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ME 240: Introduction to Engineering Materials CHAPTER 8 Chapter 8

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ME 240: Introduction to Engineering Materials CHAPTER 8 Chapter 8. Failure 8.1 ME 240: Introduction to Engineering Materials FRACTU RE The Fundamentals Fracture = separation of body into two or more pieces due to application of static stress Tensile, Compressive Shear or torsional Lets talk about the tensile loading of materials Modes of fracture Chapter 8. Failure DUCTILE BRITTLE 8.2 ME 240: Introduction to Engineering Materials Back to the The tensile test Chapter 8. Failure 8.3 ME 240: Introduction to Engineering Materials Ductile fracture in copper nucleating around inclusions Chapter 8. Failure 8.4 ME 240: Introduction to Engineering Materials Cup and Cone fracture Chapter 8. Failure Brittle fracture 8.5 ME 240: Introduction to Engineering Materials Transgranular vs. intergranular fracture (cleavage fracture) Chapter 8. Failure 8.6 ME 240: Introduction to Engineering Materials Professor Inglis (1913) The birth of the term ‘’stress concentration’’ Large structures Stress trajectories y x Chapter 8. Failure 8.7 ME 240: Introduction to Engineering Materials Griffith (1920) – application of Inglis to cracks and defects ~ 100 microns in diameter Bent to a strain of 7.5% i.e. 5000MPa. Normal strength of glass = 100-200MPa. HE PROVED HIS POINT!! Crack-free silica (glass) fiber Chapter 8. Failure 8.8 ME 240: Introduction to Engineering Materials Ellipse 2b But radius of curvature = ? t = b2/a For circular hole Chapter 8. Failure Stress concentration factor b= a = r ? Kt = 1 + 2 (1/1)1/2) = 3 8.9 ME 240: Introduction to Engineering Materials Stress concentration factor vs. Specimen Geometry/configuration So what happens if a crack intersects a hole? Chapter 8. Failure 8.10 ME 240: Introduction to Engineering Materials Griffith and his Energy criterion Crack propagates when favorable, i.e. system reduces its total energy ? Relaxed material behind crack = Elastic strain energy released Crack having surface energy (?s) a ?? 2 a G? ?Gc ?2? ?R ?cons tan t E Chapter 8. Failure 8.11 ME 240: Introduction to Engineering Materials What about ductile materials But for v. ductile materials ?p >>> ?s Define the strain energy release rate Gc (IRWIN) Hence Chapter 8. Failure 8.12 ME 240: Introduction to Engineering Materials Modes of fracture Chapter 8. Failure 8.13 ME 240: Introduction to Engineering Materials Stress intensity factor AND Chapter 8. Failure = 8.14 ME 240: Introduction to Engineering Materials What about ductile materials ? consider ?y (i.e. y means direction not yield) Plastic zone Chapter 8. Failure 8.15 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.16 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.17 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.18 ME 240: Introduction to Engineering Materials Plane strain fracture toughness Chapter 8. Failure To be plane strain 8.19 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.20 ME 240: Introduction to Engineering Materials Design using fracture mechanics Example: Compare the critical flaw sizes in the following metals subjected to tensile stress 1500MPa and K = 1.12 ???a. KIc (MPa.m1/2) Critical flaw size (microns) 7000 Al 250 280 Steel 50 0.45 Zirconia(ZrO2) 2 16 Toughened Zirconia 12 SOLUTION Where Y = 1.12. Substitute values Chapter 8. Failure 8.21 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.22 ME 240: Introduction to Engineering Materials This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of KIc, Y, and the largest value of a in the material. This requires that we solve for ? from Equation (8.7). Thus c Therefore, fracture will not occur because this specimen will tolerate a stress of 1380 MPa (199,500 psi) before fracture, which is greater than the applied stress of 1030 MPa (150,000 psi). Chapter 8. Failure 8.23 ME 240: Introduction to Engineering Materials IMPACT TESTING Tensile test vs. real life failures Impact energy measured or notch toughness Chapter 8. Failure 8.24 ME 240: Introduction to Engineering Materials How do we specify a ductile-brittle transition temperature (DBTT)?? Also HCP Not all materials exhibit DBTT Chapter 8. Failure 8.25 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.26 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.27 ME 240: Introduction to Engineering Materials FATIGUE Failure under repeated cyclic loading Definitions Mean stress Range of stress Stress amplitude For Reversed cycle fatigue R = -1 Chapter 8. Failure 8.28 ME 240: Introduction to Engineering Materials How do you practically make these fatigue measurements ? Chapter 8. Failure 8.29 ME 240: Introduction to Engineering Materials Endurance limit or Fatigue limit vs. fatigue life e.g. Al e.g. steels & Ti Alloys e.g. 35-60% of TS High cycle vs. low cycle fatigue Chapter 8. Failure 8.30 ME 240: Introduction to Engineering Materials How does a fatigue crack form and propagate? STAGE II ? Nf = Ni + Initiation Np Propagation Chapter 8. Failure 8.31 ME 240: Introduction to Engineering Materials Striations Beachmarks Chapter 