Chapter 3: The Structure of Crystalline Solids

Chapter 3: The Structure of Crystalline Solids

Properties of engineering materials Chapter 3 - materials Metals (ferrous and non ferrous) Ceramics

Polymers Composites Advanced (biomaterials, semi conductors) Chapter 3 - 2 Properties

Optical Electrical Thermal Mechanical Magnetic Deteriorative Chapter 3 - 3 The Structure of Crystalline Solids

- Crystalline : atoms/ crystals -Non crystalline or amorphous -Solidification? Chapter 3 - 4 Packing of crystals: Non dense =random Dense = regular Chapter 3 - 5

Unit cell and lattice parameters Unit cell : small repeated entities It is the basic structural unit or building block of the crystal structure Its geometry is defined in terms of 6 lattice parameters: 3 edges length (a, b, c) 3 interaxial angles Chapter 3 - 6

Unit cell parameters Chapter 3 - 7 Chapter 3 - 8 Metallic crystals 3 crystal structures: Face-Centered Cubic FCC Body-Centered Cubic BCC Hexagonal close-packed HCP

Characterization of Crystal structure: Number of atoms per unit cell Coordination number (Number of nearest neighbor or touching atoms) Atomic Packing Factor (APF) : what fraction of the cube is occupied by the atoms Chapter 3 - 9 The Body-Centered Cubic Crystal Structure Coordination no. = ? Number of atoms per unit

cell = ? ? what fraction of the cube is occupied by the atoms or Atomic packing factor Chapter 3 - 10 The Body-Centered Cubic Crystal Structure Chapter 3 - 11

APF = 0.68 solve ? Relation between r and a Chapter 3 - 12 Solution Close-packed directions: length = 4R = 3a atoms volume 4

3 ( 3a/4) 2 unit cell atom 3 APF = volume 3 a unit cell

Chapter 3 - 13 The Face-Centered Cubic Crystal Structure Coordination no. = ? Number of atoms per unit cell = ? ? what fraction of the cube is occupied by the atoms or Atomic packing factor Chapter 3 - 14

The Face-Centered Cubic Crystal Structure Chapter 3 - 15 APF = 0.74 solve ? Relation between r and a Chapter 3 - 16 Solution

Close-packed directions: length = 4R = 2a atoms volume 4 ( 2a/4)3 4 unit cell atom 3 APF =

volume a3 unit cell Chapter 3 - 17 The Hexagonal Close-Packed Crystal Structure Chapter 3 - 18 HCP

The coordination number and APF for HCP crystal structure are same as for FCC: 12 and 0.74, respectively. HCP metal includes: cadmium, magnesium, titanium, and zinc. Chapter 3 - 19 Examples Chapter 3 - 20

Polymorphism Chapter 3 - 21 Crystallographic Points, Directions, and Planes Chapter 3 - 22 POINT COORDINATES Point position specified in terms

of its coordinates as fractional multiples of the unit cell edge lengths ( i.e., in terms of X a b and c) Z 1,1,1 0,0,0

Y Chapter 3 - 23 Solve point coordinates (2,3,5,9) Chapter 3 - 24 solution Chapter 3 - 25

Crystallographic directions 1- A vector of convenient length is positioned such that it passes through the origin of the coordinate system. Any vector may be translated throughout the crystal lattice without alteration, if parallelism is maintained. 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. 3. These three numbers are multiplied or divided by a common factor to reduce them to the smallest integer. 4. The three indices, not separated by commas, are enclosed

as [uvw]. The u, v, and w integers correspond to the reduced projections along x, y, and z axes, respectively. Chapter 3 - 26 Example Chapter 3 - 27 Solution Chapter 3 - 28

Z [0 1 0] [1 0 1] Y X Chapter 3 - 29 Planes ( miller 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. 3. The reciprocals of these numbers are taken. A plane that parallels an axis may be considered to have an infinite intercept, and, therefore, a zero

index. 4. If necessary, these three numbers are changed to the set of smallest Chapter 3 - 30 integers by multiplication or division by a common factor. example Chapter 3 - 31 Solution

Chapter 3 - 32 Example : solve Z Y X Chapter 3 - 33 solve z

c z a c a x

b y x b y Chapter 3 - 34

Crystallographic Planes z example 1. Intercepts 2. Reciprocals 3. Reduction a 1

1/1 1 1 4. Miller Indices (110) example 1. Intercepts

2. Reciprocals 3. Reduction 4. Miller Indices a 1/2 1/ 2

0 2 0 b 1 1/1 1 1 c

1/ 0 0 c b a x y

z b c 1/ 1/ 0 0 c

a (100) b x Chapter 3 - 35 y

Crystallographic Planes z example 1. Intercepts 2. Reciprocals 3. Reduction 4. Miller Indices

a 1/2 1/ 2 6 (634) b 1 1/1 1

3 c c 3/4 1/ 4/3 4 a x

b y Chapter 3 - 36 Chapter 3 - 37

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