. , 1:3,.“ $1-5 . gr 5 II 4 t T ‘ II‘ . . _.\rth\. 1 -‘ , . \3 A... .. é. , . - ,rh .. x? n , 3; axtxv. . 2. 1 ‘ .11 ¢ ‘ , ‘ V . . m# a... n l ‘ 1n..!1!t.: . . ERMiJX I V ,1. Eu. 1.. .5 . . ‘ is}... :1, . ....:.x-..f . . . .z. . . 5. .35... , . ., ( . , t ”it...“ , . . t {$96.41 K..A.:L..9:hi§. m... 3.39.. “Raga. 5» n all. T :11!- 1 . i . 4.... W3 . _ . .btJF..r.s!I1.. 3 17w .1 I 3 3i ‘ .1 .H L as.» :zwmwnmgf 9: 5 ct “fix .1 7%... xv... k Luv. is {lfiz .I: . .t. , it! ,. .9 311.. LIBRARY Michigan State University This is to certify that the dissertation entitled The Analysis Of Dynamic Stress And Plastic Wave Propagation In The Taylor Impact Test presented by Gong Song has been accepted towards fulfillment of the requirements for the PhD degree in Mechanical Engineering Major Professor’s Signature FZé. /0! 2.00; Date MSU is an Afiinnative Action/Equal Opportunity Institution ---.- —-~—.—.¢.-.-.—--.-.-.--o¢- PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DWQ, (HE-ill :21, @007 2/05 p:IClRC/DateDue.indd-p.1 The Analysis Of Dynamic Stress And Plastic Wave Propagation In The Taylor Impact Test By Gong Song A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mechanical Engineering 2006 ABSTRACT The Analysis of Dynamic Stress and Plastic Wave Propagation in the Taylor Impact Test By Gong Song This thesis investigates the propagation of plastic wave in a projectile of the Taylor Impact Test. The theories of plastic wave propagation, RI theory and RD theory, are used to analyze the Taylor Impact Test. A complete method (theoretical investiga- tion, experimental and measurement and numerical analysis) of computing the dynamic stresses in the projectile is developed. The yielding dynamic stress and the shape curves of the projectile after impact are used to verify our theoretical investigation. Three phases model of the Taylor Impact Test, a new constitutive equation (RI theory) and the Modified Malvern model (RD theory) in the Taylor Impact Test are proposed. The determination of the maximum dynamic stress and the characteristics of the plastic wave propagation are included. In order to verify the proposed constitutive equation and model, the Taylor Impact Tests are performed with three different materials (Aluminum, Copper and Steel) and impact velocities, and numerical analysis is carried out. The experimental and numerical analysis results indicate that the conclusions from our theoretical investigation match experimental and numerical analysis results well. Dedicated to my father, mother, wife Xiao—Cui and son Guan-Long ACKNOWLEDGEMENT I would like to acknowledge my sincere thanks to all my committee members for their serving on my committee and making contribution to this investigation. In particular, I would like to express my deepest gratitude to Prof. D. Liu who advises me during the whole course of this investigation. Also, I would like to express my deepest thanks to Professor Zhou, Dr.Tsai, Professor Kwon and Professor Lee from whom I have received valuable advice and suggestions. The pioneers in the area of plastic wave propagation are gratefully acknowledged for their prominent theoretical work and excellent experimental results. I wish to thank many professors of the University of Alabama who helped me in my student career in one way or another. Finally, I want to express my deep thanks to everyone who has helped my in my study and life. iv TABLE OF CONTENTS LIST OF TABLES viii LIST OF FIGURES x CHAPTER 1 INTRODUCTION 1 1.1 Literature Review 2 1.1.1. Development of Theories 2 A. Strain Rate Independent Theory 2 B. Strain Rate Dependent Theory 4 C. Plasticity-Viscoplasticity Theory 6 D. Stress-Strain Relations 7 1.1.2 Development of Experimental Techniques 7 A. Experimental Measurements 7 B. Split Hopkinson’s Pressure Bar And Plate Impact 9 C. Taylor Impact Test 9 1.1.3 Development of Computational Methods 12 1.2 Statement of Problem 14 1.3 Organization 16 CHAPTER 2 FUNDAMENT EQUATIONS FOR PLASTIC WAVE PROPAGATION ----- 18 2.1 Material Deformation And Plastic Wave Propagation 18 2.2 Theories Of Plastic Wave Propagation 20 2.2.1 Strain Rate Independent Theory 20 2.2.2 Strain Rate Dependent Theory 20 2.2.3 Comparison Of the Two Theories 21 2.3 Equation Of One-Dimensional Plastic Wave 22 2.4 An Solution Of Plastic Wave Equation Of RI Theory 23 CHAPTER 3 APPLICATION OF PLASTIC WAVE PROPAGATION THEORY IN TAYLOR IMPACT TEST 27 3.1 Three Phases Model 28 3.2 Theory 30 3.2.1 First Phase 30 3.2.2 The Shape Curve Of the Projectile 34 3.2.3 The Shape Curve Of the Projectile In the First Phase 36 V 3.2.4 Plastic Wave Propagation In the Second Phase 38 3.2.4.1 Condition Of Constant Velocity Propagation 38 3.2.4.2 An Solution Of Plastic Wave Equation 39 3.2.4.3 The Shape Curve Of the Projectile In the Second Phase 41 3.3 Three Taylor Impact Tests 43 3.4 Computation And Discussion 45 3.4.1 Material Aluminum 6061-T6511 45 3.4.2 Material Copper 145-Hard-H02 51 3.4.3 Material Steel C1045 56 3.5 Results 61 3.6 Tables 62 CHAPTER 4 ANALYSIS FOR DYNAMIC MODELS IN TAYLOR TEST 74 4.1 Analysis For Power Law Plasticity Model 74 4.2 Analysis For Hartig Model 77 4.3 Analysis For Rate Sensitive Power Law Plasticity 79 4.4 Analysis For Johnson-Cook Model 81 4.5 Analysis For Malvern Model 85 4.6 Analysis For Strain Rate Dependent Plasticity 87 4.7 Analysis For Power Law 88 4.8 Analysis For Logarithmic Law 89 CHAPTER 5 THE NUMERICAL ANALYSIS FOR PLASTICITY WAVE PROPAGATION OF RD THEORY 90 5.1 Plastic Wave Equation 01' RD Theory And Its Plateau 90 5.2 The Characteristic Equation 92 5.3 Numerical Analysis Methods 95 5.3.1 Eulerian Method 96 5.3.2 Hartree Method 97 CHAPTER 6 THE DYNAMIC STRESS COMPUTATION OF PROJECTILE ALUMINUM 6061-T6511 IN THE TAYLOR IMPACT TEST 100 6.1 The Computation of Dynamic Stress In the First Phase By J ohnson-Cook Model 100 6.2 The Computation of Dynamic Stress In the Second Phase 102 6.2.1 Modified Malvern Model 102 6.2.2 Eulerian Method 103 6.2.3 Hartree Method 105 6.2.3.1 Using Arbitrary Mesh Point To Calculate Dynamic Stresses ----—-- 107 vi 6.2.3.2 Using Specific Mesh Point To Calculate Dynamic Stresses 109 CHAPTER 7 CONCLUSIONS AND FUTURE WORKS 115 7.1 Conclusion 115 7.2 Taylor Formula And New Formula 116 7.3 Future Works 119 APPENDIX A 122 APPENDIX B 126 APPENDIX C 128 APPENDIX D 131 APPENDIX E 135 APPENDIX F 141 APPENDIX G 146 APPENDIX H 154 APPENDIX 1 156 APPENDIX .1 169 APPENDIX K 170 APPENDIX L 171 BIBLIOGRAPHY 185 vii LIST OF TABLE 1. Table - Experimental Data-1 Aluminum 606l-T6511 62 2. Table - Experimental Data-2 Copper 145-Hard -H02 64 3. Table - Experimental Data-3 Steel C1045 66 4. Table 3.4.1.1 -Measured Diameter vs. Computed Diameter (Aluminum 6061- T651 1) 66 5. Table 3.4.1.1.1 —Stress vs. Strain (Aluminum 6061-T6511) 67 6. Table 3.4.1.2 —Stress, Strain, Plastic Wave Velocity vs. Diameter (Aluminum 6061-T6511) 67 7. Table- Lagrangian Diagram Data —1 (Aluminum 6061-T6511) 68 8. Table 3.4.2.1 —Measured Diameter vs. Computed Diameter (Copper 145—Hard- H02) 69 9. Table 3.4.2.1.1 -Strain vs. Distance (Copper 145-Hard-H02) 69 10. Table 3.4.2.2 —Stress, Strain, Plastic Wave Velocity vs. Diameter (Copper 145- Hard-H02) ‘ 70 11. Table- Lagrangian Diagram Data —2 (Copper-145-Hard-02 71 12. Table 3.4.3.1 —Measures Diameter vs. Computed Diameter (Steel C1045) -- 72 13. Table 3.4.3.1.1 —Strain vs. Distance (Steel C1045) 72 14. Table 3.4.3.2 —Stress, Strain, Plastic Wave Velocity vs. Diameter (Steel C1045) 73 15. Table- Lagrangian Diagram Data —3 (Steel C1045) 73 16. Table 6.] —Stress, Strain vs. Distance in the First Phase (Aluminum6061- T6511) 10] 17. Table 6.2 —Eulerian Method 104 viii 18' 19 er 18. Table 6.3 —Initial Values for Eulerian Method 107 19. Table 6.4 -Computed Results From Eulerian Method (1) 108 20. Table 6.5 —Computed Results From Hartree Method (2) 111 21. Table 6.6 —Computed Results From Modified Malvern Model vs. J ohnson- Cook Model 113 ix LIST OF FIGURES 1. Figure 2.1 -Impact Diagram 22 2. Figure 3.1 -Before Impact 31 3. Figure 3.2 -At the Moment of Impact 31 4. Figure 3.3 -After Impact 32 5. Figure 3.4 -Cross Section of Projectile 32 6. Figure. Projectile-l-Aluminum 6061-T6511 44 5. Figure. Projectile-Z-Copper 145-Hard -H02 44 7. Figure. Projectile-3-Steel C1045 I 45 8. Figure 3.4.1.1 —Deformed Diameter vs. Distance (Aluminum 6061-T6511)-- 46 9. Figure 3.4.1.1.1 —Stress vs. Distance in First Phase (Aluminum 6061-T6511)— 47 10. Figure 3.4.1.2 -Diameter vs. Stress (Aluminum 6061-T6511) 49 11. Figure 3.4.1.3 -Stress vs. Strain (Aluminum 6061-T6511) 50 12. Figure 3.4.1.4 -Diameter vs. Plastic Wave Velocity (Aluminum 6061-T6511)— 50 13. Figure 3.4.1.5 -Lagrangian Diagram (Aluminum 6061-T6511) 51 14. Figure 3.4.2.1 —Deformed Diameter vs. Diameter in First Phase (Copper 145- Hard-H02) 52 15. Figure 3.4.2.2 -Stress vs. Strain (Copper 145-Hard-H02) 55 16. Figure 3.4.2.3 -Diameter vs. Plastic Wave Velocity (Copper 145-Hard-H02)-—- 55 17. Figure 3.4.2.4 -Diameter vs. Stress (Copper 145-Hard-H02) 56 18. Figure 3.4.2.5 -Lagrangian Diagram (Copper 145-Hard-H02) 56 19. Figure 3.4.3.1 —Deformed Diameter vs. Distance in First Phase (Steel C1045)- 58 20. Figure 3.4.3.2 -Diameter vs. Plastic Wave Velocity (Steel C1045) 60 21. Figure 3.4.3.3 —Diameter vs. Stress (Steel C1045) 60 22. Figure 3.4.3.4 -Stress vs. Strain (Steel C1045) 60 23. Figure 3.4.3.5 -Lagrangian Diagram (Steel C1045) 61 24. Figure 5.1 -Hartree Method Diagram 99 25. Figure 6.1 -Stress vs. Distance in First Phase 101 26. Figure 6.2 -Stress vs. Distance in Second Phase (Eulerian Method) ------ 104 27. Figure 6.3 -Stress vs. Distance in Second Phase (Hartree Method) --------- 111 28. Figure 6.4 -Stress vs. Distance in Projectile 112 29. Figure 6.5 -Strain Rate Distribution Diagram 113 30. Figure 6.6 -Stress vs. Distance in Projectile From Modified Malvern Model And Johnson Cook Model 114 xi CHAPTER 1 INTRODUCTION Impact dynamics is one important area in engineering practice. It covers subjects from aerospace and automotive crashworthiness to human body health and military applications. In aerospace, research in impact dynamics is performed to develop techniques capable of predicting the dynamic behavior of aerospace structures and to devise advanced control techniques aimed to improve the performance of the structures. There is currently an increasing demand for dynamic, stress analyses; material behavior; and constitutive equations of metal materials. For example, rotor containment design is not yet an exact science and it currently depends mainly on empirical methods and extensive testing. The requirement to produce a safe design while at the same time reducing development cost and time in order to be competitive in the aero-engine market is of paramount importance. When engineers use the supercomputer to simulate non-linear numerical modeling of an aircrafi frame to investigate the distribution of the dynamic stress in the frame, this work needs the help of impact dynamics. In the automotive industry, because of the rapid growth of crashworthiness as a stand-along, engineering subject; there is a need for impact dynamics devoted to the crash behavior of structures and materials, body structures, and energy-absorbing systems subject to sudden dynamic loading. This includes side-impact protection, analysis of frontal collisions, structural crashworthiness simulation of the rear-end collision of small cars, prediction of energy absorption capability of composite materials, and more. In bio-mechanics, people use the research results of impact dynamics to investigate the fracture in long bones, head injury and protection, injury mechanisms and criteria for the human feet and ankle under axial impact loads, and the simulation of head/neck dynamic responses under impact loading. These research activities are now called impact bio-mechanics. Another application of impact dynamics is in the military. The modern battlefield is an increasingly dynamic and complex environment that requires weapons with increased agility and speed with strong ability to penetrate armor plate. These requirements provide challenges to the warhead or penetrator design. One challenge is that the engagement conditions translate to high, strain rates in the order of 106 US. In such an environment, the ability for a material to not fracture is challenged. In the case of a warhead/ penetrator, its ability to perform its function ( kill the target ) is compromised if a material fails. In order to design such a warhead/penetrator with material that can maintain its integrity under such loading conditions, the failure process and dynamic stress distribution, and the dynamic yield strength of a material must be understood. This is exactly the task of impact dynamics. 1.1 Literature Reviews 1.1.1 Development of Theories A. Strain Rate Independent Theory ( RI Theory) In the 19305, as one of the first pioneers in the study of plastic wave propagation, Donnell [1] introduced the first scheme for treating the wave propagation in a media beyond the Hooke’s law. Although the focus of Donnell’s research was not directly aimed at plastic wave propagation, a formula expressing the velocity of wave propagation 2 was developed Cp: [1.99. (1.1) pda where 0' is stress, 8 is strain, p is density, and C p is the velocity of a plastic wave. The systematic studies of plastic, wave propagation from the aspects of theory and testing of metallic materials, such as copper and altuninum, began in 1940. The studies were motivated in light of developments of armor-piercing shells and plates, and were performed independently by von Karman and Duwez [2], Taylor [3] and Rakhmatulin [4]. The difference among their studies was essentially based on the choice of coordinate systems. Von Karman used a Lagrangian coordinate system while Taylor used an Eulerian, coordinate system. Today their studies are known as the classical plastic wave propagation theory by some researchers or the strain rate independent (RD theory by most researchers. There were two assumptions in the RI theory. 1) The stress 0' was only a fimction of strain 8 , i.e. 0' = 0(8). 2) The curve of stress versus strain was concave toward the strain. Under these assumptions, the following one-dimensional, wave equation was formulated: flzu _ d0“ é’zu d2 d8 &2 p (1.2) where u = u(x,t) is displacement, x is coordinate and t is time. After the strain rate independent (RI) theory was proposed, some studies were conducted by Duwez and Clark [5], Campell [6], Johnson, Wood and Clark [7] from 1946 through 1953 to further examine the plastic wave propagation. Their studies could 3 be divided into two groups. The first group verified the existence of the plastic wave through the comparison between theoretical analyses and experimental investigations. The second group investigated the physical nature of plastic wave propagation, as in the relationships of wave velocity vs. time and stress vs. time. All these studies concluded that plastic wave propagation did exist and the RI theory was basically correct, although some predictions fi'om the RI theory did not totally match with the experimental results at the impacted end. B. Strain Rate Dependent Theory ( RD Theory) By the time the RI theory was proposed, the strain rate dependent (RD) theory though not aiming at plastic wave propagation had already been established. In 1909, Ludwik [8] proposed a logarithmic fimction among the stress, strain, and strain rate 0' = 01(8) + K 1n 6" where 0' 1 (8 ) is the stress of strain 8 when strain rate 5‘ is unity. The factor K was a function of strain. Prandtl [8] reached the same conclusion with the use of a physical theory of plastic flow. Meanwhile, a power law was proposed by Prandtl 0' = 01(8)?" where 11 could be a function of stress. In 1950, Sokolovsky [9] and Malvern [10] also proposed the strain rate dependent theory for studying plastic wave propagation. They suggested that the effects of visco- plasticity should be considered in the study of plastic wave propagation. Their theories could be expressed by the following stress, strain, and strain-rate relation 0' = 018,55) Based on Newton’s second law, the equation for plastic wave propagation could be written as é’zu flaé’zu é’a fisu ,0—2—= 2+ - 2 0’} 58% fiao’ko’t ( 1.3) where 5? is strain rate. Malvern proposed the following stress-strain rate relation: 0' = 0'0 + ké ( 1.4) where 0'0 represents for the static stress-strain relation and k is a constant. With the use of the equation of motion and the strain-displacement relation €2sz (1.5) 036 07 fit 0'} where V is particle velocity, the following constitutive relationship could be obtained: Eé=d+k[a—f(g)] (1.7) where E is Young’s modulus, f (6‘) is a strain function obtained from a hardened aluminum alloy, if is strain rate, and d" is stress rate. Sokolovsky proposed the following stress-strain-strain-rate relation to model dynamic behavior of materials: a=0(a)+ln(1+bé) (1.8) where 0(8) is a quasi-static stress-strain relation, dot denotes time and b is a constant. Substituting Equation (1.8) into Equations (1.5) and (1.6), yields Eé=d’+g(0’,a) (1.9) where g(0’, 6') is a stress-strain ftmction. Although the two types of theories, the RI theory and the RD theory, were 5 proposed for studying plastic wave propagation, either the RI theory or the RD theory alone could not explain all experimental results. In fact, the experimental results seemed to indicate that strain rate should not be neglected completely and there existed a strain plateau around the impacted surface [9,10,11]. The RD theory could not explain the existence of the plateau, while RI theory could. The RI theory agreed with the experi- mental results in the region away from the impact surface while the RD theory did not. C. Plasticity-Viscoplasticity Theory From 1962 through 1968, Simmons, Hauser and Dom [18], and Lubliner and Valathur [19] proposed a generalized plasticity-viscoplasticity theory. According to the theory, the strain rate independent theory (RI) could be classified as the plasticity theory and the strain rate dependent theory (RD) could be classified as the viscoplasticity theory. That is, the strain rate independent theory (RI) did not consider the viscous effect of the materials while the strain rate dependent theory (RD) did. For viscid ductile materials, the RD theory agreed with experimental results better than the RI theory. For less viscid ductile materials, the RI theory agreed with experimental results better than the RD theory. This plasticity-viscoplasticity theory was supported by the experiments from Kolsky and Dough [14] and Lindholrn [15]. For pure aluminum and pure copper, the RD theory agreed with the experiments better than the RI theory. For aluminum alloys, the RI theory agreed with the experimental results better than the RD theory. A model was proposed by Lublinear [19] as follows: é=p(0',8)o"+q(0',8) (1.10) where the function p(0’, 8) and q(0', 8) governed the instantaneous and non- instantaneous response respectively. D. Stress-Strain Relations The relationship between the stress and the strain in the impacted materials was always a primary interest in the study of plastic wave propagation. In 1968, Bell [20] proposed the following stress-strain relation based on nearly 600 experiments covering 2200 individual tests and 27 types of solids reported in the literature over a period of 45 years ’ 1 or) m where BO is a dimensionless universal constant, i.e. BO = 0.028 , 21(0) is the shear modulus at zero point; r = 1, 2, 3, is the finite deformation mode, and 5b is the parabola intercept upon the strain abscissa of the particulardeformation mode of interest (8,, = 0 for the initial finite deformation mode). T is testing temperature and T m is melting temperature. 1.1.2 Development of Experimental Techniques A. Experimental Measurements In order to further identify the above usefulness of the RI theory and the RD theory, Bell [11-—13] developed a diffraction grating method to measure the strain in rods made of aluminum and copper impacted by a constant velocity. His findings could be summarized as follows: 1 ) The RI theory could not completely agree with the experimental results in the zone close to the impact end. However, it could be applied to the rod in the zone away from the impact end. On the contrary, the RD theory could be applied to the rod in the zone close to the impact end but away fi'om it. 2 ) After impact, a non-dispersive shock wave was inaugurated in a location approximately equal to one-quarter of the diameter of the rods and then developed into a dispersive, plastic-wave front. 