1

2 Introduction to Materials Science & EngineeringCourse Objective... Introduce fundamental concepts in Materials Science You will learn about: • material structure • how structure dictates properties • how processing can change structure This course will help you to: • use materials properly • realize new design opportunities with materials 2 2

3 Chapter 1 - IntroductionWhat is materials science? Why should we know about it? Materials drive our society Stone Age Bronze Age Iron Age Now? Silicon Age? Polymer Age? 3 3

4 Example – Hip Implant With age or certain illnesses joints deteriorate. Particularly those with large loads (such as hip). Adapted from Fig , Callister 7e. 4 4

5 Example – Hip Implant Requirements mechanical strength (many cycles)good lubricity biocompatibility Adapted from Fig , Callister 7e. 5 5

6 Example – Hip Implant Adapted from Fig , Callister 7e. 6 6

7 Hip Implant Key problems to overcomefixation agent to hold acetabular cup cup lubrication material femoral stem – fixing agent (“glue”) must avoid any debris in cup Ball Acetabular Cup and Liner Femoral Stem Adapted from chapter-opening photograph, Chapter 22, Callister 7e. 7 7

8 Example – Develop New Types of PolymersCommodity plastics – large volume ca. $0.50 / lb Ex. Polyethylene Polypropylene Polystyrene etc.  Engineering Resins – small volume > $1.00 / lb Ex. Polycarbonate Nylon Polysulfone etc.   Can polypropylene be “upgraded” to properties (and price) near those of engineering resins? 8 8

9 Structure, Processing, & Properties• Properties depend on structure ex: hardness vs structure of steel (d) 30 m 6 00 5 00 (c) 4 m Data obtained from Figs (a) and with 4 wt% C composition, and from Fig and associated discussion, Callister & Rethwisch 8e. Micrographs adapted from (a) Fig. 10.19; (b) Fig. 9.30;(c) Fig ; and (d) Fig , Callister & Rethwisch 8e. 4 00 (b) 30 m (a) 30 m Hardness (BHN) 3 00 2 00 100 0.01 0.1 1 10 100 1000 Cooling Rate (ºC/s) • Processing can change structure ex: structure vs cooling rate of steel 9 9

10 Types of Materials Metals:Strong, ductile High thermal & electrical conductivity Opaque, reflective. Polymers/plastics: Covalent bonding  sharing of e’s Soft, ductile, low strength, low density Thermal & electrical insulators Optically translucent or transparent. Ceramics: ionic bonding (refractory) – compounds of metallic & non-metallic elements (oxides, carbides, nitrides, sulfides) Brittle, glassy, elastic Non-conducting (insulators) Metals have high thermal & electrical conductivity because valence electrons are free to roam 10 10

11 The Materials Selection Process1. Pick Application Determine required Properties Properties: mechanical, electrical, thermal, magnetic, optical, deteriorative. 2. Properties Identify candidate Material(s) Material: structure, composition. 3. Material Identify required Processing Processing: changes structure and overall shape ex: casting, sintering, vapor deposition, doping forming, joining, annealing. 11 11

12 ELECTRICAL • Electrical Resistivity of Copper: Resistivity,  T (ºC)-200 -100 Cu at%Ni Cu at%Ni deformed Cu at%Ni 1 2 3 4 5 6 Resistivity,  (10-8 Ohm-m) Cu at%Ni “Pure” Cu Adapted from Fig. 18.8, Callister & Rethwisch 8e. (Fig adapted from: J.O. Linde, Ann Physik 5, 219 (1932); and C.A. Wert and R.M. Thomson, Physics of Solids, 2nd edition, McGraw-Hill Company, New York, 1970.) • Adding “impurity” atoms to Cu increases resistivity. • Deforming Cu increases resistivity. 12 12

13 THERMAL • Space Shuttle Tiles: • Thermal Conductivity of Copper:-- Silica fiber insulation offers low heat conduction. • Thermal Conductivity of Copper: -- It decreases when you add zinc! Adapted from chapter-opening photograph, Chapter 17, Callister & Rethwisch 3e. (Courtesy of Lockheed Missiles and Space Company, Inc.) Composition (wt% Zinc) Thermal Conductivity (W/m-K) 400 300 200 100 10 20 30 40 100 m Adapted from Fig. 19.4W, Callister 6e. (Courtesy of Lockheed Aerospace Ceramics Systems, Sunnyvale, CA) (Note: "W" denotes fig. is on CD-ROM.) Adapted from Fig. 19.4, Callister & Rethwisch 8e. (Fig is adapted from Metals Handbook: Properties and Selection: Nonferrous alloys and Pure Metals, Vol. 2, 9th ed., H. Baker, (Managing Editor), American Society for Metals, 1979, p. 315.) 13 13

14 MAGNETIC • Magnetic Permeability • Magnetic Storage: vs. Composition:-- Recording medium is magnetized by recording head. • Magnetic Permeability vs. Composition: -- Adding 3 atomic % Si makes Fe a better recording medium! Magnetic Field Magnetization Fe+3%Si Fe Adapted from C.R. Barrett, W.D. Nix, and A.S. Tetelman, The Principles of Engineering Materials, Fig. 1-7(a), p. 9, Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey. Fig , Callister & Rethwisch 8e. 14 14

15 OPTICAL • Transmittance:-- Aluminum oxide may be transparent, translucent, or opaque depending on the material structure. single crystal polycrystal: low porosity high porosity Adapted from Fig. 1.2, Callister & Rethwisch 8e. (Specimen preparation, P.A. Lessing; photo by S. Tanner.) 15 15

16 DETERIORATIVE • Stress & Saltwater... • Heat treatment: slows-- causes cracks! • Heat treatment: slows crack speed in salt water! “held at 160ºC for 1 hr before testing” increasing load crack speed (m/s) “as-is” 10 -10 -8 Alloy 7178 tested in saturated aqueous NaCl solution at 23ºC Adapted from Fig (b), R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials" (4th ed.), p. 505, John Wiley and Sons, (Original source: Markus O. Speidel, Brown Boveri Co.) Adapted from chapter-opening photograph, Chapter 16, Callister & Rethwisch 3e. (from Marine Corrosion, Causes, and Prevention, John Wiley and Sons, Inc., 1975.) 4 m -- material: 7150-T651 Al "alloy" (Zn,Cu,Mg,Zr) Adapted from Fig , Callister & Rethwisch 8e. (Provided courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 16 16

17 SUMMARY Course Goals: • Use the right material for the job.• Understand the relation between properties, structure, and processing. • Recognize new design opportunities offered by materials selection. 17 17

18 Chapter 2: Atomic Structure & Interatomic BondingISSUES TO ADDRESS... • What promotes bonding? • What types of bonds are there? • What properties are inferred from bonding? 18 18

19 Atomic Structure (Freshman Chem.)atom – electrons – x kg  protons neutrons atomic number = # of protons in nucleus of atom = # of electrons of neutral species A [=] atomic mass unit = amu = 1/12 mass of 12C   Atomic wt = wt of x 1023 molecules or atoms   1 amu/atom = 1g/mol C H etc. } 1.67 x kg 19 19

20 number of neutrons = N number of protons = ZAVAGADRO’S NUMBER = x = NA ATOMIC OR MOLECULAR WEIGHT = NA x WEIGHT PER ATOM. number of neutrons = N number of protons = Z A= Z + N (2.1) 20

21 Atomic Structure Valence electrons determine all of the following properties Chemical Electrical Thermal Optical 21 21

22 BOHR ATOM 22

23 WAVE MECHANICAL MODEL OF ATOM23

24 Electronic Structure Electrons have wavelike and particulate properties. This means that electrons are in orbitals defined by a probability. Each orbital at discrete energy level is determined by quantum numbers.   Quantum # Designation n = principal (energy level-shell) K, L, M, N, O (1, 2, 3, etc.) l = subsidiary (orbitals) s, p, d, f (0, 1, 2, 3,…, n -1) ml = magnetic 1, 3, 5, 7 (-l to +l) ms = spin ½, -½ 24 24

25 Electron Energy StatesElectrons... • have discrete energy states • tend to occupy lowest available energy state. 1s 2s 2p K-shell n = 1 L-shell n = 2 3s 3p M-shell n = 3 3d 4s 4p 4d Energy N-shell n = 4 Adapted from Fig. 2.4, Callister & Rethwisch 8e. 25 25

26 SURVEY OF ELEMENTS • Most elements: Electron configuration not stable.... 1s 2 2s 2p 6 3s 3p 3d 10 4s 4p Atomic # 18 36 Element 1 Hydrogen Helium 3 Lithium 4 Beryllium 5 Boron Carbon Neon 11 Sodium 12 Magnesium 13 Aluminum Argon Krypton Adapted from Table 2.2, Callister & Rethwisch 8e. • Why? Valence (outer) shell usually not filled completely. 26 26

27 Electron ConfigurationsValence electrons – those in unfilled shells Filled shells more stable Valence electrons are most available for bonding and tend to control the chemical properties example: C (atomic number = 6) 1s2 2s2 2p2 valence electrons 27 27

28 Electronic Configurations26 1s2 2s2 2p6 3s2 3p6 3d 6 4s2 valence electrons ex: Fe - atomic # = 1s 2s 2p K-shell n = 1 L-shell n = 2 3s 3p M-shell n = 3 3d 4s 4p 4d Energy N-shell n = 4 Adapted from Fig. 2.4, Callister & Rethwisch 8e. 28 28

29 The Periodic Table • Columns: Similar Valence Structure give up 1e-inert gases accept 1e- accept 2e- O Se Te Po At I Br He Ne Ar Kr Xe Rn F Cl S Li Be H Na Mg Ba Cs Ra Fr Ca K Sc Sr Rb Y Adapted from Fig. 2.6, Callister & Rethwisch 8e. Electropositive elements: Readily give up electrons to become + ions. Electronegative elements: Readily acquire electrons to become - ions. 29 29

30 Electronegativity • Ranges from 0.7 to 4.0,• Large values: tendency to acquire electrons. Smaller electronegativity Larger electronegativity Adapted from Fig. 2.7, Callister & Rethwisch 8e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University. 30 30

31 Ionic bond – metal + nonmetaldonates accepts electrons electrons   Dissimilar electronegativities   ex: MgO Mg 1s2 2s2 2p6 3s O 1s2 2s2 2p4 [Ne] 3s2  Mg2+ 1s2 2s2 2p O2- 1s2 2s2 2p6 [Ne] [Ne] 31 31

