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Analytical and Numerical Evaluation of Subsurface Stresses in Anisotropic (Single-Crystal) Contacts

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PAGE 1

ANALYTICAL AND NUMERICAL EVALUATION OF SUBSURFACE STRESSES IN ANISOTROPIC (SINGLE-CRYSTAL) CONTACTS By ERIK C. KNUDSEN A THESIS PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLOR IDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE UNIVERSITY OF FLORIDA 2003

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Copyright 2003 by Erik C. Knudsen

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This thesis is dedicated to my grandmothe r, Madeline Rosenthal. I will forever miss her and always cherish the memories.

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ACKNOWLEDGMENTS I thank my advisor, Nagaraj Arakere, for his support and guidance while I did this research. Greg Swanson, Greg Duke, Gilda Batista, and Jeff Rayburn were a big help when I was at the Marshall Space Flight Center. My labmates (Jeff, Shadab, and Srikant) were an invaluable resource. My parents have been a source of wisdom and comfort all my life and none of this would have been possible without them. iv

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TABLE OF CONTENTS Page ACKNOWLEDGMENTS.................................................................................................iv LIST OF TABLES............................................................................................................vii LIST OF FIGURES.........................................................................................................viii ABSTRACT......................................................................................................................xii CHAPTER 1 INTRODUCTION........................................................................................................1 Research Objectives......................................................................................................1 Introduction to Nickel-Base Superalloys......................................................................3 Coordinate Transformations.........................................................................................4 The [110] Orientation............................................................................................6 The [213] Orientation............................................................................................9 2 ANALYTICAL SOLUTION......................................................................................14 3 FINITE ELEMENT FORMULATION-APPLIED STRESS MODEL......................22 4 RESULTS OF THE APPLIED-STRESS MODEL....................................................33 Comparison of Contour Plots: Applied-Stress Model and Analytical Solution.........34 Numerical Comparison of the Analytical and Applied-Stress Models......................39 5 CONTACT MODEL..................................................................................................42 Examination of the Normal Stresses at the Surface....................................................44 Comparison of Contact Model Stresses to the Analytical Solution...........................48 6 DEFORMATION MECHANISMS............................................................................56 7 LITERATURE REVIEW...........................................................................................67 8 CONCLUSIONS........................................................................................................73 9 RECOMMENDATIONS FOR FUTURE WORK.....................................................74 v

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LIST OF REFERENCES...................................................................................................75 BIOGRAPHICAL SKETCH.............................................................................................77 vi

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LIST OF TABLES Table page 1-1 Direction cosines........................................................................................................9 1-2 Direction cosines for the [110] orientation................................................................9 5-1 Comparison of Hertzian and numerical contact pressures and patch sizes..............47 6-1 Slip Systems.............................................................................................................60 7-1 Number of elements used.........................................................................................68 vii

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LIST OF FIGURES Figure page 1-1 Visual description of the contact problem..................................................................2 1-2 Damper and blade in contact......................................................................................2 1-3 Profile of the contact problem....................................................................................2 1-4 Damaged turbine blade...............................................................................................3 1-5 Specimen and material coordinate axes.....................................................................5 1-6 Aligned coordinate systems.......................................................................................6 1-7 Rotation of specimen coordinate axes about the z axis..............................................7 1-8 Two-dimensional view of rotation (z axis is into the page).......................................8 1-9 Schematic of the [213] orientation...........................................................................10 1-10 First rotation about y, y axis..................................................................................11 1-11 Second rotation.........................................................................................................12 1-12 Second rotation about z axis...................................................................................12 2-1 Profile view of contact problem with normal and tangential tractions....................15 2-2 Elliptic distribution...................................................................................................16 2-3 Applied forces used to derive the equilibrium equations.........................................16 3-1 Cylinder in contact with slab....................................................................................23 3-2 Face-on view of contact problem with applied load shown.....................................23 3-3 Elastic modulus of PWA 1480 as a function of crystal orientation at room temperature...............................................................................................................25 3-4 Contact patch dimensions.........................................................................................25 viii

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3-5 FE model; applied stress (no contact elements).......................................................26 3-6 Constraints................................................................................................................28 3-7 Elliptic contact load....................................................................................................28 3-8 Applied pressure.......................................................................................................29 3-9 Visualization of crystallographic orientation...........................................................32 4-1 Results were taken from the elements at the midplane............................................33 4-2 Schematic of casting coordinate system...................................................................34 4-3 Contour plots of the x stress (case 0)......................................................................35 4-4 Contour plots of the y stress (case 0)......................................................................35 4-5 Contour plots of the z stress (case 0)......................................................................36 4-6 Contour plots of the xy stress (case 0).....................................................................36 4-7 Contour plots of the x stress (case 33)....................................................................37 4-8 Contour plots of the y stress (case 33)....................................................................37 4-9 Contour plots of the z stress (case 33)....................................................................38 4-10 Contour plots of the xy stress (case 33)...................................................................38 4-11 Nodes used to extract results....................................................................................39 4-12 x stress vs depth......................................................................................................40 4-13 y stress vs depth......................................................................................................40 4-14 z stress vs depth......................................................................................................40 4-15 xy stress vs depth.....................................................................................................41 4-16 yz stress vs depth.....................................................................................................41 4-17 xz stress vs depth.....................................................................................................41 5-1 Contact model..........................................................................................................42 5-2 Densely meshed regions of the contact model.........................................................43 5-3 Contact and target elements.....................................................................................43 ix

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5-4 Nodes used to obtain the contact patch and peak pressure at the surface................45 5-5 Y-stress for case 17..................................................................................................45 5-6 Y-stress for case 21..................................................................................................46 5-7 Y-stress for case 25..................................................................................................46 5-8 Y-stress for case 29..................................................................................................47 5-9 Dimensions of the brick element used in the contact model....................................49 5-10 Sigma X comparison case 17...................................................................................49 5-11 Sigma X comparison case 25...................................................................................50 5-12 Sigma Y comparison case 17...................................................................................50 5-13 Sigma Y comparison case 25...................................................................................51 5-14 Sigma Z comparison case 17....................................................................................51 5-15 Sigma Z comparison case 25....................................................................................52 5-16 Tau XY comparison case 17....................................................................................52 5-17 Tau XY comparison case 25....................................................................................53 5-18 Sigma YZ comparison case 17.................................................................................53 5-19 Tau YZ comparison case 25.....................................................................................54 5-20 Tau XZ comparison case 17.....................................................................................54 5-21 Tau XZ comparison case 25.....................................................................................55 6-1 Example slip system.................................................................................................57 6-2 Four octahedral planes and slip directions...............................................................57 6-3 Cube slip planes and slip directions.........................................................................58 6-4 Tau 1 contour plot. .................................................................................................61 6-5 Tau 2 contour plot....................................................................................................61 6-6 Tau 3 contour plot....................................................................................................62 6-7 Tau 4 contour plot....................................................................................................62 x

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6-8 Tau 5 contour plot....................................................................................................63 6-9 Tau 6 contour plot....................................................................................................63 6-10 Tau 7 contour plot....................................................................................................64 6-11 Tau 8 contour plot....................................................................................................64 6-12 Tau 9 contour plot....................................................................................................65 6-13 Tau 10 contour plot..................................................................................................65 6-14 Tau 11 contour plot..................................................................................................66 6-15 Tau 12 contour plot..................................................................................................66 xi

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Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy ANALYTICAL AND NUMERICAL EVALUATION OF SUBSURFACE STRESSES IN ANISOTROPIC (SINGLE-CRYSTAL) CONTACTS By Erik C. Knudsen December 2003 Chair: Nagaraj K. Arakere Major Department: Mechanical and Aerospace Engineering The Space Shuttle main engine alternate high pressure-fuel turbo pump (SSME AHPFTP) turbine blades commonly fail from subsurface contact stresses imparted by a friction-damping device designed to reduce vibratory stresses. Fretting-fatigue failure is a major cause of failure in turbine components. Fretting occurs when components such as blade dampers and dovetail joints are subjected to vibration, which results in contact damage. When combined with a mean stress in the component, fretting can lead to a reduction in high cycle fatigue (HCF) life. Fretting (along with wear and corrosion) initiates cracks where the blade and damper are in contact. Although the turbo pump is only operating for ten minutes, it is cycling at a very high rate, which leads to rapid accumulation of fatigue cycles. The Marshall Space Flight Center (MSFC) has developed finite element (FE) models to investigate this problem. These models have very fine meshes and also incorporate nonlinearities such as friction. The sophistication of these models is xii

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expensive from a computational standpoint. Since the blades are made of anisotropic material, the properties of the blade change with the crystallographic orientation of the material. Manufacturing tolerances limit the precise orientation of the material, thus there are numerous orientations to examine. When one factors in the time it takes to run the models and also examine the results, many hours would have to be put in to give a thorough analysis. To speed the analysis up and also make things a bit simpler, it would be helpful to avoid such large FE models (that take hours to solve and many more hours to analyze). Ideally, one should build a contact model, to get a true sense of the stresses and contact patch size. If possible, much less complicated models should be built and compared to larger contact models. If there is reasonable agreement between the two, then the less sophisticated FE models can be used outright and the results arrived at much faster. This was done here. Results for a simple, applied-stress, FE model are compared to those for the contact model. Concurrently, analytical (or closed-form solutions) are also being developed for this problem. Getting results using analytical methods is almost always faster and easier then using numerical approaches. Numerical (or FE) results are also compared to those for analytical solutions. Results for numerical vs analytical approaches have fairly good agreement with the normal stress components, but the shear stresses do differ. The most probable source of that disagreement lies in the mesh density of the FE models. xiii

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CHAPTER 1 INTRODUCTION Extensive work has been done in the non-conformal classical area of contact mechanics of elastic solids dating back to Hertzs seminal work in the area. Johnson (1985) gives a complete discussion of his results. Relatively little work has been focused on contact mechanics of non-conformal anisotropic solids. Interest in this field is increasing because of applications of single crystal materials used in high temperature environments. Research Objectives The primary goal of the research presented is to examine the subsurface stresses in face-centered cubic (FCC) single crystal (SC), orthotropic, turbine blade damper and dovetail attachment contacts using the finite element (FE) method. Develop a simple FE model that has no contact elements (applied-stress problem) and compare the results to the more complex contact FE model. Develop a contact FE model that captures the contact patch dimensions and stresses in the blade, for fully anisotropic contacts; and compare the results to Hertzian solutions. Develop an analytical solution for anisotropic subsurface stress fields. Compare results of both the contact and applied-stress models to analytical solutions; and determine if the analytical approach is better than the FE approach for looking at the subsurface. Figures 1-1 and 1-2 show what the contact problem looks like. The damper shown in Figure 1-1 is touching the axial face of the turbine blade. Notice that the damper is upside-down in Figure 1-2. Figure 1-3 shows a profile view of the contact problem. 1

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2 Essentially, the damper and blade are modeled as a half-cylinder (damper) touching a rectangular slab (blade). The damper touches the blade platform at two separate locations. Only the circled location in Figure 1-3 is analyzed here, although this analysis could be applied to other parts of the damper. The resulting contact load is one that has a normal and tangential component. The work presented in this thesis is for normal loading only. Ultimately, cracks form and damage the blade as shown in Figure 1-4. Figure 1-1. Visual description of the contact problem Figure 1-2. Damper and blade in contact Figure 1-3. Profile of the contact problem

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3 Figure 1-4. Damaged turbine blade Source: N. Arakere, personal correspondence, 2003 Introduction to Nickel-Base Superalloys The material chosen for the blade is a precipitation strengthened single-crystal Nickel-base superalloy, Pratt and Whitney (PWA) 1480/1493. Generally speaking, when working at high temperatures and high tensile stresses where creep dictates the mode of failure, it is desirable to have a material with a close packed structure, such as FCC (which happens to be the structure of this alloy); a high melting temperature; and coarse grains (for isotropic materials). The close-packed structure makes diffusion more difficult. Since grain boundaries can be the weak link at high temperatures, it is a good idea to either eliminate them or make sure one does not have too fine a grain size (Hummel 1998). The blades in the SSME AHPFTP are made from a monograin material. The absence of grain boundaries gives improved creep resistance (which is necessary since the blades operate at high temperatures). Also, with no grain boundaries, there is no need for elements, such as Boron, that are added to increase grain boundary strength, but can also lower the melting point of the alloy (Donachie and Donachie 2002). Nickel-base superalloys can be used at a higher fraction of their melting point than just about any other alloy. Also, many elements can be dissolved into Nickel. This means that lots of additives can be used for hot corrosion and oxidation resistance, solid

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4 solution strengthening and ` formation (precipitation hardening). For example, the addition of Al provides increased oxidation resistance and also forms the compound Ni3Al (`). Chromium and Tungsten are usually added as solid solution strengtheners and Cr also helps combat corrosion (Donachie and Donachie 2002). The precipitation hardening (`) comes from heat-treating an alloy that has limited solid solubility and decreasing solid solubility with decreasing temperature. The heat treatment forms a fine dispersion of precipitates (small particles) that inhibit the motion of dislocations and make the material stronger. The highest strength is achieved when many small, round, closely spaced particles are coherently spread throughout the alloy (Hummel 1998). Since these alloys have only one grain, they are orientation dependent. The orientation of the crystal lattice plays a pivotal role in the properties of the material. Typically the blades are cast in the <001>, or low modulus, direction. This orientation gives the blades great thermal-mechanical fatigue resistance. It should be pointed out that one cannot cast the blades perfectly. Casting tolerances allow a 15o deviation from the stacking line (the line between the <001> direction and the airfoil stacking line) (Arakere and Swanson 2000). Coordinate Transformations To account for the orientation dependency of these materials, one must understand how to transform from one coordinate system to another. From there quantities such as the resolved shear stress (RSS) on different slip planes can be determined. Let us first begin with a few definitions. Figure 1-7 shows the material coordinate and specimen coordinate systems. The primed axes (x y z) are defined as the specimen coordinate system. The unprimed axes (x y z) are defined as the material (crystal) coordinate system. Coordinate

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5 transformations are necessary because material properties are functions of the orientation. This is not the case for an isotropic material. Since this material has cubic symmetry, there are three independent elastic constants as well as a tensor matrix that relates stress to strain (Dame and Stouffer 1996). y <010> y x x <100> z <001> z Figure 1-5. Specimen and material coordinate axes One way to relate one coordinate system to another is through the use of direction cosines. Once one has the cosines of the angles between the specimen and material axes, the proper coordinate systems can be constructed in whatever FE code one chooses. The model is initially built in the global (x y z) coordinate system. The material is cast with a certain crystallographic orientation which must be accounted for to obtain the proper results (stresses, strains, etc.). The orientation of the material is always given, and from there direction cosines of the angles between the material and specimen coordinate system can be determined using methods described by Lekhnitskii (1963) and also Magnan (2002).

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6 The [110] Orientation This example will deal with the [110] orientation. Let us align the x axis with the [110] direction. To do that imagine the specimen axes (they are primed) are initially aligned with the material, unprimed, coordinate system. z y x z <001> x <100> y <010> Figure 1-6. Aligned coordinate systems The orientation of the x axis is known (defined as [110]). One way to find the y and z axes is to arbitrarily define the y axis (it just has to be normal to the x axis, i.e. the dot product between the two will be zero). Now that the x axis is defined and the y axis is arbitrarily chosen, the z axis is obtained by taking the cross product of the x and y axis. The downside of employing such a method is that there is an infinite number of y vectors normal to the x axis. Thus, there are no unique vectors, or directions, that define where the y and z axes could wind up. But, the method used here does not involve an arbitrary assignment of the y and z axes. Continuing on, the y axis will also be rotated about the z axis and has the direction [-110].

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7 <-110> z y x<110> z <001> x <100> y <010> Figure 1-7. Rotation of specimen coordinate axes about the z axis Notice in Figure 1-7 the orientation (or coordinates relative to the material coordinate system) of the y vector was not arbitrarily picked. Both the specimen and material coordinate systems are orthogonal (i.e. the angle between the x, y, z axes are 90o). The entire specimen (primed) coordinate system was rotated about the z axis. The next step is to write the coordinates of the new axes with respect to the original axes. They are as follows z z yxyyxx'cossin'sincos' (1-1) Putting Equation (1-1) into matrix form

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8 zyxzyx1000cossin0sincos''' (1-2) y x y 1 x 1 Figure 1-8. Two-dimensional view of rotation (z axis is into the page) Trigonometry gives that = 45 degrees (measured counter-clockwise) from the (x y z) axes. o451tan (1-3) No further transformations are needed here to obtain the direction cosines for this case. It is convenient to enter the direction cosines into a table.

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9 Table 1-1. Direction cosines x y z x 1 1 1 y 2 2 2 z 3 3 3 Where 1 is the cosine of the angle between the x and x axes, 1 is the angle between the x and y axes, etc. Thus, Table 1-1 becomes Table 1-2. Direction cosines for the [110] orientation x y z x 0.707 0.707 0 y -0.707 0.707 0 z 0 0 1 The [213] Orientation The last example involved only one rotation. Frequently, one will have to perform two or even three rotations to get the desired crystal orientation. This example will look at the [213] orientation. Let us look at Figure 1-6 again. Once more, the x axis will be aligned with the [213] direction as shown in Figure 1-9. To do this transformation correctly, two rotations will be executed. First a rotation about the y-y axis by degrees is performed. Then a second rotation about the z axis by degrees takes place and the transformation is complete.

