MME 2001 MATERIALS SCIENCE

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1 MME 2001 MATERIALS SCIENCE

2 crystal structures X tal structure Coord. # Atoms/ unit cell a=f(r) APF % SC 6 1 2R 52 BCC 8 2 4R/ 3 68 FCC R 2 74 HCP R 74

3 Theoretical Density, knowing the crystal structure of a metallic solid we can compute its theoretical density Density = = = Mass of atoms in unit cell Total volume of unit cell n A V C N A A g/atom N A where n = number of atoms/unit cell A = atomic weight (g/mol) V C = Volume of unit cell = a 3 for cubic N A = Avogadro s number (6.023 x atoms/mol)

4 Theoretical Density, - BCC R atoms unit cell = a x10 23 volume unit cell a Ex: Cr (BCC) A = g/mol R = nm n = 2 (2 atoms/unit cell) a = 4R/3 = nm g mol = 7.18 g/cm 3 measured atoms mol 7.19 g/cm 3

5 Theoretical Density, - FCC R atoms unit cell = a x10 23 volume unit cell a Ex: Al (FCC) A = g/mol R = nm n = 4 (4 atoms/unit cell) a = 2R 2 = nm g mol measured = g/cm g/cm 3 atoms mol

6 Theoretical density estimation for Cu number of atoms per unit cell in FCC: 4 atomic weight Acu: 63.5 g/mol R = nm Vunitcell = a 3 = (2R 2) 3 = 16R 3 2 Measured: 8.94 g/cm 3

7 Densities of Material Classes metals > ceramics > polymers Why? (g/cm 3 ) 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 Platinum Gold, W Tantalum Silver, Mo Cu,Ni Steels Tin, Zinc Titanium Aluminum Magnesium Metals/ Alloys Zirconia Al oxide Diamond Si nitride Glass - soda Concrete Silicon G raphite PTFE Silicone PVC PET PC H DPE, PS PP, LDPE Graphite/ Ceramics/ Polymers Semicond Glass fibers GFRE* Carbon fibers CFRE * A ramid fibers AFRE * Wood Composites/ fibers

8 polycrystalline materials Most crystalline solids are composed of a collection of many small crystals or grains : polycrystalline. a c b d solidification of a polycrystalline material; (a) crystallite nuclei (b) their growth (c) Solidification over! (d) grain structure under the microscope; dark lines are the grain boundaries.

9 Single crystals the periodic arrangement extends throughout! All unit cells have the same orientation. Single crystals exist in nature + may also be produced artificially (crystal growth!) If a single crystal grows without any external constraint, the crystal will assume a regular geometric shape having flat faces, as with some of the gemstones. a garnet single crystal found in Tongbei, Fujian Province, China

10 anisotropy The physical properties of single crystals of some substances depend on the crystallographic direction in which measurements are taken. For example, the elastic modulus, the electrical conductivity, and the index of refraction may have different values in the [100] and [111] directions. This directionality of properties is termed anisotropy, and it is associated with the variance of atomic or ionic spacing with crystallographic direction.

11 anisotropy Substances in which measured properties are independent of the direction of measurement are isotropic. The extent and magnitude of anisotropic effects in crystalline materials are functions of the symmetry of the crystal structure; the degree of anisotropy increases with decreasing structural symmetry triclinic structures normally are highly anisotropic.

12 anisotropy For many polycrystalline materials, crystallographic orientations of the individual grains are random. even though each grain may be anisotropic, a specimen composed of the grain aggregate behaves isotropically. Also, the magnitude of a measured property represents some average of the directional values. Sometimes the grains in polycrystalline materials have a preferential crystallographic orientation, in which case the material is said to have a texture

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

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

15 Single vs Polycrystals Single Crystals Properties vary with direction: anisotropic. Example: E in BCC iron: Polycrystals Properties may/may not vary with direction. If grains are randomly oriented: isotropic If grains are textured anisotropic. E (diagonal) = 273 GPa E (edge) = 125 GPa 200 mm E (poly-fe) = 210 GPa

16 Polymorphism / allotropy Two or more distinct crystal structures for the same material (allotropy/polymorphism) modification of the density + physical properties Titanium:, -Ti carbon graphite is the stable polymorph at ambient conditions, whereas diamond is formed at extremely high pressures. liquid BCC FCC BCC iron system 1538ºC -Fe 1394ºC -Fe 912ºC -Fe

17 crystallography!

