1 IIII mmmm pppp eeee rrrr ffff eeee cccc tttt iiii oooo nnnn ssss IIII nnnn S S S S oooo llll iiii dddd ssss Chapter 5.

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1 1 IIII mmmm pppp eeee rrrr ffff eeee cccc tttt iiii oooo nnnn ssss IIII nnnn S S S S oooo llll iiii dddd ssss Chapter 5

2 2 Learning Goals  Learn The Forms of Defects in Solids Use metals as Prototypical Example  How the number and type of defects Can be varied and controlled  How defects affect material properties  Determine if “Defects” are Desireable UNdesirable

3 3 Classes of Imperfections  POINT Defects Atomic Vacancies Interstitial atoms Substitutional atoms  LINE Defects (Plane Edge) Dislocations  Area Defects Grain Boundaries –Usually 3-D HRTEM image of SrTiO3 Grain Boundary * ULTRAMICROSCOPY, vol 86 (2001) pp 303-318

4 4  Point defects - Imperfections, such as vacancies, that are located typically at one (in some cases a few) sites in the crystal.  Extended defects - Defects that involve several atoms/ions and thus occur over a finite volume of the crystalline material (e.g., dislocations, stacking faults, etc.).  Vacancy - An atom or an ion missing from its regular crystallographic site.  Interstitial defect - A point defect produced when an atom is placed into the crystal at a site that is normally not a lattice point.  Substitutional defect - A point defect produced when an atom is removed from a regular lattice point and replaced with a different atom, usually of a different size. Section 5.1 Point Defects

5 5  Vacancy  MISSING atom at Lattice Site  Self-Interstitial  “Extra” Atom “Squeezed into the Lattice Structure

6 6 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.1 Point defects: (a) vacancy, (b) interstitial atom, (c) small substitutional atom, (d) large substitutional atom, (e) Frenkel defect, (f) Schottky defect. All of these defects disrupt the perfect arrangement of the surrounding atoms.

7 7 Point Defect Concentration  Equilibrium Defect Concentration Varies With Temperature  k = 1.38x10 -23 J/at-K 8.62x10 -5 eV/at-K  N  Every Lattice Site

8 8 Measure Activation Energy  Recall The Defect Density Eqn  Take the ln of Eqn  This form of a Negative Exponential is called an Arrhenius Relation Svante Arrhenius: 1859-1927, Chem Nobel 1903  This of the form

9 9 Measure Activation Energy cont  Meausure N D /N vs T  Find the Activation Energy from the Slope  RePlot in Linear Form y = mx + b NDND N T exponential dependence 1/T N N D ln 1 -Q D /k slope

10 10 Vacancy Concentration Exmpl  In Defect Density Rln Q D Can Take Two forms Q v  Vacancies Q i  Interstitials  Consider a Q v Case Copper at 1000 C Q v = 0.9 eV/at A Cu = 63.5 g/mol  = 8400 kg/cu-m  Find the Vacancy Density First Find N in atoms per cu-m 

11 11 Vacancy Concentration cont  Since Units Chk:  At 180C (Pizza Oven) The Vacancy Rate  98 pptr  Now apply the Arrhenius Relation @1000 ºC   275 ppm Vacancy Rate

12 12 Calculate the concentration of vacancies in copper at room temperature (25 o C). What temperature will be needed to heat treat copper such that the concentration of vacancies produced will be 1000 times more than the equilibrium concentration of vacancies at room temperature? Assume that 20,000 cal are required to produce a mole of vacancies in copper. Example 5.1 SOLUTION The lattice parameter of FCC copper is 0.36151 nm. The basis is 1, therefore, the number of copper atoms, or lattice points, per cm 3 is: Example 5.1 The Effect of Temperature on Vacancy Concentrations

13 13 Example 5.1 SOLUTION (Continued) At room temperature, T = 25 + 273 = 298 K: We could do this by heating the copper to a temperature at which this number of vacancies forms:

14 14 Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87 g/cm 3. The lattice parameter of the iron is 2.866  10 -8 cm. Example 5.2 SOLUTION The expected theoretical density of iron can be calculated from the lattice parameter and the atomic mass. Example 5.2 Vacancy Concentrations in Iron

15 15 Example 5.2 SOLUTION (Continued) Let’s calculate the number of iron atoms and vacancies that would be present in each unit cell for the required density of 7.87 g/cm 3 : Or, there should be 2.00 – 1.9971 = 0.0029 vacancies per unit cell. The number of vacancies per cm 3 is:

