Chapter 3_Crystal structures_Filled

9/21/2023 3 Materials and Atomic Arrangement • periodic array of atoms or molecules • atoms pack in periodic , 3D arrays Crystalline materials... -metals -many ceramics -some polymers • atoms have no periodic packing Noncrystalline materials... -complex structures -rapid cooling crystalline SiO 2 noncrystalline SiO 2 " Amorphous " = Noncrystalline Adapted from Fig. 3.25(b), Callister 10e. Adapted from Fig. 3.25(a), Callister 10e. Si Oxygen • typical of: • occurs for: MSE220 Engineering Materials 3

9/21/2023 6 Unit Cell Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. Fig. 3.1, Callister 10e. Ex: face-centered cubic (FCC) crystal structure: a)Hard sphere unit cell representation b) Reduced-sphere unit cell representation MSE220 Engineering Materials 6

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9/21/2023 12 A sites B B B B B B B C sites C C C A B B sites • In the FCC structure, stacking of closed packed planes in ABCABC sequence • 2D Projection • FCC Unit Cell Face-Centered Cubic (FCC) A B C MSE220 Engineering Materials 12 Stacking Sequence Referenced to an FCC Unit Cell. Close-Packed Plane B B B B B B B B sites C C C A C C C A

9/21/2023 15 Face-Centered Cubic (FCC) Adapted from Fig. 3.1, Callister 10e. • Coordination number: MSE220 Engineering Materials 15 A sites B B B B B B B C sites C C C A B B sites • In closed-packed A plane: 6 • In closed-packed B plane: 3 • In closed-packed C plane: 3 • Coordination # = 6+3+3=12

9/21/2023 18 Body-Centered Cubic (BCC) a APF = 4 3 p ( 3 a /4) 3 2 atoms unit cell atom volume a 3 unit cell volume length = 4 R = Close-packed directions: 3 a • APF (BCC)= 0.68 a R Adapted from Fig. 3.2, Callister 10e. a 2 a 3 MSE220 Engineering Materials 18

9/21/2023 21 Theoretical Density, 𝝆 • Density = ? = 𝑀??? ?? ????? 𝑖? ??𝑖? ???? ????? ?????? ?? ??𝑖? ???? MSE220 Engineering Materials 21 ? = ?𝐴 𝑉 ? 𝑁 ? ? = number of atoms associated with each unit cell 𝐴 = atomic weight 𝑉 ? = volume of the unit cell 𝑁 ? = Avogadro’s number ( 6.022 × 10 23 atoms/mol)

9/21/2023 22 MSE220 Engineering Materials 22 Theoretical Density, 𝝆 • Example: Chromium (Cr) has an atomic radius of 0.125 nm, an BCC crystal structure, Compute its theoretical density, and compare the answer with its measured density ? = ?𝐴 𝑉 ? 𝑁 ? ? = 2 A=52 g/mol 𝑁 ? = 6.022 × 10 23 𝑉 ? = ? 3 ? = 4𝑅 3 = 0.2887?? • ? ?ℎ = 7.18?/?? 3 • ? ?? = 7.19?/?? 3

9/21/2023 23 Densities of Material Classes r metals > r ceramics > r polymers Why? Data from Table B1, Callister 10e. r (g/cm ) 3 Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers 1 2 20 30 B ased on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). 10 3 4 5 0.3 0.4 0.5 Magnesium Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum Gold, W Platinum G raphite Silicon Glass -soda Concrete Si nitride Diamond Al oxide Zirconia HDPE, PS PP, LDPE PC PTFE PET PVC Silicone Wood AFRE * CFRE * GFRE* Glass fibers Carbon fibers A ramid fibers Metals have... • close -packing (metallic bonding) • often large atomic masses Ceramics have... • less dense packing • often lighter elements Polymers have... • low packing density (often amorphous) • lighter elements (C,H,O) Composites have... • intermediate values In general MSE220 Engineering Materials 23

