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Unformatted text preview: Chapter 1 Introduction The introduction of the course is broken into two sections. The rst section describes the importance of friction and wear to today's engineer and introduces some of the basic concepts that will be pursued later in more detail. The second section brie y outlines the history of tribology so that we understand where we are coming from to better understand where we are going. 1.1 What is Tribology? The mechanics of friction and wear covers a portion of a larger eld known as tribology. Tribology is the SCIENCE and TECHNOLOGY of interacting surfaces in relative motion and the practices related thereto. The word was o cially coined and de ned by the Jost Report (1966) in 1966. It is derived from the Greek root tribos which means rubbing. Tribology includes friction, wear, and lubrication. Engineers are speci cally interested in the control of friction, wear, and lubrication which is why it is important that the word technology is included in the de nition of tribology. Next the three terms friction, wear, and lubrication are de ned. 1.1.1 Friction Consider the block of weight W resting on the inclined plane in Figure 1.1. As the plane is tilted to an angle with the horizontal the weight can be resolved into components perpendicular to the plane, N , and parallel to the plane, F . If is less than a certain value say, s, the block does not move. It is inferred that the plane resists the motion of the block that is driven by the component of the weight parallel to the plane. The force resisting the motion is due to friction between the plane and block. As is increased past s the block begins to move because the tangential force due to the weight, F , overcomes the frictional force. It can be shown experimentally that s is approximately independent of the size, shape, and weight of the block. The ratio of F to N at sliding is, tan s , which is called the coe cient of friction. The frictional resistance to motion is equal to the coe cient of friction times the compressive normal force between two bodies. The coe cient of friction depends on the two materials and 1 Figure 1.1: Inclined plane experiment used to de ne the coe cient of sliding friction . Physical Situation Rubber on cardboard (try it) 0.5 - 0.8 Brake material on brake drum 1.2 Dry tire on dry road 1 Wet tire on wet road 0.2 Copper on steel, dry 0.7 Ice on wood 0.05 Table 1.1: Typical Coe cients of Friction (Bowden and Tabor, 1982) . in general 0:05 < < 1. The idea that the resistance to motion caused by friction is F = N is called the Amontons-Coulomb Law of sliding friction. This law is not like Newton's Laws such as F = ma. The more we study the friction law the more complex it becomes. In fact we cannot fully explain the friction law as evidenced by the fact that it is di cult or even impossible to estimate for two materials without performing an experiment. As a point of reference, is tabulated for several everyday circumstances in Table 1.1. 1.1.2 Wear The rubbing together of two bodies can cause material removal or weight loss from one or both of the bodies. This phenomenon is called wear. Wear is a very complex process. It is much more complex even than friction. The complexity of wear is exempli ed in Table 1.2 where wear rate is shown with for several material combinations. Radical di erences in wear rate occur over relatively small ranges of the coe cient of friction. Note that wear is calculated from wear rate by multiplying by the distance traveled. Materials Wear rate cm3 10;12 cm 1 Mild steel on mild steel 0.62 157,000 2 60/40 leaded brass 0.24 24,000 3 PTFE (Te on) 0.18 2,000 4 Stellite 0.60 320 5 Ferritic stainless steel 0.53 270 6 Polyethylene 0.65 30 7 Tungsten carbide on itself 0.35 2 Table 1.2: Friction and wear from pin on ring tests. Rings are hardened tool steel except in tests 1 and 7 (Halling, 1983) . The load is 400g and the speed is 180cm=sec. 1.1.3 Lubrication Lubrication is the e ect of a third body on the contacting bodies. The third body may be a lubricating oil or a chemically formed layer from one or both of the contacting bodies (oxides). In general the coe cient of friction in the presence of lubrication is reduced so that 0:001 < < 0:1. 1.2 Why Study Tribology? Tribology, or friction, wear, and lubrication, have several consequences in mechanical components and modern day life. Most consequences of friction and wear are thought of as negative such as power consumption and the cause of mechanical failure. However, there are also some positive bene ts of friction and wear. It is estimated that 20% of the power consumed in automobiles is used in overcoming friction while friction accounts for 10% of the power consumption in airplane piston engines and 1.5-2% in modern turbojets. Friction also leads to heat build-up which can cause the deterioration of components due to thermo-mechanical fatigue. Understanding friction is the rst step towards reducing friction through clever design, the use of low friction materials and the proper use of lubricating oils and greases. Friction has many bene ts such as the interaction between the tire and the road and the shoe and the oor without which we would not be able to travel. Friction serves as the inherent connecting mechanism in knots, nails, and the nut and bolt assembly. It has some secondary bene ts such as the interaction between the ber and matrix in composites and damping which may reduce deleterious e ects due to resonance. 1.3 What is Required? The understanding of tribology requires knowledge from the disciplines of mechanics, materials science, mechanical engineering, and chemistry. In this course emphasis will be placed on the tools required from mechanics and mechanical engineering so that they can be investigated in depth. The topics to be considered include the following. 1. The manner in which surfaces that are brought into contact deform. Knowledge about this deformation leads to the calculation of contact pressure and the subsurface stress eld which governs yielding and failure. 2. The contact induced deformation is controlled by forces, material properties, and the shape and contour of the undeformed surfaces. Later we will use ideas from statistics to characterized rough surfaces. 3. The ability of surfaces to adhere to each other and the \strength" of interfaces play an important role in friction and wear. 4. We will discuss the mechanisms by which energy is lost during the deformation of surfaces. Much of this energy goes into heat which increases the temperature of the bodies. We will develop models for the calculation of the temperature rise. 5. Finally the role of surface lms in friction and wear will be discussed. This may include lubricants as well as chemical reactants formed on the surface by the interaction between the contacting bodies. The chemical reactions may be brought on in part by the frictionally induced rise in temperature. 1.4 History of Tribology Most of the information in this section is taken from the book by Frank Bowden and David Tabor (1982) entitled Friction: An Introduction to Tribology . It is repeated here to give you a feel for how little is actually known to this point about friction. Man has known about the deleterious e ects of friction for some two thousand years as evidenced by a mural painting (circa 1900 B.C.) which shows a collossus pulled along on a sledge while one man pours lubricating oil in its path. Ancient chariots have been found with grease remains on their axles that contain metal rather than wood contacting surfaces. However, no work of scienti c nature was undertaken until Leonardo da Vinci. Leonardo da Vinci deduced the two basic laws of friction using an experiment very similar to Figure 1.2. These laws are that the friction force is proportional to the weight of the block and independent of the block's shape. This work was mainly forgotten until the time of Newton and the onset of classical mechanics. The original scienti c investigations of friction were conducted by the French. Amontons (1663-1705) rediscovered the laws of friction originally derived by Leonardo da Vinci. Amontons knew that the surfaces that he worked with were rough and attributed friction to this roughness. He explained the energy loss that caused friction as that taken up by lifting one surface over the roughness of another and the energy required to bend and/or break the peaks of the surface roughness. This view of roughness induced friction was generally accepted for the next century. An English scientist, John Theophilus Desaguliers (1683-1744), took a very di erent view of friction. Desaguliers recognized that sometimes smoother surfaces lead to higher Figure 1.2: Leonardo da Vinci's Experiments. Figure 1.3: Ideal rough surface. friction which could not be explained with the surface roughness theory. He measured adhesive forces between bodies in contact. He recognized that this adhesion played a part in friction but could not see how it could be used to explain the laws of friction. As we will see, this shortcoming of the adhesive friction ideas was not overcome until we were able to consider the deformation of rough surfaces in contact. Charles Augustin Coulomb (1736-1806) provided a classical study of friction. He used ideas from work to derive the laws of friction. In particular he derived the equation for the coe cient of friction as = tan (1.1) where the value for is taken from the surface roughness and the energy is taken up by lifting one body over the hills (Figure 1.3). There are some shortcomings of Coulomb's theory of friction. First of all, what value of should be used for a rough surface that has hills of varying slopes? Second, John Leslie (1766-1832) asked about how energy is dissipated. Shouldn't all of the energy that is used to lift one body over the other be recovered when the body slides back down? In fact, this question remains unanswered. Third, Coulomb did not recognize the role that the surface roughness plays in the real area of contact. He pictured the real area of contact continually increasing as the bodies approached each other without considering the e ect of surface deformation. This surface deformation has relatively recently been used to explain the role of adhesion in friction. 1.5 Tribology and Modern Science I will end the introduction by giving hints at what sort of tools and theories are available to the modern tribologist. Perhaps this will serve as motivation. 1. Surface roughness measurements can be performed at resolutions approaching 0:005 m. 2. Failure mode analysis can be used to predict the life and reliability of components. These analyses can predict contact pressure, both dry and lubricated, subsurface stress elds, and contact temperatures due to frictional heating. 3. The growth of cracks initiated both at the surface and below the surface of contacting bodies can be calculated. 4. Magnetic bearings are being developed for applications. 5. The tribology of the magnetic head/disk interface is a very active eld. As ying heights decrease, tribological issues are becoming more and more important to further technological advances. 6. The use of hard coatings in tribological components is being heavily investigated It is hoped that we will discuss these issues in depth at some point in the semester. By in depth, it is meant that we will attempt to understand some of these topics from a mechanics point of view. Chapter 2 Two-Dimensional Elasticity First consider the contact of nonconforming surfaces. Nonconforming bodies touch at only one point or along a line when they are brought in contact without deformation. Examples include ball bearings, roller bearings, and gear teeth. A schematic of nonconforming contacting bodies is shown in Figure 2.1. The contact size is much smaller than other dimensions of the bodies. The stresses induced by the contact loading are highly concentrated in the vicinity of the contact. These stresses do not strongly depend on the overall shape of the bodies. For certain contact loads the stresses in the bodies remain below the yield stress of the material and the strains are small compared to unity. Under these assumptions the stresses and strains are related by the isotropic Hooke's Law. It is also assumed for now that the loads are applied slowly so that the deformation is static. Restricting ourselves to these assumptions allows the use of the wealth of information known about the linear elastic half-space. 