPLDE,DRPLDT,STRAN,DSTRAN, 2 TIME,DTIME,TEMP,DTEMP,PREDEF,DPRED,CMNAME,NDI,NSHR,NTENS, 3 NSTATV,PROPS,NPROPS,COORDS,DROT,PNEWDT,CELENT, 4 DFGRD0,DFGRD1,NOEL,NPT,KSLAY,KSPT,KSTEP,KINC) C INCLUDE ’ABA_PARAM.INC’ C CHARACTER*80 CMNAME DIMENSION STRESS(NTENS),STATEV(NSTATV), 1 DDSDDE(NTENS,NTENS),DDSDDT(NTENS),DRPLDE(NTENS), 2 STRAN(NTENS),DSTRAN(NTENS),TIME(2),PREDEF(1),DPRED(1), 3 PROPS(NPROPS),COORDS(3),DROT(3,3), 4 DFGRD0(3,3),DFGRD1(3,3) C C LOCAL ARRAYS C --------------------------------------------------------- C EELAS - ELASTIC STRAINS C EPLAS - PLASTIC STRAINS C AKPHA - SHIFT TENSOR C FLOW - DIRECTION OF PLASTIC FLOW C OLDS - STRESS AT START OF INCREMENT C OLDPL - PLASTIC STRAINS AT START OF INCREMENT C --------------------------------------------------------- C DIMENSION EELAS(6),EPLAS(6),ALPHA(6),FLOW(6),OLDS(6),OLDPL(6), 1 HARD(3) C DOUBLE PRECISION I1 C PARAMETER (ZERO=0.D0,ONE=1.D0,TWO=2.D0,THREE=3.D0,SIX=6.D0, 1 ENUMAX=.4999D0,NEWTON=100,TOLER=1.0D-6) C C ----------------------------------------------------------- C UMAT FOR ISOTROPIC ELASTICITY AND DRUCKER-PRAGER PLASTICITY C WITH MULTILINEAR ISOTROPIC AND KINEMATIC HARDENING C CANNOT BE USED FOR PLANE STRESS C ----------------------------------------------------------- C PROPS(1) - E C PROPS(2) - NU C PROPS(3) - A, DRUCKER-PRAGER CONSTANT, LET A=ZERO FOR C CLASSICAL VON MISES PLASTICITY C PROPS(4) - BETA, SCALAR PARAMETER TO DETERMINE C PERCENTAGE OF EACH TYPE OF HARDENING C 0.0 <= BETA <= 1.0 C PROPS(5..) - SYIELD VS PLASTIC STRAIN DATA C -----------------------------------------------------------

233

IF (NDI.NE.3) THEN WRITE(6,1) 1 FORMAT(//,30X,’***ERROR - THIS UMAT MAY ONLY BE USED FOR ’, $ ’ELEMENTS WITH THREE DIRECT STRESS COMPONENTS’) ENDIF C C ELASTIC PROPERTIES C EMOD=PROPS(1) ENU=MIN(PROPS(2),ENUMAX) EBULK3=EMOD/(ONE-TWO*ENU) EG2=EMOD/(ONE+ENU) EG=EG2/TWO EG3=THREE*EG ELAM=(EBULK3-EG2)/THREE EBULK=EBULK3/THREE C CONSTITUTIVE MODEL PROPERTIES A=PROPS(3) BETA=PROPS(4) C READ INITIAL YIELD STRESS SYINIT=PROPS(5) C C ELASTIC STIFFNESS C DO K1=1,NDI DO K2=1,NDI DDSDDE(K2,K1)=ELAM END DO DDSDDE(K1,K1)=EG2+ELAM END DO DO K1=NDI+1,NTENS DDSDDE(K1,K1)=EG END DO C RECOVER ELASTIC AND PLASTIC STRAINS AND AND SHIFT TENSOR AND C ROTATE NOTE: USE CODE 1 FOR (TENSOR) STRESS, CODE 2 FOR C (ENGINEERING) STRAIN C CALL ROTSIG(STATEV( 1),DROT,EELAS,2,NDI,NSHR) CALL ROTSIG(STATEV( NTENS+1),DROT,EPLAS,2,NDI,NSHR) CALL ROTSIG(STATEV(2*NTENS+1),DROT,ALPHA,1,NDI,NSHR) EQPLAS=STATEV(3*NTENS+1) C C SAVE STRESS AND PLASTIC STRAINS AND C CALCULATE PREDICTOR STRESS AND ELASTIC STRAIN C DO K1=1,NTENS OLDS(K1)=STRESS(K1) OLDPL(K1)=EPLAS(K1) EELAS(K1)=EELAS(K1)+DSTRAN(K1) DO K2=1,NTENS STRESS(K2)=STRESS(K2)+DDSDDE(K2,K1)*DSTRAN(K1) END DO