9/21/2023 24 Ceramic Structures MSE220 Engineering Materials 24

9/21/2023 25 Calculate the IPF of MgO (NaCl Structure) FCC Structure MSE220 Engineering Materials 25 Structure? Radius of O is 0.14nm Radius of Mg is 0.072nm ? = 2? 𝑀? 2+ + 2? 𝑂 2− = 0.424?? 𝑉 ????? = 4 × 4 3 ?? 𝑀? 2+ 3 + 4 × 4 3 ?? 𝑂 2− 3 = 0.0522?? 3 𝑉 ??𝑖????? = ? 3 = 0.076?? 3 𝐼𝑃𝐹 = 𝑉 ????? 𝑉 ??𝑖????? = 0.685?? 3

9/21/2023 26 Conceptual Question MSE220 Engineering Materials 26 • What is the difference between crystal structure and crystal system? • A crystal structure is described by both the geometry of the unit cell, and atomic arrangements within • A crystal system is described only in terms of the unit cell geometry.

9/21/2023 27 Polymorphism • Two or more distinct crystal structures for the same material = allotropy /polymorphism • Titanium: ➢  phase (HCP, below 882℃ ) ➢  -Ti phase (BCP, a bove 882℃ ) • Carbon: ➢ graphite (ambient condition) ➢ Diamond (high pressure) BCC FCC BCC 1538ºC 1394ºC 912ºC  - Fe  - Fe  - Fe liquid iron system MSE220 Engineering Materials 27

9/21/2023 28 Lattice Position, Direction, and Planes MSE220 Engineering Materials only in a cubic crystal Not in a tetragonal 28

9/21/2023 29 Crystallographic Point MSE220 Engineering Materials 29 • Crystallographic points are commonly used to describe the location of atoms/impurities within the unit cell (location of C atom in Fe BCC structure to create steel) • Point locations described by fractional q, r, and s of the unit cell edge length a, b, and c (q, r, and s are less than 1) • We denote any point P using no punctuation marks as : q r s

9/21/2023 30 Crystallographic Point (Example) MSE220 Engineering Materials 30 • For the unit cell shown in the accompanying sketch (a), locate the point having indices 1 4 1 1 2 0.12 nm 0.46 nm 0.20 nm Position of point P is (0.12, 0.46, 0.20)nm

9/21/2023 31 Crystallographic Direction MSE220 Engineering Materials 31 1) Find the location of start and end point based on a, b, c 2) Tail point coordinate are subtracted from the head point and normalized ? 2 −? 1 ? ? 2 −? 1 ? ? 2 −? 1 ? • It is often convenient to describe directions in a crystal structure (direction of slips during plastic deformations) • Here is the algorithm to find crystal direction: 3) If necessary, these three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values . ? = ? ? 2 −? 1 ? ? = ? ? 2 −? 1 ? ? = ? ? 2 −? 1 ? 4) Write it square bracket with NO commas [ uvw ]

9/21/2023 32 Crystallographic Direction MSE220 Engineering Materials 32 • Example: Determine the indices for the direction shown in the figure. −221 = [ ത 221] Use overbar to denote negative index

9/21/2023 33 Crystallographic Direction MSE220 Engineering Materials 33 • Example: Determine the indices for the direction shown in the figure. Answer: [ ത 412]

9/21/2023 34 Family of Direction MSE220 Engineering Materials 34 • Sometimes, we denote a family of directions that are crystallographic ally equivalent (have same atomic spacing along direction) • Families of directions are denoted with angle brackets

9/21/2023 35 Crystalline Planes • Miller indices is used to identify crystal orientation and directions using h, k and l • For plane orientation, it uses () and {} • For direction, it uses [] and < > • All parallel planes have same Miller indices. • Procedures for Miller indices: 1) If plane intersects origin, choose a new origin of adjacent cell 2) Determine distance to intercept plane by traveling along each axis from origin 3) Take the reciprocal of each quantity. 4) Multiply the set by the least common denominator 5) Enclose the set with the appropriate brackets 6) Negative quantities are indicated with an over score above the number MSE220 Engineering Materials 35

9/21/2023 36 Crystallographic Planes z x y a b c 4. Miller Indices (110) Example a b c z x y a b c 4. Miller Indices ( 2 00) 1. Intercepts 1 1  2. Reciprocals 1/1 1/1 1/  1 1 0 3. Multiply by LCD 1 1 0 1. Intercepts 1/2   2. Reciprocals 1/½ 1/  1/  2 0 0 3. Multiply by LCD 2 0 0 Example a b c MSE220 Engineering Materials 36