2.1 Coordinate Systems and Sign Conventions The nomenclature and sign conventions used in these notes for stresses and displacements in Cartesian and cylindrical coordinates are described in Figure 2.2. The loading is assumed to be in plane strain so that the plane of the page is viewed as a crosssection of an in nitely long cylinder. The x-axis is tangential to the half-space and the z-axis is directed into the half-space. The objective is the calculation of the stresses and displacements everywhere in the body when the body is loaded by known tractions are by contact with another body. For the coordinate systems shown in Figure 2.2, the transformation equations for changing coordinate systems are given next: x = ;r sin z = r cos r2 = x2 + z2 = ;tan;1 x z ur = ;u sin + w cos u = ; u cos ; w sin u = ;ur sin ; u cos w = ur cos ; u sin 7 (2.1) (2.2) Figure 2.1: Non-conforming contact. r = x sin = r x cos =( x 2 2 = r sin z = r cos = ; ( z cos + z sin 2 2 ; 2 xz sin cos +2 xz sin cos ; z ) sin cos ; xz (cos2 ; sin2 ) x xz + 2 2 + cos2 + 2 r sin cos + sin2 ; 2 r sin cos ; ) sin cos ; xz (cos2 ; sin2 ) 2 + z cos2 ; 2 xz sin cos r = x sin = r r x cos 2 + z sin 2 = ( x ; z ) sin cos ; x = r sin z = r cos 2 2 (2.3) + 2 xz sin cos xz (cos 2 ; sin2 ) + cos2 + 2 r sin cos + sin2 ; 2 r sin cos = ; ( r ; ) sin cos ; r (cos2 ; sin2 ) (2.4) Here u is the displacement in the x-direction, w is the displacement in the zdirection, ur is the displacement in the r-direction, and u is the displacement in the -direction. The 's and 's refer to normal stresses and strains while the 's and 's refer to shear stresses and strains. The above transformation equations are derived in most books on mechanics of materials and elasticity. xz Figure 2.2: Elastic half-space (plane strain) showing Cartesian and cylindrical coordinate systems. 2.2 Governing Equations The equations governing the stresses, strains, and displacements are derived from considerations of local equilibrium, the strain displacement relationships, Hooke's Law, and compatibility of strains. These equations are given without derivation here. See any elasticity or advanced structural analysis for the details of derivation.(For example, Timoshenko and Goodier, 1970) The equilibrium equations are written as @ x + @ xz = 0 @x @z @ xz + @ z = 0 (2.5) @x @z The strain displacement equations can be derived from geometrical considerations as @u x = @x @w z = @z @u + @w (2.6) xz = 2 xz = @z @x The plane strain condition is applied as y which implies that = yz yz = = yx yx = 0 = 0 (2.7) = ( x + z) (2.8) where is called Poisson's ratio. Utilizing Equation 2.8 in Hooke's Law yields 1 2 x = (1 ; ) x ; (1 + ) z ] E 1 2 z = (1 ; ) z ; (1 + ) x ] E 2(1 + ) 1 (2.9) xz = xz = G E xz where E and G are called the, modulus of elasticity and the shear modulus respectively. Since both stress and strain are second order tensors the elastic constants E G and are not independent and their relationship can be seen in Equation 2.9. The three inplane strains are calculated from two displacements and are not completely independent of each other. This compatibility condition is given as @ 2 x + @ 2 z = @ 2 xz (2.10) @z2 @x2 @x@z Substitution of the strain-displacement relationships Equations 2.6 into Equation 2.10 shows that it is identically satis ed by a compatible strain eld. y 2.3 The Airy Stress Function It would be nice to use the equilibrium equations (Equations 2.5) alone to calculate stress distributions. However, these are only two conditions for three stress components ( x z and xz ) and nding the stresses is statically indeterminate. That is, we must impose the additional constraint that the strains calculated from the stresses are compatible so that displacements can in turn be calculated. Compatibility is ensured through the use of the Airy stress function. Equilibrium is satis ed for a su ciently continuous Airy stress function from which the stresses are calculated as @2 @2 @2 (2.11) x= z= xz = ; @z2 @x2 @x@z Substituting Equation 2.11 into Equations 2.9 gives the strains in terms of . Substituting the strains into the compatibility condition (Equation 2.10) shows that if r4 where r2 (r2 ) = 0 (2.12) @2 + @2 @x2 @z2 then the compatibility condition is satis ed. In summary, if satis es Equation 2.12 then the stresses which satisfy equilibrium and compatibility can be calculated using Equations 2.11. r2 If r2 = 0, then is called harmonic. If r4 = 0, then is called bi-harmonic. Note that if r2 = 0 thenr4 = 0 r4(x ) = 0 r4(z ) = 0 andr4(r2 ) = 0: It is sometimes convenient to work in cylindrical coordinates. The r2 operator is invariant so that a function which is harmonic in Cartesian coordinates is also harmonic in cylindrical coordinates. The r2 operator can be transformed from Cartesian coordinates to cylindrical coordinates resulting in @2 + 1 @ + 1 @2 @r2 r @r r2 @ 2 One could substitute Equations 2.11 into Equations 2.3 to show that 2 @ 1@ ) 1 @ + 1 @2 = @ 2 (2.13) r =; ( r= 2 @ 2 r @r r @r @r r @ satis es equilibrium and compatibility provided r4 = 0. For cylindrical coordinates equilibrium is given by @ r + 1@ r + r ; = 0 @r r @ r 1@ + @ r + 2 r = 0 (2.14) r@ @r r in the r and directions respectively. The strain-displacement relations are @ur 1 @ur + @u ; u (2.15) = ur + 1 @u r= r = 2r = @r r r@ r@ @r r and Hooke's Law is 1 2 r = (1 ; ) r ; (1 + ) ] E 1 = E (1 ; 2 ) ; (1 + ) r ] 1 = 2(1 + ) (2.16) r = Gr E r Several Airy Stress Functions and the corresponding stresses and displacements are given on a handout in class. r2 2.4 The Uniform Tension and Shear Airy Stress Functions In this section two of the simpler Airy stress functions are examined. This process will demonstrate the calculation of stresses through di erentiation of the Airy stress function. It will also be shown that these stresses cause compatible strains which can be integrated to give displacements. Consider = T z2 2 Di erentiating (Equations 2.11) yields x=T z = 0 xz = 0 which corresponds to uniform uniaxial tension in the x-direction (Figure 2.3). The corresponding strains are @u = 1 ; 2 T x = @x E @w = ; (1 + ) T z = @z E @u + @w = 0 xz = @z @x The strain x can be integrated to give 2 u = 1 ; Tx + f (z) E where f is an as yet unknown function of z. The strain z can be integrated to give + w = ; (1E ) Tz + g(x) where g is an as yet unknown function of x. Substituting the displacements u and w into the shear strain xz yields g0(x) + f 0(z) = 0: This equation can only be satis ed by g0(x) = ; f 0(z) = C where C is an arbitrary constant. Integration of and substitution yields 1 ; 2 Tx + A ; Cz u = E + w = ; (1 E ) Tz + B + Cx where A and B are constants. The terms containing A, B , and C correspond to rigid body motion: the rigid body displacement in the x-direction (A) the rigid body displacement in the z-direction (B) and the rigid body rotation counter-clockwise about the origin (C). The rigid body displacement terms do not generate any strain or stress and can be added when convenient. For instance, they are sometimes used to satisfy displacement boundary conditions. The stress function = ; Sxz leads to stresses x=0 z=0 xz = S: which corresponds to uniform shear (Figure 2.3). Figure 2.3: Uniform tension and Shear. 2.5 Boundary Conditions Solving plane strain elasticity problems involves nding stresses, strains, and displacements which are related through the equations of the preceding sections that satisfy the boundary conditions. For half-space problems, boundary conditions must be speci ed on the surface and the stresses must decay appropriately for large values of r in the half-space. The problem is well-posed (it has a unique solution) if a boundary condition is imposed in both the x and z directions at every location on the surface. Tractions can be speci ed, displacements can be speci ed, or one traction and one displacement can be speci ed for mixed conditions 2.4. 2.6 Cylindrical Coordinate Stress Distributions 2.6.1 The Hole in an In nite Sheet The discussion of contact to follow relies on familiarity with the stress components and how they vary from point to point when using cylindrical coordinates. It is instructive to solve some problems to illustrate stresses in cylindrical coordinates. A simple classical problem that serves this point is that of an in nite elastic containing a circular hole and loaded at in nity by a uniform tension (Figure 2.5). The procedure also provides an exercise for how candidate Airy stress functions are chosen for a particular problem. This solution also illustrates the rapid decay of stresses near a local disturbance which is a useful concept that will appear in contact problems. The boundary conditions required for the circular hole problem are that the surface of the hole is traction free and that a uniform uniaxial tension eld exists as r ! 1. Figure 2.4: Mixed boundary conditions. Figure 2.5: The hole in a uniform tension eld. These conditions are given in equation form as r r r r = = = = = 0 r=a 0< <2 0 r=a 0< <2 2 T sin r!1 T cos2 r!1 T sin cos r!1 (2.17) (2.18) (2.19) (2.20) (2.21) (2.22) The problem can be treated through the superposition of two problems as illustrated in Figure 2.5. One problem, far eld problem, is that of an in nite solid without a circular hole that is loaded by tension at in nity. The remaining problem, the local problem, is that of an in nite solid containing a circular hole that is not loaded at in nity. The circular hole is loaded by tractions that are opposite those calculated in the far eld problem so that the resulting tractions on the circular hole boundary are zero. Utilizing the half-angle trigonometric identities, the conditions at r ! 1 can be written as T (1 ; cos 2 ) r!1 (2.23) r = 2 T sin 2 r!1 (2.24) r = 2 r!1 (2.25) = T (1 + cos 2 ) 2 Consideration of the Table shows that these conditions are satis ed by the Airy stress function T 2 2 (2.26) F = (r + r cos 2 ) 2 where the subscript F denotes F as the far eld Airy stress function. This is the correct Airy stress function in cylindrical coordinates for the in nite solid without a hole loaded by tension at in nity. The disturbance due to the circular hole is a local one which can be treated by superposition. That is we want a local stress distribution corresponding to tractions applied to the circular hole given by T T r=a (2.27) r = ; sin 2 r = ; (1 ; cos 2 ) 2 2 In addition the stresses corresponding to the local problem should go to zero at in nity. Examining the Table for candidate solutions that give the right dependence for r and r leads to the following candidate Airy stress function ! T Aa2 log r + Ba2 cos 2 + Ca4 cos 2 (2.28) L= 4 r2 where A, B , and C are as yet unknown dimensionless constants and the subscript L refers to the local problem. Equation (2.27) can be used to solve for the constants by realizing that the equations must be satis ed for arbitrary . The resulting Airy stress function for a circular hole of radius a disturbing a uniform tension eld (Figure 2.5) is obtained by superposing F and L resulting in ! T r2 + r2 cos 2 ; 2a2 log r ; 2a2 cos 2 + a4 cos 2 =4 (2.29) r2 Consideration of the resulting stresses is interesting. First of all note that the normal stress acting parallel to the hole, (a ) is not zero. The general form of is ! T 1 + cos 2 + a2 + 3 a4 cos 2 (2.30) (r ) = 2 r2 r4 Along the surface of the hole it is given by (a ) = T (1 + 2 cos 2 ) (2.31) resulting in the famed stress concentration factor of 3 at = 0 and compression of T at = =2 3 =2. Also note the localized nature of the disturbance by considering as a function of r along the z-axis where = x and r = z ! a2 + 3 a4 (r 0) = T 1 + 2z2 2z4 (2.32) This decays from 3 at z = a to 1.22 at z = 2a. The e ect of the hole is localized! 