Materials Science for Engineering Technology (MET 161-3) PDF

Materials Science for Engineering Technology (MET 161-3) Crystalline and non-crystalline materials Dr. Sandeepa Lakshad [email protected] 1  Course Conte...

Materials Science for Engineering Technology (MET 161-3) Crystalline and non-crystalline materials Dr. Sandeepa Lakshad [email protected] 1  Course Content 1. Introduction (Evolution of engineering materials, Classification of materials.) 2. Structures of Materials (Atomic Structure, Bonding forces and energies, Primary interatomic bonds and secondary bonding) 3. Crystalline and non-crystalline materials 4. Phase diagram and microstructure 5. Electrical and Optical Properties of Materials (Conductors, semiconductors, and insulators.) 6. Mechanical Properties of Materials (Tensile, compression, impact energy, fracture toughness.) 7. Mechanical behavior of Materials (Stress-strain behavior, Elastic and plastic properties of materials, dislocations and strengthening mechanisms.) 8. Introduction to Failure Analysis and Prevention Fundamentals of fracture (Fractures Types (ductile, brittle, fatigue and creep), Corrosion, Nondestructive Testing and Techniques for failure analysis and Prevention.) 9. Selection of Engineering Materials 2 (Characterization of Materials, Design and safety factors) 3. Crystalline and non-crystalline materials 3 Short-Range order vs. Long-Range order No Order: In monoatomic gases or plasma, atoms or ions have no orderly arrangement. 4 Short-Range Order: A material displays short-range order if the special arrangement of the atoms extends only to the atom’s nearest neighbors. 5 Long-Range Order: The special atomic arrangement extends over much larger length scales, larger than 100 nm. 6 Classification of materials based on the type of atomic order: Liquid Crystals Amorphous Materials Crystalline Materials Monoatomic Gases Short and Long-range Only short-range Short and long-range No order order in small order order [Ex: Ar] volumes [Ex: Glass] [Ex: Metals] [Ex: LCD polymers] Single Crystal Materials Polycrystalline Materials 7 Crystalline Materials The atoms, ions, or molecules in crystalline materials form a regular, repeating pattern in three dimensions/ two dimensions/ one dimension. 8 Single Crystal Materials If a crystalline material consists of only one large crystal, we refer to it as a single crystal. Single crystals are useful in many electronic and optical applications. 9 Poly Crystalline Materials A polycrystalline material is composed of many small crystals with varying orientations in space. These smaller crystals are known as grains. The borders between crystals, where the crystals are in misalignment, are known as grain boundaries. 10 https://doi.org/10.1038/ncomms3811 HRTEM images 11 12 Liquid Crystals Liquid crystals are polymeric materials that have a special type of order. Liquid crystal polymers behave as amorphous materials (liquid-like) in one state. Upon exposure to external stimuli such as electric fields or temperature changes, certain polymer molecules align and form small crystalline regions. 13 Properties of Crystalline and amorphous materials Crystalline Materials Amorphous Materials Repetitive atomic Random atomic arrangement structure Melting point Glass transition temperature Definite geometrical Irregular shape shape 14 15 Lattice, basis, Unit cells, and Crystal Structures Crystalline Structure: Ordered arrangement of atoms, ions, or molecules. 16  A lattice is a collection of points called lattice points, arranged in a periodic pattern. In other words, lattice means a three-dimensional array of points coinciding with atom positions. A lattice is a purely mathematical construct and is infinite in extent. A group of one or more atoms located in a particular way with respect to each other and associated with each lattice point is known as the basis. We obtain a crystal structure by placing the atoms of the basis on every lattice point. 17 (a) A one-dimensional lattice. The lattice points are separated by an equal distance. (b) A basis of one atom. (c) A crystal structure is formed by placing the basis of (b) on every lattice point in (a). (d) A crystal structure is formed by placing a basis of two atoms of different types on the lattice in (a). 18  When describing crystalline structures, atoms are often represented as solid spheres with well-defined diameters. This atomic hard-sphere model depicts spheres of nearest- neighbor atoms touching one another. 19 Unit Cells The atomic order in crystalline solids indicates that small groups of atoms form a repetitive pattern. Thus, it is often convenient to subdivide crystal structures into small repeat entities called unit cells when describing them. 20 21 22  Three-dimensional arrangements of lattice points are known as the Bravais lattices. There are 14 Bravais lattices. These 14 Bravais lattices are grouped into 7 crystal systems. 