11.1 Crystal Structure, Phase Diagrams, and Material Processing

11.1 Crystal Structure, Phase Diagrams, and Material Processing

Exam Focus: Materials science on the FE Other Disciplines exam tests crystal lattice geometry, theoretical density calculations using atomic radius and unit cell volume, X-ray diffraction via Bragg's Law, quantitative phase diagram evaluation using the Lever Rule on binary systems (especially (\text{Fe-Fe}_3\text{C})), equilibrium vs non-equilibrium heat treatment microstructures, and polymer structure classification.

Unit Cells, Atomic Packing Factor, and Theoretical Density

Metals and many ceramics solidify into repeating three-dimensional atomic arrangements known as crystal lattices. The smallest repeating structural unit that reflects the full symmetry of the crystal lattice is the unit cell. The FE Reference Handbook highlights three primary metallic crystal structures: Body-Centered Cubic (BCC), Face-Centered Cubic (FCC), and Hexagonal Close-Packed (HCP).

Lattice Characteristics and Geometry

• Simple Cubic (SC): Contains 8 corner atoms, each shared among 8 neighboring unit cells, giving (n = 8 \times (1/8) = 1) atom per unit cell. Atoms touch along the cube edge: (a = 2r).
• Body-Centered Cubic (BCC): Contains 8 corner atoms plus 1 central atom, giving (n = 8 \times (1/8) + 1 = 2) atoms per unit cell. Atoms touch along the body diagonal: (\sqrt{3}a = 4r \implies a = \frac{4r}{\sqrt{3}}). Coordination number is 8. Examples include (\alpha)-iron, chromium, tungsten, molybdenum, and vanadium.
• Face-Centered Cubic (FCC): Contains 8 corner atoms plus 6 face-centered atoms (each shared by 2 cells), giving (n = 8 \times (1/8) + 6 \times (1/2) = 4) atoms per unit cell. Atoms touch along the face diagonal: (\sqrt{2}a = 4r \implies a = 2\sqrt{2}r). Coordination number is 12. Examples include (\gamma)-iron, aluminum, copper, gold, silver, nickel, and platinum.
• Hexagonal Close-Packed (HCP): Features top and bottom hexagonal planes with 6 corner atoms, 1 center atom, plus 3 interior atoms between planes, yielding (n = 6) atoms per hexagonal unit cell (or 2 atoms for the primitive cell). Ideal height-to-edge ratio is (c/a = 1.633). Coordination number is 12. Examples include titanium, zinc, magnesium, and zirconium.

Atomic Packing Factor (APF)

The Atomic Packing Factor (APF) represents the fraction of solid sphere volume occupied by atoms within a unit cell:

$\text{APF} = \frac{V_{\text{atoms}}}{V_{\text{unit cell}}} = \frac{n \cdot \left(\frac{4}{3}\pi r^3\right)}{V_c}$

Theoretical Mass Density Equation

The theoretical mass density ((\rho)) of a crystalline material is determined from its unit cell parameters:

$\rho = \frac{n \cdot A}{V_c \cdot N_A}$

where:

• (n) = number of atoms per unit cell
• (A) = atomic weight of the element ((\text{g/mol}) or (\text{kg/kmol}))
• (V_c) = unit cell volume ((a^3) for cubic systems, (\text{cm}^3) or (\text{m}^3))
• (N_A) = Avogadro's number ((6.022 \times 10^{23} \text{ atoms/mol}))

Worked Example: Theoretical Density of FCC Copper

Problem: Face-Centered Cubic (FCC) copper ((\text{Cu})) has an atomic radius (r = 0.1278 \text{ nm}) ((1.278 \times 10^{-8} \text{ cm})) and an atomic weight (A = 63.55 \text{ g/mol}). Calculate its theoretical mass density in (\text{g/cm}^3).

Solution:

1. Identify atoms per cell: For FCC, (n = 4).
2. Calculate lattice parameter (a): $a = 2\sqrt{2} r = 2\sqrt{2} (1.278 \times 10^{-8} \text{ cm}) = 3.6148 \times 10^{-8} \text{ cm}$
3. Calculate unit cell volume (V_c = a^3): $V_c = (3.6148 \times 10^{-8} \text{ cm})^3 = 4.7236 \times 10^{-23} \text{ cm}^3$
4. Compute density (\rho): $\rho = \frac{4 \text{ atoms} \times 63.55 \text{ g/mol}}{(4.7236 \times 10^{-23} \text{ cm}^3) \times (6.022 \times 10^{23} \text{ atoms/mol})} = \frac{254.2}{28.4456} \approx 8.936 \text{ g/cm}^3$

Bragg's Law and X-Ray Diffraction

X-ray diffraction (XRD) is the primary technique used to determine crystallographic structures, lattice parameters, and interplanar spacings. When a monochromatic X-ray beam strikes parallel crystallographic planes specified by Miller indices ((hkl)), constructive interference occurs only when the path length difference between reflected waves equals an integer number of wavelengths.