234

END DO C C CALCULATE EQUIVALENT VON MISES STRESS C SMISES=(STRESS(1)-ALPHA(1)-STRESS(2)+ALPHA(2))**2 1 +(STRESS(2)-ALPHA(2)-STRESS(3)+ALPHA(3))**2 1 +(STRESS(3)-ALPHA(3)-STRESS(1)+ALPHA(1))**2 DO K1=NDI+1,NTENS SMISES=SMISES+SIX*(STRESS(K1)-ALPHA(K1))**2 END DO SMISES=SQRT(SMISES/TWO) SHYDRO=(STRESS(1)+STRESS(2)+STRESS(3))/THREE I1=SHYDRO*THREE C CALCULATE EQUIVALENT DRUCKER-PRAGER STRESS C EQUATION (5.10) DPYIEL=SMISES+A*I1 C C GET YIELD STRESS FROM THE SPECIFIED HARDENING CURVE C NVALUE=(NPROPS-4)/2 CALL UHARD(SYIELD,HARD,EQPLAS,EQPLASRT,TIME,DTIME,TEMP, 1 DTEMP,NOEL,NPT,LAYER,KSPT,KSTEP,KINC,CMNAME,NSTATV, 2 STATEV,NUMFIELDV,PREDEF,DPRED,NVALUE,PROPS) C C CALCULATE PREVIOUS YIELD VALUE AND DETERMINE IF C ACTIVELY YIELDING C C STORE PREVIOUS TABLE VALUE FOR SYIELD SYT1=SYIELD SYPREV=SYINIT+(SYT1-SYINIT)*BETA IF (DPYIEL.GT.(ONE+TOLER)*SYPREV) THEN C C ACTIVELY YIELDING C SEPARATE THE HYDROSTATIC FROM THE DEVIATORIC STRESS C CALCULATE THE FLOW DIRECTION C EQUATION (5.14) C DO K1=1,NDI FLOW(K1)=(STRESS(K1)-ALPHA(K1)-SHYDRO)/SMISES END DO DO K1=NDI+1,NTENS FLOW(K1)=(STRESS(K1)-ALPHA(K1))/SMISES END DO C C SOLVE FOR NEW YIELD STRESS C AND EQUIVALENT PLASTIC STRAIN INCREMENT USING NEWTON ITERATION C SYIELD=SYPREV DEQPL=ZERO DO KEWTON=1,NEWTON C EQUATION (5.121) RHS=SMISES-SYPREV-(THREE*EBULK3*A*A+EG3)*DEQPL