3 ) The dispersive, plastic wave developed into two wave fronts. The first wave front was associated with the deviatoric component of the dynamic stress and was called the shear wave. The second wave front was associated with the hydrostatic component of the dynamic stress and was called the longitudinal wave. The longitudinal wave was developed much more slowly than the shear wave. 4 ) The deviatoric stress was about 2/3 of the dynamic stress while the hydrostatic stress was about 1/3 of the dynamic stress. 5 ) The initial non-dispersive shock wave first developed the shear wave and then the longitudinal wave. The shear wave and the longitudinal wave each contained approximately one-half the initial kinetic energy. 6) The von Karman critical velocity was found to be about 48.5 ft/sec for aluminum and 73 fi/sec for copper. In 1964, Kolsky and Dough [14] studied wave propagation in short bars made of pure aluminum, pure copper, and an aluminum alloy. They found no appreciable strain rate dependence for the aluminum alloy. For pure aluminum and pure copper, the strain rate dependent theory had reasonable agreement. In 1964, Lindholm [15], in a series of tests in which short specimens (aspect ratio of length to diameter ranging from 0.2 to 2.0) made of pure aluminum were subjected to strain cycling at various strain rates, showed that the RD theory agreed better with the experimental results than the RI theory. These experiments indicated that both the aspect ratio and material characteristics played important roles in the strain rate effect. Malvern and Huffington [16], Gillich and Ewing [17], and Bell [13] conducted many experiments attempting to directly measure the profiles of strain vs. time and velocity vs. time. They found that the velocity of the plastic wave propagation was a constant for each strain and the strain rate did not affect the velocity of plastic wave propagation very much. That is to say, the influence of strain rate could be neglected in the zone away from the impact end. B. Split Hopkinson’s Pressure Bar and Plate Impact Split Hopkinson’s pressure bars (SHPB) were used in the studying of plastic wave propagation investigation. Kames [21] and Bertholf and Kames [22] showed that the length-to-diameter ratio of the specimens used in the SHPB tests should be around 0.3, and that both the faces of the specimens should be lubricated to reduce friction. In the 19705, the experimental investigation of plastic wave propagation turned to develop the plate impact technique. Because the longitudinal dimension of plate specimen, i.e. the thickness, was small plate specimens were more suitable for the investigation of strain rate effect than thicker specimens used in SPHB. C1ifion[23,24] found that high strain rates could cause thermal disturbance in the plate specimens due to the elevated temperature around the impacted surface, hence, the effect of strain rate could influence the plastic wave propagation. This conclusion seemed to agree with Bell’s conclusion [12,13]. C. Taylor Impact Test In the 19403, Taylor devised a method to estimate the dynamic strength of materials. Due to World War II, this work was not published until the late 1940s [32,33]. Now this test is called Taylor Impact Test, or Taylor test. The Taylor Test consisted of 9 firing a solid cylinder of the material against a rigid target. The dynamic flow stress of the cylinder could be estimated from the deformed cylinder by measuring the overall length of the un-deformed section. One of the assumptions in the Taylor Test was that the rear of the cylinder undergoes constant deceleration. Today, this test has being investigated by many people and military labs. For example, Wilkins and Guinan in 1973 [34] used computer simulation to compute the dynamic stress of the material and found the final position of the plastic wave front inside the cylinder differed from the one resulting from the surface deformation evaluation of the cylinder used by Taylor. In 1985, Kuscher [35] experimentally proved the theory of Taylor, using a laser velocity interferometer, the so-called VISAR. The front-side movement of the cylinder and the plastic-wave-front movement can be seen, along with the velocity, which decreases exponentially. The propagation of the elastic wave traveling in between the plastic wave and the rear end of the cylinder is also proved. Now the Taylor test has not been used for its original purpose of obtaining dynamic-yield stresses of the materials. It has come into its own usage for checking constitutive equations by comparing the shapes of the deformed cylinders with the predictions from the constitutive equations. Since the 19905, the investigation about the Taylor impact test has made progress in theoretical model, numerical 'simulation and experimental methods. In theoretical investigation, J .W.House, J .C.Lewis, P.P.Gills and L.L.Wilson [36] in 1995 modified the Jones and co-authors’ works. This paper introduced approximation for the J ones' model. The author used the experiments to verify their results in aluminum, copper, and steel. In 1997 and 1998, W.K.Rule, S.E.Jones, J .A.Drinkard and L.L.Wilson [37,3 8] revised the J ohnson-Cook model in the term of strain rate. This 10 model was used in four materials (aluminum, copper, iron, and steel). In the papers, they proposed a theoretical analysis for the linearity and several important parameters. These parameters were used to determine the state of stress at strain rates exceeding 104 (US). A numerical integration method was used to obtain the parameters for the revised J ohnson-Cook model. Essentially, this method considered the effect of thermal energy and the concept optimization. In 2001, Guoxing Lu, Bin Wang, and Tieguang Zhang [39,40] modified the original Taylor impact model for porous materials. Calculations were made for porous materials with a relative density that was a linear function of compressive strain. The final length of the projectile after impact was plotted against the density, impact velocity, and the dynamic yielding stress. The experimental results with bronze and iron were presented for the static and dynamic stresses of porous metals. R.L.Woodward, N.M.Burman, and B.J.Baxter in 1994 [41] presented the experimental method of using Taylor impact test with a Hopkinson bar as a target to record the load/time data for an aluminum alloy. The elastic deformation and plastic deformation were checked. In 1999, C.G.Lamontage, G.N.Manuelpillai, E.A.Taylor, and R.C.Tennyson [42] reported the results of a hypervelocity (5km/s) oblique impact test (35 and 45 degrees) which were performed on carbon fiber/Peek composite specimens using the light gas gun. In 2001, E.A.Taylor [43] investigated the numerical simulation of the impact of hollow shaped charge jet projectiles onto stuffed, Whipple-bumper shielding using AUTODYN-ZD and AUTODYN-3D. A total of 56 simulations were carried out. J .D.Yatteau, G.W.Recht, R.H.Zemow, and K.T.Edquist in 2001 [44] described the experiments to record deformation profiles versus time in metallic rods subjected to transverse ballistic impact (materials: tungsten alloy, alloy steel, titanium and copper). ll The deformation profiles were analyzed to determine transverse plastic deformation wave , speeds. The rod materials were also subjected to Taylor impact tests to measure the static tensile and dynamic yielding strengths and longitudinal wave speeds. In 2004, I.Rohr, H.Nahme, and K.Thomas [45] described the behavior of 35NiCrMoV109 high strength steel over a wide range of strain rates for numerical simulations of dynamic events. The low, strain-rate, tensile tests and a numerical stress state were carried out. These data were linked to high, strain-rate data from a novel modified Taylor impact test using VISAR technique and data from planar, plate-impact tests. A material model and an equation of state were developed. The enhanced model was made by comparing the measured and calculated VISAR signals. 1.1.3 Development of Computational Methods In the 19805, finite element methods and other computer-based methods, such as LS-DYNA and Pam & Trade, were developed to study structural crashworthiness. The models of the materials included in the software packages consisting of both the RD theory and the RI theory. For instance, the J ohnson-Cook model was a model of the RD theory while the power law plasticity model was a model of the RI theory. Many other models are given in Appendix C. Nicholas [25] in 1981 pointed out that there appeared to be no theoretical basis for the RD models. In other words, the model was not unique. Besides, there was no basis to claim that other models could not be proposed for the investigation of plastic wave propagation. It was not expected to propose a model based on some experimental results to model all effects due to strain rate. Even if one model could be generally accepted, the methodology for determining the empirical constants might not be straightforward. 12 In 1987, Jones [26] proposed a two-phase theory, which was later developed into a three-phase theory [27], to study Taylor’s test. According to Jones, the deformation of copper in the Taylor test could be characterized by three phases of behavior. In the beginning, denoted by phase I, the behavior was dominated by a shock-wave front. This was followed by a steady motion of plastic wave front, denoted by phase II. In phase III, the deceleration of the plastic wave front occurred. Jones’ study showed that the plastic wave front entered phase II from phase I about 10 us after the impact of a 30—caliber, copper rod by 4340 steel anvils. His study also showed that there was a decay of the function of strain rate vs. time. Jones’ study further supported Bell’s conclusions that the strain rate dependent theory did not seem to be applicable for the zone beyond the first diameter of the rods from the impact ends. In 1991, Khan [28] and Hsiao used electrical resistance strain gages to investigate both small-amplitude and large-amplitude plastic wave propagation in fully annealed 1100 aluminum subjected to axi-symmetric free projectile impact. The results showed that the dynamic behavior of the material could be described by the strain rate independent theory for large-amplitude plastic wave propagation if elastic strain and strain rate sensitivity were negligible. On the contrary, the material response appeared to be strain rate sensitive for small-amplitude plastic wave propagation if elastic strain was not negligible when compared to plastic strain. In 1997, Maudlin, Foster, and Jones [29] used a continuum mechanics code to analyze the plastic wave propagation in the Taylor test. They found that the use of the strain rate dependent J ohnson-Cook model [29], the MTS flow stress mode [29], and the thermodynamic response with the Mie-Grunisen equation-of—state [29] could not capture 13 the plastic wave propagation in annealed and hardened copper. They also found the velocity of the plastic wave propagation became slow as the strain decreased. In 1998, Wright [30] used the three-dimensional, nonlinear continuum to investi- gate the plastic wave propagation in a circular rod with the effects of radial inertia and radial shear. His investigation showed that the RI theory (von Karman-Taylor theory) should hold eventually at a large distance from the impact end of a bar where the radial effects had decayed essentially to zero. This distance could be several tens of radii from the impact end before the von Karman-Taylor solution became a good approximation. Alternatively, if the axial loading at the impact end was slow enough and the wave front could be distributed over the time axis, the radial shear was always small and the von Karman-Taylor theory (the RI theory) would be valid throughout the analysis. In 2000, Rusinek and Klepaczko [31] reported the critical impact velocity value in the steel sheet ’ with the help of finite element code ABAQUS. Their conclusion was consistent with von Karman-Taylor’s theory and Bell’s experiments. 1.2 Statement of Problem From the foregoing review, it can be seen that impact dynamics is a discipline of studying material responses under impact loading. Being different from the static loading, dynamic loading such as impact loading causes instantaneous wave propagation. Hence, wave propagation is a branch of this discipline for studying waves, elastic as well as plastic, propagating in impacted materials such as rods and plates. When the impact velocity becomes high, the elastic wave propagation is usually neglected and only plastic wave propagation is considered. Although the study of plastic wave propagation has made significant progress in the past six decades, some fundamental problems still 14 remain open and hinder the development of impact dynamics to some extent. This deserves some attention. The purpose of this thesis is to study the plastic wave propagation and constitutive equation of the materials in the Taylor test. Taylor impact tests are specially designed to exploit the inertia of a test specimen to produce very high loading rates that is used to study a large deformation, high strain ( 50 % ) and high strain rate behavior of materials. It consisted of firing a short rod of the material against a rigid anvil or wall and making measurements on the rod before and after. During the deformation of the specimen, the plastic deformation and fracture will take place and a plastic wave will propagate. So the Taylor test is suited very well to investigate and study constitutive equation of materials, the nonlinear response of materi- als and structures, fundamental theoretical analyses and dynamic materials characteriza- tion. Typical applications include armor and anti-armor, penetration mechanics, physical security, demilitarization, safety studies, accident post-mortem analyses, containment studies, transient loads, structural response analyses, and computational modeling of materials. The approach is first to use the two types of plastic-wave theory, the strain rate independent theory ( RI ) and the strain rate dependent theory ( RD ) to investigate the plastic wave propagation. The theoretical results from the plastic wave propagation are then used to investigate the distributions of the dynamic stress and dynamic strain of the projectiles in the Taylor test. 15 1.3 Organization This thesis starts with a historical literature review on plastic wave propagation in Chapter 1. It is followed by the investigation of the fundamental equations of plastic wave propagation in Chapter 2. In Chapter 2, a constitutive equation is proposed based on VISAR experimental results. Using this constitutive equation, an analytical solution of one-dimensional, plastic wave equation derived from RI theory is obtained under the condition of constant velocity impact. Using this solution, the propagation of the plastic wave can be described. Based on the theoretical results obtained in Chapter 2, the model of three phases theory in the Taylor test is proposed in Chapter 3. Using this model, the propagation of the plastic wave of four kinds of material ( 6061-T6 Aluminum, 145-Hard-H02 Copper, 6061-T6511 Aluminum, and C1045 Steel ) in the Taylor tests for different impact velocities are investigated and the dynamic stress, dynamic strain and the velocity of the propagation of the plastic wave are calculated. According to the three phases model, there exists a mushroom shapes in the projectiles after impact. This conclusion is verified by the four experimental results. In Chapter 4, eight constitutive equations from RI theory and RD theory are used to investigate the propagation of the plastic wave in the Taylor tests. The results indicate that the conclusions from these constitutive equations cannot match the experimental phenomena well. In Chapter 5, the one-dimensional plastic wave of RD theory is investigated. The investigation includes the plateau of RD theory, the characteristic equation of the plastic 16 wave propagation and numerical analyses ( Eulerian method and Hartree method ) along the characteristic curves. The results indicate that there exists a plateau for RD theory. The propagation of the plastic wave can be computed by the numerical methods along the characteristic curves. It is shown in Chapter 6 that the Modified Molvern model is used to numerically compute the propagation of the plastic wave ( RD theory) in the Taylor test for material 606l-T6511. Both the Eulerian method and the Hartree method are used, separately. The computed results indicate that the Eulerian method is simple but not accurate; the Hartree method is complicated but possesses better convergence. Finally, in Chapter 7, what was done in this investigation is summarized and some recommendations for fixture works are made. 17 CHAPTER 2 FUNDAMENTAL EQUATIONS FOR PLASTIC WAVE PROPAGATION 2.1 Material Deformation and Plastic Wave Propagation The impact process is a very complex phenomenon. One of the important characteristics of the impact process is its short duration. Experimental results have indicated that the impacted object during the impact process is deformed both elastically and plastically and with thermal energy change. In this investigation, the thermal energy is not considered. Experimental results have also shown that the elastic deformation occurs prior to the plastic deformation under hi gh-velocity impact. The amount of the deformation is dependent on the amount of energy absorbed by the impacted object. Corresponding to the elastic deformation (linear as well as non-linear), two types of elastic waves ( linear and non-linear ) are formed and propagate in the impacted object. From the elastic theory for isotropic materials, there are two types of waves in solids: longitudinal (or dilatational) wave and transverse (or distortional, or shear) waves. The longitudinal waves are those in which the particles move in the same direction as the wave propagation. The transverse waves are those in which the particles move in the direction normal to the wave propagation. In the meanwhile, elastic waves may propagate on the surface of a solid. Two types of surface waves, the Rayleigh wave and Love wave, have been investigated extensively. Rayleigh waves decay exponentially with the depth from the surface to the interior of the medium and move only in two dimensions. They propagate more slowly than other types of elastic waves. Love waves are induced by material mismatch, e. g. a layer of material that possesses one set of 18 physical constants overlies another layer of the material that possesses a different set of physical constants. As the intensity of the impact force increases, the material is driven beyond its elastic limit and becomes a plastic deformation, resulting in two waves propagating in the solid: an elastic wave and a much slower but more intensive plastic wave. In addition, if the characteristics of the medium are such that the velocity of propagation of large disturbances is greater than that of smaller ones, the stress pulse develops a steeper and steeper front when propagating through the medium. In this way, a shock wave is formed. Shock waves can also be formed if the impact velocity exceeds the wave velocity in the material. In this thesis, only plastic wave propagation is considered. A plastic wave is formed and propagated in the impacted object following the plastic deformation. The plastic wave is a kind of nonlinear wave. Compared with the plastic deformation from high-velocity impact, the magnitude of elastic deformation is much smaller. Hence the elastic waves can be neglected in the investigating of wave propagation under high-velocity impacts. In this thesis, the high-velocity impact is defined as the impact velocity which results in plastic deformation. Historically, most research activities on plastic wave propagation dealt with long bars. According to experimental [13] and theoretical investigations [1,2], the plastic waves were not formed at the impacted face; instead they were inaugurated at a distance away from the impacted face. The plastic wave propagation causes plastic deformation and sharp changes of stress and strain in the impacted object. The stress and the strain are called dynamic stress and dynamic strain in this investigation. The propagation can be described by a governing equation based on assumptions concerning the relationship 19 among stress, strain and strain rate. There are two major approaches for the propagation of plastic wave: the strain rate independent theory (RI) and the strain rate dependent theory (RD). 