Volume 08 - Mechanical Testing and Evaluation Part 1 docx

151 377 0
Volume 08 - Mechanical Testing and Evaluation Part 1 docx

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

ASM INTERNATIONAL ® Publication Information and Contributors Introduction Mechanical Testing and Evaluation was published in 2000 as Volume of the ASM Handbook The Volume was prepared under the direction of the ASM Handbook Committee Volume Coordinator The Volume Coordinators were Howard Kuhn, Concurrent Technologies Corporation and Dana Medlin, The Timkin Company Authors and Contributors • LAMET UFRGS • John A Bailey North Carolina State University • John Barsom Barsom Consulting Limited • Peter Blau Oak Ridge National Laboratory • Kenneth Budinski Eastman Kodak Company • Vispi Bulsara Cummins Engine Company • Leif Carlsson Florida Atlantic University • Norm Carroll Applied Test Systems, Inc • Srinivasan Chandrasekar Purdue University • P Dadras Wright State University • K.L DeVries Univerisity of Utah • George E Dieter University of Maryland • James Earthman University of California, Irvine • Horatio Espinosa Northwestern University • Henry Fairman MQS Inspection, Inc • Thomas Farris Purdue University • Andrew R Fee Consultant • William W Gerberich University of Minnesota • Jeffrey Gibeling University of California, Irvine • Amos Gilat The Ohio State University • Peter P Gillis University of Kentucky • William Glaeser Battelle Memorial Institute • Blythe Gore Northwestern University • George (Rusty) Gray III Los Alamos National Laboratory • Amitava Guha Brush Wellman Inc • Gary Halford NASA Glenn Research Center at Lewis Field • John Harding Oxford University • Jeffrey Hawk Albany Research Center • Jennifer Hay Applied Nano Metrics, Inc • Robert Hayes Metals Technology Inc • K.H Herter University of Stuttgart • John (Tim) M Holt Alpha Consultants and Engineering • Joel House Air Force Research Laboratory • Roger Hurst Institute for Advanced Materials (The Netherlands) • Ian Hutchings University of Cambridge • Michael Jenkins University of Wyoming • Steve Johnson Georgia Institute of Technology • Serope Kalpakjian Illinois Institute of Technology • Y Katz University of Minnesota • Kevin M Kit University of Tennessee • Brian Klotz The General Motors Corporation • Howard A Kuhn Concurrent Technologies Corporation • John Landes University of Tennessee • Bradley Lerch NASA Glenn Research Center at Lewis Field • Peter Liaw University of Tennessee • John Magee Carpenter Technology Corporation • Frank N Mandigo Olin Corporation • Michael McGaw McGaw Technology, Inc • Lothar Meyer Technische Universität Chemnitz • James Miller Oak Ridge National Laboratory • Farghalli A Mohamed University of California • William C Mohr Edison Welding Institute • Charles A Moyer The Timken Company (retired) • Yukitaka Murakami Kyushu University • Sia Nemat-Nasser University of California, San Diego • Vitali Nesterenko University of California, San Digeo • Todd M Osman U.S Steel • F Otremba University of Stuttgart • T Ozkai Olin Brass Japan, Inc • George M Pharr The University of Tennessee • Paul Phillips The University of Tennessee • Martin Prager Welding Research Council and Materials Properties Council • Lisa Pruitt University of California, Berkeley • George Quinn National Institute of Standards and Technology • J.H Rantala Institute for Advanced Materials (The Netherlands) • Suran Rao Applied Research Laboratories • W Ren Air Force Research Laboratory • Gopal Revankar Deere & Company • Robert Ritchie University of California, Berkeley • Roxanne Robinson American Association for Laboratory Accreditation • E Roos University of Stuttgart • Clayton Rudd Pennsylvania State University • Jonathan Salem NASA Glenn Research Center at Lewis Field • P Sandberg Outokupa Copper • Ashok Saxena Georgia Institute of Technology • Eugene Shapiro Olin Corporation • Ralph S Shoberg R.S Technologies Ltd • M.E Stevenson The University of Alabama • Ghatu Subhash Michigan Technological University • Ed Tobolski Wilson Instruments • N Tymiak University of Minnesota • George Vander Voort Buehler Ltd • Howard R Voorhees Materials Technology Corporation • Robert Walsh Florida State University • Robert Waterhouse University of Nottingham • Mark Weaver University of Alabama • Dale Wilson The Johns Hopkins University • David Woodford Materials Performance Analysis, Inc • Dan Zhao Johnson Controls,Inc Reviewers • Julie Bannantine Consultant • Raymond Bayer Tribology Corporation • Peter Blau Oak Ridge National Laboratory • Toni Brugger Carpenter Technology Corporation • Prabir Chaudhury Concurrent Technologies Corporation • Richard Cook National Casein of California • Dennis Damico Lord Corporation Chemical Products Division • Craig Darragh The Timken Corporation • Mahmoud Demeri Ford Research Laboratories • Dez Demianczuk LTV Steel • George E Dieter University of Maryland • James Earthman University of California/Irvine • Kathy Faber Northwestern University • Henry Fairman MQS Inspection, Inc • Gerard Gary Polytechnique Institute (France) • Thomas Gibbons Consultant • William Glaeser Battelle Memorial Institute • Jennifer Hay Applied Nano Metrics, Inc • Robert Hayes Metals Technology, Inc • David Heberling Southwestern Ohio Steel • Kent Johnson Engineering Systems Inc • Brian Klotz The General Motors Corporation • Howard A Kuhn Concurrent Technologies Corporation • Lonnie Kuntz Mar-Test, Inc • David Lambert Air Force Research Laboratory • John Landes University of Tennessee • Iain LeMay Metallurgical Consulting Services • John Lewandowski Case Western Reserve University Fig Linear distribution of shear stress in torsion of a round bar Mt, applied torque Fig Linear distribution of normal stress in bending Mb, bending moment The shear stress distribution in torsion is given by (Ref 15): τ = Mt r/J (Eq 10) where τ is the shear stress, Mt is the applied torque, r is measured from the axis, and J is the polar moment of inertia (second moment about the axis of rotation) of the bar cross section The influence of J in design is discussed later in “Shape Design” in this article More detailed descriptions of rotational shear (i.e., torsion) are provided in the article “Fundamental Aspects of Torsional Loading” in this Volume The normal stress distribution in bending is given by (Ref 16): σ = Mbz/I (Eq 11) where σ is the longitudinal stress in the bar, Mb is the bending moment, z is measured from the neutral axis, and I is the moment of inertia (second moment about the bending axis) of the bar cross section The role of I in design is also discussed later in “Shape Design.” More detailed descriptions of bending stress and strain behavior are provided in the article “Stress-Strain Behavior in Bending” in this Volume Design for Strength in Torsion or Bending Design of bars under torsional or bending moments is based on preventing the maximum surface stresses from exceeding the failure limit of the material For example, in torsion of a round bar, τmax occurs at r = D/2 (where D is the bar diameter) and must satisfy the design condition: τmax = Mt D/2J < τf (Eq 12) where τf is the failure shear strength of the material at failure This value may be the shear yield strength, τo, or the ultimate shear strength, τu Similarly, in bending, the maximum normal stress, σmax, occurs at z = H/2 (where H is the thickness of a symmetrical beam) and must satisfy the design condition: σmax = Mb H/2I < σf (Eq 13) where σf may be the tensile yield strength, σo, or the ultimate tensile strength, σu, for failure on the convex side of the bar Compressive strengths apply to failure on the concave side of the bar In the same way that Eq is used for design of bars under tensile loading, Eq 12 and 13 can be used to determine the maximum torque, Mt, or bending moment, Mb, that can be transmitted by a specific bar geometry (D and J, or H and I) and material (τf or σf) These equations can also be used to determine the geometric parameters required to transmit a specified torque or bending moment with a given bar material Alternatively, Eq 12 and 13 can be used to select materials having the proper values of τf or σf to transmit a specified torque or bending moment in a bar of given geometry As with the case of simple tension, the design equations for torsion and bending can be modified to include the additional criteria of minimum weight or minimum cost The material parameters for minimum weight or minimum cost in this case depend on the geometric parameters that are fixed and those that are variable For a beam with the width undefined, the design parameters are (σf/ρ) and (σf/ρc), while for beams in which the height is undefined, the design parameters are (σf1/2/ρ) and (σf1/2/cρ) (Ref 1, 7) Design