2.6.2 Rigid Circular Inclusion The Airy stress function corresponding to the uniaxial tension eld in an in nite body disturbed by a rigid circular inclusion of radius a is given by ! T r2 + r2 cos 2 + 2(1 ; 2 )a2 log r + 2 a2 cos 2 ; a4 cos 2 (2.33) =4 3;4 3 ; 4 r2 The procedure for obtaining this solution is analogous to the circular hole problem with the traction boundary conditions at r = a replaced by ur = 0 u =0 r=a and the process is assigned as homework. 2.7 Closure Insert Dundurs table as the next 4 pages in your notes. (2.34) Chapter 3 Distributed Loading of a Plane Strain Half-Space In Chapter 2 the equations governing the deformation of an elastic body were given along with some sample Airy Stress Functions. In this Chapter solutions are given for particular loadings of the elastic half-space. These solutions will be used to nd the Hertz contact stress distribution in Chapter 4. 3.1 Normal Point Load Consider a half-space loaded by a normal point load of magnitude P per unit length on its surface at the origin (Figure 3.1). The solution to this problem is given by the Airy stress function = Ar sin where A is a constant to be determined. The stresses are given by (using di erentiation or Dundurs' Table) cos =0 (3.1) r = 2A r =0 r which satisfy the traction free conditions at the surface of the half-space. Notice that the stresses decay as 1=r as r approaches in nity and are singular or approach in nity as 1=r as r approaches zero. The constant A is determined by considering the equilibrium of a semi-circular disk cut from the half-space (Figure 3.1). The tractions on the surface of this disk are directed radially and are of magnitude r rd per unit thickness of the disk. The horizontal and vertical components of this traction are resolved by multiplying by sin and cos respectively. Horizontal equilibrium is identically satis ed as demonstrated in the following: Z Z Z ; ;2 r sin rd = ; ;2 2A cos sin rd = ; ;2 2Acos sin d = 0 r 2 2 2 17 P P ? ? -x ? z / r r Figure 3.1: Normal point load on a half-space and the disk used for the equilibrium calculation. Vertical equilibrium requires Z2 Z 2 cos Z2 ; ; r cos rd = ; ; 2A r cos rd = ; ; 2A cos2 d = P 2 2 2 which yields A = ;P Notice that A is independent of the size of the disk as must be the case. This is why the stresses had to behave as 1=r. Moment equilibrium of the disk is identically satis ed since the applied loading and all of the tractions pass through the origin. Substituting for A into Equations (3.1) gives 2P cos = ; P r sin =0 (3.2) r=; r =0 r Drawing a circle of radius d tangent to the surface at r = 0 leads to the relationship r = dcos on the boundary of the circle. This implies that on this circle the stresses are of constant magnitude with r = 2P= d. This relationship is illustrated beautifully in photoelastic fringe patterns. This fact will be used to solve the Brazilian disk problem later. Note that r is not perpendicular to this surface. Later we will calculate stresses and displacements due to distributed normal loads by superposing point loads. This superposition must be done in Cartesian coordinates. The equations of Chapter 2 can be used to write the stresses in Cartesian coordinates as 2P cos sin2 = ; 2P x2 z 2 x = r sin = ; r (x2 + z2 )2 z= z3 cos2 = ; 2P cos cos2 = ; 2P (x2 + z2 )2 r r 2 = ; r sin cos = 2P cos sin cos = ; 2P (x2 xz z2 )2 xz r + The surface displacements can be written as (see the Dundurs' Table and note that u = ;ur , for x < 0, u = +ur , for x > 0, w = ;u , for x < 0, and w = +u , for x > 0) )(1 u = ; (1 ; 22E + ) P sgn(x) w = ; 1 ;E 2P log jxj + C1 where it is assumed that the surface does not tilt and that the line of symmetry (x=0) does not move laterally. The overbar indicates a surface value so that u u(x 0). The function sgn(x) is called the sign or signal of x and it is -1 for x < 0 and +1 for x > 0 and log jxj is the natural logarithm of x. The tangential displacement is directed inward. The logarithm term causes some trouble because we cannot take the logarithm of a length. However, there is no length scale in the problem that we can use to make the argument of the logarithm dimensionless. Later, when we consider distributed loadings, we will be able to make the argument of the logarithm dimensionless. The logarithm term and the free vertical rigid body displacement are consequences of the two dimensional problems that we are considering. It will be shown that these issues can be circumvented by working with the gradient of the displacement. However, the two dimensional solutions will always have a rigid body displacement that cannot be speci ed unless problems with nite geometry are solved. 2 3.2 Tangential Point Load Consider a half-space loaded by a tangential point load of magnitude Q per unit length on its surface at the origin (Figure 3.2). The solution to this problem is given by the Airy stress function = ; Q r cos which can be veri ed using the procedure of the previous section. The stresses are 2Q sin =0 (3.3) r= r =0 r Again the equations of Chapter 2 can be used to write the stresses in Cartesian coordinates as 2Q sin sin2 = ; 2Q x3 2 x = r sin = r (x2 + z2 )2 z= 2 cos2 = 2Q sin cos2 = ; 2Q (x2 xz z2 )2 r r + Q - Q - -x ? z / r r Figure 3.2: Tangential point load on a half-space and semi-circle used for the equilibrium calculation 2 = ; r sin cos = ; 2Q sin sin cos = ; 2Q 2 x z 2 2 r (x + z ) The surface displacements can be written as (see the Dundurs' Table) 2 u = ; 1 ;E 2Q log jxj + C1 )(1 w = (1 ; 22E + ) Qsgn(x) xz 3.3 Distributed Tractions The stresses and surface displacements for distributed tractions can be found through the superposition of point loads (see Figure 3.3). For tractions distributed from b to a the stresses are given by 2z Z a p(s)(x ; s)2ds ; 2 Z a q(s)(x ; s)3 ds x = ; b (x ; s)2 + z 2 ]2 b (x ; s)2 + z 2 ]2 2 Z a p(s)ds ( )(x s 2z3 Z a ; 2z b (qxs; s);+ )zds]2 (3.4) z = ; 2 + z 2 ]2 2 2 b (x ; s) 2 Z a p(s)(x ; s)ds ; 2z Z a q(s)(x ; s)2 ds = ; 2z (3.5) xz b (x ; s)2 + z 2 ]2 b (x ; s)2 + z 2 ]2 These equations can be used to evaluate the full eld stress distribution for known distributions of p and q. However, the integrations may be di cult in closed form. For constant p and q, the integration can be done. One could then get the solution for traction distributions by replacing the distribution with an approximate piecewise ? z b a -x Figure 3.3: Distributed normal and shear tractions on a half-space. constant distribution (Figure 3.3). This procedure is an example of a large class of methods called the boundary integral method. The integration for the pressure and shear distributions could be performed on the Airy stress function itself. The stresses could then be calculated by di erentiation. When this sort of integration is done, the solutions for the point loads are called Green's functions. The point loads could also be superposed to develop the solution for a concentrated moment applied to the surface or a small region of concentrated pressure. This region of concentrated pressure is useful in the modeling of microindentation fracture of ceramic materials. The concentrated pressure is used to model the plastic deformation region directly underneath the apex of the indenter. The constant and the logarithm in the surface displacements can be removed by di erentiation. This results in @ u = ; (1 ; 2 )(1 + ) p(x) ; 2(1 ; 2 ) Z a q(s) ds @x E E b x;s @ w = ; 2(1 ; 2 ) Z a p(s) ds + (1 ; 2 )(1 + ) q(x) (3.6) @x E b x;s E Contact problems can now be solved by converting Equation (3.6) into an integral equation for the unknown surface pressure in terms of the undeformed shape of the bodies. That is the subject of Chapter 4. This technique will be used to show that the pressure distribution induced by contacting cylinders when aligned parallel to their axes is elliptical. Equations (3.4) cannot be integrated easily for this elliptical pressure. However, we can use some simple complex variable techniques to calculate the stress elds for this case. -x ? z -x ? z Figure 3.4: The constant pressure and shear patches 3.4 Constant Pressure and Traction Distribution Over a Patch Consider a patch of constant pressure, p, distributed from ;a to a (Figure 3.4). The integrations in Equations (3.4) and (3.6) can be conducted leading to p 2( ; ) + (sin 2 ; sin 2 )] z (x z ) = 1 2 1 2 2 p 2( ; ) ; (sin 2 ; sin 2 )] x (x z ) = 1 2 1 2 2 p (cos 2 ; cos 2 ) 1 2 xz = 2 8 > ;a x < ;a < )(1 u = ; (1 ; 2 E + )p > x ;a < x < a :a a<x 2 (3.7) w = ; (1 ;E )p f(x + a) log( x + a )2 ; (x ; a) log( x ; a )2g a a 2 where the substitutions r1 2 = (x a)2 + z2 , z = r1 2 sin 1 2 and x a = r1 2 cos 1 2 have been made and the overbar implies that the quantity is calculated on the surface. The constants of integration were used to impose that the origin does not translate or rotate. The distribution of w is given in Figure 3.5. This solution has the amazing property that the tangential stress at the surface is equal to the applied surface pressure. Since any distribution of surface pressure can be thought of as a distribution of constant pressure patches, this result is true for any 2D frictionless contact. We will see that this has tremendous implications to the failure mode of 2D contacts. The normal displacement of the surface corresponding to this contact is shown in Figure 3.5. A rigid indenter of this surface pro le would generate the constant pressure patch contact pressure. 2 Displacement (wE/p(1-ν )) 2.0 1.5 1.0 0.5 0.0 -0.5 -2.0 -1.0 0.0 Position (x/a) 1.0 2.0 Figure 3.5: The surface displacement caused by a patch of constant pressure. It is also constructive to consider the subsurface stress distribution due to this load. The stresses along the line of symmetry, x = 0, are shown in Figure 3.6. Note that x and z are equal at the surface. For ductile materials, the maximum shear stress leads to rst yield. Along x = 0, this is represented by max = 1=2( x ; z ) which is also shown. The rst yield occurs below the surface. For the tangentially loaded patch (Figure 3.4) q (cos 2 ; cos 2 ) z= 1 2 2 q 4 log r1 ; (cos 2 ; cos 2 )] x= 1 2 2 r2 q 2( ; ) ; (sin 2 ; sin 2 )] 1 2 1 2 xz = 2 2 u = ; (1 ;E )q f(x + a) log( x + a )2 ; (x ; a) log( x ; a )2g a a 8 x < ;a < (1 ; 2 )(1 + )p > ;a ;a < x < a w= (3.8) >x E :a a<x The tangential stress at the surface x(x 0) is sketched in Figure 3.7. This stress has logarithmic asymptotes at x = a. The high tensile stresses at the surface at x = ;a could cause cracking in brittle materials. This in nite stresses violate some of the assumptions of linear elasticity but we do learn from the approximate linear theory that these stresses are indeed high. In fact, the material will yield at the edge of the shear patch unless the shear can be relieved in some way. The shear must approach zero near the edge of the shear patch in a continuous way. We will keep this in mind later when we discuss friction laws. Stress (σ/p) -0.5 0.0 -1.0 0.5 σx 0.5 τmax 1.0 σz Depth (z/a) 1.5 2.0 2.5 3.0 Figure 3.6: Subsurface stress along x = 0 caused by a patch of constant pressure. Note that the maximum value of max = p= 0:318p at z = a. Stress (σx/q) 2.5 0.0 -2.5 -3.0 -2.0 -1.0 0.0 1.0 Position (x/a) 2.0 3.0 Figure 3.7: The tangential surface stress due to the constant shear patch. Chapter 4 Singular Integral Equations and Hertz Contact The solution of contact problems can be reduced to a Cauchy Singular Integral Equation of the rst kind. In this chapter solution techniques for this equation are described. The solution technique is then applied to contact problems. 