23 24 Simple Cubic (SC) Ex: Po 25 Body-Centered Cubic (BCC) Ex: Na, Li 26 27 Face Centered Cubic (FCC) Ex: Cu, Al 28 29 30 Number of atoms per unit cell Each unit cell contains a specific number of lattice points. When counting the number of lattice points belonging to each unit cell, we must recognize that, lattice points may be shared by more than one unit cell. 31 A lattice point at a corner of one unit cell is shared by seven adjacent unit cells (thus a total of eight cells); only one-eighth of each corner belongs to one particular cell. 32 Example: Determine the number of lattice points per unit cell in SC, BCC, and FCC crystal systems. If only one atom is located at each lattice point, calculate the number of atoms per unit cell. 33 34 Atomic Radius vs. Lattice Parameter (relationship between the apparent size of the atom and the size of the unit cell) By determining the length of the direction relative to the lattice parameters and adding the number of atomic radii along that direction, we can establish the desired relationship. 35 Example: Determining the Relationship between Atomic Radius and Lattice Parameters of SC, BCC, and FCC. 36 37 Calculate the atomic radius in cm for the following: (a) BCC metal with a = 0.3294 nm; and (b) FCC metal with a = 4.0862 Å. 38 Packing Factor (Atomic Packing Factor) Fraction of space occupied by atoms, assuming that the atoms are hard spheres. Volume of atoms in a unit cell Packing Factor = Volume of unit cell Number of atoms per unit cell ×Volume of an atom Packing Factor = Volume of unit cell 39 Example: Calculate the packing factor for FCC, SC unit cell. 40 Density The theoretical density of a material can be calculated using the properties of the crystal structure. Number of atoms per unit cell ×Atomic mass Denity = Volume of unit cell×Avogadro constant 41 Example: Determine the density of BCC iron which has lattice parameter of 0.2866 nm. 42 Crystallographic Points, Directions, and Planes When dealing with crystalline materials, specifying a particular point within a unit cell, a crystallographic direction, or some crystallographic plane of atoms often becomes necessary. 43 POINT COORDINATES (specifying a lattice position within a unit cell) The position of a lattice can be described using three coordinates, one for each axis (x, y, and z.) In this case, we have labeled them as Px, Py, and Pz. To specify the coordinates, we use three fractional indices: q, r, and s. These indices are expressed as multiples of the unit cell lengths a, b, and c. 44  𝑃! = 𝑞𝑎 𝑃" = 𝑟𝑏 𝑃# = 𝑠𝑐 Point P: 𝑞 𝑟 𝑠 45 Example: For the unit cell shown in the accompanying $ $ sketch, locate the point having indices 1 % % 46 Example: Specify indices for all lattice points of the unit cell. 47 CRYSTALLOGRAPHIC DIRECTIONS (line directed between two points or a vector) 1. First, a right-handed x-y-z coordinate system is constructed (its origin may be located at a unit cell corner for convenience). 2. The coordinates of two points that lie on the direction vector (referenced to the coordinate system) are determined. 3. Tail point coordinates are subtracted from head point components (that is, x2 - x1, y2 - y1, and z2 - z1). 4. These coordinate differences are then normalized in terms of (i.e., divided by) their respective a, b, and c lattice parameters, which yields a set of three numbers. 5. If necessary, these three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values. 6. The three resulting indices, not separated by commas, are enclosed in square brackets. 48 In summary, the 𝑢, 𝑣, & 𝑤 indices are determined using the following equations: 𝑥& − 𝑥$ 𝑢=𝑛 𝑎 𝑦& − 𝑦$ 𝑣=𝑛 𝑏 𝑧& − 𝑧$ 𝑤=𝑛 𝑐 49 Determine the indices for the direction shown in the accompanying figure. 50 CRYSTALLOGRAPHIC PLANES (Miller indices) The orientations of planes for a crystal structure are represented in a similar manner. The procedure used to determine the h, k, and l index numbers is as follows: 51 1. When a plane passes through the chosen origin, we have two options. We can either construct another parallel plane within the unit cell through an appropriate translation or establish a new origin at the corner of another unit cell. 2. At this stage, the crystallographic plane either intersects or runs parallel to the three axes. We determine the coordinates for the intersection of the crystallographic plane with each of the axes, referenced to the origin of the coordinate system. These coordinates for the x, y, and z axes will be denoted as A, B, and C, respectively. 3. The reciprocals of these numbers are taken. A plane that parallels an axis is considered to have an infinite intercept and, therefore, a zero index. 4. The reciprocals of the intercepts are then normalized in terms of (i.e., multiplied by) their respective a, b, and c lattice parameters. That is, 5. If necessary, these three numbers are changed to the set of smallest integers by multiplication or by division by a common factor. 6. Finally, the integer indices, not separated by commas, are enclosed within parentheses, thus: (hkl). The h, k, and l integers correspond to the normalized intercept reciprocals referenced to the x, y, and z axes, respectively. 52 7. In summary, the h, k, and l indices may be determined using the following equations: In summary, the ℎ, 𝑘, & 𝑙 indices are determined using the following equations: 𝑛𝑎 ℎ= 𝐴 𝑛𝑏 𝑘= 𝐵 𝑛𝑐 𝑙= 𝐶 53 Determine the Miller indices for the plane shown in the accompanying sketch. 54 Thank You! 55