for Stiffness in Torsion or Bending Elastic deflection of a bar under bending moments and elastic twisting of the bar under torsion may lead to additional design limits For example, in bending a simply supported beam (Fig 7), the deflection at the center point is (Ref 16): δ = FL3/48EI (Eq 14) The beam stiffness is determined by the material parameter, E, and the geometry parameters, L and I Beam design to meet deflection limitations can be accomplished by proper material specification or by geometric specifications Similarly, torsional rotation of a round bar is given by (Ref 15): θ = MtL/GJ (Eq 15) where Mt is the applied torque, L is the length of the bar, J is the polar moment of inertia, and G is the shear modulus of elasticity Design for torsional stiffness involves selection of the bar dimensions as well as the material via its elastic property, G, which is related to other elastic properties by (Ref 15): G = E/2(1 + ν) (Eq 16) Shape Design Of particular interest in design of bars under torsion or beams under bending are the moments of inertia, J and I Since the stress distributions in each case are linear and reach a maximum at the surface, as shown in Fig and 9, the best use of material is accomplished by distributing it near the surfaces rather than at the center For example, the value of J for a solid circular cross section is: J = πD4/32 (Eq 17) By removing material from the central region of the bar, which is under little stress, the torsional load carrying capacity is reduced slightly, but the area (and therefore weight) is reduced more significantly Referring to Fig 10, if one-half of the inner material is removed (i.e., inside diameter Di = D/ 50%, but the value of J is reduced by just 25% ), the weight is reduced by Fig 10 Comparison of polar moment of inertia, J, for (a) solid and (b) hollow round bars A, crosssectional area; D, diameter Similarly, in bending, efficient material use is accomplished by placing material at the upper and lower surfaces, as in the shape of an I-beam To illustrate this, consider that the moment of inertia for a rectangle is given by: I = bH3/12 (Eq 18) where b is the width and H is the height of the rectangle Obviously, increasing H has a much larger effect on I than increasing b For example, as shown in Fig 11, a rectangle that is units thick and units long has a value of I that is times larger in the vertical orientation compared to the horizontal orientation However, if the same amount of material is rearranged into an I-beam configuration, the value of I increases further by a factor of nearly Thus, weight savings in design for torsion and bending can be accomplished not only by selecting materials having high strength-to-weight ratios, but also by careful attention to the distribution of material in the component Fig 11 Comparison of moment of inertia, I, for three different arrangements of the same amount of material Mechanical Testing Testing of materials for shear yield strength, τo, to be used in Eq 12, or for tensile yield strength, σo, for use in Eq 13, can be accomplished through a tension test If the material is isotropic and homogeneous, the shear yield strength can also be calculated from the tensile yield strength (Ref 15): τo = σo/ (Eq 19) Fracture strengths, however, must be measured in torsion and bend tests because the mechanisms or modes of fracture may be different from those in tension testing Tensile yield strength, σo, and shear yield strength, τo, can also be derived directly from bending and torsion tests using Eq 12 and 13 Accuracy will be limited, however, because in both cases yielding occurs initially at the outside surfaces, so the effect on the measured loads is not as easily detectable as in the tension test, where yielding occurs simultaneously across the entire section Torsion and bend tests are particularly useful in evaluating materials that have been given surface treatments such as carburizing or shot peening to increase the strength of the surfaces and improve their resistance to the high stresses at the surface generated by torsional or bending moments Generally, the metallurgical structures of such surfaces occur in a thin layer and cannot be produced easily in bulk form for measurement by tension testing Then, Eq 12 can be used to determine the strength, τf, of the surface material in torsion, and Eq 13 