32 Electrons in different shells32

33 Electrons in Sodium and ChlorineTABLE 2.2 / P 25 3s1 3s2 3p5 33

34 Ionic Bonding - + • Occurs between + and - ions.• Requires electron transfer. • Large difference in electronegativity required. • Example: NaCl Na (metal) unstable Cl (nonmetal) electron + - Coulombic Attraction Na (cation) stable Cl (anion) 34 34

35 35

36 FORCES AND ENERGIES 36

37 37

38 38

39 Bonding Forces and Energies2.13 Calculate the force of attraction between a K+ and an O2- ion the centers of which are separated by a distance of r0 =1.5 nm. Solution The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation 2.8, which is just 39

40 (Z1 and Z2) are +1 and -2, respectively, Z1 = 1 and Z2 = 2, thenThe constant A in this expression is defined in footnote 3. Since the valences of the K+ and O2- ions (Z1 and Z2) are +1 and -2, respectively, Z1 = 1 and Z2 = 2, then 40

41 =2.05  N 41

42 IONIC FORCE / P 31 FOOT-NOTEF= (Z1 *Z2 * e^2)/(4*π*ε0*r^2); e= *10^(-19) COULOMBS ; ε0 = 8.85 * 10^(-12 ) Z1, Z2 = VALENCIES OF IONS 42

43 Ionic Bonding Energy – minimum energy most stable r A B EN = EA + ER =Energy balance of attractive and repulsive terms r A n B EN = EA + ER =   Attractive energy EA Net energy EN Repulsive energy ER Interatomic separation r Adapted from Fig. 2.8(b), Callister & Rethwisch 8e. 43 43

44 Examples: Ionic Bonding• Predominant bonding in Ceramics NaCl MgO Give up electrons Acquire electrons CaF 2 CsCl Adapted from Fig. 2.7, Callister & Rethwisch 8e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University. 44 44

45 Covalent Bonding similar electronegativity  share electronsbonds determined by valence – s & p orbitals dominate bonding Example: CH4 shared electrons from carbon atom from hydrogen atoms H C CH 4 C: has 4 valence e-, needs 4 more H: has 1 valence e-, needs 1 more Electronegativities are comparable. Adapted from Fig. 2.10, Callister & Rethwisch 8e. 45 45

46 Primary Bonding Metallic Bond -- delocalized as electron cloud Ionic-Covalent Mixed Bonding % ionic character =   where XA & XB are Pauling electronegativities %) 100 ( x Ex: MgO XMg = XO = 3.5 46 46

47 METALLIC BONDING 47

48 SECONDARY BONDING + - Arises from interaction between dipoles• Fluctuating dipoles asymmetric electron clouds + - secondary bonding H 2 ex: liquid H Adapted from Fig. 2.13, Callister & Rethwisch 8e. • Permanent dipoles-molecule induced + - -general case: secondary bonding Adapted from Fig. 2.15, Callister & Rethwisch 8e. Cl Cl -ex: liquid HCl secondary H H bonding secondary bonding -ex: polymer secondary bonding 48 48

49 Summary: Bonding Type Bond Energy Comments Ionic Large!Nondirectional (ceramics) Covalent Variable Directional (semiconductors, ceramics polymer chains) large-Diamond small-Bismuth Metallic Variable large-Tungsten Nondirectional (metals) small-Mercury Secondary smallest Directional inter-chain (polymer) inter-molecular 49 49

50 Properties From Bonding: Tm• Bond length, r • Melting Temperature, Tm r o Energy r • Bond energy, Eo Eo = “bond energy” Energy r o unstretched length smaller Tm larger Tm Tm is larger if Eo is larger. 50 50

51 Properties From Bonding : • Coefficient of thermal expansion,   L length, o unheated, T 1 heated, T 2 coeff. thermal expansion  L =  ( T - T ) 2 1 L o •  ~ symmetric at ro r o smaller  larger  Energy unstretched length Eo  is larger if Eo is smaller. 51 51

52 Summary: Primary BondsCeramics Large bond energy large Tm large E small  (Ionic & covalent bonding): Metals (Metallic bonding): Variable bond energy moderate Tm moderate E moderate  Polymers (Covalent & Secondary): Directional Properties Secondary bonding dominates small Tm small E large  secondary bonding 52 52

53 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 53 53

54 Chapter 3: The Structure of Crystalline SolidsISSUES TO ADDRESS... • How do atoms assemble into solid structures? • How does the density of a material depend on its structure? • When do material properties vary with the sample (i.e., part) orientation? 54 54

55 Energy and Packing • Non dense, random packingtypical neighbor bond length bond energy • Dense, ordered packing Energy r typical neighbor bond length bond energy Dense, ordered packed structures tend to have lower energies. 55 55

56 Materials and Packing Si Oxygen Crystalline materials...• atoms pack in periodic, 3D arrays • typical of: -metals -many ceramics -some polymers crystalline SiO2 Adapted from Fig. 3.23(a), Callister & Rethwisch 8e. Si Oxygen Noncrystalline materials... • atoms have no periodic packing • occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline noncrystalline SiO2 Adapted from Fig. 3.23(b), Callister & Rethwisch 8e. 56 56

57 Metallic Crystal StructuresHow can we stack metal atoms to minimize empty space? 2-dimensions vs. Now stack these 2-D layers to make 3-D structures 57 57

58 Metallic Crystal Structures• Tend to be densely packed. • Reasons for dense packing: - Typically, only one element is present, so all atomic radii are the same. - Metallic bonding is not directional. - Nearest neighbor distances tend to be small in order to lower bond energy. - Electron cloud shields cores from each other • Have the simplest crystal structures. We will examine three such structures... 58 58

59 Simple Cubic Structure (SC)• Rare due to low packing density (only Po has this structure) • Close-packed directions are cube edges. • Coordination # = 6 (# nearest neighbors) Click once on image to start animation (Courtesy P.M. Anderson) 59 59

60 Atomic Packing Factor (APF)Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres • APF for a simple cubic structure = 0.52 Adapted from Fig. 3.24, Callister & Rethwisch 8e. close-packed directions a R=0.5a contains 8 x 1/8 = 1 atom/unit cell atom volume atoms unit cell 4 3  (0.5a) 1 APF = 3 a unit cell volume 60 60

61 Body Centered Cubic Structure (BCC)• Atoms touch each other along cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. ex: Cr, W, Fe (), Tantalum, Molybdenum • Coordination # = 8 Adapted from Fig. 3.2, Callister & Rethwisch 8e. Click once on image to start animation (Courtesy P.M. Anderson) 2 atoms/unit cell: 1 center + 8 corners x 1/8 61 61

62 Atomic Packing Factor: BCC• APF for a body-centered cubic structure = 0.68 a R a 3 a a 2 length = 4R = Close-packed directions: 3 a Adapted from Fig. 3.2(a), Callister & Rethwisch 8e. APF = 4 3  ( a/4 ) 2 atoms unit cell atom volume a 62 62

63 Face Centered Cubic Structure (FCC)• Atoms touch each other along face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag • Coordination # = 12 Adapted from Fig. 3.1, Callister & Rethwisch 8e. Click once on image to start animation 4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8 (Courtesy P.M. Anderson) 63 63

64 Atomic Packing Factor: FCC• APF for a face-centered cubic structure = 0.74 a 2 a maximum achievable APF Close-packed directions: length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell Adapted from Fig. 3.1(a), Callister & Rethwisch 8e. APF = 4 3  ( 2 a/4 ) atoms unit cell atom volume a 64 64

65 FCC Stacking Sequence • ABCABC... Stacking Sequence • 2D ProjectionA sites B C sites A B sites C A C A A B C • FCC Unit Cell 65 65

66 Hexagonal Close-Packed Structure (HCP)• ABAB... Stacking Sequence • 3D Projection • 2D Projection c a A sites B sites Bottom layer Middle layer Top layer Adapted from Fig. 3.3(a), Callister & Rethwisch 8e. • Coordination # = 12 6 atoms/unit cell • APF = 0.74 ex: Cd, Mg, Ti, Zn • c/a = 1.633 66 66

67 Theoretical Density,  Density =  = n A  = VC NACell Unit of Volume Total in Atoms Mass Density =  = VC NA n A  = where n = number of atoms/unit cell A = atomic weight VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = x 1023 atoms/mol 67 67

68 Theoretical Density,   = a R Ex: Cr (BCC) A = 52.00 g/molR = nm n = 2 atoms/unit cell a = 4R/ 3 = nm Adapted from Fig. 3.2(a), Callister & Rethwisch 8e.  = a3 52.00 2 atoms unit cell mol g volume 6.022 x 1023 theoretical = 7.18 g/cm3 actual = 7.19 g/cm3 68 68

69 Densities of Material ClassesIn general Graphite/  metals  ceramics  polymers Metals/ Composites/ > > Ceramics/ Polymers Alloys fibers Semicond 30 Why? B ased on data in Table B1, Callister Magnesium Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum Gold, W Platinum Metals have... • close-packing (metallic bonding) • often large atomic masses 2 *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers 10 in an epoxy matrix). G raphite Silicon Glass - soda Concrete Si nitride Diamond Al oxide Zirconia Ceramics have... • less dense packing • often lighter elements 5 3 4 (g/cm ) 3 Wood AFRE * CFRE GFRE* Glass fibers Carbon fibers A ramid fibers H DPE, PS PP, LDPE PC PTFE PET PVC Silicone Polymers have... • low packing density (often amorphous) • lighter elements (C,H,O)  2 1 Composites have... • intermediate values 0.5 0.4 0.3 Data from Table B.1, Callister & Rethwisch, 8e. 69 69

70 Crystals as Building Blocks• Some engineering applications require single crystals: -- diamond single crystals for abrasives -- turbine blades Fig. 8.33(c), Callister & Rethwisch 8e. (Fig. 8.33(c) courtesy of Pratt and Whitney). (Courtesy Martin Deakins, GE Superabrasives, Worthington, OH. Used with permission.) • Properties of crystalline materials often related to crystal structure. -- Ex: Quartz fractures more easily along some crystal planes than others. (Courtesy P.M. Anderson) 70 70

71 Polycrystals Anisotropic• Most engineering materials are polycrystals. Adapted from Fig. K, color inset pages of Callister 5e. (Fig. K is courtesy of Paul E. Danielson, Teledyne Wah Chang Albany) 1 mm Isotropic • Nb-Hf-W plate with an electron beam weld. • Each "grain" is a single crystal. • If grains are randomly oriented, overall component properties are not directional. • Grain sizes typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers). 71 71

