Upload
others
View
0
Download
0
Embed Size (px)
Citation preview
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 = 0.010 in................................................................................ 61 Figure 4.17. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB
Tests with ρ = 0.020 in................................................................................ 61 Figure 4.18. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB
Tests with ρ = 0.040 in................................................................................ 62 Figure 4.19. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB
Tests with ρ = 0.080 in................................................................................ 62 Figure 4.20. 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.
................................................................................................................... 170 Figure 7.20. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
................................................................................................................... 171 Figure 7.21. Third 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.
................................................................................................................... 177 Figure 7.26. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
................................................................................................................... 178 Figure 7.27. Third 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
xxi
(b) Material Constant in Mowbray’s Equation
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
xxii
Ro Initial Radius of Spherical Void
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
xxiii
(b) Scalar Multiplier of Plastic Strain
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
xxiv
σult Ultimate Tensile Strength
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
xxv
VLCF Very Low Cycle Fatigue
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
Lo
g S
cale
∆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 ≈ 1.1, indicating that the material will likely cyclically soften. A
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)
0.2% Offset Yield Strength,
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)
T rue 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
aσσ−+=
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Specimen No.
Figure 4.16. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.010 in.
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Specimen No.
Figure 4.17. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.020 in.
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Specimen No.
Figure 4.18. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.040 in.
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Specimen No.
Figure 4.19. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.080 in.
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Specimen No.
Figure 4.20. Composite Load-Gage Displacement Plot for 2024-T851 L Direction NRB Tests with ρ = 0.120 in.
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)
Gage Displacement, v (in)
Load
, P (
N)
Gage Displacement, v (mm)
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)
Gage Displacement (in.)
Load
, P (
N)
Gage Displacement (mm)
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)
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)
Gage Displacement (in.)
Load
, P (
N)
Gage Displacement (mm)
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)
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 ≈ 1.4, which indicates that the material will likely cyclically harden.
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)
0.2% Offset Yield Strength,
ksi (MPa)
Ultimate Tensile Strength, ksi (MPa)
True Fracture Strain
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)
0.2% Offset Yield Strength,
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, andijΙ 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+= pijqplij 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
ijpleqij n εσεσεσ &
v&& +
Ι=
3, (5.25)
which can be written as
87
qeffppleqij p εσεεσ &&& +−= . (5.26)
Finally, substituting the expression for pleqij εσ & 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
ijpnijqprijnij Kn Ι∆−∆−=
++εεµσσ
112
v, (5.28)
ijpnijqplij n Ι∆+∆=∆
+εεε
3
11
v, (5.29)
88
( ) 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=ijij nnvv
and effijij n 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
apprpleqys
preff
q , (5.45)
and
( )
µεσσ
ε39
32 +
−−=∆
Ka
apprpleqys
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
ijqijpplij n
vεεε ∆+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
Sij
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
ijplij 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
yield surface at all times. For the isotropic case, the consistency condition is formed by
taking the time derivative of Equation (5.53)
94
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
ijijpleq QH σε &=
3
2. (5.67)
The elastic predictor stress rate, prijσ& is
klijklprij 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, ijijprijij 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
ijplijij 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
( )ijprijijijij QQQHQ µγσγ 2
32 −= & . (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
( ) ijijijijklijklijprijij QHQQHQQDQQ
=
−−− βγβγγσ32
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
Sij 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
ijpleqij 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
appr
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
appr
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)
Plastic Strain, εpl
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 ≈ 0.15).
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)
Plastic Strain, εpl
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)
Plastic Strain, εpl
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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
Distance Across Neck (in.)
Distance Across Neck (mm)
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.
Uniform Displacement
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)
Distance Across Bottom Symmetry Plane (in.)
Distance Across Bottom Symmetry Plane (mm)
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)
Distance Across Bottom Symmetry Plane (in.)
Distance Across Bottom Symmetry Plane (mm)
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
Distance Across Bottom Symmetry Plane (in.)
Distance Across Bottom Symmetry Plane (mm)
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
Uniform Displacement
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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)
Distance Across Neck (in.)
Distance Across Neck (mm)
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
Distance Across Neck (in.)
Distance Across Neck (mm)
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), andpleqε 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)
Distance Across Fillet Section (in.)
Distance Across Fillet Section (mm)
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
)
Distance Across Fillet Section (in.)
Distance Across Fillet Section (mm)
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
Distance Across Fillet Section (in.)
Distance Across Neck (mm)
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
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,
145
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)
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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
Drucker-Prager UMAT, β = 0.5
Drucker-Prager UMAT, β = 0.0
-24000
-16000
-8000
0
8000
16000
24000
-0.15 -0.1 -0.05 0 0.05 0.1 0.15Lo
ad, P
(lb
s)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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
Drucker-Prager FEA, a = 0.041
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
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Figure 7.8. Load-Gage Displacement Results for NRB with ρ = 0.010 in.
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
Drucker-Prager FEA, a = 0.029
Drucker-Prager FEA, a = 0.041
0
5000
10000
15000
20000
25000
0 0.05 0.1 0.15 0.2Lo
ad, P
(lb
s)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Figure 7.9. Load-Gage Displacement Results for NRB with ρ = 0.020 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Figure 7.10. Load-Gage Displacement Results for NRB with ρ = 0.040 in.
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
Drucker-Prager FEA, a = 0.029
Drucker-Prager FEA, a = 0.041
0
5000
10000
15000
20000
25000
0 0.05 0.1 0.15 0.2Lo
ad, P
(lb
s)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Figure 7.11. Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
0.5 in. Gage Length
Figure 7.12. Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
1st Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
2nd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
3rd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
4th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
5th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.57
Figure 7.17. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
9th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.57
Figure 7.18. Ninth Cycle Load-Gage Displacement Results for NRB with ρ = 0.040 in.