can be used to determine the strength, σf, of surface material in bending This approach is particularly useful for determining fracture strengths of the surface materials During bend or torsion testing of surface treated materials, however, certain precautions must be taken First, the orientation of the bending and shear stresses in the test specimens must be in the same orientation with respect to the material microstructure as occurs in the actual components Second, variations in material strength beneath the surface must be considered in comparison with the linear distributions of stress in bending and torsion shown in Fig and Even though the surface material strength may be greater than the stress at the surface (such as in a case hardened part), away from the surface at the neutral axis of the bar the strength may decrease more rapidly than the applied stress, as illustrated in Fig 12 (Ref 17) Then failure will occur beneath the surface where the stress exceeds the local strength of the material Clearly, it is important to understand the strength distribution in a material in relation to the stress distribution acting on the material to ensure that the product design prevents such insidious failures Details on torsion and bend testing are covered in separate articles in this Volume Fig 12 Stress and strength distributions in a bar under torsion or bending Failure occurs beneath the surface if the material strength decreases more rapidly than the stress in the material Source: Ref 17 Equation 14 for deflection of a beam also suggests an alternative method for measuring the elastic modulus, E, of a material For given geometry parameters, measured data pairs for force, F, and deflection, δ, lead to a calculation of E Alternatively, resonant frequency methods can be used to measure E in the beam geometry shown in Fig The natural frequency of vibration of a beam is related to its geometry, density, and elastic modulus To carry out the measurement, the beam is vibrated by a transducer connected to a frequency generator producing sinusoidal waves By varying the frequency of the generator, the natural frequency is determined when resonance occurs Then the elastic modulus can be calculated from (Ref 18): E = Cf2 ρL4/H2 (Eq 20) where f is the resonant frequency, ρ is the material density, L is the beam length, H is the beam thickness, and C is a constant The shear modulus, G, can also be measured by sonic and resonant frequency methods in torsion Generally, resonant and sonic methods of measuring elastic properties are more accurate and easier to perform than direct measurement of stress and strain in a tension or torsion test References cited in this section Cambridge Engineering Selector, Granta Design Ltd., Cambridge, UK, 1998 M.F Ashby, Materials Selection for Mechanical Design, 2nd ed., Butterworth-Heinemann, 1999 15 G.E Dieter, Mechanical Metallurgy, 2nd ed., McGraw Hill, 1976, p 49–50, 79–80, 379, 381, 385 16 J.H Faupel and F.E Fisher, Engineering Design, John Wiley & Sons, 1981, p 102, 113, 230–235, 802 17 D.J Wulpi, Understanding How Components Fail, ASM International, 1966, p 27 18 T Baumeistes, Ed., Marks' Mechanical Engineers' Handbook, 6th ed., McGraw-Hill, 1958, p 5–106 Overview of Mechanical Properties and Testing for Design Howard A Kuhn, Concurrent Technologies Corporation Shear Loading The torsion test described above subjects material to shear stress in a rotational mode, but shear stresses also occur in a translational mode Such modes can be found, for example, in the end connectors of tie bars Design Examples In an actual tie bar, the loads at each end of the bar in Fig would be applied through a pin connector or a threaded coupling, for example, as shown in Fig 13(a) If the tie bar were a composite material (which cannot be threaded or drilled), the loads would most likely be applied through an adhesive lap joint, as shown in Fig 13(b) Fig 13 Methods of connecting loads (a) A load is applied to the tie bar in Fig by a pin/eye connector and a threaded connector (b) End loads are applied to a composite bar by adhesives In these and all other types of connectors, shear stresses occur in the transition from the connection to the straight section of the bar For example, Fig 14(a) shows a pin through the eye of the tie bar and the forces applied to the tie bar through a clevis The