4.1 Cauchy Singular Integral Equation and Solution The Cauchy Singular Integral Equation has the form Z a F (s)ds = g(x) ;a < x < a (4.1) ;a x ; s where g(x) is known and F (s) is unknown. Note that both the left and right-hand-sides of Equation (4.1) are functions of x and s is a dummy variable of integration. The equation holds for ;a < x < a, corresponding to the contact length, say, and F (x) is zero outside this region, as the contact pressure is zero outside of contact. The solution of Equation (4.1) for F (x) is written as Z a pa2 ; s2 g (s)ds 1p 1 + C] ;a < x < a (4.2) F (x) = ; 2 2 2 x;s a ; x ;a where C is a free constant. The free constant comes from there being a non-zero solution to the homogeneous form of Equation (4.1)(g(x) = 0). The approach used is that an equation similar to Equation (4.1) is derived for contact problems where the contact pressure plays the role of F and the shape of the bodies plays the role of g. Equation (4.2) is then written for the pressure in terms of the shape of the contacting bodies. Some integrals that are useful to this process are given next: Za ds p 2 2 = psgn(x)2 H (jxj ; a) I0 ;a (x ; s) a ; s x2 ; a 25 10 I0(x,a) 5 0 -5 -10 -3.0 -2.0 -1.0 0.0 1.0 Position (x/a) 2.0 3.0 Figure 4.1: Sketch of I0 (x a). Za sds p = I1 ;a (x ; s) a2 ; s2 Za ( Za p p ; ;a (x ;xs; sa)2ds s2 + x ;a (x ; s)dsa2 ; s2 = ; + xI0 ) ; Za s2 ds p I2 = xI1 = ; x + x2 I0 ;a (x ; s) a2 ; s2 Za s3ds p = ; 2 a2 + xI2 = ; 2 a2 ; x2 + x2I0 I3 ;a (x ; s) a2 ; s2 Integrals involving higher powers of s can be derived by repeatedly applying the above procedure. The integral I0 is sketched in Figure 4.1. Recall that for the normal loading of a half-space in plane strain (Figure 3.3), the normal displacement of the surface (w(x)) is related to the contact pressure (p(x)) through 2(1 ; 2 ) Z a p(s) log jx ; sjds (4.3) w(x) = C2 + C1 x ; w0(x) = ; E ;a where the contants C1 and C2 de ne the rigid body rotation and translation of the punch respectively. The logarithm term and the constant translation in Equation (4.3) are inconvenient. Therefore it is convenient to di erentiate for the displacement gradient as 2 Z a 0 C1 ; dw0(x) C1 ; w0 = ; 2(1 ; ) px(s)ds (4.4) dx E ;a ; s Comparing Equations (4.4) and (4.1) reveals that the contact pressure, p(s), plays the (C ; role of the function F (s) and that ; E2(11; w0)(x) plays the role of g(x). The constant C 2 in Equation (4.2) is used to satisfy force equilibrium while C1 is used to satisfy moment equilibrium. 0 4.2 Trigonometric Substitution An alternative scheme for solving Cauchy Singular Integral equations relies on a trigonometric change of variable in Equation (4.4). Let x = a cos s = a cos dx = ;a sin d ds = ;a sin d Equation (4.4) is written in the transformed variables as 2 Z 1 p( d C1 + a sin dw0 = ; 2(1 ; ) cos ) sincos d E 0 ; This substitution is useful due to the following identity. Z cos m d = ; sin m sin 0 cos ; cos Now, let 1 1 X pm cos m dw0( ) = X w sin m p( ) = m sin d m=0 (4.5) m=1 After some manipulation, Equation (4.5) can be written as 1 1 2 X X wm sin m + C1a sin = 2(1 ; )a pm sin m E m=1 m=1 (4.6) which can be solved for pm as E pm = 2(1Ewm2 )a ) m > 1 (4.7) p1 = 2(1 ; 2 )a (C1a + w1) m = 1 ; Equilibrium requires that p1 = 2Pd (4.8) p0 = Pa a2 where P is the normal load per unit length in the y;direction applied a distance d from the center of contact. Equation (4.8) xes C1. 4.3 Rigid Flat Frictionless Punch Consider the loading of the half-space by a rigid at frictionless punch by a force per unit length P (Figure 4.2). For the special case of d = 0, the surface remains at under the indenter so 0 w = z = constant and w0 0 Equation (4.4) is written as 2(1 ; 2) Z a p(s)ds = 0 ; E ;a x ; s ;a < x < a (4.9) P ? - x ? z Figure 4.2: Rigid at punch loading a half-space. Global equilibrium requires Za p(s)ds = P (4.10) Comparing Equations (4.9) and (4.1) yields p(x) = ; 12 p 2C 2 a ;x Equation (4.10) requires that C = ;P and p(x) = p P 2 a2 ; x The pressure distribution is sketched in Figure 4.3. The symmetric stress distribution corresponds to d = 0, since Za Pd xp(x)dx = 0 ;a ;a Alternatively, using the method of Section 4.2, wm = 0 m > 0 Now and p0 = Pa d = 0 ) C1 = 0 pm = 0 m > 0 p p(x) = sin0 = p P 2 a2 ; x which is the same result as that of the previous paragraph and again is shown in Figure 4.3. Pressure (pa/P) 4.0 3.0 2.0 1.0 -0.0 -3.0 -2.0 -1.0 0.0 1.0 Position (x/a) 2.0 3.0 Figure 4.3: Pressure distribution under rigid at punch. 4.4 Hertz Contact of Cylinders In this section the contact stress distribution for the frictionless contact of two long cylinders along a line parallel to their axes is derived (Figure 4.4). This is a special case of the contact of two ellipsoidal bodies rst solved by Hertz in 1881. The origin is placed at the point where the cylinders rst come into contact. At rst contact the separation of the cylinders is q q 2 2 w0 = z1 + z2 z1 = R1 ; R1 ; x2 z2 = R2 ; R2 ; x2 (4.11) If the contact length is small compared to the size of the cylinders, a Equation (4.11) can be approximated as R, then 1 x2 1 w0 = x2 ( 2R + 2R ) = 2R 1 2 where 1/R = 1/R1 + 1/R2. The loads cause the cylinders to approach each other a distance = 1 + 2 . The cylinders must deform to cancel the interpenetration. This is written in equation form as w1 + w2 = ; w0 or Di erentiating gives x2 w1 + w2 = ; 2R @ w1 + @ w2 = ; x @x @x R Figure 4.4: Long cylinders brought into contact by a load per unit length P. Substituting into Equation (4.4) gives Z a p(s)ds = 2E x R ;a x ; s ;a < x < a 2 2 where 1=E (1 ; 1 )=E1 + (1 ; 2 )=E2. This equation can be inverted for p(x) using Equation (4.2) yielding Z a spa2 ; s2 ds 1E p 1 p(x) = ; 2 R 2 2 + C] ;a < x < a a ; x ;a x ; s ;a < x < a p(x) = ; 21 E p 21 2 a2 I1 ; I3 + C ] R a ;x 2 ;a < x < a (4.12) p(x) = ; 21 E p 21 2 (x2 ; a ) + C ] R a ;x 2 The constant C is used to ensure that the pressure is continuous everywhere by making the coe cient of the radical in Equation (4.12) zero so that C = ; a2 =2. This results in E p p(x) = 2R a2 ; x2 This pressure distribution is shown in Figure 4.5. The remaining unknown is the contact length a which is used to ensure global equilibrium so that Za p(x)dx = P ;a Za E p a2 ; x2dx = P ;a 2R s E a2 = P ) a = 2 PR (4.13) 4R E Pressure (p/p0) 1.00 0.75 0.50 0.25 -0.00 -1.0 0.0 Position (x/a) 1.0 Figure 4.5: Hertz line contact pressure distribution. and the maximum contact pressure is pmax = p(0) s pmax = PE R (4.14) Again, the technique of Section 4.2 can be used noting that the separation can be written 2 2 a2 w = w1 + w2 = C0 ; 2x 2 = C0 ; a cos 2 ; 4R R 4R There is some confusion here caused by the subscripts on w referring to the surface displacements of each body and the subscripts on w referring to the terms in the Fourier series. Now a2 Ea w2 = ; 2R ) p2 = ; 4R(1 ; 2 ) p0 = Pa Substituting, P ; Ea cos 2 (1; 2 p( ) = a 4Rsin ) (4.15) The stress must be continuous at the edge of contact, x = a, at which points = 0 and sin = 0. Thus the numerator in Equation (4.15) must be zero at = 0 leading to P = Ea a 4R(1 ; 2 ) Solving for a and substituting into Equation (4.15) leads to p( ) = Pa 1 ; cos 2 sin which can be written in terms of x as Pp p(x) = 2a2 a2 ; x2 which is identical to the result of the previous paragraph. It is interesting to note that the maximum pressure distribution varies as the square root of the load rather than linearly with the load. This is because the contact length increases with the applied load resulting in an increased area that bears the load. The most important thing to remember is that the Hertz contact problem is nonlinear. It would be interesting at this stage to substitute the contact pressure into Equations 3.4 and perform the resulting integrations for the stress distribution. Unfortunately, the integrations are di cult. In fact, the performance of these integrations has been the subject of recent publications including Sack eld and Hills (1983a, 1983b, 1983c). The symmetry of the problem requires that initial yielding occurs along the z-axis. Since xz = 0 along the z-axis the stresses x and z are principal stresses and the maximum shear stress is one-half of their di erence. These stresses along x = 0 are, see Johnson (1985) page 102, a pmax p2 + 2z2 ; 2z] x (0 z ) = ; a a2 + z2 pmaxa z (0 z ) = ; p 2 a + z2 1( ; ) max 2 x z z2 ; z] (4.16) max = ;pmax p 2 a a + z2 The maximum shear stress occurs below the surface at a depth of z 0:78a and has a value of max = 0:3pmax . These stress distributions are shown in Figure 4.6 4.5 The Westergaard Stress Functions The Westergaard stress functions can be used to get full eld representations of the Hertz contact stress eld. The Westergaard stress functions are outlined in this section. The Westergaard method yields a succinct solution to linear elastic contact crack problems. Sun and Farris (1989) have shown that the superposition of the Westergaard Functions with a uniform uniaxial stress eld provides a complete solution to the equations of two-dimensional linear elasticity for the in nite plane. The following is a transcription of their presentation. 4.5.1 Introduction to Westergaard Stress Functions The Westergaard method (Westergaard, 1939) provides a powerful technique for solving the problem of an in nite linear elastic plane containing a crack or array of cracks or contact of half planes. It can be used to succinctly write the stress and displacement -1.0 Stress (σ/p0) -0.5 0.0 0.5 σx 0.5 τmax 1.0 σz Depth (z/a) 1.5 2.0 2.5 3.0 Figure 4.6: Stress below the surface for Hertz contact. elds throughout the body. This property makes the Westergaard method particularly attractive for use as a teaching tool in fracture mechanics and contact mechanics courses. It has been suggested that the Westergaard method solves only a restricted class of crack problems (Sih, 1966) and that this restriction (Eftis and Liebowitz, 1972) is due to an oversight (MacGregor, 1935) on which Westergaard based his formulation. Here it is shown that the removal of this oversight leads only to a uniform uniaxial state of stress. Therefore, the superposition of the Westergaard Functions with a stress function that produces a uniform uniaxial stress provides a complete solution to the equations of elasticity. 4.5.2 Formulation of the Westergaard Stress Functions The problems under investigation involve two-dimensional linear isotropic elasticity. Figure 4.7 show the coordinate system that is used for the Westergaard stress functions. The stresses and displacements are written in terms of the two Kolosov-Muskhelishvili (Muskhelishvili, 1954) potential functions as 0 x + y = 4Re (z )] 00 0 (4.17) y ; x + 2i xy = 2 z (z ) + (z )] 0 (z ) ; (z ) 2G(u + iv) = (z) ; z where = 3 ; 4 for plane strain and (3 ; )=(1 + ) for plane stress, is Poisson's p ratio, and G is the shear modulus. Also z = x + iy where i = ;1, the prime denotes Figure 4.7: Coordinate system for Westergaard stress functions. di erentiation with respect to z, and the bar denotes complex conjugation. Normally is used rather than but we have reserved for the Airy stress function. The Airy stress function is written in terms of the Kolosov-Muskhelishvili potential functions as (x y) = Re z (z) + ^(z)] (4.18) where d ^=dz . The problem geometry is an in nite plane containing a straight crack or a system of straight cracks along the x-axis or a half-plane subjected to surface loading. The loading consists of tractions applied to the crack faces, applied stresses at in nity, or loads applied in the plane which can be treated by superposition. Symmetric and skew-symmetric loadings (normal pressure and shear tractions for contact problems) are treated separately and the total loading is found by superposition of the two solutions. 