72 Single vs PolycrystalsE (diagonal) = 273 GPa E (edge) = 125 GPa • Single Crystals -Properties vary with direction: anisotropic. Data from Table 3.3, Callister & Rethwisch 8e. (Source of data is R.W. Hertzberg, Deformation and Fracture Mechanics of Engineering Materials, 3rd ed., John Wiley and Sons, 1989.) -Example: the modulus of elasticity (E) in BCC iron: • Polycrystals -Properties may/may not vary with direction. -If grains are randomly oriented: isotropic. (Epoly iron = 210 GPa) -If grains are textured, anisotropic. 200 m Adapted from Fig. 4.14(b), Callister & Rethwisch 8e. (Fig. 4.14(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].) 72 72

73 Polymorphism Two or more distinct crystal structures for the same material (allotropy/polymorphism)     titanium   , -Ti carbon diamond, graphite BCC FCC 1538ºC 1394ºC 912ºC -Fe -Fe -Fe liquid iron system 73 73

74 Crystal Systems Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. Fig. 3.4, Callister & Rethwisch 8e. 7 crystal systems 14 crystal lattices a, b, and c are the lattice constants 74 74

75 Point Coordinates z Point coordinates for unit cell center are cx y a b c 000 111 Point coordinates for unit cell center are a/2, b/2, c/ ½ ½ ½ Point coordinates for unit cell corner are 111 Translation: integer multiple of lattice constants  identical position in another unit cell z 2c y b b 75 75

76 Crystallographic Directionsz Algorithm 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a, b, and c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvw] y x ex: 1, 0, ½ Lecture 2 ended here => 2, 0, 1 => [ 201 ] -1, 1, 1 where overbar represents a negative index [ 111 ] => families of directions 76 76

77 Linear Density 3.5 nm a 2 LD  Linear Density of Atoms  LD = [110]Number of atoms Unit length of direction vector a [110] ex: linear density of Al in [110] direction  a = nm # atoms length 1 3.5 nm a 2 LD   Adapted from Fig. 3.1(a), Callister & Rethwisch 8e. 77 77

78 HCP Crystallographic Directions- a3 a1 a2 z Algorithm 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a1, a2, a3, or c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvtw] dashed red lines indicate projections onto a1 and a2 axes a1 a2 a3 -a3 2 a 1 Adapted from Fig. 3.8(a), Callister & Rethwisch 8e. [ 1120 ] ex: ½, ½, -1, => 78 78

79 HCP Crystallographic DirectionsHexagonal Crystals 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows. Fig. 3.8(a), Callister & Rethwisch 8e. - a3 a1 a2 z  ' w t v u ) ( + - 2 3 1 ] uvtw [  79 79

80 Crystallographic PlanesAdapted from Fig. 3.10, Callister & Rethwisch 8e. 80 80

81 Crystallographic PlanesMiller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices. Algorithm  1.  Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl) 81 81

82 Crystallographic Planesz x y a b c example a b c Intercepts  Reciprocals 1/ / / Reduction Miller Indices (110) example a b c z x y a b c Intercepts 1/   Reciprocals 1/½ 1/ 1/ Reduction Miller Indices (100) 82 82

83 Crystallographic Planesz x y a b c example a b c Intercepts 1/ /4 Reciprocals 1/½ 1/ /¾ /3 Reduction Miller Indices (634) (001) (010), Family of Planes {hkl} (100), (001), Ex: {100} = (100), 83 83

84 Crystallographic Planes (HCP)In hexagonal unit cells the same idea is used a2 a3 a1 z example a a a c Intercepts  -1 1 Reciprocals / -1 1 Reduction -1 1 Miller-Bravais Indices (1011) Adapted from Fig. 3.8(b), Callister & Rethwisch 8e. 84 84

85 Crystallographic PlanesWe want to examine the atomic packing of crystallographic planes Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important. Draw (100) and (111) crystallographic planes for Fe. b) Calculate the planar density for each of these planes. 85 85

86 Virtual Materials Science & Engineering (VMSE)• VMSE is a tool to visualize materials science topics such as crystallography and polymer structures in three dimensions • Available in Student Companion Site at and in WileyPLUS 86

87 VMSE: Metallic Crystal Structures & Crystallography Module• VMSE allows you to view crystal structures, directions, planes, etc. and manipulate them in three dimensions 87

88 Unit Cells for Metals FCC Structure HCP Structure• VMSE allows you to view the unit cells and manipulate them in three dimensions • Below are examples of actual VMSE screen shots FCC Structure HCP Structure 88

89 VMSE: Crystallographic Planes ExercisesAdditional practice on indexing crystallographic planes 89

90 Planar Density of (100) IronSolution:  At T < 912ºC iron has the BCC structure. 2D repeat unit R 3 4 a  (100) Radius of iron R = nm Adapted from Fig. 3.2(c), Callister & Rethwisch 8e. = Planar Density = a 2 1 atoms 2D repeat unit nm2 12.1 m2 = 1.2 x 1019 R 3 4 area 90 90

91 Planar Density of (111) IronSolution (cont):  (111) plane 1 atom in plane/ unit surface cell 2 a atoms in plane atoms above plane atoms below plane 2D repeat unit 3 h  a 2 3 2 R 16 4 a ah area        1 = nm2 atoms 7.0 m2 0.70 x 1019 3 2 R 16 Planar Density = 2D repeat unit area 91 91

92 VMSE Planar Atomic Arrangements• VMSE allows you to view planar arrangements and rotate them in 3 dimensions BCC (110) Plane 92

93 X-Ray Diffraction Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation. Can’t resolve spacings   Spacing is the distance between parallel planes of atoms.   93 93

94 X-Rays to Determine Crystal Structure• Incoming X-rays diffract from crystal planes. Adapted from Fig. 3.20, Callister & Rethwisch 8e. reflections must be in phase for a detectable signal spacing between planes d incoming X-rays outgoing X-rays detector   extra distance travelled by wave “2” “1” “2” X-ray intensity (from detector)  c d  n  2 sin Measurement of critical angle, c, allows computation of planar spacing, d. 94 94

95 X-Ray Diffraction Patternz x y a b c z x y a b c z x y a b c (110) (211) Intensity (relative) (200) Diffraction angle 2 Diffraction pattern for polycrystalline -iron (BCC) Adapted from Fig. 3.22, Callister 8e. 95 95

96 SUMMARY • Atoms may assemble into crystalline or amorphous structures.• Common metallic crystal structures are FCC, BCC, and HCP. Coordination number and atomic packing factor are the same for both FCC and HCP crystal structures. • We can predict the density of a material, provided we know the atomic weight, atomic radius, and crystal geometry (e.g., FCC, BCC, HCP). • Crystallographic points, directions and planes are specified in terms of indexing schemes. Crystallographic directions and planes are related to atomic linear densities and planar densities. 96 96

97 SUMMARY • Materials can be single crystals or polycrystalline.Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains. • Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy). • X-ray diffraction is used for crystal structure and interplanar spacing determinations. 97 97

98 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 98 98

99 Chapter 4: Imperfections in SolidsISSUES TO ADDRESS... • What are the solidification mechanisms? • What types of defects arise in solids? • Can the number and type of defects be varied and controlled? • How do defects affect material properties? • Are defects undesirable? 99 99

100 Imperfections in SolidsSolidification- result of casting of molten material 2 steps Nuclei form Nuclei grow to form crystals – grain structure Start with a molten material – all liquid nuclei liquid crystals growing Adapted from Fig. 4.14(b), Callister & Rethwisch 8e. grain structure Crystals grow until they meet each other 100 100

101 Polycrystalline MaterialsGrain Boundaries regions between crystals transition from lattice of one region to that of the other slightly disordered low density in grain boundaries high mobility high diffusivity high chemical reactivity Adapted from Fig. 4.7, Callister & Rethwisch 8e. 101 101

102 Solidification Grains can be - equiaxed (roughly same size in all directions) - columnar (elongated grains) ~ 8 cm heat flow Shell of equiaxed grains due to rapid cooling (greater T) near wall Columnar in area with less undercooling Adapted from Fig. 5.17, Callister & Rethwisch 3e. Grain Refiner - added to make smaller, more uniform, equiaxed grains. 102 102

103 Imperfections in SolidsThere is no such thing as a perfect crystal. What are these imperfections? Why are they important? Many of the important properties of materials are due to the presence of imperfections. 103 103

104 Types of Imperfections• Vacancy atoms • Interstitial atoms • Substitutional atoms Point defects • Dislocations Line defects • Grain Boundaries Area defects 104 104

105 Point Defects in Metals• Vacancies: -vacant atomic sites in a structure. Vacancy distortion of planes • Self-Interstitials: -"extra" atoms positioned between atomic sites. self- interstitial distortion of planes 105 105

106 Equilibrium Concentration: Point Defects• Equilibrium concentration varies with temperature! No. of defects Activation energy N   Q  v v  exp   No. of potential N  k T  defect sites Temperature Boltzmann's constant -23 (1.38 x 10 J/atom-K) -5 (8.62 x 10 eV/atom-K) Each lattice site is a potential vacancy site 106 106

107 Measuring Activation Energy  N v = exp  Q k T     • We can get Qv from an experiment. • Measure this... N v T exponential dependence! defect concentration • Replot it... 1/ T N v ln - Q /k slope 107 107

108 Estimating Vacancy Concentration• Find the equil. # of vacancies in 1 m3 of Cu at 1000C. • Given:  = 8.4 g / cm 3 A = 63.5 g/mol Cu Q = 0.9 eV/atom N = 6.02 x 1023 atoms/mol v A = 2.7 x 10-4 8.62 x 10-5 eV/atom-K 0.9 eV/atom 1273 K  N v  exp  Q k T      For 1 m3 , N = N A Cu  x 1 m3 = 8.0 x 1028 sites • Answer: N v = (2.7 x 10-4)(8.0 x 1028) sites = 2.2 x 1025 vacancies 108 108

109 Observing Equilibrium Vacancy Conc.• Low energy electron microscope view of a (110) surface of NiAl. • Increasing temperature causes surface island of atoms to grow. • Why? The equil. vacancy conc. increases via atom motion from the crystal to the surface, where they join the island. Click once on image to start animation Reprinted with permission from Nature (K.F. McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies in Solids and the Stability of Surface Morphology", Nature, Vol. 412, pp (2001). Image is 5.75 m by 5.75 m.) Copyright (2001) Macmillan Publishers, Ltd. I sland grows/shrinks to maintain equil. vancancy conc. in the bulk. 109 109