Bragg's Law Formula

$n \lambda = 2 d_{hkl} \sin \theta$

where:

• (n) = order of diffraction (an integer, typically (n = 1))
• (\lambda) = wavelength of incident X-ray radiation ((\text{nm}) or (\text{\AA}))
• (d_{hkl}) = interplanar spacing between adjacent parallel ((hkl)) planes
• (\theta) = Bragg diffraction angle (half of the total reflection angle (2\theta))

Interplanar Spacing for Cubic Crystals

For any cubic crystal system (SC, BCC, or FCC) with lattice parameter (a), the interplanar spacing (d_{hkl}) for a specific set of planes ((hkl)) is given by:

$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$

Worked Example: X-Ray Diffraction Angle for BCC Iron

Problem: Monochromatic X-rays with wavelength (\lambda = 0.1542 \text{ nm}) ((\text{Cu } K_\alpha) radiation) reflect off the ((110)) planes of a Body-Centered Cubic (BCC) iron crystal with lattice parameter (a = 0.2866 \text{ nm}). Assuming first-order diffraction ((n = 1)), calculate the diffraction angle (\theta) and the detector position (2\theta).

Solution:

1. Compute interplanar spacing (d_{110}): $d_{110} = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{0.2866}{\sqrt{2}} = \frac{0.2866}{1.4142} \approx 0.20265 \text{ nm}$
2. Apply Bragg's Law to solve for (\sin \theta): $\sin \theta = \frac{n \lambda}{2 d_{110}} = \frac{1 \times 0.1542 \text{ nm}}{2 \times 0.20265 \text{ nm}} = \frac{0.1542}{0.4053} \approx 0.38046$
3. Calculate (\theta) and (2\theta): $\theta = \arcsin(0.38046) \approx 22.36^\circ$ $2\theta = 2 \times 22.36^\circ = 44.72^\circ$

Phase Diagrams, Gibbs Phase Rule, and the Lever Rule

A phase diagram maps the equilibrium microstructures and phases present in a material system as a function of temperature, pressure, and chemical composition.

Gibbs Phase Rule

The number of degrees of freedom ((F)) in an equilibrium system is governed by Gibbs Phase Rule:

$P + F = C + N$

For binary systems under constant atmospheric pressure ((N = 1)), the reduced Gibbs Phase Rule simplifies to:

$P + F = C + 1$

where (P) is the number of coexisting phases, (C) is the number of chemical components, and (F) is the degrees of freedom (number of state variables like temperature and concentration that can be independently varied without altering the number of phases present).

Boundary Lines on Phase Diagrams

• Liquidus Line: The boundary above which the system is entirely liquid.
• Solidus Line: The boundary below which the system is entirely solid.
• Solvus Line: The boundary defining the limit of solid solubility between two solid phases.

The Lever Rule Equation

To determine the mass fraction of coexisting phases in a two-phase region (e.g., (\alpha + \beta)) at temperature (T), construct a horizontal tie-line extending across the two-phase region from phase boundary (C_\alpha) to phase boundary (C_\beta). For an overall alloy composition (C_0):

$W_\alpha = \frac{C_\beta - C_0}{C_\beta - C_\alpha}$

$W_\beta = \frac{C_0 - C_\alpha}{C_\beta - C_\alpha} = 1 - W_\alpha$

Note that the mass fraction of phase (\alpha) is proportional to the length of the tie-line segment opposite to phase (\alpha).

Binary Invariant Reactions

An invariant reaction occurs at a specific temperature and composition where three phases coexist in equilibrium ((F = 0)):

The Iron-Carbon ((\text{Fe-Fe}_3\text{C})) System and Heat Treatment Processes

The iron-carbon system forms the cornerstone of steel metallurgy. Steel is defined as an iron-carbon alloy containing less than (2.14 \text{ wt% C}), whereas cast iron contains between (2.14 \text{ wt% C}) and (6.70 \text{ wt% C}).