235

$ -SYIELD+A*I1+SYT1 DEQPL=DEQPL+RHS/(THREE*EBULK3*A*A+EG3+HARD(1)) EQPLA2=EQPLAS+DEQPL CALL UHARD(SYIELD,HARD,EQPLA2,EQPLASRT,TIME,DTIME $ ,TEMP,DTEMP,NOEL,NPT,LAYER,KSPT,KSTEP,KINC,CMNAME $ ,NSTATV,STATEV,NUMFIELDV,PREDEF,DPRED,NVALUE,PROPS) C IF(ABS(RHS).LT.TOLER*SYPREV) GOTO 10 END DO C C WRITE WARNING MESSAGE TO THE .MSG FILE C WRITE(7,2) NEWTON 2 FORMAT(//,30X,’***WARNING - PLASTICITY ALGORITHM DID NOT ’, 1 ’CONVERGE AFTER ’,I3,’ ITERATIONS’) 10 CONTINUE C C STORE THE CURRENT TABLE YIELD VALUE SYT2=SYIELD C UPDATE THE EQUIVALENT PLASTIC STRAIN EQPLAS=EQPLAS+DEQPL C C SOLVE FOR UPDATED YIELD STRESS C EQUATION (5.104) SYIELD=SYINIT+(SYT2-SYINIT)*BETA C C CALCULATE NEEDED CONSTANTS C EQUATIONS (5.42) THROUGH (5.45) P=(-EBULK*A*SYIELD-SHYDRO*EG+EBULK*A*SMISES)/(EBULK3*A*A+EG) Q=(EBULK3*A*A*SMISES+SYIELD*EG-A*I1*EG)/(EBULK3*A*A+EG) DEP=(A*(SMISES-SYIELD+A*I1))/(EBULK3*A*A+EG) DEQ=(SMISES-SYIELD+A*I1)/(THREE*EBULK3*A*A+EG3) C C UPDATE STRESS, ELASTIC AND PLASTIC STRAINS AND SHIFT TENSOR C DO K1=1,NDI C EQUATION (5.105) ALPHA(K1)=ALPHA(K1)+(SYT2-SYT1)*FLOW(K1)*(ONE-BETA) C CHANGE IN PLASTIC STRAIN FROM EQUATION (5.47) EPLAS(K1)=EPLAS(K1)+DEP/THREE+THREE*DEQ*FLOW(K1)/TWO EELAS(K1)=EELAS(K1)-(DEP/THREE+THREE*DEQ*FLOW(K1)/TWO) C EQUATION (5.48) STRESS(K1)=ALPHA(K1)-P+Q*FLOW(K1) END DO DO K1=NDI+1,NTENS C EQUATION (5.105) ALPHA(K1)=ALPHA(K1)+(SYT2-SYT1)*FLOW(K1)*(ONE-BETA) C CHANGE IN PLASTIC STRAIN FROM EQUATION (5.47) EPLAS(K1)=EPLAS(K1)+THREE*DEQ*FLOW(K1) EELAS(K1)=EELAS(K1)-THREE*DEQ*FLOW(K1) C EQUATION (5.48) STRESS(K1)=ALPHA(K1)+Q*FLOW(K1)

236

END DO C C CALCULATE THE PLASTIC DISSIPATION C SPD=ZERO DO K1=1,NTENS SPD=SPD+(STRESS(K1)+OLDS(K1))*(EPLAS(K1)-OLDPL(K1))/TWO END DO C C FORMULATE THE JACOBIAN (MATERIAL TANGENT) C FIRST CALCULATE THE EFFECTIVE MODULI C EFFG=EG*SYIELD/SMISES EFFG2=TWO*EFFG EFFG3=THREE*EFFG EFFLAM=(EBULK3-EFFG2)/THREE EFFHRD=EG3*HARD(1)/(EG3+HARD(1))-EFFG3 DO K1=1,NDI DO K2=1,NDI DDSDDE(K2,K1)=EFFLAM END DO DDSDDE(K1,K1)=EFFG2+EFFLAM END DO DO K1=NDI+1,NTENS DDSDDE(K1,K1)=EFFG END DO DO K1=1,NTENS DO K2=1,NTENS DDSDDE(K2,K1)=DDSDDE(K2,K1)+EFFHRD*FLOW(K2)*FLOW(K1) END DO END DO ENDIF C C STORE ELASTIC STRAINS, PLASTIC STRAINS AND SHIFT TENSOR C IN STATE VARIABLE ARRAY C DO K1=1,NTENS STATEV(K1)=EELAS(K1) STATEV(K1+NTENS)=EPLAS(K1) STATEV(K1+2*NTENS)=ALPHA(K1) END DO STATEV(3*NTENS+1)=EQPLAS C RETURN END C C