4.5.3 Symmetric Loading (Mode I) If the loading is distributed symmetrically with respect to the y-axis or there is normal pressure only for the contact problem, then xy = 0 along y = 0. Examination of Equation 4.17 shows that the zero shear condition is satis ed by Im z 00 (z) + 0 (z)] = 0 y=0 (4.19) Theorem 4.1 Theorem: The most general solution to Equation (4.19) is given by 0 (z ) + z 00 (z ) + A = 0 where A is a real constant. Proof: Consider (4.20) 0 (z ) + z 00 (z ) = ;A(z ) (4.21) where A(z) is analytic and bounded for y 6= 0 (since 0 (z ) + z 00 (z ) is analytic and bounded for y 6= 0). Substituting Equation (4.21 into Equation (4.19) yields Im A(z)] = 0 y = 0: (4.22) Let A(z) = a1 (x y) + ia2 (x y). Since r2 a2 = 0, a2 (x 0) = 0, and a2 is bounded elsewhere, a2 0 by Liouville's Theorem (Churchill, Brown, and Verhey, 1976). The Cauchy-Riemann conditions (Churchill, Brown, and Verhey, 1976) in turn yield a1 = constant. Therefore, A(z) = real constant. Equation (4.20) allows the stresses and displacements to be written in terms of one potential function and a real constant. This reduced form appropriate for symmetric problems is written as = = = 2Gu = 2Gv = (x y) = x y xy 2Re 0] ; 2yIm 00 ] + A 2Re 0] + 2yIm 00] ; A ;2yRe 00 ] ( ; 1)Re ] ; 2yIm 0] + Ax ( + 1)Im ] ; 2yRe 0 ] ; Ay Re 2^] + yIm 2 ] ; A Re z2 ] 2 (4.23) where d ^=dz . Next, the stresses, displacements, and Airy stress function are written in terms of the Westergaard function ZI (z) by substituting 0 (z ) = 1 Z (z ) + A] (4.24) 2 I ^ and de ning dZI =dz ZI . These de nitions yield = Re ZI ] ; yIm ZI0 ] + 2A = Re ZI ] + yIm ZI0 ] = ;yRe ZI0 ] ^ 2Gu = ; 1 Re ZI ] ; yIm ZI ] + 2 ^ 2Gv = + 1 Im ZI ] ; yRe ZI ] + 2 ^ ^ (x y) = Re Z I ] + yIm ZI ] + Ay2 x y xy + 1 Ax 2 ; 3 Ay 2 (4.25) Equations (4.25 correspond to Westergaard's Eqnuations (4-6) and (9-10) (Westergaard, 1939) if A is taken as zero. Since A is constant the Westergaard method for symmetric loadings is complete with the exception of a constant uniaxial stress x = 2A which acts parallel to the line of the crack or parallel to the surface in contact problems. The stress eld caused by the constant A does not contain any singularities. The appropriate choice for A can be found by recognizing that 1 A = 2 x(x 0) ; y (x 0)] (4.26) In particular, if the far- eld loading is biaxial tension, A = 0. Sanford (1979) pointed out that for nite bodies A could be replaced with a power series. The power series can be used in a collocation technique to solve plane nite body crack and contact problems. 4.5.4 Skew-Symmetric Loading Mode II] For loadings distributed skew-symmetrically to the y-axis (or shear contact tractions alone) y = 0 along y = 0. Equations (4.17) are used to write this condition as Re 2 0(z) + z 00 (z) + 0(z)] = 0 y=0 (4.27) A procedure analogous to that of the previous section shows that the most general solution to Equation (4.27) is given by 0 (z ) + 2 0 (z ) + z 00 (z ) + iB =0 (4.28) where B is a real constant. Again the stresses and displacements can be written in terms of one potential function and a real constant. The stresses and displacements are put into the Westergaard form by de ning ZII = 2i 0 ; B (4.29) ^ and ZII =dz ZII . These de nitions give 0 = 2Im ZII ] + yRe ZII ] 0 = ;yRe ZII ] 0 = Re ZII ] ; yIm ZII ] ^ 2Gu = + 1 Im ZII ] + yRe ZII ] + + 1 By 2 2 ; 1 Re Z ] ; yIm Z ] ; + 1 Bx ^II 2Gv = ; II 2 2 ^II ] (x y) = ;yRe Z x y xy (4.30) The terms due to B represent a rigid body rotation. If the rigid body rotation is eliminated, Equations (4.30) correspond to the stresses and displacements due to the skew-symmetric Westergaard function. 4.5.5 Superposition and the Airy Stress Function The solution to the plane elasticity problem of an in nite sheet containing a straight crack or array of straight cracks along the x-axis (or contact of a half-plane whose boundary lies along the x-axis) is given completely by ^ ^ ^ = Re Z I ] + yIm ZI ] ; yRe ZII ] + Ay2 (4.31) The functions ZI and ZII were rst given by Westergaard and the term Ay2 induces a uniform uniaxial stress. Previously, de Wit (1977) added additional terms to the Airy stress function of Westergaard, Equation (4.31), to construct the uniaxial tension eld. However, some of de Wit's terms are unnecessary and he does not show that the additional terms make the Westergaard formulation complete. The inclusion of the constants A in Equation (4.24) and B in Equation (4.28) makes the Airy stress function, Equation (4.31), identical to Westergaard's Airy Stress Function with the superposition of a uniaxial tension eld. The Westergaard method provides the complete stress and displacement elds for in nite plane crack problems. The method as introduced by Westergaard (1939) omits only a simple uniform stress and should not be perceived as restrictive. 4.5.6 Subsurface Hertz Contact Stress The Westergaard stress functions can be used in conjunction with a simple FORTRAN program to evaluate the subsurface stress eld induced by Hertz contact. For frictionless Hertz contact, the appropriate Westergaard stress function is P p ZI = ; 2a2 ( a2 ; z2 + iz) (4.32) where P is the contact force per unit length and a is the half contact length. Equations (4.25) can be used to verify that the boundary conditions along y = 0 are satis ed. The branch cut on the radical in Equation (4.32) is chosen so that ZI ! 0 as z ! 0. Contours of the in-plane maximum shear stress s x; y 2 2 + xy (4.33) = max 2 are shown in Figure 4.8. The maximum in-plane shear stress is about 0:3p0 where p0 is the maximum contact pressure and it occurs on the y-axis at a depth of about y 0:78a. If the contacting cylinders are loaded tangentially as well as normally and caused to slide over each other then a shear stress will exist at the surface. This shear stress is equal in magnitude to the coe cient of friction multiplied by the normal contact pressure. The shear stress acts to oppose the tangential motion of each cylinder. Thus if the top cylinder moves from the left to the right relative to the bottom cylinder then the tangential traction on the bottom cylinder is given by Pp (4.34) q(x) = 2a2 a2 ; x2 0.0 0.0 0.5 2 4 Position (x/a) 1.0 1.5 1 3 6 5 2 1 7 0.5 3 8 6 5 1.0 Level tauma 4 9 8 7 6 5 4 0.299 0.265363 0.231725 0.198088 0.16445 0.130812 3 2 1 4 7 Depth (y/a) 2.0 0.097175 0.063537 0.0299 7 1.5 6 2.0 Figure 4.8: Stress contours of 5 4 max =pmax for Hertz contact The Westergaard stress function that yields these surface tractions is P p ZII = ; 2a2 ( a2 ; z2 + iz) (4.35) as can be veri ed using Equations (4.30). Contours of the in-plane maximum shear stress for the combined shear and normal tractions are shown in Figures 4.9 and 4.10 for = 0:1 and = 0:4 respectively. The indenter is sliding over the surface from left to right. Notice that increasing the coe cient of friction increases the maximum shear stress while changing its location to be nearer the surface and o of the y-axis towards the leading edge of contact. For the higher value of , max occurs on the surface. Position (x/a) -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 5 4 0.0 5 6 4 2 Level tauma 7 6 Depth (y/a) 0.5 2 9 3 8 7 6 5 5 4 9 6 8 1.0 4 8 5 7 7 1.5 6 6 2.0 0.301 0.267278 0.233555 0.199833 0.16611 0.132388 3 2 1 8 3 0.098665 0.064942 0.031220 5 2.5 4 5 4 3.0 Figure 4.9: Stress contours of max =pmax for frictional Hertz contact with = 0:1, the load is sliding from left to right. Position (x/a) -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 9 5 0.0 8 Level tauma 7 34 0.5 Depth (y/a) 9 6 2 1 1.0 8 5 7 1.5 3 2.0 6 4 7 9 8 6 7 6 0.360543 0.321166 0.281789 0.242412 5 4 5 3 2 1 0.203034 0.163657 0.12428 0.084903 0.045525 2 5 4 2.5 4 3.0 3 Figure 4.10: Stress contours of max =pmax for frictional Hertz contact with = 0:4, the load is sliding from left to right. Chapter 5 The Boundary Element Method The boundary element method is a general numerical scheme for the solution of partial di erential equations such as the equations of elasticity. The boundary element approach 9] requires only that the boundary of the body to be modeled be discretized. On the other hand, the nite element method, which now must be considered the more traditional technique for engineering analysis, requires that the complete domain be discretized. This is ine cient when considering problems like contact in which the region of interest is much smaller than the overall body. The boundary element approach actually uses this to its advantage by naturally calculating temperatures, stresses, and displacements on the boundary. I am interested in how well the boundary element method really works. One way to nd out is by trying it out on some problems of interest. You are given access to a boundary element program (bemprg) that my student Scott Hucker is developing which solves both two-dimensional elasticity and steady state heat conduction problems. The operation of the program is described in another handout. Operating bemprg as a black box would be a learning experience because you can get numbers for problems that are more di cult than those solved on the blackboard in class. However, knowing something about what is in the black box is very important to you as students. The next section discusses some of the insides of the black box for steady state heat conduction problems. 5.1 Steady State Heat Conduction Let T (x y) (we used f in class) be the steady-state temperature distribution in the body with the boundary ;. The Fourier Law of heat conduction and conservation of energy can be used to show that r2T (x) = 0 in (5.1) where T is the unknown, it may represent the steady-state temperature distribution, x is the vector representing position, x = (x y), represents the domain of interest (Figure 5.1, and the Laplacian operator r2 @ 2 =@x2 + @ 2 =@y2. Equation (5.1) can 41 Figure 5.1: Geometry used for boundary element formulation have essential boundary conditions T =T on ;1 (5.2) and natural boundary conditions q = k @T = q on ;2 (5.3) @n where n is the normal to the boundary ; = ;1 + ;2 , the bars indicate known values, and k is the thermal conductivity. It is also possible to have mixed conditions which involve combinations of T and q such as convective heat conduction. The rst step in the boundary element procedure is to multiply Equation 5.1 by an as yet arbitrary weighting function T and integrate over the body. For now we just require that T be smooth enough that its second partial derivatives have meaning. Equation 5.1 now becomes R 2 (r T )T d = 0 (5.4) Now we use Green's Theorem (my famous Japanese professor, Toshio Mura, always claims that this is the only math that he knows) R 2 R R (5.5) (r T )T d = rT rT d + ; @T T d; @n which reduces Equation 5.4 R R (rT rT )d + ; @T T = 0 @n Now use Equation 5.5 with T replaced by T and vice versa to yield R R R T (r2T )d + ; @T T d; ; ; @T Td; = 0 (5.6) @n @n Figure 5.2: The delta function Remember that the boundary value problem de ned by Equations 5.1-5.3 is fully contained in this equation. The next question is the following: What is a convenient choice for T ? The weight function T is chosen as the solution to the fundamental equation (for the in nite domain). The fundamental equation is written as r2T (x ) + (x ) = 0 (5.7) where is the Dirac delta function representing a source of heat at the source point . The delta function is plotted schematically in Figure 5.2. Its property of importance to us is its sifting property described by Zb f (x) (x ; )dx = f ( ) a< <b a The fundamental solution to Equation (5.7) is T = ; 21 log r (5.8) where r is the distance from the source point to the observation point x (r2 = (x ; )2 + (y ; )2). This is shown by noting that r2T is zero every except where r = 0 and that Z Z @T 1 Z rd = ;1 2 r T d = ; @n = ; 2 ; r Note that , ;, , and ; are de ned in Figure 5.1. Substitution of Equation 5.7 into Equation 5.6 leads to 1Z 1Z ciT (x) + k T (x)q (x )d; = k T (x )q(x)d; (5.9) ; ; where ci is a constant equal to 1/2 for a smooth boundary and 1 in the interior. Figure 5.3: Geometry used for constant boundary elements It is important to note that Equation (5.9) is an exact representation of Equation (5.1) and the accompanying boundary conditions so that no approximation has been made to this point. The approximation in the Boundary Element Method begins with the discretization of Equation (5.9) used for numerical integration. 