Crystal structure, defects and mechanical properties of metals

Why it matters

Every steel connecting rod, aluminium piston and magnesium gearbox housing behaves the way it does because of how its atoms are stacked and how that stacking is imperfect. Crystal structure fixes density, stiffness and how easily a metal can be formed; defects – above all dislocations and grain boundaries – decide yield strength, ductility and how heat treatment and cold work change them. Reading a tensile-test curve correctly is the first step in every design and every manufacturing calculation that follows.

Key ideas

Crystal structures. Metals solidify as crystals: a unit cell repeated in three dimensions. Three structures cover almost all engineering metals.

• BCC (body-centred cubic) – atoms at the 8 corners plus one at the body centre: 2 atoms per cell, atoms touch along the body diagonal so a = 4r/√3, coordination number 8, atomic packing factor (APF) 0.68. Examples: α-iron (ferrite, below 912 °C), Cr, Mo, W, V.
• FCC (face-centred cubic) – corners plus the 6 face centres: 4 atoms per cell, atoms touch along the face diagonal so a = 2√2·r, coordination number 12, APF 0.74 (close-packed). Examples: γ-iron (austenite, 912–1394 °C), Al, Cu, Ni, Ag, Au.
• HCP (hexagonal close-packed) – 6 atoms per hexagonal cell, coordination number 12, APF 0.74, ideal c/a = 1.633. Examples: Mg, Zn, Ti (α, room temperature), Co. Iron is allotropic (polymorphic): BCC → FCC at 912 °C, FCC → BCC (δ-iron) at 1394 °C. This change is what makes steels heat-treatable.

Slip and ductility. Plastic deformation happens by dislocations gliding on close-packed planes along close-packed directions (a slip system). FCC has 12 slip systems on truly close-packed {111} planes, so it stays ductile even at cryogenic temperature. BCC has many potential systems (up to 48) but no truly close-packed plane, so dislocation glide needs a higher, strongly temperature-dependent stress – BCC steels show a ductile-to-brittle transition. HCP has only 3 easy basal systems at room temperature, so Mg and Zn are hard to cold-form.

Defects.

• Point defects (0-D): vacancies, self-interstitials, substitutional and interstitial solute atoms. Carbon in iron is interstitial; Ni in Cu is substitutional. Solute atoms distort the lattice and impede dislocations (solid-solution strengthening). Vacancies enable diffusion.
• Line defects (1-D): edge, screw and mixed dislocations, described by the Burgers vector b. Because slip happens one dislocation step at a time, real metals yield at roughly 1/100 to 1/1000 of the theoretical shear strength of a perfect crystal.
• Planar defects (2-D): grain boundaries, twin boundaries, stacking faults, free surfaces. Grain boundaries block dislocations, so finer grains mean higher yield strength (Hall–Petch).
• Volume defects (3-D): voids, porosity, inclusions – usually from casting or powder processing.

Strengthening mechanisms all work by obstructing dislocation motion: grain refinement, solid solution, cold work (strain hardening – dislocations tangle with each other), and precipitation/dispersion hardening. Grain refinement is the only one that raises strength and toughness together.

Mechanical properties from the tensile test. Engineering stress σ = F/A₀, engineering strain ε = ΔL/L₀. The initial straight line gives Young's modulus E (about 200 GPa for steel, 70 GPa for aluminium – a structure-insensitive property set by bonding). Yield strength (or 0.2 % proof stress where there is no sharp yield point), ultimate tensile strength (maximum engineering stress, where necking begins), percentage elongation and reduction in area (ductility), modulus of resilience (elastic energy per volume) and toughness (area under the whole curve) are structure-sensitive. Hardness (Brinell, Vickers, Rockwell) is resistance to plastic indentation and correlates roughly with UTS. Fatigue strength and creep resistance describe behaviour under cyclic load and at high temperature.

Formulas

ρ = n·A / (V_c·N_A) ρ = theoretical density (g/cm³ or kg/m³), n = atoms per unit cell (BCC 2, FCC 4, HCP 6), A = atomic mass (g/mol), V_c = unit-cell volume (cm³), N_A = 6.022 × 10²³ mol⁻¹.

a_BCC = 4r / √3 a_FCC = 2√2·r a = lattice parameter (m), r = atomic radius (m). Hard-sphere model.