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10 Figure 1-6. Reprinted z y x z <001> x <100> y <010> x <213> 2 1 3 z <001> x <100> y <010> Figure 1-9. Schematic of the [213] orientation In Figure 1-11 one can see the first rotation from the projection (denoted by the dashed line) of the [213] vector into the x-z plane. Since the dimensions of the projection are known, the angle can be readily determined.

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11 01309.5623tan23tan (1-4) y,y <010> x <100> z x: projection of x into x-z plane z < 001 > Figure 1-10. First rotation about y, y axis For clarity, the rotation is redrawn in Figure 1-10. Writing the coordinates of the transformed coordinates in terms of the original set of axes we have zyxzyxcos0sin010sin0cos'''''' (1-5) There is just one more step which entails a rotation about the z axis by an amount of degrees. Figure 1-11 illustrates this.

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12 y <010> 3 x <213> 1 x <100> 2 z <001> Figure 1-11. Second rotation To make this second rotation clear, it is redrawn in Figure 1-12. Figure 1-12. Second rotation about z axis Once more this transformation is put in matrix form ''''''1000cossin0sincos'''zyxzyx (1-6) From looking at Figure 1-11, the value of is

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13 01505.15131tan131tan (1-7) The only thing left to do is solve for the table of direction cosines. cos0sin010sin0cos1000cossin0sincos333222111 (1-8) Finally, we note that counterclockwise rotations are positive and clockwise rotations are negative. Following this convention, theta is negative and phi is positive. Substituting and solving for the table of direction cosines 555.00832.0222.0964.0148.0802.0267.0535.0333222111 (1-9) Of course one could perform a third rotation about the x axis, perhaps by an angle If that were to happen, we would have cos0sin010sin0cos1000cossin0sincoscossin0sincos0001333222111 (1-10)

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CHAPTER 2 ANALYTICAL SOLUTION A closed-form solution to this contact problem was formulated by Dr. Arakere (University of Florida) which utilizes complex potential stress functions developed by Lekhnitskii (1963). This is a two-dimensional method with a generalized plain strain condition. The load and resulting stress functions have no z-dependency. The shape of the contact patch was assumed and the applied load equations were taken from Hertzs work for line-contact problems (such as a cylinder touching a cylinder) with isotropic materials. The subsurface stresses can be solved for any general traction distribution over the surface. In chapter two it was stated that the normal pressure distribution for Herzian (non-conformal) anisotropic bodies is the same as for isotropic ones. From Hertz the normal load distribution is 221)(axpxNo (2-1) For a tangential load we have 221)(axpxTof (2-2) Where po is the peak pressure, a is the contact half-width, and f is the coefficient of friction. In Figure 1-3, a profile view of the contact problem was shown. That figure is reprinted in Figure 2-1 with a normal and tangential traction applied. If both materials are isotropic, then a solution exists for this contact problem.. Essentially all one needs are 14

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15 the material properties and applied loads. But since the blade is made out of an orthotropic material, there is no closed-form solution available. P Q Figure 2-1. Profile view of contact problem with normal and tangential tractions Once again, there is only a normal load applied to the top of the cylinder (i.e. Q = 0). Hertz has proved, for non-conforming bodies in contact, that the normal pressure distribution on the block above is elliptic, Figure 2-2, for isotropic materials (Johnson 1985). In his paper of anisotropic contact problems, Willis (1966), states that this same distribution can be applied when anisotropic materials are used. What follows is an overview of Lehknitiskiis work applied to a non-conformal contact problem such as a cylinder touching a flat substrate. Equations that determine the subsurface stresses are developed with the main assumption is knowing the contact patch dimensions beforehand.

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16 Figure 2-2. Elliptic distribution Although Hertzs equations are used to describe the normal contact pressure, the analytical solution is derived in such a fashion that any loading can be applied to the surface of the anisotropic substrate (Figure 2-3). Where Hertzs equations are used can be seen in Equations 2-27 and 2-28. Lekhnitskii presents a stress function approach, F(x,y) and (x,y), to obtain the subsurface stresses. The stress functions are used to satisfy the equations of equilibrium which come from Figure 2-3. Surface Traction x Anisotropic Half-Space y Figure 2-3. Applied forces used to derive the equilibrium equations Equation 2-3 shows the relations for equilibrium. U is the applied body force. Notice that if no body forces are present Equation 2-3 becomes homogeneous. 0xUxyxyx

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17 0yUxyyxy (2-3) 0yyzxzx The relations listed in Equation 2-4 are the stress functions used to satisfy the equilibrium equations. UyFx22 U x Fy 22 yxFxy2 (2-4) xyz yxz This approach to solving these kinds of problems was initially developed by Airy in 1882 when he was solving two-dimensional stress problems. A good primer to the concept of stress function can be found in Timoshenko (1982). The stress functions must satisfy the above equations (which become homogeneous when there are no body forces). Lekhnitksii integrates three stress-strain equations that come from the generalized Hookes law. The aforementioned integration gives rise to three displacement equations. These equations must also satisfy the leftover three equations from the generalized Hookes law. They are substituted into the remaining three stress-strain relations that give rise to another set of three equations where the coefficients of the left and right-hand sides are equated. Those equations are differentiated to eliminate certain variables from the generalized displacement equations that will be determined in the future. The result of that differentiation is listed in Equation 2-5 and Equation 2-6. yxUyxUxULFL212112261622221234)()()( (2-5)

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18 yUxUBALFL)()(225152414353423 (2-6) L2, L3, L4 are differential operators of the second, third, and fourth orders. A and B describe the bending in the x-z and y-z planes. is the rotation about the z-axis. In the special case of plane deformation; A = B = = 0. Equation (2-6) can be written as follows F = F` + Fo (2-7) = ` + o (2-8) where F` and ` are the general solutions to a system of homogeneous differential equations. They are homogenous because of the absence of body forces. L4F` + L3` = 0 (2-9) L3F` + L2` = 0 (2-10) To obtain the solution to the set of differential equations above, one of the functions, such as `, is eliminated. The next step is to apply the operator L2 on Equation 2-9 and L3 on Equation 2-10 and subtract what remains. The result is an equation of the sixth order (L4L2-L32)F` = 0 (2-11) To get the solution for `, the exact same procedure is followed. Since Equation (2-11) is of the sixth order, it can be broken down into six linear operators of the first order D6 D5 D4 D3 D2 D1F` = 0 2-12) Where Dk = (/y)-k/x and k are the roots to the characteristic equation 0)()().(2324lll (2-13)

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19 where l255224555 (2-14) l315314562254624 (2-15) l4114216321266 222622 (2-16) and ijaijai3aj3a33 (2-17) The term aij are the elastic constants for this orthotropic material. See Equation 3-13. Since Equation 2-12 is of the sixth order, it can be resolved into the integration of six equations of the first order D1F` = 2, D22 = 3, D33 = 4, D44 = 5, D55 = 6, D66 = 0 (2-18) In order to integrate these first order equations, it is assumed the general integral is a function of the argument x + y. For example, the integral for 6 is denoted by fV6(x + 6y). Following this convention, 5 satisfies the nonhomogeneous equation D55 = fV6(x + 6y) (2-19) Integrating (2-19) the following expression is obtained 5 = f5IV(x + 5y) + fV6(x + 6y)/ 6 5 (2-20) Using the same method 4, 3, 2, and F` will be obtained. Using a change of notation, expressions for F` and ` are determined 61)(`kkkyxFF (2-21)

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20 61)(`kkkyx (2-22) Lekhnitskii goes on to prove four theorems that are necessary for the theory governing a body bounded by a cylindrical surface. He shows that a certain set of four equations cannot have real roots and that ultimately k is always complex or imaginary. Now the stress functions take the following form oFzFzFzFF )()()(Re2332211 (2-23) ozFzFzF)(`1)(`)`(Re233322211 (2-24) Where 1I31I22 2I32 I22 3I33 I43 (2-25) Introducing the complex variable zk (k = 1, 2), new functions are introduced )(`)(kkkkzFz )(`1)(33333zFz 2,1 k (2-26) dzxxNizzzaa)(21)()()('33'2'1 (2-27) dxzxxTizzzaa)(21)()()('333'22'11 (2-28) (2-29) 0)()()('3'22'11zzz Equations 2-30 are the resulting stress functions. x2Re12'1z()22'2z()323'3z() (2-30) y2Re'1z()'2z()3'3z() (2-31)

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21 xy2Re1'1z()2'2z()33'3z() (2-32) xz2Re11'1z()22'2z()3'3z() (2-33) yz2Re1'1z()2'2z()'3z() (2-34) z1a33 a13xa23ya34yza35xza36xy (2-35) zixiy (2-36)

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CHAPTER 3 FINITE ELEMENT FORMULATION-APPLIED STRESS MODEL Our first goal with this model is to evaluate the subsurface stresses in the substrate by decoupling the contact mechanics from the stress analysis. To do this, an anisotropic substrate is modeled in 3-D, subject to normal loading over a contact width chosen arbitrarily. One of the difficulties in trying to get reliable results is that FE models need very fine meshes and must account for nonlinearities such as friction and plasticity. The numerical modeling procedure would be simplified considerably if the contact problem is reduced to a stress analysis problem. This can be done if the size of the contact patch and the normal contact pressure distribution is determined from Hertzian assumptions. These models are less complex than contact models and will run faster. They are easier to troubleshoot and modify should there be a problem or design change. Hertz developed closed-form solutions for isotropic contact problems. His theory is grounded on four assumptions: the bodies in contact are continuous and non-conforming, the strains are small, each solid is considered to be an elastic half space, and there is no friction on the surfaces (Johnson 1985). These assumptions were built into the models to analyze the anisotropic material. The applied stress model is a rectangular slab (composed of the superalloy) with an applied load. The load was determined from the Hertzian assumptions stated above. Since the contact is modeled as a cylinder touching a rectangular slab, this is a line-contact problem. Hertz determined the contact patch dimensions, maximum contact 22

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23 pressure, and profile of the contact load (which happens to be elliptic). It has been shown that the pressure distribution for contact problems of a similar type (non-conforming) involving anisotropic materials of is also parabolic (Willis 1966). The true contact patch size and load distribution were not known when these models were first being built. The idea is to build a simple model and compare it a more sophisticated one that uses contact elements to see how well Hertzs equations will predict the load distribution and contact patch size for the anisotropic material. Figure 3-1. Cylinder in contact with slab P Figure 3-2. Face-on view of contact problem with applied load shown Figure 3-2 shows the non-conformal contact problem of a cylinder on a flat substrate and in Figure 3-4 shows a close-up view of the contact region. Hertz (1881)

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24 determined this contact width and also the maximum contact pressure. The equations are as follows: 12111RRR (3-1) where R is the equivalent radius of curvature. 2/1222121*11EEE (3-2) 2/1*4EPRa (3-3) aPpo2 (3-4) Where P is the load per unit length, a is the contact half-width, and po is the maximum contact pressure. The subscripts 1 and 2 refer to the two bodies that are in contact with each other. The elastic modulii, E1 and E2, can be easily defined if the elastic bodies are isotropic. The problem inherent here is the contact of isotropic indenter (damper) with anisotropic substrate (blade). For the anisotropic substrate the modulus will vary as a function of crystallographic orientation. There is no procedure to determine the effective elastic modulus, E*, for an anisotropic contact, since E1 and E2 vary with the orientation of the material. See Figure 2-3; the modulus is 126 GPa for the [001] orientation and increases to 310 GPa for the [-111] orientation (Sims et al. 1987).

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25 Figure 3-3. Elastic modulus of PWA 1480 as a function of crystal orientation at room temperature Source: Modified from Donachie and Donachie, 2002 Figure 3-4. Contact patch dimensions The contact dimension, 2a, is known here and is given by Equation 3-3. The contact takes place over an area that is rectangular (width x length). The width is 2a and the length is 0.5 inches as shown in Figure 3-3. Figure 3-5 shows the FE model used for the applied/load stress conditions. The model was built using ANSYS FE software and is composed of solid-45 elements. These are 8-node elements used for 3-D models. This element has three degrees of freedom at

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26 each node and also has the capability to analyze creep, swelling, large deflection and large strains (ANSYS 1999). Figure 3-5. FE model; applied stress (no contact elements) Since a plane stress or plane strain condition cannot be enforced with this material, a 3-D model must be built. The analytical solution (discussed in chapter two) uses a generalized plane strain condition that generates equations for the subsurface stresses that are functions of x and y only. When Lekhnitskii (1963) developed the closed-form solution used in this analysis, his boundary conditions were prescribed on an infinitely long cylinder. This is not a plain strain condition is the conventional sense since an orthotropic material is involved. The orthotropy induces shear coupling in the z-direction.

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27 One cannot introduce infinitely large elastic bodies into a numerical analysis. The dimensions used in the FE model drawn in Figure 3-5 were sufficiently large so the rectangular slab could be treated as an elastic half space. That is, the stresses are uniform throughout the thickness (z dimension) of the material. In most non-conformal contact problems, the contact patch is small compared to the dimensions of the bodies in contact. Consequently, stress gradients in the region affected by the contact are very large, necessitating an extremely refined FE mesh to accurately capture the subsurface stresses. The densely meshed region in Figure 3-5 is where the contact would take place. It is in this region that the results were extracted and where the stresses are the highest. A coarser mesh is used for the rest of the substrate since the stresses are not as high and to reduce the overall size of the model. The less elements there are the better, since the model will run faster on the computer. The base of the model, or slab, is constrained so the block cannot move in the normal, or y-direction. It is important to allow the elements to have some flexibility, thus one does not want to over-constrain the model. That is why only two of the bottom corners of the slab were constrained in the x-direction. This ensures that the model cannot rotate about the y axis, or have any rigid body motion. Now that the model is properly constrained, only entering the material properties and applying the loads remain. The load was applied using a macro (algorithm written in a computer language such as FORTRAN). The Hertzian contact pressure distribution for normal, frictionless, loading is shown above in Figure 3-7. The loading curve takes the form of an ellipse when the normal load

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28 P is applied to the cylinder. This distribution was applied over the densely meshed region in Figure 3-8. Figure 3-6. Constraints Figure 3-7 Elliptic contact load

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29 Figure 3-8. Applied pressure 2/1221)(axpxpo (3-6) With the geometry and loading in hand, the model was drawn and then meshed with the solid-45 elements. Finally, the material properties had to be entered along with the crystal orientation of the material. Determining the elastic constants in the specimen coordinate system can be achieved by using the following equations taken from Lekhnitskii (1963). The relationship between the stresses in the material and specimen coordinate systems is: }']{[}{ Q (3-7) where xyzxyzzyx and (3-8) xyzxyzzyx'''''''

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30 Recall that 1 is the cosine of the angle between the x and x axis, 1 is the cosine of the angle between the x and y axes, etc. )()()()()()()()()(222222222122113312332332211122113312332332211122113312332332211123123232221123123232221123123232221Q (3-9) The strain vector transformations are similar. The [ matrix is different because there is a factor of two in the relationship between engineering and tensor shear strain components(i.e. xy = 2 xy, zx = 2 zx, and yz = 2 yz). ]Q }]{'[}'{ Q (3-10) )()()(222)()()(222)()()(222122113312332332211122113312332332211122113312332332211123123232221123123232221123123232221Q (3-11) Hookes law for a homogeneous anisotropic body using Cartesian coordinates is: ija (3-12) Since this superalloy is FCC and has cubic symmetry, there are only three elastic independent elastic constants needed to define the shear modulus, elastic modulus and Poisson ratio.

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31 444444111212121112121211000000000000000000000000aaaaaaaaaaaaija (3-13) yyxyxxyxyzxxEEaGaEa 124411,1,1 (3-14) The equation below relates the stresses to the strains in the material specimen coordinate system: '' ija' (3-16) The elastic constants can be transformed from one coordinate system to another by: )6......,,2,1,(6161jiQQamnnjmimnTQaQaijij (3-17) With these relationships in hand, the stresses, strains, and elastic constants can be determined in either the material or specimen coordinate systems. Consider a block of the Ni-based superalloy shown in the Figure 3-9. The desired orientation of the crystal lattice is <110> with the x axis aligned with the <110> direction. One way to look at the problem is to imagine a cylindrical piece of material, shown in the Figure 3-8, cut from the block. That cylinder now has the desired material properties in the <110> direction. Using equation 3-17 the elastic constants in any orientation can be determined if one has the direction cosines between the specimen and material coordinate system.