2.2 Theories of Plastic Wave Propagation 2.2.1 Strain Rate Independent Theory (RI) The strain rate independent theory assumes that a stress 0' is only a function of strain 8 when plastic wave propagates in a solid. The relation between stress and strain can be written as 0' = 0'( 8) and the curve between them has been formed to be concave toward the axis of strain for most solids. The equation of plastic wave propagation can be derived from Newton’s second law. Without the loss of generality, only one-dimensional motion is considered. The governing equation can then be expressed as all; _d0'd¢: _____ 2.1 Pa, daéx ( ) where 0' is stress, 8 is strain, p is mass density, t is time, u is displacement, and x is displacement. 2.2.2 Strain Rate Dependent Theory (RD) Two strain rate dependent theories were proposed independently by Solovsky [9] in 1949 and by Malvern [10] in 1950. These theories consider stress as a function of strain 8 and strain rate 53 , i,e. a = 0(6, 6") From Newton’s second law, without loss of generality, the following one-dimensional motion in x direction can be formatted: 20 flzu 070' 015' é’o as" P—7=——+—.—— (2.2) 07 fig dt 58 dc 2.2.3 Comparison of the Two Theories As mentioned above, the problems involving plastic wave propagation in metals and ductile materials can be analyzed based on two types of hypotheses: the strain rate independent theory (RI) and the strain rate dependent theory (RD). From the point of view of Simmons [18] and Lubliner [19], the strain rate independent theory is associated with plasticity theory while the strain rate dependent theory is associated with visco- plastic theory. For less viscid ductile materials, the strain rate independent theory seems to agree with experimental results, e.g., the strain rate independent theory agreed better with the experimental results for aluminum alloy in the experiments conducted by Kolsky and Dutch [14] and Lindholm [15]. For viscid ductile materials, the strain rate dependent theory seems to agree better with the experimental results for aluminum and copper. Because the viscosity of materials is dependent on temperature and the elevated temperature mainly appears around the impacted face under the high velocity impact [22], the effect of strain rate (viscosity effect) should only influence the plastic wave propagation around the impacted face. This result is strongly supported by Bell’s experiment [13]. From Bell’s experiment, the strain rate independent theory of plastic wave propagation was feasible in annealed aluminum and copper with a distance approximately equal to the diameter from the impacted surface. Further experiments [4] showed that this conclusion is applicable for zinc, lead, magnesium, nickel, and 70-30 or- brass. In addition, Jones’ experimental result [26] also concluded that the strain rate 21 dependent theory was not applicable after the first diameter from the impact face if the ratio of length to diameter was greater than 5. It should be noted that the RD theory is referred to 0' = 0(6‘, if) and the wave equation. Many models based on the RD theory have been proposed. However, they cannot completely represent the RD theory. From the RD theory, the stress is a function of strain and strain rate, Equation (2.2) tells us that 0' the effect of strain rate influences the plastic wave propagation by term of —7. If the 8 . . . 0' . strain rate 15 a constant, say cg , then 0' = 0'( 8,68). Thls means that —7 rs zero and 8 Equation (2.2) will become Equation (2.1). That is, the RI theory is a special case of the RD theory. 2.3 Equation of One-Dimensional Plastic Wave Assume that a bar extending from x = 0 to x = L . The endpoint, x = 0 , of the bar is suddenly impacted with a constant velocity V0 , as shown in Figure 2.1. The relation 0' = 0'( 8, 6") holds only for the initial deformation of the material of the rod beyond the elastic limit. The lateral contraction of the material and the inertia effects are neglected. ~sno---.-r---------.-o-.-----c-----a---o-t---o-----I-~-uq k x0 L Figure 2.1 Impact Diagram 22 . 0’11 . u Recall Equation (2.2) and use 6‘ = -— and 8 = —— , then 03c écét 0’2u 0’0' 0‘2u 0’0 0’3u p = + . 02 as a2 as 030 (2.3) where u is the displacement function of an element in the longitudinal direction, i .e, u = u(x,t). If the strain rate is not considered, the equation of plastic wave propagation, Equation (2.3), can be reduced to Equation (2.1) p321: _ d0 fizu d2 d8 &2 This is a plastic-wave equation of RI theory in Lagrangian coordinate system. This equation was first proposed by von Karman in 1942. It is usually called the von Karman plastic-wave equation. When the deformation of the materials is within the elastic range, £13 d5 equation = E , where E is Young’s modulus. Equation (2.1) becomes a linear elastic wave 3211 521.! P7 = E —2' 07 0c Apparently, the linear, elastic-wave equation is a special case of a nonlinear, plastic-wave equation. 2.4 A Solution of the One-Dimensional Plastic-Wave Equation When a bar with length L is impacted by a constant velocity V0 at x = 0 , the constitutive equation of a material (the relation between the stress 0' and the strain 8) can be expressed as 0 = 0(8). In 1985, Kuscher used his VISAR test to prove that the plastic wave propagates exponentially in the projectile. Based on this experimental 23 result, a constitutional equation is proposed as follows: BC; ’73—] (2.4) where C1 and C2 are constants. Based on Newton’s second law, the plastic wave equation can be written as follows: é’zu _ d0 é’zu _ 2.5 Pa, (150,2 ( ) and subjected to the following boundary conditions: u(x09t0) = hd; u(L,tp)=0. where u(x,t) is the displacement function, x is position, t is time, t p is the plastic wave propagation time and p is the density. x0 and to are constants. Using the . . X . . . substitution, 5 = — , then 8 can be consrdered as a function of 5 , 1.6.8 = f (5) . 1 According to the experimental results obtained by Bell, a plastic wave is inaugurated at a distance, i.e. at x = x0 > 0 ( t = to > 0). So, the computation of displacement should begin from x = x0. That is, the coordinate original point is set at x = x0. It yields u(x,t) = Egg—’90 = lgféwy =ti§fdn (2.6) 24 where 77=-)t:, dy=td77. Take the derivative for the above Equation, it gives duékééléfi. 01 0.5011 tag’ é’zu_ 0104 ag_1 , are :25 Kim "'7’ 5 3;: lf(n)d77 — 0(5) 0 02 2 , if," = i—f (5) (2.8) Take Equations (2.7) and (2.8) into Equation (2.5), then 2 I r p§f(5)=d0f(5) (2.9) t dgt From Equation (2.9), we have f'(5)=0 (210) Or 2 =ld_‘7 (2.11) pda From (2.10), we know that f (5) = constant, Cl . That is, the strain is a constant. From the experimental results, this solution is in the interval of 0 S x < x0. From Equations (2.11) and (2.4), we have that 1 2(5-01) :2 =—pe C2 (2.12) ,0 25 ———> 8 = C1 + C2 lné Taking Equation (2.13) into Equation (2.14), then we have Mm u(x,t) = 1.1—510 =t1fftn)dn = t 130, + 021117711177 = Clx + C,x(1n-’tf — 1) ( 2.14) where 0[ Figure 3.2 At the Moment of Impact 31 Third Phase Second Phase First Phase \ x L Rigid Wall \\/§<\\\ Figure 3.3 After Impact Rigid Wall \\\\\ \ Figure 3.4 Cross Section of Projectile 32 "i In order to use Equation (3.2) to describe the propagation of the plastic wave, t h x . must be known. Notice ( t + — ) in Equation (3.2) is a concept of t1me. Let C1 x 2' = t +-—. Thenwehave Cl u(x,t): V07, OS 15th (3.5) . . . . . X . That is, 2' can be d1vrded into two parts: deformation of the material, —— and pure t1me, 01 t . As we know, the impact process is very complicated. Here, it is assumed that the impact process can be divided into two sub-processes: pure-impact process and deformation process. We define that the pure-impact process is the process of impact taking place. The time of the pure impact process takes time t In this sub-process, no deformation pure ' takes place. The deformation process is the process of deformation of the material. In this sub-process, only the deformation of the material takes place; impact does not take place. It takes time tdeform' Because the pure impact process takes place first, and then the deformation process begins. The time of the impact process should be equal to the sum of the times of the pure impact process and the deformation process. And during the impact, the total time, to , can be calculated as follows: ’0 = tpure + (deform (3'6) Because in the first phase the velocity of the particle is a constant, that is to say, a 33 particle moves uniformly for a distance 11 with velocity V0 from the impact end. According to Newton’s first law, the sum of the total forces applied to the particle in the first phase should be zero. That is to say, when the particle moves uniformly for a distance h , the impact force is always applied to the particle. According to Bell’s experiments [13], beyond h , the velocity of particle is not a constant. So, it can be concluded that the impact force is no longer applied to the particle. Based on this analysis, the time of material deformation, th, is equal to tdeform and can be calculated as follows: th = tdeform : 7 (3'7) 0 . h . . that 1S, after ’1 = — , the deformation of material ends. 0 From the above analysis, we give two conditions at x = h and t = th as follows: 1401.0.) = hd (3.8) 011 ——h,t =V 3.9 a( h) 0 ( ) where hd is the length of h after impact. 3.2.2 The Shape Curve of the Projectile Suppose the radius of the projectile is r0 or the diameter is D1 before impact. Taking a polar coordinate at the centerline of the projectile, x is along the centerline, shown in Figure 3.4. In order to describe the shape curve of the projectile after impact, we must relate r with x , that is, to find r = r(x). In order to do so, let us use the plastic incompressibility [Zukas,25] 34 1111 .\'o in d8, +d£g+d8x =0 (3.10) where 8, and ‘96 are the strain along r and 9, and 8x is the plastic strain along x axis. Noticing that for the centerline of the geometry, the transverse strain increment (1'8, and 0'89 are equal [Maudlin, 29], d5, = are, (3.11) From Equation (3.11), Equation (3.10) can be written as 2d8, = -d8x Taking integration for the above equation, we have 8 8, =-——"+Cq 2 where Cq is an integrative constant. Assume when 5r = 0, Cq = 0. Then, we have 8, = ——" (3.12) 8,. =— (3.13) combining Equation (3.12) and Equation (3.13), we have fl——lg (314) r 2 x . Equation (3.14) gives the relationship between x and r . Now we are going to use Equation (3. 14) to get the shape curve of the projectile in the plastic region after impact. 35 3.2.3 The Shape Curve of the Projectile In the First Phase Since in this phase the strain is a constant, we take integration for both sides of Equation (3.14) dr 1 _ : __gx r dr 1011 :> — = ——— r 2 0: :> idlnr = —li@-dx r0 20% u(x,t) :> lnr—lnro = — +u(0,t) According to the von Karman boundary condition, when x = 0 and t = 0, it gives u(0,0) = 0. So we have _u(x,t) r=BOe 2 (3.15) where BO is a constant. From Equation (3.15), we have _u(x,t) r = 306 2 IE _ 2 — Boe Suppose that after impact R0 is the radius of the projectile at 2' = 0 and x = 0, and Rh is the radius of the projectile at 2' = t* , then we have T = 0, r = R0; T : t*, r = Rh; 36 Take the above conditions back into Equation (3.15), then we have Rh = 308 2 (3.17) Solving the above equations, it gives 1: 2 R 1 =— ——°— (3.18) V0 Rh Sowehave IQ, r=ROe 2 , Osr.<_t*, Osxsh (3.19) Notice that the movement of the particle is uniform in this region, we have t*_ x2 n—Rfl _i h hVO ”Rh Then, Equation (3.19) can be rewritten as 2 r=ROeh Rh, OSxSh (3.20) The shape curve of the projectile in the first sub-region can be described by Equation (3.20) after impact. Taking a derivative twice for x to Equation (3.20), it gives R H "“ r=11 _R_OR(ln 0 728) h Rh Because the second derivative is greater than zero, the shape of the projectile is 37 concave. Using the rule of plastic incompressibility, I? can be calculated from Equation (3.20) and hd erozh = 7r! R58 h thx 0 Solve this equation, it gives Zlnfl R h=— 2 0 hd (3.20-1) 2 R ln[1 — i 111(1)] 2 R0 R0 3.2.4 Plastic Wave Propagation in the Second Phase 3.2.4.1 Condition of Constant Velocity Propagation At x = h and t = th , the plastic wave enters the second phase from the first phase. Consider the propagation of the plastic wave is constant velocity during the plastic wave propagates in the second phase, that is, the velocity of the plastic wave, C , satisfies C: x p t (3.21) According to von Karman [2] and Taylor’s derivations [3], the velocity of the plastic wave in Equation (3.1) should satisfy _1:I_<: pd8 C2 p (3.22) Combing Equation (3.21) and Equation (3.22), it gives the condition of constant velocity propagation 38 f1 x2 _ld0 (3.23) t2 _ ,0 d8 3.2.4.2 A Solution of Plastic Wave Equation In Chapter 2, a constitutive equation is proposed as follows when the plastic wave propagates in the second phase: 2(8+C1) a = _Ae C2 (3.24) 2 And the solution of Equation (3.1) is that x u(x,t) = Clx + C2x(ln—t— — 1)+ C3t (3.25) where C1, C2 and C3 are constants and will be determined by boundary conditions. In order to determine three constants, three boundary conditions are needed. Because we have two boundary conditions already, Equations (3.8) and (3.9), another one is needed. At the interface of the second phase and the third phase, we have a boundary condition as follows: u(L—Le,tp)=(L—LO)—(h—hd) (3.26) Suppose that after impact the time when the plastic wave reaches the interface of the second and third phase is t p. The time when the elastic wave propagates in the projectile and reaches the interface after reflecting at the back end is t e . Then t e should be equal to z‘p approximately, that is 1, =1 (3.27) E If the velocity of the elastic wave propagation is Ce , Ce = — , where pe is \i p the density of projectile and E is Young’s modulus, then it is easy to calculate te as follows: = e (3.28) So, from Equations (3.27) and (3.28), we have I" + I" t = e (3.29) ‘u m Now, we assume the displacement function u(x, t) and the velocity of the particles are continuous when the plastic wave enters the second phase from the first phase. Then we have the following three boundary conditions, Equations (3.8), (3.9) and (3.26), to solve the three integrative constants in Equation (3.25). Taking the three boundary conditions into Equation (3.25), then we have h th L—Le CltL—L.)+Cz(L—L.)(In 1 —1>+C.t.=(L—Lo)—(h-hd) P (3.32) 40 Write the above three equations as a matrix form h h(ln-}—l——l) 1,, ’11 Cl ha, 0 _ti 1 C2 = V0 2-1 G. (L—Lo)— dr = —%(C1 + C2 In?) Since r0 is the radius of the projectile before impact, then after impact, the radius r can be written as r = r0 + roldr| = r0(1+‘dri) 41 C C x =r 1+—‘-+ 2ln— o( 2 1) C C x :3 r=r 1+—‘—+—iln— 3.34 0( 2 2 t) ( ) or D=D1(1+£1-+§lln£) (3.35) 2 2 t where D1 is the diameter of the projectile before impact. D is the diameter of the projectile after impact. The shape curve of the projectile in the second phase can be described by Equation(3.34) or Equation (3.44) after impact. Taking derivative twice for roC2 2 Because the second derivative is less than 2x x to Equation (3.34), it gives r" = zero, the shape of the projectile is convex. Using Equation (3.20), Equation (3.34) and considering the third phase of the projectile, the shape curve of the whole projectile can be drawn as follows: _5 n59 First Phase: r = Roe h Rh , 0 S x S h SecondPhase: =r0(l+%+%ln§),thSL—Le,thStStP or D: Dl(l+fl+gln£) 2 2 t Third Phase: r = r0, L — Le S x S L The above formulas indicate that in the first phase, the shape curve of the projectile is concave. In the second phase, the shape curve is convex, like the mushroom. In the third phase, the shape curve is a straight line. 42 3.3 Three Taylor Impact Tests Three Taylor impact tests were carried out to verify the theoretical analysis at the Impact Dynamics Lab of the Department of Mechanical Engineering, Michigan State University. The test data and figures of the projectiles are shown in Figure. Projectile-l- Aluminum 606l-T6511, Figure. Projectile-Z-Copper l45-Hard-H02 and Figure. Projectile—3-Stee1 C1045. The measured test data are shown in Table-Experimental Data- l-Aluminum 606l-T6511, Table-Experimental Data-2-Copper 145-Hard-H02 and Table- Experimental Data 3-Stee1 C1045. All tables can be seen in Section 3.5. Three figures only Show the plastic region of the projectile after impact. The parameters of projectiles can be seen in Section 3.4. From the test data and three figures, there exist two phases in every projectile. For the projectile of Aluminum 6061-T6511, the first phase’s length is 0.299 cm. The length of the second phase is 3.4282 cm. The length of the third phase is 4.12 cm. In the first phase, the shape curve of the projectile afler impact is concave. In the second phase, the shape curve of the projectile after impact is convex. For the projectile of Copper l45- hard-HOZ, the first phase’s length is 0.32 cm. The length of the second phase is 2.60 cm. The length of the third phase is 6.788 cm. In the first phase, the shape curve of the projectile after impact is concave. In the second phase, the shape curve of the projectile after impact is convex. For the projectile of Steel C1045, the first phase’s length is 0.24 cm. The length of the second phase is 2.1532 cm. The length of the third phase is 4.422 cm. In the first phase, the shape curve of the projectile after impact is concave. In the second phase, the 43 shape curve of the projectile afler impact is convex. These three tests support firmly our theoretical model and investigation. Experimental Data (Aluminum 6061 -T6511 ) 13.6 , A 13.4- 5 ".3 5 128 - ‘9» o ' h. g 12.6 ”a... 5 12'4 ’ “.W 12.2 . . o 10 20 30 4o 50 . Distance (mm ) Figure. 3-Projectile-1 —Aluminum 6061-T6511 Experimental Data (Copper 145 -Hard-H02 ) A 13.2 L E E 13 ~ E 12.8 ~ 0 00 g 12.6 T .- a 12.4 - ”a... 12.2 . , , 0 10 20 30 40 Distance ( mm ) Figure.3-Projectile—2 -Copper 145-Hard-H02 44 Experimental Data (Steel C1045 ) 13.3 __________--_,__-,__.,___-W-- *- 13.2 4, 13.1 -° 13 .ON 12.9 \ 12.8 1 ”a. 12.7 ‘s O 12.6 « o 12.5 Diameter( mm ) 0 5 10 15 2O 25 Distance ( mm ) Figure.3-Projectile-3- Steel C1045 3.4 Computation And Discussion 3.4.1 Material: Aluminum 6061-T6511 The physical and measured parameters, geometrical parameters, and computed parameters of the projectile are as follows: Density: p=2700 kg/ m3 Static Yield Stress: 05 = 275 MPa Young Modulus: E = 69 x 109 N /m2 Elastic Wave Velocity: C, = 5055m/s Impact Velocity: V0 = 93m/ 8 Original Length: L = 7.8232011 Final Length: L0 = 7.5727cm L p = 3.42820m Le = 4121:»: Projectile Diameter: D1 = 122936011 Measured h = 0.275cm , Computed h = 0.299cm 11,, = 0.2565cm, 211,, = 1.3195cm, 2RO :1 335cm AL = (L-L0)—(h-hd)= 0.269cm ,, = (L+ Le)/Ce = 23.63x 10—63 1,, = h / V0 = 29.55 x 10—63 45 In the first phase, the radius of projectile is obtained from Equation (3.20) —0375 r = 0.69688 h , O S x S 0.299 cm Table 3.4.1.1 shows the measured values of r and theoretically computed values of r . Figure 3.4.1.1 shows the theoretical values and the actual values. The series 1 represents the computed theoretical curve and the series 2 represents the measured curve. In order to calculate the dynamic stress in this phase, the J ohnson-Cook model is used. According to analysis in the first phase, the velocity of the particle is a constant. This is then an adiabatic phase. The thermal effect can be omitted. Because the strain is a constant, 0.148, the effect of the strain rate is not considered. So the J ohnson-Cook model can be written a 0 = (A + B8") , where A = 275 MPa, B = 426 MPa and n = 0.34. From the computed strain and the above model, the dynamic stress in the first phase of Aluminum 6061-T6511 can be computed as shown in Table 3.4.1.1.1. The relationship of stress 0 1 and x is shown in Figure 3.4.1.1.1 P Deformed Diameter vs. Distance (Aluminum 6061-T6511 ) 1.34 " 1.335 1 5 1L1. 4,_,,”_.Hg : 1.33 I o Series1 ‘ 3 O I ; . g 1.325 - ’ l} [EEBESK a \ x E 1.32 , O O O ' .