pin, Fig 14(b), is subjected to shear stresses along the dotted planes between the eye of the tie bar and the clevis Shear failure may occur along these planes On the face view of the eye on the tie bar, Fig 14(c), shear also acts along the dotted planes from the hole to the end of the bar, potentially leading to tear out of the end of the conductor On the threaded end of Fig 13(a), transmission of load from a threaded connector to the tie bar also is carried by shear stresses at the root of the threads, shown by the dotted lines in Fig 15 In the lap joint shown in Fig 13(b), shear would occur along the adhesive interfaces Concentrated shear stresses, similar to those shown in Fig 14(b) and 14(c), also occur in many machine elements, such as keys and keyways in drive shafts, shear pins, and splined couplings These applications are critical to the safe operation of machinery and require robust design methods using accurate material properties Fig 14 Shear stresses in load end connectors (a) A clevis and pin connected to the eye in Fig 13(a) (b) Shear planes in the pin (c) Shear planes in the eye Fig 15 Shear planes in a threaded end connector In all of these examples, the average shear stress acting on the shear planes is: τ = F/As (Eq 21) where F is the load transmitted and As is the total area of the shear planes For design against failure, this shear stress must be less than the shear strength of the material Mechanical Testing Measurement of the shear yield strength of the material can be extracted from a tension test by using Eq 19 if the material has an isotropic and homogeneous microstructure The torsion test described previously also can be used to determine the shear strength of the material However, the torsion test measures the shear strength of the material in rotational shear, but the applications shown in Fig 13, 14, and 15 involve linear shear along a plane through the material or along an interface Linear shear behavior is affected significantly by anisotropy of the microstructure of the material, and specialized tests have been developed to determine the linear shear yield and fracture strength of materials More details on torsion loading are in the article “Shear, Torsion, and Multiaxial Testing” in this Volume Overview of Mechanical Properties and Testing for Design Howard A Kuhn, Concurrent Technologies Corporation Complex Stresses The previous sections describe the relatively simple uniform and linear stress distributions occurring during tension, compression, torsion, bending, and shear In all of these cases, one primary stress occurred Nonelementary shapes, however, lead to nonuniform, nonlinear, and multiaxial stresses, and these complex stress states must be considered in a complete analysis for product design Constitutive Relations To consider the effects of combined stresses in design, constitutive equations are required that give the relationships between multiple stresses and strains A one-dimensional example of a constitutive relation is the combination of Eq and 7, describing the elastic connection between uniaxial stress and strain Extending this behavior to three dimensions, if σx is applied in the x-direction, the strain εx = σx/E occurs, and that same stress will generate transverse strains: εy = εz = -νεx = -νσx / E (Eq 22) This leads to the complete three-dimensional expressions relating elastic stresses and strains (Ref 15): εx = (σx - νσy - νσz)/E (Eq 23a) εy = (σy - νσz - νσx)/E (Eq 23b) εz = (σz - νσx - νσy)/E (Eq 23c) A complete description of material behavior requires a description of yielding under multiaxial stresses Equations 2, 12, 13, and 21 express the conditions for yielding under the action of a single tensile or shear stress For three-dimensional stresses, a commonly used criterion that includes the effect of all stresses acting at a point in a material is: [(σx - σy)2 + (σy - σz)2 + (σz - σx)2 + 6τxy2 (Eq 24) + 6τyz2 + 6τzx2] 1/2/ = σo where σo is the yield strength of the material in simple tension Equation 24, known as the von Mises yield criterion (Ref 19), relates all of the stresses acting in a material to its yield strength Note that, in a torsion test where normal stresses are zero and only one shear stress is applied, Eq 24 reduces to Eq 19, which enables shear yield strength values for design to be determined from a simple tension test