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32 ANSYS allows one to input values of the stiffness (units of psi, or Pa) or compliance matrix (units of 1/psi or 1/Pa). Once those values are entered into ANSYS-the finite element model now has the material properties in the desired orientation. <110> y y x < 010 > x <100> <001>, z, z Figure 3-9. Visualization of crystallographic orientation

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CHAPTER 4 RESULTS OF THE APPLIED-STRESS MODEL Contour plots of the normal and shear stresses were obtained from the densely meshed region at the midplane of the model (see Figure 4-1). To make the analysis simpler, the analytical solution does not account for the boundary conditions on the free ends, or surfaces. Since the analytical solution uses a generalized plane strain condition, it is important to extract the results from nodes that are not near the edges, or faces, of the model. Figure 4-1. Results were taken from the elements at the midplane How the material orientations are selected is best seen in Figure 4-2. The cones represent variations in the primary orientation. The inner and outer cones represent a 7.5 degree and fifteen degree variation, respectively. The red dots indicate some possible positions of the z-axis (33 total) that come from the intersection of the x-z and y-z planes. The third plane is the theta () plane, or a rotation about the z axis. Since this material 33

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34 exhibits cubic symmetry, theta only has to vary between 0 and 80 degrees. Following the convention of Arakere and Swanson (2000), theta is broken up into nine degree increments (0, 10, 20, 30, 40, 50, 60, 70, 80). Multiplying the possible values for theta by the 33 primary axis cases, there is a total of 297 material orientations. Figure 4-2. Schematic of casting coordinate system Source: G. Duke, personal correspondence, 2002 Case 0 corresponds to the angles , and all being equal to zero as shown in Figure 4-2. Another example would be case 17. Here, = 15o, and are zero degrees. Two sets of results will be presented here: case 0 and case 33. Case 33 is not on the chart but = 30o, and are zero degrees. Comparison of Contour Plots: Applied-Stress Model and Analytical Solution The first set of results is a contour plot for case 0. This is where the material and specimen coordinate systems are coincident with each other. Right away, one notices the

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35 symmetry between all of the plots. The ANSYS plots have the black background. The analytical plots for the corresponding stress are placed next to the ANSYS contour plots. A B Figure 4-3. Contour plots of the x stress (case 0). A) ANSYS contour plot. B) Analytical contour plot A B Figure 4-4. Contour plots of the y stress (case 0). A) ANSYS contour plot. B) Analytical contour plot

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36 A B Figure 4-5. Contour plots of the z stress (case 0). A.) ANSYS contour plot. B) Analytical contour plot A B Figure 4-6. Contour plots of the xy stress (case 0). A). ANSYS contour plot. B). Analytical contour plot

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37 The next set of results is for case 33. Again, the two plots look very similar from a graphical standpoint. A B Figure 4-7. Contour plots of the x stress (case 33). A). ANSYS contour plot. B). Analytical contour plot A B Figure 4-8. Contour plots of the y stress (case 33). A.) ANSYS contour plot. B). Analytical contour plot

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38 A B Figure 4-9. Contour plots of the z stress (case 33). A.) ANSYS contour plot. B). Analytical contour plot A B Figure 4-10. Contour plots of the xy stress (case 33). A.) ANSYS contour plot. B). Analytical contour plot

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39 Numerical Comparison of the Analytical and Applied-Stress Models Contour plots establish a pattern of how the stresses vary, but it is a good idea to also look at the numbers and compare the magnitudes of the stresses. The following plots are for case 33. These results were taken from the elements at the midplane of the model in the densely meshed region (see Figure 4-11). In Figures 4-16 and 4-17, the shear stresses between the two models do not agree very well. The likely cause will be discussed in chapter four. results were obtained from these nodes densely meshed region y x z midplane Figure 4-11. Nodes used to extract results

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40 Sigma X vs Depth -3.00E+05-2.50E+05-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+0000.0050.010.015Depth into Material (in.)psi Sigma X ANSYS Sigma X Analytical Figure 4-12. x stress vs depth Sigma Y vs Depth -3.00E+05-2.50E+05-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+0000.0050.010.015Depth into Material (in.)psi Sigma Y ANSYS Sigma Y Analytical Figure 4-13. y stress vs depth Sigma Z vs Depth-2.50E+05-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+0000.0050.010.015Depth into Material (in.)psi Sigma Z ANSYS Sigma Z Analytical Figure 4-14. z stress vs depth

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41 Tau XY vs Depth-50000500010000150002000000.0050.010.015 Depth into Material (in.)psi Tau XY ANSYS Tau XY Analytical Figure 4-15. xy stress vs depth Tau YZ vs Depth-250-200-150-100-50000.0050.010.015Dpeth into Material (in.)psi Sigma YZ ANSYS Tau YZ Analytical Figure 4-16. yz stress vs depth Tau XZ vs Depth020040060080010001200140000.0050.010.015Depth into Material (in.)psi Sigma XZ ANSYS Tau XZ Analytical Figure 4-17. xz stress vs depth

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CHAPTER 5 CONTACT MODEL Figure 5-1. Contact model The contact model is a half-cylinder resting on a rectangular block. The same assumptions are in place for the applied stress model: no friction and normal loading only. The volumes are comprised of SOLID-45 elements. CONTA174 and TARGE170 elements were used to establish the contact between the half-cylinder and block. The primary aim of running the contact model is to determine the dimensions of the contact patch (the contact half-width, a). The applied stress and analytical solutions relied on an assumed value for that number. The contact model will take much longer to run on the 42

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43 computer versus the applied stress or analytical solution, but the computations are necessary to get a handle on the contact patch size. Figure 5-2. Densely meshed regions of the contact model In Figure 5-2 above one can see the densely meshed region where the contact elements were placed on the block and where the target elements were applied on the half-cylinder. Figure 5-3 shows the resulting contact pair created in ANSYS. Target Elements Contact Elements Figure 5-3. Contact and target elements

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44 Notice the curved shape of the target elements. That is because they are attached to the curved profile of the half-cylinder. In order to create contact models in ANSYS, a contact pair of elements must be created-a contact element and a target element. ANSYS has general guidelines as to what line, surface, or volume these elements should be applied. Perhaps the most critical feature is the mesh size. For example, a large target element size and very fine contact element will not work. The sizes of the contact and target elements should be fairly close to one another. It is possible to get a solution to converge, but the results will most likely be incorrect. That is why there is a densely meshed region in both the bottom part of the half-cylinder and on the surface of the block shown in Figures 5-2. Another important characteristic of this contact model is that displacement boundary conditions had to be used instead of applied forces. Applied displacements are more stable than applied forces in ANSYS. Examination of the Normal Stresses at the Surface The analysis was run for four cases: case 17, 21, 25, and 29. To obtain the size of the contact patch and peak pressure at the surface, the same procedure in chapter three was followed. A group of nodes selected at the midplane of the densely meshed region of the block (see Figure 5-4). Below are plots of the stress in the y-direction for the selected nodes. Notice that they have an elliptic profile. The normal contact pressure should be elliptical (even though orthotropic materials are involved). If the correct distribution is not apparent, then the size of the elements may be too big and additional refinement is needed so the proper load is applied to the surfaces of the bodies in contact.

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45 z x y results were obtained from these nodes densely meshed region midplane Figure 5-4. Nodes used to obtain the contact patch and peak pressure at the surface Case 17 Sigma Y-250000-200000-150000-100000-500000500000510152025milspsi Sigma Y Figure 5-5. Y-stress for case 17

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46 Case 21 Sigma Y-250000-200000-150000-100000-500000500000510152025milspsi Sigma Y Figure 5-6. Y-stress for case 21 Case 25 Sigma Y-250000-200000-150000-100000-500000500000510152025milspsi Sigma Y Figure 5-7. Y-stress for case 25

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47 Case 29 Sigma Y-250000-200000-150000-100000-500000500000510152025milspsi Sigma Y Figure 5-8. Y-stress for case 29 Table 5-1 lists the contact patch (taken from the above plots) and also the peak pressures from the four cases that were run in ANSYS. These numbers are compared to peak pressure and contact patch when Equations 3-2, 3-3, and 3-4 are used. The cylinder was composed of an isotropic steel (E = 30 x 106 psi and = 0.3), the effective radius, R, was 0.1 in. Table 5-1. Comparison of Hertzian and numerical contact pressures and patch sizes Case17212529 ANSYS Contact Half-Width (in.)0.0030.0030.0020.002ANSYS Peak Pressure (psi)-195000-206000-194770-200240 Using Equations 3-2, 3-3, and 3-4 the peak (Hertzian) contact pressure and contact half-width are 180100 psi and 0.002774 inches (from the 79 lbf normal load). Looking at the table above it would appear the true contact half-width is between 0.002 and 0.003 inches. A finer mesh would help to determine if that is the case. The consequence of a better mesh (more elements) is computational cost. ANSYS does have a limit on the number of elements that can be used. Approaching that limit would tax

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48 ANSYS and the abilities of the computer to the point that one run could take many hours to finish. Comparison of Contact Model Stresses to the Analytical Solution The following plots compare stresses between the contact model and analytical solution taken at the midplane straight down (-y direction) into the rectangular block. The stresses between the applied stress model and analytical solution were already compared and the results below are similar in that the x, y z, and xy stresses match well but the remaining two shear components do not. It should be pointed out that the xz and yz are not always zero for the analytical solution. As the orientation changes, those two stress components will not be zero. For case 21, xz is nonzero and in case 29 yz is nonzero. Even though Lehnitkskii (1963) uses what he refers to as a generalized plane strain condition, there is still shear coupling in the z direction. The elements in the densely meshed regions of both models had finer meshes in the x and y directions than the z direction. Figure 5-9 illustrates the size of the brick element used in the contact model. Brick elements of a slightly smaller size were used in the applied stress model (the z-dimension was 0.005 in.). One can see that there is a large difference between the length in the x and y directions and the z direction of the brick element. The difference between the z dimensions was only 0.002 inches, but that is enough to throw off at least one of the shear stress components. For example, the xy component from the contact model does not agree with the analytical solution. The issue here is probably one of convergence. One way that might help is to use a finer mesh. Four of the six stress components from the applied stress model match up very with the analytical solution and that is most likely due to the smaller element size that model uses.

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49 Figure 5-9. Dimensions of the brick element used in the contact model The reason why these elements had such dimensions is computational cost. A smaller dimension in the z direction would have resulted in more elements. Ideally one would like to have many cube elements or small tetrahedrons. But that would have resulted in contact models having a large number of elements. A model that big would take a very long time to solve and the results files (where the stresses are stored) also would have been large. Sigma X vs Depth (case 17)-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+00051015milspsi Sigma X ANSYS Analytical Figure 5-10. Sigma X comparison case 17

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50 Sigma X vs Depth (case 25)-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+00051015milspsi Sigma X ANSYS Analytical Figure 5-11. Sigma X comparison case 25 Sigma Y vs Depth (case 17)-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+00051015milspsi Sigma Y ANSYS Analytical Figure 5-12. Sigma Y comparison case 17

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51 Sigma Y vs Depth (case 25)-2.00E+05-1.50E+05-1.00E+05-5.00E+040.00E+00051015milspsi Sigma Y ANSYS Analytical Figure 5-13. Sigma Y comparison case 25 Sigma Z vs Depth (case 17)-1.60E+05-1.40E+05-1.20E+05-1.00E+05-8.00E+04-6.00E+04-4.00E+04-2.00E+040.00E+00051015milspsi Sigma Z ANSYS Analytical Figure 5-14. Sigma Z comparison case 17

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52 Sigma Z vs Depth (case 25)-1.60E+05-1.40E+05-1.20E+05-1.00E+05-8.00E+04-6.00E+04-4.00E+04-2.00E+040.00E+00051015milspsi Sigma Z ANSYS Analytical Figure 5-15. Sigma Z comparison case 25 Tau XY vs Depth (case 17)-60000-50000-40000-30000-20000-1000001000020000051015milspsi Tau XY ANSYS Analytical Figure 5-16. Tau XY comparison case 17

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53 Tau XY vs Depth (case 25)-40000-35000-30000-25000-20000-15000-10000-50000051015milspsi Tau XY ANSYS Analytical Figure 5-17. Tau XY comparison case 25 Tau YZ vs Depth (case 17)-2000-10000100020003000051015milspsi Tau YZ ANSYS Analytical Figure 5-18. Sigma YZ comparison case 17

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54 Tau YZ vs Depth (case 25)-1000-500050010001500200025003000051015milspsi Tau YZ ANSYS Analytical Figure 5-19. Tau YZ comparison case 25 Tau XZ vs Depth (case 17)-500005000100001500020000051015milspsi Tau XZ ANSYS Analytical Figure 5-20. Tau XZ comparison case 17

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55 Tau XZ vs Depth (case 25)-500005000100001500020000051015milspsi Tau XZ ANSYS Analytical Figure 5-21. Tau XZ comparison case 25

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CHAPTER 6 DEFORMATION MECHANISMS Plastic deformation in metals results from a phenomenon called slip-that is the movement of dislocations throughout the material. Typically slip occurs over planes with high atomic density in the close-packed directions. When slip occurs atoms are being forced to relocate to new lattice positions within the crystal structure. Atoms located on densely packed planes will have a shorter distance to travel when compared to atoms situated on other, less densely packed, planes. This is not always the case, and at higher temperatures dislocations can move along planes with lower atomic densities because of two things: cross slip and climb. Cross slip (phenomenon that entails dislocations moving onto an intersecting slip plane) is a thermally activated process. Climb (also a thermally activated process) is associated with diffusion (movement of atoms into vacancies, etc) (Hummel1998). Since the material in question is a Nickel based superalloy, the planes with the highest atomic density are the octahedral planes. A slip system requires the definition of the slip plane and the slip direction. For example, (111)[0 1] is a slip system where (111) is the slip plane and [0 1] is the slip direction. The slip direction always lies in the slip plane (Figure 6-2). The aforementioned notation comes from the Miller indices. Let us look at the (111)[0 1] system graphically (Figure 6-1). The (111) plane is encased by the cube and the vector <0-11> is the slip direction drawn from the origin. With the four octahedral planes there are 12 close-packed directions the planes can move and also another set of 12 directions in the non-close-packed slip direction. Thus, 56

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57 there are 24 slip systems on the octahedral planes. At higher temperatures, slip can occur on the cube planes. These planes are not as densely packed as the octahedral planes and they, along with the octahedral planes, are shown below in Figures 6-2 and 6-3. Counting the cube slip systems, there are a total of 30 slip systems (Dame and Stouffer 1996). Figure 6-1. Example slip system Figure 6-2. Four octahedral planes and slip directions. Source: Modified from Dame and Stouffer 1996

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58 Figure 6-3. Cube slip planes and slip directions. Source: Modified from Dame and Stouffer 1996 With the 30 slip systems defined one can now establish equations to acquire local stresses and strains within the material. From Dame and Stouffer (1996) we have ,11001110110101111010110101111011001101111011001110110111001101111010110161121110987654321yzzxxyzzyyxx (6-1)

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59 yzzxxyzzyyxx112211121121211112121121211112112211211112112211121121112211211112121121231242322212019181716151413 (6-2) The six slip systems below are for the cube slip planes. yzzxxyzzyyxx11000011000010100010100001100001100021302928272625 (6-3) 1-30 are the local shear stresses associated with that slip system. For example, 1 is associated with slip system number one. If any of these values (1-30) are above a threshold stress called the critical resolved shear stress, then that slip system will become active. It is possible for more than one slip system to become active depending on the orientation and loading of the material. It might be of some use to investigate how the local shear stresses (1-30) vary in the material. In order to calculate these local shear stresses, all six stress components

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60 (x, y, z, xy,zx, yz) are needed for the equations listed above. Then at each node all of the local shear stresses can be tabulated. Table 6-1. Slip Systems Slip Number Slip Plane Slip Direction Octahedral Slip a/2<110>{111} 12 Primary Slip Directions 1 [ 1 1 1] [1 0 -1] 2 [ 1 1 1] [0 -1 1] 3 [ 1 1 1] [1 -1 0] 4 [-1 1 -1] [1 0 -1] 5 [-1 1 -1] [1 1 0] 6 [-1 1 -1] [0 1 1] 7 [1 -1 -1] [1 1 0] 8 [1 -1 -1] [0 -1 1] 9 [1 -1 -1] [1 0 1] 10 [-1 -1 1] [0 1 1] 11 [-1 -1 1] [1 0 1] 12 [-1 -1 1] [1 -1 0] Octahedral Slip a/2<112>{111} 12 Secondary Slip Directions 13 [ 1 1 1] [-1 2 -1] 14 [ 1 1 1] [2 -1 -1] 15 [ 1 1 1] [-1 -1 2] 16 [-1 1 -1] [1 2 1] 17 [-1 1 -1] [1 -1 -2] 18 [-1 1 -1] [-2 -1 1] 19 [1 -1 -1] [-1 1 -2] 20 [1 -1 -1] [2 1 1] 21 [1 -1 -1] [-1 -2 1] 22 [-1 -1 1] [-2 1 -1] 23 [-1 -1 1] [1 -2 -1] 24 [-1 -1 1] [1 1 2] Cube Slip a/2<110>{100} 6 Cube Slip Directions 25 [1 0 0] [0 1 1] 26 [1 0 0] [0 1 -1] 27 [0 1 0] [1 0 1] 28 [0 1 0] [1 0 -1] 29 [0 0 1] [1 1 0] 30 [0 0 1] [-1 1 0] Source: Modified from Dame and Stouffer 1996