- 1.315 . l o 01 02 03 Distance ( cm ) Figure 3.4.1.1 Deformed Diameter vs. Distance (Aluminum) 46 Stress vs. Distance in the First Phase Aluminum 6061-1'8511 51 0 508 506 i 504 502 « o 500 498 O . Q Q o 496 O 494 . Wis , 0 0.5 1 1.5 2 2.5 3 Stress ( MPa ) Distance ( mm ) Figure 3.4.1.1.1 Stress vs. Distance in the First Phase (Aluminum) In Figure 3.4.1.1, the experimental and computed results indicate that the radius of the projectile is convex. This result matches our theoretical analysis. In the first phase, the strain is approximately 0.148 except near to impact end. The stress is about 498 MPa. It indicates that the strain is a constant in the first phase. This result matches our theoretical investigation well. In the second phase, according to Equation (3.33) and the above parameters of the projectile, we have 0.275 —1.5613 29.56 Cw 0.2565 0 —0.0093 1 C2 3: 0.0093 3.7032 —10.56 23.63 63 0.269 Solve the above matrix equation. Then C1 = 0.1321 , C2 = 0.0426,u/sec, C3 = 0.0097cm/ysec Dynamic Yielding Stress 0': 2700 x 0.0426 0' = 2 e 2x(0.l321) 0-0426 x104 = 283Mpa 47 From Equations (3.24), (3.25) and (3.35), we have: a) Plastic Dynamic Stress 0': 2(a—0.1321) 0'=14.91x(e 00426 )(Pa) b) Plastic Wave Propagation Velocity C p : £+O.1321 C p = e 0-0426 (cm/,usec) c) Displacement Function u(x,t) : u(x,t) = 0.132126 + 0.0426x(1n C, — 1) + 0.0097: (cm) (1) Plastic Strain 8: a = 0.1321+ 0.04261n C], e) Particle Velocity v: v = —0.0426Cp + 0.0097 (cm/ psec) 1') Diameter of Projectile: D =1.22936(1+ 0.1321+ 0.0426 1n Cp) (cm) 0.2756m S x S 3.7032cm Table 3.4.1.2 gives the values of the strain 8 , the stress 0' , C p and the diameter in the second phase from 0.764cm S D S 0.9626cm. Figure 3.4.1.2, Figure 3.4.1.3, Figure 3.4.1.4 and Figure 3.4.1.5 show the relationship among variables. Figure 3.4.1.2 indicates the relationship between the diameter and the stress. From this figure, the stress decreases gradually along the projectile. In the second phase, 48 Figure 3.4.1.3 indicates that the dynamic stress decays as the strain decreases. When the strain is very small, the stress can be expressed approximately by a straight line. Figure 3.4.1.4 indicates that in the second phase the shape of the projectile is like a mushroom after impact. The reason for the mushroom shape is due to the velocity of the plastic wave propagation. However, if the velocity of the plastic wave is higher ( about above 4000 m/s ), the shape of projectile can be approximately replaced by a straight line. This explains why Taylor could use a straight line to substitute the curve of the projectile and good agreement was obtained. Table 3.4.1.2 gives the value when the static stress is 283 MPa. It matches well the given static stress value of Aluminum 6061—T651 1, 275 MPa. The Lagrangian Diagram of the plastic wave propagation is shown in Figure 3.4.1.5. The data for Lagrangian Diagram is shown in Table Lagrangian Diagram Data-1. This one describes the propagation of the plastic wave in the projectile after impact using the displacement and time as variables. From Table 3.4.1.1.1, the impact velocity and Equation (3.3), 61:0.0628 X 10.6 cm/s. Diameter vs. Stress ( Aluminum 6061 -T6511 ) A 600 a Q s 4009 / I «umw“” m 200 — g m 0 T g . T . 1.22 1.24 1.26 1.28 1.3 1.32 1.34 . Diameter ( cm ) > Figure 3.4.1.2 Diameter vs. Stress (Aluminum) 49 Stress vs. Strain ( Aluminum 6061 -T6511 ) 500 400 - 0 300 ooomouee”°” Stress ( MPa ) N O O 100 1 O 0.05 0.1 0.15 ‘ Strain V Figure 3.4.1.3 Stress vs. Strain (Aluminum) Diameter vs. Plastic Wave Velocity (Aluminum 6061-T6511 ) is: 1.32 - 1.3 - 1.28 9 ,o0 . 1.26 « ° | 1.24 1.22 Diameter ( cm ) 0 5000 10000 Plastic Wave Velocity ( mls) Figure 3.4.1.4 Diameter vs. Plastic Wave Velocity (Aluminum) 50 Lagrangian Diagram Aluminum 6061 -T6511 0') (D O O . _._1_. I O Time(us) N a. o o o O N h Distance ( cm) Figure 3.4.1.5 Lagrangian Diagram (Aluminum) 3.4.2 Material: Copper 145-Hard-H02 The physical and measured parameters, geometrical parameters and computed parameters of the projectile are as follows: Density: p = 8960kg / m3 Static Yield Stress: as = 32 MPa Young Modulus: E = 110 ><109 N / m2 Elastic Wave Velocity: Ce = 3504m/ s Impact Velocity: VO = 49.56m/s Original Length: L = 9525cm Final Length: L0 = 9.4425cm L p = 2.60cm Le = 6.788cm Projectile Diameter: 01 =1.22936cm Measured h = 0.335cm; Computed h = 0.363cm hd = 0.320cm 2R,, =1.3025cm 2R0 = 1.3178cm AL = (L—L0)—(h-hd)= 0.1105cm 1p =(L+ [.8)/Ce =4656x10"6s 1,, = h/VO = 71.23x10‘6s 51 In the first phase, the diameter of projectile can be obtained using Equation (3.20): 0.005): D=1.3177Se h , OSxS0.3356'm Table 3.4.2.1 shows the measured values of D and theoretically computed values of D. This table indicates the measured diameter matches the computed diameter very well. Figure 3.4.2.1 is drawn from Table 3.4.2.1. The series 2 represents the computed theoretical curve and the series 1 represents the measured curve. In Figure 3.4.2.1, the experimental and computed results indicate that the radius of the projectile is convex. This result matches our theoretical analysis. In this phase, the strain is approximately 0.12 except near the impact end. It indicates that the strain is a constant in the first phase. This result matches our theoretical investigation well. Deformed Diameter vs. Distance In First Phase ( Copper 145 -Hard-H02 ) l l A 1.32 + 1 5 1.315 ~ 0. . m_ 1" .I l O Series1 l % 1'31 4 I‘ PI Seriesz g 1.305 - . , o 13 . 0 e e I O 0.1 0.2 0.3 0.4 Distance ( cm ) ‘ Figure 3.4.2.1 Deformed Diameter vs. Distance in First Phase (Copper) In the second phase, according to Equation (3.33) and the above parameters, we have 52 0.353 —2.226 71.23 CI 0.325 0 —0.004956 1 C2 = 0.004956 2.737 —10.49 46.56 C3 0.1105 Solve the above matrix, then C1 = 0.0420, C2 = 0.0229 (ysec/cm) C3 = 0.0051 (cm/psec); Dynamic Yielding Stress: 2.400420) 0.0229 X104 :33 (MP0) 8960 x 0.0229 0' = 2 e From Equations (3.24), (3.25) and (3.35), we have: a) Plastic Dynamic Stress 0': 2(5—00420) 0:102.59x(e 0-0229 ) (Pa) b) Plastic Wave Propagation Velocity C p : 5-00420 C p = e 0-0229 (cm/ ,usec) c) Displacement Function u(x,t): u(x,t) = 0.0420x + 0.0229x(ln Cp — 1) + 0.0051t ( cm/ ,usec ) (1) Plastic Strain 8: 8 = 0.0420 + 0.0229ln C p e) Particle Velocity v: v = —0.0229Cp + 0.0051( cm/ysec ) 53 0.0420 + 0.0229 2 1‘) Diameter of Projectile: D = 1.22936(1 + In C p) (cm) 0.325 S x S 2.925 cm Table 3.4.2.2 gives the values of the strain 8 the stress 0' , C p and the diameter in the second phase from 1.22936 cm S D S 1.3025 cm. In the second phase, from Figure 3.4.2.2, the maximum strain is 0.1194. In the first phase, the strain value is about 0.12, this indicates the strain is continuous from the first phase to the second phase. The relationship between the stress and the strain is shown in Figure 3.4.2.2. From this figure, the stress decreases gradually along the projectile. In the second phase, Figure 3.4.2.3 indicates that the dynamic stress decays as the plastic wave propagates, the shape of the projectile is like a mushroom after impact, too. In Figure 3.4.2.4, the dynamic stress, the dynamic strain decay sharply after the propagation of the plastic wave around the interface of the first phase and the second phase. The computed static stress of Copper 145-Hard-H02 is 33 MPa. It matches the given static yielding stress of the material very well. When the propagation of the plastic wave in the projectile stops, its velocity is 1597 m/s. As the propagation of the plastic wave, the decaying will gradually become slow. Figure 3.4.2.5 shows the Lagrangian Diagram of the plastic wave propagation. The data for Lagrangian Diagram are shown in Table Lagrangian Diagram Data-1. From Table 3.4.2.1.1, the impact velocity and Equation (3.3), 61:0.0413 x 10—6 cm/s. 54 Stress vs. Strain ( Copper 145-Hard-H02 ) A 30000 . a 25000 - . 5 20000 - 8 2; 15000 - ‘f 8 10000 - ’ 1: 5000 .o’ m 0 {WI 0 0.05 0.1 0.15 Strain Figure 3.4.2.2 Stress vs. Strain (Copper) W ' ' ' ' ' 'l 1 Diameter vs. Plastlc Wave Velocity | p ( Copper 145-Hard-H02 ) i . 1.32 * ‘ .3 OD 000’“... i 126 . ‘124 i I 1.22 T . , . 1 3 0 50000 100000 150000 200000 250000 1 Plastic Wave Velocity ( rnls ) _j Figure 3.4.2.3 Diameter vs. Plastic Wave Velocity (Copper) Diameter ( cm ) 55 Diameter vs. Stress (Copper 145-Hard-HOZ ) 15000 A o If c 10000 9 E o m g 5000 j *' e (D O 7W. o 1.22 1.24 1.26 1.28 1.3 1.32 Diameter ( cm ) Figure 3.4.2.4 Diameter vs. Stress (Copper) Lagrangian Diagram 1 Copper-145-Hard-H02 30 0) '320 0° 0 510 ’ (i: «0’ CW . Distance ( cm ) Figure 3.4.2.5 Lagrangian Diagram (Copper) 3.4.3 Material: Steel C1045 The physical and measured parameters, geometrical parameters and computed parameters of the projectile are as follows: 56 Density: p = 7870kg / m3 Static Yield Stress: a, = 585MPa Young Modulus: E = 200 x 109 N / m2 Elastic Wave Velocity: C, = 5041m/s Impact Velocity: V0 = 57.85m/s Original Length: L = 7.8232cm Final Length: L0 = 7.683cm LP = 2.1532cm Le = 5.44cm Projectile Diameter: 01 =1.22936cm Measured h = 0.240cm; Computed h = 0.223cm hd = 0.211cm 2R,, =.125925cm 2R0 = 1.26925cm AL=(L—LO)—(h—hd)=0.10120m tp =(L+ Le)/Ce =26.31x10’6s :h = h/V0 = 39.76x10_(’s In the first phase, the diameter of projectile is obtained using Equation (3.20): 0.0079x D=1.269258 h , Table 3.4.3.1 shows the measured values of D and theoretically computed values of D. Figure 3.4.3.1 indicates the measured diameter matches the computed diameter very well. The series 2 represents the computed theoretical curve and the series 1 represents the measured curve. In Figure 3.4.3.1, the experimental and computed results indicate that the radius of the projectile is convex. In this phase, the strain is approxima- tely 0.052 except near to impact end. It indicates that the strain is approximately a constant in the first phase. 57 OSxSO.Z4cm Deformed Diameter vs. Distance In First Phase ( SteelC1045) 1 1 . E 1.27 -.-' 1 3 I ,__. 1 E 1265 "I. 1 O Series11! 1 .. 1.26 0 .‘~ _ 1,-*§erie$2, g 3 I ’ 1 o 1.255 , i 0 0.1 0.2 0.3 l Distance (cm ) Figure 3.4.3.1 Deformed Diameter vs. Distance in the First Phase (Steel) In the second phase, according to Equation (3.33) and the above parameters, we have 0.23 —1.415 39.76 CIl [ 0.211 0 —0.005785 1 C2 = 0.005785 2.3832 —8.1064 26.31 C3 0.1212 Solving the above matrix, then C1 = 0.1151; C2 =0.0384ysec/cm; C3 = 0.0060 cm/ysec. Dynamic Yielding Stress: 259-112 0' = 7870 x20'0384e 0384 x104 = 613 (MPa) From Equations (25), (24), (30), (43) and (26) in Section 2, we have: a) Plastic Wave Propagation Velocity C p : 8 -0. 1 15 1 Cp =e 0-0384 (cm/,usec) b) Plastic Dynamic Stress 0': 209-01151) 0'=15.11><(e 003“ )(Pa) 58 c) Displacement Function u(x,t): u(x,t) = 0.1151x + 0.0384x(1n Cp — 1) + 0.006t ( cm ) (1) Plastic Strain 8: a = 0.1151 + 0.03841n C, e) Particle Velocity V: v=—0.1151Cp +0.006( cm/ysec) 1‘) Diameter of Projectile: 0.1151 0.0384 + D = 1.22936(1+ 2 lnCp) (cm) 0.2116m S x S 2.36420m Table 3.4.3.2 gives the values of the strain 8 the stress 0' , C p and the diameter in the second phase from 1.22936 cm S D S 1 1.25925 cm. For this test, although the impact velocity is 59 m/s, its maximum strain is much less than the maximum one in Aluminum 6061-T6511. When the plastic wave stops propagating, its velocity is 499 m/s. The computed static yielding stress is 612 MPa and the given static yielding stress is 585 MPa. These data match very well. The Lagrangian Diagram of the plastic wave propagation is shown in Figure Lagrangian Diagram ( Steel C1045 ). The data for Lagrangian Diagram are shown in Table Lagrangian Diagram Data-3. From Figure 3.4.3.1, Figure 3.4.3.2 and Figure 3.4.3.3 indicate the relationships of the diameter vs. the velocity of the plastic wave, the diameter vs. stress and the stress vs. strain. From Table 3.4.3.1.], the impact velocity and Equation (3.3),61 =0.1 l 125 x 10"6 cm/s. 59 Diameter vs. Plastic Wave Velocity ( SteelClo45) A 1.27 E 1.25 ‘ ...0 g 1.24 - g 1.23 e 0 1.22 . 0 500 1000 1 500 2000 Plastic Wave Velocity ( mls) Figure 3.4.3.2 Diameter vs. Plastic Wave velocity (Steel) Diameter vs. Stress ( Steel C1045) ‘ “l 625 « w m a . O ' 605 1 I g . 1.22 1.23 1.24 1.25 1.26 1.27 Diameter ( cm ) Figure 3.4.3.3 Diameter vs. Stress (Steel) Stress vs. Strain ( Steel C1045) N 625 i § 620 /’ z 615 .. ‘ m 610 : i g 605 1 "’ 600 0 0.02 0.04 0.06 Strain Figure 3.4.3.4 Stress vs. Strain (Steel) 60 Lagrangian Diagram Distance ( cm ) Steel C1045 { j i x 25 1 "‘ 20 - o 0') ‘ 3 15 « .’ l I g 10i .. . i j ._ , e r l— 5 . i 1 O .,..__Q . 0 0.5 1 15 2 1 l I Figure 3.4.3.5 Lagrangian Diagram (Steel) 3.5 Results 1. Three test results indicate that the propagation of the plastic wave can be divided into two phases. The shape curve of the projectile in the first phase is concave, and the shape curve of the projectile in the second phase is convex. The strain is a constant in the first phase. This conclusion is approximate near the impact end. 2. When using our model to determine the dynamic stress of the projectiles, the results are as follows: Static Yield Computed Yield Material Stress ( MPa) Discrepancies ( MPa ) Aluminum 6061-T6511 275 283.92 3.24% Cg)per 145-Hard-H02 32 33 3.125% Steel C1045 585 612 4.6% This table indicates that the model can be used to calculate the dynamic stress of the projectile. 61 3. Many models proposed by other people cannot predict that the shape curve of the projectile in the second phase is convex, as in J ohnson-Cook model. These models can be used in the first phase. 4. Using the model proposed by the author, the dynamic stress, dynamic strain, the velocity of the propagation of the plastic wave and the velocity of the particle can be calculated. 5. In order to verify our theoretical investigation, three of the Taylor tests are carried out for Copper 145-Hard-H02, Aluminum 6061-T6511, and Steel C1045. The experimental results show good agreement between the shape curves of the projectiles predicted by the constitutive equation and experimental results. 6. It should be pointed out that the model proposed by the author does not consider the effect of the strain rate of the propagation of the plastic wave in the projectile. So, the model is not very accurate around the interface between the first phase and the second phase. This will be improved in the next chapters. 3.6 Tables Table-Experimental Data-l-Aluminum 6061-T651 Distance x( mm) Diameter ( mm) 0 13.3525 0.2 1.3275 0.4 1.3275 0.6 13.27 0.8 13.2425 1 13.2075 1.2 13.2175 1.4 13.195 1.6 13.22 1.8 13.1975 2 13.22 2.2 13.23 62 2.4 13.2075 2.6 13.195 2.8 13.1875 3 13.1855 3.2 13.1845 3.4 13.1825 3.6 13.165 3.8 13.1475 4 13.1525 4.2 13.145 4.4 13.145 4.6 13.123 4.8 13.1375 5 13.13 5.2 13.12 5.4 13.12 5.6 13.1 5.8 13.1075 6 13.095 6.2 13.0975 6.4 13.0825 6.6 13.09 6.8 13.0975 7 13.095 7.2 13.08 7.4 13.0875 7.6 13.0925 7.8 13.0875 8.2 13.0875 8.4 13.06 8.6 13.0725 8.8 13.0675 9 13.07 9.2 13.0725 9.4 13.0625 9.6 13.0625 9.8 13.055 10 13.055 11 13.0225 12 12.9975 13 12.9475 14 12.9225 15 12.8725 16 12.8325 63 17 12.8025 18 12.7525 19 12.7225 20 12.7075 21 12.6625 22 12.6375 23 12.5975 24 12.6125 25 12.5875 26 12.555 27 12.5275 28 12.5 29 12.4625 30 12.4575 31 12.4225 32 12.38 33 12.365 34 12.3375 35 12.3325 36 12.3425 37 12.35 38 12.34 39 12.35 40 12.34 41 12.34 42 12.3425 43 12.3375 Table-Experimental Data-Z-Copper 145-Hard-H02 Distance x ( mm) Diameter ( mm) 0 13.1775 0.5 13.12 1 13.035 1.5 13.04 2 13.0075 2.5 13.025 3 13.0225 3.5 13.0275 4 13.0125 4.5 13.0225 5 13.005 64 5.5 12.9975 6 13 6.5 12.99 7 12.975 7.5 12.9825 8 12.9725 8.5 12.965 9 12.9725 9.5 12.96 10 12.96 10.5 12.9425 11 12.945 11.5 12.935 12 12.9125 12.5 12.905 13 12.8875 13.5 12.8725 14 12.8525 14.5 12.8225 15 12.815 15.5 12.7875 16 12.7625 16.5 12.74 17 12.725 17.5 12.7125 18 12.6925 19 12.65 20 12.6725 21 12.6 22 12.5625 23 12.5425 24 12.4975 25 12.505 26 12.485 27 12.4825 28 12.4475 29 12.44 30 12.425 31 12.415 32 12.3945 33 12.415 34 12.4025 65 Table-Experimental Data-3- Steel C1045 Distance x( mm) Diameter ( mm) 0 12.6925 0.5 12.6725 1 12.6 1.5 12.5975 2 12.5925 2.5 12.5925 3 12.58 3.5 12.5925 4 12.58 4.5 12.59 5 12.585 5.5 12.5925 6 12.5825 6.5 12.595 7 12.5925 7.5 12.59 8 12.585 8.5 12.595 9 12.585 9.5 12.565 10 12.5575 11 12.5475 12 12.545 Table 3.4.1.1 Measured Diameter vs. Computed Diameter (Aluminum) Distance x Measured Diameter Computed Diameter cm cm cm 0.00 1.33525 1.33525 0.34 1.3275 1.333 0.69 1.3256 1.3314 1.03 1.321 1.3294 1.38 1.3206 1.3275 1.72 1.3208 1.3256 2.06 1.322 1.3236 2.14 1.32075 1.3217 2.75 1.3191 1.3191 66 Table 3.4.1.1.] Stress vs. Strain (Aluminum) x (mm) 8p, 0p1(MPa) 0.00 0.1723 509.33 0.34 0.1596 503.31 0.69 0.1565 501.79 1.03 0.1491 498.09 1.38 0.1484 497.74 1.72 0.1487 497.74 1.72 0.1487 497.88 2.06 0.1507 498.90 2.41 0.1486 497.84 2.75 0.1459 496.45 Table 3.4.1.2 Stress, Strain, Plastic Wave Velocity vs. Diameter (Aluminum) D(cm) C p (m / s) a 0'( MPa) 1.31875 13644 0.1454 813 1.31855 13568 0.1451 806 1.31825 13410 0.1446 795 1.3165 12557 0.1418 731 1.31525 11953 0.1397 689 1.3145 11621 0.1385 667 1.31375 11298 0.1373 646 1.312 10554 0.1344 600 1.31 9790 0.1312 556 1 .3095 9608 0.1304 546 1.30925 9519 0.13 541 1 .309 9430 0.1296 536 1.30875 9341 0.1292 531 1 .3085 9233 0.1287 525 1.30825 9147 0.1283 521 1 .30725 8809 0.1267 504 1.30625 8485 0.1251 488 1 .3055 8249 0.1239 477 1 .30225 7284 0.1 186 434 1.29975 6616 0.1145 408 1 .29475 5470 0.1064 368 1 .29225 4968 0.1023 357 1.28725 4108 0.0942 331 1.28325 3527 0.0877 319 1 .28025 3143 0.0828 308 1 .27525 2599 0.0747 303 1 .27225 2184 0.0673 297 1 .27075 1840 0.0673 297 1 .26625 1671 0.06 293 1.26375 1435 0.0559 291 1 .25975 1408 0.0494 289.76 67 1 .25925 1382 0.0486 289.55 1 .25875 1220 0.0478 289.34 1.2555 1 100 0.0425 288 1 .25275 1050 0.0381 290 1.25 990 0.0336 289.5 1.24625 858 0.0275 286.6 1 .24575 842 0.0267 285.96 1 .24225 736 0.021 285.46 1.238 626 0.0141 284.28 1.2365 590 0.0116 284.18 1 .23375 450 0 283.92 Table Lagrangian Diagram Data-l Aluminum 6011-T6511 Distance (cm) Time (gs) 0.28 0.299305 0.3 0.322824 0.34 0.369525 0.36 0.414603 0.38 0.448642 0.4 0.480192 0.42 0.516859 0.48 0.605525 0.5 0.646663 0.52 0.67253 0.56 0.8 0.6 0.87146 0.62 0.907893 0.66 0.974601 0.82 1 .231046 0.86 1 .345642 0.94 1 .520544 0.98 1 .624938 1 .1 2.031394 1 .2 2.407222 1 .3 3.077652 1 .4 3.600823 1 .5 4.553734 1 .6 5.545927 1 .7 6.508423 1 .8 8.130081 1 .9 9.481038 2 10.48768 2.1 12.7815 2.2 14.55027 ON 00 2.3 17.3716 2.4 18.41903 2.5 19.516 2.6 22.6087 2.7 25.71429 2.8 29.19708 2.9 34.23849 3 36.01441 3.1 41.77898 3.2 49.68944 3.3 53.83361 3.4 60.71429 Table 3.4.2.] Measure Diameter vs. Computed Diameter (Copper) Distance x Measured Diameter Computed Diameter (cm) (cm (cm) 0 1.31775 1.31775 0.044125 1.3152 1.3158 0.08825 1.3115 1.3139 0.13238 1.3038 1.3120 0.1765 1.3016 1.3101 0.22063 1.3025 1.3082 0.26475 1.3020 1.3063 0.30887 1.3023 1.3044 0.353 1.3025 1.3025 Table 3.4.2.1.1 Strain vs. Distance (Copper) Distance x Strain (cm) 0.00 0.1438 0.044125 0.1396 0.08825 0.1336 0.13238 0.1211 0.1765 0.1175 0.22063 0.11898 0.26475 0.118175 0.30887 0.11866 0.353 0.1189 69 Table 3.4.2.2 Stress, Strain, Plastic Wave Velocity vs. Diameter (Copper) D(cm) C p (m / s) 6‘ 0'( MPa) 1 .30275 293682 0.1 194 34699 1.30125 264461 0.117 28144 1 .3005 249866 0.1 157 25094 1.3 241288 0.1149 23400 1 .299 225005 0.1 133 20349 1.2975 202617 0.1 109 16534 1.2973 198241 0.1 104 15796 1.2965 188121 0.1092 14257 1 .295 169403 0.1068 1 1567 1.2943 160754 0.1056 10419 1.294 157971 0.1052 10063 1.2935 151822 0.1043 9505 1 .29125 129788 0.1007 6803 1 .28875 108512 0.0966 4766 1 .2853 84601 0.0909 2809 1.2815 64817 0.0848 1722 1 .27625 44719 0.0763 836 1 .2725 34261 0.0702 504 1 .26925 27182 0.0649 330 1.265 20111 0.058 195 1.26275 17116 0.0543 150 1.26 13875 0.0495 110 1 .2563 10770 0.0437 79 1 .2543 9365 0.0405 68 1 .2498 6809 0.0332 51 1 .249 6462 0.032 49 1.2485 6213 0.0311 48 1 .2448 4759 0.025 42 1.244 4516 0.0238 41 1 .2425 4067 0.0214 39 1.24 3400 0.0173 37 1.235 2387 0.0092 35 1 .23 1668 0.001 34 1 .2294 1597 0 33 70 Table-Lagrangian Diagram Data-2 Copper—145—Hard—02 Distance ( cm) Time (,us) 0.35 0.011966 0.4 0.015178 0.5 0.019989 0.6 0.024835 0.65 0.028841 0.7 0.034474 0.8 0.04009 0.85 0.044875 1 0.05859 1.05 0.06482 1.1 0.069098 1.15 0.074785 1.2 0.091289 1.3 0.117634 1.4 0.161584 1.5 0.224692 1.6 0.345252 1.7 0.476097 1.8 0.631889 1.9 0.896311 2 1.103205 2.1 1.402431 2.2 1.906908 2.3 2.290837 2.4 3.268864 2.5 3.587315 2.6 3.862725 2.7 5.205321 2.8 5.687589 2.9 6.538895 3 8.04721 3.1 11.77364 3.2 17.2043 3.3 18.54975 71 Table 3.4.3.] Measured Diameter vs. Computed Diameter (Steel) Distance x Measured Diameter Computed Diameter ( cm) ( cm) 0 1.26925 1.26925 0.02875 1.267 1.268 0.0575 1 .266 1.2667 0.08625 1 .264 1 .2655 0.115 1.263 1.2642 0.14375 1.261 1.263 0.1725 1.26 1.2617 0.20125 1.259 1.2605 0.23 1.25925 1.25925 Table 3.4.3.1.1 Strain vs. Distance (Steel) Distance x ( cm) 0.00 0.06489 0.02875 0.0612 0.0575 0.0596 0.08625 0.0563 0.1 15 0.0547 0.14375 0.0515 0.1725 0.0498 0.20125 0.0482 0.23 0.0486 72 Table 3.4.3.2 Stress, Strain, Plastic Wave Velocity vs. Diameter (Steel) D(cm) C p (m / s) 6‘ 0"( MPa) 1.25925 1010.3 0.0486 630.99 1.25908 1010 0.0484 630.79 1.25905 1009.9 0.0483 630.69 1.259 1008.8 0.0482 630.60 1.2583 1008.7 0.0471 630.06 12.5825 1008.6 0.047 629.56 1.2582 1008.5 0.0469 629.47 1.258 1008.2 0.0466 629.38 1.2575 1007.4 0.0458 628.42 1.256 1005 0.0433 626.41 1.2557 1003.3 0.0417 625.26 1.2547 1002.8 0.0412 624.92 1.2545 1001.7 0.0401 624.19 1.253 1000.1 0.0385 623.22 1 .2522 999 0.0372 622.48 1 .252 998 0.0368 622.27 1.25 995 0.0336 620.69 1.2485 993 0.0311 619.63 1.247 987 0.0257 617.76 1.24725 986 0.0251 617 1.22936 499 0 612 Table Lagrangian Diagram Data-3 Steel C1045 Distance (cm) Time (,us) 0.2 1.1293 0.4 2.2779 0.6 3.5356 0.85 5.5087 1 6.7613 1.2 8.47457 1.3 9.5658 1.5 11.5119 1.7 14.20217 1.8 17.0778 1.9 19.40756 73 CHAPTER 4 ANALYSIS FOR PLASTIC DYNAMIC MODELS IN THE TAYLOR TEST Today, the Taylor test has not ofien been used for its original purpose of obtaining dynamic, yield stress of materials. It has come into its own purpose in recent years for checking constitutive equations by comparing the shapes of projectiles with predictions. From the experimental results, there are three main characteristics in the Taylor test. The first one is that in the second zone of specimens the shapes appear to be like mushroom. The second one is that as the propagation of the plastic wave the strain and the velocity of the plastic wave will decrease. The third one is when the propagation of the plastic wave stops the strain is zero but the velocity of the plastic wave propagation is not zero, either. The constitutive equation proposed in this thesis has predicted the three characteristics. In this chapter, we will use models proposed by other persons to verify whether or not these models can predict the three characteristics in the specimens of the Taylor test. 4.1 Analysis for Power Law Plasticity Model This is an isotropic plasticity model with rate effect, which uses a power law hardening rule. Elastoplastic behavior with isotropic hardening is provided by this model. The yield stress is a function of plastic strain and can be written as 0' = k(se + 8)" where 8e is the elastic strain to yield , 6' is the effective plastic strain, k is strength 74 coefficient and n < l is hardening exponent. When this model is used to describe the propagation of plastic wave in the second zone of the specimen, the procedures of solving the equation of plastic wave in Appendix -D is followed. Based on this model, it yields T(a) = _c_i_0'_ = nk(se + a)"—1 d8 Let 8: f (C ), where Cp — is the velocity of the plastic wave. Then :I 717161.11: nk[8e + 7831,11“ (4.1 1 Taking Equation (4.1) back into Equation (D-S), it yields 2 _ k C "—1 pCp—n [86+f( p)] 1 zpnCp—_1_ 8e (4.2) According to von Karman’s solution in Appendix - D, the following equations can be achieved: The displacement function 2 1 10C —— u(x, t)- kp 1 — 8e]de COP A A = A11Cp 12t — X88 _t(A11CpO 12 — Cpoge) (4.3) 1 1 n+1 . . .. . where A“ = ———(— '0‘)” A12: and C p0 1s an1n1t1al plastic wave n+1 nk n— propagation velocity. 