01 introduction to_mechanical_metallurgy

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01 introduction to_mechanical_metallurgy

  1. 1. Chapter 1Introduction to mechanicalmetallurgySubjects of interest • Introduction to mechanical metallurgy • Strength of materials – Basic assumptions • Elastic and plastic behaviour • Average stress and strain • Tensile deformation of ductile metals • Ductile vs brittle behaviour • What constitute failure? • Concept of stress and the types of stresses • Units of stress and other quantities Suranaree University of Technology Tapany Udomphol May-Aug 2007
  2. 2. Objectives • This chapter provides a background of continuum description of stress and strain and extends it to the defect mechanisms of flow and fracture of metals. • Elastic and plastic behaviours of metals are highlighted and factors influencing failure in metals are finally addressed.Suranaree University of Technology Tapany Udomphol May-Aug 2007
  3. 3. IntroductionMechanical metallurgy : Response of metals to forces or loads.Mechanical assessment of Forming of metals intoMaterials useful shapes• Structural materials • Forging, rolling, extrusion.• Machine, aircraft, ship, car etc • drawing, machining, etc We need to know We need to know conditions limiting values of which of load and temperature to materials in service can minimise the forces that are withstand without needed to deform metal failure. without failure. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  4. 4. Strength of materials Strength of materials deals with relationships between; • internal resisting forces • deformation • external loads , which act on some part of a body (member) in equilibrium.• In equilibrium condition, if there are the Internal force External forceexternal forces acting on the member,there will be the internal forces resisting ∫ σdA Pthe action of the external loads.• The internal resisting forces are usually expressed by thestress acting over a certain area, so that the internal force is theintegral of the stress times the differential area over which it acts. P = ∫ σdA …Eq.1 Suranaree University of Technology Tapany Udomphol May-Aug 2007
  5. 5. Assumptions in strength of materialsThe body (member) is Remark:• Continuous: No voids or empty spaces. Anisotropic is when• Homogeneous: Has identical properties at all the body has points. property that varies with orientation.• Isotropic: Has similar properties in all directions or orientation.Macroscopic scale, engineering materials suchas steel, cast iron, aluminium seems to becontinuous, homogeneous and isotropic.Microscopic scale, metals are made up of anaggregate of crystal grains having differentproperties in different crystallographic directions. However, these crystal grains are very small, and therefore the properties are homogenous Microstructure of in the macroscopic scale. low carbon steel Suranaree University of Technology Tapany Udomphol May-Aug 2007
  6. 6. Elastic and plastic behaviour• All solid materials can be deformed stress Maximumwhen subjected to external load. tensile strength Elastic• In elastic region, stress is proportional to limit Fracturestrain. This follows Hook’s law up to theelastic limit . The material now has elasticbehaviour. σmax• At the elastic limit, when the load is Elastic Plasticremoved, the material will change back toits original shape. strain• Beyond the elastic limit, material Load and extension curve ofpermanently deformed or the material has under uniaxial tensile loading.undergone plastic deformation. Note: elastic deformations in metals are relatively small in comparison to plastic deformations. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  7. 7. Average stress and strain No load applied Load applied Lo+ δ Lo Lo P• A uniform cylindrical bar which is The external load P is balanced bysubjected to an axial tensile load P. the internal resisting force, givingThe original gauge length Lo has the equilibrium equation;undergone a slight increase in Internal Externallength to Lo+ δ with a slight P = ∫ σdA force forcedecrease in diameter. ∫σdA P• The average elastic strain e is If the stress σ is uniform over the areathe ratio of change in length to the A, σ = constant, thenoriginal length. P = σ ∫ dA = σA δ∆L L − Lo…Eq.2 e= = = P Lo Lo Lo The average stress σ = …Eq.3 A Suranaree University of Technology Tapany Udomphol May-Aug 2007