9/21/2023 37 Crystallographic Planes MSE220 Engineering Materials 37 z x y a b c · · · x y z Example: 4. Miller Indices (634) 1. Intercepts 1 /2 1 3/4 2. Reciprocals 2 1 4/3 3. Multiply by LCD 6 3 4

9/21/2023 38 Crystallographic Planes MSE220 Engineering Materials 38 Example: find the orientation of these two planes

9/21/2023 39 Crystallographic Planes MSE220 Engineering Materials 39 Example: Draw the orientation of plane (2,1,2) and (2, ത 1 ,2) (2,1,2) (2, ത 1, 2)

9/21/2023 40 Crystallographic Planes MSE220 Engineering Materials 40 • Family of Plane: Given any plane in a lattice, there is a infinite set of parallel lattice planes (or family of planes) that are equally spaced from each other. • All planes that are crystallographically equivalent (have the same atomic packing) – indicated by indices in braces • Ex: {100} = (100), (010), (001),( ത 1 00), (0 ത 1 0), (00 ത 1 )

9/21/2023 41 Crystallographic Planes …a set of parallel planes MSE220 Engineering Materials 41

9/21/2023 42 Miller Index MSE220 Engineering Materials 42

9/21/2023 43 Intercepts →  ½  Plane → (0 2 0) • (010) and (020) are parallel plane, are they in the same plane family? Conceptual Question 43 No, suppose we consider a simple cubic crystal, then alternate (020) planes will not have any atoms in them

9/21/2023 44 Crystallographic Planes (HCP) MSE220 Engineering Materials 44 a 2 a 3 a 1 z • For hexagonal unit cells a similar procedure is used ➢ In the hexagonal unit cell, instead of 3 axis, we have to find the intercept of 4 axis of Miller-Bravais Indices h, k, i , and l . ? 1 ? 2 ? 3 𝑧 4. Miller-Bravais Indices ( 10 ത 11 ) 1. Intercepts 1 ∞ -1 1 2. Reciprocals 1 0 -1 1 3. Multiply by LCD 1 0 -1 1

9/21/2023 45 Crystallographic Planes (HCP) MSE220 Engineering Materials 45 • Example:

9/21/2023 46 Linear and Planar Densities MSE220 Engineering Materials 46 • Linear density (LD) is defined as the number of atoms per unit length whose centers lie on the direction vector for a specific crystallographic direction 𝐿𝐷 = ?????? ?? ????? ???????? ?? ?𝑖????𝑖?? ?????? ?????ℎ ?? ?𝑖????𝑖?? ?????? 𝐿𝐷 110 = 2 4𝑅 = 1 2𝑅 ? −1 • Example: find the linear density of [110] in FCC

9/21/2023 47 Linear and Planar Densities MSE220 Engineering Materials 47 • planar density (PD) is taken as the number of atoms per unit area that are centered on a particular crystallographic plane 𝑃𝐷 = ?????? ?? ????? ???????? ?? ? ????? ???? ?? ????? • Example: find the planar density of (110) within an FCC unit cell 𝑃𝐷 = 2 (4𝑅)(2 2𝑅) = 1 4𝑅 2 2 a 2 a

9/21/2023 48 Linear and Planar Densities MSE220 Engineering Materials 48 Example: Find the Planar Density of (100) Iron at T < 912℃ Solution: At T < 912℃ iron has the BCC structure with Radius of iron R = 0.1241 nm (100) R 3 3 4 a = Adapted from Fig. 3.2(c), Callister 10e. 2D repeat unit 𝑃𝐷 = 1 (4 3𝑅/3) 2 = 12.1 ????? ?? 2

9/21/2023 49 Anisotropy MSE220 Engineering Materials 49 • The physical properties of single crystals of some substances depend on the crystallographic direction in which measurements are taken. • For example, the elastic modulus , and the electrical conductivity , may have different values in the [100] and [111] directions. • This directionality of properties is termed anisotropy , associated with the variance of atomic spacing with crystallographic direction. • Substances in which measured properties are independent of the direction of measurement are isotropic . • Degree of anisotropy increases with decreasing structural symmetry • Triclinic structures normally are highly anisotropic.