75 If the procedures in Chapter 3 are followed, the radius of the specimen, r , in the second zone is 2 1 r=MU+EKiE __1_ )n—1 — 8e]} The second derivative of r about x is 2—n 1 (3_") p )"b-i(x);7] (4.4) (n — 1)2 nkt2 7| 4 r": rol: where r0 is the initial radius of a specimen. Now we check three main characteristics in this model one by one: A. It can be seen that after impact, the radius of the specimen will become a curve. But because r" is greater then zero, the shape of the specimen does not appear to be mushroom. This does not match the first characteristic. B. As the plastic wave propagates, if the strain decreases, the velocity of the plastic wave will increase. This one does not match the second characteristic. C. When the strain is zero, the velocity of the plastic wave is not zero. This one matches the third characteristic. Example: Suppose in the Taylor Test, a specimen is made from Aluminum 1 100, with parameters as follows: Aluminum 1100, k = 0.598, n = 0.216, p = 2710kg / m3 Then, from the foregoing formulas, the displacement function, strain and the radius and the second derivative of the specimen in the second zone can be obtained as follows: 76 a = 1.682 x 10‘7 C132“ u(x,t) = —4.789 x 10’6C131-55t + 1.4789 x 10‘6C131'55t r = r0[(1+ 8.41 x 10'8(5-)2-55] X 8 t2.55 ) x455 r": r0(76.13 x 10— In this example, because r" > 0 then the contour of the specimen is not like a mushroom. Summarizing, this model only matches one characteristic of the plastic wave propagation in the Taylor test. The other two characteristics cannot be matched. 4.2 Analysis for Hartig Model This model was proposed by Ernst K. Harti g in 1893. It was used to investigate the nonlinearity in small deformation and tangent modulus of metals, especially in cast iron, for both tension and compression. It can be written as E —b 0'=—01—e g b( ) where E 0 is zero stress modulus and b > 0 is a constant. Similarly, if the same procedures are followed in Appendix - D, based on this model it yields d0 —b£ T = — = E = E (<9) d8 09 0 e‘bf(Cp) :5 T(.9) = E0e_bf(CP) (4.5) Taking Equation (4.5) back into Equation (D-S) in Appendix - D, it yields 77 P 1111—5—63] 0 => 8=f(C,,) = ——b—— According to von Karman’s solution in Appendix - D, the following equations can be achieved: The displacement function Cp big—(5,2)] u(x,t) =t j ———0—de . 8,, b i 1 2 = 431117336 — Cpt) + guano}, - 1) — Cp0(1nCp0 —1)t]} 0 The radius of the specimen in the second zone is _ __1_ 202 r-ro[1 2b1n( E0 )1 The second derivative of r about x is r" = r0 bx2 Now we check three main characteristics in this model: A. It can be seen that after impact, the radius of the specimen will become a curve. But r" is greater then zero ( b > 0 ), the shape of the specimen does not appear to be mushroom. This does not match the first characteristic. B. Under the condition of the strain absolute values being taken, if the strain decreases, the velocity of the plastic wave will decrease. This one matches the second 78 characteristic. When the strain is zero, the velocity of the plastic wave is not zero. This one does not match the third characteristic. So it can be concluded that because r" is greater then zero, the shape of the specimen does not appear to be mushroom. This is a characteristic not matching the experimental results. 4.3 Analysis for Rate Sensitive Power Law Plasticity This one models strain rate sensitive material with power law hardening. This model follows a constitutive relationship of the form: a = kam 6" where k,m,n are material constants and m < 1 and n < 1. When this model is used to describe the propagation of plastic wave in the second zone of the specimen, the strain rate must be set to a constant. The procedures of solving the equation of plastic wave in Appendix - D are followed. Based on this model, it yields d . _ T(£) = _0' = mkgnam 1 d8 Define 8 = f(Cp), it yields pC; = mk.é".e’"'1 (4.6) Taking Equation (4.6) hack into Equation (D-S) in Appendix - D, it yields 2 1 PCP )E mké‘" 8=f(Cp)=( 79 According to von Karman’s solution in Appendix - D, the following equations can be achieved: The displacement fimction: u(x,t) = A11t(C:12 — C562 ) 1 m—l —‘ m+1 ( p )m‘l, 412: m+1mkgn m—l where A11 = The radius of the specimen in the second zone is 2 r=r0{l+l( pCp 2 mké‘" __L )m"1} The second derivative of r about x is l 4—2m 3 — m —— —— r"=r0[( ), ‘1, 214100511 (m — 1) mks t where r0 is the radius of a specimen. Because r" is greater than zero, the shape of the specimen does not appear to be a mushroom. Example: Take A356 Aluminum as an example with the material parameters being m = 0.7 ,k = 0.002Mpa and n = 0.32. At the same time suppose 8‘ = 106 US , then from the foregoing formulas, the displacement function, strain and the radius and the second derivative of the specimen in the second zone can be obtained as follows: _ —6.67 .c: .. 1310,, , u(x,t) = —30.17t(C;5-67 — 93367). r = r0(1+ 65.67C136'67), 80 2 r": r0(3359 x t0'3x78'67) In this example, because 7'" > 0 then the contour of the specimen is not like a mushroom, it does not match the first characteristic. As the plastic wave propagates, if the strain decreases, the velocity of the plastic wave will increase. This one does not match the second characteristic. When the strain is zero, the velocity of the plastic wave is infinite. This one does not match the third characteristic. So, there are not characteristics to match the ones of the propagation of the plastic wave in the Taylor test. 4.4 Analysis for Johnson-Cook Model The J ohnson-Cook model is mainly used for problems where the strain rates vary over a large range and adiabatic temperature increases due to plastic heating cause material softening in most metals. Typical applications include explosive metal forming, ballistic penetration and impact. Without considering thermal effect, this model can be expressed a = (A + Ba")(1+ Clné) where A ,B ,C are material constants and n < 1. When this model is used to calculate the propagation of the plastic wave in the Taylor Test, the procedures in Appendix - D are followed and the strain rate is supposed to be a constant. From this model, it yields T(a) = d_0' = Bn(1+ C1n.é).9”—1 d8 Let 6‘ = f(Cp),thus 81 pC}, = 1‘5 = Bn(1+ Clné)£"_1 d8 8C2 Bn(1+ Clné) 1 => 8:80,.) =1 11* (4.7) According to von Karman’s solution in Appendix - D, the following equations can be achieved: The displacement function u(x,t) = A1 1t(CI’,412 — C262 ) 1 n—1[ P ],,—__1‘, A12:n+1 n+1nB(1+C1né) n—l where A1 1 = The radius of the specimen in the second zone 1 _ 1 PC?) ;__ r_r0{1+2[Bn(1+Clné)] 1} The second derivative of r about x is 1 4—2n 213300 "-1 1 (3 - n) p (n — 1)2 Bn(l + C1né)t r": r(){ where r0 is the radius of a specimen. Because r" is greater then zero, the shape of the specimen does not appear to be a mushroom. In the next four examples, four materials are specimens in the Taylor tests and the J ohnson-Cook model is used to compute the plastic wave propaga- tion, where set 5‘ = 1061/s. 82 Example] Amco Iron, A =175MPa, B = 380MPa, C = 0.06, n = 0.32 p = 7890kg / m3 Computed results as follows: a = 121576cg2-94, u(x,t) = —626301(C,;‘-94 — C1394), r = r0(1+ 60788C132'94), r": r0(704147 x x'4'94t2'941). Example 2 OFHC Copper, A = 89MPa, B = 292 MPa, C = 0.025, n = 0.31 ,0 = 8330kg / m3 Computed results as follows: .9 = 81658C52-89, u(x,t) = —43010t(C;1’898 —- (33898), r = r0(1+ 40828C132'89), r": r0 (458992 x x74‘89t2'898). Example 3 4340 Steel, A = 792MPa, B = 510MPa, C = 0.014, n = 0.26 83 p: 7830kg / m3 Computed results as follows: a = 29695C132'7, u(x,t) = —174401(C,;‘-7 — 3,”), r = r0(1+14847C;2'7), r": r0(148326.53 x x‘4'7t2'7). Example 4 7039 Aluminum, A = 336MPa, B = 343MPa, C = 0.01, n = 0.41 p = 2720kg / m3 Computed results as follows: a = 2429621C53-38, u(x,t) = —1016650t(C;2'39 — ,3339), r = r0(1+ 1214811C133'38), r": r0(17984548 x {53813-3891 The foregoing formulas and the common characteristics of the above four examples show that: A. Although the material is different, the radius of the specimen will become a curve in the second zone. But because 7'" s are greater then zero, the shapes of the 84 specimens do not appear to be mushroom. This does not match the first characteristic. B. As the plastic wave propagates, if the strain decreases, the velocity of the plastic wave will increase. This one does not match the second characteristic. C. When the strain is zero, the velocity of the plastic wave is infinite. This one does not match the third characteristic. In his paper, Maudlin also proves that the J ohnson-Cook model cannot describe the propagation of the plastic wave in the Taylor Test. This conclusion matches our one. 4.5 Analysis for Malvern Model One of the earliest models used to describe rate-dependent material behavior is due to Malvern in 1951. The model can be written as 0' = (71(8) + aln(1+ bé) where 01(8) is the static stress and a and b are constants which describe the rate sensitivity. The linear form of the Malvern model is 0' = 01(8) + ké where k is a material constant. For Harden Aluminum, the linear form of Malvern is that 0' = (20000 — 19) + 10000006 8 Suppose a specimen in the Taylor Test is made from Harden Aluminum. The linear Malvern model is used to describe the propagation of the plastic wave in the second zone, then it yields 85 d5 6'2 Let £=f(Cp),then 2 -1219 pCp — d8 52 Taking the above equation back into Equation (D-5) in Appendix - D, it yields 10 1 6=f(C,,)= —— (4.8) pCp From this formula, if the velocity of the plastic wave propagation gradually decreases, the strain 8 will increase. Apparently, this one does not mach the experimental results, that is, the second characteristic. When the strain is zero, the velocity of the plastic wave propagation is not a constant. This one does not match the third characteristic of the Taylor test. According to von Karman’s solution in Appendix, the following equations can be achieved: 1 0 The displacement function u(x,t) = t — ln—p— p CpO The radius of the specimen in the second zone is r=r0(1+-1— EL) 2 ,0 CP The second derivative of r is r": r0(1+ -1— End) 2 p 86 where r0 is the radius of a specimen. Because r" is greater then zero, the shape of the specimen does not appear to be a mushroom. This one does not match the first charac- teristic of the Taylor test. 4.6 Analysis for Strain Rate Dependent Plasticity In this model, the yield stress is defined as 0' = 0'1 (6') + E p8 where E p is plastic tangent modulus and is given in terms of Young’s modulus and the tangent modulus. 01(8) is the stress caused by the strain rate. do T =—=E (8) d3 p If this model is used in the Taylor test, then first define 6' = f (C p ) and it yields 2_d0'_ pCp_—CE_E (4.9) .0 Because E p is a constant, the plastic wave velocity will be a constant and the strain is a constant. According to this model, the shape of the specimen will be a straight line. For example, for 4140 Steel, p = 7850kg / m3 and E p = 22 x 105 Pa. If the procedures in Appendix- D are followed, then one can calculate that the strain is 8 = 0.190’ and r = r0(1 + 0.09350). The shape of the specimen is a straight line. These conclusions totally do not match the experimental results. 87 4.7 Analysis for Power Law The Power Law can be written as 0' = 01(8)8n where n < l is a material constant and 0'1 (8) is quasi-static stress. Set .6" be a constant. Based on this model, it yields 712:}? Let .9 = f(Cp), it yields 71f(Cp )]=—-8" (4.10) Taking the above equation back into Equation (D-S) in Appendix -D, it yields 2 _§0-18n pC= ‘0 0'19 2 60' Define inv(— ——)is an inverse function of —(8), then g" 019 PCp 8=f(C,,)= T,-,W(.— ) (411) According to von Karman’s solution in Appendix - D, the following equations can be achieved: “(x t)=tC.1T inv(p:-—_p -——)de So it can be seen that the expressions of strain and displacement depend on the expressions of the material model. Only the model is given for a material; the Power 88 Law can be used in the Taylor Test. 4.8 Analysis for Logarithmic Law Logarithmic law can be expressed in the form 0' = 01(8) + k 1n .9 where k is amaterial constant and 01(8) is quasi-static stress. Set 6‘ be a do 30' constant. Based on this model T (8) = —- = --—l , let 8 = f (C p ) , it d8 38 . a welds: 71f (Cp )1 = g 070' ' Define 72m, (pC; ) is an inverse function of 271(8) , then 8 8=f(C,,)=72..(pCf,) (4.12) Taking Equation (1) back into Equation (D-5) in Appendix, it yields C 2 P pC p u(x,t) =t I Env(—_n_)de C W 6' So it can be seen that the expressions of strain and displacement depend on the expressions of the material model. Only the model is given for a material; the Power Law can be used in the Taylor Test. 89 CHAPTER 5 THE NUMERICAL ANALYSIS FOR PLASTIC WAVE PROPAGATION OF RD THEORY In Chapter 3, RI theory is used to investigate the plastic wave propagation in the Taylor test. That is, the effect of strain rate is not considered. If the strain rate is considered, RD theory is applied to investigate the plastic wave propagation in the Taylor test. It is very difficult to get the analytical solution of the plastic wave equation of RD theory, so numerical methods are used. When numerical methods are used, the numerical computation is along the characteristic curve of the plastic wave. The plateau of RD theory, the characteristic equation, characteristic curves and Eulerian method and Hartree methods are discussed. The Eulerian method is an explicit numerical method and the Hartree method is an implicit numerical method. 5.1 Plastic Wave Equation Of RD Theory And Its Plateau Shown in Figure 2.1, assume a rod with length L is subjected to impact loading with a constant velocity V0. The relationship among stress, strain and strain rate is 0' = 0(8, 8) , where 0' is stress, 8 is strain and 8‘ is strain rate. The transverse effect is neglected. 0' = 0(8, 6‘) holds for plastic deformation of the material. The boundary conditions are u(0,t) = Vot and u(L,t ) = O , where t p is the time of the plastic wave reaching the other end of the rod. From Newton’s Second Law, the plastic wave equation with the boundary conditions is given as follows: 90 421.6288 48 012 6851’- 6601261 (5.1) B.C: u(0,t) = VOt and u(L,t )= 0 where p is the mass density, 1‘ is time, u = u(x,t)is displacement function; 6' = — According to experimental results, the plastic wave propagation in the Taylor test can be divided into two phases. In the first phase, the strain is a constant. Physically, it is called plateau. It can be seen that there is a linear solution with the following form: u(x,t) = V0(t + 11) (5.2) 0 with a constant, k0 , satisfying Equation (5.1). And the first boundary condition is satisfied at x = 0. From this solution, the strain 8 and the velocity of the particle v are obtained as follows: c9='03l—=£/9 (5.2.1) dc 0 fix v:--=V 5.2.2 07 O ( ) Equation (5.2.2) is the first solution to Equation (5.1). For this solution, the strain is a constant and the velocity of the particle is equal to the impact velocity V0. Physically, the constant strain is the plateau. That velocity of the particle being equal to V0 means the plateau is near the impact end. So it can be concluded that there exists a plateau and this plateau is near to the impact end. Because the first solution to Equation (5.2) is a 91 linear solution, and it always satisfies the plastic wave propagation equation, this linear solution is independent of the constitutive equation of the material, 0' = 0'( 8, 5‘) or 070' 0' E— and 27. This means that no matter what kinds of models have been proposed, the 8 8 plateau always exists. That is, RD theory can predict the existence of the plateau and the plateau is near the impact end. Basedon this conclusion, RD theory and RI theory are the same in the first phase. Because the linear solution or the plateau is near the impact end, it gives boundary conditions for the second phase as follows: k0 8(xO,t0)=']I:—O' (5.11) 0 V(X0,t0) = V0 (5.32) where x0 is the coordinate of the plateau and to is the time of the plateau ending. 5.2 The Characteristic Equation In order to investigate the propagation of the plastic wave in the second phase, the characteristic equation of the plastic wave equation should be gotten first and the numerical method applied to the characteristic curves. Recall that the equation of motion @_60 ___ 5.4 pg, dc ( ) 92 During the propagation of plastic wave, v and 0' are function of x and t, then the increments of v and 0' can be written as dv = fldx + fldt 5x 0‘! d0 =€gdx + ZJ—dt 5x 01 The following strain-velocity relation should be satisfied: 2-3: dc d (5.5) (5.6) (5.7) In the RD theory, 0' = 0(8, 8). Taking the derivative of the stress 0' with respect to t , it yields 22 _ 43.3 . @224: d 58 d 5'6" d 60' 0' Define T = — and T1 = —T, the above becomes 55 0’19 0"_0'_ 0"_0' 0'16 00' 0"_8' 0'19 ———+ =T 01 07.901 07.901 01 or ———— —— Substituting Equation ( 5.8 ) into Equation ( 5.7 ), it yields 0'1)_1§0' T016 03c Ta T01 61 1 56 T, 53 3 ___—___ ___ dcTél T01 Gather Equations (5.4), (5.5), (5.6) and (5.9) together 93 +Tl— 0'19 62‘ (5.8) (5.9) 0'12 100'=0 01 péc dxfl+dtfl=dv dc 0‘! dx2q+dtéq=d0 0x 01 0’1» 10’0'_ 71019 18. 