  8. 8. Average stress and strain • In general, the stress is not uniform over the area A average stress. • On a microscopic scale, metals consist of more than one phase and therefore give rise to nonuniformity of stress. stress Maximum tensile strength• Below the elastic limit, Hook’s law Elasticcan be applied, so that the average limit Fracturestress is proportional to the averagestrain, σ …Eq.4 σmax =E ε The constant E is the modulus of elasticity or Young’s modulus. strain Load and extention curve of under uniaxial tensile loading Suranaree University of Technology Tapany Udomphol May-Aug 2007
  9. 9. Tensile deformation of ductile metal• In tension test, specimen is subjected to axial tensile loaduntil fracture. Load and extension are measured and expressedas stress and strain, see Fig.• The initial linear portion of the curve (OA) is the elastic region,following Hook’s law. D• Point A is the elastic limit (the greatest stress that the metal Bcan withstand without undergone permanent or plastic Elastic A’ Adeformation. A’ is the proportional limit where the curve limitdeviates from linearity.• The slope of the linear portion is the modulus of elasticity.• Point B is the yield strength, defined as the stress which willproduce a small amount of strain equal to 0.002 (OC). C• As the plastic deformationincreases, the metalbecomes stronger (strainhardening) until reaching themaximum load, givingultimate tensile strength D.• Beyond D, metal necks(reduce in x-section). Loadneeded to continuedeformation drops off tillfailure. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  10. 10. Ductile and brittle behaviour• Completely brittle materials i.e.,ceramics would fracture almost atthe elastic limit. stress stress• Brittle metals such as cast ironshow small amounts of plasticitybefore failure. strain strain• For engineering materials, Completely brittle Slightly brittle with small amount ofadequate ductility is important ductilitybecause it allows the materials toredistribute localised stresses. • Fracture behaviour of metal (ductile or brittle) also depends on some conditions, i.e., temperature, tension or compression, state of stresses, strain rate and embrittling agent. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  11. 11. What constitute fracture? Three general ways that cause failures in structural members and machine elements:1) Excessive elastic deformation stress2) Yielding or excessive plastic deformation Plastic energy Elastic energy3) Fracture strain Types of Load/forces Good structural components in Nature of Materials service without Structural design failure Suranaree University of Technology Tapany Udomphol May-Aug 2007
  12. 12. Excessive elastic deformation • Failure due to excessive elastic deformation are controlled by the modulus of elasticity not the strength of the materials. Two general types of excessive elastic deformation: 1) Excessive elastic deformation under stress condition of stable equilibrium. Elastic EX: too much deflection in a shaft can cause energy wear in bearing and damage other parts. strain 2) Excessive elastic deformation under condition of unstable equilibrium. EX: A sudden deflection or buckling of a slender column. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  13. 13. Excessive plastic deformation • Excessive plastic deformation occurs when the elastic limit is exceeded yielding. • Yielding produce permanent change of shape and can cause fracture. stress EX: When component changed in shape, it cannot function properly any longer. Plastic Elastic energy energy• Failure by excessive plastic straindeformation is controlled by the yieldstrength of the materials. Even in morecomplex loading conditions, yield strengthis still a significant parameter. Large deformation that cause failure in a pipe Suranaree University of Technology Tapany Udomphol May-Aug 2007
  14. 14. Fracture Metals fail by fracture in three general ways: 1) Sudden brittle fracture Ex : some ductile metal such as plain carbon steel will undergo ductile to brittle transition with decreasing temperature, increasing rate of loading and triaxial state of stresses. 2) Fatigue, or progressive fracture Most fracture in machine parts are due to fatigue (subjected to alternating stresses). Failure is due to localised tensile stress at spot or notch or stress concentration. 3) Delayed fracture Ex: Stress-rupture failure, which occurs when a metal has been statistically loaded at an elevated temperature. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  15. 