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61 Once again, nodes from the densely meshed region at the midplane of the rectangular slab were used to obtain the required stresses. Local shear stresses from the primary planes (1-12) are given below for cases 17 and 25. In all cases the peak stresses occurred a few mils below the surface, and then decrease as a function of depth (-y direction) into the material. Also, in comparing the local shear stresses one notices how similar the contours are between the two cases. bb bb A B Figure 6-4. Tau 1 contour plot. A.) Case 17. B.) Case 25 cc cc A B Figure 6-5. Tau 2 contour plot. A.) Case 17. B.) Case 25

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62 dd dd A B Figure 6-6. Tau 3 contour plot. A.) Case 17. B.) Case 25 ee ee A B Figure 6-7. Tau 4 contour plot. A.) Case 17. B.) Case 25

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63 ff ff A B Figure 6-8. Tau 5 contour plot. A.) Case 17. B.) Case 25 gg gg A B Figure 6-9. Tau 6 contour plot. A.) Case 17. B.) Case 25

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64 hh hh A B Figure 6-10. Tau 7 contour plot. A.) Case 17. B.) Case 25 ii ii A B Figure 6-11. Tau 8 contour plot. A.) Case 17. B.) Case 25

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65 jj jj A B Figure 6-12. Tau 9 contour plot. A.) Case 17. B.) Case 25 kk kk A B Figure 6-13. Tau 10 contour plot. A.) Case 17. B.) Case 25

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66 ll ll A B Figure 6-14. Tau 11 contour plot. A.) Case 17. B.) Case 25 mm mm A B Figure 6-15. Tau 12 contour plot. A.) Case 17. B.) Case 25

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CHAPTER 7 LITERATURE REVIEW This section gives a discussion of the limited work done in the area of anisotropic contact mechanics. Relevant work dealing with analytical and numerical solutions is highlighted. Sinclair and Beisheim (2002) developed a procedure to analyze stresses of dovetail attachments in turbine blades using a submodeling routine. These stresses are difficult to obtain because of the nature of the contact problem (conforming) and also the large magnitude and gradient of the contact stresses. These parameters require a high number of elements to achieve accurate results. A submodeling routine was implemented in order to get converged peak stresses without considerable computational cost. The first step is to run a global analysis that includes the highly stressed contact region. Three grids labeled coarse, medium, and fine are used to mesh the model. The FE model is run with each of the three grids and the results from each grid are compared to check for convergence. The stress component in question, such as a peak stress, is said to have converged if maxmaxmaxfmf < s (7-1) where s is the error level (0.01 is excellent, 0.05 is good, and 0.1 is satisfactory). The superscripts f and m denote fine and course grids. The next step is the submodeling routine. This is where the highly stressed region, or submodel, is broken out and analyzed separately using the information 67

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68 obtained from the global analysis. Distances, or boundaries, away from peak stress component are chosen such that the stress component in question has converged. Then the submodel is divided up with its own coarse, medium, and fine grids. Displacements were used to check for convergence since they converge faster than stresses. Boundary displacements are said to be converging if miciuu > fimiuu (7-2) where i denotes the node in question. Cubic spline equations were fit to the nodal displacements. This was done so even more displacements could be ascertained between nodes for all of the grids. Using the fitted splines, the three grids are run and checked to see if the submodel boundary conditions have converged using two expressions that calculate the boundary condition error and the discretization error. The latter is built in to all FE analysis. Even though Sinclair and Beisheim (2002) claim that their numerical approach requires modest computing resources, a very high number of elements was used for their analysis. Table 7-1. Number of elements used GridSpatialContactTotalGlobal Course21274782614Medium14990194316993Fine1088707765116635Outer Ring Course5400540Medium414004140Fine34147034147Submodel Coarse9216115210368Medium73728460878336Fine58982418432608256 Source: Modified from Sinclair and Beisheim (2002)

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69 On the average, contact models in this work, composed of roughly 30,000 elements, took approximately two hours for the computer to complete its run for each crystal orientation. Over 600,000 elements were used in the submodeling routine to get accurate results in the contact region. Only had normal tractions applied to the model, just like the models used in this analysis. In chapter three the applied stress results were compared to analytical solution. The normal stress components agreed fairly well, but the shear stresses did not. A potential source of this problem was the size of the elements used, or the mesh size. Smaller elements, and consequently bigger FE models, are needed to get the correct stresses-as shown in Table 7-1. Using that many elements resulted in a total estimated error of 2.7%. When looking at contact problems (such as those involving indenters) the contact patch is a function of the material properties, geometry, and load. In most cases, one has a handle on what the geometry of the indenter and half-space is and also what external force is being applied. When isotropic materials are used the contact patch is only dependant on two material properties: such as the elastic modulus and the poisson ratio. Anisotropic materials are different in that the contact patch is related to an equivalent, or composite, modulus since the material properties are direction dependent. Vlassak et al. (2003) looked at calculating the indentation modulus of anisotropic materials. Analytical solutions were developed for indenters of arbitrary shape being pressed into anisotropic half-space. A theorem stating that the solution to this contact problem is one that maximizes the load on the indenter for a certain depth. One of the principle assumptions was to assume the contact patch had a certain shape, then the best approximate solution exists in the Rayleigh-Ritz sense. For axisymmetric indenters, a

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70 limited family of Greens functions, using the first term in the Fourier expansion, can be used to obtain a much simpler solution. This solution is denoted as the equivalent isotropic solution. First, the Greens function used to obtain displacements at the surface of the half-space is defined. The next step is to determine the load on the punch, or indenter. With the load function available, it can be differentiated and the modulus of the half-space is found. The Greens function gives the displacement in the direction of a load applied to the half-space. rhyyyywmkmk1281)( (7-3) Equation 7-4 is the force that will establish the elliptical contact area. AobyaxdxdyyxupAF22221),()( (7-4) where ),(1ebpo and u(x,y) is the displacement from the indenter. Now the solution for a conical indenter is produced so the modulus for this case can be found. For a conical indenter the displacement is 22),(yxCUyxu (7-5) where U is the rigid body displacement and C is the cotangent of the cone angle. Using Equation 7-4 the force on the indenter is now

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71 AbyaxdxdyyxCUAF2222221)( (7-6) After a change of variables and also switching to polar coordinates the force is ),()(2),,(2eeECaUabaF (7-7) The resulting contact stiffness for a conical indenter is 4/12)1)(,(12eeAdUdFS (7-8) where U is the approach of the two bodies and A = ab. The modulus of the half-space is 4/12)1)(,(1eeMeqv (7-9) A less complicated expression for conical indenters can be found by assuming the contact patch is circular. Now the orientation angle, is arbitrary and the eccentricity drops out oohdh)(),0( (7-10) ho is the first term of the Fourier expansion of the function h(), and modulus now becomes oeqvhM1 (7-11) which is the same solution for a flat circular punch for an isotropic half-space; the equivalent isotropic solution.

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72 The modulus in Equation 7-11 is compared to the modulii one gets from when a sphere and cone are the indenters and the Greens function is not truncated using a Fourier expansion. When the modulii are plotted against crystal orientation, the modulii of the cone and sphere do not vary much (the difference between the two is less than 0.03%). Using the equivalent isotropic solution, the difference is approximately 0.1%. Green and Zerna (1954) looked at the two-dimensional contact problem for anisotropic materials. Their work is somewhat confusing because they retain the notation used for isotropic materials. Their solution is not applicable for coupled in-plane and antiplane deformation. Fan and Keer (1994) re-examine the two-dimensional contact problem using the analytic function continuation approach. They arrive at a compact solution that may be applied to other problems such as examining subsurface stresses in contact problems involving sliding and sticking. Willis (1966) examined the Hertzian contact for anisotropic half-spaces. He assumed the contact patch is elliptical and the contact pressure will be of the form Hertz used when he was looking at this problem for isotropic materials. Willis does not use potential theory to solve this problem. He used Fourier transforms and the solution lies in solving contour integrals. Evaluating those integrals can be done numerically.

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CHAPTER 8 CONCLUSIONS The 2-D analytical solution developed for subsurface stresses in anisotropic half-spaces is an efficient and practical method for performing design iterations. Contact dimensions and pressure distributions obtained by Hertzian isotropic assumptions are a good approximation for anisotropic contacts also. The 3-D FE model, or applied stress model, for stress analysis in the anisotropic substrate (single crystal), using contact dimensions and pressure profile from Hertzian solutions is an excellent representation of the contact problem. This is an accurate and efficient technique for determining subsurface stresses for anisotropic contacts. The 3-D contact model for anisotropic substrates, while numerically intensive, is the best way to solve problems involving complex geometries. Just like the applied stress model, effects of crystal orientation were included. Additional mesh refinement might produce more accurate stress solutions in the edge of contact. Overall, most of the results had excellent agreement with the applied stress model and analytical solution. Contour plots for resolved shear stresses on slip planes are presented in the contact region. These results are required for determining fatigue life (crack initiation) in the substrate. 73

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CHAPTER 9 RECOMMENDATIONS FOR FUTURE WORK Effects of contact friction and tangential forces must be incorporated in the FE models. The analytical solution can already account for tangential traction forces of the contact. Relationships between shear stress amplitude on the slip planes and cycles to fatigue crack nucleation need to be determined. To effectively model the fatigue damage process, the ability to evaluate plastic stress and strain amplitudes on the contact is required. This will entail developing constitutive relations for crystal plasticity, which will prove to be a complex endeavor. 74

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LIST OF REFERENCES ANSYS Elements Reference, ANSYS Release 5.6; ANSYS, Inc. November 1999. Arakere N, Swanson G. Effect of crystal orientation on analysis of single-crystal, Nickel-base turbine blade superalloys, National Aeronautics and Space Administration Technical Publication 2000; Feb: 6-10. Beisheim JR, Sinclair GB. Three-dimensional finite element analysis of dovetail attachments. American Society of Mechanical Engineers Turbo Expo. 2002 Jun 3-6; Amsterdam, the Netherlands. Dame L, Stouffer D. Inelastic deformation of metals. New York: John Wiley and Sons; 1996. p. 3-46, 387-435. Donachie M, Donachie S. Superalloys. Materials Park, Ohio: American Society for Metals International, 2002. p. 25-39. Fan H, Keer LM. Two-dimensional contact on an anisotropic half-space. Transactions of the American Society of Mechanical Engineers 1994; 61: 250-255. Green AE, Zerna W. Theoretical Elasticity. Oxford, UK: Clarendon Press; 1954. Hummel R. Understanding materials science. New York: Springer; 1998. p.74-101, 102-122. Johnson KL. Contact mechanics. Cambridge, UK: Cambridge University Press; 1985. p. 84-106 Lekhnitskii SG. Theory of elasticity of an anisotropic body, San Francisco: Holden-Day; 1963. Magnan S. Three-dimensional stress fields and slip systems for single crystal superalloy notched specimens, Masters Thesis, University of Florida, Gainesville 2002, p. 41-52. Sims CT, Stoloff NS, Hegel WC, editors. Superalloys II. New York: John Wiley and Sons; 1987. Timoshenko SP, Goodier JN. Theory of elasticity. Tokyo: McGraw-Hill; 1983. p. 15-34. 75

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76 Vlassak JJ, Ciavarella M, Barber JR, Wang X. The indentation modulus of elastically anisotropic materials for indenters of arbitrary shape. Journal of the Mechanics and Physics Solids 2003;51: p. 1701-1721. Willis JR. Hertzian contact of anisotropic bodies. Journal of the Mechanics and Physics Solids 1966; 14: 163-176.

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BIOGRAPHICAL SKETCH Born on February 18, 1979 in West Chester, Pennsylvania I moved to Naples, Florida when I was four. I graduated with a B.S. in mechanical engineering from Villanova University and am now seeking a doctorate in mechanical engineering. I plan on becoming a professor someday. 77


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Title: Analytical and Numerical Evaluation of Subsurface Stresses in Anisotropic (Single-Crystal) Contacts
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Copyright Date: 2008

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Permanent Link: http://ufdc.ufl.edu/UFE0002849/00001

Material Information

Title: Analytical and Numerical Evaluation of Subsurface Stresses in Anisotropic (Single-Crystal) Contacts
Physical Description: Mixed Material
Copyright Date: 2008

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Holding Location: University of Florida
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ANALYTICAL AND NUMERICAL EVALUATION OF SUBSURFACE STRESSES
IN ANISOTROPIC (SINGLE-CRYSTAL) CONTACTS
















By

ERIK C. KNUDSEN


A THESIS PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF SCIENCE

UNIVERSITY OF FLORIDA


2003

































Copyright 2003

by

Erik C. Knudsen



























This thesis is dedicated to my grandmother, Madeline Rosenthal. I will forever
miss her and always cherish the memories.















ACKNOWLEDGMENTS

I thank my advisor, Nagaraj Arakere, for his support and guidance while I did this

research. Greg Swanson, Greg Duke, Gilda Batista, and JeffRayburn were a big help

when I was at the Marshall Space Flight Center. My labmates (Jeff, Shadab, and Srikant)

were an invaluable resource. My parents have been a source of wisdom and comfort all

my life and none of this would have been possible without them.
















TABLE OF CONTENTS
Page

A C K N O W L E D G M E N T S ................................................................................................. iv

LIST OF TABLES ............. ................... ........... .. .. ............... .. vii

LIST O F FIG U RE S ......................................................... ......... .. ............. viii

ABSTRACT .............. ..................... .......... .............. xii

CHAPTER

1 IN TR OD U CTION ............................................... .. ......................... ..

R e search O bje ctiv e s............... ................................ .................. .............. ............. .....
Introduction to N ickel-Base Superalloys.................................. ........................ 3
C oordinate Transform nations ............................................................................. 4
The [110] O orientation ................................................. .... .. ...............
The [213] O orientation ................................................... ........ ...............

2 A N A L Y TIC A L SO LU TIO N ........................................................... .....................14

3 FINITE ELEMENT FORMULATION-APPLIED STRESS MODEL......................22

4 RESULTS OF THE APPLIED-STRESS MODEL ..............................................33

Comparison of Contour Plots: Applied-Stress Model and Analytical Solution.........34
Numerical Comparison of the Analytical and Applied-Stress Models ...................39

5 C O N T A C T M O D E L ......................................................................... ...................42

Examination of the Normal Stresses at the Surface................................ ........ 44
Comparison of Contact Model Stresses to the Analytical Solution .........................48

6 DEFORM ATION M ECHANISM S.................................. .................................... 56

7 L ITE R A TU R E R E V IE W ........................................ ............................................67

8 CON CLU SION S .................................. .. .......... .. .............73

9 RECOMMENDATIONS FOR FUTURE WORK ............................................... 74



v









L IST O F R E F E R E N C E S ...................................... .................................... ....................75

B IO G R A PH IC A L SK E TCH ...................................................................... ..................77
















LIST OF TABLES

Table p

1-1 D direction cosines .............. ...................... ........................... .... .... 9

1-2 Direction cosines for the [110] orientation ............................................................9

5-1 Comparison of Hertzian and numerical contact pressures and patch sizes.............47

6 -1 S lip S y stem s ....................................................... ................ 6 0

7-1 N um ber of elem ents used ........................................................................... .... ... 68
















LIST OF FIGURES

Figure page

1-1 Visual description of the contact problem ............................................. ...............2

1-2 D am per and blade in contact................................ ....... ................ ............... 2

1-3 Profile of the contact problem .......................... .................. ...............

1-4 D am aged turbine blade............................................................... ....................... 3

1-5 Specim en and m material coordinate axes .......................................... ............... 5

1-6 A ligned coordinate system s ............................................... ............................ 6

1-7 Rotation of specimen coordinate axes about the z axis.........................................7

1-8 Two-dimensional view of rotation (z axis is into the page).......................... ........8

1-9 Schem atic of the [213] orientation .............................................. ........ ............... 10

1-10 First rotation about y, y" axis ...................... ............................... .....................

