Crystal structure, defects and mechanical properties of metals

Why it matters

Whether a shaft bends or snaps, whether a sheet can be deep drawn, and why a cold-rolled bar is stronger than an annealed one all trace back to how atoms are packed and how imperfections in that packing move. A mechatronics engineer choosing between an aluminium bracket, a steel shaft or a titanium linkage is really choosing a crystal structure and a defect population. These ideas also underpin every later topic in this subject: heat treatment, forming and machining all work by changing or exploiting defects.

Key ideas

Crystal structure. In a crystalline metal the atoms sit on a repeating three-dimensional lattice. The smallest repeating block is the unit cell, described by its edge length (lattice parameter) a. Treating atoms as hard spheres of radius r, three structures cover almost all engineering metals:

• Body-centred cubic (BCC): atoms at the 8 corners and one at the body centre. Atoms touch along the body diagonal, so √3·a = 4r. Effective atoms per cell n = 8 × 1/8 + 1 = 2. Coordination number 8. Atomic packing factor (APF) 0.68. Examples: α-iron (ferrite) at room temperature, Cr, Mo, W, V.
• Face-centred cubic (FCC): corners plus the centre of each face. Atoms touch along the face diagonal, so √2·a = 4r. n = 8 × 1/8 + 6 × 1/2 = 4. Coordination number 12. APF 0.74 (the maximum for equal spheres). Examples: Al, Cu, Ni, Au, Ag, γ-iron (austenite).
• Hexagonal close-packed (HCP): close-packed layers stacked ABAB. n = 6 per hexagonal prism, coordination number 12, APF 0.74, ideal c/a = 1.633. Examples: Mg, Zn, Ti (α), Co.
• Simple cubic (n = 1, a = 2r, APF 0.52) is a teaching reference only (polonium).

FCC and HCP have the same packing density; they differ only in stacking sequence (ABCABC versus ABAB). Iron is polymorphic (allotropic): BCC up to 912 °C, FCC from 912 to 1394 °C, BCC (δ) again up to melting at 1538 °C. That change is what makes steel heat-treatable.

Slip and ductility. Plastic deformation happens by dislocations gliding on close-packed planes in close-packed directions; a plane plus a direction is a slip system. FCC has 12 slip systems on {111}⟨110⟩ that operate easily at all temperatures, so FCC metals stay ductile even at cryogenic temperatures. BCC has 12 or more nominal systems on {110}⟨111⟩ (48 counting {112} and {123}), but its planes are not truly close-packed, so slip needs more stress and is strongly temperature-sensitive; BCC steels show a ductile-to-brittle transition. HCP has only 3 easy basal systems, so it is the least ductile at room temperature.

Defects. Real crystals are never perfect, and the useful properties come from the imperfections:

• Point defects (zero-dimensional): vacancies, self-interstitials, substitutional and interstitial solute atoms. Vacancies control diffusion; carbon in iron is an interstitial solute.
• Line defects (one-dimensional): dislocations, edge (Burgers vector perpendicular to the line) and screw (Burgers vector parallel to the line). Moving a dislocation shears the crystal one atom spacing at a time, so real yield strengths are 100–1000 times lower than the theoretical strength of a perfect crystal.
• Surface (planar) defects: grain boundaries, twin boundaries, stacking faults, free surfaces.
• Volume defects: pores, cracks, inclusions, second-phase particles.

Strengthening mechanisms. Every practical way of strengthening a metal makes dislocation motion harder: grain refinement (Hall–Petch), solid-solution strengthening, strain (work) hardening by raising dislocation density in cold work, and precipitation or dispersion hardening. Grain refinement is the only one that raises strength and toughness together; the others usually cost ductility.

Mechanical properties from the tensile test. Engineering stress uses the original area, engineering strain the original length. The linear region gives Young's modulus E (a property of atomic bonding, almost unaffected by heat treatment or cold work). Yield strength (or 0.2 % proof stress when there is no sharp yield point) marks the onset of plastic flow; ultimate tensile strength (UTS) is the maximum engineering stress, where necking begins. Ductility is reported as percentage elongation or percentage reduction in area. Resilience is the elastic strain energy per unit volume; toughness is the total energy absorbed to fracture (area under the curve, or impact energy in Charpy/Izod tests). Hardness is resistance to localised plastic indentation (Brinell, Vickers, Rockwell); for steels UTS roughly scales with Brinell hardness, but take the conversion factor from your data book.

Formulas

ρ = n·A / (a³·N_A)

• ρ: theoretical density (kg/m³, or g/cm³ if a is in cm and A in g/mol); n: atoms per unit cell; A: atomic mass (g/mol); a: lattice parameter (m or cm); N_A = 6.022 × 10²³ mol⁻¹. Applies to cubic crystals.