16 16 In FCC iron, carbon atoms are located at octahedral sites at the center of each edge of the unit cell (1/2, 0, 0) and at the center of the unit cell (1/2, 1/2, 1/2). In BCC iron, carbon atoms enter tetrahedral sites, such as 1/4, 1/2, 0. The lattice parameter is 0.3571 nm for FCC iron and 0.2866 nm for BCC iron. Assume that carbon atoms have a radius of 0.071 nm. (1) Would we expect a greater distortion of the crystal by an interstitial carbon atom in FCC or BCC iron? (2) What would be the atomic percentage of carbon in each type of iron if all the interstitial sites were filled? Homework: Example 5.3 Sites for Carbon in Iron

17 17 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.3 (a) The location of the ¼, ½, 0 interstitial site in BCC metals, showing the arrangement of the normal atoms and the interstitial atom (b) ½, 0, 0 site in FCC metals, (for Example 5.3). (c) Edge centers and cube centers are some of the interstitial sites in the FCC structure (Example 5.3).

18 18 Three separate samples of germanium (Ge) crystals contain small concentrations of either silicon (Si), arsenic (As), or boron (B) as dopants. Based on the valence of these elements, what type of semiconductivity is expected from these materials? Assume that these elements will occupy Ge sites. Example 5.4 SOLUTION When Si is added to Ge, silicon atoms can form four bonds with neighboring Ge atoms. As a result, there is no need to donate or accept an electron. The resultant material then does not show either ‘‘n-type’’ or ‘‘p-type’’ conductivity. When we add As, we expect n-type conductivity since each As atom brings in five valence electrons. When we add small concentrations of B to Ge we expect p- type conductivity for the resultant material, since B has a valence of 3. Example 5.4 Dopants in Germanium Semiconductor

19 19  Interstitialcy - A point defect caused when a ‘‘normal’’ atom occupies an interstitial site in the crystal.  Frenkel defect - A pair of point defects produced when an ion moves to create an interstitial site, leaving behind a vacancy.  Schottky defect - A point defect in ionically bonded materials. In order to maintain a neutral charge, a stoichiometric number of cation and anion vacancies must form.  Kr Ö ger-Vink notation - A system used to indicate point defects in materials. The main body of the notation indicates the type of defect or the element involved. Section 5.2 Other Point Defects

20 20 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.3 When a divalent cation replaces a monovalent cation, a second monovalent cation must also be removed, creating a vacancy.

21 21 Write the appropriate defect reactions for (1) incorporation of magnesium oxide (MgO) in nickel oxide (NiO), and (2) formation of a Schottky defect in alumina (Al 2 O 3 ). Example 5.5 SOLUTION 1. MgO is the guest and NiO is the host material. We will assume that Mg +2 ions will occupy Ni +2 sites and oxygen anions from MgO will occupy O -2 sites of NiO. Example 5.5 Application of the Kröger-Vink Notation 2. Thus describes one vacancy of an Al +3. Similarly, represents an oxygen ion vacancy.

22 22 Write the appropriate defect reactions for the incorporation of calcium oxide (CaO) in zirconia (ZrO 2 ) using the Kröger-Vink notation. Example 5.6 SOLUTION We will assume that Ca +2 will occupy Zr +4 sites. If we send one Ca +2 to Zr +4, the site will have an effective negative charge of -2 (instead of having a charge of +4 we have a charge of +2). We have used one Zr +4 site and site balance would require us to utilize two oxygen sites. We can send one O -2 from CaO to one of the O -2 sites in ZrO 2. The other oxygen site must be used and since mass balance must also be maintained we will have to keep this site vacant. Example 5.6 Point Defects in Stabilized Zirconia for Solid Electrolytes

23 23 Point Impurities in Solids  Two outcomes if impurity (B) added to host (A) 1.Solid solution of B in A (i.e., random dist. of point defects) 2.Solid solution of B in A plus particles of a NEW PHASE (usually for a larger amount of B) OR Substitutional alloy (e.g., Cu in Ni) Interstitial alloy (e.g., C in Fe) Second phase particle different composition (chem formula) often different structure e.g.; BCC in FCC

24 24 Composition/Concentration  Composition  Amount of impurity/solute (B) and host/solvent (A) in the SYSTEM.  Two Forms  Convert Between Forms Using A J Weight-% Where –m J = mass of constituent “J” Atom/Mol % Where –n mJ = mols of constituent “J”