110 Imperfections in Metals (i)Two outcomes if impurity (B) added to host (A): • Solid solution of B in A (i.e., random dist. of point defects) OR Substitutional solid soln. (e.g., Cu in Ni) Interstitial solid soln. (e.g., C in Fe) • Solid solution of B in A plus particles of a new phase (usually for a larger amount of B) Second phase particle -- different composition -- often different structure. 110 110

111 Imperfections in Metals (ii)Conditions for substitutional solid solution (S.S.) W. Hume – Rothery rule 1. r (atomic radius) < 15% 2. Proximity in periodic table i.e., similar electronegativities 3. Same crystal structure for pure metals 4. Valency All else being equal, a metal will have a greater tendency to dissolve a metal of higher valency than one of lower valency 111 111

112 Imperfections in Metals (iii)Application of Hume–Rothery rules – Solid Solutions 1. Would you predict more Al or Ag to dissolve in Zn? 2. More Zn or Al in Cu? Element Atomic Crystal Electro- Valence Radius Structure nega (nm) tivity Cu FCC C H O Ag FCC Al FCC Co HCP Cr BCC Fe BCC Ni FCC Pd FCC Zn HCP Table on p. 118, Callister & Rethwisch 8e. 112 112

113 Impurities in Solids Specification of composition weight percentm1 = mass of component 1 nm1 = number of moles of component 1 atom percent 113 113

114 Line Defects Dislocations: Schematic of Zinc (HCP): slip steps• are line defects, • slip between crystal planes result when dislocations move, • produce permanent (plastic) deformation. Schematic of Zinc (HCP): • before deformation • after tensile elongation slip steps 114 114

115 Imperfections in SolidsLinear Defects (Dislocations) Are one-dimensional defects around which atoms are misaligned Edge dislocation: extra half-plane of atoms inserted in a crystal structure b perpendicular () to dislocation line Screw dislocation: spiral planar ramp resulting from shear deformation b parallel () to dislocation line Burger’s vector, b: measure of lattice distortion 115 115

116 Imperfections in SolidsEdge Dislocation Fig. 4.3, Callister & Rethwisch 8e. 116 116

117 Motion of Edge Dislocation• Dislocation motion requires the successive bumping of a half plane of atoms (from left to right here). • Bonds across the slipping planes are broken and remade in succession. Atomic view of edge dislocation motion from left to right as a crystal is sheared. Click once on image to start animation (Courtesy P.M. Anderson) 117 117

118 Imperfections in SolidsScrew Dislocation Screw Dislocation b Dislocation line Burgers vector b (b) (a) Adapted from Fig. 4.4, Callister & Rethwisch 8e. 118 118

119 VMSE: Screw DislocationIn VMSE: a region of crystal containing a dislocation can be rotated in 3D dislocation motion may be animated Front View Top View VMSE Screen Shots 119

120 Edge, Screw, and Mixed DislocationsAdapted from Fig. 4.5, Callister & Rethwisch 8e. 120 120

121 Imperfections in SolidsDislocations are visible in electron micrographs Fig. 4.6, Callister & Rethwisch 8e. 121 121

122 Dislocations & Crystal Structures• Structure: close-packed planes & directions are preferred. view onto two close-packed planes. close-packed directions close-packed plane (bottom) close-packed plane (top) • Comparison among crystal structures: FCC: many close-packed planes/directions; HCP: only one plane, 3 directions; BCC: none • Specimens that were tensile tested. Mg (HCP) tensile direction Al (FCC) 122 122

123 Planar Defects in SolidsOne case is a twin boundary (plane) Essentially a reflection of atom positions across the twin plane. Stacking faults For FCC metals an error in ABCABC packing sequence Ex: ABCABABC Adapted from Fig. 4.9, Callister & Rethwisch 8e. 123 123

124 Catalysts and Surface DefectsA catalyst increases the rate of a chemical reaction without being consumed Active sites on catalysts are normally surface defects Fig. 4.10, Callister & Rethwisch 8e. Single crystals of (Ce0.5Zr0.5)O2 used in an automotive catalytic converter Fig. 4.11, Callister & Rethwisch 8e. 124 124

125 Microscopic ExaminationCrystallites (grains) and grain boundaries. Vary considerably in size. Can be quite large. ex: Large single crystal of quartz or diamond or Si ex: Aluminum light post or garbage can - see the individual grains Crystallites (grains) can be quite small (mm or less) – necessary to observe with a microscope. 125 125

126 Optical Microscopy • Useful up to 2000X magnification.• Polishing removes surface features (e.g., scratches) • Etching changes reflectance, depending on crystal orientation. 0.75mm crystallographic planes Adapted from Fig. 4.13(b) and (c), Callister & Rethwisch 8e. (Fig. 4.13(c) is courtesy of J.E. Burke, General Electric Co.) Micrograph of brass (a Cu-Zn alloy) 126 126

127 Optical Microscopy Grain boundaries... Fe-Cr alloy N = 2 n -1• are imperfections, • are more susceptible to etching, • may be revealed as dark lines, • change in crystal orientation across boundary. Fe-Cr alloy (b) grain boundary surface groove polished surface (a) Adapted from Fig. 4.14(a) and (b), Callister & Rethwisch 8e. (Fig. 4.14(b) is courtesy of L.C. Smith and C. Brady, the National Bureau of Standards, Washington, DC [now the National Institute of Standards and Technology, Gaithersburg, MD].) ASTM grain size number N = 2 n -1 number of grains/in2 at 100x magnification 127 127

128 Optical Microscopy Polarized lightmetallographic scopes often use polarized light to increase contrast Also used for transparent samples such as polymers 128 128

129 Microscopy Optical resolution ca. 10-7 m = 0.1 m = 100 nmFor higher resolution need higher frequency X-Rays? Difficult to focus. Electrons wavelengths ca. 3 pm (0.003 nm) (Magnification - 1,000,000X) Atomic resolution possible Electron beam focused by magnetic lenses. 129 129

130 Scanning Tunneling Microscopy (STM)• Atoms can be arranged and imaged! Photos produced from the work of C.P. Lutz, Zeppenfeld, and D.M. Eigler. Reprinted with permission from International Business Machines Corporation, copyright 1995. Carbon monoxide molecules arranged on a platinum (111) surface. Iron atoms arranged on a copper (111) surface. These Kanji characters represent the word “atom”. 130 130

131 Summary • Point, Line, and Area defects exist in solids.• The number and type of defects can be varied and controlled (e.g., T controls vacancy conc.) • Defects affect material properties (e.g., grain boundaries control crystal slip). • Defects may be desirable or undesirable (e.g., dislocations may be good or bad, depending on whether plastic deformation is desirable or not.) 131 131

132 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 132 132

133 Chapter 7: Deformation & Strengthening MechanismsISSUES TO ADDRESS... • Why are the number of dislocations present greatest in metals? • How are strength and dislocation motion related? • Why does heating alter strength and other properties? 133 133

134 Dislocations & Materials Classes• Metals (Cu, Al): Dislocation motion easiest - non-directional bonding - close-packed directions for slip electron cloud ion cores + • Covalent Ceramics (Si, diamond): Motion difficult - directional (angular) bonding • Ionic Ceramics (NaCl): Motion difficult - need to avoid nearest neighbors of like sign (- and +) + - 134 134

135 Dislocation Motion Dislocation motion & plastic deformationMetals - plastic deformation occurs by slip – an edge dislocation (extra half-plane of atoms) slides over adjacent plane half-planes of atoms. So we saw that above the yield stress plastic deformation occurs. But how? In a perfect single crystal for this to occur every bond connecting tow planes would have to break at once! Large energy requirement Now rather than entire plane of bonds needing to be broken at once, only the bonds along dislocation line are broken at once. If dislocations can't move, plastic deformation doesn't occur! Adapted from Fig. 7.1, Callister & Rethwisch 8e. 135 135

136 Dislocation Motion A dislocation moves along a slip plane in a slip direction perpendicular to the dislocation line The slip direction is the same as the Burgers vector direction Edge dislocation Adapted from Fig. 7.2, Callister & Rethwisch 8e. Screw dislocation 136 136

137 Deformation MechanismsSlip System Slip plane - plane on which easiest slippage occurs Highest planar densities (and large interplanar spacings) Slip directions - directions of movement Highest linear densities Adapted from Fig. 7.6, Callister & Rethwisch 8e. FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed) => total of 12 slip systems in FCC For BCC & HCP there are other slip systems. 137 137

138 Stress and Dislocation Motion• Resolved shear stress, R results from applied tensile stresses Applied tensile stress: = F/A  direction slip F A slip plane normal, ns Resolved shear stress: R = F s /A direction slip AS FS direction slip Relation between  and R = FS /AS F cos  A /  nS AS 138 138

139 Critical Resolved Shear Stress• Condition for dislocation motion: 10-4 GPa to 10-2 GPa typically • Ease of dislocation motion depends on crystallographic orientation R = 0  = 90°  R =  /2  = 45°  R = 0  = 90°   maximum at  =  = 45º 139 139

140 Single Crystal Slip Adapted from Fig. 7.9, Callister & Rethwisch 8e.140 140

141 Ex: Deformation of single crystala) Will the single crystal yield? b) If not, what stress is needed?  = 60° crss = 20.7 MPa  = 35° Adapted from Fig. 7.7, Callister & Rethwisch 8e.  = 45 MPa So the applied stress of 45 MPa will not cause the crystal to yield. 141 141

142 Ex: Deformation of single crystalWhat stress is necessary (i.e., what is the yield stress, y)? So for deformation to occur the applied stress must be greater than or equal to the yield stress 142 142

143 Slip Motion in Polycrystals 300 m • Polycrystals stronger than single crystals – grain boundaries are barriers to dislocation motion. • Slip planes & directions (, ) change from one grain to another. • R will vary from one • The grain with the largest R yields first. • Other (less favorably oriented) grains yield later. Adapted from Fig. 7.10, Callister & Rethwisch 8e. (Fig is courtesy of C. Brady, National Bureau of Standards [now the National Institute of Standards and Technology, Gaithersburg, MD].) 143 143

144 Anisotropy in y • Can be induced by rolling a polycrystalline metal- before rolling - after rolling - anisotropic since rolling affects grain orientation and shape. rolling direction Adapted from Fig. 7.11, Callister & Rethwisch 8e. (Fig is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. I, Structure, p. 140, John Wiley and Sons, New York, 1964.) 235 m - isotropic since grains are equiaxed & randomly oriented. 144 144