18 Crystallographic points, directions and planes it is often necessary to specify a particular point within a unit cell, a crystallographic direction, or some crystallographic plane of atoms. three indices are used to designate point locations, directions, and planes. The basis for determining index values is the unit cell, with a right-handed coordinate system consisting of three (x, y, and z) axes situated at one of the corners and coinciding with the unit cell edges

19 Crystallographic points directions and planes The q coordinate corresponds to the distance qa along the x axis, where a is the unit cell edge length. The respective r and s coordinates for the y and z axes are determined similarly

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

21 Location of points with specific coordinates Locate the point having coordinates: ¼, 1, ½ Move ¼ units along the x axis, 1 unit along the y axis ½ units along the z axis

22 Specification of point coordinates point coordinates for atom positions for a BCC unit cell: The BCC unit cell has 8 corner atoms and a single atom at the centre.

23 Learning check Show the point having coordinates: ½, 1, 1/3 y x

24 Learning check x 1 3 z y 1: 1 0 ½ 2: 1 ½ 0 3: ½ 0 1 4: 0 ½ 1 5: ½ 1 1 6: 0 0 ½ 7: ½ 1 0 8: ½ 0 0 9: 0 ½ 0 10: ½ ½ 1 11: 0 ½ ½ 12: 1 ½ ½

25 crystallographic directions to determine the 3 directional indices: 1. A vector of convenient length is positioned such that it passes through the origin of the coordinate system. 2. The length of the vector projection on each of the three axes is determined; these are measured in terms of the unit cell dimensions a, b, and c. [100], [110], and [111] directions within a unit cell.

26 Crystallographic directions 3. These three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values. 4. The three indices, not separated by commas, are enclosed in square brackets, thus: [uvw]. The u, v, and w integers correspond to the reduced projections along the x, y, and z axes, respectively. [100], [110], and [111] directions within a unit cell.

27 Crystallographic Directions x 1/2 1 z Algorithm ex: 1, 0, ½ 2, 0, 1 [ 201 ] ex: -1, 1, 1 [1 1 1] families of directions <uvw> y 1. Vector repositioned 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]

28 Crystallographic directions For each of the three axes, there will exist both positive and negative coordinates. -1, 1, 1 [ 111 ] where overbar represents a negative index - the [111] direction would have a negative component in the y direction. Also, changing the signs of all indices produces an - - antiparallel direction; [111] is directly opposite to - [111].

29 Determination of directional indices x y z projection 1/2 1 0 Multiply each of the above by 2 to transform these numbers to the lowest set of integers indices direction [120]

30 Crystallographic Directions Point P is located by moving along the x axis a units: 1 From this position a units parallel to the y axis: -1 Z projection is 0; so there is no z component: 0 [110] direction within a cubic unit cell

31 Hexagonal crystals a four-axis, or Miller Bravais, coordinate system is used in hexagonal crystals. The three a 1, a 2, and a 3 axes are all contained within the basal plane and are at 120 angles to one another. The z axis is perpendicular to this basal plane. Directional indices will be denoted by four indices as [uvtw]; the first three indices pertain to projections along the respective a 1, a 2, and a 3 axes in the basal plane.

32 HCP Crystallographic Directions a 3 z a 2 a 1 - Algorithm 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] a 2 ex: ½, ½, -1, 0 [ 1120 ] dashed red lines indicate projections onto a 1 and a 2 axes a 3 a 2 2 a 1 2 -a 3 a 1

33 HCP Crystallographic Directions 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows. a 3 z a 2 a 1 - [ u ' v ' w ' u v t w = = = = ] 1 ( 2 u '- v ') 3 1 ( 2 v '- u ') 3 -( u + v ) w ' [ uvtw ]

34 Construction of specific crystal directions [122] direction within a cubic unit cell Considering unit cell scales: ½, 1, 1 z Move along x axis by ½ Move parallel to the y axis by 1 Move parallel to the z axis by 1 y x

35 Construction of specific crystal directions [122] direction within a cubic unit cell z y x

36 Construction of specific crystal directions _ [122] direction within a cubic unit cell Considering unit cell scales: ½, -1, 1 z Move along x axis by ½ Move parallel to the y axis by -1 Move parallel to the z axis by 1 x y

37 Learning check Draw the directions [236] and [203] in a cubic unit cell! x: 2 2/6 1/3 y: 3 3/6 1/2 z: 6 6/6 1 x: 2 2/3 2/3 y: 0 0/3 0 z: 3 3/3 1

38 Learning check Draw the direction [211] in a cubic unit cell! 2 2/ /2-1/2 1 1/2 1/2 1/2-1/2 1-1/2 1

39 Exercise - crystal directions

40 Learning check

41 Exercise - crystal directions

42 Exercise - crystal directions

43 Learning check 1 1 1/3

44 Learning check

45 Learning check ½ 0-1

46 Learning check Click on the diagram that shows the direction [112] correctly drawn:

47 Learning check Click on the diagram that shows the [021] direction correctly drawn!

48 Learning check Which one shows the [111] direction correctly drawn.

49 Learning check Which one shows the [210] direction correctly drawn? Which one shows the [210] direction correctly drawn? Which one shows the [201] direction correctly drawn?