5.2 Constant Boundary Elements The boundary of the domain is discretized which allows the integrations in Equation (5.9) to be reduced to sums. Constant boundary elements approximate T and q as piecewise constant on the boundaries. The boundary is broken into N nodal points or nodes which are joined by N elements (Figure 5.3). Equation 5.9 is now approximated on the boundary by N N 1 T (x) + X 1 Z T (x)q (x )d; = X 1 Z T (x )q (x)d; (5.10) j j ij ij 2 i j =1 k ;j j =1 k ;j where the ;j refer to integration over each element and the subscripts ij refer to the fundamental solution being proportional to the logarithm of the distance from the source point to the observation point. Since the unknowns (T and q) are constant in each element the integrals in Equation 5.10 just depend on the fundamental solution. Equation 5.10 is written in matrix form by de ning the components of matrices as 1 Z q d; Hij = k ij ;j 1 Z T d; Gij = k ij ;j resulting in Iij ] Tj ] + Hij ] Tj ] = Gij ] qj ] (5.11) where Iij ] is the identity matrix. Equation (5.11) is rearranged placing the unknown boundary values on the lefthand-side of the equation and the known boundary values (Equations 5.2- 5.3) on the right. The rearrangement results in a system of algebraic equations for the unknown boundary values. The diagonal terms in the resulting matrix equation are evaluated exactly by integrating the fundamental solution in closed form while the o -diagonal terms are evaluated numerically using Gaussian quadrature. The closed form integrations are given as Hii = 0 L Gii = 2 (1 ; log L ) 2 where L is the length of the element. Once the equations are inverted, the temperature can be evaluated anywhere in the body using Equation (5.9) by making ci equal to one and choosing x to be away from the boundary. However, it should be noted that the boundary element method is uniquely suited to contact problems as it solves for boundary values which are normally of the greatest interest. 5.3 Elasticity and Thermoelasticity Elasticity problems can be handled in much the same way by using Navier's equations for equilibrium rather than the Laplace equation. In two dimensions, Navier's equations are two vector equations which are elliptic leading to the same type of singularity in the fundamental solution as that of Equation 5.8. The details of the fundamental solution and the boundary element equation corresponding to Equation (5.9) are slightly more lengthy than those given here, but all of the principals are the same. Thermoelasticity problems are then handled by treating the thermal strains as body forces, transformed to the surface using the divergence theorem, in the elasticity equations. Here the transformation of the body forces to the surface allows the numerical work to involve only discretization of the boundary. Some of these features have been programmed and are available to you in bemprg. Other features will be available soon. The details of the development of bemprg will have to wait for next year's handouts. 5.4 The Importance of Transient Thermal Behavior This section partially describes some of my present research and the potential role of transient boundary elements. The speci c problems of interest are those of high speed sliding contact. In particular, the program will be used to predict failure modes in brittle ceramic materials in high speed sliding applications. One long range goal of the research is the determination of whether ceramics are appropriate for applications in which high speed sliding is important. Also, the research is one step towards the prediction of damage induced by nishing of ceramics 10]. There is increasing interest in using structural ceramics such as partially stabilized zirconia and silicon nitride in structural applications. In many of these applications, such as bearings and the piston/cylinder system in engines, the possibility for high speed sliding contact exists. Ceramics show particular promise in high speed sliding applications due to their stability, high hardness, and strength at elevated temperatures. There is also signi cant use of ceramics in electronic applications. Nickel-zinc and Manganese-zinc ferrites are widely used as recording head materials in magnetic storage systems. This application requires extremely good control of the geometry, surface nish, and integrity (residual stress, cracking, and alteration of surface structure and chemistry) of the nished surface. Diamond grinding is an important manufacturing process used in the nishing of these ceramics to meet surface nish requirements. However, the grinding process has a signi cant in uence on the quality of the nished work surface 11]. A lapping machine that will be used to develop nishing schemes to produced ultra-smooth surfaces will be obtained in the near future. The use of structural ceramics in high speed sliding contact applications and the understanding of nishing of ceramics requires that the thermomechanical phenomena of high speed sliding contacts be studied. The proposed research will characterize some thermomechanical e ects at high speed sliding interfaces. These thermal e ects can be broadly classi ed into three categories: 1. The high temperatures generated in the surface/subsurface of the members as a consequence of frictional heating 2. Thermal stresses in the contacting solids, both transient and residual, induced by the frictional heating process and 3. The corresponding wear and failure modes. The high temperatures and stresses generated at high speed sliding contacts play an important role in in uencing the very friction and wear processes out of which they arise. They signi cantly alter the mechanical, magnetic, and electrical properties of the contacting surfaces 12, 13, 14, 15, 16, 17]. For instance, in some zirconium oxide ceramics, the temperatures and stresses generated can cause an improvement in the fracture toughness of the ceramic. Thermally induced stresses are particularly important in dictating crack nucleation and propagation and plastic deformation processes in ceramics 18], thereby controlling the wear and fracture resistance of the component. It is therefore critical that these e ects be well understood and quanti ed so that the tribological performance of these ceramics and the property changes occurring in them can be predicted and controlled. There have been several studies pertaining to the analysis and measurement of interface temperatures in sliding contact 19, 20, 21, 22, 23, 24, 25, 26]. Foremost amongst these are the classical papers of Blok 19], Jaeger 20], and Archard 21]. In recent years with the development of high frequency response infrared sensors, radiation measurement techniques have emerged as a powerful tool for estimating sliding contact temperatures 23, 24, 25, 26]. However, microindentation contacts, which have been studied extensively under static loading conditions, have been neglected in most of the high speed sliding temperature studies. Reliable temperature measurements, in addition to providing insight into various physical and chemical processes that can occur at sliding interfaces, provide a means for veri cation of contact temperature prediction models. The proposed research would be combined with the ongoing temperature measurements to enable the analysis of thermal stresses developed at these contacts. There has been quite a bit of research directed at the calculation of thermal and mechanical stresses induced by high speed sliding contact. In particular, see Bryant 27] and Ling 28], who describe in detail some of the past work in the eld. These techniques usually involve integral transforms which lead to integral equations which must be inverted numerically or Green's functions which must be integrated numerically to account for the distributed heat source. The author used the Green's function for the insulated thermoelastic half-space subjected to a uniformly moving heat source to calculate temperatures and stresses related to single point grinding 25, 29]. The solution showed remarkable agreement with measured temperatures. However, this type of analysis is quite ad hoc in the sense that one must start from the beginning with each new problem. The mathematical details of the transforms and the numerical details of performing the inverse transforms can be quite lengthy. On the other hand, a boundary element approach would lead to an accurate and general numerical technique for such problems. 5.5 Transient Boundary Elements The transient problems will be attacked using the fast Fourier transform which has been applied successfully to structural dynamics 30]. For example, consider the transient thermal problem where r2T = 1 @T in (5.12) @t where is the thermal di usivity, t is time and T = T (x t). Again there may be both essential boundary conditions (Equation 5.2) and natural boundary conditions (Equation 5.3) and there is also an initial condition which may be that at the initial time the body is at zero temperature, T (x 0) = 0 (5.13) The transient problem, Equation (5.12), will be transformed into a series of pseudosteady state problems using the spectral representation X^ T (x t) = Tn(x !)ei!t (5.14) n Equation (5.12) is now reduced to ^ ^ r2Tn ; i! Tn = 0 (5.15) The fundamental form of Equation (5.15) is ^ ^ r2 Tn ; i! Tn + (x ) = 0 whose solution is (5.16) r ^n = 1 K0( i ! r) (5.17) T 2 where K0 is the modi ed Bessel function of the second kind. The weighted residual technique can now be applied to Equation (5.15) to yield the boundary element equation Z Z ^n(x) + 1 Tn(x)^n(x )d; = 1 Tn (x )^n(x)d; ^ q ^ ciT q k ; k ; which is the same as the boundary element equation for the steady-state problem ^ (Equation (5.9)) remembering that there is a new fundamental solution Tn . The transient solution is performed by solving the boundary element equations at each frequency and using the fast Fourier transform to convert to the time domain (Equation 5.14). A computer program will be written that uses these techniques to solve for the temperatures and stresses in high speed sliding contact. The temperature calculations can be compared to temperatures obtained obtained experimentally using infrared sensors 25, 29]. The stress calculations can then be used to make failure predictions for brittle ceramics used in high speed sliding contacts. Chapter 6 Three-Dimensional Elasticity and Hertz Contact In this chapter the equations for three-dimensional or point contacts are derived. The approach taken is very similar to that used in two dimensional problems where the point force is superposed to yield the distributed load solutions. The point load solution is derived using Love's axisymmetric stress function. Much of the information in this handout can be found in various forms in Chapter 12 of Timoshenko and Goodier (1970) and Chapters 3 and 4 of Johnson (1985) 6.1 Love's Axisymmetric Stress Function The coordinate system used is shown in Figure 6.1. The z-axis points down, r is the projection of the distance to a point on the x ; y plane, and R is the distance to a point so that R2 = r2 + z2 . The angle is related to r R, and z through cos = z=R and sin = r=R. The displacement in the r direction is u and the displacement in the z direction is w. The stress components are de ned in Figure 6.1. Equilibrium in the r and z requires that @ r + @ rz + r ; = 0 @r @z r @ rz + @ z + rz = 0 (6.1) @r @z r These stresses can be calculated from Love's axisymmetric stress function using @ ( r2 ; @ 2 ) r = @z @r2 @ = @z ( r2 ; 1 @ ) r @r @ ((2 ; )r2 ; @ 2 ) z = @z @z2 @ ((1 ; )r2 ; @ 2 ) (6.2) rz = @r @z2 49 Figure 6.1: Three-dimensional coordinate system. It can be shown by substitution that these stresses satisfy equilibrium provided r4 = 0 where r4 r2 r2 and r2 @ 2 =@r2 + 1=r@=@r + @ 2 =@z2 . The strain-displacement relations are @u @w =u r= z= @r r @z @u + @w (6.3) rz = @z @r Now Hooke's Law and integration can be used to show that 2 @2 2Gw = 2(1 ; )r2 ; @ 2 2Gu = ; @r@z @z Since the displacements can be calculated, compatibility is satis ed. The solution of axisymmetric elasticity problems is now reduced to nding the Love's stress function that satis es all of the boundary conditions. 