145 Anisotropy in Deformation• The noncircular end view shows anisotropic deformation of rolled material. end view 3. Deformed cylinder plate thickness direction Photos courtesy of G.T. Gray III, Los Alamos National Labs. Used with permission. 1. Cylinder of tantalum machined from a rolled plate: rolling direction 2. Fire cylinder at a target. side view 145 145

146 Four Strategies for Strengthening: 1: Reduce Grain Size• Grain boundaries are barriers to slip. • Barrier "strength" increases with Increasing angle of misorientation. • Smaller grain size: more barriers to slip. • Hall-Petch Equation: Adapted from Fig. 7.14, Callister & Rethwisch 8e. (Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.) 146 146

147 Four Strategies for Strengthening: 2: Form Solid Solutions• Impurity atoms distort the lattice & generate lattice strains. • These strains can act as barriers to dislocation motion. • Smaller substitutional impurity Impurity generates local stress at A and B that opposes dislocation motion to the right. A B • Larger substitutional impurity Impurity generates local stress at C and D that opposes dislocation motion to the right. C D 147 147

148 Lattice Strains Around DislocationsDon’t move past one another – hardens material Adapted from Fig. 7.4, Callister & Rethwisch 8e. 148 148

149 Strengthening by Solid Solution AlloyingSmall impurities tend to concentrate at dislocations (regions of compressive strains) - partial cancellation of dislocation compressive strains and impurity atom tensile strains Reduce mobility of dislocations and increase strength Adapted from Fig. 7.17, Callister & Rethwisch 8e. 149 149

150 Strengthening by Solid Solution AlloyingLarge impurities tend to concentrate at dislocations (regions of tensile strains) Adapted from Fig. 7.18, Callister & Rethwisch 8e. 150 150

151 VMSE Solid-Solution Strengthening Tutorial151

152 Ex: Solid Solution Strengthening in Copper• Tensile strength & yield strength increase with wt% Ni. Tensile strength (MPa) wt.% Ni, (Concentration C) 200 300 400 10 20 30 40 50 Yield strength (MPa) wt.%Ni, (Concentration C) 60 120 180 10 20 30 40 50 Adapted from Fig. 7.16(a) and (b), Callister & Rethwisch 8e. • Empirical relation: • Alloying increases y and TS. 152 152

153 Four Strategies for Strengthening: 3: Precipitation Strengthening• Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View precipitate Top View Slipped part of slip plane Unslipped part of slip plane S spacing Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation “advances” but precipitates act as “pinning” sites with spacing S . • Result: 153 153

154 Application: Precipitation Strengthening• Internal wing structure on Boeing 767 Adapted from chapter-opening photograph, Chapter 11, Callister & Rethwisch 3e. (courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) • Aluminum is strengthened with precipitates formed by alloying. 1.5m Adapted from Fig , Callister & Rethwisch 8e. (Fig is courtesy of G.H. Narayanan and A.G. Miller, Boeing Commercial Airplane Company.) 154 154

155 Four Strategies for Strengthening: 4: Cold Work (Strain Hardening)• Deformation at room temperature (for most metals). • Common forming operations reduce the cross-sectional area: Adapted from Fig. 11.8, Callister & Rethwisch 8e. -Forging A o d force die blank -Rolling roll A o d -Drawing tensile force A o d die -Extrusion ram billet container force die holder die A o d extrusion 155 155

156 Dislocation Structures Change During Cold Working• Dislocation structure in Ti after cold working. • Dislocations entangle with one another during cold work. • Dislocation motion becomes more difficult. Fig. 4.6, Callister & Rethwisch 8e. (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.) 156 156

157 Dislocation Density Increases During Cold Workingtotal dislocation length unit volume Dislocation density = Carefully grown single crystals  ca. 103 mm-2 Deforming sample increases density  mm-2 Heat treatment reduces density  mm-2 Again it propagates through til reaches the edge • Yield stress increases as d increases: 157 157

158 Lattice Strain Interactions Between DislocationsAdapted from Fig. 7.5, Callister & Rethwisch 8e. 158 158

159 Impact of Cold Work As cold work is increased• Yield strength (y) increases. • Tensile strength (TS) increases. • Ductility (%EL or %AR) decreases. Adapted from Fig. 7.20, Callister & Rethwisch 8e. low carbon steel 159 159

160 Mechanical Property Alterations Due to Cold Working• What are the values of yield strength, tensile strength & ductility after cold working Cu? Cold Work Dd = 12.2 mm Copper Do = 15.2 mm 160

161 Mechanical Property Alterations Due to Cold Working• What are the values of yield strength, tensile strength & ductility for Cu for %CW = 35.6%? % Cold Work 100 300 500 700 Cu 20 40 60 % Cold Work 200 Cu 400 600 800 20 40 60 % Cold Work 20 40 60 Cu yield strength (MPa) tensile strength (MPa) ductility (%EL) y = 300 MPa 300 MPa 340 MPa TS = 340 MPa 7% %EL = 7% Adapted from Fig. 7.19, Callister & Rethwisch 8e. (Fig is adapted from Metals Handbook: Properties and Selection: Iron and Steels, Vol. 1, 9th ed., B. Bardes (Ed.), American Society for Metals, 1978, p. 226; and Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.), American Society for Metals, 1979, p. 276 and 327.) 161

162 Effect of Heat Treating After Cold Working• 1 hour treatment at Tanneal... decreases TS and increases %EL. • Effects of cold work are nullified! tensile strength (MPa) ductility (%EL) tensile strength ductility Recovery Recrystallization Grain Growth 600 300 400 500 60 50 40 30 20 annealing temperature (ºC) 200 100 700 • Three Annealing stages: Recovery Recrystallization Grain Growth Adapted from Fig. 7.22, Callister & Rethwisch 8e. (Fig is adapted from G. Sachs and K.R. van Horn, Practical Metallurgy, Applied Metallurgy, and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, 1940, p. 139.) 162 162

163 Three Stages During Heat Treatment: 1. RecoveryReduction of dislocation density by annihilation. • Scenario 1 Results from diffusion extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. atoms diffuse to regions of tension • Scenario 2 R 1. dislocation blocked; can’t move to the right Obstacle dislocation 3 . “Climbed” disl. can now move on new slip plane 2 . grey atoms leave by vacancy diffusion allowing disl. to “climb” 4. opposite dislocations meet and annihilate 163 163

164 Three Stages During Heat Treatment: 2. Recrystallization• New grains are formed that: -- have low dislocation densities -- are small in size -- consume and replace parent cold-worked grains. 33% cold worked brass New crystals nucleate after 3 sec. at 580C. 0.6 mm Adapted from Fig. 7.21(a),(b), Callister & Rethwisch 8e. (Fig. 7.21(a),(b) are courtesy of J.E. Burke, General Electric Company.) 164 164

165 As Recrystallization Continues…• All cold-worked grains are eventually consumed/replaced. After 4 seconds After 8 0.6 mm Adapted from Fig. 7.21(c),(d), Callister & Rethwisch 8e. (Fig. 7.21(c),(d) are courtesy of J.E. Burke, General Electric Company.) 165 165

166 Three Stages During Heat Treatment: 3. Grain Growth• At longer times, average grain size increases. -- Small grains shrink (and ultimately disappear) -- Large grains continue to grow After 8 s, 580ºC After 15 min, 0.6 mm Adapted from Fig. 7.21(d),(e), Callister & Rethwisch 8e. (Fig. 7.21(d),(e) are courtesy of J.E. Burke, General Electric Company.) • Empirical Relation: elapsed time coefficient dependent on material and T. grain diam. at time t. exponent typ. ~ 2 166 166

167 TR = recrystallization temperatureAdapted from Fig. 7.22, Callister & Rethwisch 8e. 167 167

168 Recrystallization TemperatureTR = recrystallization temperature = temperature at which recrystallization just reaches completion in 1 h. 0.3Tm < TR < 0.6Tm For a specific metal/alloy, TR depends on: %CW -- TR decreases with increasing %CW Purity of metal -- TR decreases with increasing purity 168 168

169 Diameter Reduction Procedure - ProblemA cylindrical rod of brass originally 10 mm (0.39 in) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 380 MPa (55,000 psi) and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm (0.30 in). Explain how this may be accomplished. 169 169

170 Diameter Reduction Procedure - SolutionWhat are the consequences of directly drawing to the final diameter? Brass Cold Work D f = 7.5 mm D o = 10 mm 170 170

171 Diameter Reduction Procedure – Solution (Cont.)420 540 6 For %CW = 43.8% Adapted from Fig. 7.19, Callister & Rethwisch 8e. y = 420 MPa TS = 540 MPa > 380 MPa %EL = < 15 This doesn’t satisfy criteria… what other options are possible? 171 171

172 Diameter Reduction Procedure – Solution (cont.)380 12 15 27 Adapted from Fig. 7.19, Callister & Rethwisch 8e. For TS > 380 MPa > 12 %CW For %EL > 15 < 27 %CW  our working range is limited to 12 < %CW < 27 172 172

173 Diameter Reduction Procedure – Solution (cont.)Cold work, then anneal, then cold work again For objective we need a cold work of 12 < %CW < 27 We’ll use 20 %CW Diameter after first cold work stage (but before 2nd cold work stage) is calculated as follows: So after the cold draw & anneal D02=0.335m  Intermediate diameter = 173 173

174 Diameter Reduction Procedure – SummaryStage 1: Cold work – reduce diameter from 10 mm to 8.39 mm Stage 2: Heat treat (allow recrystallization) Stage 3: Cold work – reduce diameter from 8.39 mm to 7.5 mm Therefore, all criteria satisfied Fig 7.19  174 174

175 Cold Working vs. Hot WorkingHot working  deformation above TR Cold working  deformation below TR 175 175

176 Grain Size Influences PropertiesMetals having small grains – relatively strong and tough at low temperatures Metals having large grains – good creep resistance at relatively high temperatures

177 Summary • Dislocations are observed primarily in metals and alloys.• Strength is increased by making dislocation motion difficult. • Strength of metals may be increased by: -- decreasing grain size -- solid solution strengthening -- precipitate hardening -- cold working • A cold-worked metal that is heat treated may experience recovery, recrystallization, and grain growth – its properties will be altered. 177 177

178 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 178 178

179 Chapter 9: Phase DiagramsISSUES TO ADDRESS... • When we combine two elements... what equilibrium state do we get? • In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature (T ) then... How many phases do we get? What is the composition of each phase? How much of each phase do we get? Phase B Phase A Nickel atom Copper atom 179 179