50 crystal directions

51 Exercise crystal directions

52 Exercise - crystal directions

53 Learning check sketch the following directions in a cubic unit cell

54 Learning check Determine indices for the directions shown in the cubic unit cell

55 Learning check Determine indices for the directions shown in the following cubic unit cell

56 Linear densities Linear density (LD) is the number of atoms per unit length along a particular direction linear density = # of atoms on the direction vector length of the direction vector equivalent directions have identical linear densities. [110] direction, through atoms X, Y,and Z.

57 Linear Density <110> directions in the FCC lattice have 2 atoms (2 corner atoms x ½) + 1 center atom Length of the <110> vector is 2a a [110] # atoms length LD = a = nm 2 2 a = 3.5 nm - 1 Most densely packed direction in FCC

58 Learning check The atomic radius of Fe is nm. Find the lattice parameter of Fe.

59 Crystallographic planes crystallographic planes are specified by three Miller indices as (hkl). Any two planes parallel to each other are equivalent and have identical indices. The procedure used to determine the h, k, and l indices: 1. If the plane passes through the selected origin, either another parallel plane must be constructed within the unit cell by an appropriate translation, or a new origin must be established at the corner of another unit cell. 2. At this point the crystallographic plane either intersects or parallels each of the three axes; the length of the planar intercept for each axis is determined in terms of the lattice parameters a, b, and c.

60 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 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)

61 Crystallographic planes (001) (110) (111) crystallographic planes. unique characteristic of cubic crystals: planes and directions having the same indices are perpendicular to one another; for other crystal systems there are no such simple geometrical relationships between planes and directions

62 Crystal planes intercepts on a, b, c: ¾, ½, ¼ Reciprocals 4/3, 2, 4 Clear off fractions 4, 6, 12 Reduce to smallets integers 2, 3, 6 Miller indices (2 3 6) x z y

63 Equivalent crystal planes Planes with similar indices are equivalent. Faces of the cube (100), (010) and (001). Family of planes: {100} includes all (100) combinations including those with negative indices!

64 Determination of planar (miller) indices x y z intercept -1 1/2 reciprocal of each of the above reductions direction (012)

65 Crystallographic Planes _ Construct a (011) plane within a cubic unit cell. Take the reciprocals of the indices:, -1, 1. The plane parallels the x axis, intersect the y axis at b intersects the z axis at c _ (011)

66 Crystallographic Planes z example a b c 1. Intercepts Reciprocals 1/1 1/1 1/ Reduction a c b y 4. Miller Indices (110) x z example a b c 1. Intercepts 1/2 2. Reciprocals 1/½ 1/ 1/ Reduction Miller Indices (100) x a c b y

67 Crystallographic Planes z example a b c 1. Intercepts Reciprocals 1/1 1/-1 1/ Reduction Miller Indices (110) x a z c b y example a b c 1. Intercepts 1/2 2. Reciprocals 1/ 1/1/2 1/ Reduction Miller Indices (010) x a c b y

68 Crystallographic Planes example a b c 1. Intercepts 1/2 1 3/4 2. Reciprocals 1/½ 1/1 1/¾ 2 1 4/3 c z 3. Reduction Miller Indices (634) a b y x Family of Planes {hkl} Ex: {100} = (100), (010), (001), (100), (010), (001)

69 Miller-Bravais indices for a plane in a hexagonal unit cell The plane intersects a 1 axis at a distance of a from the origin (point C). a 2 axis at a distance of a in the opposite direction! (negative) z axis at a distance of c the intersections are: 1, -1, 1 The reciprocal of these numbers are also 1, -1, 1 i= (h+k)= (1-1)=0 (hkil) = (1101)

70 Crystallographic Planes (HCP) In hexagonal unit cells the same idea is used example a 1 a 2 a 3 c 1. Intercepts Reciprocals 1 1/ z 3. Reduction a 2 4. Miller-Bravais Indices (1011) a 3 a 1

71 see you next week!