25 25  Dislocation - A line imperfection in a crystalline material.  Screw dislocation - A dislocation produced by skewing a crystal so that one atomic plane produces a spiral ramp about the dislocation.  Edge dislocation - A dislocation introduced into the crystal by adding an ‘‘extra half plane’’ of atoms.  Mixed dislocation - A dislocation that contains partly edge components and partly screw components.  Slip - Deformation of a metallic material by the movement of dislocations through the crystal. Section 5.3 Dislocations

26 26 Dislocations  First PREDICTED as defects in crystals since theoretical strength calculations (due to multibond breaking) were far too low as compared to experiments  later invention of the Transmission Electron Microscope (TEM) PROVED their Existence deformed steel (40,000X) Ti alloy (51,500X)

27 27 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.4 the perfect crystal (a) is cut and sheared one atom spacing, (b) and (c). The line along which shearing occurs is a screw dislocation. A Burgers vector b is required to close a loop of equal atom spacings around the screw dislocation.

28 28 Figure 5.5 The perfect crystal in (a) is cut and an extra plane of atoms is inserted (b). The bottom edge of the extra plane is an edge dislocation (c). A Burgers vector b is required to close a loop of equal atom spacings around the edge dislocation. (Adapted from J.D. Verhoeven, Fundamentals of Physical Metallurgy, Wiley, 1975.)

29 29 Figure 5.6 A mixed dislocation. The screw dislocation at the front face of the crystal gradually changes to an edge dislocation at the side of the crystal. (Adapted from W.T. Read, Dislocations in Crystals. McGraw-Hill, 1953.)

30 30 Figure 5.7 Schematic of slip line, slip plane, and slip (Burgers) vector for (a) an edge dislocation and (b) for a screw dislocation. (Adapted from J.D. Verhoeven, Fundamentals of Physical Metallurgy, Wiley, 1975.)

31 31

32 32 Linear Defects → Dislocations  Edge dislocation: extra half-plane of atoms linear defect moves in response to shear stress and results in bulk atomic movement –cause of slip between crystal planes when they move

33 33 Figure 5.8 (a) When a shear stress is applied to the dislocation in (a), the atoms are displaced, causing the dislocation to move one Burgers vector in the slip direction (b). Continued movement of the dislocation eventually creates a step (c), and the crystal is deformed. (Adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill, 1976.) (d) Motion of caterpillar is analogous to the motion of a dislocation.

34 34 Movement of Edge Dislocations  Dislocations Move Thru the Crystal in Response to Shear Force Result in Net atomic Movement or DEFORMATION

35 35 A sketch of a dislocation in magnesium oxide (MgO), which has the sodium chloride crystal structure and a lattice parameter of 0.396 nm, is shown in Figure 5.9. Determine the length of the Burgers vector. Example 5.7 Dislocations in Ceramic Materials Figure 5.9 An edge dislocation in MgO showing the slip direction and Burgers vector (for Example 5.7). (Adapted from W.D. Kingery, H.K. Bowen, and D.R. Uhlmann, Introduction to Ceramics, John Wiley, 1976.) for Example 5.7)

36 36 Example 5.7 SOLUTION In Figure 5.9, we begin a clockwise loop around the dislocation at point x, then move equal atom spacings to finish at point y. The vector b is the Burgers vector. Because b is a [110] direction, it must be perpendicular to {110} planes. The length of b is the distance between two adjacent (110) planes. From Equation 3-7, Note that this formula for calculating the magnitude of the Burgers vector will not work for non-cubic systems. It is better to consider the magntiude of the Burgers vector as equal to the repeat distance in the slip direction.

37 37 Calculate the length of the Burgers vector in copper. Homework: Example 5.8 Burgers Vector Calculation (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.10 (a) Burgers vector for FCC copper. (b) The atom locations on a (110) plane in a BCC unit cell (for example 5.8 and 5.9, respectively)

38 38 The planar density of the (112) plane in BCC iron is 9.94  10 14 atoms/cm 2. Calculate (1) the planar density of the (110) plane and (2) the interplanar spacings for both the (112) and (110) planes. On which plane would slip normally occur? Homework: Example 5.9 Identification of Preferred Slip Planes (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.10 (a) Burgers vector for FCC copper. (b) The atom locations on a (110) plane in a BCC unit cell (for example 5.8 and 5.9, respectively)

39 39  Etch pits - Tiny holes created at areas where dislocations meet the surface. These are used to examine the presence and number density of dislocations.  Slip line - A visible line produced at the surface of a metallic material by the presence of several thousand dislocations.  Slip band - Collection of many slip lines, often easily visible. Section 5.4 Observing Dislocations

40 40 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.11 A sketch illustrating dislocations, slip planes, and etch pit locations. (Source: Adapted from Physical Metallurgy Principles, Third Edition, by R.E. Reed-Hill and R. Abbaschian, p. 92, Figs. 4-7 and 4-8. Copyright (c) 1992 Brooks/Cole Thomson Learning. Adapted by permission.)