180 Phase Equilibria: Solubility LimitIntroduction Solutions – solid solutions, single phase Mixtures – more than one phase Adapted from Fig. 9.1, Callister 7e. Sucrose/Water Phase Diagram Pure Sugar Temperature (°C) 20 40 60 80 100 Co =Composition (wt% sugar) L (liquid solution i.e., syrup) Solubility Limit (liquid) + S (solid sugar) 4 6 8 10 Water • Solubility Limit: Max concentration for which only a single phase solution occurs. Question: What is the solubility limit at 20°C? 65 Answer: 65 wt% sugar. If Co < 65 wt% sugar: syrup If Co > 65 wt% sugar: syrup + sugar. 180 180

181 Components and Phases • Components: • Phases:  (lighter phase) The elements or compounds which are present in the mixture (e.g., Al and Cu) • Phases: The physically and chemically distinct material regions that result (e.g.,  and ). Aluminum- Copper Alloy  (lighter phase)  (darker phase) Adapted from chapter-opening photograph, Chapter 9, Callister 3e. 181 181

182 Effect of T & Composition (Co)• Changing T can change # of phases: path A to B. Changing Co can change # of phases: path B to D. B (100°C,70) 1 phase D (100°C,90) 2 phases 70 80 100 60 40 20 Temperature (°C) Co =Composition (wt% sugar) L ( liquid solution i.e., syrup) (liquid) + S (solid sugar) water- sugar system A (20°C,70) 2 phases Adapted from Fig. 9.1, Callister 7e. 182 182

183 Phase Equilibria Simple solution system (e.g., Ni-Cu solution) CrystalStructure electroneg r (nm) Ni FCC 1.9 0.1246 Cu 1.8 0.1278 Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. Ni and Cu are totally miscible in all proportions. 183 183

184 Phase Diagrams • Indicate phases as function of T, Co, and P.• For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). • 2 phases: L (liquid)  (FCC solid solution) • 3 phase fields: L + wt% Ni 20 40 60 80 100 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid) (FCC solid solution) + liquidus solidus • Phase Diagram for Cu-Ni system Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). 184 184

185 Phase Diagrams: # and types of phases• Rule 1: If we know T and Co, then we know: --the # and types of phases present. wt% Ni 20 40 60 80 100 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L + liquidus solidus Cu-Ni phase diagram • Examples: A(1100°C, 60): 1 phase:  B (1250°C,35) B (1250°C, 35): 2 phases: L +  Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991). A(1100°C,60) 185 185

186 Phase Diagrams: composition of phases• Rule 2: If we know T and Co, then we know: --the composition of each phase. wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L + liquidus solidus 30 40 50 Cu-Ni system • Examples: T A C o = 35 wt% Ni tie line 35 C o At T A = 1320°C: Only Liquid (L) C L = C o ( = 35 wt% Ni) B T 32 C L 4 C  3 At T D = 1190°C: D T Only Solid (  ) C = C o ( = 35 wt% Ni At T B = 1250°C: Both  and L Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) C L = C liquidus ( = 32 wt% Ni here) C  = C solidus ( = 43 wt% Ni here) 186 186

187 Phase Diagrams: weight fractions of phases• Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%). wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L + liquidus solidus 3 4 5 Cu-Ni system T A 35 C o 32 B D tie line R S • Examples: C o = 35 wt% Ni At T A : Only Liquid (L) W L = 100 wt%, W  = 0 At T D : Only Solid (  ) W L = 0, W = 100 wt% At T B : Both  and L = 27 wt% WL  S R + W Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) 187 187

188 The Lever Rule Tie line – connects the phases in equilibrium with each other - essentially an isotherm wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L + liquidus solidus 3 4 5 B T tie line C o S R How much of each phase? Think of it as a lever (teeter-totter) ML M R S Adapted from Fig. 9.3(b), Callister 7e. 188 188

189 Ex: Cooling in a Cu-Ni Binary• Phase diagram: Cu-Ni system. wt% Ni 20 120 130 3 4 5 110 L (liquid)  (solid) L + T(°C) A 35 C o L: 35wt%Ni Cu-Ni system • System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another;  phase field extends from 0 to 100 wt% Ni. : 46 wt% Ni L: 35 wt% Ni B 46 35 C 43 32  : 43 wt% Ni L: 32 wt% Ni D 24 36 L: 24 wt% Ni  : 36 wt% Ni E • Consider Co = 35 wt%Ni. Adapted from Fig. 9.4, Callister 7e. 189 189

190 Cored vs Equilibrium Phases• C changes as we solidify. • Cu-Ni case: First  to solidify has C = 46 wt% Ni. Last  to solidify has C = 35 wt% Ni. • Fast rate of cooling: Cored structure • Slow rate of cooling: Equilibrium structure First  to solidify: 46 wt% Ni Uniform C : 35 wt% Ni Last  < 35 wt% Ni 190 190

191 Mechanical Properties: Cu-Ni System• Effect of solid solution strengthening on: --Tensile strength (TS) --Ductility (%EL,%AR) Tensile Strength (MPa) Composition, wt% Ni Cu Ni 20 40 60 80 100 200 300 400 TS for pure Ni TS for pure Cu Elongation (%EL) Composition, wt% Ni Cu Ni 20 40 60 80 100 30 50 %EL for pure Ni for pure Cu Adapted from Fig. 9.6(a), Callister 7e. Adapted from Fig. 9.6(b), Callister 7e. --Peak as a function of Co --Min. as a function of Co 191 191

192 Binary-Eutectic Systemshas a special composition with a min. melting T. 2 components Cu-Ag system T(°C) Ex.: Cu-Ag system 1200 • 3 single phase regions L (liquid) (L,  ) 1000 • Limited solubility:  L +  L +   800 779°C   TE : mostly Cu 8.0 71.9 91.2  : mostly Ag 600 • TE : No liquid below TE    • CE : Min. melting TE 400 composition 200 20 40 60 CE 80 100 • Eutectic transition L(CE) (CE) + (CE) Co , wt% Ag Adapted from Fig. 9.7, Callister 7e. 192 192

193 EX: Pb-Sn Eutectic System (1)• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present:  +  Pb-Sn system --compositions of phases: L +   200 T(°C) 18.3 C, wt% Sn 20 60 80 100 300 L (liquid) 183°C 61.9 97.8 CO = 40 wt% Sn C = 11 wt% Sn C = 99 wt% Sn --the relative amount of each phase: W  = C - CO C - C 59 88 = 67 wt% S R+S 150 R 11 C 40 Co S 99 C W  = CO - C C - C R R+S 29 88 = 33 wt% Adapted from Fig. 9.8, Callister 7e. 193 193

194 EX: Pb-Sn Eutectic System (2)• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present:  + L Pb-Sn system --compositions of phases: L +   200 T(°C) C, wt% Sn 20 60 80 100 300 L (liquid) 183°C CO = 40 wt% Sn C = 17 wt% Sn CL = 46 wt% Sn --the relative amount of each phase: 220 17 C R 40 Co S 46 CL W  = CL - CO CL - C 6 29 = 21 wt% W L = CO - C CL - C 23 29 = 79 wt% Adapted from Fig. 9.8, Callister 7e. 194 194

195 Microstructures in Eutectic Systems: I• Co < 2 wt% Sn • Result: --at extreme ends --polycrystal of  grains i.e., only one solid phase. L +  200 T(°C) Co , wt% Sn 10 2 20 300 100  30  400 (room T solubility limit) TE (Pb-Sn System) L: Co wt% Sn  L : Co wt% Sn Adapted from Fig. 9.11, Callister 7e. 195 195

196 Microstructures in Eutectic Systems: IIL: Co wt% Sn • 2 wt% Sn < Co < 18.3 wt% Sn • Result: Initially liquid +  then  alone finally two phases  polycrystal fine -phase inclusions Pb-Sn system L +  200 T(°C) Co , wt% Sn 10 18.3 20 300 100 30  400 (sol. limit at TE) TE 2 (sol. limit at T room ) L  : Co wt% Sn   Adapted from Fig. 9.12, Callister 7e. 196 196

197 Microstructures in Eutectic Systems: III• Co = CE • Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of  and  crystals. Adapted from Fig. 9.14, Callister 7e. 160 m Micrograph of Pb-Sn eutectic microstructure Pb-Sn system L  200 T(°C) C, wt% Sn 20 60 80 100 300 L   + 183°C 40 TE L: Co wt% Sn CE 61.9 18.3 : 18.3 wt%Sn 97.8 : 97.8 wt% Sn Adapted from Fig. 9.13, Callister 7e. 197 197

198 Lamellar Eutectic StructureAdapted from Figs & 9.15, Callister 7e. 198 198

199 Microstructures in Eutectic Systems: IV• wt% Sn < Co < 61.9 wt% Sn • Result:  crystals and a eutectic microstructure WL = (1- W  ) = 50 wt% C = 18.3 wt% Sn CL = 61.9 wt% Sn S R + = • Just above TE : Adapted from Fig. 9.16, Callister 7e. Pb-Sn system L +  200 T(°C) Co, wt% Sn 20 60 80 100 300  40 TE L: Co wt% Sn L  L  18.3 61.9 S R 97.8 S R primary  eutectic  • Just below TE : C  = 18.3 wt% Sn  = 97.8 wt% Sn S R + W = = 73 wt% = 27 wt% 199 199

200 Hypoeutectic & Hypereutectic300 L T(°C) Adapted from Fig. 9.8, Callister 7e. (Fig. 9.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.) L +   200 L +   (Pb-Sn TE System)  +  100 (Figs and 9.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 175 m  hypoeutectic: Co = 50 wt% Sn Adapted from Fig. 9.17, Callister 7e. hypereutectic: (illustration only)  Adapted from Fig. 9.17, Callister 7e. (Illustration only) Co, wt% Sn 20 40 60 80 100 eutectic 61.9 eutectic: Co = 61.9 wt% Sn 160 m eutectic micro-constituent Adapted from Fig. 9.14, Callister 7e. 200 200

201 Intermetallic CompoundsAdapted from Fig. 9.20, Callister 7e. Mg2Pb Note: intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact. 201 201

202 Eutectoid & PeritecticEutectic - liquid in equilibrium with two solids L  +  cool heat intermetallic compound - cementite cool heat Eutectoid - solid phase in equation with two solid phases S S1+S3   + Fe3C (727ºC) cool heat Peritectic - liquid + solid 1  solid 2 (Fig 9.21) S1 + L S2  + L  (1493ºC) 202 202

203 Eutectoid & PeritecticPeritectic transition  + L  Cu-Zn Phase diagram Eutectoid transition   +  Adapted from Fig. 9.21, Callister 7e. 203 203