Key Constituents of the Iron-Carbon System

• Ferrite ((\alpha)-iron): Body-Centered Cubic (BCC) solid solution of carbon in iron. Stable below (912^\circ\text{C}). Maximum carbon solubility is (0.022 \text{ wt% C}) at (727^\circ\text{C}). Ferrite is soft, ductile, and magnetic.
• Austenite ((\gamma)-iron): Face-Centered Cubic (FCC) solid solution of carbon in iron. Stable between (912^\circ\text{C}) and (1394^\circ\text{C}). Maximum carbon solubility is (2.14 \text{ wt% C}) at (1147^\circ\text{C}). Austenite is non-magnetic and tough.
• Cementite ((\text{Fe}_3\text{C})): An intermetallic compound containing (6.70 \text{ wt% C}). Extremely hard, brittle, and ceramic-like.
• Pearlite: A two-phase lamellar (layered) microstructure composed of alternating plates of (\alpha)-ferrite ((88 \text{ wt%})) and cementite ((12 \text{ wt%})) formed during slow equilibrium cooling of austenite across the eutectoid temperature ((727^\circ\text{C}), (0.76 \text{ wt% C})).

Microstructural Calculations: Hypoeutectoid vs. Hypereutectoid Steels

• Hypoeutectoid Steel ((< 0.76 \text{ wt% C})): Upon cooling from austenite, proeutectoid ferrite ((\alpha')) forms first along grain boundaries until (727^\circ\text{C}), at which point remaining austenite transforms to pearlite.
• Hypereutectoid Steel ((0.76 - 2.14 \text{ wt% C})): Proeutectoid cementite forms first along grain boundaries, followed by pearlite transformation at (727^\circ\text{C}).

Worked Example: Microstructural Fractions in (0.40 \text{ wt% C}) Steel

Problem: A (0.40 \text{ wt% C}) hypoeutectoid steel is slowly cooled under equilibrium conditions to just below (727^\circ\text{C}). Given phase compositions of (C_\alpha = 0.022 \text{ wt% C}), (C_\gamma = 0.76 \text{ wt% C}), and (C_{\text{Fe}_3\text{C}} = 6.70 \text{ wt% C}):

1. Calculate total mass fraction of ferrite ((W_\alpha)) and cementite ((W_{\text{Fe}_3\text{C}})).
2. Calculate mass fraction of proeutectoid ferrite ((W_{\alpha'})) formed prior to the eutectoid reaction.

Solution:

1. Total Ferrite and Cementite Fractions: $W_\alpha = \frac{C_{\text{Fe}_3\text{C}} - C_0}{C_{\text{Fe}_3\text{C}} - C_\alpha} = \frac{6.70 - 0.40}{6.70 - 0.022} = \frac{6.30}{6.678} \approx 0.9434 \text{ (94.34\%)}$ $W_{\text{Fe}_3\text{C}} = \frac{C_0 - C_\alpha}{C_{\text{Fe}_3\text{C}} - C_\alpha} = \frac{0.40 - 0.022}{6.70 - 0.022} = \frac{0.378}{6.678} \approx 0.0566 \text{ (5.66\%)}$
2. Proeutectoid Ferrite Fraction ((W_{\alpha'})): $W_{\alpha'} = \frac{C_\gamma - C_0}{C_\gamma - C_\alpha} = \frac{0.76 - 0.40}{0.76 - 0.022} = \frac{0.36}{0.738} \approx 0.4878 \text{ (48.78\%)}$ (Pearlite fraction is (W_{\text{pearlite}} = 1 - W_{\alpha'} = 0.5122) or 51.22%).

Steel Heat Treatments

• Full Annealing: Heating steel into the single-phase austenite region followed by slow furnace cooling. Produces coarse pearlite, low strength, maximum ductility, and stress relief.
• Normalizing: Heating above the upper critical temperature followed by still-air cooling. Yields fine pearlite, refined grain structure, and higher yield strength than full annealing.
• Quenching: Rapidly cooling austenite in water, oil, or brine to suppress carbon diffusion. Traps carbon in a Body-Centered Tetragonal (BCT) lattice called Martensite. Martensite is extremely hard, strong, and brittle.
• Tempering: Reheating quenched martensitic steel to a temperature below the eutectoid line ((250^\circ\text{C} - 650^\circ\text{C})). Promotes diffusion of trapped carbon to form Tempered Martensite (fine cementite particles in a ferrite matrix), restoring ductility and impact toughness while retaining high strength.

Polymer Structure and Crystallinity

• Thermoplastics: Linear and branched polymer chains connected by weak van der Waals secondary bonds. Reversibly soften/melt above melting temperature ((T_m)) and re-harden upon cooling (e.g., polyethylene, PVC, polystyrene).
• Thermosets: Cross-linked 3D network polymer chains joined by strong covalent bonds. Do not melt upon heating; decompose permanently at elevated temperatures (e.g., epoxy, phenolic resins).
• Thermal Transitions: Amorphous regions exhibit a Glass Transition Temperature ((T_g)), below which the polymer behaves as a rigid brittle glass and above which it becomes rubbery and flexible.