Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 3 -1 ISSUES TO ADDRESS... How do atoms assemble (arrange) into solid structures? (focus on metals) How does the density of a material depend on.

Similar presentations


Presentation on theme: "Chapter 3 -1 ISSUES TO ADDRESS... How do atoms assemble (arrange) into solid structures? (focus on metals) How does the density of a material depend on."— Presentation transcript:

1 Chapter 3 -1 ISSUES TO ADDRESS... How do atoms assemble (arrange) into solid structures? (focus on metals) How does the density of a material depend on its structure? When do material properties vary with the sample (i.e., part) orientation? Structure Levels: Subatomic  now: Atomic Chapter 3: The Structure of Crystalline Solids

2 Chapter 3 -2 Non dense, random packing Dense, ordered packing Dense, ordered packed structures tend to have lower energies. Energy and Packing Energy r typical neighbor bond length typical neighbor bond energy Energy r typical neighbor bond length typical neighbor bond energy

3 Chapter 3 -3 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 Materials and Packing SiOxygen typical of: occurs for:

4 Chapter 3 -4 Crystal Systems There are 6 lattice parameters Their different combinations lead to: 7 crystal systems (shapes) 14 crystal lattices In metals: Cubic and Hexagonal Fig. 3.4, Callister 7e. Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. a, b, and c are the lattice constants

5 Chapter 3 -5 Table 3.2 Lattice Parameter Relationships and Figures Showing Unit Cell Geometries for the Seven Crystal Systems See other shapes

6 Chapter 3 -6 Metallic Crystal Structures How can we stack metal atoms to minimize empty space? 2-dimensions vs. Now stack these 2-D layers to make 3-D structures

7 Chapter 3 -7 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... Metallic Crystal Structures

8 Chapter 3 -8 Parameters in Studying Crystal Structures Number and Position of Atoms in unit cell Coordination No. No. of touching-neighbor atoms Cube edge (length) Atomic Packing Factor (APF): Fraction of volume occupied by atoms Fraction of empty space = 1 – APF APF = Total volume of atoms in the unit cell Volume of the unit cell

9 Chapter 3 -9 Rare due to low packing density (only Po* has this structure) Close-packed directions are cube edges. Coordination # = 6 (# nearest neighbors) * Po: Polonium Simple Cubic Structure (SC)

10 Chapter APF for a simple cubic structure = 0.52 APF = a  (0.5a) 3 1 atoms unit cell atom volume unit cell volume Atomic Packing Factor (APF) APF = Volume of atoms in unit cell* Volume of unit cell *assume hard spheres close-packed directions a R=0.5a contains 8 x 1/8 = 1 atom/unit cell

11 Chapter Atoms touch each other along cube diagonals. examples: Cr, Fe(  ), W, Ta, Mo* Coordination # = 8 --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. * Cr: Chromium, W: Tungsten, Ta: Tantalum, Mo: Molybdenum Body Centered Cubic Structure (BCC) No of Atoms per unit cell = 1 center + 8 corners x 1/8 =2 atoms

12 Chapter Atomic Packing Factor: BCC a APF = 4 3  (3a/4) 3 2 No. atoms unit cell atom volume a 3 unit cell volume length = 4R = Close-packed directions: 3 a APF for a body-centered cubic structure = 0.68 a R a 2 a 3

13 Chapter Atoms touch each other along face diagonals. ex: Al, Cu, Au, Pb, Ni, Pt, Ag Coordination # = 12 Face Centered Cubic Structure (FCC) No of atoms per unit cell = 6 face x 1/2 + 8 corners x 1/8 = 4 atoms --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

14 Chapter APF for a face-centered cubic structure = 0.74 Atomic Packing Factor: FCC maximum achievable APF APF = 4 3  (2a/4) 3 4 atoms unit cell atom volume a 3 unit cell volume Close-packed directions: length = 4R = 2 a a

15 Chapter A sites B B B B B BB C sites C C C A B B ABCABC... Stacking Sequence 2D Projection FCC Unit Cell FCC Stacking Sequence B B B B B BB B sites C C C A C C C A

16 Chapter Coordination # = 12 ABAB... Stacking Sequence APF = D Projection 2D Projection Hexagonal Close-Packed Structure (HCP) No. of atoms/unit cell = 6 ex: Cd, Mg, Ti, Zn c/a = c a A sites B sites A sites Bottom layer Middle layer Top layer Cd: Cadmium, Mg: Magnesium, Ti: Titanium, Zn: Zinc