204 Iron-Carbon (Fe-C) Phase Diagram• 2 important Adapted from Fig. 9.24,Callister 7e. Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L  (austenite)  +L +Fe3C  + L+Fe3C  (Fe) Co, wt% C 1148°C T(°C) 727°C = T eutectoid points -Eutectic (A): L   + Fe3C A S R 4.30 -Eutectoid (B):    + Fe3C  Result: Pearlite = alternating layers of  and Fe3C phases 120 m (Adapted from Fig. 9.27, Callister 7e.) 0.76 C eutectoid B R S Fe3C (cementite-hard)  (ferrite-soft) 204 204

205 Hypoeutectoid Steel T(°C)  L  +L  (austenite) Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L  (austenite)  +L + Fe3C  L+Fe3C  (Fe) Co , wt% C 1148°C T(°C) 727°C (Fe-C System) C0 0.76   Adapted from Figs and 9.29,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) r s w  = /( + )  (1- Adapted from Fig. 9.30,Callister 7e. proeutectoid ferrite pearlite 100 m Hypoeutectoid steel R S  w = /( + ) Fe3C (1-  205 205

206 Hypereutectoid Steel Fe3C (cementite) L  (austenite)  +L   T(°C)1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L  (austenite)  +L +Fe3C  L+Fe3C  (Fe) Co , wt%C 1148°C T(°C) (Fe-C System)   Adapted from Figs and 9.32,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) s r w Fe3C = /( + )  =(1- Adapted from Fig. 9.33,Callister 7e. proeutectoid Fe3C 60 m Hypereutectoid steel pearlite R S w  = /( + ) Fe3C (1-  Co 0.76 206 206

207 Example: Phase EquilibriaFor a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following composition of Fe3C and ferrite () the amount of carbide (cementite) in grams that forms per 100 g of steel the amount of pearlite and proeutectoid ferrite () 207 207

208 Chapter 9 – Phase EquilibriaSolution: a) composition of Fe3C and ferrite () the amount of carbide (cementite) in grams that forms per 100 g of steel CO = 0.40 wt% C C = wt% C CFe C = 6.70 wt% C 3 Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L  (austenite)  +L + Fe3C  L+Fe3C  Co , wt% C 1148°C T(°C) 727°C CO R S CFe C 3 C 208 208

209 Chapter 9 – Phase Equilibriathe amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of  just above TE Co = 0.40 wt% C C = wt% C Cpearlite = C = 0.76 wt% C Fe3C (cementite) 1600 1400 1200 1000 800 600 400 1 2 3 4 5 6 6.7 L  (austenite)  +L + Fe3C  L+Fe3C  Co , wt% C 1148°C T(°C) 727°C CO R S C C pearlite = 51.2 g proeutectoid  = 48.8 g 209 209

210 Alloying Steel with More Elements• Teutectoid changes: • Ceutectoid changes: Adapted from Fig. 9.34,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Adapted from Fig. 9.35,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T Eutectoid (°C) wt. % of alloying elements Ti Ni Mo Si W Cr Mn C eutectoid (wt%C) 210 210

211 Summary • Phase diagrams are useful tools to determine:--the number and types of phases, --the wt% of each phase, --and the composition of each phase for a given T and composition of the system. • Alloying to produce a solid solution usually --increases the tensile strength (TS) --decreases the ductility. • Binary eutectics and binary eutectoids allow for a range of microstructures. 211 211

212 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 212 212

213 Chapter 11: Metal Alloys Applications and ProcessingISSUES TO ADDRESS... • How are metal alloys classified and how are they used? • What are some of the common fabrication techniques? • How do properties vary throughout a piece of material that has been quenched, for example? • How can properties be modified by post heat treatment? 213 213

214 Taxonomy of Metals Steels Cast Irons <1.4 wt% C 3-4.5 wt% CAlloys Steels Ferrous Nonferrous Cast Irons Cu Al Mg Ti <1.4wt%C 3-4.5 wt%C Adapted from Fig. 11.1, Callister 7e. Steels <1.4 wt% C Cast Irons 3-4.5 wt% C microstructure: ferrite, graphite cementite Fe 3 C cementite 1600 1400 1200 1000 800 600 400 1 2 4 5 6 6.7 L  austenite +L +Fe3C  ferrite + L+Fe3C  (Fe) Co , wt% C Eutectic: Eutectoid: 0.76 4.30 727°C 1148°C T(°C) Adapted from Fig. 9.24,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 214 214

215 Steels Low Alloy High Alloy low carbon <0.25 wt% C Med carbonhigh carbon wt% C plain HSLA heat treatable tool austenitic stainless Name Additions none Cr,V Ni, Mo Cr, Ni Mo Cr, V, Mo, W Cr, Ni, Mo Example 1010 4310 1040 43 40 1095 4190 304 Hardenability + ++ +++ TS - + ++ EL + + - - -- ++ Uses auto bridges crank pistons wear drills high T struc. towers shafts gears applic. saws applic. sheet press. bolts wear dies turbines vessels hammers applic. furnaces increasing strength, cost, decreasing ductility blades V. corros. resistant Based on data provided in Tables 11.1(b), 11.2(b), 11.3, and 11.4, Callister 7e. 215 215

216 Refinement of Steel from OreIron Ore Coke Limestone 3CO + Fe2O3  2Fe +3CO2 C + O2 CO2 2CO CaCO3 CaO+CO2 CaO + SiO2 + Al2O3 slag purification reduction of iron ore to metal heat generation Molten iron BLAST FURNACE air layers of coke and iron ore gas refractory vessel 216 216

217 Ferrous Alloys Iron containing – Steels - cast ironsNomenclature AISI & SAE 10xx Plain Carbon Steels 11xx Plain Carbon Steels (resulfurized for machinability) 15xx Mn (10 ~ 20%) 40xx Mo (0.20 ~ 0.30%) 43xx Ni ( %), Cr ( %), Mo ( %) 44xx Mo (0.5%) where xx is wt% C x 100 example: steel – plain carbon steel with 0.60 wt% C Stainless Steel -- >11% Cr 217 217

218 Cast Iron Ferrous alloys with > 2.1 wt% Cmore commonly wt%C low melting (also brittle) so easiest to cast Cementite decomposes to ferrite + graphite Fe3C  3 Fe () + C (graphite) generally a slow process So phase diagram for this system is different (Fig 12.4) 218 218

219 Fe-C True Equilibrium Diagram1600 1400 1200 1000 800 600 400 1 2 3 4 90 L  +L  + Graphite Liquid + Graphite (Fe) Co , wt% C 0.65 740°C T(°C)  + Graphite 100 1153°C Austenite 4.2 wt% C  Graphite formation promoted by Si > 1 wt% slow cooling Cast irons have graphite Adapted from Fig. 11.2,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 219 219

220 Types of Cast Iron Gray iron graphite flakesweak & brittle under tension stronger under compression excellent vibrational dampening wear resistant Ductile iron add Mg or Ce graphite in nodules not flakes matrix often pearlite - better ductility Adapted from Fig. 11.3(a) & (b), Callister 7e. 220 220

221 Types of Cast Iron White iron <1wt% Si so harder but brittlemore cementite Malleable iron heat treat at ºC graphite in rosettes more ductile Adapted from Fig. 11.3(c) & (d), Callister 7e. 221 221

222 Production of Cast IronAdapted from Fig.11.5, Callister 7e. 222 222

223 Limitations of Ferrous AlloysRelatively high density Relatively low conductivity Poor corrosion resistance 223 223

224 Nonferrous Alloys NonFerrous Alloys • Cu Alloys • Al AlloysBrass: Zn is subst. impurity (costume jewelry, coins, corrosion resistant) Bronze : Sn, Al, Si, Ni are subst. impurity (bushings, landing gear) Cu-Be : precip. hardened for strength • Al Alloys -lower  : 2.7g/cm3 -Cu, Mg, Si, Mn, Zn additions -solid sol. or precip. strengthened (struct. aircraft parts & packaging) NonFerrous Alloys • Mg Alloys -very low  : 1.7g/cm3 -ignites easily - aircraft, missiles • Ti Alloys -lower  : 4.5g/cm3 vs 7.9 for steel -reactive at high T - space applic. • Refractory metals -high melting T -Nb, Mo, W, Ta • Noble metals -Ag, Au, Pt - oxid./corr. resistant Based on discussion and data provided in Section 11.3, Callister 7e. 224 224

225 Metal Fabrication How do we fabricate metals?Blacksmith - hammer (forged) Molding - cast Forming Operations Rough stock formed to final shape Hot working vs Cold working • T high enough for • well below Tm recrystallization • work hardening • Larger deformations • smaller deformations 225 225

226 Metal Fabrication Methods - IFORMING CASTING JOINING A o d force die blank • Forging (Hammering; Stamping) (wrenches, crankshafts) often at elev. T Adapted from Fig. 11.8, Callister 7e. roll A o d • Rolling (Hot or Cold Rolling) (I-beams, rails, sheet & plate) tensile force A o d die • Drawing (rods, wire, tubing) die must be well lubricated & clean ram billet container force die holder die A o d extrusion • Extrusion (rods, tubing) ductile metals, e.g. Cu, Al (hot) 226 226

227 Metal Fabrication Methods - IIFORMING CASTING JOINING Casting- mold is filled with metal metal melted in furnace, perhaps alloying elements added. Then cast in a mold most common, cheapest method gives good production of shapes weaker products, internal defects good option for brittle materials 227 227

228 Metal Fabrication Methods - IIFORMING CASTING JOINING • Sand Casting (large parts, e.g., auto engine blocks) trying to hold something that is hot what will withstand >1600ºC? cheap - easy to mold => sand!!! pack sand around form (pattern) of desired shape Sand molten metal 228 228

229 Metal Fabrication Methods - IIFORMING CASTING JOINING • Sand Casting (large parts, e.g., auto engine blocks) Investment Casting pattern is made from paraffin. mold made by encasing in plaster of paris melt the wax & the hollow mold is left pour in metal Sand molten metal • Investment Casting (low volume, complex shapes e.g., jewelry, turbine blades) plaster die formed around wax prototype wax 229 229

230 Metal Fabrication Methods - IIFORMING CASTING JOINING • Sand Casting (large parts, e.g., auto engine blocks) • Die Casting (high volume, low T alloys) Sand molten metal • Continuous Casting (simple slab shapes) molten solidified • Investment Casting (low volume, complex shapes e.g., jewelry, turbine blades) plaster die formed around wax prototype wax 230 230