41 41 Figure 5.12 Optical image of etch pits in silicon carbide (SiC). The etch pits correspond to intersection points of pure edge dislocations with Burgers vector a/3 and the dislocation line direction along [0001] (perpendicular to the etched surface). Lines of etch pits represent low angle grain boundaries (Courtesy of Dr. Marek Skowronski, Carnegie Mellon University.)

42 42 Figure 5.13 Electron photomicrographs of dislocations in Ti 3 Al: (a) Dislocation pileups (x26,500). (b) Micrograph at x 100 showing slip lines and grain boundaries in AI. (c) Schematic of slip bands development. (c) 2003 Brooks/Cole Publishing / Thomson Learning (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.

43 43  Plastic deformation refers to irreversible deformation or change in shape that occurs when the force or stress that caused it is removed.  Elastic deformation - Deformation that is fully recovered when the stress causing it is removed.  Dislocation density - The total length of dislocation line per cubic centimeter in a material. Section 5.5 Significance of Dislocations

44 44 Interfacial Defects  2D, Sheet-like Defects are Termed as Interfacial  Some Macro-Scale Examples Solid Surfaces (Edges) –Bonds of Surface Atoms are NOT Satisfied  Source of “Surface Energy” in Units of J/sq-m Stacking Faults – When atom-Plane Stacking Pattern is Not as Expected Phase Boundaries – InterFace Between Different Xtal Structures

45 45 Interface Def. → Grain Boundaries  Grain Boundaries are Boundaries BETWEEN crystals Produced by the solidification process, for example Have a Change In Crystal Orientation across them IMPEDE dislocation motion Generally Weaker that the Native Xtal –Typically Reduce Material Strength thru Grain-Boundary Tearing Crack Along GB

46 46 Area Defects: Grain Boundaries  Schematic Representation Note GB Angles  Metal Ingot: GB’s Follow Solidification Path ~ 8cm

47 47 Section 5.6 Schmid’s Law  Schmid’s law -The relationship between shear stress, the applied stress, and the orientation of the slip system—that is,  Critical resolved shear stress - The shear stress required to cause a dislocation to move and cause slip.

48 48 Significant of slip Slip can occur in some ceramics and polymers Useful for understanding the mechanical behavior of materials Slip explains why the strength of metals is much lower than predicted from the metalic bond Slip provides ductility of metals, so that metal is not brittle Slip is useful to control the mechanical properties of metal

49 49 (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license. Figure 5.14 (a) A resolved shear stress τ is produced on a slip system. (Note: (ø + λ) does not have to be 90°.) (b) Movement of dislocations on the slip system deforms the material. (c) Resolving the force.

50 50 Apply the Schmid’s law for a situation in which the single crystal is at an orientation so that the slip plane is perpendicular to the applied tensile stress. Example 5.10 Calculation of Resolved Shear Stress (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 15.15 When the slip plane is perpendicular to the applied stress σ, the angle λ is 90° and no shear stress is resolved.

51 51 Example 5.10 SOLUTION Suppose the slip plane is perpendicular to the applied stress σ, as in Figure 5.15. Then, ø = 0 o, λ = 90 o, cos λ = 0, and therefore  r = 0. As noted before, the angles ø and λ can but do not always add up to 90 o. Even if the applied stress  is enormous, no resolved shear stress develops along the slip direction and the dislocation cannot move. Slip cannot occur if the slip system is oriented so that either λ or ø is 90 o.