17 Chapter Polymorphism Polymorphous Material: Same Material with two or more distinct crystal structures Crystal Structure is a Property: It may vary with temperature  Allotropy /polymorphism Examples Titanium: ,  -Ti Carbon: diamond, graphite BCC FCC BCC 1538ºC 1394ºC 912ºC  - Fe  - Fe  - Fe liquid iron system Heating

18 Chapter Theoretical Density,  where n = number of atoms/unit cell A = atomic weight V C = Volume of unit cell = a 3 for cubic N A = Avogadro’s number = x atoms/mol Density =  = VC NAVC NA n An A  = Cell Unit of VolumeTotal Cell Unit in Atomsof Mass

19 Chapter Example: For Cr Known: A = g/mol and R = nm Crystal Structure: BCC For BCC: n = 2 and a = 4R/ 3 = nm Calculate: a R  = a atoms unit cell mol g unit cell volume atoms mol x Theoretical Density,   theoretical  actual = 7.18 g/cm 3 = 7.19 g/cm 3 Compare

20 Chapter Densities of Material Classes  metals >  ceramics >  polymers Why?  (g/cm ) 3 Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers Based 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) Magnesium Aluminum Steels Titanium Cu,Ni Tin, Zinc Silver, Mo Tantalum Gold, W Platinum Graphite 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 Carbonfibers Aramid 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 Water

21 Chapter Structure-Properties Materials-Structures Amorphous: Random atomic orientation Crystalline: –Single Crystal Structure: One Crystal Perfectly repeated unit cells. –Polycrystalline Many crystals Properties Anisotropy: depend on direction Isotropy: does not depend on direction

22 Chapter Some engineering applications require single crystals: Properties of crystalline materials often related to crystal structure. Ex: Quartz fractures more easily along some crystal planes than others. --diamond single crystals for abrasives Single Crystals --turbine blades

23 Chapter Single Crystals -Properties vary with direction: anisotropic. -Example: the modulus of elasticity (E) in BCC iron: Polycrystals - Properties may/may not vary with direction. -If grains are randomly oriented: isotropic. (E poly iron = 210 GPa) -If grains are textured, anisotropic. 200  m Properties for Single and Polycrystals E (diagonal) = 273 GPa E (edge) = 125 GPa

24 Chapter Most engineering materials are polycrystals. Each "grain" is a single crystal (composed of many unit cells). If grains are randomly oriented, overall component properties are not directional. Grain sizes typ. range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers). 1 mm Polycrystals Isotropic Anisotropic

25 Chapter Point Coordinates Point coordinates for unit cell center are a/2, b/2, c/2 ½ ½ ½ Point coordinates for unit cell corner are 111 Translation: integer multiple of lattice constants  identical position in another unit cell z x y a b c y z 2c2c b b

26 Chapter Crystallographic Directions Vectors or lines between two points Described by indices Algorithm (for obtaining indices) 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 (multiply by a factor) 4. Enclose in square brackets, no commas [uvw] (note: If the value is –ve put the – sign over the number) Instead of 1 and 2: you can 1. Determine coordinates of head (H) and tail (T) 2. Subtract: head – tail Then as before (steps 3 and 4) Ex 1: 1, 0, ½  2, 0, 1  [ 201 ] Ex 2: -1, 1, 1 families of directions (overbar represents a negative index) [ 111 ] => z x y

27 Chapter ex: linear density of Al in [110] direction a = nm Linear Density Linear Density of Atoms  LD = a [110] Unit length of direction vector Number of atoms # atoms length nm a2 2 LD  

28 Chapter HCP Crystallographic Directions 1. Vector repositioned (if necessary) to pass through origin. 2. Read off projections in terms of unit cell dimensions a 1, a 2, a 3, or c 3. Adjust to smallest integer values 4. Enclose in square brackets, no commas [uvtw] [ 1120 ] ex: ½, ½, -1, 0 => dashed red lines indicate projections onto a 1 and a 2 axes a1a1 a2a2 a3a3 -a3-a3 2 a 2 2 a 1 - a3a3 a1a1 a2a2 z Algorithm

29 Chapter HCP Crystallographic Directions Hexagonal Crystals –4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows.    'ww t v u )vu( +- )'u'v2( )'v'u2(  ]uvtw[]'w'v'u[  Fig. 3.8(a), Callister 7e. - a3a3 a1a1 a2a2 z

30 Chapter Crystallographic Planes

31 Chapter Crystallographic Planes Miller Indices: Reciprocals of the (three) axial intercepts for a plane, Cleared of fractions & common multiples. All parallel planes have same Miller indices. Algorithm (for obtaining Miller Indices) If the plane passes through the origin, choose another origin 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)

