MATS105 Introduction to Engineering Materials Lecture 2 (W2) Bonding in materials PDF

MATS105 Introduction to Engineering Materials Lecture 2 (W2) 4th Oct. 2024 Bonding in materials Maulik Patel [email protected] Topics covered in the module Ø Introduction to materials properties, microstructure, classifi...

MATS105 Introduction to Engineering Materials Lecture 2 (W2) 4th Oct. 2024 Bonding in materials Maulik Patel [email protected] Topics covered in the module Ø Introduction to materials properties, microstructure, classifications and selection. Ø Materials Ø Metals and alloys Ø Polymers and polymer-composites Ø Structural ceramics and glasses Ø Construction materials (cement, concrete, wood etc) Ø Mechanical properties/testing/failure of materials Ø The tensile test & deformation Ø Hardness Ø Fatigue Ø Creep Ø Impact fracture Ø Industrial alloys Ø Strengthening processes in metals Ø Heat treatment of metals Ø Industrial alloys – specification and application Learning objectives Recommended reading: Reading list ebook by Callister Chapters 1 & 2 Quiz What is the most fundamental physical unit that make up materials Quiz What are valence electrons ? Quiz What are the primary bonding types ? Atomic Structure orbital electrons: electrons: THE charge =n-1.602x10 = principal 9.11x10-31quantum kg -19 C, mass = BOHR number 1 n=3 2 ATOM K L M Nucleus: protons and neutrons n = principal quantum protons: number; describes the charge = 1.602x10-19 C total energy of the Mass = 1.67x10-27 kg electron neutrons: neutral, 1.67x10-27 kg Atomic number (Z) = # protons (electrically neutral = # electrons) Z ranges from 1 (H) to 94 (Pu) - naturally occurring elements Atomic mass number (A) = Z + N, N = # neutrons A uniquely ID’s ea. isotope of an element Electronic Configurations ex: Fe - atomic # = 26 1s2 2s2 2p6 3s2 3p6 3d 6 4s2 4d valence 4p N-shell n = 4 electrons 3d 4s Energy 3p M-shell n = 3 3s 2p L-shell n = 2 2s 1s K-shell n = 1 Adapted from Fig. 2.6, Callister & Rethwisch 9e. Electronic configurations Valence electrons – those in unfilled shells Filled shells more stable Valence electrons are most available for bonding and tend to control the chemical properties – example: C (atomic number = 6) 1s2 2s2 2p2 valence electrons The Periodic Table Columns: Similar Valence Structure inert gases give up 1e- give up 2e- accept 2e- accept 1e- give up 3e- H He Li Be O F Ne Na Mg S Cl Ar Adapted from Fig. 2.8, K Ca Sc Se Br Kr Callister & Rethwisch 9e. Rb Sr Y Te I Xe Cs Ba Po At Rn Fr Ra Electropositive elements: Electronegative elements: Readily give up electrons Readily acquire electrons to become + ions. to become - ions. Electronegativity Ranges from 0.9 to 4.1, Large values: tendency to acquire electrons. Smaller electronegativity Larger electronegativity Bonding forces and energies Ionic bonding Ionic bond – metal + nonmetal donates accepts electrons electrons Dissimilar electronegativities ex: MgO Mg 1s2 2s2 2p6 3s2 O 1s2 2s2 2p4 [Ne] 3s2 Mg2+ 1s2 2s2 2p6 O2- 1s2 2s2 2p6 [Ne] [Ne] Ionic bonding Occurs between + and - ions. Requires electron transfer. Large difference in electronegativity required. Non-directional Range between 600 to 1500 kJ/mol Ionic bonding Predominant bonding in Ceramics NaCl MgO CaF2 CsCl Give up electrons Acquire electrons Ionic bonding Covalent bonding similar electronegativity and therefore share electrons bonds determined by valence – s & p orbitals dominate bonding Example: H2 H2 Each H: has 1 valence e-, needs 1 more H H Electronegativities are the same. shared 1s electron shared 1s electron from 1st hydrogen from 2nd hydrogen atom atom Fig. 2.12, Callister & Rethwisch 9e. Covalent bonding Example: CH4 shared electrons H from carbon atom C: has 4 valence e, CH4 needs 4 more H: has 1 valence e, H C H needs 1 more shared electrons Electronegativities H from hydrogen are comparable. atoms Covalent bonding Bond energies and bond distances (lengths) in covalent bonded organic compounds vary with the atoms involved and the nature of the bond. Metallic bonding Outer electrons released and shared between all atoms in solid Primary bond for metals and their alloys Metallic bonding Atoms of pure metals are all the same size– leads to high coordination numbers C.N. = 12 Occurs in Al, Ag, Au, Cu, Ni, Pb, Pt This is a “close-packed” structure Primary bonding in most materials Metallic Bond -- delocalized as electron cloud Ionic-Covalent Mixed Bonding where XA & XB are Pauling electronegativities Ex: MgO XMg = 