1-11 S econ d rotation ................................................. ................. 12

1-12 Second rotation about z' axis ............................................................................12

2-1 Profile view of contact problem with normal and tangential tractions .................. 15

2-2 E lliptic distribution ........ ....... ............................................................ .. .... .. ... ... 16

2-3 Applied forces used to derive the equilibrium equations................. ............. ...16

3-1 C ylinder in contact w ith slab ........................................................... ......... ......23

3-2 Face-on view of contact problem with applied load shown.............................. 23

3-3 Elastic modulus of PWA 1480 as a function of crystal orientation at room
tem perature ..................................... ................................ ........... 25

3-4 C contact patch dim ensions................................................ ............................. 25









3-5 FE model; applied stress (no contact elements).................................................26

3-6 Constraints ......... ...... ........ ............................28

3-7 Elliptic contact load ......... ..... .......................... ................ .... 28

3-8 A applied pressure............ .............................................................. .......... ....... 29

3-9 Visualization of crystallographic orientation ................................ ..................... 32

4-1 Results were taken from the elements at the midplane ........................................33

4-2 Schematic of casting coordinate system.................................... ................34

4-3 Contour plots of the ox stress (case 0). ................................................ .... ........... 35

4-4 Contour plots of the Gy stress (case 0). ...................................... ...............35

4-5 Contour plots of the z stress (case 0). ................................................ .... ........... 36

4-6 Contour plots of the Txy stress (case 0). ............................................... ............... 36

4-7 Contour plots of the ox stress (case 33). ........................................ .... ........... 37

4-8 Contour plots of the Gy stress (case 33). ........................... ....................... 37

4-9 Contour plots of the cz stress (case 33). ............................................... .. ........... 38

4-10 Contour plots of the Txy stress (case 33). ........................ ....................................... 38

4-11 N odes used to extract results ........................................................... ......... ......39

4-12 ox stress vs depth .............. ............ ........ ............... .. .. ........... 40

4-13 Gy stress vs depth ......... .... ........................... ............. 40

4-14 ,z stress vs depth .......................................... ............. .... ....... 40

4-15 Txy stress vs depth ........................................ ......... .............. .. 41

4-16 Tyz stress vs depth ........................................ ........ .............. .. 41

4-17 Txz stress vs depth ............................... ..... .. ... ......... .............. .. 41

5-1 C contact m odel ................................................................42

5-2 Densely meshed regions of the contact model .................................... ..................43

5-3 Contact and target elem ents .............................................................................. 43









5-4 Nodes used to obtain the contact patch and peak pressure at the surface ................45

5-5 Y -stress for case 17 .......................................... ................... ........ 45

5-6 Y -stress for case 21 ............................................ ................. ........ 46

5-7 Y -stress for case 25 .......................................... ................... ........ 46

5-8 Y -stress for case 29 .......................................... ................... ........ 47

5-9 Dimensions of the brick element used in the contact model.............. .......... 49

5-10 Sigm a X com prison case 17 ............................................................................49

5-11 Sigm a X com prison case 25 ............................................................................50

5-12 Sigm a Y com prison case 17 ............................................................................50

5-13 Sigm a Y com prison case 25 ............................................................................51

5-14 Sigm a Z com prison case 17......................................................... ............... 51

5-15 Sigm a Z com prison case 25 ............................... ........................ ............... 52

5-16 Tau XY com prison case 17 ............................................................................. 52

5-17 Tau XY com prison case 25 .............................................................................. 53

5-18 Sigm a Y Z com prison case 17........................................... .......................... 53

5-19 Tau Y Z com prison case 25 ......... ............... .................................. ............... 54

5-20 Tau X Z com prison case 17.......................................................... ............... 54

5-2 1 T au X Z com prison case 25 ......................................................................... ... ... 55

6-1 Exam ple slip system ....................................................... ................. 57

6-2 Four octahedral planes and slip directions. ................................... ............... 57

6-3 Cube slip planes and slip directions ....................................................................... 58

6-4 Tau 1 contour plot. ....................................... ... .... ........ ......... 61

6-5 Tau 2 contour plot. ......................................... ... .... ........ ......... 61

6-6 Tau 3 contour plot. ......................................... ... .... ........ ......... 62

6-7 Tau 4 contour plot. ......................................... ... .... ........ ......... 62









6-8 Tau 5 contour plot. ......................................... ... .... ........ ......... 63

6-9 Tau 6 contour plot. ......................................... ... .... ........ ......... 63

6-10 Tau 7 contour plot. ......................................... ... .... ........ ......... 64

6-11 Tau 8 contour plot. ......................................... ... .... ........ ......... 64

6-12 Tau 9 contour plot. ......................................... ... .... ........ ......... 65

6-13 Tau 10 contour plot. ....................................... ... .... ........ ..... .... 65

6-14 Tau 11 contour plot. ....................................... ... .... ........ ..... .... 66

6-15 Tau 12 contour plot. ....................................... ... .... ........ ..... .... 66















Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy

ANALYTICAL AND NUMERICAL EVALUATION OF SUBSURFACE STRESSES
IN ANISOTROPIC (SINGLE-CRYSTAL) CONTACTS

By

Erik C. Knudsen

December 2003

Chair: Nagaraj K. Arakere
Major Department: Mechanical and Aerospace Engineering

The Space Shuttle main engine alternate high pressure-fuel turbo pump (SSME

AHPFTP) turbine blades commonly fail from subsurface contact stresses imparted by a

friction-damping device designed to reduce vibratory stresses. Fretting-fatigue failure is a

major cause of failure in turbine components. Fretting occurs when components such as

blade dampers and dovetail joints are subjected to vibration, which results in contact

damage. When combined with a mean stress in the component, fretting can lead to a

reduction in high cycle fatigue (HCF) life. Fretting (along with wear and corrosion)

initiates cracks where the blade and damper are in contact. Although the turbo pump is

only operating for ten minutes, it is cycling at a very high rate, which leads to rapid

accumulation of fatigue cycles.

The Marshall Space Flight Center (MSFC) has developed finite element (FE)

models to investigate this problem. These models have very fine meshes and also

incorporate nonlinearities such as friction. The sophistication of these models is









expensive from a computational standpoint. Since the blades are made of anisotropic

material, the properties of the blade change with the crystallographic orientation of the

material. Manufacturing tolerances limit the precise orientation of the material, thus

there are numerous orientations to examine. When one factors in the time it takes to run

the models and also examine the results, many hours would have to be put in to give a

thorough analysis.

To speed the analysis up and also make things a bit simpler, it would be helpful to

avoid such large FE models (that take hours to solve and many more hours to analyze).

Ideally, one should build a contact model, to get a true sense of the stresses and contact

patch size. If possible, much less complicated models should be built and compared to

larger contact models. If there is reasonable agreement between the two, then the less

sophisticated FE models can be used outright and the results arrived at much faster. This

was done here. Results for a simple, applied-stress, FE model are compared to those for

the contact model.

Concurrently, analytical (or closed-form solutions) are also being developed for

this problem. Getting results using analytical methods is almost always faster and easier

then using numerical approaches. Numerical (or FE) results are also compared to those

for analytical solutions. Results for numerical vs analytical approaches have fairly good

agreement with the normal stress components, but the shear stresses do differ. The most

probable source of that disagreement lies in the mesh density of the FE models.














CHAPTER 1
INTRODUCTION

Extensive work has been done in the non-conformal classical area of contact

mechanics of elastic solids dating back to Hertz's seminal work in the area. Johnson

(1985) gives a complete discussion of his results. Relatively little work has been focused

on contact mechanics of non-conformal anisotropic solids. Interest in this field is

increasing because of applications of single crystal materials used in high temperature

environments.

Research Objectives

The primary goal of the research presented is to examine the subsurface stresses

in face-centered cubic (FCC) single crystal (SC), orthotropic, turbine blade damper and

dovetail attachment contacts using the finite element (FE) method.

* Develop a simple FE model that has no contact elements (applied-stress problem)
and compare the results to the more complex contact FE model.

* Develop a contact FE model that captures the contact patch dimensions and stresses
in the blade, for fully anisotropic contacts; and compare the results to Hertzian
solutions.

* Develop an analytical solution for anisotropic subsurface stress fields.

* Compare results of both the contact and applied-stress models to analytical
solutions; and determine if the analytical approach is better than the FE approach
for looking at the subsurface.

Figures 1-1 and 1-2 show what the contact problem looks like. The damper shown

in Figure 1-1 is touching the axial face of the turbine blade. Notice that the damper is

upside-down in Figure 1-2. Figure 1-3 shows a profile view of the contact problem.









Essentially, the damper and blade are modeled as a half-cylinder (damper) touching a

rectangular slab (blade). The damper touches the blade platform at two separate

locations. Only the circled location in Figure 1-3 is analyzed here, although this analysis

could be applied to other parts of the damper. The resulting contact load is one that has a

normal and tangential component. The work presented in this thesis is for normal

loading only. Ultimately, cracks form and damage the blade as shown in Figure 1-4.

Blade Coordinate System


Figure 1-1. Visual description of the contact problem


Figure 1-2. Damper and blade in contact


Figure 1-3. Profile of the contact problem














,,4a- I

i~iGISZ~iil~c1ft ,~'i~m~ir-.s


Figure 1-4. Damaged turbine blade Source: N. Arakere, personal correspondence, 2003

Introduction to Nickel-Base Superalloys

The material chosen for the blade is a precipitation strengthened single-crystal

Nickel-base superalloy, Pratt and Whitney (PWA) 1480/1493. Generally speaking, when

working at high temperatures and high tensile stresses where creep dictates the mode of

failure, it is desirable to have a material with a close packed structure, such as FCC

(which happens to be the structure of this alloy); a high melting temperature; and coarse

grains (for isotropic materials). The close-packed structure makes diffusion more

difficult. Since grain boundaries can be the weak link at high temperatures, it is a good

idea to either eliminate them or make sure one does not have too fine a grain size

(Hummel 1998).

The blades in the SSME AHPFTP are made from a monograin material. The

absence of grain boundaries gives improved creep resistance (which is necessary since

the blades operate at high temperatures). Also, with no grain boundaries, there is no need

for elements, such as Boron, that are added to increase grain boundary strength, but can

also lower the melting point of the alloy (Donachie and Donachie 2002).

Nickel-base superalloys can be used at a higher fraction of their melting point

than just about any other alloy. Also, many elements can be dissolved into Nickel. This

means that lots of additives can be used for hot corrosion and oxidation resistance, solid









solution strengthening and y' formation (precipitation hardening). For example, the

addition of Al provides increased oxidation resistance and also forms the compound

Ni3Al (y'). Chromium and Tungsten are usually added as solid solution strengtheners

and Cr also helps combat corrosion (Donachie and Donachie 2002).

The precipitation hardening (y') comes from heat-treating an alloy that has limited

solid solubility and decreasing solid solubility with decreasing temperature. The heat

treatment forms a fine dispersion of precipitates (small particles) that inhibit the motion

of dislocations and make the material stronger. The highest strength is achieved when

many small, round, closely spaced particles are coherently spread throughout the alloy

(Hummel 1998). Since these alloys have only one grain, they are orientation dependent.

The orientation of the crystal lattice plays a pivotal role in the properties of the material.

Typically the blades are cast in the <001>, or low modulus, direction. This orientation

gives the blades great thermal-mechanical fatigue resistance. It should be pointed out

that one cannot cast the blades perfectly. Casting tolerances allow a 15 deviation from

the stacking line (the line between the <001> direction and the airfoil stacking line)

(Arakere and Swanson 2000).

Coordinate Transformations

To account for the orientation dependency of these materials, one must understand

how to transform from one coordinate system to another. From there quantities such as

the resolved shear stress (RSS) on different slip planes can be determined. Let us first

begin with a few definitions. Figure 1-7 shows the material coordinate and specimen

coordinate systems.

The primed axes (x' y' z') are defined as the specimen coordinate system. The

unprimed axes (x y z) are defined as the material (crystal) coordinate system. Coordinate









transformations are necessary because material properties are functions of the orientation.

This is not the case for an isotropic material. Since this material has cubic symmetry,

there are three independent elastic constants as well as a tensor matrix that relates stress

to strain (Dame and Stouffer 1996).

y <010>

y







x<100>



z <001> z




Figure 1-5. Specimen and material coordinate axes

One way to relate one coordinate system to another is through the use of direction

cosines. Once one has the cosines of the angles between the specimen and material axes,

the proper coordinate systems can be constructed in whatever FE code one chooses. The

model is initially built in the global (x y z) coordinate system. The material is cast with a

certain crystallographic orientation which must be accounted for to obtain the proper

results (stresses, strains, etc.). The orientation of the material is always given, and from

there direction cosines of the angles between the material and specimen coordinate

system can be determined using methods described by Lekhnitskii (1963) and also

Magnan (2002).









The r1101 Orientation

This example will deal with the [110] orientation. Let us align the x' axis with

the [110] direction. To do that imagine the specimen axes (they are primed) are initially

aligned with the material, unprimed, coordinate system.

y <010>

y'




/ /--------------

z' x <100>



z <001>


Figure 1-6. Aligned coordinate systems

The orientation of the x' axis is known (defined as [110]). One way to find the y'

and z' axes is to arbitrarily define the y' axis (it just has to be normal to the x' axis, i.e.

the dot product between the two will be zero). Now that the x' axis is defined and the y'

axis is arbitrarily chosen, the z' axis is obtained by taking the cross product of the x' and

y' axis. The downside of employing such a method is that there is an infinite number of

y' vectors normal to the x' axis. Thus, there are no unique vectors, or directions, that

define where the y' and z' axes could wind up. But, the method used here does not

involve an arbitrary assignment of the y' and z' axes.

Continuing on, the y' axis will also be rotated about the z axis and has the

direction [-110].

















<-110>
0

y


y <010>


x'<110>


0


x <100>


z <001>


Figure 1-7. Rotation of specimen coordinate axes about the z axis

Notice in Figure 1-7 the orientation (or coordinates relative to the material

coordinate system) of the y' vector was not arbitrarily picked. Both the specimen and

material coordinate systems are orthogonal (i.e. the angle between the x, y, z axes are

900). The entire specimen (primed) coordinate system was rotated about the z axis.

The next step is to write the coordinates of the new axes with respect to the

original axes. They are as follows

x'= xcos O + y sin
y'= -xsin0 + ycosO (1-1)
Z'= z

Putting Equation (1-1) into matrix form










x' cosO sinO 0 x
y' = -sin0 cosO 0 y
z' 0 0 1 z


y x'


-- x

1


Figure 1-8. Two-dimensional view of rotation (z axis is into the page)


Trigonometry gives that 0

(x y z) axes.


45 degrees (measured counter-clockwise) from the


tan(O)=


S= 450


(1-3)


No further transformations are needed here to obtain the direction cosines for this

case. It is convenient to enter the direction cosines into a table.


(1-2)







~--I









Table 1-1. Direction cosines

x y z

x' a 13Pi y

y' a2 12 72

z' 13 03 73





Where at is the cosine of the angle between the x' and x axes, P1 is the angle

between the x' and y axes, etc. Thus, Table 1-1 becomes



Table 1-2. Direction cosines for the [110] orientation
x y z

x' 0.707 0.707 0

y' -0.707 0.707 0

z' 0 0 1



The [213] Orientation

The last example involved only one rotation. Frequently, one will have to

perform two or even three rotations to get the desired crystal orientation. This example

will look at the [213] orientation. Let us look at Figure 1-6 again. Once more, the x' axis

will be aligned with the [213] direction as shown in Figure 1-9.

To do this transformation correctly, two rotations will be executed. First a

rotation about the y-y" axis by 0 degrees is performed. Then a second rotation about the

z' axis by (p degrees takes place and the transformation is complete.








y <010>


P --W


x <100>


z <001>
Figure 1-6. Reprinted


y <010>


x <100>


z <001>


Figure 1-9. Schematic of the [213] orientation
In Figure 1-11 one can see the first rotation from the projection (denoted by the

dashed line) of the [213] vector into the x-z plane. Since the dimensions of the projection

are known, the angle 0 can be readily determined.


zoo'


'"'-~r.:
~...
~









3
tan = -
2


S= tan' (1-4)
2



0 = 56.3090

y,y" <010>







x<100>



z


x": projection ofx' into x-z plane
z <001>




Figure 1-10. First rotation about y, y" axis

For clarity, the rotation is redrawn in Figure 1-10. Writing the coordinates of the

transformed coordinates in terms of the original set of axes we have

x" cos 0 sinO x
y' = 0 1 0 y (1-5)
z"1 -sinO 0 cosOI z

There is just one more step which entails a rotation about the z" axis by an

amount of p degrees. Figure 1-11 illustrates this.










y <010>


x <100>


2


Figure 1-11. Second rotation

To make this second rotation clear, it is redrawn in Figure 1-12.


y,y


x": x' projected into x-z plane


Figure 1-12. Second rotation about z' axis

Once more this transformation is put in matrix form



y' = sin cos 0 y"
z' 0 0 1 z"


From looking at Figure 1-11, the value of p is


z <001


y








z',z"


(1-6)










tan l



= tan (1-7)


=15.5050



The only thing left to do is solve for the table of direction cosines.

La A, / 0 cos sin 0 cos 0 0 sin0
a2 A 72 = -sine cos 0 0 1 0 (1-8)
a3 P3 73 0 0 1 --sin 0 cos6O

Finally, we note that counterclockwise rotations are positive and clockwise

rotations are negative. Following this convention, theta is negative and phi is positive.

Substituting and solving for the table of direction cosines



a2 2 2 = 0.148 0.964 0.222 (1-9)
a,3 p3 73 0.832 0 0.555

Of course one could perform a third rotation about the x' axis, perhaps by an

angle y. If that were to happen, we would have

al A 1 0 0 cos sino 0 cosO 0 sino
a2 = c sinin/ -sin 0 cos 0 0 1 0 (1-10)
a3 /73 73 0 -sin/ cosi/l 0 0 1 -sinO 0 cosO














CHAPTER 2
ANALYTICAL SOLUTION

A closed-form solution to this contact problem was formulated by Dr. Arakere

(University of Florida) which utilizes complex potential stress functions developed by

Lekhnitskii (1963). This is a two-dimensional method with a generalized plain strain

condition. The load and resulting stress functions have no z-dependency. The shape of

the contact patch was assumed and the applied load equations were taken from Hertz's

work for line-contact problems (such as a cylinder touching a cylinder) with isotropic

materials. The subsurface stresses can be solved for any general traction distribution over

the surface. In chapter two it was stated that the normal pressure distribution for Herzian

(non-conformal) anisotropic bodies is the same as for isotropic ones.