APF = n·(4/3)·π·r³ / a³

• r: atomic radius; a: lattice parameter (same length unit). BCC: a = 4r/√3, APF = √3·π/8 = 0.68. FCC: a = 2√2·r, APF = π/(3√2) = 0.74.

σ_y = σ₀ + k·d^(−1/2) (Hall–Petch)

• σ_y: yield strength (MPa); σ₀: friction stress (MPa); k: locking parameter (MPa·m^0.5); d: average grain diameter (m). Valid for grain sizes from about 1 µm upward; material constants come from tests or your data book.

σ = F / A₀, ε = ΔL / L₀, E = σ / ε (elastic region only)

• F: load (N); A₀: original area (m²); ΔL, L₀: extension and gauge length (m); E: Young's modulus (Pa).

σ_T = σ·(1 + ε), ε_T = ln(1 + ε)

• True stress and true strain; valid up to the onset of necking (uniform deformation, constant volume).

U_r = σ_y² / (2E)

• Modulus of resilience (J/m³), elastic strain energy per unit volume up to yield.

Worked examples

Example 1 (standard): density of iron. α-iron is BCC with a = 0.2866 nm and A = 55.85 g/mol. Find its theoretical density and atomic radius.

1. BCC: n = 2.
2. a = 0.2866 nm = 0.2866 × 10⁻⁷ cm, so a³ = 2.354 × 10⁻²³ cm³.
3. ρ = n·A / (a³·N_A) = (2 × 55.85) / (2.354 × 10⁻²³ × 6.022 × 10²³) = 111.70 / 14.18 g/cm³.
4. ρ = 7.88 g/cm³ (7880 kg/m³; the measured value is 7.87 g/cm³).
5. Radius: r = √3·a/4 = 1.732 × 0.2866 / 4 = 0.124 nm.

Example 2 (GATE level): Hall–Petch from two tests. Annealed samples of a low-carbon steel give σ_y = 150 MPa at grain size 64 µm and σ_y = 250 MPa at 16 µm. Find σ₀, k, and the yield strength if thermo-mechanical processing refines the grain to 4 µm.

1. d₁ = 64 × 10⁻⁶ m, so d₁^(−1/2) = 1/(8 × 10⁻³) = 125 m^(−1/2). d₂ = 16 × 10⁻⁶ m, so d₂^(−1/2) = 250 m^(−1/2).
2. Subtract the two Hall–Petch equations: 250 − 150 = k·(250 − 125), so k = 100/125 = 0.8 MPa·m^0.5.
3. σ₀ = 150 − 0.8 × 125 = 50 MPa.
4. At d = 4 × 10⁻⁶ m: d^(−1/2) = 500 m^(−1/2).
5. σ_y = 50 + 0.8 × 500 = 450 MPa. Refining the grain 16 times tripled the yield strength.

Example 3 (true stress). At maximum load a specimen has engineering stress 400 MPa and engineering strain 0.20. Then σ_T = 400 × 1.20 = 480 MPa and ε_T = ln 1.20 = 0.182.

Common mistakes

• Using a = 4r/√2 for BCC or a = 4r/√3 for FCC. Atoms touch along the body diagonal in BCC and the face diagonal in FCC.
• Counting 9 atoms in a BCC cell or 14 in FCC: corner atoms count 1/8 and face atoms 1/2.
• Mixing nm with cm in the density formula. Convert a to cm (1 nm = 10⁻⁷ cm) to get g/cm³ directly.
• Saying dislocations "make metals weaker". They make plastic flow possible; adding more dislocations (cold work) or obstacles to them makes the metal stronger.
• Assuming heat treatment changes Young's modulus. It changes yield strength, UTS and hardness, but E barely moves.
• Applying σ_T = σ(1 + ε) after necking, where deformation is no longer uniform.
• Saying grain refinement trades ductility for strength. It raises both strength and toughness.

For GATE ME

Expect one-mark conceptual questions on atoms per cell, coordination number, packing factor, which metals are FCC/BCC/HCP, classification of defects (point, line, surface) and the slip-system argument for ductility. Two-mark numericals ask for theoretical density from lattice parameter, the lattice parameter from atomic radius, Hall–Petch constants from two data points, or true stress and strain from tensile-test data. Practise unit conversion in the density formula until it is automatic, and be able to sketch and label an engineering stress–strain curve.

Quick check

1. How many atoms are there per unit cell in FCC, and what is its APF?
2. Copper is FCC with r = 0.128 nm and A = 63.5 g/mol. What is its theoretical density?
3. Is a dislocation a point, line or surface defect?
4. Which strengthening mechanism improves both strength and toughness?
5. Why are FCC metals ductile even at very low temperatures?

Answers: 1. 4 atoms, 0.74. 2. a = 2√2 × 0.128 = 0.362 nm, ρ ≈ 8.89 g/cm³. 3. Line defect. 4. Grain refinement. 5. They have 12 close-packed slip systems that operate at low stress at all temperatures, so they show no ductile-to-brittle transition.