1.3 XO = 3.5 " − (3.5−1.3)2 % % ionic character = $$1 − e 4 ' x (100%) = 70.2% ionic ' # & Bonding types & classification of materials Type Bond Energy Comments Ionic Large! Nondirectional (ceramics) Covalent Variable Directional (semiconductors, ceramics large-Diamond polymer chains) small-Bismuth Metallic Variable large-Tungsten Nondirectional (metals) small-Mercury Secondary smallest Directional inter-chain (polymer) inter-molecular Bonding types Strong primary/chemical Bonds – Ionic Bonding- transfer of electrons – Covalent Bonding- sharing of electrons – Metallic Bonding- delocalized electrons Weak Secondary Bonds- van der Waals – Molecular Dipoles – Induced Dipoles – Hydrogen Bridge Properties from Bonding: Melting point Bond length, r Melting Temperature, Tm Energy r Bond energy, Eo r o r Energy smaller Tm unstretched length larger Tm r o r Eo = Tm is larger if Eo is larger. “bond energy” Properties from Bonding: Thermal expansion Coefficient of thermal expansion, α length, L o coeff. thermal expansion unheated, T 1 ΔL ΔL Lo = α (T2 -T 1 ) heated, T 2 α ~ symmetric at ro Energy unstretched length r o α is larger if Eo is smaller. r larger α Eo Eo smaller α Atomic and ionic radii Energy and packing Non dense, random packing Energy typical neighbor bond length typical neighbor r bond energy Dense, ordered packing Energy typical neighbor bond length typical neighbor r bond energy Dense, ordered packed structures tend to have lower energies. Classes of materials One type of classification based largely on bonding types – Metals – Ceramics – Polymers Another type of classification based on ‘atomic order’ – Crystalline – Noncrystalline – Semicrystalline  MATS105 Introduction to Engineering Materials Lecture 2 (W2) 4th Oct. 2024 Crystal Structures Maulik Patel [email protected] Learning objectives Chapter 3 How do atoms assemble into solid structures? How does the density of a material depend on its structure? When do material properties vary with the crystal orientation? How we name planes and directions in crystals ? Classes of materials One type of classification based largely on bonding types – Metals – Ceramics – Polymers Another type of classification based on ‘atomic order’ – Crystalline – Noncrystalline – Semicrystalline Types of atomic arrangements Classification based on atomic order Materials and packing Crystalline materials... atoms pack in periodic, 3D arrays typical of: -metals -many ceramics -some polymers crystalline SiO2 Si Oxygen Noncrystalline materials... atoms have no periodic packing occurs for: -complex structures -rapid cooling "Amorphous" = Noncrystalline noncrystalline SiO2 Lattice & Basis Unit cell-2D Space lattice (3D) Unit Cell Lattice, Basis, Unit cells & crystal structures § Lattice: A collection of points that divide space into smaller equally sized segments. § Basis: A group of atoms associated with a lattice point (same as motif). § Unit cell: A subdivision of the lattice that still retains the overall characteristics of the entire lattice. § Crystal structure: The arrangement of the atoms in a material into a regular repeatable lattice. The structure is fully described by a lattice and a basis. § Crystallography: The formal study of the arrangements of atoms in solids. 14 Bravais Lattices 7 crystal systems 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 Metallic crystal structures 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 Visualizing crystal structures Inorganic Crystal Structure Database VESTA Crystalmaker Simple cubic structure (SC) Rare due to low packing density (only Po has this structure) Close-packed directions are cube edges. Coordination # = (# nearest neighbors) Simple cubic structure (SC): Atomic Packing Factor (APF) Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres APF for a simple cubic structure = 0.52 volume atoms atom a 4 π 3 unit cell 1 (0.5a) 3 R = 0.5a APF = a3 volume unit cell close-packed directions contains 8 x 1/8 = 1 atom/unit cell Body Centres Cubic structure (BCC) Atoms touch each other along cube diagonals. Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. ex: Cr, W, Fe (α), Tantalum, Molybdenum Coordination # = 2 atoms/unit cell: 1 center + 8 corners x 1/8 BCC Atomic Packing Factor APF for a body-centered cubic structure = 0.68 3a a 2a Close-packed directions: R length = 4R = 3 a a atoms volume 4 π ( 3 a/4 ) 3 unit cell 2 3 atom APF = volume a3 unit cell Face Centred Cubic structure (FCC) Atoms touch each other along face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. ex: Al, Cu, Au, Pb, Ni, Pt, Ag Coordination # = 4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8 FCC Atomic Packing Factor APF for a face-centered cubic structure = 0.74 maximum achievable APF Close-packed directions: length = 4R = 2 a 2a Atoms per unit cell: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell a atoms 4 volume unit cell 4 π ( 2 a/4 ) 3 3 atom APF = volume a3 unit cell Relation between atomic radius and lattice parameter Characteristics of metallic crystal structures 6 Theoretical density Density (Number of atoms/cell ‘n’) (Atomic mass ‘A’) (ρ) (Volume of unit cell ‘Vc’) (Avogadro’s number ‘NA’) VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = 6.022 x 1023 atoms/mol Theoretical density Ex: Cr (BCC) A = 52.00 g/mol R = 0.125 nm n = 2 atoms/unit cell R a = 4R/ 3 = 0.2887 nm a atoms g unit cell 2 52.00 mol ρtheoretical = 7.18 g/cm3 ρ= a 3 6.022 x 1023 ρactual = 7.19 g/cm3 volume atoms unit cell mol Density of various classes of materials In general ρ ρ ρ metals > ceramics > polymers Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond Why? 30 Based on data in Table B1, Callister Platinum Metals have... 20 Gold, W Tantalum *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced close-packing Epoxy composites (values based on 60% volume fraction of aligned fibers (metallic bonding) 10 Silver, Mo Cu,Ni in an epoxy matrix). often large atomic masses Steels Tin, Zinc Zirconia Ceramics have... 5 (g/cm3 ) Titanium 4 Al oxide less dense packing 3 Diamond Si nitride Aluminum often lighter elements Glass -soda Concrete Glass fibers Silicon PTFE GFRE* 2 Carbon fibers Polymers have... Magnesium Graphite Silicone CFRE* ρ PVC Aramid fibers AFRE* low packing density 1 PET PC HDPE, PS (often amorphous) PP, LDPE lighter elements (C,H,O) 0.5 Wood Composites have... 0.4 0.3 intermediate values Data from Callister & Rethwisch, 9e. Allotropic or Polymorphic transformations § Allotropy: The characteristic of an element being able to exist in more than one crystal structure depending on temperature and pressure. § Polymorphism: Compounds exhibiting more than one type of crystal structure. § Ceramic materials, such as silica (SiO2) and zirconia (ZrO2), are polymorphic. Polymorphism in ZrO2 and Yttrium Stabilised Zirconia (YSZ) Ceramic made from pure ZrO2 will fracture as the temperature is lowered and as ZrO2 transforms from the tetragonal to monoclinic form because of volume expansion. Pol Duwez et al., J. Electrochem. Soc. 98 (1951) 356 Polymorphism in Fe Two or more distinct crystal structures for the same material (allotropy/polymorphism) Titanium (α, β –Ti) iron system liquid Carbon, diamond, graphite 1538°C BCC δ-Fe 1394°C FCC γ-Fe 912°C BCC α-Fe Crystal Structures Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal. z 7 crystal systems ! 14 crystal lattices # y c " a a, b, and c are the lattice constants b x Point co-ordinates Point co-ordinates Point co-ordinates Point co-ordinates Algorithm 1. Determine coordinates of vector tail, pt. 1: x1, y1, & z1; and vector head, pt. 2: x2, y2, z & z2. 2. Tail point coordinates subtracted from head point coordinates. 3. Normalize coordinate differences in terms pt. 2 of lattice parameters a, b, and c: pt. 1: head y x 2 − x1 y 2 − y1 z2 − z1 tail a b c x 4. Adjust to smallest integer values. 5. Enclose in square brackets, no commas ex: [uvw] pt. 1 x1 = 0, y1 = 0, z1 = 0 pt. 2 x2 = a, y2 = 0, z2 = c/2 => 1, 0, 1/2 => 2, 0, 1 a−0 0−0 c 2−0 => [ 201 ] a b c Crystallographic directions z pt. 2 Example 2: head pt. 1 x1 = a, y1 = b/2, z1 = 0 pt. 2 x2 = -a, y2 = b, z2 = c −a − a b−b 2 c−0 y a b c pt. 1: x tail => -2, 1/2, 1 Multiplying by 2 to eliminate the fraction -4, 1, 2 => [ 412 ] where the overbar represents a negative index families of directions Crystallographic directions Crystallographic directions Which is the [1-10] crystallographic direction Crystallographic Planes Adapted from Fig. 3.11, Callister & Rethwisch 9e. Crystallographic Planes 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. When selecting a new origin, the following procedure is suggested: – If the crystallographic plane that intersects the origin lies in one of the unit cell faces, move the origin one unit cell distance parallel the the axis. – If the crystallographic plane that intersects the origin passes through one of the unit cell axes, move the origin one unit cell distance parallel to either of the to other axes. – For