Equations 23a, 23b, 23c, and 24 are the constitutive relations for elastic deformation and yielding of an isotropic material More complex constitutive relations have been developed for anisotropic materials, composites, and rate dependent materials For failure by fracture, no simple criterion, such as Eq 24, exists to express the effects of multiaxial stresses on failure Qualitative results from Eq 23a, 23b, 23c, and 24 lead to insights on the relationships between combined stresses and strains For example, in a plane-strain tension test (Fig 16), a groove across the face of the specimen will lead to stress σx = F/tW where F is the applied load, t is the groove thickness, and W is the specimen width Under the action of this stress, the material in the groove will tend to contract in the thickness direction, z, and in the width direction, y There is no constraint preventing the material from contracting in the thickness, so σz = In the width direction, however, the bulk material (thickness, T, much greater than t) on each side of the reduced section prevents material from contracting That is, εy = Then, from Eq 23a, 23b, and 23c: εy = = (σy - νσx)/E (Eq 25a) or σy = νσx (Eq 25b) That is, a stress is generated in the y-direction because the natural tendency for the material to contract is constrained The strain in the thickness direction will be: εz = -ν(σx + σy)/E = -ν(1 + ν)σx/E (Eq 26) which is slightly larger than the strain that would have occurred with no constraint in the y-direction More complex interactions occur between three-dimensional stresses and strains in more complex geometries, and these interactions are influenced strongly by the Poisson ratio, ν Fig 16 Plane strain tension test specimen, showing the strains and stresses in the gage section The multiaxial stresses in this example have an effect on yielding The applied stress, σx, combined with the constraint of transverse strain in the y-direction, generates a stress in the y-direction given by Eq 25a and 25b The applied stress σx required to cause failure by yielding can be found by substituting the stresses σy = νσx and σz = into Eq 24, giving: (σx - νσx)2 + (νσx)2 + (-σx)2 = 2σo2 (Eq 27a) or σx = σo/(1 - ν + ν2)1/2 (Eq 27b) Substituting the value of ν = 0.5 for material undergoing plastic deformation, σx = 1.15 σo; that is, because of the constraint to contraction in the y-direction of the geometry shown in Fig 16, the stress required to yield the material is 15% higher than the axial stress required to yield the material if it were in simple tension Following are further examples of basic design stress conditions that involve nonlinear stress distributions and multiaxial stresses Stress Concentrations In the transition from the pin connector to the uniform cylindrical part of the tie bar (Fig 13a), the irregular geometry leads to concentrations of stress around the hole in the tie bar The load applied by the pin is distributed as a nonuniform stress across the eye cross section, as shown in Fig 17 Although the average stress on the cross section equals the load divided by the area, peak values of the stress occur at the inside of the eye Fig 17 Stress distribution on a cross section through the eye shown in Fig 13(a) Such concentrations of stress occur in all geometric irregularities such as fillet radii, notches, and holes A simplistic but useful expression for the stress concentration is given by (Ref 19): σmax = kt σa ≈ (1 + 2a/b) σa (Eq 28) where kt is the stress concentration factor, σa is the average stress, a is the dimension of the geometric irregularity perpendicular to the applied load, and b is the dimension parallel to the load (Fig 18) In Fig 17, for example, a = b, so the stress concentration factor is For very small cracks perpendicular to the load, a is much greater than b so the stress concentration becomes a/b Stress concentration factors for a wide range of practical geometries have been developed through extensive experimentation and analysis (Ref 16, 20) Failures usually initiate at these points of stress concentration and must be considered in all product designs Fig 18 Stress concentration around a flaw Source: Ref 19 Pressure Vessels Multiaxial stresses occur in shells subjected to internal or external pressure Examples include tanks containing fluids, vessels containing a high-pressure and