32 Chapter Crystallographic Planes z x y a b c 4. Miller Indices (110) Example 2a b c z x y a b c 4. Miller Indices (100) 1. Intercepts 1 1  2. Reciprocals 1/1 1/1 1/  Integers Intercepts 1/2   2. Reciprocals 1/½ 1/  1/  Integers Example 1a b c

33 Chapter Crystallographic Planes z x y a b c 4. Miller Indices (634) Example 3 1. Intercepts 1/2 1 3/4 a b c 2. Reciprocals 1/½ 1/1 1/¾ 21 4/3 3. Integers 63 4 (001)(010), Family of Planes {hkl} (100),(010),(001),Ex: {100} = (100),

34 Chapter Crystallographic Planes (HCP) In hexagonal unit cells the same idea is used example a 1 a 2 a 3 c 4. Miller-Bravais Indices(1011) 1. Intercepts 1  1 2. Reciprocals 1 1/  Integers1 0 1 a2a2 a3a3 a1a1 z

35 Chapter Planar Density Concept The atomic packing of crystallographic planes No. of atoms per area of a plane Importance Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important. Example a)Draw (100) and (111) crystallographic planes for Fe (BCC). b) Calculate the planar density for each of these planes.

36 Chapter Planar Density of (100) Iron Solution: At T < 912  C iron has the BCC structure. (100) Radius of iron R = nm R 3 34 a  2D repeat unit = Planar Density = a 2 1 atoms 2D repeat unit = nm 2 atoms 12.1 m2m2 atoms = 1.2 x R 3 34 area 2D repeat unit

37 Chapter Planar Density of (111) Iron Solution (cont): (111) plane 1 atom in plane/ unit surface cell R 3 16 R a3ah2area           atoms in plane atoms above plane atoms below plane ah 2 3  a 2 2D repeat unit 1 = = nm 2 atoms 7.0 m2m2 atoms 0.70 x R 3 16 Planar Density = atoms 2D repeat unit area 2D repeat unit

38 Chapter Determination of Crystal Structure Wave: has an amplitude and wave length ( ) Diffraction: phase relationship between two (or more) waves that have been scattered. Scattering: –In-phase Constructive –Out-of-phase Destructive Detecting scattered waves after encountering an object: Identify positions of atoms مثل الرؤية البصرية Diffraction occurs when a wave encounters a series of regularly spaced atoms – IF: 1)The obstacles are capable of scattering the wave 2)Spacing between them nearly = wave length source Wave length depend on the source (i.e. material of the anode).

39 Chapter X-Ray Diffraction Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation. Can’t resolve spacings 

40 Chapter Constructive versus Destructive Diffraction

41 Chapter 3 - Constructive X-Ray Diffraction 41

42 Chapter 3 - X-Ray Experiment 42

43 Chapter X-Rays to Determine Crystal Structure  c Constructive diffraction occurs when the second wave travels a distance = n From Geometry extra traveled distance = 2 d sin  Then Incoming X-rays diffract from crystal planes. reflections must be in phase for a detectable signal Interplaner spacing spacing between planes d incoming X-rays outgoing X-rays detector   extra distance traveled by wave 2 “1” “2” “1” “2” d d  n 2 sin  c In-phase X-ray intensity  Constructive

44 Chapter Calculations d = a for plane (h k l) (h 2 +k 2 +l 2 ) 0.5 Atoms in other planes cause extra scattering For validity of Bragg’s law For BCC h+k+l = even For FCC h, k, l all even or odd Sample calculations Given the plane indices and Crystal structure, get a (from C.S. and R) and d (from indices) Then with wavelength get angle or wavelength d d  n 2 sin  c Measurement of critical angle,  c, allows computation of planar spacing, d. Bragg’s Law, n: order of reflection

45 Chapter 3 - Solved Problem Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium when monochromatic radiation of wavelength nm is used. BCC d = a for plane (h k l) (h 2 +k 2 +l 2 ) 0.5 For (310) Brag’s Law

46 Chapter X-Ray Diffraction Pattern (110) (200) (211) z x y a b c Diffraction angle 2  Diffraction pattern for polycrystalline  -iron (BCC) Intensity (relative) z x y a b c z x y a b c

47 Chapter Atoms may assemble into crystalline or amorphous 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). SUMMARY 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. 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.

48 Chapter Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy). 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 interplanar spacing determinations.


Download ppt "Chapter 3 -1 ISSUES TO ADDRESS... How do atoms assemble (arrange) into solid structures? (focus on metals) How does the density of a material depend on."

Similar presentations


Ads by Google