707" rd and write the above equation in the form of matrix 0‘1» ' 1 "— r — 1 __ a 0 0 ,0 0 Q: dv 2) dx dt 0 0 5:; ___ d0 (5.10) 1 o o —— 751 1 LEE _ Lot. The characteristic equation of the plastic wave equation can be obtained by letting determinant of the coefficient matrix of Equation(5.10) be equa1 to zero. From the above matrix, we obtain the characteristic equation dx 100' __=_ /__ (5.11) dt p06 As we known, when A5? = b and det(A) = 0 , where A is a matrix, 55 is a column vector and b is a vector, then a necessary condition for the existence of finite solution for 55 is when I) is substituted for any column of A the resulting determinant must also 94 vanish [Ames,36]. Taking the right column in Equation (5.10) into 1St column of the matrix, it gives the stress increment along the characteristic curves 078 d0=pdev+715dL Ipr>0; (5.12) d - d aéd C 0 0'——,0Cp v+Tlg t, If p< . (5.13) where Cp = d—x. dt If one considers the equation —— = 29- , then dx dv = —d8 = de8 dt Take it into Equation (5.12) and (5.13), then it yields 028‘ do = 2d.9+ T —dt 5.14 do = —,0C2d£+ T1 @dt (5.15) P 0 Equations (5.14) and (5.15) describe differential relationships for a variety of variables of the plastic wave propagation and provide the theoretical bases to use numerical methods to compute the variables of the plastic wave propagation along the characteristic curves. 5.3 Numerical Analysis Methods There are many numerical methods which can be used to compute variables in plastic wave propagation. Here only two methods are discussed. The first method is an 95 explicit numerical method, it called Eulerian method. The second method is an implicit numerical method, it is called Hartree method. 5.3.1 Eulerian Method From Equation (5.14), the increment of the stress along the characteristic curve, if Cp > 0, can be written as do = pcjdg + T1 gar: In order to use numerical method along the characteristic curve, we use (10' z A0: 0'(i+1)—0'(i) for point (i + l) and point i , where i = O,1,2,3- ° - , and notice that 15 ~93. _ .._ .. __.. _8(i+1)-£(i)_8(i)-8(i—1) 01 dt ~ [ At (1)]At — A80) —[.9(z +1) 5(1)] — At At _ a(i+1)+a(i—1) — At then it gives Eulerian method as follows: x(i+1)= x(i) + Cp(i)At (5.16) 00' + 1) = 0(1) + 72C: (0141 + 1) — 401+ 711015“ + "if“ " 1)1 (5.17) . . 1 Cp(i+1)={ (“MD—0(1) 2 (5.18) 10160 + 1) - 80)] This method is very simple. If the initial values are given, then the computation 96 can go step by step along the characteristic curve. However, it asks that before computing variables at point (i + 1) , the values at point i must be known. Especially, the strain at point (i + 1) must be given. But the convenience of this method is not very good. 5.3.2 Hartree Method Hartree method is an implicit method along the characteristic curve. This method can be introduced as follows. Suppose that a fixed rectangle grid is imposed on the integration with distance interval Ax and time interval At , as shown in Figure 5.1. The relations along the characteristics arethat d0+pC12,ds+T1—g£dt=0 2 55‘ d da— d8+T——dt=0 an pCp 1a Define Ga =pC2, Ha, =7j—Zf, Gfl=—pC; and Hfl=7]-§-,thenonehas d0+Gada+Hadt=0 (5.19) d0+Gfld8+Hfldt=0 (5.20) Suppose that solution is known at the mesh points and R, S, are equally spaced along the next line, t = (j + 1)At . Draw the a and fl characteristics from Equation (5.19) and (5.20) through R, back to their intersection with first line t = jAt. These two points 97 of intersection are unknown. We wish to calculate O'R, 8R, xp and xQ. The Hartree’s computational form is as follows: 1 x, — x, = §[CP(R) + Cp(P)]At (5.21 ) l xR - xQ = E[-C,,(12) — Cp(Q)]At (5.22) (a). — ap)+%1pC§(R)+pC§(P)1(sR —ap)+ 1 A6 A6 _ 71781030055 T.(P)-§ 0), can be written as 00 + 1) = 0(1) + pC,:(i)Aa(z') + Twig-(i) As from the Modified Malvem model, T = k , then the above equation becomes 0(i+1) = 0R, (i) + pC:(i)Ag(i) + ki—fh‘) (6.3) Take Equation (6.2) into Equation (6.3), it gives 218(0-C1 )1 0'(i + 1) = 8%— {9 C2 }+ ké(i) + pCf,(1)A.e(i) + k%(i) (6.4) From Malvern experimental results, the physical parameter k is given k = 1 x 106 Pa sec. Suppose average strain rate during the plastic wave propagation is a constant, 5‘ = 100 l / sec. Equation (6.4) becomes 2180' )-C1 )1 , _ pC2 C2 . . 2 . . . 0'(l +1) — ——2 {e }+ k8(l) + pCp(1)[£(z +1)— 8(1)] ( 6.5 ) From test data in Chapter 3 and the above equation, the computed results are shown in Figure 6.2 and Table 6.2. 103 Stress vs Duistance in the Second Phase ( Eilerian Method ) 600 ...-.....-...-.._. - _ O A 400 . .0 O O ”O O 200. e e g .° 0 e . 0 r m ~ — e — A 2 -200 9° . 14 20 30 40 a O m -400 . O -500 - _ ..._____._..-._~__L ______ Distance ( mm ) Figure 6.2 Stress vs. Distance in the Second Phase ( Eulerian Method ) Table 6.2 Eulerian Method Distance ( mm ) 0'( MPa) 2.8 0 3.4 441.71 3.9 -314.45 4.2 182.58 5 142.01 5.6 -238.64 6.3 257.8 6.4 401.15 9.4 -129.57 11 -179.7 13 -499.85 16 120.35 18 233.3 20 114.47 22 230.61 25 281.39 26 341.94 28 359.87 30 355.01 32 368.51 33 361.39 34 368.13 It can been seen that results from the Eulerian method are not accurate. The convergence is not good from x = 0 to x = 18mm. Afier that distance, convergence 104 tends to be better. From x = 0 to x = 18 mm, the stresses computed from the Modified Malvern model are less than the stresses computed fiom the RI theory. Gradually, the stresses computed from the RD theory are greater than the stresses computed from the RI theory. It indicates if the plastic wave propagates very fast, this method cannot describe the propagation of the plastic wave well. Only when the strain is very small, the propagation of the plastic wave can be described by this method. 6.2.3 Hartree Method When the Modified Malvern model is used to the Hartree method, the Hartree method from Chapter 5 can written as Notice that XR— ( ( k 2 A8( R) A£( P) xp = %[CP(R) + Cp(P)]At — x9 = ér—CAR) — Cp(Q)]At 0‘1 — 0.) +%1pC;(R)+pC:(P)1(a. — a.) + k _[Aé_(_R) Aé(P) ———]A—t _ 2 At A 0. —aQ)+%1—pCZ(R)—pcz (0)109. - eQ)+ k _[_A_g'(R) Aé(Q) ——]A-t _ 2 At At At At IAt {EKER —8A )+(8A _EPH :kgk—8P_£P—£R k =—8-8 At At ] At[R P] 105 (6.5) {101. —e‘.)1 k Aé [58' k S' '11, ——R+— At=——£—£ Imlary 2[At( ) At(Q)] Atl R p] Then the Hartree method can be written as l x, — x, = -—2-[CP(R) + Cp(P)]At 1 xR _ xQ = §[—Cp(R) _ Cp (Q)]At (a. — 0,) + -;-[pC§(R) + pC:(P)1(aR — 8p) + glen. — 6P] = o (a. —aQ)+-;—[—pC;(R)—pC;(Q)J(aR whim —eQ] = 0 (6.6) Because Equation (6.6) is a non-linear system of equations, when applying Equation (6.6) to calculate the dynamic stresses, the initial values of x P , xQ , C P (R) , C p (P) , C P (Q) , 8p , 8Q , 0'}; and O'Q are very important. There are two methods to apply Equation (6.6) to calculate the dynamic stresses in the second phase. The first method is to use arbitrary mesh points to calculate the dynamic stresses. Since the length of the projectile is not very long, there is not enough space to get many test data. In order to get other data, the interpolation method must be used. However, the errors from the interpolation functions affect the accuracy of thecomputation. The second method is to use the particular mesh points to calculate the dynamic stresses. That is, the points calculated from RI theory’s values are used as mesh points to compute the dynamic stresses. Next, Equation (6.6) is used to compute the stresses and strains of Aluminum 6061-T6511 in the second phase of the Taylor Test step by step. 106 6.2.3.1 Using Arbitrary Mesh Point to Calculate Dynamic Stresses t For the next computation, At = 2% = 1.1815 #3 1. Get the values of RI theory at different mesh points, 5 points, shown in Table 6.3. Table 6.3 Initial Values for Hartree Method xR (mm) 0R, (Mpa) 6‘ CI) (m/s) 0 530.79 0.1454 9355 8.6 398.66 0.1269 6391 17 302.36 0.0828 2612 25.5 287.68 0.0478 1281 34.28 283.92 0 560 2. Using different values of xto get the interpolation functions of 0R1: 8 and C p about x. Here Lagrangian interpolation method is used. 0R1 (x) = 0.00415(x — 8.6)(x — l7)(x - 25.5)(x - 34.3) - 0.00127x (x — 255)(x — 34.3) — 0.0l44x(x — 8.6)(x — 17)(x — 34.3) — 0.0087x (x — 8.6)(x — 17)(x - 34.3) — 0.0021x(x — 8.6)(x — 17)(x — 25.5) ( 6.7 ) £(x) = 1.147 x 10_6(x — 8.6)(x — 17)(x — 255)(x — 34.3) — 4.038 x 10-6x )(x — 17)(x — 255)(x — 34.3) + 3.94 x 10‘6x(x — 8.6)(x — 17)(x — 34.3) — — 1.48 x 10‘6x(x — 8.6)(x - 17)(x — 34.3) (6.8 ) 5p (x) = 0.073(x — 8.6)(x — l7)(x — 255)(x — 34.3) — O.2x(x — 17)(x - 255) (x — 34.3) + O.124x(x - 8.6)(x — 255)(x — 34.3) - 0.039x(x — 8.6)(x — 17) (x — 34.3) + 0.043x(x — 8.6)(x — 17)(x — 255) (6.9) 107 3. Using the below equation to get initial values xp = xR — Cp(A)At xQ -_- xR + Cp(A)At (6.10) 4. Using x p and xQ computed from Equation (6.10) to compute interpolation values O'P, Hg, 8}), EQ, C p and CQof Equations (6.8), (6.9) and (6.10). The computed results are shown in Table 6.4. Table 6.4 Computed Results From Hartree Method (1) XP XQ XR CPU”) Cp(Q) CPU?) 5,» £9 6,; 0P 09 0;; mm mm mm m / s m / s m / 3 MP0 MPa MPa 7.55 9.65 8.6 6799 3743 6391 0.1388 0.1130 0.1452 414.65 383.00 -6.173 13.91 20.09 17.00 3743 1824 2612 0.1130 0.0550 0.0442 329.00 286.00 9.727 23.98 27.00 25.5 1335 1216 1281 0.000 0.000 0.000 279.00 282.00 279.00 From the above values for three points, if not enough mesh points are obtained, themethod is not very accuracte. During the plastic wave propagation, the plastic wave propagates initially very fast and then decreases gradually. So the interval of time must be taken very small for the initial plastic wave propagation. However, the increase of the number of the interval of time causes the accumulation of computation error. On the other hand, when the Hartree method is used, the initial values must be local values near the true values of xp, xQ, CP(R), Cp(P) , CP(Q) , 8p, 8Q, JP and O'Q. That is to say, the interpolation should be avoided. Otherwise, the computed values cannot converge to the true values. In order to use the Hartree method well, the initial values 108 should be given or obtained by simple computation as much as possible. 6.2.3.2 Using Specific Mesh Point to Calculate Dynamic Stresses From the above computation, the Hartree method is very sensitive to the errors from computing xp, xQ, CP(R), CP(P) , CP(Q) , 81;, 8Q, 0P and O'Q. To improve the accuracy of computed results, the mesh points are placed the measured values xp, xQ. The values of C p (P) , C P (Q) , 0p and (IQ are from computed RI theory’s values. The interval of time is taken differently for every mesh point. The following flow chart tells us how to compute necessary data for the Hartree Method. The computed results are shown in Table 6.5 The following calculation begins in the second phase. The following values should be known: xP’an 8P9 5Q, UP,O'Q, Cp(P)’ Cp(Q)’ At, k The following values are from the measurements of a projectile: XP , x Q The following values are from the calculation: ( ) — -————D’” _ D“ a, x p — Do Do Using the rule of plastic incompressibility to calculate 6p , 5Q: 8r (xQ) : 8P : —28r (x13) 109 The values of are from the computation of RI theory: UPaOQ, Cp(P) a Cp(Q) The computation of x R : xk = (xp :xQ) The computation of velocity of plastic wave at point R: CPU”) + Cp(Q) 2 The computation of time interval: _ 2(xP - xQ) ' c,(P) + Cp(R) CP(R)= Material constant k from Malvern experimental result. Using Hartree method finds the following values: 5R , 0' R Strain rate can be calculated as follows: . 5P _ 3R . 8R _' 69 or 8R = At Repeat the above steps for next point. The relationship of the stress JR and x is shown in Figure 6.3. The distribution of the dynamic stress in the projectile is shown in Figure 6.4. Now we discuss the propagation of the plastic wave from the first phase to the second phase with the help of Figure 6.4. 110 Table 6.5 Computed Results From Hartree Method (2) A! x1» x9 xR C,.(P) 6,19) 9.00 3,. 3Q a, 0p 0Q 0R ><10‘2 mm mm mm m/s m/s m/s MP0 MP0 MP0 #SCC 2.80 3.40 3.10 9355 8470 8912 0.1454 0.1405 0.1430 813 704 1171 6.372 3.40 5.00 4.20 8470 7732 8100 0.1405 0.1361 0.1384 704 625 975 9.875 5.00 6.30 5.65 7732 6716 7224 0.1361 0.1292 0.1329 625 531 940 17.99 6.30 9.40 7.85 6716 6182 8912 0.1292 0.1251 0.1272 531 487 740 48.00 9.40 14.00 11.70 6182 4224 5203 0.1251 0.1064 0.1151 487 368 491 88.40 14.00 16.00 15.00 4224 2885 3554 0.1064 0.0877 0.099 368 318 433 56.30 16.00 24.50 18.25 2885 1907 2396 0.0877 0.0673 0.0842 318 296 472.3 166.9 20.00 24.50 22.25 1907 1303 1605 0.0673 0.0486 0.0592 296 288 460 280.4 24.50 28.00 26.25 1303 959 1131 0.0486 0.0336 0.0416 288 285 415.8 309.5 28.00 32.00 30.00 959 644 801 0.0336 0.0141 0.0254 285 284 386.5 499.1 32.00 34.00 33.50 1335 1216 602 0.0141 0.0000 0.014 284 283 384 332.2 ' + *1”? . . f M1 1 Stress vs Distance In Second Phase 1 (Aluminum 6061-T6511 ) 1 1 1 1 A 1500 "1 1 1 a 1 1 ‘ 2 10001 e. 1 1; 5001 O o 1 ~ 3 ‘ 9 ’ ° 0 e e ' 1 a 1 1 i 0 +--- ———— '1' rfi 1 ‘ 0 10 20 30 40 1 Distance ( mm ) (Aluminum 6061-T6511) 111 Figure 6.3 Stress vs. Distance in Second Phase ._ _ _ _. _ AAA ___2. _.. ___A_”..._ _..1 Stress vs Distance In Projectlle (Aluminum 6061-T6511 ) 1400 12004 . 10004 .. 800+ 600« 400. ’ O ’ ’ 0 s . 200 0 . 4 0 1o 20 30 40 Stress ( MPa ) i 1 1 1 1 . 1 1 1 1 1 1 Distance ( mm ) 1 Figure 6.4 Stress vs. Distance in Projectile (Aluminum 6061-T651 1) 1. When the projectile heats the rigid wall around the impact end of the projectile, x _<_ 0.69 mm, the stress changes a little. But after x > 0.69 mm, the strain and the stress do not change. Basically, they are constants. For strain, 8 is about 0.15 or so. For stress, 0' = 498 MPa. They indicate that when using the first solutions of the plastic wave equations ( RD theory or R1 theory) to solve the dynamic stresses in the first phase, the theory can explain experimental results well. At the same time, it indicates that the theory can be applied a distance from the impact end. If we check other Taylor tests in Chapter 3, then it can be discovered that above conclusions are true. 2. When the plastic wave propagates from the first phase entering the second phase, the dynamic stress increases suddenly. This one takes place at the interface between the first phase and the second phase. The interface is about at x = 2.75mm. The dynamic stress value is about the twice of the value of the dynamic stress in the first phase. This tells us that the maximum dynamic stress value takes place at the interface between the first phase and the second phase. 3. As the plastic wave propagates from the first phase to the second phase, the 112 values of the dynamic stresses gradually decay, that is, the effect of the strain rate plays an import role in this region. After a distance from the interface, x = 12 mm, the change of the stress gradually is steady. The effect of the strain rate plays a less important role. This means that RD theory can be replaced by RI theory. This conclusion matches one of Bell’s tests. The distribution of the strain rate along the projectile is shown in Figure 6.5 and the third column in Table 6.6. Strain Rate Distrbution Diagram (Aluminum 6061-T6511 ) 1 1 .. 1 § 40000 . 1 ‘; 30000 - 1 a 20000 ~ 0. 1 “2 10000 . . ° 1 E 0 .e 9 ° 9 .L 1‘5 0 10 20 30 40 1 1 Distance ( mm ) E 1 Figure 6.5 Strain Rate Distribution Diagram Table 6.6 Computed Results From Modified Malvern Model vs. Johnson-Cook Model Distance Strain Strain Rate Stress ( MPa) Stress ( MPa ) (mm) ( Us) New RD Model Johnson-Cook Model 0 0.1723 0 509.33 509.33 0.343 0.1596 0 503.31 503.31 0.6875 0.1565 0 501.79 501 .79 1.03125 0.1491 0 498.09 498.09 1.375 0.1484 0 497.74 497.74 1.71875 0.1487 0 497.88 497.88 2.0625 0.1486 0 497 .84 497.84 2.40625 0.1486 0 497.84 497.84 2.75 0.1459 0 496.45 496.45 3.10 0.1430 35650 1171 572.71 4.20 0.1384 21265 975 564.21 5.65 0.1329 17780 940 560.77 7.85 0.1272 4166 740 547.1 11.70 0.1151 0 491 485.12 15.00 0.099 13140 43 0 535.77 18.25 0.0842 2097 472 510.54 22.25 0.0592 2889 460 489.23 26.25 0.04163 2252 415 468.12 30.00 0.0254 1642 386 441.3 33.50 0.014 4214 384 421.71 113 4. When both the Johnson-Cook model and the Modified Malvern model are used to compute the dynamic stress’s distribution along the projectile, the computed results are shown in Table 6.6 and Figure 6.6. From the Figure 6.6, the ways of the stress decay are different for two models. The Johnson-Cook model is not sensitive to the change of the stress. The maximum stress computed from the Modified Malvern model is about twice of the value computed from the Johnson-Cook model. 1 Stress vs. Distance in the Projectile 1 New RD Model and Johnson-Cook Model 1 ( Nuninum 6061-T6511 ) l 1:1000 °o 54~~F ,.,__‘ ‘ g 800. o 1814611718011116661 1 .— a Johnson-Cook Model‘ 1 3 SW C O n o o o a 8 i 1* 7 ’ ' i J 1 to 400 B 8 g 8 1 200 1 o c 1 o 5 1o 15 20 25 30 35 4o 1 Distance (mm ) Figure 6.6 Stres vs. Distance in the Projectile From Modified Malvern Model and Johnson-Cook Model 114 CHAPTER 7 CONCLUSIONS AND FUTURE WORKS 7.1 Conclusion In this thesis, the propagation of the plastic wave and its applications in the Taylor Impact Test are discussed. The shape curves and the dynamic yielding stress of the projectile are used to verify our theoretical investigation. A complete method to calculate the dynamic yielding stress, strain, and the velocity of the plastic wave in the projectile is developed. Our investigation indicates that according to the shape curve, the projectile can be divided into three phases. Afier impact, the shape curve of the first phase is concave, the one of the second phase is convex and the third one is a straight line. When the plastic wave propagates in the projectile, the first phase can be described by a linear solution of the plastic wave equation (RI theory). It is an adiabatic process. The strain and wave velocity are constants. The second phase can be described by a nonlinear solution of the plastic wave equation. In this phase, the velocity of the plastic wave, the strain and the strain rate are not constants. In order to get the nonlinear solution of the equation of the plastic wave, a constitutive equation (R1) is proposed. Based on this constitutive equation, the shape curve of the projectile is convex can be predicted, and the dynamic yielding stress of the projectile can be calculated. Other models cannot predict that the shape curve of the projectile in the second phase is convex. 115 In order to consider the effect of the strain rate in the propagation of the plastic wave, the Modified Malvern model is proposed. This model, the characteristic method and the numerical method are used to analyze the propagation of the plastic wave. Our investigation indicates that with the propagation of the plastic wave the effect of the strain rate gradually decreases. So, at a distance from the impacted end, the RD theory can be replaced by the RI theory. This conclusion matches the Bell’s test result. The von Karman solution is a characteristic equation of the plastic wave equation. The RD theory can predict the existence of the plateau around the impacted end. The existence of the plateau is independent of the model of the materials. As an application of the theory, the Taylor Impact Tests with three different materials and impact velocities are carried out. Using the constitutive equation proposed in this investigation to compute the dynamic stress in the projectiles gets good results. The experimental results indicate that the shape curves of projectiles after impact match the conclusions from our theoretical results very well. The theoretical analyses in the three materials indicate that there exists the maximum dynamic stress around the interface between the first phase and the second phase. 7.2 Taylor Formula and New Formula When deriving his formula, Taylor did not use the theory of plastic wave propagation. So he thought that his formula could be used to calculate the dynamic yielding strength of a material roughly. In this thesis, the theory of plastic wave propagation is used to calculate the dynamic yielding strength of a material. The result is better than the one fiom Taylor’s formula. But more parameters are needed. Comparison between two formulas are shown as follows: 116 Taylor’s Formula New Formula 1. Objective: find Gd), (6‘) 1. Objective: find 0'05, (8, 6") 0d (x) and é(x) 2. Test: V0,L,L0,Le,H,p 2. Test:V0,L,LO,Le,p,h,hd,Rd,Rh,k,C1,C2 where k is a material constant from Malvern model, C1 and C2 are integration constants. 