high-temperature chemical reaction (such as in petroleum refining), dirigibles, and submarine pressure hulls Proper design of such pressure vessels must account for the multiaxial stresses A spherical shell containing pressure, for example, has equal stresses in any two perpendicular directions tangent to the sphere (Fig 19), given by (Ref 22): σθ = σφ = pD/4t (Eq 29) where p is the internal pressure, D is the spherical diameter, and t is the wall thickness Note that for D much greater than t, these stresses are much larger than the pressure, and stress through the thickness σt ≈ Fig 19 Stresses in the wall of a spherical pressure vessel A cylindrical pressure vessel wall will have a stress in the circumferential, or hoop, direction (Fig 20), given by (Ref 16): σθ = pD/2t (Eq 30a) and a stress in the axial direction given by: σz = pD/4t (Eq 30b) Fig 20 Stresses in the wall of a cylindrical pressure vessel Failure by yielding of a spherical pressure vessel can be found by substituting Eq 29 into Eq 24, giving: pmax = 4σot/D (Eq 31) For the cylindrical pressure vessel, substituting Eq 30a and 30b into Eq 24 gives: pmax = (4 / )σot/D (Eq 32) In all pressure vessel applications, the stresses in the wall, given by Eq 29, 30a, and 30b, may be further complicated by stress concentrations due to holes for inlet and outlet piping In cases where the wall thickness is much less than the diameter, the stresses given by Eq 29, 30a, and 30b are nearly uniform through the thickness For pressure containment where the wall thickness is a significant fraction of the diameter, such as in metalworking dies, the stress distribution is nonuniform with peak values at the inner surface (Ref 16) In these cases as well, Eq 24 is used to evaluate the initiation of failure by yielding Bearing Loads Close examination of the pin end connector in Fig 13(a) shows that the pin rests in a circular hole at the end of the tie bar For ease of attachment to the connector, the pin diameter is smaller than the hole diameter As a result, the load transmitted from the pin to the inside surface of the hole occurs over a small area of contact, leading to high local pressures, as shown in Fig 21 The pressure distribution in bearing contacts of this type is generally elliptical because the greatest elastic deformation occurs at the center of the contact zone Figure 22 shows this more clearly in the contact pressure between a flat plate and a cylinder Similar types of bearing loads and the accompanying elliptical pressure distributions occur in journal bearings, ball and roller bearings, gear tooth contacts, and railroad wheel/rail contacts The nature of the contact pressure distributions and the internal distribution of stresses resulting from these loads have special implications regarding material testing and selection Fig 21 Contact stresses between a pin and eye Fig 22 Flattening and contact pressure distribution between a roller and flat plate In roller-on-roller contacts, the elliptical pressure distributions in Fig 21 and 22 have peak pressures given by (Ref 21): pmax = (Eq 33a) where P is the load per unit length of contact and Δ [1/(1/D1 + 1/D2)] [(1 - ν12)/E1 + (1 - ν22)/E2] (Eq 33b) where D1 and D2 are the diameters of the rollers, and ν and E are the elastic properties of the materials in contact The width, w, of the contact zone between the rollers is given by: w= (Eq 34) In addition to the normal contact pressure shown as σz in Fig 22, elastic deformation in the contact zone also generates σx parallel to the surface and σy parallel to the axes of the rollers (perpendicular to the view in Fig 21) In this case, σx has the same magnitude and elliptical distribution as σz Beneath the surface in Fig 22, all of the stresses decrease because the contact load is spread over a larger area Stress analysis leads to the stress distributions (Ref 21): σz = - w/2Δζ (Eq 35a) σx = - (ζ - z′)2 w/2Δζ (Eq 35b) σy = - ν(ζ - z′) w/Δζ (Eq 35c) where z is the distance beneath the contact surface, z′ = 2z/w, and ζ = These stresses are plotted in Fig 23, which shows that σx decreases more rapidly than σz The von Mises stress σvm is also plotted, which shows a maximum at a distance 0.4w beneath the surface (where w is the contact width calculated from Eq 34) Subsurface failure as a result of these stresses is common in rolling contact configurations such as