6.2 The Boussinesq Solution In this section the derivation of the stresses and displacements for the normal point loading of an elastic half-space will be given. This is known as the Boussinesq solution. The procedure superposes the fundamental Love's stress function in a few di erent ways. 6.2.1 The Point Load in an In nite Solid Next the solution will be given for the point force in an in nite solid (Figure 6.2). It can be shown by di erentiation that the Love's stress function = B (r2 + z2 )1=2 is biharmonic and that the components of stress are given by 2 = B (1 ; 22 )3z 2 ; 2 3r z 5=2 ] r (r2 + z ) = (r + z2 ) = B (1 ; 22 )3z 2 (r2 + z ) = (1 ; 2 )z + 3z3 ] z = ;B 2 (r + z2 )3=2 (r2 + z2 )5=2 (1 ; 2 )r + 3rz2 ] (6.4) rz = ;B 2 (r + z2 )3=2 (r2 + z2 )5=2 These stresses behave as 1=R2 and the stress function is a candidate for the point force stress function. Consider the force due to the stresses on a spherical hole cut-out around the location of p the force. The z-component of this force on an annular element of area dA r2 + z2 d 2 r is p 2 2 dFz = ;( z cosp + rz sin )dA Using cos = z= r + z and sin = r= r2 + z2 one can show that the total force on the sphere is Z =2 P = ;2 dFz = 8 (1 ; 2 )B ! B = 8 (1P 2 ) ; 0 Along z=0 the normal traction is zero ( z (r 0) = 0) but the shear traction is given by P 1;2 1 (6.5) rz (r 0) = ; 8 1 ; r2 which is inversely proportional to the square of the distance from the load. We must come up with a solution to cancel these tractions before we have the solution for the point force on the half space. 6.2.2 The Doublet Operation and the Center of Compression The doublet is formed by superposing two point forces (Figure 6.3). The forces are brought together with a separation of the distance d. The distance d is made to approach zero while the magnitude of the force P approaches in nity such the strength of the doublet given by A = Bd is nite. The stresses are then given by 2 @ = ;A @z (1 ; 22 )3z 2 ; 2 3r z 5=2 ] r (r2 + z ) = (r + z2 ) = ;A @ (1 ; 22 )3z 2 @z (r2 + z ) = @ (1 ; 2 )z + 3z3 ] z = A @z (r2 + z2 )3=2 (r2 + z2 )5=2 @ (1 ; 2 )r + 3rz2 ] (6.6) rz = A @z (r2 + z2 )3=2 (r2 + z2 )5=2 Figure 6.2: The point load in an in nite body. Figure 6.3: The three-dimensional force doublet. These stresses are now written in spherical coordinates (Figure 6.4) using the transformation equations 2 2 R = r sin + z cos + 2 rz sin cos 2 2 R = ( r ; z ) sin cos ; rz (sin ; cos ) r z sin = p 2r 2 = R cos = p 2z 2 = R r +z r +z Carrying the substitutions through leads to 2(1 + )A sin2 + 2(2 ; ) cos2 ] R=; R3 1+ 2(1 + )A sin cos R =; R3 Figure 6.4: Angle used to transform the doublet stress into spherical coordinates. The solution for the center of compression (Figure 6.4) is written by superposing the above doublet with one directed along r by setting = ; =2 and one directed out of the page by setting = =2. The resulting stresses are R 2 = ; 4(1 ; 3 )A R R = 0: The solution for a spherical body containing a spherical hole subjected to internal and external pressure (Figure 6.5)can now be obtained. Equilibrium in the R direction can be written (noting that the two hoop stresses are equal) as 31] d R + 2 ; 2 =0 dR R R R t Rd R + t= R 2 dR Taking the radial stress as the superposition of the center of pressure with a uniform radial stress such that C R = 2 +D R and solving for C and D by using the pressure boundary conditions leads to pob3 (R3 ; a3) + pia3 (b3 ; R3) R= R3(a3 ; b3 ) R3(a3 ; b3 ) pob3 (2R3 + a3) ; pia3 (2R3 + b3 ) t= 2R3(a3 ; b3 ) 2R3(a3 ; b3 ) (6.7) Figure 6.5: Hollow sphere subjected to internal and external pressure. 6.2.3 Superposition of the Center of Compression and the Point Load The spherical components of stress due to the center of compression ( R = A=R3 ;A=(2R3 )) can now be transformed into cylindrical coordinates using t = 2 2 + t cos2 = A r ; z =2 R5 z 2 ; r 2 =2 2 2 z = R cos + t sin = A R5 1 3A rz rz = ( R ; t ) sin 2 = 2 2 R5 = t = ; A3 2R Now the center of compression is distributed over the negative z-axis from z = 0 to z = ;1 with a strength A1 leading to Z0 r2 ; (z ; )2=2 d = Z 1 A r2 ; u2=2 du A1 r2 + (z ; )2]5=2 1 2 r= (r + u2)5=2 ;1 z A1 ( 1 ; z ; z ) r= 2 r 2 r 2 R R3 A1 z A1 r (6.8) z= rz = 3 2R 2 R3 = ; A1 ( 12 ; 2z ) 2 r rR Notice that the shear stress on z = 0 behave as 1=r2 which is the same behavior exhibited by the shear stress due to the point load in the in nite solid (Equation 6.5). r = R sin 2 Therefore, the solution for the point load on the half-space is given by Equations 6.4 added to Equations 6.8 with A1 = 2B (1 ; 2 ) and this solution is summarized in the next subsection. 6.2.4 Normal Point Load on a Half-Space The displacements and stresses for the normal point loading of an elastic half-space is are known as Boussinesq's solution. The Cartesian components of displacement and stress are xz u = 4 PG R3 ; (1 ; 2 ) R(Rx+ z) ] yz ) v = 4 PG R3 ; (1(; 2 zy ] R R+ ) z2 ; w = 4 PG R3 + 2(1R ) ] P f 1 ; 2 (1 ; z ) x2 ; y2 + zy2 ] ; 3zx2 g x= 2 r2 R r2 R3 R5 P f 1 ; 2 (1 ; z ) y2 ; x2 + zx2 ] ; 3zy2 g y= 2 r2 R r2 R3 R5 z3 = ; 3P R5 z 2 P f 1 ; 2 (1 ; z ) xy ; xyz ] ; 3 xyz g xy = 2 r2 R r 2 R3 R5 3P xz2 3P yz2 xz = ; yz = ; 2 R5 2 R5 The cylindrical components for the Boussinesq solution are given as rz ; ur = 4 PG R3 ; (1 ; 2 ) RRr z ] z2 ; uz = 4 PG R3 + 2(1R ) ] P z zr2 = ; 2 1 ;22 (1 ; R ) ; 3R5 ] r r = ; P (1 ; 2 ) 12 (1 ; z ) z3 ] 2 r R R 3 3P z z=; 2 R5 3P rz2 rz = ; 2 R5 (6.9) Figure 6.6: Point loading of an elastic half-space (Boussinesq's Problem). 6.3 Pressure applied to circular regions The surface displacements are given by the relatively simple relations ur = ; (1 + 2)(1 ; 2 ) P E r 2 uz = 1 ;E P r Contact problems require the surface displacement in terms of distributed pressure loadings. The surface displacement due to distributed pressure is calculated by integration where the point at which the displacement is calculated is taken as the origin. The normal displacement at B due the pressure applied to the element at C is (Figure 6.7) 2 uz = 1 ;E p(ss ) sdsd where s is distance from B to C. Notice that the s in the Jacobian cancels the s in the denominator of the Boussinesq solution. Calculating the displacement due to the entire pressure distribution is accomplished by 1 ; 2 Z p(s )dsd uz = E A where A is the circular area to which the pressure is applied. Special cases of the results are given next. The details of the integration are not reproduced from class. 6.3.1 Uniform pressure For a uniform pressure distribution applied to a circle of radius a, 2 uz = 4(1 ; E )p0a E (r=a) r<a Figure 6.7: Pressure applied to circular region. Figure 6.8: Normal displacement due to uniform pressure distribution. 4(1 ; 2 )p0r E (a=r) ; (1 ; a2=r2)K (a=r)] r>a uz = E where K and E are the complete elliptic integrals of the rst and second kind respectively. This displacement is sketched in Figure 6.8. The radial surface displacement is )(1 ur = ; (1 ; 22E + ) pr r<a 2 )(1 ur = ; (1 ; 22E + ) p a r>a r This last equation is the same as that for a point load with magnitude P = p a2 which implies that the radial displacement outside of contact is independent of the pressure distribution. 6.3.2 Hertz Pressure For an ellipsoidal pressure distribution applied to a circle of radius a such that p(r) = q p0 1 ; r2=a2, 2 uz = 4 1 ; p0 (2a2 ; r2) r<a (6.10) E a 2 p0 uz = 1 ; 2a (2a2 ; r2) sin;1 (a=r) + r2 (a=r)(1 ; a2=r2)1=2 ] r>a E Note that Johnson has a sign error for the last term. Also for the radial displacement 2 r2 )(1 r<a ur = ; (1 ; 23E + ) a p0 1 ; (1 ; a2 )3=2 ] r 2 )(1 ur = ; (1 ; 23E + ) p0 a r>a r 6.4 Hertz Point Contact for Spheres Consider the spheres being brought into contact in Figure 6.9. The load is P, the total approach is , and the radius of contact is a. Geometric considerations very similar to those for the contacting cylinders in chapter 4 reveal that the sum of the displacements for the two spheres should satisfy uz1 + uz2 = ; 21 r2 (6.11) R where 1/R = 1/R1 + 1/R2 . That is a pressure distribution which gives a constant plus r2 term is needed to cancel the potential interpenetration of the spheres. Comparing Equations 6.10 and 6.11 reveals that the contacting spheres induce an ellipsoidal pressure distribution and p0 (2a2 ; r2) = ; 1 r2 4 aE 2R This equation must be valid for any r < a requiring 0 a = 2 pER and = 2 ap0 E Global equilibrium requires Za P = p(r)2 rdr = 2 p0 a2 3 0 Finally (6.12) a = ( 3PR )1=3 4E Figure 6.9: Hertz contact of spheres. 2 (6.13) = ( 169P 2 )1=3 RE 2 p0 = ( 6PE 2 )1=3 (6.14) 3R These equations describe Hertz contact for spheres. Notice that these results are nonlinear and that the maximum pressure increases as the load raised to the 1=3 power. The equations in Section 6.3.2 can be di erentiated to yield the tangential surface strains. These can be combined with Hooke's Law and the contact pressure to calculate the surface stresses as 1 ; 2 a2 1 ; (1 ; r2=a2 )3=2 ] ; (1 ; r2=a2 )1=2 g (6.15) r = p0 f 3 r2 2 (6.16) = ;p0 f 1 ; 2 a2 1 ; (1 ; r2=a2)3=2 ] + 2 (1 ; r2=a2 )1=2 g 3 r 2 2 1=2 (6.17) z = ;p0 (1 ; r =a ) inside the contact patch (r < a) and (1 ; 2 )a2 (6.18) r = ; = p0 3r2 outside the contact patch (r > a). These stresses are shown in Figure 6.10. Notice that the radial stress is tensile outside the circle and that it reaches its maximum value at r = a. This is the maximum tensile stress in the whole body. The stresses along the z-axis (r = 0) can be calculated by rst evaluating the stress due to a ring of point force along r = r and integrating from r = 0 to r = a. For example, using Equation 6.9 and noting that the area of a thin annulus is dA = 2 rdr leads to p 3p0z3 Z a r a2 ; r2 dr z (0 z ) = ; a 0 (z2 + r2)5=2 0.50 0.25 σr Stress (σ/p0) -0.00 -0.25 σθ -0.50 -0.75 σz -1.00 -2.0 -1.0 0.0 Position (r/a) 1.0 2.0 Figure 6.10: Surface stresses induced by circular point contact ( = 0:3) Making the substitution a2 ; r2 = u2, using integration by parts, some integrals in the APPENDIX, and substitution of limits leads to 2 (0 z) = ;p0 z2 a a2 z + ( ) z tan;1 a ] + 1 a2 (0 z) = p0 ;(1 + ) 1 ; a r (0 z ) = z 2 z2 + a2 (6.19) (6.20) Symmetry dictates that the maximum shear stress in the body occurs along r = 0. Manipulation of the above equations leads to " # 1 ( (0 z) ; (0 z)) = p (1 + )(1 ; z tan;1 a ) ; 3 a2 max = r 0 2 z a z 2 z2 + a2 For = 0:3 the maximum shear stress is about 0:31p0 and occurs at a depth of approximately z 0:48a. The stresses are plotted for = 0:3 in Figure 6.11. 6.5 Elliptical Contact When solids that have unequal curvatures in two directions are brought into contact the contours of constant separation are ellipses. When the principal curvatures of the two bodies are aligned, the axes of the ellipse are correspond to these directions. Placing the x and y axes in the directions of principal curvature, the equation comparable to Equation (6.11) is 1 1 (6.21) uz1 + uz2 = ; 2R0 x2 ; 2R00 y2 Stress (σ/p0) -0.5 0.0 -1.0 0.5 τmax 0.5 σr σz Depth (z/a) 1.0 1.5 2.0 2.5 3.0 Figure 6.11: Subsurface stresses induced by circular point contact ( = 0:3) where 1 = 1 + 1 1 = 1 + 1 0 R0 00 00 R 0 R1 R00 R1 R2 2 and Ri0 are the curvatures in the x direction and Ri00 are the curvatures in the y direction. The contact area is an ellipse adn the resulting pressure distribution is semiellipsoidal given by s 2 2 p(x y) = p0 1 ; x2 ; y2 a b and the actual calculation of a and b is cumbersome. However, for mildly elliptical contacts (Greenwood, 1985), the contact can be approximated as circular with p 3PRe 1=3 4E with and p0 given by Equations (6.13- 6.14). c ab Re p R0 R00 (6.22) Chapter 7 Introduction to Plasticity In this Chapter we will introduce plasticity in order to consider elastic-plastic contact. There are a few plasticity books of general interest which include Engineering Plasticity by W. Johnson and P.B. Mellor 32] and The Mathematical Theory of Plasticity by R. Hill 33]. These should be consulted if the reader has further interest. 