From Hertz the normal load distribution is

N(x)= po 1 x a2 (2-1)

For a tangential load we have

T(x) = p, po 1- x/a (2-2)

Where po is the peak pressure, a is the contact half-width, and [f is the coefficient of

friction.

In Figure 1-3, a profile view of the contact problem was shown. That figure is

reprinted in Figure 2-1 with a normal and tangential traction applied. If both materials are

isotropic, then a solution exists for this contact problem.. Essentially all one needs are









the material properties and applied loads. But since the blade is made out of an

orthotropic material, there is no closed-form solution available.




p- Q





















Figure 2-1. Profile view of contact problem with normal and tangential tractions

Once again, there is only a normal load applied to the top of the cylinder (i.e.

Q = 0). Hertz has proved, for non-conforming bodies in contact, that the normal pressure

distribution on the block above is elliptic, Figure 2-2, for isotropic materials (Johnson

1985). In his paper of anisotropic contact problems, Willis (1966), states that this same

distribution can be applied when anisotropic materials are used. What follows is an

overview of Lehknitiskii's work applied to a non-conformal contact problem such as a

cylinder touching a flat substrate. Equations that determine the subsurface stresses are

developed with the main assumption is knowing the contact patch dimensions

beforehand.






















Figure 2-2. Elliptic distribution

Although Hertz's equations are used to describe the normal contact pressure, the

analytical solution is derived in such a fashion that any loading can be applied to the

surface of the anisotropic substrate (Figure 2-3). Where Hertz's equations are used can

be seen in Equations 2-27 and 2-28.

Lekhnitskii presents a stress function approach, F(x,y) and y(x,y), to obtain the

subsurface stresses. The stress functions are used to satisfy the equations of equilibrium

which come from Figure 2-3.

Surface Traction





Anisotropic Half-Space


Figure 2-3. Applied forces used to derive the equilibrium equations

Equation 2-3 shows the relations for equilibrium. U is the applied body force. Notice

that if no body forces are present Equation 2-3 becomes homogeneous.

a0
x 8) a x






17


+ o 0 (2-3)
y





The relations listed in Equation 2-4 are the stress functions used to satisfy the

equilibrium equations.

2 F 02F 2F
x = -+U 2 = +U (2-4)
By 2 a2 X axB

all a'i
yz z y

This approach to solving these kinds of problems was initially developed by Airy

in 1882 when he was solving two-dimensional stress problems. A good primer to the

concept of stress function can be found in Timoshenko (1982). The stress functions must

satisfy the above equations (which become homogeneous when there are no body forces).

Lekhnitksii integrates three stress-strain equations that come from the generalized

Hooke's law. The aforementioned integration gives rise to three displacement equations.

These equations must also satisfy the leftover three equations from the generalized

Hooke's law. They are substituted into the remaining three stress-strain relations that

give rise to another set of three equations where the coefficients of the left and right-hand

sides are equated. Those equations are differentiated to eliminate certain variables from

the generalized displacement equations that will be determined in the future. The result

of that differentiation is listed in Equation 2-5 and Equation 2-6.

"2U 02U 82U
L4FL = + L3 12 22) a2 + ( 6 )c (A 11 + 12) (2-5)
~sC2 ax----@









OU OU
L3F+L, = -20+ Aa34 Ba35 +(f,1 + 9, ) (,15 + f2,) (2-6)


L2, L3, L4 are differential operators of the second, third, and fourth orders. A and B

describe the bending in the x-z and y-z planes. 0 is the rotation about the z-axis. In the

special case of plane deformation; A = B = 0 = 0. Equation (2-6) can be written as

follows

F = F + Fo (2-7)

y = y + Vo (2-8)

where F' and y' are the general solutions to a system of homogeneous differential

equations. They are homogenous because of the absence of body forces.

L4F' + L3Y' = 0 (2-9)

L3F' + L2y' = 0 (2-10)

To obtain the solution to the set of differential equations above, one of the

functions, such as y', is eliminated. The next step is to apply the operator L2 on Equation

2-9 and L3 on Equation 2-10 and subtract what remains. The result is an equation of the

sixth order

(L4L2-L32)F = 0 (2-11)

To get the solution for y', the exact same procedure is followed. Since Equation

(2-11) is of the sixth order, it can be broken down into six linear operators of the first

order

D6 D5 D4 D3 D2 D1F' = 0 2-12)

Where Dk = (a/ay)-Ltka/ax and [tk are the roots to the characteristic equation


2( 3 (/()= 0 (2-13)









where


12(11) =5512 2.-45-1 + P55 (2-14)


13(1) = 15-31 -(114 + 356).2 +(1325+ 146) B- 24 (2-15)


14() = 11.4 21316.3 +(2.P12+ P66)2 -22-P261 + P22 (2-16)

and

ai3aj3
Pij = aij -
a33 (2-17)

The term aj are the elastic constants for this orthotropic material. See Equation

3-13.

Since Equation 2-12 is of the sixth order, it can be resolved into the integration of

six equations of the first order

D1F' = (2, D2(P2 = (P3, D3(P3 = (p4, D4(4 = (P5, D5(P5 = (6, D6(6 = 0 (2-18)

In order to integrate these first order equations, it is assumed the general integral

is a function of the argument x + [ty. For example, the integral for (p6 is denoted by

fV6(x + t6y). Following this convention, ps5 satisfies the nonhomogeneous equation

D5(p5 = fV6(x + t6y) (2-19)

Integrating (2-19) the following expression is obtained

(p5 = fsIV(x + k5sy) + fV(x + X 6y)/ 1-6 [15 (2-20)

Using the same method (p4, (p3, (P2, and F' will be obtained. Using a change of notation,

expressions for F' and y' are determined

6
F'= FFk(x +ky) (2-21)
k=1










(2-22)


6
'k= Yk +k Y)
k=l


Lekhnitskii goes on to prove four theorems that are necessary for the theory

governing a body bounded by a cylindrical surface. He shows that a certain set of four

equations cannot have real roots and that ultimately tlk is always complex or imaginary.

Now the stress functions take the following form

F = 2 Re[F,(zi) + F,(z,) + F (z3)] + F (2-23)


(2-24)


I= 2Re F(z,)+ A2F'2 (z,) + F'3() + Yo
A3


Where


-I3(tl) -I3(112) -I3(3)
1 = (2 = (3 = 1(3)
12G12) 12(1 2) 14( 3)


(2-25)


Introducing the complex variable zk (k = 1, 2), new functions are introduced

1
k (zk)= F'k(zk) 3(z3)= F'3(z3) k= 1,2 (
A3





A (z)+ A2(z)+ 3 (z)= 0--- (
2m J x- z
/0/1 (Z) + //2 (Z)+ 303 (Z) = -- (
2;d x-z

A10(z)+., 202(z)+ +3(z) =( 0 (=

Equations 2-30-35 are the resulting stress functions.

cx= 2.Re (112. 1(z) + 12 .2(z) + 132 3 ,3'. )) (


Ty = 2.Re(1(z) + 2(z 2) + )3 '3', ))


2-26)



2-27)


Z-28)

2-29)




Z-30)


(2-31)






21


txy = 2-Re(Gi'1(z) + -12'2(z) + 3 1'3' )) (2-32)


Txz = 2-Re(Gl l 'lI(z) + 2-2 '2(z) + ,-'3 '3' ) (2-33)


tyz = 2-Re( l4'(z) + 2'2() + (z)) (2-34)

-1
7z -- (al3-x+ a23'Sy + a34-yz + a35'cxz + a36.'xy)
a33 (2-35)

zi-= x+ rli-y (2-36)














CHAPTER 3
FINITE ELEMENT FORMULATION-APPLIED STRESS MODEL

Our first goal with this model is to evaluate the subsurface stresses in the substrate

by decoupling the contact mechanics from the stress analysis. To do this, an anisotropic

substrate is modeled in 3-D, subject to normal loading over a contact width chosen

arbitrarily.

One of the difficulties in trying to get reliable results is that FE models need very

fine meshes and must account for nonlinearities such as friction and plasticity. The

numerical modeling procedure would be simplified considerably if the contact problem is

reduced to a stress analysis problem. This can be done if the size of the contact patch and

the normal contact pressure distribution is determined from Hertzian assumptions. These

models are less complex than contact models and will run faster. They are easier to

troubleshoot and modify should there be a problem or design change.

Hertz developed closed-form solutions for isotropic contact problems. His theory

is grounded on four assumptions: the bodies in contact are continuous and non-

conforming, the strains are small, each solid is considered to be an elastic half space, and

there is no friction on the surfaces (Johnson 1985).

These assumptions were built into the models to analyze the anisotropic material.

The applied stress model is a rectangular slab (composed of the superalloy) with an

applied load. The load was determined from the Hertzian assumptions stated above.

Since the contact is modeled as a cylinder touching a rectangular slab, this is a line-

contact problem. Hertz determined the contact patch dimensions, maximum contact








pressure, and profile of the contact load (which happens to be elliptic). It has been shown

that the pressure distribution for contact problems of a similar type (non-conforming)

involving anisotropic materials of is also parabolic (Willis 1966).

The true contact patch size and load distribution were not known when these

models were first being built. The idea is to build a simple model and compare it a more

sophisticated one that uses contact elements to see how well Hertz's equations will

predict the load distribution and contact patch size for the anisotropic material.


Figure 3-1. Cylinder in contact with slab


I P


Figure 3-2. Face-on view of contact problem with applied load shown

Figure 3-2 shows the non-conformal contact problem of a cylinder on a flat

substrate and in Figure 3-4 shows a close-up view of the contact region. Hertz (1881)









determined this contact width and also the maximum contact pressure. The equations are

as follows:

fi-1
R K + 1 (3-1)
Ri R2

where R is the equivalent radius of curvature.


E* = + v2 (3-2)










2P
a E (3-3)




S= 2P (3-4)




Where P is the load per unit length, a is the contact half-width, and po is the

maximum contact pressure. The subscripts 1 and 2 refer to the two bodies that are in

contact with each other.

The elastic modulii, El and E2, can be easily defined if the elastic bodies are

isotropic. The problem inherent here is the contact of isotropic indenter (damper) with

anisotropic substrate (blade). For the anisotropic substrate the modulus will vary as a

function of crystallographic orientation. There is no procedure to determine the effective

elastic modulus, E*, for an anisotropic contact, since El and E2 vary with the orientation

of the material. See Figure 2-3; the modulus is 126 GPa for the [001] orientation and

increases to 310 GPa for the [-111] orientation (Sims et al. 1987).




















%.-\ "\ \ \ ) 32.1
0 s Io Iis a6 30 :s 40 50


Figure 3-3. Elastic modulus of PWA 1480 as a function of crystal orientation at room
temperature Source: Modified from Donachie and Donachie, 2002











2a




Figure 3-4. Contact patch dimensions

The contact dimension, 2a, is known here and is given by Equation 3-3. The

contact takes place over an area that is rectangular (width x length). The width is 2a and

the length is 0.5 inches as shown in Figure 3-3.

Figure 3-5 shows the FE model used for the applied/load stress conditions. The

model was built using ANSYS FE software and is composed of solid-45 elements. These

are 8-node elements used for 3-D models. This element has three degrees of freedom at









each node and also has the capability to analyze creep, swelling, large deflection and

large strains (ANSYS 1999).


Figure 3-5. FE model; applied stress (no contact elements)

Since a plane stress or plane strain condition cannot be enforced with this material,

a 3-D model must be built. The analytical solution (discussed in chapter two) uses a

'generalized plane strain' condition that generates equations for the subsurface stresses

that are functions ofx and y only. When Lekhnitskii (1963) developed the closed-form

solution used in this analysis, his boundary conditions were prescribed on an infinitely

long cylinder. This is not a plain strain condition is the conventional sense since an

orthotropic material is involved. The orthotropy induces shear coupling in the z-

direction.









One cannot introduce infinitely large elastic bodies into a numerical analysis. The

dimensions used in the FE model drawn in Figure 3-5 were sufficiently large so the

rectangular slab could be treated as an elastic half space. That is, the stresses are uniform

throughout the thickness (z dimension) of the material.

In most non-conformal contact problems, the contact patch is small compared to

the dimensions of the bodies in contact. Consequently, stress gradients in the region

affected by the contact are very large, necessitating an extremely refined FE mesh to

accurately capture the subsurface stresses. The densely meshed region in Figure 3-5 is

where the contact would take place. It is in this region that the results were extracted and

where the stresses are the highest. A coarser mesh is used for the rest of the substrate

since the stresses are not as high and to reduce the overall size of the model. The less

elements there are the better, since the model will run faster on the computer.

The base of the model, or slab, is constrained so the block cannot move in the

normal, or y-direction. It is important to allow the elements to have some flexibility, thus

one does not want to over-constrain the model. That is why only two of the bottom

corners of the slab were constrained in the x-direction. This ensures that the model

cannot rotate about the y axis, or have any rigid body motion.

Now that the model is properly constrained, only entering the material properties

and applying the loads remain. The load was applied using a macro (algorithm written in

a computer language such as FORTRAN).

The Hertzian contact pressure distribution for normal, frictionless, loading is shown

above in Figure 3-7. The loading curve takes the form of an ellipse when the normal load









P is applied to the cylinder. This distribution was applied over the densely meshed region

in Figure 3-8.


Figure 3-6. Constraints


Figure 3-7 Elliptic contact load














111111
*^J~j kl ''il
17-P I







Figure 3-8. Applied pressure


(x)= p 1a (3-6)


With the geometry and loading in hand, the model was drawn and then meshed

with the solid-45 elements. Finally, the material properties had to be entered along with

the crystal orientation of the material.

Determining the elastic constants in the specimen coordinate system can be

achieved by using the following equations taken from Lekhnitskii (1963). The

relationship between the stresses in the material and specimen coordinate systems is:

{u} = [Q]{cr'} (3-7)

where

anx 073x
7'Y '7 '

S. and (3-8)

TZ 7zx
7T T X
TX)? TX)?








Recall that aL is the cosine of the angle between the x' and x axis, 01 is the cosine

of the angle between the x' and y axes, etc.

a2 32 2aaa2 2aaa3 2a2a1
#12 2 32 2#3#2 2#1A3 2f2A1
7= 2 Y22 32 27372 2yl73 2721
ArY A22 A73 (3l273 + 3 2) (A213 +A3y1) (A1/2 + )2y1))
ryl, 7,2a2 ,3a3 (Y203 + y3a2) (Y1a3 + y3 1) (y1a2 +2C)
a 1 a2l2 a3 3A (a2A3 + 1a3 2) (a1i3 +a3)1) (a1f2 +a2-A1)


The strain vector transformations are similar. The [Q] matrix is different because

there is a factor of two in the relationship between engineering and tensor shear strain

components(i.e. yxy = 2 yxy, yzx = 2 yzx, and 7yz = 2 yyz).


{'} =[Q ]{e} (3-10)


a12 a2 a a3a2 a1a3 a2a1

2 2 2
Q= 7 2 73 7Y32 7173 7271 (3-11)
2Ay, 2027, 2A373 (y073 + 73y2) (A,y3 + 37y,) (Ay2 + P7y,)
2y a, 22a,2 273a3 (y2a3 + y3a2) (y1a3 + y3a) (1,a2 + ya )
2aAP, 2a2,2 2a3A3 (a2,3 + a 2) (afl + a3/1) (af2 + a2/,)


Hooke's law for a homogeneous anisotropic body using Cartesian coordinates is:

{} = [a ]o} (3-12)


Since this superalloy is FCC and has cubic symmetry, there are only three elastic

independent elastic constants needed to define the shear modulus, elastic modulus and

Poisson ratio.









a11 a12 a12 0 0 0
a12 a11 a12 0 0 0
[, aL12z a12 a111 0 0 0
a = a(3-13)
0 0 0 a44 0 0
0 0 0 0 a44 0
0 0 0 0 0 a44


1 1 vyx vV
all = a44 a12 (3-14)
Ex Gyz Ex Eyy

The equation below relates the stresses to the strains in the material specimen

coordinate system:

{}c = L[a'J{ 1' (3-16)

The elastic constants can be transformed from one coordinate system to another

by:

6 6
[a ]= T [airk 1=iria^ a, 3-1
m=l n=l (3-17)
(i, j=l,2,......,6)

With these relationships in hand, the stresses, strains, and elastic constants can be

determined in either the material or specimen coordinate systems.

Consider a block of the Ni-based superalloy shown in the Figure 3-9. The desired

orientation of the crystal lattice is <110> with the x' axis aligned with the <110>

direction. One way to look at the problem is to imagine a cylindrical piece of material,

shown in the Figure 3-8, cut from the block. That cylinder now has the desired material

properties in the <110> direction.

Using equation 3-17 the elastic constants in any orientation can be determined if

one has the direction cosines between the specimen and material coordinate system.