3. Analysis: 3. Analysis: 2C1 V L- —‘ ad, = p M H)L 1) cay(RI)=%e C2 2(L— L0)ln-I; 2) adymD) = ody(R1)+ ké \\\\ 5 ‘1\\\ \ L 2 Le _4 h _ V L ‘ LP L0 '/ 1.,r—-—> / g 1/ L, / x 7 M - H . \Rh Z>R0 A Ln Using the rule of plastic incompressibility, h can be calculated from hd and the 1 17 measured data h= hd ___511) R0 The computed results and measured results are as follows Table 7.1 Values of h From Computed Results and Measured Results Computed Measured Material Results (cm) Results (cm) Discrepancies Aluminum 6061-T6511 0.299 0.275 8.7% Copper l45-Hard-H02 0.363 0.353 2.8% Steel C1045 0.223 0.211 2.9% The computation of dynamic stress in the first phase is by the J ohnson-Cook model. From the measured strain and the model, the dynamic stress in the first phase of Aluminum 6061-T6511 can be computed. In order to compute the dynamic stress in the second phase, the modified Malvern model is proposed and used to calculate the dynamic stress in the second phase. This model can be written 2(6-C1) 2[6 C2 ]+ké 0:0R,(s)+ké= When the Modified Malvern model is used to the Hartree method, the Hartree method can be written as 118 1 201 -xp = 516,,110 + (2,110)].0 1 xR — xQ = §[_Cp(R) " Cp(Q)]At 1 k (0R _ 0P) + §[pC;(R) +pC;(P)](£R — 8P) +—A7[8R - Sp] = O 1 k (0R — 4Q1+ 5117930?) —pC,%cot, 8:0 Figure D-l 134 V Appendix E: Taylor’s Test A technique determining the dynamic yield stress of materials was developed by Taylor. The method involves impacting a right circular cylinder onto a rigid target at a constant velocity V0 , as shown in Figure E-l , and measuring the post-impact deformation of the cylinder. The cylinder has a length L0 + H and is assumed to be rigid strain-hardening, i.e. elastic strains are neglected. One-dimensional plastic wave propagation analysis was developed to identify the yield stress. In Taylor dynamic study, the material was assumed to be rate independent, i.e. 0' = 0(8). Figure E-2 shows the cylinder at some point of impact. The deformed surface is propagating away from the rigid target at a plastic wave velocity C p . The deformed portion has an instantaneous length L , cross sectional area A , density p and stress 0'. The particle velocity behind the plastic wave is assumed to be at rest, i.e. v = 0. The undeformed portion travels with an instantaneous velocity v0 , which is less than V0. It has an instantaneous length h , cross-sectional area A0 and density p0. The stress in the undeformed portion is assumed to be at yielding point O'y. Based on the law of conservation of mass, the flows of mass into the interface I from both portions are equal to each other, i.e. p(Cp — v)A = p0(Cp — v0)A0 pCpA = p0(Cp — v0)A0 135 Assume p = p0, it yields The decreasing velocity of the underformed portion is dh Based on material incompressibility, the strain directly behind the plastic wave front can be calculated, i.e. Vv01_Vv01 L0 Vvol A A A0 where Vvol is the volume of the material and L is the length of the deformed portion. Combining Equation (E3) and Equation (E. 1), the following relation can be established g=—v°—- (13.4) Cp —VO Consider an element with length dx and unit cross-sectional area 1 ahead of the plastic wave front, as shown in Figure E-2. Assume the velocity of the initial state of the element is v00; the velocity of the final state of the element is v01. It takes time dt for the element to deform from the initial state to the final state. From Newton’s law F dt = mdv th = podx- 1-(v01 — v00) Assuming the initial state of the element is v00 = v0 and the final state of the element is at rest, v01 = 0 , the above equation can be rewritten as 136 (0' — 0'}, )dt = podx(—v00) d (a—ay)=—p0voj:— (E5) In analogy to Equation (E2), the following equation an be established for the element d 3:246], —v0) (E.6) Taking Equation (E.6) into Equation (E.5), it yields J-O'y =p0(Cp—v0)v0 (E7) The equation of motion of the undeformed portion is dv 150210177: = ay A0 £9— - 0' (E 8) '00 dt y ' Recall Equation (E.2), dh dt = —— ( E.9) Cp —' V0 With substitution of Equation (E9) into Equation (E.8), it yields dh Potho = -Uy(-——) Cp — V0 0' — (Cp — 120)de = 151—’1 (13.10) p0 h Assuming C p is a constant, and taking integration from V0 to 0 and from L0 to H, where L0 is the original length, then we have 137 1 a —E(Cp — V0)2 9V0: $111]? 80 V—02-+V0Cp=:y—ln—Ii (EH) 2 p0 L0 where specific — V0 is used because the velocity is negative in x direction. It was further assumed by Taylor that the undeformed portion of the specimen decelerates at a uniform rate. Then it can be seen that the plastic wave propagates a distance (L1 — H), where L1 is the final length of the cylinder and H is the final length of the undeformed portion, as shown in Figure E-3. The plastic wave propagating time is t: (E.12) If it is assumed that the time of deceleration is uniform, it is approximately equal to 1:2(L0—L‘) (E.13) V0 From Equation (BIZ) and Equation (E.13), it gives p : EM ( 13.14 ) 2 (Lo — L1) Taking Equation (E.14) into Equation (E.13), the yield stress obtained from Taylor’s study because O'y _ (L0 ‘ H) 1 poi/02 2(L0 - L1) 1n(_L_Q_) H (E.15) The yield stress in the above equation only involves measurements of the undeformed portion of the cylinder and the impact velocity. 138 V A \VNWA L0+H Figure E-l Schematic of the rod before impact test. k 1. 0',Cp,v, A0,0'y,p0,v0 V0 A,p \\\\\\\\\\ Figure E-2 Schematic of the rod during impact test. 139 fi-—*——m——t_ \\\\\ Figure E-3- Schematic of the rod at the end of impact test. 140 Appendix F: Formulation of Plastic Wave Propagation The investigation of a plastic wave propagating in continuum media can be based on interfacial conditions. It based on the laws of the conservation of mass, conservation of momentum and conservation of energy. In the analysis, the stress, strain, density, particle velocity, plastic wave velocity and internal energy are taken as state variables. The laws of conservation of mass, momentum and energy are combined with equation of state and constitutive equation to constitute a complete set of equations. As shown in Figure F-l , there is an interface I to which a plastic wave propagates from the right end of the media. Assume that behind the plastic wave front, the cross sectional area is A , the density is p, the stress is 0' , the particle velocity is v , the velocity of the plastic wave is C p and the specific energy density is I E . Ahead of the plastic wave front, the cross sectional area is A0 , the density is p0 , the stress is 0'0( yield stress ), the particle velocity is v0 and the specific energy density is [130° The velocities are referred to an Eulerian coordinate system fixed in space. 1. Conservation of Mass At the interface I , the conservation of mass requires that the rate of mass flowing into the plastic wave front equals the rate of mass exiting the plastic wave front. Hence pA(Cp—v)=pOAO(Cp—v0) (F.1) If the particles entering the plastic wave front are rest immediately afier the plastic deformation, i.e. v = 0 , Equation (F-l) can be rewritten as 141 pACp = p0A0(Cp — v0) (Taylor-1) However, if v0 = 0 is imported, Equation (F-l) will become pA(Cp — v) = pOAOCp (MFJ-l) 2. Conservation of Momentum The law of conservation of momentum requires that the change of a mass entering the plastic wave front in an increment of time At equals the impulse imparted to the mass over the same time At, i.e. F At = mAv. Consider a free element ahead of the plastic wave front. The mass of the element is Amo = p0 A0(Cp — v0)At . Before entering the plastic wave front, the particle velocity of the element is v0. After entering the plastic wave front, the particle velocity of the element is v. The force acting on the element is (O'A + 0'0 A0) . Thus, from Newton’s law FAt = Am0(v — v0) 2) (0A + 00A0)At = pOAO(Cp — vO)At(v — v0) :> (0'A+0'0A0)=,00AO(Cp —v0)(v—v0) (F-2) In MFJ’s analysis, let v0 = 0 and consider the sign of 0'0 is negative, then Equation (F-2) becomes 0A — 0'vo = 100140va (MFJ-2) In the Taylor’s formulation, since v = 0 and A = A0 = 1, and then consider 142 the sign of v0 is negative, Equation (F-2) becomes 0' — 0'0 = ,00(Cp - v0)v0 (Taylor-2) 3. Conservation of Energy The law of conservation of energy requires that the energy before entering the interface I and the work done by the acting forces is equal to the energy after exiting the interface. For an element with a mass of AmO = p0 AO(Cp — v0)At , it yields 122010,. — 120140411150 + %P0(Cp — 4140440,. - v.02 + (UAvAt + 00 onAt) = = [pomp — 1201404015 + imam, — 4140446,. — 02 P0(Cp " V0)A0150 + %P0(Cp — V0)A0(Cp " "0)2 + (0A + 00A0)V = = p016, — vo>AoIE +§po(cp — V0)A0(Cp — 02 v + (C, —vo)2 _ (C, —v)2 A0 p0(Cp — V0) 2 2 (F-3 ) Eliminating v by using combing Equation (F-l), Equation (F -2) with Equation (F -3), it gives 1 1 1 IE—IEO=E(U+JO)(—_;) (F-4) .00 In MFJ analysis, let v0 = 0 and A = A0. Combing Equation (F-3) with Equation (MF J -1) and Equation (MFJ-2) and consider the sign of 0'0 is negative, then 143 1 IE _IEO =-2-[Cf,-(Cp—v)’]+(%—-:§) (MN-3) In Taylor’s formulation, the conservation of energy is not considered. Collect Equations (F-l), (F-2) and (F-4), a complete set of equations for formulation plastic wave propagation is pA(C,, — V) = pvo(C,, - v0) (*) (0A + (IO/10) = p0A0(Cp — v0)(v — v0) (**) 1 1 l IE_IEO=§(0'+0'0)(—“—) (***) P0 P In the above equations, there are 10 variables. However, A , A0 , p0 , v0 and I 50 are usually a given and measurable. If so, there are 5 variables in the equations, two more equations are needed. In general, an equation of state and a constitutive equation are combined with 5 original equations to solve 5 variables. 144 / Rigid Wall Projectile \\\\\\ \\\ \ A0, 1’03 00,100 ‘—' x I A, v, 0', ,0 I Figure F-l 145 A Appendix G Maudlin-Foster-Jones Model In Maudlin-Foster-J ones ( MF J )model, the conservation relationships ( jump conditions ) in conjunction with postulated material constitutive behavior are applied to study steady plastic wave propagation. Strength constitutive behavior and an equation of state ( EOS ) are combined with the jump equations. For example, the von Mises yield surface, Johnson-Cook ( J C ) or MTS (Mechanical Threshold Stress) flow stress models, and Mie—Grunisen EOS are used as constitutive equations in the paper by Maudlin- Foster-Jones. Obviously, the choice of constitutive equation is not unique, but the main technical points were illustrated in their paper. The following one is a summary from the Maudlin-F oster—J ones paper. The details can be referred to the article. 1. Preliminaries Consider an axi-symmetric cylinder with radius of r0 , as shown in Figure G-l. An area strain is defined as a — A0 _ A ( G 1 ) A A ' where A0 is the initial undeformed area, while A is current deformed area. Taking differentiation of Equation (G.1), it gives A A volumetric strain increment referenced to the initial configuration can be defined as 146 d5, =54? (G3) V where Vvol is the current volume, Vv is the initial volume. 8v is the volume strain. An increment of small Lagrangian strain is defined as 1 im ij where F}!- is the deformation gradient, i.e. 2. Conservation Relationships (Jump Conditions) As shown in Figure G-2, assume that behind the plastic wave the cross-sectional area is A , the density is ,0, the specific internal energy density is I E , the stress is 0' and the velocity of the particle is v. Ahead of the plastic wave, the cross-sectional area is A0 , the density is p0 , the specific internal energy density is I EO , the stress is 0'0 and the velocity of the particle is v0. The velocity of plastic wave is C p . From the law of conservation of mass, we have pOAOCp = pA(Cp — v) (6.5) From the law of conservation of momentum, we have From the law of the conservation of energy 147 1 115‘150=51C;_(Cp“1’)21+—:—--2—Q (G7) .00 Equation (G.5) can be rearranged to yield A (pon —1)Cp = —v (6.8) The expression within parenthesis can be rewritten as “‘ .0vo A0 P0 A0 -—————1=-—————1+——l 6.9 M A(p ) (A ) ( ) l 1 With use of p = —-— and p0 = V—, it yields vol v _L _ _L &_1: VV VVOI : VVOI _ VV : AVVOI : 8v (G10) 10 _L V. Vv Vv __zwzmzl4-8 (3.11 L ( ) where 8 is one dimensional engineering strain, L is the deformed length of cylinder and AL is the increment of L0 . Equation (G.9) becomes __p0A0—1=g+ag,+g, (6.12) pA and the conservation of mass equation, i.e. Equation (G.5), becomes C —— v (G13) 1” 8(8v+1)+8v ' Similar algebraic manipulations will yield conservation of momentum in the 148 following form 0 m— 0 = —p0CpV (6.14) The conservation of mass equation can then be used to eliminate the particle velocity v in Equation (G. 14), it gives 0' 00 - 1+ 8 2 = —p C (G.15) [8v(1+8)+8] 0 p .no' .._- The wave velocity C p can then be eliminated from Equation (GM) and the following '11 can be achieved 0' [0'0 ][.sv(l+r:)+6‘]=—,00v2 (G.16) 1+ 8 It is the equation for the constant particle velocity in terms of stress and strain across the wave front. Finally, the conservation of mass and conservation of momentum equations can be used to rewrite the conservation of energy equation, 1 0'1 p0(IE ‘150)=§[5v(5+1)+8][0'0 er] (G.17) in terms of stress and strain. It is a form of the Hugoniot equation. Equations (G.15), (G.16) and (G.17) in general require an equation of state and a constitutive equation to completely specify the curves in stress and strain space. This problem will be solved in the next section. 3. Constitutive Relationships Specification of a yield surface provides an equation relating the deviatoric stress 149 Sij to a yield stress or flow stress 0' f. MFJ model uses von Mises surface and a flow stress model that is functionally dependent on equivalent plastic strain 8 , strain rate 8p , and temperature T: 3 . —2-S,-jsj,- =a}(5p,gp,T) (G.18) In Maudlin-F oster-J ones’ paper, two flow stress models are investigated; the Johnson-Cook model and the MTS model. Here only the Johnson-Cook model is illustrated. The J ohnson-Cook model has the following form: 7} n ' T" m of =[0'y +ng][1+Cln8p][l_(T;—77) ] (G.19) The first factor in Equation (G.19) represents strain hardening with 0'}, interpreted as the initial yield stress, B is a strain-hardening coefficient, and n is a strain-hardening exponent. The second factor accounts for strain-rate hardening effects with C being a strain-rate hardening coefficient. The last factor represents thermal softening relative to room temperature I; , decreasing to zero as the melting point T m is realized. The quantity m is a softening exponent. If follow von Karman and Duwez and Kolsky by relating the plastic wave velocity to the partial derivative of the stress with respect to strain suggested by a method of characteristics solution of the equation motion for the uniaxial stress problem, then it gives C; = 51.7% (G20) where p0 is the initial density and E H is the small Lagrangian strain given by Equation 150 (G.4). For the Johnson-Cook model, the plastic wave velocity can be calculated as follows: 3076‘ n—l T‘7i =B 1+C1 1— &% n4 1 net (3”,. )m] When we use the energy jump equation to combine with a Mie-Grunisen EOS and a von Mises yield function, then we obtain an analogous Hugoniot result: 1 5(5A+8v+548v)0y+150l’0+ . 1 + 1 ; (1+82) F(1+#) 0.1::— %(EA + 8v + 8248v) +—~—————1 ——[—2— F 0' + k +k 2+k 3 1—— 1.0+,” 3 f (11“ 21“ 3/1 X 2.11)] 1 1 (Hwn+r0+m $094 + 5v + 5431») (G21) If the volumetric strain now assumed to be small, then we have that: l l 2 “is/40'), '1' IEOPO +F(—3‘0f — ‘9ka) a :_ 8 (G22) A 1 l __.______.+__ 2(1+8A) F where k1 is a linear coefficient, k2 is a quadratic coefficient, k3 is a cubic coefficient, fl—l. F is Grunisen coefficient, #0 = 151 0 L L ////«/ ////n// G-l Figure 152 STATE VARIABLES C10 ——> 0'1,V1,p1,51,11 00,V0=O’100’50110 Figure G-2 Variables relative to the plastic wave front. Variables with subscript O are up stream of the plastic wave and subscripts l are downstream. 153 Appendix-H Name Model a0. 0770. Material Contents 58 68" Hartig a=%(l_e-bg) Ee—bs 0 X b>0 Logarithmic a = a, (8) + k ln 8“ 301 k X X W ‘5; 3' Power Law 0' = 01(£)én n 50‘ no‘én—i X X (.96 MalverrgModel a = (20000 __1_0_) oo'l k Alummum k = 10 1/ s 5 86 +10000008 Power Law _ n n-l 0 Aluminum 1100 k = 0.598 , Plasticity " ’ We + 5) "We + a) n = 0.216 Strain Rate a = E p£+ 00(8) 55 500 4140 Steel X Dependent ———p— 72;: Plasticity as 10h“5°“'C°°k a = (A + Be" )* B8"_1 (1 + Cln 5‘) ,4 + 35" X X Model . ___'___ (1 + C 1n 5) g , Rate Sens1t1ve 0’ = k8m 6:11 mkén gm—l nkgmén_l A356. k = 0.002 , Power Law Alummum m = 0,7 , Plasticity n = 0.32 154 ._-:r.~:=q ’) .—‘ l . J ohnson-Cook Model Material C n A(MPa) B(MPa) OFHC 89 292 0.0025 031 Catridge 1 17 505 0.009 0.42 Brass Nickle 200 163 648 0.006 0.33 Armco Iron 175 380 0.06 0.32 Carpenter 290 336 0.055 0.4 Elertrc Iron 1006 Steel 350 275 0.022 0.36 2024-T351 265 426 0.015 0.34 Aluminum 7038 336 343 0.01 0.41 Aluminum 4340 Steel 792 510 0.014 0.26 S-7 Tool 154 476 0.012 . 0.18 Steel Tungsten 1056 176 0.016 0.12 Depleted 1079 11 19 0.007 0.25 Uranium Tantalum 140 300 X 0.3 155 Appendix I Elastic Wave in The Taylor Test In the Taylor test [2], a compressive elastic wave and a compressive plastic wave occur in rod when it strikes onto a rigid wall. The elastic wave moves into the rod faster than the plastic wave and reflects from the free end of the rod as a tensile wave. The reflected elastic wave collides with the ongoing plastic wave somewhere in the rod. When they meet, the tensile elastic wave reflects from the plastic wave front again as a compressive wave and moves toward the free end of the rod. This back and forth process continues until the propagation of the elastic wave stops as the effect of interior material friction force. In the absence of body forces, the elastic wave propagation can be expressed by the following wave equation, which is based on Newton second law and Hooke’s law Eazue=p€iu£, OSxSL,t20; (I-l) 03.2 9032 where x is the coordinate, t is time, ue (x,t) is elastic displacement, L is the length of the rod. E is Young’s modulus and pe is density. There are three kinds of boundary condition in the elastic wave propagation of the Taylor test. The first kind is associated with the impacted end between the rod and the rigid wall, the second kind is the free boundary at the free end of the rod and the third kind is located right at the plastic wave front. 1. Impact Boundary If the impact velocity of the rod onto the rigid wall is V0 strikes the rigid wall, the boundary condition and the initial conditions can be expressed as follows: 156 ue(0,t) = 0, ue (x,0) = 0, 0716 (x,0) = V0 0’1 2. Free Boundary At the free end of the rod, there is no strain. The boundary condition can be given by 0% (LJ) _ d1: 0 When a compressive elastic wave propagates into the free end of the rod, it comes out as a tensile wave. Kuscher [3 5] used a laser velocity interferometer, the so-called VISAR, to measure the wave propagation in Taylor’s test for a 9820K steel rod. The propagation of the plastic wave front can be calculated based on the experimental technique. He also presented a Lagrangian diagram, i.e. x — tplot, to express the one- dimensional wave propagation phenomenon. Kuscher found that the velovity of the plastic wave decreases exponentially along the rod which had an aspect ratio L/D=10 and an impact velocity V0 = 314m / S. 3. Elasto-plastic Boundary When the tensile elastic wave meets with the compressive plastic wave, an elasto- plastic boundary is created. The boundary essentially divides the rod into two regions, an elastic region and a plastic region. The investigation of the elasto-plastic boundary before the plastic wave can be based on the laws of the conservation of mass and conservation of momentum. As shown in Figure I-l, there is an elasto-plastic boundary I . Assume that 157 behind the plastic wave front, the cross sectional area is A p , the density is pp , the stress is 0' , the particle velocity is vp , the velocity of the plastic wave is C p while they are Ae , pe , O'e and ve ahead of the plastic wave fiont, respectively. At the elasto-plastic boundary, the conservation of mass requires that the rate of mass flowing into the plastic wave front equals the rate of mass exiting the plastic wave front. Hence ppAp(Cp _vp)=peAe(Cp —ve) (1'2) The law of conservation of momentum requires that the change of a mass entering the plastic wave front in an increment of time At equals the impulse imparted to the mass over the same time At, i.e. FAt = mAv Consider a free element ahead of the plastic wave front. The mass of the element is Ame = ,oeAe(Cp — ve)At Before entering the plastic wave front, the particle velocity of the element is ve. After entering the plastic wave front, the particle velocity of the element is vp. The force acting on the element is (O'Ap — 0',, A6). Thus, the forgoing equation becomes (O'Ap — aeAe)At = peAe(Cp — ve)At(vp — ve) 2) (CAP —0'eAe)=,0eAe(Cp —ve)(vp —ve) (1-3) Rewrite Equations ( I-2 ), ( I-3 ) and set A p = A (3, then become pp v. = C. 76". -vp> “-41 158 0',, =0'—pe(Cp—ve)(vp—ve) (1-5) Equation ( I-5 ) is a formula of the stress. According to Hooke’s law, the strain at the elasto-plastic boundary should be 0-10 (C -V )(V -V) 8e= e p e p 8 (151) E At the elasto-plastic boundary, before the elastic wave meets the plastic wave, the elastic displacement should be zero. Then, it gives u,,(L —l,t = 0) = 0 where uer is the displacement function of the reflective elastic wave, 1 is the distance from the free end to the elasto-plastic boundary and tis the time of the propagation of the reflective elastic wave from the elasto-plastic boundary to the free end. Hence, when the elastic wave is reflected at the elasto-plastic boundary, the following conditions can be obtained as follows: ue,(L —l,t = 0) = 0, die, (x,t = 0) _ 0’1 pp Cp ————-(Cp —vp), e dter(x,t '—" O) _ a—pe(Cp — ve)(vp — ve) dc E ’ where L—leS L. 4. The Propagation of the Elastic Wave in the Rod The equation of the elastic wave propagation with boundary conditions and initial conditions from the impacted end toward the free end in the rod can be written as 159 52.. 