169
Representative load-gage displacement data from selected cycles of the LCF tests
of the NRB with ρ = 0.080 are plotted along with NRB low cycle fatigue finite element
solutions in Figures 7.19 through 7.24. A β of 0.52 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 7% 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 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)
Gage Displacement, v (in.)
Load
, P (
N)
1st Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure 7.19. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
2nd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure 7.20. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
3rd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure 7.21. Third Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
4th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure 7.22. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
5th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure 7.23. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
10th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.52
Figure 7.24. Tenth Cycle Load-Gage Displacement Results for NRB with ρ = 0.080 in.
176
Representative load-gage displacement data from selected cycles of the LCF tests
of the NRB with ρ = 0.120 are plotted along with NRB low cycle fatigue finite element
solutions in Figures 7.25 through 7.30. A β of 0.50 was used for the von Mises FEA and
a β of 0.23 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 7% 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 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)
Gage Displacement, v (in.)
Load
, P (
N)
1st Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure 7.25. First Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
2nd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure 7.26. Second Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
3rd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure 7.27. Third Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
4th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure 7.28. Fourth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
5th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure 7.29. Fifth Cycle Load-Gage Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Load
, P (
N)
10th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Load
, P (
N)
Gage Displacement, v (mm)
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)
0.2% Offset Yield Strength,
ksi (MPa)
Ultimate Tensile Strength, ksi (MPa)
True Fracture Strain
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
Gage Diameter, in. (mm)
Gage Length, in. (mm)
Young's Modulus,
103 ksi (GPa)
0.2% Offset Yield Strength,
ksi (MPa)
Ultimate Tensile Strength, ksi (MPa)
True Fracture Strain
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
Gage Diameter, in. (mm)
Gage Length, in. (mm)
Young's Modulus,
103 ksi (GPa)
0.2% Offset Yield Strength,
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
Gage Diameter, in. (mm)
Gage Length, in. (mm)
Young's Modulus,
103 ksi (GPa)
0.2% Offset Yield Strength,
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
Gage Diameter, in. (mm)
Gage Length, in. (mm)
Young's Modulus,
103 ksi (GPa)
Upper Yield Strength, ksi
(MPa)
0.2% Offset Yield Strength,
ksi (MPa)
Ultimate Tensile Strength, ksi (MPa)
True Fracture Strain
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
Gage Diameter, in. (mm)
Gage Length, in. (mm)
Young's Modulus,
103 ksi (GPa)
Upper Yield Strength, ksi
(MPa)
0.2% Offset Yield Strength,
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
1st Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
5th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Lo
ad
, P
(N
)
6th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
7th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
8th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
9th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Lo
ad
, P
(N
)
1st Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
Figure B.10. First Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
Figure B.11. Second 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
Loa
d, P
(lb
s)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
Figure B.12. Third Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
Figure B.13. Fourth 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
Loa
d, P
(lb
s)
Gage Displacement, v (in.)
Loa
d, P
(N
)
5th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
Figure B.14. Fifth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
10th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Lo
ad
, P
(N
)
15th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
20th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
1st Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
Figure B.18. First Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
Figure B.19. Second 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
Figure B.20. Third Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
Figure B.21. Fourth 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
Lo
ad
, P
(lb
s)
Gage Displacement, v (in.)
Lo
ad
, P
(N
)
5th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
Figure B.22. Fifth Cycle Load-Gage Displacement Plot for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
10th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
20th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
30th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
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
Figure F.14. Medium Mesh FEM of NRB with ρ = 0.010 in.
258
Figure F.15. Medium Mesh FEM of NRB with ρ = 0.020 in.
259
Figure F.16. Medium Mesh FEM of NRB with ρ = 0.040 in.
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
Figure F.20. Medium Mesh FEM of NRB with ρ = 0.080 in.
263
Figure F.21. Medium Mesh FEM of NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
1st Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
5th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
6th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
7th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
8th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
9th Cycle, ρ = 0.040 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
1st Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure G.10. First Cycle Load-Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure G.11. Second 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 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure G.12. Third Cycle Load-Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure G.13. Fourth 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 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
5th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.27
β = 0.52
Figure G.14. Fifth Cycle Load-Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
6th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.52
Figure G.15. Sixth 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 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
7th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.52
Figure G.16. Seventh Cycle Load-Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
8th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.52
Figure G.17. Eighth 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 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
9th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.52
Figure G.18. Ninth Cycle Load-Displacement Results for NRB with ρ = 0.080 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
10th Cycle, ρ = 0.080 in.
Gage Displacement, v (mm)
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
1st Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.20. First Cycle Load-Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
2nd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.21. Second Cycle Load-Displacement Results for NRB with ρ = 0.120 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
3rd Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.22. Third Cycle Load-Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
4th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.23. Fourth Cycle Load-Displacement Results for NRB with ρ = 0.120 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
5th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.24. Fifth Cycle Load-Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
6th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.25. Sixth Cycle Load-Displacement Results for NRB with ρ = 0.120 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
7th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.26. Seventh Cycle Load-Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
8th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.27. Eighth Cycle Load-Displacement Results for NRB with ρ = 0.120 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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
9th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
β = 0.23
β = 0.50
Figure G.28. Ninth Cycle Load-Displacement Results for NRB with ρ = 0.120 in.
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)
Gage Displacement, v (in.)
Loa
d, P
(N
)
10th Cycle, ρ = 0.120 in.
Gage Displacement, v (mm)
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
Engineering.