52 52 We wish to produce a rod composed of a single crystal of pure aluminum, which has a critical resolved shear stress of 148 psi. We would like to orient the rod in such a manner that, when an axial stress of 500 psi is applied, the rod deforms by slip in a 45 o direction to the axis of the rod and actuates a sensor that detects the overload. Design the rod and a method by which it might be produced. Example 5.11 SOLUTION Dislocations begin to move when the resolved shear stress τ r equals the critical resolved shear stress, 148 psi. From Schmid’s law: τ r = σ cos λ cos ø ; or 148 psi = (500 psi ) cos λ cos ø Example 5.11 Design of a Single Crystal Casting Process

53 53 Example 5.11 SOLUTION (Continued) Because we wish slip to occur at a 45 o angle to the axis of the rod, λ = 45 o, and: Therefore, we must produce a rod that is oriented such that λ = 45 o and ø = 65.2 o. Note that ø and λ do not add to 90 o. We might do this by a solidification process. We would orient a seed crystal of solid aluminum at the bottom of a mold. Liquid aluminum could be introduced into the mold. The liquid begins to solidify from the starting crystal and a single crystal rod of the proper orientation is produced.

54 54  Critical Resolved Shear Stress: The shear stress required to cause a dislocation to move and cause slip.  Number of Slip Systems  Cross-slip - A change in the slip system of a dislocation. Section 5.7 Influence of Crystal Structure

55 55

56 56 A single crystal of magnesium (Mg), which has a HCP crystal structure, can be stretched into a ribbon-like shape four to six times its original length. However, polycrystalline Mg and other metals with a HCP structure show limited ductilities. Use the values of critical resolved shear stress for metals with different crystal structures and the nature of deformation in polycrystalline materials to explain this observation. Example 5.12 Ductility of HCP Metal Single Crystals and Polycrystalline Materials

57 57

58 58 Example 5.12 SOLUTION From Table 4-2, we note that for HCP metals such as Mg, the critical resolved shear stress is low (50–100 psi). We also note that slip in HCP metals will occur readily on the basal plane—the primary slip plane. When a single crystal is deformed, assuming the basal plane is suitably oriented with applied stress, a very large deformation can occur. This explains why single crystal Mg can be stretched into a ribbon four to six times the original size. When we have a polycrystalline Mg, the deformation is not as simple. Each crystal must deform such that the strain developed in any one crystal is accommodated by its neighbors. In HCP metals, there are no intersecting slip systems, thus dislocations cannot glide over from one slip plane in one crystal (grain) onto another slip plane in a neighboring crystal. As a result, polycrystalline HCP metals such as Mg show limited ductility.

59 59 Optical Microscopy  Since Most Solid Materials are Opaque, MicroScope Uses REFLECTED Light These METALLOGRAHPIC MScopes do NOT have a CONDENSOR Lens

60 60 Optical MicroScopy cont  The Resolution, Z  The Magnification, M Where –  Light Wavelength   550 nm For “White” Light (Green Ctr) –NA  Numerical Aperture for the OBJECTIVE Lens   0.9 for a Very High Quality Lens  Typical Values Z  375 nm –Objects Smaller than This Cannot be observed –Objects Closer Together than This Cannot Be Separated M true  200

61 61 Optical MicroScopy cont.2  Sample Preparation grind and polish surface until flat and shiny sometimes use chemical etch use light microscope different orientations → different contrast take photos, do analysis –e.g. Grain Sizing

62 62 Optical MicroScopy cont.3  Grain Boundaries are imperfections, with high surface energy are more susceptible to etching; may be revealed as –dark lines –change of direction in a polycrystal  ASTM E-112 Grain Size Number, n Where –N  grain/inch 2 Fe-Cr alloy microscope grain boundary surface groove polished surface

63 63 Electron Microscopy  For much greater resolution, use BEAM OF ELECTRONS rather that light radiation  Transmission Electron Microscopy (TEM): very high magnifications! contrast from different diffraction conditions very thin samples needed for transmission  Scanning Electron Microscopy (SEM): surface scanned, TV-like depth of field possible

64 64 Atomic Force MicroScopy  Also called Scanning Probe Microscopy (SPM) tiny probe with tiny tip rasters surface topographical map on atomic scale  Will See on SEM & AFM on CE&A Tour Polymer

65 65 Section 5.8 Surface Defects  Surface defects - Imperfections, such as grain boundaries, that form a two-dimensional plane within the crystal.  Hall-Petch equation - The relationship between yield strength and grain size in a metallic material—that is,  ASTM grain size number (n) - A measure of the size of the grains in a crystalline material obtained by counting the number of grains per square inch a magnification  100.  Small angle grain boundary - An array of dislocations causing a small misorientation of the crystal across the surface of the imperfection.

66 66 Figure 5.16 (a) The atoms near the boundaries of the three grains do not have an equilibrium spacing or arrangement. (b) Grains and grain boundaries in a stainless steel sample. (Courtesy Dr. A. Deardo.)