Fig 22 Fig 23 Stresses beneath the contact zone in Fig 22 Finite Element Analysis For complex stress problems that are not easily solved using the basic loading situations (or their combinations) described previously, finite element analysis is commonly used Over the past four decades, the finite element method has evolved from a structural analysis tool to a set of refined commercial programs that are invaluable aids in modern design Much of this utility is a result of the availability of more rapid and powerful computers, improved numerical methods for solving the equations, and enhanced computer graphics for display of the results Finite element analysis reduces the infinite number of points in a prescribed product geometry to a finite number of elements bound by a finite number of points The behavior of each element in a finite-element grid is governed by the same principles of equilibrium as in the tensile, compressive, torsional, and bend loadings described previously in this article, and by the material behavior described through constitutive equations (Eq 23a, 23b, 23c, and 24) The beauty of finite element analysis for design is that it allows consideration of complicated geometries and any type of material for which the properties are known Finite element analysis methods thus assist in the specification of geometry and material, which are the two primary objectives of design for mechanical applications Evaluating small changes in geometry or variations in material properties via finite element modeling permits design optimization In the application of finite element analysis (Ref 22, 23), first the product geometry is described by a collection of two-dimensional or three-dimensional elements, such as those shown in Fig 24 The choice of element type depends on the problem geometry and loading From displacements of the nodal points in each element, the deformations and strains in the element are defined Then the material constitutive equations (e.g., Eq 23a, 23b, and 23c) relate these strains to the stresses in each element Each node is connected to two or more elements, so there is interdependence between neighboring elements Fig 24 Examples of element geometries used for finite element models The next step is to define the boundary conditions (externally applied loads at the appropriate nodes) and constraints (identification of the nodes that are constrained from moving in at least one direction) This is a very important key to development of an accurate model and is based on mechanical insight of the person doing the modeling Energy methods are used to determine the interdependent nodal displacements (and stresses and strains) that minimize the total energy of the system The resulting myriad of linear algebraic equations are then solved for all of the nodal displacements such that the stresses and deformations are compatible from element to element and are consistent with the externally applied loads and nodal constraints As an example, Fig 25 shows a strip under tension and containing a hole, which was shown previously to cause a concentration of stress The elements used for the analysis are quadrilateral to conform to the overall geometry of the part (Ref 24) In the vicinity of the hole, the elements are distorted to conform to the shape of the hole In addition, the number of elements per unit area is increased in the vicinity of the hole because the stresses change rapidly Taking advantage of symmetry, one-quarter of the geometry is considered to reduce the number of elements, nodes, and equations to be solved ... 19 74? ?19 80) • W.L Mankins (19 9 4 -1 997) (Member 19 8 9-) • J.L McCall (19 82) (Member, 19 77? ?19 82) • W.J Merten (19 27? ?19 30) (Member, 19 23? ?19 33) • D.L Olson (19 9 0 -1 992) (Member 19 8 2 -1 988, 19 8 9 -1 992)... (19 66? ?19 68) (Member, 19 61? ? ?19 70) • J.F Harper (19 23? ?19 26) (Member, 19 23? ?19 26) • C.H Herty, Jr (19 34? ?19 36) (Member, 19 30? ?19 36) • D.D Huffman (19 8 6 -1 990) (Member 19 8 2-) • J.B Johnson (19 48? ?19 51) ... (Member 19 8 1- 1 986) • E.O Dixon (19 52? ?19 54) (Member, 19 47? ?19 55) • R.L Dowdell (19 38? ?19 39) (Member, 19 35? ?19 39) • M.M Gauthier (Chair 19 9 7 -1 998; Member 19 9 0-) • J.P Gill (19 37) (Member, 19 34? ?19 37)

Ngày đăng: 10/08/2014, 12:22

Từ khóa liên quan

Tài liệu cùng người dùng

Tài liệu liên quan