ANSYS allows one to input values of the stiffness (units of psi, or Pa) or compliance

matrix (units of 1/psi or 1/Pa). Once those values are entered into ANSYS-the finite

element model now has the material properties in the desired orientation.




<110>
y

v A x


<010>


<100>


<001>, z, z'


Figure 3-9. Visualization of crystallographic orientation














CHAPTER 4
RESULTS OF THE APPLIED-STRESS MODEL

Contour plots of the normal and shear stresses were obtained from the densely

meshed region at the midplane of the model (see Figure 4-1). To make the analysis

simpler, the analytical solution does not account for the boundary conditions on the free

ends, or surfaces. Since the analytical solution uses a generalized plane strain condition,

it is important to extract the results from nodes that are not near the edges, or faces, of the

model.

midplane of
Densely meshed region the model













Figure 4-1. Results were taken from the elements at the midplane

How the material orientations are selected is best seen in Figure 4-2. The cones

represent variations in the primary orientation. The inner and outer cones represent a 7.5

degree and fifteen degree variation, respectively. The red dots indicate some possible

positions of the z-axis (33 total) that come from the intersection of the x-z and y-z planes.

The third plane is the theta (0) plane, or a rotation about the z axis. Since this material









exhibits cubic symmetry, theta only has to vary between 0 and 80 degrees. Following the

convention of Arakere and Swanson (2000), theta is broken up into nine degree

increments (0, 10, 20, 30, 40, 50, 60, 70, 80). Multiplying the possible values for theta by

the 33 primary axis cases, there is a total of 297 material orientations.

z



29^^ j- 2 27l2n





Casing Coo e System







correspondence, 2002
i1"-'- .^^ "-







9 Plane
Casting Comdinate System



Figure 4-2. Schematic of casting coordinate system Source: G. Duke, personal
correspondence, 2002

Case 0 corresponds to the angles A, y, and 0 all being equal to zero as shown in

Figure 4-2. Another example would be case 17. Here, A = 15, y and 0 are zero degrees.

Two sets of results will be presented here: case 0 and case 33. Case 33 is not on the chart

but A = 30, y and 0 are zero degrees.

Comparison of Contour Plots: Applied-Stress Model and Analytical Solution

The first set of results is a contour plot for case 0. This is where the material and

specimen coordinate systems are coincident with each other. Right away, one notices the










symmetry between all of the plots. The ANSYS plots have the black background. The

analytical plots for the corresponding stress are placed next to the ANSYS contour plots.



xx















A B

Figure 4-3. Contour plots of the ox stress (case 0). A) ANSYS contour plot. B)
Analytical contour plot


A B

Figure 4-4. Contour plots of the Gy stress (case 0). A) ANSYS contour plot. B)
Analytical contour plot


























Figure 4-5. Contour plots of the cz stress (case 0). A.) ANSYS contour plot. B)
Analytical contour plot


Figure 4-6. Contour plots of the Txy stress (case 0). A). ANSYS contour plot. B).
Analytical contour plot









The next set of results is for case 33. Again, the two plots look very similar from

a graphical standpoint.


Figure 4-7. Contour plots of the cx stress (case 33). A). ANSYS contour plot. B).
Analytical contour plot


Figure 4-8. Contour plots of the Gy stress (case 33). A.) ANSYS contour plot. B).
Analytical contour plot





























A B


Figure 4-9. Contour plots of the cz stress (case 33). A.) ANSYS contour plot. B).
Analytical contour plot


































A B


Figure 4-10. Contour plots of the Txy stress (case 33). A.) ANSYS contour plot. B).
Analytical contour plot


I00 IOOU


rmo-











Numerical Comparison of the Analytical and Applied-Stress Models

Contour plots establish a pattern of how the stresses vary, but it is a good idea to

also look at the numbers and compare the magnitudes of the stresses. The following plots

are for case 33. These results were taken from the elements at the midplane of the model

in the densely meshed region (see Figure 4-11). In Figures 4-16 and 4-17, the shear

stresses between the two models do not agree very well. The likely cause will be

discussed in chapter four.








results were obtained
from these nodes


densely meshed
region y


mx






z


midplane


Figure 4-11. Nodes used to extract results












Sigma X vs Depth


00 1 0(








Depth into Material (in.)
Depth into Material (in.)


p11


SCZgn-m3 r lS S '
m- g .rnj$,$


Figure 4-12. cx stress vs depth





Sigma Y vs Depth


' C"J5 L J 1






~_--U--


Depth into Material (in.)



Figure 4-13. oy stress vs depth


Sigma Z vs Depth


0005 001 0i









Depth into Material (in.)


---Sigma ZANSYS
-m-Sigma Z Analytical


Figure 4-14. cz stress vs depth


0.00E+00

-5.00E+04

-1.00E+05

,. -1.50E+05
-2.00E+05

-2.50E+05

-3.00E+05


0.00E+00
-5.00E+04
-1.00E+05
' -1.50E+05
-2.00E+05
-2.50E+05
-3.00E+05


0.OOE+00

-5.00E+04

-1.00E+05

S-1.50E+05

-2.00E+05 i

-2.50E+05













Tau XY vs Depth


--Tau XY ANSYS
---Tau XY Analytical


Depth into Material (in.)



Figure 4-15. Txy stress vs depth


1400
1200
1000
.- 800
-
S600
400
200
0


Tyz stress vs depth


Tau XZ vs Depth


0 0.005 0.01
Depth into Material (in.)


---Sigma XZANSYS
-m--Tau XZ Analytical


0.015


Figure 4-17. Txz stress vs depth


20000

15000

10000
CL
5000

01

-5000


) 005 0 01 0 C


Tau YZ vs Depth

0
0005 001 0O':I
-50

S-100 --Sigma YZANSYS
a -150 ---Tau YZ Analytical

-200

-250
Dpeth into Material (in.)


Figure 4-16.


~4--~___














CHAPTER 5
CONTACT MODEL


Figure 5-1. Contact model

The contact model is a half-cylinder resting on a rectangular block. The same

assumptions are in place for the applied stress model: no friction and normal loading

only. The volumes are comprised of SOLID-45 elements. CONTA174 and TARGE170

elements were used to establish the contact between the half-cylinder and block. The

primary aim of running the contact model is to determine the dimensions of the contact

patch (the contact half-width, a). The applied stress and analytical solutions relied on an

assumed value for that number. The contact model will take much longer to run on the









computer versus the applied stress or analytical solution, but the computations are

necessary to get a handle on the contact patch size.


Figure 5-2. Densely meshed regions of the contact model

In Figure 5-2 above one can see the densely meshed region where the contact

elements were placed on the block and where the target elements were applied on the

half-cylinder. Figure 5-3 shows the resulting contact pair created in ANSYS.


Figure 5-3. Contact and target elements









Notice the curved shape of the target elements. That is because they are attached to the

curved profile of the half-cylinder.

In order to create contact models in ANSYS, a contact pair of elements must be

created-a contact element and a target element. ANSYS has general guidelines as to

what line, surface, or volume these elements should be applied. Perhaps the most critical

feature is the mesh size. For example, a large target element size and very fine contact

element will not work. The sizes of the contact and target elements should be fairly close

to one another. It is possible to get a solution to converge, but the results will most likely

be incorrect. That is why there is a densely meshed region in both the bottom part of the

half-cylinder and on the surface of the block shown in Figures 5-2.

Another important characteristic of this contact model is that displacement

boundary conditions had to be used instead of applied forces. Applied displacements are

more stable than applied forces in ANSYS.

Examination of the Normal Stresses at the Surface

The analysis was run for four cases: case 17, 21, 25, and 29. To obtain the size of

the contact patch and peak pressure at the surface, the same procedure in chapter three

was followed. A group of nodes selected at the midplane of the densely meshed region

of the block (see Figure 5-4). Below are plots of the stress in the y-direction for the

selected nodes. Notice that they have an elliptic profile.

The normal contact pressure should be elliptical (even though orthotropic

materials are involved). If the correct distribution is not apparent, then the size of the

elements may be too big and additional refinement is needed so the proper load is applied

to the surfaces of the bodies in contact.













results were obtained from
these nodes


densely meshed
region


midplane


Figure 5-4. Nodes used to obtain the contact patch and peak pressure at the surface











Case 17 Sigma Y


- Sigma Y


mils


Figure 5-5. Y-stress for case 17


50000

0

-50000

CL -100000

-150000

-200000

-250000


15 20


5







46






Case 21 Sigma Y


50000

0
-
-50000

CL -100000

-150000

-200000

-250000


--Sigma Y


mils


Figure 5-6. Y-stress for case 21


Case 25 Sigma Y


50000

0

-50000

C -100000

-150000

-200000

-250000


-*-Sigma Y


mils


Figure 5-7. Y-stress for case 25


15 20


) 5 10 15 20



^


I










Case 29 Sigma Y

50000

0
-50000 5 10 15 20 25
-50000 5

-100000 Sigma Y
-150000

-200000
-250000
mils


Figure 5-8. Y-stress for case 29

Table 5-1 lists the contact patch (taken from the above plots) and also the peak

pressures from the four cases that were run in ANSYS. These numbers are compared to

peak pressure and contact patch when Equations 3-2, 3-3, and 3-4 are used. The cylinder

was composed of an isotropic steel (E = 30 x 106 psi and v = 0.3), the effective radius, R,

was 0.1 in.

Table 5-1. Comparison of Hertzian and numerical contact pressures and patch sizes
Case 17 21 25 29
ANSYS Contact Half-Width (in.) 0.003 0.003 0.002 0.002
ANSYS Peak Pressure (psi) -195000 -206000 -194770 -200240


Using Equations 3-2, 3-3, and 3-4 the peak (Hertzian) contact pressure and

contact half-width are 180100 psi and 0.002774 inches (from the 79 lbf normal load).

Looking at the table above it would appear the true contact half-width is between 0.002

and 0.003 inches. A finer mesh would help to determine if that is the case. The

consequence of a better mesh (more elements) is computational cost. ANSYS does have

a limit on the number of elements that can be used. Approaching that limit would tax









ANSYS and the abilities of the computer to the point that one run could take many hours

to finish.

Comparison of Contact Model Stresses to the Analytical Solution

The following plots compare stresses between the contact model and analytical

solution taken at the midplane straight down (-y direction) into the rectangular block.

The stresses between the applied stress model and analytical solution were already

compared and the results below are similar in that the cx, y oz, and Txy stresses match

well but the remaining two shear components do not. It should be pointed out that the Txz

and Tyz are not always zero for the analytical solution. As the orientation changes, those

two stress components will not be zero. For case 21, Txz is nonzero and in case 29 Tyz is

nonzero. Even though Lehnitkskii (1963) uses what he refers to as a generalized plane

strain condition, there is still shear coupling in the z direction.

The elements in the densely meshed regions of both models had finer meshes in

the x and y directions than the z direction. Figure 5-9 illustrates the size of the brick

element used in the contact model. Brick elements of a slightly smaller size were used in

the applied stress model (the z-dimension was 0.005 in.). One can see that there is a large

difference between the length in the x and y directions and the z direction of the brick

element.

The difference between the z dimensions was only 0.002 inches, but that is

enough to throw off at least one of the shear stress components. For example, the Txy

component from the contact model does not agree with the analytical solution. The issue

here is probably one of convergence. One way that might help is to use a finer mesh.

Four of the six stress components from the applied stress model match up very with the

analytical solution and that is most likely due to the smaller element size that model uses.


















0001 in
0. 001 in.

-4


0.07n.
0.007 m.


Ly






x


z
z


-.001 in.-


Figure 5-9. Dimensions of the brick element used in the contact model

The reason why these elements had such dimensions is computational cost. A

smaller dimension in the z direction would have resulted in more elements. Ideally one

would like to have many cube elements or small tetrahedrons. But that would have

resulted in contact models having a large number of elements. A model that big would

take a very long time to solve and the results files (where the stresses are stored) also

would have been large.


Sigma X vs Depth (case 17)


0.00E+00

-5.00E+04

'C -1.00E+05

-1.50E+05

-2.00E+05


---Sigma XANSYS
- Analytical


mils


Figure 5-10. Sigma X comparison case 17







50






Sigma X vs Depth (case 25)


0.00E+00

-5.00E+04

C. -1.00E+05

-1.50E+05

-2.00E+05


- Sigma XANSYS
---Analytical


mils


Figure 5-11. Sigma X comparison case 25


Sigma Yvs Depth (case 17)


--Sigma Y ANSYS
-- Analytical


mils


Figure 5-12. Sigma Y comparison case 17


0.00E+00

-5.00E+04

S-1.00E+05

-1.50E+05

-2.00E+05


5 10






51


Sigma Y vs Depth (case 25)


O.OOE+00

-5.00E+04

'i. -1.00E+05

-1.50E+05

-2.00E+05


- Sigma Y ANSYS
-Analytical


mils


Figure 5-13. Sigma Y comparison case 25


Sigma Z vs Depth (case 17)


0.00E+00
-2.00E+04
-4.00E+04
-6.00E+04
i -8.00E+04
-1.00E+05
-1.20E+05
-1.40E+05
-1.60E+05


-- Sigma ZANSYS
-- Analytical


mils


Figure 5-14. Sigma Z comparison case 17


) 5 10 1
-s


-


S5 1


-r







52



Sigma Z vs Depth (case 25)


O.OOE+00
-2.00E+04
-4.00E+04
-6.00E+04
'. -8.00E+04
-1.00E+05
-1.20E+05
-1.40E+05
-1.60E+05


-- Sigma ZANSYS
-- Analytical


mils


Figure 5-15. Sigma Z comparison case 25


Tau XY vs Depth (case 17)


--Tau XY ANSYS
-- Analytical


mils


Figure 5-16. Tau XY comparison case 17


S5 1




-/____


20000
10000
0
-10000 1
' -20000
-30000
-40000
-50000
-60000


5 0--~


*







53



Tau XY vs Depth (case 25)


0
-5000
-10000
-15000
'. -20000
-25000
-30000
-35000
-40000
mils



Figure 5-17. Tau XY comparison case 25


Tau YZ vs Depth (case 17)


3000

2000

1000

0

-1000

-2000


---Tau XY ANSYS
-- Analytical


-- Tau YZ ANSYS
-- Analytical


mils


Figure 5-18. Sigma YZ comparison case 17


1) 5







54



Tau YZ vs Depth (case 25)


---Tau YZANSYS
-- Analytical


mils



Figure 5-19. Tau YZ comparison case 25


Tau XZ vs Depth (case 17)


- Tau XZ ANSYS
- Analytical


mils


Figure 5-20. Tau XZ comparison case 17


3000
2500
2000
1500
'. 1000
500
0
-500
-1000


5 1


20000

15000

10000

5000

0

-5000


S1 10











Tau XZ vs Depth (case 25)


--Tau XZANSYS
--Analytical


mils


Figure 5-21. Tau XZ comparison case 25


20000

15000

10000

5000

0

-5000


10 1














CHAPTER 6
DEFORMATION MECHANISMS

Plastic deformation in metals results from a phenomenon called slip-that is the

movement of dislocations throughout the material. Typically slip occurs over planes with

high atomic density in the close-packed directions. When slip occurs atoms are being

forced to relocate to new lattice positions within the crystal structure. Atoms located on

densely packed planes will have a shorter distance to travel when compared to atoms

situated on other, less densely packed, planes. This is not always the case, and at higher

temperatures dislocations can move along planes with lower atomic densities because of

two things: cross slip and climb. Cross slip (phenomenon that entails dislocations

moving onto an intersecting slip plane) is a thermally activated process. Climb (also a

thermally activated process) is associated with diffusion (movement of atoms into

vacancies, etc) (Hummel 1998).

Since the material in question is a Nickel based superalloy, the planes with the

highest atomic density are the octahedral planes. A slip system requires the definition of

the slip plane and the slip direction. For example, (111)[0-1 1] is a slip system where

(111) is the slip plane and [0 -1 1] is the slip direction. The slip direction always lies in

the slip plane (Figure 6-2). The aforementioned notation comes from the Miller indices.

Let us look at the (111)[0 -1 1] system graphically (Figure 6-1). The (111) plane is

encased by the cube and the vector <0-11> is the slip direction drawn from the origin.

With the four octahedral planes there are 12 close-packed directions the planes can

move and also another set of 12 directions in the non-close-packed slip direction. Thus,








57



there are 24 slip systems on the octahedral planes. At higher temperatures, slip can occur


on the cube planes. These planes are not as densely packed as the octahedral planes and


they, along with the octahedral planes, are shown below in Figures 6-2 and 6-3.


Counting the cube slip systems, there are a total of 30 slip systems (Dame and Stouffer


1996).