all other cases, move the origin one unit cell distance parallel to any of the three unit cell axes 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) Crystallographic Planes z example a b c 1. Intercepts c 1 1 ∞ 2. Reciprocals 1/1 1/1 1/∞ 1 1 0 3. Reduction 1 1 0 y a b 4. Miller Indices (110) x z example a b c 1. Intercepts 1/2 ∞ ∞ c 2. Reciprocals 1/½ 1/∞ 1/∞ 2 0 0 3. Reduction 2 0 0 y 4. Miller Indices (100) a b x Crystallographic Planes Construct a (101) crystallographic plane  Crystallographic Planes Adapted from Fig. 3.11, Callister & Rethwisch 9e. Single v/s Polycrystalline Some engineering applications require single crystals: -- diamond single -- turbine blades crystals for abrasives Fig. 8.34(c), Callister & (Courtesy Martin Deakins, Rethwisch 9e. GE Superabrasives, Worthington, (courtesy of Pratt and Whitney) OH. Used with permission.) Properties of crystalline materials often related to crystal structure. -- Ex: Quartz fractures more easily along some crystal planes than others. (Courtesy P.M. Anderson) Polycrystalline Anisotropic Most engineering materials are polycrystals. Fig. K, color inset pages of Callister 5e. (Courtesy of Paul E. Danielson, Teledyne Wah Chang Albany) 1 mm Nb-Hf-W plate with an electron beam weld. Isotropic Each "grain" is a single crystal. If grains are randomly oriented, overall component properties are not directional. Grain sizes typically range from 1 nm to 2 cm (i.e., from a few to millions of atomic layers). Single v/s Polycrystalline Single Crystals E (diagonal) = 273 GPa Data from Table 3.4, -Properties vary with Callister & Rethwisch 9e. (Source of data is R.W. direction: anisotropic. Hertzberg, Deformation and Fracture Mechanics of -Example: the modulus Engineering Materials, 3rd ed., John Wiley and Sons, 1989.) of elasticity (E) in BCC iron: E (edge) = 125 GPa Polycrystals -Properties may/may not 200 μm Adapted from Fig. 4.15(b), Callister & vary with direction. Rethwisch 9e. [Fig. 4.15(b) is courtesy of -If grains are randomly L.C. Smith and C. Brady, the National Bureau of oriented: isotropic. Standards, Washington, DC (now the National Institute of (Epoly iron = 210 GPa) Standards and Technology, Gaithersburg, MD).] -If grains are textured, anisotropic.  Problem set Bonding and crystal structures 1. Compute the %IC of the interatomic bond for each of the following compounds: MgO, GaP, CsF, CdS, and FeO. 2. (a) Calculate %IC of the interatomic bonds for the intermetallic compound Al6Mn. (b) On the basis of this result what type of interatomic bonding would you expect to be found in Al6Mn? 3. What type(s) of bonding would be expected for each of the following materials: solid xenon, calcium fluoride (CaF2), bronze, cadmium telluride (CdTe), rubber, and tungsten? 4. Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and compare its theoretical density with the experimental value found inside the front cover of the book. 5. Calculate the radius of a palladium (Pd) atom, given that Pd has an FCC crystal structure, a density of 12.0 g/cm3, and an atomic weight of 106.4 g/mol. 6. Niobium (Nb) has an atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it has an FCC or a BCC crystal structure. 7. What are the indices for the two planes drawn in the following sketch? 8. Determine the Miller indices for the planes shown in the following unit cell: 9. The metal rhodium (Rh) has an FCC crystal structure. If the angle of diffraction for the (311) set of planes occurs at 36.12° (first-order reflection) when monochromatic x-radiation having a wavelength of 0.0711 nm is used, compute the following: (a) the interplanar spacing for this set of planes and (b) the atomic radius for a Rh atom. 10.The following table lists diffraction angles for the first four peaks (first-order) of the x-ray diffraction pattern for platinum (Pt), which has an FCC crystal structure; monochromatic x- radiation having a wavelength of 0.0711 nm was used. Plane Indices Diffraction Angle (2theta) (111) 18.06° (200) 20.88° (220) 26.66° (311) 31.37° (a) Determine the interplanar spacing for each of the peaks. (b) For each peak, determine the atomic radius for Pt, and compare these with the value presented in Table 3.1. 11. Within a cubic unit cell, sketch the following directions: (a) (e) [-1 1 -1] (d) [3-1 3] (h) 12.Determine the indices for the A and D directions shown in the following cubic unit cell: 13. Sketch within a cubic unit cell the following planes: (a) (1 0 -1); (b) (3 -1 3); (c) (2 -1 2)