67 67 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.17 The effect of grain size on the yield strength of steel at room temperature.

68 68 The yield strength of mild steel with an average grain size of 0.05 mm is 20,000 psi. The yield stress of the same steel with a grain size of 0.007 mm is 40,000 psi. What will be the average grain size of the same steel with a yield stress of 30,000 psi? Assume the Hall-Petch equation is valid and that changes in the observed yield stress are due to changes in dislocation density. Example 5.13 SOLUTION Example 5.13 Design of a Mild Steel Thus, for a grain size of 0.05 mm the yield stress is 20  6.895 MPa = 137.9 MPa. (Note:1,000 psi = 6.895 MPa). Using the Hall-Petch equation

69 69 Example 5.13 SOLUTION (Continued) For the grain size of 0.007 mm, the yield stress is 40  6.895 MPa = 275.8 MPa. Therefore, again using the Hall- Petch equation: Solving these two equations K = 18.43 MPa-mm 1/2, and σ 0 = 55.5 MPa. Now we have the Hall-Petch equation as σ y = 55.5 + 18.43 d -1/2 If we want a yield stress of 30,000 psi or 30  6.895 = 206.9 MPa, the grain size will be 0.0148 mm.

70 70 Figure 5.18 Microstructure of palladium (x 100). (From ASM Handbook, Vol. 9, Metallography and Microstructure (1985), ASM International, Materials Park, OH 44073.)

71 71 Suppose we count 16 grains per square inch in a photomicrograph taken at magnification  250. What is the ASTM grain size number? Example 5.14 SOLUTION If we count 16 grains per square inch at magnification  250, then at magnification  100 we must have: N = (250/100) 2 (16) = 100 grains/in. 2 = 2 n-1 Log 100 = (n – 1) log 2 2 = (n – 1)(0.301) n = 7.64 Example 5.14 Calculation of ASTM Grain Size Number

72 72 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.19 The small angle grain boundary is produced by an array of dislocations, causing an angular mismatch θ between lattices on either side of the boundary.

73 73 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.20 Application of a stress to the perfect crystal (a) may cause a displacement of the atoms, (b) causing the formation of a twin. Note that the crystal has deformed as a result of twinning.

74 74 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure 5.20 (c) A micrograph of twins within a grain of brass (x250).

75 75

76 76 Figure 5.21 Domains in ferroelectric barium titanate. (Courtesy of Dr. Rodney Roseman, University of Cincinnati.) Similar domain structures occur in ferromagnetic and ferrimagnetic materials.

77 77  Effect on Mechanical Properties via Control of the Slip Process  Strain Hardening  Solid-Solution Strengthening  Grain-Size Strengthening  Effects on Electrical, Optical, and Magnetic Properties Section 5.9 Importance of Defects

78 78 (c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license. Figure 5.22 If the dislocation at point A moves to the left, it is blocked by the point defect. If the dislocation moves to the right, it interacts with the disturbed lattice near the second dislocation at point B. If the dislocation moves farther to the right, it is blocked by a grain boundary.

79 79 We would like to produce a bracket to hold ceramic bricks in place in a heat-treating furnace. The bracket should be strong, should possess some ductility so that it bends rather than fractures if overloaded, and should maintain most of its strength up to 600 o C. Design the material for this bracket, considering the various crystal imperfections as the strengthening mechanism. Example 5.15 SOLUTION In order to serve up to 600 o C, the bracket should not be produced from a polymer material. Instead, a metal or ceramic would be considered. Example 5.15 Design/Materials Selection for a Stable Structure

80 80 Example 5.15 SOLUTION (Continued) In order to have some ductility, dislocations must move and cause slip. Because slip in ceramics is difficult, the bracket should be produced from a metallic material. We might add carbon to the iron as interstitial atoms or substitute vanadium atoms for iron atoms at normal lattice points. These point defects continue to interfere with dislocation movement and help to keep the strength stable. Of course, other design requirements may be important as well. For example, the steel bracket may deteriorate by oxidation or may react with the ceramic brick.

81 81 Figure 5.23 Microstructure of iron, for Problem 4-54 (x500). (From ASM Handbook, Vol. 9, Metallography and Microstructure (1985), ASM International, Materials Park, OH 44073.)

82 82 Figure 5.24 The microstructure of BMT ceramics obtained by compaction and sintering of BMT powders. (Courtesy of H. Shirey.)