9/21/2023 50 X-Ray Diffraction • X-rays are a form of electromagnetic radiation that have high energies and short wavelengths. • When a beam of X-rays impinges on a solid material, the crystalline atoms cause that beam to diffract into many specific directions. • Diffracted beam provides useful information concerning the structure of the material. • By measuring the angles and intensities of these diffracted beams, a crystallographer can produce a three-dimensional picture of the density of electrons within the crystal and all other properties. MSE220 Engineering Materials 50

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9/21/2023 53 X-Ray Diffraction Constructive : Successful diffraction Destructive: Failed diffraction MSE220 Engineering Materials 53 When incident waves are shifted by an integer multiple of wavelength ?𝜆 When incident waves are shifted by an integer multiple of wavelength (? + 1 2 )𝜆

9/21/2023 54 X-Ray Diffraction • At the specific angle, all the scattered waves from parallel crystal planes have constructive interference which results in diffraction . • Bragg’s law: ➢ the relation between the spacing of atomic planes in crystals and the angles of incidence at which these planes produce the most intense reflections of X-Ray (diffraction) ?𝜆 = 2? ℎ?? ?𝑖?𝜃 Adapted from Fig. 3.21, Callister 10e . reflections must be in phase for a detectable signal spacing between planes d q l q extra distance travelled by wave “ 2 ” MSE220 Engineering Materials 54 ? = integer 𝜆 = wave length 2𝜃 = angle of diffraction ? ℎ?? = interplanar spacing ? ℎ?? = ? ℎ 2 + ? 2 + ? 2 ? = unit cell edge length

9/21/2023 55 X-Ray Diffraction Pattern (110) (200) (211) z x y a b c Diffraction angle 2 q Diffraction pattern for polycrystalline  -iron (BCC) Intensity (relative) z x y a b c z x y a b c MSE220 Engineering Materials 55 How we can get these planes?

9/21/2023 56 X-Ray Diffraction Pattern MSE220 Engineering Materials 57 One form of question: ?𝜆 = 2? ℎ?? ?𝑖?𝜃 ? ℎ?? = ? ℎ 2 + ? 2 + ? 2 ? ??? = 4𝑅 3

9/21/2023 57 X-Ray Diffraction Pattern MSE220 Engineering Materials 58 ?𝜆 = 2? ℎ?? ?𝑖?𝜃 ? ℎ?? = ? ℎ 2 + ? 2 + ? 2 ? ??? = 4𝑅 3

9/21/2023 58 X-Ray Diffraction Pattern z x y a b c If we don’t know that what type of crystal we are dealing with: 1- Fcc: h, k, l need to be all even or all odd 2- Bcc: h+k+l=even Intensity (relative) z x y a b c z x y a b c MSE220 Engineering Materials 59

9/21/2023 59 X-Ray Diffraction Pattern MSE220 Engineering Materials 60 Another form of question: https://www.youtube.com/watch?v=6IzrOWIw3zQ

9/21/2023 60 X-Ray Diffraction Pattern MSE220 Engineering Materials 61

9/21/2023 61 X-Ray Diffraction Pattern MSE220 Engineering Materials 62

9/21/2023 62 X-Ray Diffraction Pattern MSE220 Engineering Materials 64

9/21/2023 63 65 ?𝜆 = 2? ℎ?? ?𝑖?𝜃 ? ℎ?? = ? ℎ 2 + ? 2 + ? 2 ? ??? = 4𝑅 3 Adapted from Fig. 3.21, Callister 10e . reflections must be in phase for a detectable signal spacing between planes d q l q extra distance travelled by wave “ 2 ”

9/21/2023 64 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. MSE220 Engineering Materials 66

9/21/2023 65 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. • X-ray diffraction is used for crystal structure and inter-planar spacing determinations. MSE220 Engineering Materials 67

9/21/2023 66 Home work • Practice these questions on the back of your textbook. 10 th edition’s Questions and Problems 3.3, 3.5, 3.7, 3.9, 3.15, 3.22, 3.43,3.45, 3.46, 3.47, 3.52, 3.58, 3.59, 3.60 • Read your text book Chapter 3 MSE220 Engineering Materials 68