Figure 6-1. Example slip system


DIi.- I
010 eoP-Yd]i.z


Pla 2
P-nay 0, t, T'
Secondaiv T11, T17 t11


Pl, 3
010 iD .-
Sec..day Oe


010


Figure 6-2. Four octahedral planes and slip directions. Source: Modified from Dame and
Stouffer 1996












11W Flamel



PHM

001010
100


001
010 Plane 3







S100











1 1001 1
Figure 6-3. Cube slip planes and slip directions. Source: Modified from Dame and
Stouffer 1996



With the 30 slip systems defined one can now establish equations to acquire local

stresses and strains within the material. From Dame and Stouffer (1996) we have




S1 0 -1 1 0 -1
S0 -1 1 -1 1 01 1
S1 1 0 0 1 -1
4 -1 0 1 1 0 -1 co
5 -1 1 0 0 -1 -1 O
r6 1 0 1 -1 -1 -1 0 ^ (6-1)
r7 -6 1 -1 0 0 -1 -1 ,y

T8 0 1 -1 -1 1 0 o"
r9 1 0 -1 -1 0 -1 yz
r10 0 -1 1 -1 -1 0

11 -1 0 1 -1 0 -1
12 -1 1 0 0 1 -1






59


13 1 2 -1 1 -2 1
14 2 -1 -1 1 1 -2
15 -1 -1 2 -2 1 1
16 -1 2 -1 -1 -2 -1 o,
17 -1 -1 2 2 1 -1 o

718 1 2 -1 -1 -1 1 2 2)
I (6-2)
-19 3= -1 -1 2 2 -1 1 C ,X
-20 2 -1 1 -1 -1 -2 o
21 -1 2 -1 -1 2 1 o-
z22 2 -1 -1 1 -1 2
23 -1 2 -1 1 2 -1
z24 1 -1 2 -2 -1 -1




The six slip systems below are for the cube slip planes.

'25 0 0 0 1 1 0 oxx
S26 0 0 1 1 0
S27 0 0 0 1 0 1 C (
'28 V2 0 0 0 1 0 -1 x
29 0 0 0 0 1 1 O-
30 0 0 0 0 -1 1 OC


cI-'c30 are the local shear stresses associated with that slip system. For example, TI is

associated with slip system number one.

If any of these values ('l-C30) are above a threshold stress called the critical

resolved shear stress, then that slip system will become active. It is possible for more

than one slip system to become active depending on the orientation and loading of the

material. It might be of some use to investigate how the local shear stresses (--'C30) vary

in the material. In order to calculate these local shear stresses, all six stress components









(Ox, O, Oz, C xy,'-zx, Tyz) are needed for the equations listed above. Then at each node all of

the local shear stresses can be tabulated.


Table 6-1. Slip Systems
Slip Number Slip Plane
Octahedral Slip a/2<110>{111}
1 [1 1 1]
2 [ 1 1]
3 [1 1 1]
4 [-1 1-1]
5 [-1 1-1]
6 [-1 1 -1]
7 [1 -1 -1]
8 [1 -1 -1]
9 [1 -1 -1]
10 [-1 -1 1]
11 r-1 -1 1]
12 [-1 -1 1]
Octahedral Slip a/2<112>{111}
13 [1 1 1]
14 [1 1 1]
15 [1 1 1]
16 [-1 1 -1]
17 [-1 1 -1]
18 [-1 1 -1]
19 [1 -1 -1]
20 [1 -1 -1]
21 [1 -1 -1]
22 [-1 -1 1]
23 [-1 -1 1]
24 [-1 -1 1]
Cube Slip a/2<110>{100}
25 [1 0 0]
26 [1 0 0]
27 [0 1 0]
28 [0 1 0]
29 [0 0 1]
30 [0 0 1]
Source: Modified from Dame and Stouffer 1996


L


Slip Direction
12 Primary Slip Directions
[1 0 -1]
[0 -1 1]
[1 -1 0]
[1 0 -1]
[1 1 0]
[0 1 1]
[1 1 0]
[0 -1 1]
[1 0 1]
[0 1 1]
[1 0 1]
[1 -1 0]
12 Secondary Slip Directions
[-1 2 -1]
[2 -1 -1]
[-1 -1 2]
[1 2 1]
[1 -1 -2]
[-2 -1 1]
[-1 1 -2]
[2 1 1]
[-1 -2 1]
[-2 1 -1]
[1 -2 -1]
[1 1 2]
6 Cube Slip Directions
[0 1 1]
[0 1 -1]
[1 0 1]
[1 0 -1]
[1 1 0]
r-1 1 0]







61


Once again, nodes from the densely meshed region at the midplane of the

rectangular slab were used to obtain the required stresses. Local shear stresses from the

primary planes ('1-'c12) are given below for cases 17 and 25. In all cases the peak stresses

occurred a few mils below the surface, and then decrease as a function of depth (-y

direction) into the material. Also, in comparing the local shear stresses one notices how

similar the contours are between the two cases.


5 10 15


A

Figure 6-4. Tau 1 contour plot. A.)


0 5 10 15 20


A

Figure 6-5. Tau 2 contour plot. A.) Case


Case 17. B.) Case 25


5 10 15 20


17. B.) Case 25



























2



0
0 5 10 15 20

dd
A


Figure 6-6. Tau 3 contour plot. A.) Case 17.



8



62
1-

44
4
1.






0
0 5 10 15 20

ee
A


Figure 6-7. Tau 4 contour plot. A.) Case 17.


4



a



0
0 5

dd




B.) Case 25



8



6



4



2



0
0 5

ee




B.) Case 25


10 15 20



B












4






4 -



0 5 10 15 20
ff
A

Figure 6-8. Tau 5 contour plot. A.) Case 17.










:1 1


0 5 10 15 20

gg
A

Figure 6-9. Tau 6 contour plot. A.) Case 17.


0 5
ff


B.) Case 25




8-





4-






0 5

gg


B.) Case 25


10 15 20


B


10 15 20


B






64



8 -











0 0
0 5 10 15 20 0 5
hh hh
A

Figure 6-10. Tau 7 contour plot. A.) Case 17. B.) Case 25


4.0




a-
0 5 10 15 20


A

Figure 6-11. Tau 8 contour plot. A.) Case 17.


10 15 20


B


4-
4.






0 5




B.) Case 25


15 20













8- 0 8



6 6



4 4
4


2 1- 2



0 0 0
0 5 10 15 20 0 5

ii ii
A


Figure 6-12. Tau 9 contour plot. A.) Case 17. B.) Case 25



8


6-











o 0
0 10 15 20 0 5

kk kk
A


Figure 6-13. Tau 10 contour plot. A.) Case 17. B.) Case 25


10 15 20



B


15 20













8



6



41



21



0
0 5 10 15 20

11
A


Figure 6-14. Tau 11 contour plot. A.) Case 17.




















0 5 10 15 20

mm
A


Figure 6-15. Tau 12 contour plot. A.) Case 17.


8



6



4 4



2



0 0
0 5


















2-
11























0

0 5

mm



B.) Case 25


10 15 20



B















CHAPTER 7
LITERATURE REVIEW

This section gives a discussion of the limited work done in the area of anisotropic

contact mechanics. Relevant work dealing with analytical and numerical solutions is

highlighted.

Sinclair and Beisheim (2002) developed a procedure to analyze stresses of

dovetail attachments in turbine blades using a submodeling routine. These stresses are

difficult to obtain because of the nature of the contact problem (conforming) and also the

large magnitude and gradient of the contact stresses. These parameters require a high

number of elements to achieve accurate results.

A submodeling routine was implemented in order to get converged peak stresses

without considerable computational cost. The first step is to run a global analysis that

includes the highly stressed contact region. Three grids labeled coarse, medium, and fine

are used to mesh the model. The FE model is run with each of the three grids and the

results from each grid are compared to check for convergence. The stress component in

question, such as a peak stress, is said to have converged if

f max -mmax < (71)
< Es (7-1)
0" max
where Es is the error level (0.01 is excellent, 0.05 is good, and 0.1 is satisfactory). The

superscripts f and m denote fine and course grids.

The next step is the submodeling routine. This is where the highly stressed

region, or submodel, is broken out and analyzed separately using the information









obtained from the global analysis. Distances, or boundaries, away from peak stress

component are chosen such that the stress component in question has converged. Then

the submodel is divided up with its own coarse, medium, and fine grids.

Displacements were used to check for convergence since they converge faster

than stresses. Boundary displacements are said to be converging if

Uc --_Um > zm -Ulf r (7-2)
uf tU, U1i (7-2)

where i denotes the node in question.

Cubic spline equations were fit to the nodal displacements. This was done so

even more displacements could be ascertained between nodes for all of the grids. Using

the fitted splines, the three grids are run and checked to see if the submodel boundary

conditions have converged using two expressions that calculate the boundary condition

error and the discretization error. The latter is built in to all FE analysis.

Even though Sinclair and Beisheim (2002) claim that their numerical approach

requires 'modest' computing resources, a very high number of elements was used for

their analysis.

Table 7-1. Number of elements used
Grid Spatial Contact Total
Global Course 2127 478 2614
Medium 14990 1943 16993
Fine 108870 7765 116635

Outer Ring Course 540 0 540
Medium 4140 0 4140
Fine 34147 0 34147

Submodel Coarse 9216 1152 10368
Medium 73728 4608 78336
Fine 589824 18432 608256
Source: Modified from Sinclair and Beisheim (2002)









On the average, contact models in this work, composed of roughly 30,000

elements, took approximately two hours for the computer to complete its run for each

crystal orientation. Over 600,000 elements were used in the submodeling routine to get

accurate results in the contact region. Only had normal tractions applied to the model,

just like the models used in this analysis. In chapter three the applied stress results were

compared to analytical solution. The normal stress components agreed fairly well, but

the shear stresses did not. A potential source of this problem was the size of the elements

used, or the mesh size. Smaller elements, and consequently bigger FE models, are

needed to get the correct stresses-as shown in Table 7-1. Using that many elements

resulted in a total estimated error of 2.7%.

When looking at contact problems (such as those involving indenters) the contact

patch is a function of the material properties, geometry, and load. In most cases, one has

a handle on what the geometry of the indenter and half-space is and also what external

force is being applied. When isotropic materials are used the contact patch is only

dependant on two material properties: such as the elastic modulus and the poisson ratio.

Anisotropic materials are different in that the contact patch is related to an equivalent, or

composite, modulus since the material properties are direction dependent.

Vlassak et al. (2003) looked at calculating the indentation modulus of anisotropic

materials. Analytical solutions were developed for indenters of arbitrary shape being

pressed into anisotropic half-space. A theorem stating that the solution to this contact

problem is one that maximizes the load on the indenter for a certain depth. One of the

principle assumptions was to assume the contact patch had a certain shape, then the best

approximate solution exists in the Rayleigh-Ritz sense. For axisymmetric indenters, a









limited family of Green's functions, using the first term in the Fourier expansion, can be

used to obtain a much simpler solution. This solution is denoted as the "equivalent

isotropic solution."

First, the Green's function used to obtain displacements at the surface of the half-

space is defined. The next step is to determine the load on the punch, or indenter. With

the load function available, it can be differentiated and the modulus of the half-space is

found.

The Green's function gives the displacement in the direction of a load applied to

the half-space.


w(y) = I-, akk :1 a (7-3)



Equation 7-4 is the force that will establish the elliptical contact area.

F(A) u(x, y)dxdy (74)

a2 b2


1
where po = and u(x,y) is the displacement from the indenter.
7;iba(e, p)

Now the solution for a conical indenter is produced so the modulus for this case

can be found.

For a conical indenter the displacement is

u(x,y) =U- Cx2 + y (7-5)

where U is the rigid body displacement and C is the cotangent of the cone angle.

Using Equation 7-4 the force on the indenter is now









U -C-Cx + y2 dxdy
F(A) JJf (7-6)

r a2 b2

After a change of variables and also switching to polar coordinates the force is

2Ua -Ca2E(e)
F(a, b, () = (7-7)


The resulting contact stiffness for a conical indenter is
sdF 21
S -= 1(7-8)
dU 4 a(e,p)(1-e2)1/4

where U is the approach of the two bodies and A = nab.

The modulus of the half-space is

eqv = 1/4 (7-9)
a(e, p)(1 e2)1

A less complicated expression for conical indenters can be found by assuming the

contact patch is circular. Now the orientation angle, p, is arbitrary and the eccentricity

drops out


a(O,() = h(O)dO = 7h (7-10)
0

ho is the first term of the Fourier expansion of the function h(O), and modulus

now becomes


qM (7-11)
zho

which is the same solution for a flat circular punch for an isotropic half-space; the

equivalent isotropic solution.









The modulus in Equation 7-11 is compared to the modulii one gets from when a

sphere and cone are the indenters and the Green's function is not truncated using a

Fourier expansion.

When the modulii are plotted against crystal orientation, the modulii of the cone

and sphere do not vary much (the difference between the two is less than 0.03%). Using

the equivalent isotropic solution, the difference is approximately 0.1%.

Green and Zerna (1954) looked at the two-dimensional contact problem for

anisotropic materials. Their work is somewhat confusing because they retain the notation

used for isotropic materials. Their solution is not applicable for coupled in-plane and

antiplane deformation. Fan and Keer (1994) re-examine the two-dimensional contact

problem using the analytic function continuation approach. They arrive at a compact

solution that may be applied to other problems such as examining subsurface stresses in

contact problems involving sliding and sticking.

Willis (1966) examined the Hertzian contact for anisotropic half-spaces. He

assumed the contact patch is elliptical and the contact pressure will be of the form Hertz

used when he was looking at this problem for isotropic materials. Willis does not use

potential theory to solve this problem. He used Fourier transforms and the solution lies in

solving contour integrals. Evaluating those integrals can be done numerically.














CHAPTER 8
CONCLUSIONS

The 2-D analytical solution developed for subsurface stresses in anisotropic half-

spaces is an efficient and practical method for performing design iterations. Contact

dimensions and pressure distributions obtained by Hertzian isotropic assumptions are a

good approximation for anisotropic contacts also.

The 3-D FE model, or applied stress model, for stress analysis in the anisotropic

substrate (single crystal), using contact dimensions and pressure profile from Hertzian

solutions is an excellent representation of the contact problem. This is an accurate and

efficient technique for determining subsurface stresses for anisotropic contacts.

The 3-D contact model for anisotropic substrates, while numerically intensive, is

the best way to solve problems involving complex geometries. Just like the applied stress

model, effects of crystal orientation were included. Additional mesh refinement might

produce more accurate stress solutions in the edge of contact. Overall, most of the results

had excellent agreement with the applied stress model and analytical solution.

Contour plots for resolved shear stresses on slip planes are presented in the

contact region. These results are required for determining fatigue life (crack initiation) in

the substrate.















CHAPTER 9
RECOMMENDATIONS FOR FUTURE WORK

Effects of contact friction and tangential forces must be incorporated in the FE

models. The analytical solution can already account for tangential traction forces of the

contact.

Relationships between shear stress amplitude on the slip planes and cycles to

fatigue crack nucleation need to be determined. To effectively model the fatigue damage

process, the ability to evaluate plastic stress and strain amplitudes on the contact is

required. This will entail developing constitutive relations for crystal plasticity, which

will prove to be a complex endeavor.
















LIST OF REFERENCES


ANSYS Elements Reference, ANSYS Release 5.6; ANSYS, Inc. November 1999.

Arakere N, Swanson G. Effect of crystal orientation on analysis of single-crystal, Nickel-
base turbine blade superalloys, National Aeronautics and Space Administration Technical
Publication 2000; Feb: 6-10.

Beisheim JR, Sinclair GB. Three-dimensional finite element analysis of dovetail
attachments. American Society of Mechanical Engineers Turbo Expo. 2002 Jun 3-6;
Amsterdam, the Netherlands.

Dame L, Stouffer D. Inelastic deformation of metals. New York: John Wiley and Sons;
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Donachie M, Donachie S. Superalloys. Materials Park, Ohio: American Society for
Metals International, 2002. p. 25-39.

Fan H, Keer LM. Two-dimensional contact on an anisotropic half-space. Transactions of
the American Society of Mechanical Engineers 1994; 61: 250-255.

Green AE, Zerna W. Theoretical Elasticity. Oxford, UK: Clarendon Press; 1954.

Hummel R. Understanding materials science. New York: Springer; 1998. p.74-101, 102-
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Johnson KL. Contact mechanics. Cambridge, UK: Cambridge University Press; 1985. p.
84-106

Lekhnitskii SG. Theory of elasticity of an anisotropic body, San Francisco: Holden-Day;
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Magnan S. Three-dimensional stress fields and slip systems for single crystal superalloy
notched specimens, Master's Thesis, University of Florida, Gainesville 2002, p. 41-52.

Sims CT, StoloffNS, Hegel WC, editors. Superalloys II. New York: John Wiley and
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Timoshenko SP, Goodier JN. Theory of elasticity. Tokyo: McGraw-Hill; 1983. p. 15-34.






76


Vlassak JJ, Ciavarella M, Barber JR, Wang X. The indentation modulus of elastically
anisotropic materials for indenters of arbitrary shape. Journal of the Mechanics and
Physics Solids 2003;51: p. 1701-1721.

Willis JR. Hertzian contact of anisotropic bodies. Journal of the Mechanics and Physics
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BIOGRAPHICAL SKETCH

Born on February 18, 1979 in West Chester, Pennsylvania I moved to Naples,

Florida when I was four. I graduated with a B.S. in mechanical engineering from

Villanova University and am now seeking a doctorate in mechanical engineering. I plan

on becoming a professor someday.