237

SUBROUTINE UHARD(SYIELD,HARD,EQPLAS,EQPLASRT,TIME,DTIME,TEMP, $ DTEMP,NOEL,NPT,LAYER,KSPT,KSTEP,KINC,CMNAME,NSTATV,STATEV $ ,NUMFIELDV,PREDEF,DPRED,NVALUE,PROPS) C INCLUDE ’ABA_PARAM.INC’ C CHARACTER*80 CMNAME DIMENSION HARD(3), STATEV(NSTATV),TIME(*), 1 PREDEF(NUMFIELDV),DPRED(*),PROPS(*) C DIMENSION TABLE(2,NVALUE) C PARAMETER(ZERO=0.D0) C C FILL IN SY VS EQPLAS TABLE FROM PROPS ARRAY I=5 DO K2=1,NVALUE TABLE(1,K2)=PROPS(I) TABLE(2,K2)=PROPS(I+1) I=I+2 END DO C SET YIELD STRESS TO LAST VALUE OF TABLE, HARDENING TO ZERO SYIELD=TABLE(1,NVALUE) HARD(1)=ZERO C C IF MORE THAN ONE ENTRY, SEARCH TABLE C IF(NVALUE.GT.1) THEN DO K1=1,NVALUE-1 EQPL1=TABLE(2,K1+1) IF(EQPLAS.LT.EQPL1) THEN EQPL0=TABLE(2,K1) IF(EQPL1.LE.EQPL0) THEN WRITE(7,1) 1 FORMAT(//,30X,’***ERROR - PLASTIC STRAIN MUST BE ’, $ ’ENTERED IN ASCENDING ORDER’) CALL XIT ENDIF C CURRENT YIELD STRESS AND HARDENING DEQPLS=EQPL1-EQPL0 SYIEL0=TABLE(1,K1) SYIEL1=TABLE(1,K1+1) DSYIEL=SYIEL1-SYIEL0 HARD(1)=DSYIEL/DEQPLS SYIELD=SYIEL0+(EQPLAS-EQPL0)*HARD(1) GOTO 10 ENDIF END DO 10 CONTINUE ENDIF RETURN END

238

APPENDIX D – ABAQUS MATERIAL DATA TABLES

239

Table D.1. Effective Stress versus Equivalent Plastic Strain ABAQUS Input Table for 2024-T851

Equivalent Plastic Strain

Effective Stress, psi

0.00000 59000 0.00011 60013 0.00016 61118 0.00021 61682 0.00025 62792 0.00032 63896 0.00038 64448 0.00044 65000 0.00050 65553 0.00058 66107 0.00069 66673 0.00082 67238 0.00100 67803 0.00131 68373 0.00183 68958 0.00272 69571 0.00367 70134 0.00464 70479 0.00656 71117 0.00848 71700 0.01039 72234 0.01231 72733 0.01423 73170 0.01616 73652 0.01808 74057 0.02000 74500 0.02286 75096 0.02476 75467 0.02667 75750 0.02858 76065 0.03049 76336 0.03240 76555 0.03526 76812 0.04000 77055 0.04500 77282 0.05000 77486 0.05500 77671 0.06000 77840 0.06500 77996 0.07000 78141 0.07500 78275 0.08000 78402

240

0.08500 78521 0.09000 78633 0.09500 78739 0.10000 78840 0.10500 78937 0.11000 79028 0.11500 79116 0.12000 79201 0.12500 79281 0.13000 79359 0.13500 79434 0.14000 79506 0.14500 79576 0.15000 79644 0.15500 79709 0.16000 79772 0.16500 79834 0.17000 79893 0.17500 79951 0.18000 80007 0.18500 80062 0.19000 80116 0.19500 80168 0.20000 80219 0.30000 81036 0.40000 81621 0.50000 82077 0.60000 82452 0.70000 82771 0.80000 83047 0.90000 83292 1.00000 83512

241

Table D.2. Effective Stress versus Equivalent Plastic Strain ABAQUS Input Table for IN100

Equivalent Plastic Strain

Effective Stress, psi

0.00000 162500 0.00378 166620 0.01124 166620 0.01511 168757 0.01898 169928 0.02285 171129 0.02671 172245 0.03057 173366 0.03439 174314 0.04139 180983 0.05963 195348 0.07780 210081 0.09562 224514 0.11314 238033 0.13093 250936 1.00000 976172

242

APPENDIX E – SCRIPT FILE

243

/NODE FOOT- U1/,/MAXIMUM/{ /^$/d /NODE/,/NOTE/d /MAXIMUM/d p } /TOTAL NUMBER OF VAR/d /TOTAL/!d /TOTALS/d /TOTAL CPU TIME/d /TOTAL TIME/d #/TOTAL/p