92.. Edrzezpeaze, osst,120; (1-6) B.C: ue(0,t)=0, Mw dc I.C: ue(x,0) = 0, €53ng 2 VO Now using the method of separation of variables to solve Equation ( I-6 ). Set ue (x,t) = X (x)T (t) and take it back the equation of the elastic wave, it yields Pe ,where a = — X"(X) = T"(t) X(x) a2T(t) The right side and the left side is equal each other, if and only if the above equation is equal to a constant — 2.. Then X"(x): 7"(t) :_ X(x) 2227(1) From the above equation, it yields X"(x)+/1.X(x)=0 (I-7) T"(x)+/1a2T(t) = 0 (I-8) From the boundary conditions in Equation ( I-6 ), it gives u,(0,t) = X(0)T(t) = 0 die(L.t) 03c and = X'(x)T(t) = 0 So, it yields 160 (1-9) X(0) = 0 X'(L) = 0 Combine Equation ( I-7 ) and Equation ( I-9 ), it gives X"(x)+ 21X(x) = 0 X(O)=O (1-10) X'(L)=0 Assume the solution of Equation ( I-lO ) is X(x) = Acosflx + Bsinflx, where )82 = 2. Using the boundary conditions, it gives A = 0, and BflCOSflL = 0 ’17! So, flzfi, n =1,2,3... 727: 2 or 2: — , n=1,2,3--- (1-11) (2L) In this way, we obtain a series of eigenvalues, A." , and the responding function, X n 7271' 2 = —- , n = l,2,3-~ 4.. 12L) . 117! X" (x) = B" Slnfix, n =1,2,3--- (I—12) Combine Equation ( I-7 ) and Equation ( 1-1 1 ), another equation about the time- dependent part can be obtained 2 2 2 1' a n 72' 7;: (11+77‘YLU) = 0 161 Apparently, the solution of the above equation is 7;, (t) = C; cos ”5? + D. sinmmt n (1-13) From ue (x,t) = X (x)T (t) , Equation ( 1-2 ) and ( l-13 ), a solution can be obtained nn'at . mrat . n7: un(x,t) = (CH cos 2L + D" srn According to the principle of superposition, the solution of Equation ( I-6 ) is given 00 mrat . n7rat . n7: ue(x,t) = Z (Cn cos 2L +Dn sm =1 n Using the initial condition, ue (x,0) = 0, then u,(x,0) = E C, shun—’3 = 0 n=1 2L that is, C" = 0. Using the initial condition again, L that is, D" = fV—Ojsinfl-dx = 8LVO mm 0 2L nzzrza mt 1— cos—— ( 2 ) The solution of the elastic wave can be written as . 1272' srn—x (I-14) 2L 0° . mrat ue(x,t)= 2 D" s1n n=l 162 Take D" into Equation ( I-14 ), then ue(x,t)= 023 8LZ°( "7r ' "flatsinll—q (1-15) n=ln27[ a 2 2L 2L From this equation, at any time the velocity of the particle, the strain and the stress can be obtained: . oo 4VO n7: . mzat mzx The stra1n: 88 = Z ——(1 — cos—) sm cos—— n=1n7ra 2 2L 2L 00 4V nit . mz'at mzx The stress: 06 = E 2 —0(1 — cos———) s1n cos—— n=1n7ra 2 2L 2L . . co 4V0 117: mrat . m: The veloc1ty of the particle: Ve = Z —(1 - 008—) COS sm—x n=l [172' 2 2L 2L 5. The Propagation of the Reflective Elastic Wave in the Rod As we known, when the elastic wave is reflected at the free end, it propagates toward the plastic wave. The reflected elastic wave collides with the ongoing plastic wave somewhere in the rod. When they meet, the tensile elastic wave reflects from the plastic wave front again as a compressive wave and moves toward the free end of the rod. This back and forth process continues until the propagation of the elastic wave stops as the effect of interior material friction force. Here, the simplest case is considered, that is, when the plastic wave stops propagating, the elastic wave come back from the free end and is reflected at the elasto-plastic boundary. That is to say, the elastic wave is only reflected once during the propagation of the plastic wave. 80, the equation of the reflection elastic wave with boundary conditions and the initial conditions can be given as follows 163 E%= 225%, L-leSL,t20; (I-l6) B.C: uer(L—l,t)=0, M=Q 03c ale, x,0 ,0 I.C: ——;£t—)—=Cp——£(Cp-vp), ale, (x,0) _ 0_ pe(Cp _ ve)(vp — ve) 03c E ' where x is the coordinate, t is the time, uer (x,t) is the reflective elastic displacement. Set uer (x,t) = X r (x)T, (t) and take it back the equation of the elastic wave, it X3002 T."(t> Mew-_- \[Z X,(x) a2];(t) , E The right side and the left side is equal each other, if and only if the above yields equation is equal to a constant - a. Then X."(x) : m.) _ Xr(x) 027;“) _ From the above equation, it yields X,"(x)+aX,(x)=0 (1-17) T,"(x)+aa27;(t)=0 (I-18) From the boundary conditions, it gives 3,,(1. — 1.0 = X..(L — 040) = 0 4mm a. and = X.' 4], 1) Apparently, the solution of the above equation is that . n7rat mrat 1 + D," sm 2] T. t =C' cos m() 7‘71 2] ( 1-24 ) From ue, (x,t) = X, (x)T, (t) , Equation ( I-23 ) and ( I-24 ), a solution can be obtained mrat . n um (x,t) = (Cm cos 2, + D," s1n zz'at . n7! 2] )srnEI—[x—(L—lfl According to the principle of superposition, the solution is given 00 nzrat . n u,, (x,t) = 2 (Cm cos 2] + D,” s1n 1 n: frat 717: ' — — L—l 2, )Sln211x ( )1 Using the initial conditions, (31.2.0.0) pp 7 : CP __p_(CP _Vp) and M92: U_pe(Cp‘Ve)(Vp_Ve) dc E then o'iter(x,0) : E (_KECm Sinnflat +mra Drn cosnflatfiinflix—(L _1)] 0’1 n=1 2] 21 21 2] 21 — EwD sinfl[x—(L—1)]—C —&(C —-v) ":12! m 21 P pg p p 4[C,, - Z—NC, — v, )] L Sinmflx — (L — 1)] dx . 9 :>D= rn 1 ””0 L—l 166 81[Cp — gimp — vp)] p 1272' = e (1 — cos—) ( 1-25 ) nzzrza 2 W = E (Cm Sn + Drn 03c n=l co 0'— C —V V —V _ Z Cm MCOS-’Z£x= pe( p e)( p e) n=1 2] 2] E => Cm =—i[U—pe(Cp —v 8],—xv vein cos ””[x (L mdx nflE L— 81 . nn Cm: ———-—-[0' pe(Cp —ve)(v —ve)]sm— (I-26) ”272,2E P 2 Now assume that pp = pe and Ve = 0, then 81 . n7: Cm: n2 7:2E ————(0' peC pvp)sm7 (1-27 ) D 81v” (1 n” ) (128) m - —cos— - nzrrza 2 The solution of the reflective elastic wave can be written as ue,(x, t)- — Z 1,(Cn coanlt + Dml sin ”gatbinlz—Er—[x— (L— 1)] (I-29) n21 From Equation ( I-29 ), the strain, the stress and the velocity of the particle can be obtained: Strain: flue (x t) 00 mrat . n7rat 8, =——'——’——= Cm cos +D, sm cos—x- L— I a a} £13? 2, n ) ”t( )1 Stress: 167 00 t . 1‘ Ge, 2 E Z fl(Cm cosnm + Dm smmm )cosfl[x—(L—l)] n=1 2] 21 21 21 The velocity of the particle: 00 mm mrat 1171511 1171 v = Z——C sin +D cos sin—x— L—l 9’ ”=1 21 ( ’” 21 ’" 21 ) 21[ ( H /Rigid Wall Projectile \\\ \ \\\\ Figure H 168 _.._.__—t'\" " I Appendix J O O The volume of the element can be written as A Vznrzx Take derivative for both sides of the above equation, then dV 2 2x7zrdr + mzdx Consider the incomprehensibility of plasticity, it gives dV = 2x71'rdr + flrzdx = O :> 2x71rdr + 7zr2dx = 0 Zxflrdr mzdx _ :> + O nrzx zzrzx 2dr dx :> —— + — = 0 r x d8, +dgx +1189 = 0 d8, —23, +d59 = O — d8, + dag = 0 d8, = deg 169 Appendix K Other Models’ Shape Curve Name Model Material Constants Shape of E g Projectile 5’6 078' Hartig a : f—(l —e'b‘) Ee—bs 0 X b > 0 Concave Logarithmic 0' = a, (g) + k In 1," 0‘10] k X X Undefined Law _ ‘7 as a Power Law 0' = 0,1 (£)é" é" €31 n 01 én—l X X Undefined 36 $2115” a = (20000-19) E k “Air?" concave . 8 36 k =1061/S “000000; :pwerLaw a = k ( 5e + 8),: nk(6‘e + £)n—I 0 Alumi- k = 0598 Concave astICIty num , 1100 n = 0.216 Strain Rate a- = E p1: + 0'0 (5') of (300 4140 X Straight Dependent —£ E'— Steel Line Plasticity 5‘5 Johnson-Cook a: (A + 35")" Be""(l+Cln é) A + Ba" X X Concave Model . . ' (1 + C In a) a Rate Sensitive 0' ___ k Em 6111 m k E'n gm—l n k 6m én- A356. k = 0,002 Concave Power Law Aluml- , Plasticity num m = 0] , n = 0.32 Johnson — Cook Model Material A(MPa) B(MPa) C n Shape of Projectile OFHC 89 292 0.0025 0.31 Concave Catridge Brass 117 505 0.009 0.42 Concave Nickle 200 163 648 0.006 0.33 Concave Arrnco Iron 175 380 0.06 0.32 Concave Carpenter Elertrc 290 336 0.055 0.4 Concave Iron 1006 Steel 350 275 0.022 0.36 Concave 2024-T351 265 426 0.015 0.34 Concave Aluminum 7038 Aluminum 336 343 0.01 0.41 Concave 4340 Steel 792 510 0.014 0.26 Concave S-7 Tool Steel 154 476 0.012 0.18 Concave Tungsten 1056 176 0.016 0.12 Concave Depleted Uranium 1079 1119 0.007 0.25 Concave Tantalum 140 300 X 0.3 Concave 170 Appendix L Computations of Dynamic Yielding Stress, Plastic Wave Velocity and Taylor Formula Computation Results 1. Computations of Dynamic Yielding Stress Consider that a cylindrical projectile with length L and radius r0 impacts a rigid wall at a constant velocity V0 , shown in Figure 1,Figure 2 and Figure 3. Plastic Region: a. First Phase: In this phase, the density is ppl . The shape of the projectile is concave in transverse direction afler impact. Before impact, the length is h. After impact, the length is hd . The plastic dispersive wave would be inaugurated at x = h . b. Second Phase: In this phase, the density is p. The shape of the projectile is convex in transverse direction after impact. Physically, that is the so-called shape of a mushroom. Before impact, the length is L — Le — h. After impact, the length is LP' Elastic Region: 0. Third Phase: In this phase, the elastic wave oscillates and disappears after impact as friction in the material. The density is pe. Before impact, the length is Le. After impact, the length is still Le. The shape of the projectile remains a straight line. 171 Rigid Wall V0 // 0 ¢ L Figure 1 Before Impact Rigid Wall V0 Second Pb: F irst Phase / / L — L, — h —+ h7‘ Third Phase Le / Elastic Region /'/' *—— Plastic Region—>l Figure 2 At the Moment of Impact Third Phase Second Phase First Phase L\ 1 Le p Rigid Wall / \\/§<\\\ Figure 3 After Impact 1 72 1. Computation Steps 1.1 V0, E , p, L should be known. V0 ----- Impact Velocity; p---- Density of Material; E --- Young Modulus; L - -- The Length of Projectile Before Impact. 1.2 Measure LO, Le, hd , ZR], and 2R0 very precisely. L0 ----- The Length of Projectile After Impact; Le ---- The Length of the Third Phase; hd ---- The Length of the First Phase After Impact; 2R h ----- The Small End Diameter of the First Phase After Impact; 2R0 ---- The large End Diameter of the First Phase After Impact. 1.3 Calculating h from hd: 2m?- h=- 0 hd 2 mn—Eg4mgin R0 R0 h ---- The Length of the First Phase Before Impact. 1.4 th can be calculated as follows: 1 = — 12 V0 t h ---- The Time of Plastic Wave Propagation in the First Phase. 173 1.5 t p can be calculated as follows t p ---- The Time of Plastic Wave Propagation in the Second Phase. 1.6 Using the following matrix to find Cl, C 2 , C3: h h(ln—h——1) 1,, th Cl hd O -‘t!1— 1 C2 = V0 11.2 C. (L‘Lo)—(h—hd) L—Le (L—Le)(ln e—1) p tp I L. 1.7 The dynamic yielding strength of the material can be obtained: 2ICII 0.:pCZeC2 2 See Explanation of Dynamic Model in Figure 5. 174 2. Example Aluminum 6061-T6511 Density: p=2700 kg/m3 Static Yield Stress: as = 275MPa Young Modulus: E = 69 x 109 N /m2 Elastic Wave Velocity: Ce = SOSSm/ 3 Impact Velocity: V0 = 93m/s Original Length: L = 7.8232cm Final Length: L0 = 7.512cm L p = 3.4282cm Le = 4.12cm Projectile Diameter: D1 =1.22936cm Measured h = 0.275cm , Computed h = 0.299cm 12,, = 0.2565cm 2R,l = 1.3195cm 2R0 =.1 335cm AL = (1. — L0)—(h —hd) = 0.927cm 1,, =(L+ Le)/Ce =23.63x10‘6s 1,, = h/V0 = 2955x10—6S 1. h = 0.26cm h(ln V0 — 1) = —1.476cm 1,, = £— = —O—'2—6— : 27.9psec V0 0.0093 L — Le = 3.70320m tp = 23.63115 L — L (L — Le)[ln( e —1)]= —10.56cm p (L — L0) — (h — hd ) = 0.5447 — 0.26 = 0.2847cm 175 0.26 — 1.476 27.95 C1 0.2565 0 —0.0093 1 C2 = 0.0093 3.7032 — 10.56 23.63 C3 0.2847 C1=0.063l, C2 =0.0163, C3 =0.00950m/ys Dynamic Yielding Stress 0': 2700 0 0163 M0063” 0' = x2 ° 8 0-0163 x104 = 499Mpa h = 0.2650m 11(an0 -1)= —1.505cm/,usec th = ——h— = 28.49psec 0 L — Le = 3.70326m tp = 23.63115 (L —- Le)[ln(L " Le — 1)] = —10.56cm ’12 (L — L0) — (h — hd ) = 0.5447 — 0.265 = 0.2797cm I 0.265 — 1.505 28.49“?)1 ”0.2565" 0 —o.0093 1 C2 = 0.0093 L3.7032 —lO.56 23.634 _0.2797J C1 = 0.0874, C2 = 0.0255, C3 = 0.0095cm/w Dynamic Yielding Stress 0': 240.0874) 8 0-0255 x104=326Mpa 2700 x 0.0255 0' = 2 176 h = 0.27cm h(ln V0 — 1) = —1.533cm/,usec h t}, = 7 = 29.03psec o L — Le = 3.70320m tp = 23.63115 (L — L.)[1n(L ’ Le —1)]= —10.56cm P (L - L0) — (h — 11d ) = 0.5447 — 0.27 = 0.2747cm 0.27 -— 1.533 29.03 C1 0.2565 0 -0.0093 1 C2 = 0.0093 3.7032 —10.56 23.63 C3 0.2747 C1 = 0.1107, 02 = 0.0343, 03 = 0.0096cm/113 Dynamic Yielding Stress 0': 2x(0.1_1_0_7) 0' = 2700 2003439 003‘” x104 2 294Mpa h = 0.2750m h(anO — l) = —l.56l3cm/,usec h 1,, = -V— : 29.56psec 0 L — Le = 3.70320m 1,, = 23.63115 L —L, (L — Le)[ln( —1)]= —10.56cm P (L — L0) — (h — hd) = 0.5447 — 0.275 = 0.269cm 177 0.275 — 1.5613 29.56 C1 0.2565 0 —0.0093 1 C2 = 0.0093 3.7032 —10.56 23.63 C3 0.269 C1=0.1321, C2 =0.0426, C3 =0.001010m/,us Dynamic Yielding Stress 0': 2x(0.l321) C 004” x104=283Mpa 2700 x 0.0426 0' = 2 h = 0.280m h(ln VO — 1) = —1.589cm/,usec th = £- = 30.101,usec 0 L — Le = 3.70320m tp = 23.63113 L—Le tp (L — L0) — (h — h, ) = 0.5447 — 0.28 = 0.2647cm (L — Le)[ln( —1)]= —10.56cm 0.28 —1.589 30.101 C] 102565 0 -0.0093 1 C2 = 0.0093 3.7032 —10.56 23.63 C3 0.2647 C1 = 0.1543, C2 = 0.0509, C3 = 0.0098cm/ 113 Dynamic Yielding Stress 0': 2x(0.1543) a = 2700x 0.05098 0050, x10. __z 295Mpa 2 178 h = 0.29cm h(1n V0 — 1) = —1.64650m/ysec th =§—=31.182psec 0 L — L, = 3.70320m 1,, = 23.63/.15 (L — L.)11n8d 1 I 2101 pC _2_._ I I "”— I I 2 2 I I I I l I I I r > A C1 _, 5d 8 8 II. Computations of Plastic Wave Velocity Plastic Wave Velocity: (8-C1) _ C2 Cp—e C E 8 _ (327% 9C2 :> Cp=e — C] 6C2 £1 __8. :> ecszp-eC2 8=8d+C1 £1— £d+C1 8C2 xCp—e C2 5d :> Cp-eCZ If 8d =O,then Cp(8d =O)=1 _91 If 8=O,then Cp(8=0)=e C2 51 .. Cp(8d:0)= 8C2 C,(g=0) .-c, e C2 E—Cl .21 id— '“C—l Ed—Cl . 3 C2 _e 2 XCp(8=O):eC2 xe C2 :6 C2 C,(a,=0) 1 183 111. Taylor Formula Computation Results 0.61: pV02(L _ Le) 2(L — L,)1n(2L—) e Gong Song’s Calculated Results Material Static Yield Computed Dynamic Discrepancies Stress(MPa) Yielding Stress (MPa) Aluminum 6061- 275 283 3.24% T651] Copper 145-Hard- 32 33 3.125% H02 Steel C1045 585 604 3.25% Taylor’s Calculated Results Material Static Yield Computed Dynamic Discrepancies Stress(MPa) Yielding Stress (MPa) Aluminum 6061- 275 225 —18% T6511 Copper l45-Hard- 32 1099 3200% H02 Steel C1045 585 628 7.3% 184 Bibliography 1. L. H. Donnell, “ Longitudinal Wave Transmission and Impact ” Trans. ASME, Vol.52, 1930, pp. 153 — 167. 2. T. von Karman and P. Duwez, “ The Propagation of Plastic Deformation In Solids ”, J .Appl. Physics, Vol.21, 1950, pp. 987 — 994. 3. G. 1. Taylor, “ The Plastic Wave In A Wire Extended By An Impact Load ”. The Scientific Papers of Sir Geoffrey Taylor, Cambridge University Press, 195 8, pp. 467 -— 479. 4. J. Bell, MECHANICS OF SOLIDS, Volume I, “ The Experimental Foundation of Solid Mechanics ”, Springer _Verlag, 1984. 5. P. Duwez and D. S. Clark, “ An Experiment Study of the Propagation of Plastic Deformation Under Condition of Longitudinal Impact”, Proc.Am.Soc.Test.Mat., Vol.47, 1949, pp. 502 — 532. O\ . J. D. Campell, “ An Investigation of the Plastic Behavior of Metal Rods Subjected to Longitudinal Impact”, Journal of the Mechanics and Physics of Solids, Vol.1, 1953, pp.113. 7. J. E. Johnson, D. S. Wood and D. S. Clark, “ Dynamic Stress-Strain Relations for Annealed 2S Aluminum under Compression Impact”. 8. L.Efron, “ Longitudinal Plastic Wave Propagation in Annealed Aluminum Bars ”, main Library/ 107242 THS, Michigan State University. 9. V.V. Sokolovsky, “ The Propagation of Elastic-Viscous-Plastic Waves In Bars ”, 185 10. 11. 12. 13. 14. 15. 16. 17. 18 Prikl. Mat. i Mek., Vol. 12, pp. 261 — 286, 1948 ( Russian ). Translation All — T6, Brown University, 1949. L. E. Malvern, “ Plastic Wave Propagation In a Bar of Material Exhibiting a Strain Rate Effect”, Quart. Appl. Math., Vol. 8, 1950, pp. 405 — 11. J .F. Bell, “ Diffraction Grating Strain Gauge ”, Proc. SESA, Vol. 1, 1959, pp. 51-64. J. F. Bell, “ Propagation of Large Amplitude Waves in Annealed Aluminum ”, J. Appl. Phys. Vol.31, No.1, 1960, pp. 277 — 282. J. F. Bell, “ Study of Initial Conditions in Constant Velocity Impact”, J. Appl. Phys., Vol. 31, N02,1960, pp. 2188 — 2195. H. Kolsky and L. S. Dough, “ Experimental Studies in Plastic Wave Propagation ”, J. Mech. Phys. Solids, Vol. 10, July/Sept. 1962, pp. 195 — 223. US. Lindholm, “ Some Experiments With the Split Hopkinson Pressure Bar ”, Technical Report No.1, Contact No. DA-23—072-ORD-1674, Sw RI Project No. 02 — 1102, Southwest Research Institute, San Antonio, Texas, March 1964. Malvern, L.E. ( 1965 ), in N.J.Huffington, Jr( Ed.), Behaviour of Materilas Under Dnamic Loading, ASME, New York, p.81. W.J.Gillich and W.O.Ewing, “ Mechanics of Solids ”, Volume I, The Experimental Foundations of Solid Mechanics, pp. 630, Author: James Bell, 1984. J. A. Simmons, F. Hauser and J .E. Dom, “ Mathematical Theories of Plastic Deformation Under Impulsive Loading ”, University. Calif. Pub. Engng 5, 177 — 230 ( 1962 ). 186 19. J. Lubliner and M. Valathur, “ Some Wave-Propagation Problems In Plastic- Viscoplastic Materials.” Int.J. Solids Structure, 1969, Vol.5, pp.1275 — 1298, Pergamon Press. Printed In Great Britain. 20. J. Bell, “ An Experimental Study of Instability Phenomena In the Initial of Plastic Waves In Long Rods.” Symposium, Held in San Antonia, Texas, Septrnber 6 — 8, 1967., Printed by Springer — Verlay New York Inc, 1968. t/ 21. R. J. Clifton, “ Plastic Waves: Theory and Experiment.” in Nemeat- Nasser, 5., ( ed), Mechanics Today, 1, Pergamon, New York. 22. R]. Clifton, “ Some recent Developments in Plate Impact Experiments.” in Varley, E., ( ed. ), Propagation of Shock Waves in Solids, ASME, New York, 27 — 40. 2000. 23. C. H. Kames, ( 1968 ) “ The Plate Impact Configuration for Determining Mechanical properties of materials at High Strain rates.” in Lindholm, U.S., ( ed), Mechanical Behavior of material Under Dynamic Loads, Spring- Verlay, Berlin, 270 — 293. 24. ED. Bertholf and CH. Kames, ( 1970 ) “ Two-Dimensional Analysis of the Split Hopkinson Pressure Bar System.” J. Mech.Phys. Solids, 23, 1. 25. J. A. Zukas, T. Nicholas, H.F. Swift, L. B. Greszczuk and D. R. Curran, “ IMPACT DYNAMICS ”, A Wiley-Interscience Publication, JOHN WILEY & SONS, 1982. 26. SE. Jones, P.P. Gillis , J .C. Foster, JR., and L. L. Wilson, “ A One Dimensional, Two-Phase Flow Model for Taylor Impact Specimens.” Recent Advances In Impact Dynamics of Engineering Structures, ASME, AMD — Vol. 105, AD — Vol.17. 1989. 27 . S.E. Jones, P.J. Maudlin and J .C. Foster, JR., “ An Engineering Analysis of Plastic Wave Propagation in the Taylor Test.” Int. J. Engng. Vol. 19, No.2, pp. 95-106,1997. 187 28. 29. 31. 32. 33. AS. Khan and C. Hsiao, “ Behavior of Fully Annealed and As-Received Polycrystalline 1100 Aluminum During Propagation of Small and large Amplitude Plastic Waves." International Journal of Plasticity, Vol.7, pp. 773 — 748, 1991. P.J. Maudlin, J .C. Foster, JR., and SE. Jones, “ A Continuum Mechanics Code Analysis of Steady Plastic Wave Propagation in the Taylor Test.” Int. J. Impact Engng Vol. 19, No.3. . T.W. Wright, “ Axial Plastic Waves In A Rod.” International Journal of Plasticity, 14: ( 1 —3 ) 25 —42,1998. A. Rusinek and Klepaczko J R., “ A Numerical Study on the Wave Propagation in Tensile and Performance Test.” Journal De Physique IV, 10: (p9 ) 653 — 658, Sept. 2000. G.I.Taylor, “ The testing of materials at high rates of loading” J .Inst.Civi Eng.26 ( 1946 ) 486-519. G.I.Taylor, “ The use of flat ended projectiles for determining yield stress. 1: Theoretical considerations ” Proc.R.Soc.Lond.A194 ( 1948 ) 289-299. 34. M.L.Wilkins and Michael W. Guinan, “ Impact of Cylinders on 3 Rigid Boundary ”, 35. J .Appl.Phys., Vol.44, No.3, Marcg 1973. Kuscher, G. ( 1985 ), “ Nicht-lineare Ausbreitung elasto-plastischer Wellen in Staben,” Emst-Mach-Inst. (EMI), Berjcht 2/85, Freigurg, West germany. English see Reference [25]. 36. William F .Ames, “ Numerical Methods For partial Differential Equation”, Academic Press, New York, 1997. 188 37. G.Thomas Mase and George.E.Mase, “ CONTINUUM MECHANICS for ENGINEER”, Secong Edition, CRC Press LLC, 1999. 38. R.C.Batra and J .B.Stenens, “ Adiabatic Shear Banding in Axisymmetric Impact and Penetration Problems”, Computer Methods in Applied Mechanics and Engineering, 151 3-4 ): 325 — 342, Jan 20, 1998. 189 11111110111111111111111111