8. Failure 8.32 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.33 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.34 ME 240: Introduction to Engineering Materials One of failure analysis goals = prediction of fatigue life of component knowing service constraint and conducting Lab tests Ignores crack initiation and fracture times Chapter 8. Failure 8.35 ME 240: Introduction to Engineering Materials Can make extrapolations To obtain log A Chapter 8. Failure 8.36 ME 240: Introduction to Engineering Materials QUESTION Eqn. 8.26 Chapter 8. Failure 8.37 ME 240: Introduction to Engineering Materials Other effects: a) Mean stress b)stress concentrations c) Surface treatments Chapter 8. Failure 8.38 ME 240: Introduction to Engineering Materials CREEP Time dependent and permanent deformation of materials when subjected to load or stress (significant at T = 0.4Tm) ? = f (T, t, ?) Lead pipes deforming under their own weight Chapter 8. Failure 8.39 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.40 ME 240: Introduction to Engineering Materials Effect of temperature and stress Chapter 8. Failure 8.41 ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.42 ME 240: Introduction to Engineering Materials Data extrapolation methods – e.g. prolonged exposures (years) Perform creep tests in excess of T and shorter time but at same stress level Chapter 8. Failure 8.43 ME 260: Introduction to Engineering Materials CHAPTER 9 Chapter 9. PHASE DIAGRAMS 9.1 ME 260: Introduction to Engineering Materials Definitions System, e.g. Ag - Cu system Component of a system, e.g. Cu and Ag Solute and solvent Solubility limit Chapter 9. PHASE DIAGRAMS 9.2 ME 260: Introduction to Engineering Materials Phase = homogeneous portion of a system that has uniform physical and chemical characteristics E.g. – pure elements - syrup and sugar - Ice and water - FCC and BCC steel (polymorphs) Single phase system = Homogeneous system Multi phase system = Heterogeneous system or mixtures MICROSTRUCTURE Single phase vs. two-phase Phase equilibrium = stable configuration with lowest free-energy Any change in T, Comp. Pressure = increase in free energy and away from equilibrium And move to another state Chapter 9. PHASE DIAGRAMS 9.3 ME 260: Introduction to Engineering Materials BINARY ‘ISOMORPHOUS’ SYSTEMS Phase amount ? Lever Rule Chapter 9. PHASE DIAGRAMS 9.4 ME 260: Introduction to Engineering Materials The Lever Rule R S W’L W’? W’? x S= W’L x R Weight W’? = W’L R/S W’? + W’L = W’LR/S + W’L = W’L (R/S+1) = W’L((R+S)/S) Phase amount ? Lever Rule Hence, W’? / (W’? +W’L) = W? = W’L R/S W’L((R+S)/S) Therefore, W? = R/(R+S) Chapter 9. PHASE DIAGRAMS 9.5 ME 260: Introduction to Engineering Materials Volume fraction Volumes of respective phases in the alloy Similarly, Chapter 9. PHASE DIAGRAMS 9.6 ME 260: Introduction to Engineering Materials Microstructural Developments – Equilibrium cooling Chapter 9. PHASE DIAGRAMS 9.7 ME 260: Introduction to Engineering Materials What happens to the same alloy if it was cooled in a non-equilibrium manner (i.e. fast) ? Segregation CORING Chapter 9. PHASE DIAGRAMS 9.8 ME 260: Introduction to Engineering Materials Gibb’s Phase Rule 1876-78 : Prof. Of Mathematical Physics at Yale Number of non-compositional variables Number of phases present Number of degrees of freedom Number of components in system For single phase molten alloy containing 2 metals P = 1, C= 2, N = 1 Therefore, F = 2 Temperature and composition i.e. both temperature and comp. Can be varied independently without introducing a second phase. Chapter 9. PHASE DIAGRAMS 9.9 ME 260: Introduction to Engineering Materials For two-phase field Therefore F = 1. ? P=2 But why do we need to know all this?? Spot non-equilibrium phase formation Chapter 9. PHASE DIAGRAMS 9.10 ME 260: Introduction to Engineering Materials Effect on Mechanical properties Chapter 9. PHASE DIAGRAMS 9.11 ME 260: Introduction to Engineering Materials Binary Eutectic System Solidus and Solvus lines Solidus line at TE = Eutectic isotherm Solid solubility Invarient point Terminal solid solutions Single phase regions separated by 2-phase regions Eutectic reaction Chapter 9. PHASE DIAGRAMS 9.12 ME 260: Introduction to Engineering Materials DEVELOPMENT OF MICROSTRUCTURE IN EUTECTIC ALLOYS Chapter 9. PHASE DIAGRAMS 9.13 ME 260: Introduction to Engineering Materials Chapter 9. PHASE DIAGRAMS 9.14 ME 260: Introduction to Engineering Materials Eutectic reaction Tin rich Chapter 9. PHASE DIAGRAMS Pb rich 9.15 ME 260: Introduction to Engineering Materials Primary ? + Eutectic microconstituents Chapter 9. PHASE DIAGRAMS 9.16 ME 260: Introduction to Engineering Materials Can use the Lever Rule again EXAMPLE Chapter 9. PHASE DIAGRAMS 9.17 ME 260: Introduction to Engineering Materials Intermediate solid solutions Chapter 9. PHASE DIAGRAMS 9.18 ME 260: Introduction to Engineering Materials INTERMETALLICS, e.g. Mg2Pb Mg-Mg2Pb system + Mg2Pb–Pb system Chapter 9. PHASE DIAGRAMS 9.19 ME 260: Introduction to Engineering Materials THE IRON-CARBON SYSTEM Steel and cast iron Iron 912oC 1394oC Ferrite ? (BCC)?Austenite ? (FCC)? Ferrite ? (BCC) All steels and cast irons have compositions