7.1 The Stress-Strain Diagram The stress-strain diagram is a material property which is usually constructed from a tensile test (Figure 7.1). The stress is taken as = P=a and the engineering strain is = L=L where P is the load, A is the cross-sectional area, and L is the specimen length. In the elastic region of the diagram in which < Y where Y is the yield stress, the stress is directly proportional to strain, = E . If the specimen is unloaded the stress and strain would follow the same path taken during loading. The energy stored in the specimen is the area under the stress-strain curve times the volume of the specimen and unloading along the same path as loading implies that all of the energy put into the specimen during loading is recovered during unloading. That is why this portion of the curve is referred to as elastic. Once the applied stress exceeds the yield stress, the diagram moves into the plastic region. Typically, ductile materials strain harden which implies that the yield stress increases with increased plastic deformation. If the load is reduced, the material unloads on a line nearly parallel to the original elastic curve. It will also reload along this line. The plastic strain, P , is de ned as the permanent strain left after unloading to zero. The elastic strain, E , is the di erence between the total strain and the plastic strain or E+ P = Residual stress is de ned as any stress that remains in the body after the load is removed. Some machining and nishing operations leave near surface compressive residual stresses acting parallel to the surface which must be overcome by subsequent stresses to cause tensile yielding. The development of this residual stresses is due to 63 Figure 7.1: The tensile test and stress-strain diagram. the occurrence of plastic deformation during the loading process. Plastic deformation occurs when the state of stress at a point satis es the yield criteria. 7.2 Yield Criteria Two yield criteria that have proven to be useful for ductile materials are the von-Mises and Tresca criteria. The von-Mises yield criteria is written 1 J2 6 ( 1 ; 2 )2 + ( 2 ; 3 )2 + ( 3 ; 1 )2] = k2 = Y 2=3 where J2 is the second invariant of the stress tensor, 1 , 2, and 3 are the principal stresses, k is the yield stress in pure shear, and Y is the yield stress in uniaxial tension. The Tresca yield criteria is written maxfj 1 ; 2 j j 2 ; 3j j 3 ; jsigma1jg = 2k = Y shear If the same yield stress in p is used in each criteria then the tensile yield stress in the Tresca condition is 2= 3 1:155 times the tensile yield stress in the von-Mises condition. For plane stress ( 3 = 0), the yield criteria are shown in Figure 7.2. These yield criteria are rst met below the surface in normally loaded contact problems. The deformation in the regions below the surface where the yield criteria are satis ed are governed by ow rules rather than Hooke's Law. 7.3 Flow Rules As a rst step it is convenient to separate the volumetric strains from the shape change strains. Let the mean stress be de ned as 1 m = ( x + y + z) 3 Figure 7.2: The von-Mises and Tresca yield criteria for plane stress. and the volumetric strain be de ned as m = x+ y+ z= m 3 K = 1 ;E2 K where the material property K is called the bulk modulus. De ning the deviatoric stress components as 0 x = x; m 0 y = y; m 0 z = 0 z; m 0 0 = xy yz = yz zx = zx Now Hooke's Law can be written in the convenient form 0 ij = 2G + 1 ; 2 m ij ij E Here ij = 1 when i = j and ij = 0 when i 6= j . Reuss assumed that the plastic strain increment is proportional to the instantaneous stress deviator so that d P = d P = d P = d xyP = d y zP = d z xP = d y x z 0 0 0 xy0 y z0 z x0 x y z or 0 d P = d ij d ij Now the total strain increment can be written as xy d ij = d P + d E ij ij d ij = 0 d ij d + 2G + 1 ; 2 ij E 0 m ij Noting that plastic strain is volume conserving so that d P +d P +d P =0 x y z and de ning the deviatoric strain increment as d 0ij = d ij ; 1 ij m 3 leads to 0 d ij 0 d 0ij = ij d + 2G d m = dKm 0 0 2 ij ij = 2k where repeated indices imply a summation and the last equation is a representation of the von-Mises yield criteria. These three equations uniquely de ne a stress increment for loading. The solution of a plasticity problem involving these equations is di cult. The solution for rolling of a thin strip was developed in class. 7.4 Hardness Testing Hardness testing is very useful in tribology. The book by Mott provides an introduction 34] while the volume edited by Blau and Lawn 35] provides more up to date information. In the Brinell test a hardened steel ball is pressed into the surface of the test specimen while in Vickers or Knoop indentation a diamond pyramid is pressed into the surface. The hardness, H , is de ned as the load divided by the indentation area (Figure 7.3). That is H=P A For most materials it turns out the hardness is about three times the yield stress in a simple compression test. This result is consistent with calculations performed using slip-line eld theory. 7.5 Elastic-Plastic Contact Hardness testing could be classi ed as plastic contact whereas lighter loaders or larger indenters lead to constrained plastic deformation below the surface (Figure 7.4). This problem is di cult because the elastic-plastic boundary is not known a priori and must be calculated as part of the solution process. There is an enormous amount of research directed at using the nite element method to address these problems and we heard a couple of projects on this subject. Figure 7.3: Hardness testing. Figure 7.4: Elastic-plastic contact. The plastic deformation will lead to residual stresses. Melan's Theorem states that if any time-independent distribution of residual stresses can be found which, together with the elastic stresses due to the load, constitutes a system of stresses within the elastic limit, then the system will shakedown so that no additional plastic deformation occurs. For rolling cylinders, the shakedown limit for loads below which shakedown will occur is Ps = 1:66Py where Py is the load at which rst yield occurs. 7.6 Slip-Line Fields Slip-line elds provide a means to get upper and lower bound solutions to problems of rigid-plastic deformation. Chapter 8 Characterization of Surface Roughness The discussion of surface roughness and the techniques used to quantify it are discussed here. Some general work in this area includes Rough Surfaces by T.R. Thomas 36] and and the article by J.A. Greenwood, \A Uni ed Theory of Surface Roughness" 37]. We are now changing our scale of interest. We calculated contact stresses to discover what was happening inside the body such as the location of rst yield or cracking. Now we are going to focus on the surface of the bodies. The lms attached to a metal surface are shown in Figure 8.1 38]. One convenient manner of characterizing this surface is to measure its surface roughness. 8.1 De nition of Surface Roughness Consider a surface pro le whose height is given as a function of position z(x). The datum is chosen so that ZL z(x)dx = 0 0 The average roughness Ra is de ned as 1 Z L jzjdx Ra = L 0 where the absolute value implies that peaks and valleys have the same contribution. The root-mean-square roughness or standard deviation is de ned by ZL 2 = 1 z2 dx L 0 The RMS roughness is always greater than the average roughness so that > Ra and 1:2 69 Figure 8.1: Surface lm attached to a metal surface. Process RMS Roughness (Microns) Grinding 0.8 0.4 Fine Grinding 0.25 Polishing 0.1 Super Finishing 0.025 0.01 Table 8.1: Surface roughness for various nishing processes Figure 8.2: A rough surface. for most surfaces. Typical RMS values for nishing processes are given in Table 8.1. Of course widely di erent surfaces could give the same Ra and RMS values. The type of statistical quantity needed will depend on the application. 8.2 Probability and the Bearing Area Curve The bearing area curve 8.3 Typical Rough Surfaces One project dealt with setting-up a Talysurf in the lab and making measurements concerning the e ects of various nishing techniques on the quality of surface nish. 8.4 Contact of Regular Wavy Surfaces 8.5 The Greenwood-Williamson Model Chapter 9 Friction Theories This chapter discusses the development of friction theories in a very general way. 9.1 The Adhesive Theory of Friction The adhesive theory of friction is discussed with the aid of Figure 9.1. 9.2 Abrasive Friction 9.3 Surface Energy and Friction 9.4 Experimental Data and Friction 73 Figure 9.1: The rough surface and adhesive friction. Chapter 10 Frictional Contact This chapter discusses frictionally induced stresses. 10.1 Sliding Contact Consider an indenter drawn across the surface of an elastic half-space (Figure 10.1). 10.2 Oscillating Shear Force and Fretting 75 Figure 10.1: Sliding indenter with friction. Chapter 11 Frictional Heating This chapter discusses the development of heat caused by friction and the corresponding temperatures and thermal stress. 11.1 Thermoelasticity The equations of coupled thermoelasticity can be summarized in the following form. kr2 T = cE @T + 3K T0(_x + _y + _z ) @t 1 ; ( + )] + (T ; T ) (11.1) y z 0 x= E x 1 ; ( + )] + (T ; T ) (11.2) y= z x 0 E y 1 ; ( + )] + (T ; T ) (11.3) x y 0 z= E z where k is thermal conductivity, T is temperature with T0 being the referenc temperature, is mass density, cE is the speci c heat at constant deformation, K is the bulk modulus, is the thermal expansion coe cient, and the dot implies partial di erentiation with respect to time. The remaining equations concerning Hooke's Law for shear stress and strain, the strain-displacement equations, and equilibrium are unchanged. In most practical situations, the energy dissipated due to elastic volume expansion is small and the equation govering the temperature distribution can be written as kr2 T = cE @T (11.4) @t ZZ 11.2 Transient Heat Conduction Consider an in nite body subjected to a point source of heat (Figure 11.1). 77 Figure 11.1: The in nite solid subjected to a point source of heat. 11.3 The Moving Heat Source 11.4 Contact of Bodies at Di erent Temperatures 11.5 Thermoelastic Instability Chapter 12 Wear Models This chapter discusses the development of wear models. The models can be categorized into 1. Adhesive Wear 2. Abrasive Wear 3. Fatigue Wear 4. Corrisive Wear 12.1 Adhesive Wear Archard's theory of adhesive wear can be derived from rough contact (Figure 12.1). 12.2 12.3 12.4 12.5 Abrasive Wear Fatigue Wear Corrisive Wear Experimental Wear Data 79 Figure 12.1: Archar's adhesive wear model. Chapter 13 Contact Fatigue and Fracture Contact fatigue occurs during repeated contacts. It includes plastic deformation and fracture. 13.1 Residual Stress The bodies may have reached elastic shakedown. (Figure 13.1). 13.2 Fracture 13.3 Fatigue Crack Propagation 81 Figure 13.1: Repeated contact leading to elastic shakedown. Chapter 14 Lubrication Lubrication is very important in most contact application. 14.1 Hydrodynamic Lubrication Reynolds lubrication theory is illustrated in (Figure 14.1). 14.2 Elasto-Hydrodynamic Lubrication 14.3 Lubricant Properties 83 Figure 14.1: The hydrodynamic lm in hydrodynamic lubricaton. 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Mellor, Engineering Plasticity, Ellis-Horwood, West Sussex, England (1983). 33] R. Hill, The Mathematical Theory of Plasticity, Oxford, Oxford, England (1985). 34] , B.W. Mott, Micro-Indentation Hardness Testing, Butterworths, London (1956). 35] P.J. Blau and B.R. Lawn (Ed), Microindentation Techniques in Materials Science and Engineering, ASTM STP 889 (1985). 36] T.R. Thomas (Ed), Rough Surfaces, Longman, London (1982). 37] \A Uni ed Theory of Surface Roughness," Proc. R. Soc. Lond., A393, (1984) 133-157. 38] E. Rabinowicz, Friction and Wear of Materials, Wiley, New York (1965) p 71. Integral Table Some useful integrals are given in this APPENDIX. Z du 1 = z tan;1 u 2 + z2 z Z udu u = 2z2 (u2 + z2 ) + 21 3 tan;1 u 2 + z 2 )2 (u z z Z u2 du = ; 2(u2 u z2 ) + 21z tan;1 u (u2 + z2 )2 + z Z du = 2p u 2 (z2 ; u2)3=2 z z2 ; u (1) (2) (3) (4) ...
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