15. Working stress To prevent structural members or machine elements from failure,such members should be used under a stress level that is lower thanits yield stress σo. This stress level is called the working stress σw. According to American Society of Mechanical Engineering (ASME), the working stress σw may be considered as either the yield strength σo or the tensile strength σu divided by a number called safety factor. σo σu …Eq.5 σw = or σw = …Eq.6 No Nu Where σw = working stress σo = yield strength σu = tensile strength No = safety factor based on yield strength Nu = safety factor based on tensile strength Safety factor depends on loading/service conditions, consequences, etc. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  16. 16. Concept of stress and the type ofstresses Definition: stress is force per unit area. There are two kinds of external forces which may act on a body; 1) Surface forces : forces distributed over the surface of the body, i.e., hydrostatic pressure 2) Body forces : forces distributed over the volume of the body, i.e., gravitational force, magnetic force, centrifugal force, thermal stress.Suranaree University of Technology Tapany Udomphol May-Aug 2007
  17. 17. Stress at a point • Consider a body having forces F1…F4 acting on it, see Fig (a). The body is cut by a plane passing through point O. If one half is removed and replaced by an equivalent force F, acting on the x-sectional area A to remain the static equilibrium, see Fig (b). • We can resolve F into component normal to the plane Fn and tangential to the plane Fs. (a) Equilibrium of an arbitrary body. • The concept of stress at a point is shrinking the area A into infinitesimal dimensions. Fn Fs…Eq.7 σ = lim and τ = lim …Eq.8 ∆A→0 A ∆A→0 A Note: σ and τ depend on the orientation of the plane (b) Force acting passing through P and will vary from point to point. on parts. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  18. 18. The total stress can be resolved into two components; 1) Normal stress σ perpendicular to A. 2) Shear stress τ lying in the plane of the area. • The force F makes an angle θ with the normal z to the plane x-y of the area A. • The plane containing the normal z and F intersects the plane A along a dash line that makes an angle φ with the y axis.The normal stress F The shear stress in the planeis given by σ= cos θ …Eq.9 A acting along OC has the magnitude. F F τ = sin θ …Eq.10 A F x direction τ = sin θ sin φ …Eq.11 A F y direction τ = sin θ cos φ …Eq.12 Resolutions of total stress into its A components. Suranaree University of Technology Tapany Udomphol May-Aug 2007
  19. 19. Concept of strain and types of strain The average linear strain was defined as the ratio of the change in length to the original length. δ∆L L − Lo Where e = average linear strain e= = = Lo Lo Lo δ = deformation. …Eq.2Strain at a point is the ratio of the deformation to the gaugelength as the gauge length 0.True strain or natural strain is the strain as the change inlinear dimension divided by the instantaneous value of thedimension. Lf dL Lf ε=∫ = ln …Eq.13 Lo L Lo Suranaree University of Technology Tapany Udomphol May-Aug 2007
  20. 20. Shear strain • Shear strain is the angular change in a right angle. • The angle at A, which is originally 90o, is decreased by a small amount θ when the shear stress is applied. • The shear strain γ will be given by a a θ γ = = tan θ = θ h h …Eq.14 A Shear strainSuranaree University of Technology Tapany Udomphol May-Aug 2007
  21. 21. Units of stress and other quantities • Following SI unitLength metre (m)Mass kilogram (kg) Frequency (s-1, Hz)Time second (s) Force Newton (N)Electric current ampere (A) Stress (N.m-2, Pa)Temperature Kelvin (K) Strain dimensionlessAmount of substance mole (mol)Luminous intensity candela (cd) Suranaree University of Technology Tapany Udomphol May-Aug 2007
  22. 22. Example: The shear stress required to nucleate a grain boundary crack in high-temperature deformation has been estimated to be 1  3πγ b G  2 τ =  8(1 − ν ) L     Where γb is the grain boundary surface energy ~ 2 J.m-2, G is shear modulus, 75 GPa, L is the grain boundary sliding distance, assume = grain diameter 0.01 mm, and the Poisson’s ratio ν = 0.3.Checking the unit 1  Nm N  1 2 1  2 × 2  3π × 2 × 75 × 10 9 2 m  =N  = N τ = 2 2 m  = 15.89 × 10 7 N .m −2τ =  4 m   8(1 − 0.3) × 10 − 2 × 10 −3     m    m 2   τ = 158.9 MN .m −2 = 158.9 MPa   Suranaree University of Technology Tapany Udomphol May-Aug 2007
  23. 23. References • Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition, McGraw-Hill, ISBN 0-07-100406-8. • Sanford, R.J., Principles of fracture mechanics, 2003, Prentice Hall, ISBN 0-13-192992-1. • www.indiamart.com. • www.technion.ac.il/.../ Figures/Light-steel.jpg • www.uvi.edu Suranaree University of Technology Tapany Udomphol May-Aug 2007

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