244

APPENDIX F – FINITE ELEMENT MODEL MESHES

245

Figure F.1. Coarse Mesh FEM of the 0.25 in. Diameter Smooth Tensile Bar

246

Figure F.2. Medium Mesh FEM of the 0.25 in. Diameter Smooth Tensile Bar

247

Figure F.3. Fine Mesh FEM of the 0.25 in. Diameter Smooth Tensile Bar

248

Figure F.4. Coarse Mesh FEM of the 0.35 in. Diameter Smooth Tensile Bar

249

Figure F.5. Medium Mesh FEM of the 0.35 in. Diameter Smooth Tensile Bar

250

Figure F.6. Fine Mesh FEM of the 0.35 in. Diameter Smooth Tensile Bar

251

Figure F.7. Coarse Mesh FEM of Smooth Compression Cylinder

252

Figure F.8. Medium Mesh FEM of Smooth Compression Cylinder

253

Figure F.9. Fine Mesh FEM of Smooth Compression Cylinder

254

Figure F.10. Medium Mesh FEM of NRB with ρ = 0.005 in.

255

Figure F.11. Coarse Mesh FEM in the Notch Region of NRB with ρ = 0.005 in.

Figure F.12. Medium Mesh FEM in the Notch Region of NRB with ρ = 0.005 in.

256

Figure F.13. Fine Mesh FEM in the Notch Region of NRB with ρ = 0.005 in.

257


258


259


260

Figure F.17. Coarse Mesh FEM in the Notch Region of NRB with ρ = 0.040 in.

Figure F.18. Medium Mesh FEM in the Notch Region of NRB with ρ = 0.040 in.

261

Figure F.19. Fine Mesh FEM in the Notch Region of NRB with ρ = 0.040 in.

262


263


264

Figure F.22. Coarse Mesh FEM of Equal-Arm Bend Specimen

265

Figure F.23. Medium Mesh FEM of Equal-Arm Bend Specimen

266

Figure F.24. Fine Mesh FEM of Equal-Arm Bend Specimen

267

Figure F.25. Medium Mesh FEM in the Fillet Region of Equal-Arm Bend Specimen

Figure F.26. Fine Mesh FEM in the Fillet Region of Equal-Arm Bend Specimen

268

APPENDIX G – NRB LOW CYCLE FATIGUE FEA PLOTS

269

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.1. First Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.2. Second Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

270

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.3. Third Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.4. Fourth Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

271

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.5. Fifth Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.6. Sixth Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

272

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.7. Seventh Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.8. Eighth Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

273

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 403

von Mises FEA

Drucker-Prager FEA

-30000

-20000

-10000

0

10000

20000

30000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.57

Figure G.9. Ninth Cycle Load-Displacement Results for NRB with ρ = 0.040 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.52


274

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.52


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.52


275

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.52


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.27

β = 0.52


276

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.52


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.52


277

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.52


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.52


278

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 803

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.52

Figure G.19. Tenth Cycle Load-Displacement Results for NRB with ρ = 0.080 in.

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


279

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


280

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


281

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


282

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50


283

-6000

-4000

-2000

0

2000

4000

6000

-0.004 -0.002 0 0.002 0.004

Specimen 125

von Mises FEA

Drucker-Prager FEA

-20000

-10000

0

10000

20000

-0.1 -0.05 0 0.05 0.1

Loa

d, P

(lb

s)


Loa

d, P

(N

)



β = 0.23

β = 0.50

Figure G.29. Tenth Cycle Load-Displacement Results for NRB with ρ = 0.120 in.

284

VITA

Phillip A. Allen was born in Nashville, Tennessee, on May 23, 1975. He attended

elementary and high school at Columbia Academy in Columbia, Tennessee, and

graduated valedictorian in May 1993. The following August he attended Freed-

Hardeman University and studied pre-engineering until May 1996. In August 1996, he

entered the mechanical engineering program at The University of Memphis. In

December 1998, he graduated Summa Cum Laude with a Bachelor of Science in

Mechanical Engineering. He entered Tennessee Technological University in January

1999 and graduated with a Master of Science Degree in Mechanical Engineering in

December 2000. He entered the Doctoral program at Tennessee Technological

University in January 2001 and is a candidate for the Doctor of Philosophy Degree in

Mechanical Engineering.

an abstract of a dissertation hydrostatic stress...

Documents