231 Metal Fabrication Methods - IIIFORMING CASTING JOINING • Powder Metallurgy (materials w/low ductility) • Welding (when one large part is impractical) • Heat affected zone: (region in which the microstructure has been changed). Adapted from Fig. 11.9, Callister 7e. (Fig from Iron Castings Handbook, C.F. Walton and T.J. Opar (Ed.), 1981.) piece 1 piece 2 fused base metal filler metal (melted) base metal (melted) unaffected heat affected zone pressure heat point contact at low T densification by diffusion at higher T area contact densify 231 231

232 Thermal Processing of MetalsAnnealing: Heat to Tanneal, then cool slowly. Types of Annealing • Stress Relief: Reduce stress caused by: -plastic deformation -nonuniform cooling -phase transform. • Spheroidize (steels): Make very soft steels for good machining. Heat just below TE & hold for 15-25 h. • Full Anneal (steels): Make soft steels for good forming by heating to get  , then cool in furnace to get coarse P. • Process Anneal: Negate effect of cold working by (recovery/ recrystallization) • Normalize (steels): Deform steel with large grains, then normalize to make grains small. Based on discussion in Section 11.7, Callister 7e. 232 232

233 Heat Treatments Annealing Quenching Tempered Martensite a) b) c) TE800 Austenite (stable) a) b) Annealing TE T(°C) A Quenching P 600 Tempered Martensite B 400 A 100% Adapted from Fig , Callister 7e. 50% 0% c) 0% 200 M + A 50% M + A 90% -1 3 5 10 10 10 10 time (s) 233 233

234 Hardenability--Steels• Ability to form martensite • Jominy end quench test to measure hardenability. 24°C water specimen (heated to  phase field) flat ground Rockwell C hardness tests Adapted from Fig , Callister 7e. (Fig adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1978.) • Hardness versus distance from the quenched end. Hardness, HRC Distance from quenched end Adapted from Fig , Callister 7e. 234 234

235 Why Hardness Changes W/Position• The cooling rate varies with position. 60 Martensite Martensite + Pearlite Fine Pearlite Pearlite Hardness, HRC 40 20 distance from quenched end (in) 1 2 3 600 400 200 A  M  P 0.1 1 10 100 1000 T(°C) M(start) Time (s) 0% 100% M(finish) Adapted from Fig , Callister 7e. (Fig adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 376.) 235 235

236 Hardenability vs Alloy CompositionCooling rate (°C/s) Hardness, HRC 20 40 60 10 30 50 Distance from quenched end (mm) 2 100 3 4140 8640 5140 1040 80 %M 4340 • Jominy end quench results, C = 0.4 wt% C Adapted from Fig , Callister 7e. (Fig adapted from figure furnished courtesy Republic Steel Corporation.) • "Alloy Steels" (4140, 4340, 5140, 8640) --contain Ni, Cr, Mo (0.2 to 2wt%) --these elements shift the "nose". --martensite is easier to form. T(°C) 10 -1 3 5 200 400 600 800 Time (s) M(start) M(90%) shift from A to B due to alloying B A TE 236 236

237 Quenching Medium & Geometry• Effect of quenching medium: Medium air oil water Severity of Quench low moderate high Hardness low moderate high • Effect of geometry: When surface-to-volume ratio increases: --cooling rate increases --hardness increases Position center surface Cooling rate low high Hardness 237 237

238 Precipitation Hardening• Particles impede dislocations. • Ex: Al-Cu system • Procedure: 10 20 30 40 50 wt% Cu L +L    +L 300 400 500 600 700 (Al) T(°C) composition range needed for precipitation hardening CuAl2 A --Pt A: solution heat treat (get  solid solution) B Pt B --Pt B: quench to room temp. C --Pt C: reheat to nucleate small  crystals within  crystals. Other precipitation systems: • Cu-Be • Cu-Sn • Mg-Al Adapted from Fig , Callister 7e. (Fig adapted from J.L. Murray, International Metals Review 30, p.5, 1985.) Temp. Time Pt A (sol’n heat treat) Pt C (precipitate  Adapted from Fig , Callister 7e. 238 238

239 Precipitate Effect on TS, %EL• 2014 Al Alloy: • TS peaks with precipitation time. • Increasing T accelerates process. • %EL reaches minimum with precipitation time. %EL (2 in sample) 10 20 30 1min 1h 1day 1mo 1yr 204°C 149 °C precipitation heat treat time precipitation heat treat time tensile strength (MPa) 200 300 400 100 1min 1h 1day 1mo 1yr 204°C non-equil. solid solution many small precipitates “aged” fewer large “overaged” 149°C Adapted from Fig (a) and (b), Callister 7e. (Fig adapted from Metals Handbook: Properties and Selection: Nonferrous Alloys and Pure Metals, Vol. 2, 9th ed., H. Baker (Managing Ed.), American Society for Metals, p. 41.) 239 239

240 Metal Alloy Crystal StuctureAlloys substitutional alloys can be ordered or disordered disordered solid solution ordered - periodic substitution example: CuAu FCC Cu Au 240 240

241 Metal Alloy Crystal StuctureInterstitial alloys (compounds) one metal much larger than the other smaller metal goes in ordered way into interstitial “holes” in the structure of larger metal Ex: Cementite – Fe3C 241 241

242 Metal Alloy Crystal StuctureConsider FCC structure what types of holes are there? Octahedron - octahedral site = OH Tetrahedron - tetrahedral site = TD 242 242

243 Metal Alloy Crystal StuctureInterstitials such as H, N, B, C FCC has 4 atoms per unit cell OH sites 4 OH sites TD sites 8 TD sites metal atoms 243 243

244 Summary • Steels: increase TS, Hardness (and cost) by adding--C (low alloy steels) --Cr, V, Ni, Mo, W (high alloy steels) --ductility usually decreases w/additions. • Non-ferrous: --Cu, Al, Ti, Mg, Refractory, and noble metals. • Fabrication techniques: --forming, casting, joining. • Hardenability --increases with alloy content. • Precipitation hardening --effective means to increase strength in Al, Cu, and Mg alloys. 244 244

245 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: 245 245

246 CHAPTER 14: POLYMER STRUCTURESISSUES TO ADDRESS... • What are the basic microstructural features? • How are polymer properties effected by molecular weight? • How do polymeric crystals accommodate the polymer chain? 246 246

247 Chapter 14 – Polymers What is a polymer? Poly mer many repeat unit C HPolyethylene (PE) Cl C H Polyvinyl chloride (PVC) H Polypropylene (PP) C CH3 Adapted from Fig. 14.2, Callister 7e. 247 247

248 Ancient Polymer HistoryOriginally natural polymers were used Wood – Rubber Cotton – Wool Leather – Silk Oldest known uses Rubber balls used by Incas Noah used pitch (a natural polymer) for the ark 248 248

249 Polymer Composition Most polymers are hydrocarbons– i.e. made up of H and C Saturated hydrocarbons Each carbon bonded to four other atoms CnH2n+2 249 249

250 250 250

251 Unsaturated HydrocarbonsDouble & triple bonds relatively reactive – can form new bonds Double bond – ethylene or ethene - CnH2n 4-bonds, but only 3 atoms bound to C’s Triple bond – acetylene or ethyne - CnH2n-2 251 251

252 Isomerism Isomerism two compounds with same chemical formula can have quite different structures Ex: C8H18 n-octane 2-methyl-4-ethyl pentane (isooctane)  252 252

253 Chemistry of Polymers Free radical polymerizationInitiator: example - benzoyl peroxide 253 253

254 Chemistry of Polymers Note: polyethylene is just a long HCAdapted from Fig. 14.1, Callister 7e. Polymer- can have various lengths depending on number of repeat units Note: polyethylene is just a long HC - paraffin is short polyethylene 254 254

255 Bulk or Commodity PolymersRelatively few polymers responsible for virtually all polymers sold – these are the bulk or commodity polymers 255 255

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257 257 257

258 MOLECULAR WEIGHT • Molecular weight, Mi: Mass of a mole of chains.Lower M higher M Simple for small molecules All the same size Number of grams/mole Polymers – distribution of chain sizes Mw is more sensitive to higher molecular weights Adapted from Fig. 14.4, Callister 7e. 258 258

259 Molecular Weight CalculationExample: average mass of a class 259

260 Degree of Polymerization, nn = number of repeat units per chain ni = 6 Chain fraction mol. wt of repeat unit i 260

261 End to End Distance, r Adapted from Fig. 14.6, Callister 7e. 261

262 Molecular Structures • Covalent chain configurations and strength:B ranched Cross-Linked Network Linear secondary bonding Direction of increasing strength Adapted from Fig. 14.7, Callister 7e. 262

263 Polymers – Molecular ShapeConformation – Molecular orientation can be changed by rotation around the bonds note: no bond breaking needed Adapted from Fig. 14.5, Callister 7e. 263

264 Polymers – Molecular ShapeConfigurations – to change must break bonds Stereoisomerism mirror plane 264

265 Tacticity Tacticity – stereoregularity of chainisotactic – all R groups on same side of chain syndiotactic – R groups alternate sides atactic – R groups random 265

266 cis/trans Isomerism cis trans cis-isoprene (natural rubber)bulky groups on same side of chain trans trans-isoprene (gutta percha) bulky groups on opposite sides of chain 266

267 Copolymers two or more monomers polymerized togetherAdapted from Fig. 14.9, Callister 7e. two or more monomers polymerized together random – A and B randomly vary in chain alternating – A and B alternate in polymer chain block – large blocks of A alternate with large blocks of B graft – chains of B grafted on to A backbone A – B – random alternating block graft 267

268 Polymer CrystallinityAdapted from Fig , Callister 7e. Ex: polyethylene unit cell Crystals must contain the polymer chains in some way Chain folded structure Adapted from Fig , Callister 7e. 10 nm 268

269 Polymer CrystallinityPolymers rarely 100% crystalline Too difficult to get all those chains aligned crystalline region • % Crystallinity: % of material that is crystalline. -- TS and E often increase with % crystallinity. -- Annealing causes crystalline regions to grow. % crystallinity increases. amorphous region Adapted from Fig , Callister 6e. (Fig is from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, John Wiley and Sons, Inc., 1965.) 269

270 Polymer Crystal Forms Single crystals – only if slow careful growthAdapted from Fig , Callister 7e. 270

271 Polymer Crystal Forms Spherulites – fast growth – forms lamellar (layered) structures Spherulite surface Nucleation site Adapted from Fig , Callister 7e. 271

272 Spherulites – crossed polarizersMaltese cross Adapted from Fig , Callister 7e. 272