APF = n·(4/3)·π·r³ / V_c Dimensionless. BCC 0.68, FCC and HCP 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^½), d = average grain diameter (m). Valid for grain sizes down to roughly tens of nanometres.

σ = F / A₀ ε = ΔL / L₀ E = σ / ε (elastic region only) F in N, A₀ original area in m², σ and E in Pa, ε dimensionless.

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).

BHN = 2P / [π·D·(D − √(D² − d²))] P = load (kgf), D = ball diameter (mm), d = indentation diameter (mm); BHN in kgf/mm².

Worked examples

Example 1 (standard) – density of α-iron. Given: BCC iron, r = 0.124 nm, A = 55.85 g/mol, N_A = 6.022 × 10²³ mol⁻¹.

1. Lattice parameter: a = 4r/√3 = 4 × 0.124/1.732 = 0.28637 nm = 2.8637 × 10⁻⁸ cm.
2. Cell volume: V_c = a³ = 2.3484 × 10⁻²³ cm³.
3. Density: ρ = n·A/(V_c·N_A) = (2 × 55.85)/(2.3484 × 10⁻²³ × 6.022 × 10²³) = 111.70/14.142.
4. ρ = 7.90 g/cm³ (7900 kg/m³), close to the measured 7.87 g/cm³.

Example 2 (GATE level) – Hall–Petch from two data points. Given: a low-carbon steel has σ_y = 200 MPa at d = 64 µm and σ_y = 300 MPa at d = 16 µm. Find the yield strength at d = 4 µm.

1. Compute d^(−1/2): for 64 × 10⁻⁶ m, d^(−1/2) = 1/0.008 = 125 m^(−½); for 16 × 10⁻⁶ m, 1/0.004 = 250 m^(−½); for 4 × 10⁻⁶ m, 1/0.002 = 500 m^(−½).
2. Slope: k = Δσ_y / Δ(d^(−1/2)) = (300 − 200)/(250 − 125) = 0.8 MPa·m^½.
3. Intercept: σ₀ = 200 − 0.8 × 125 = 100 MPa.
4. At d = 4 µm: σ_y = 100 + 0.8 × 500 = 500 MPa. Note that halving d does not double strength; it multiplies the grain-boundary term by √2.

Example 3 – true stress at UTS. A specimen reaches its maximum load at an engineering stress of 450 MPa and engineering strain of 0.20. True stress = 450 × 1.20 = 540 MPa; true strain = ln 1.20 = 0.182.

Common mistakes

• Using a = 2r for BCC or FCC. Atoms touch along the body diagonal (BCC) or face diagonal (FCC), not along the cube edge (that is simple cubic only).
• Counting atoms per cell wrongly: a corner atom is shared by 8 cells, a face atom by 2.
• Forgetting to convert nm to cm when density is wanted in g/cm³ (1 nm = 10⁻⁷ cm).
• Saying "FCC is ductile because it has more slip systems than BCC". BCC has as many or more; FCC wins because its slip planes are truly close-packed and the needed stress is low and insensitive to temperature.
• Treating E as structure-sensitive. Heat treatment and cold work change yield strength and hardness, not Young's modulus.
• Using engineering stress after necking or using σ_T = σ(1 + ε) beyond the UTS point.
• In Hall–Petch, using d instead of d^(−1/2), or mixing µm and m between the two data points.

For GATE ME

Expect short conceptual questions on atoms per cell, coordination number, APF and which metal has which structure; classification of defects; and which mechanisms strengthen a metal. Numericals are usually density from lattice data, Hall–Petch with two data points, engineering versus true stress/strain, resilience or toughness from a stress–strain curve, and Brinell hardness from indentation size. Practise moving between r and a for each lattice and keeping units consistent.

Quick check

1. How many atoms belong to one FCC unit cell, and what is its APF?
2. Is a grain boundary a point, line or planar defect?
3. A tensile specimen shows engineering strain 0.1 at 300 MPa. What is the true stress?
4. Which property is least affected by cold working: yield strength, hardness or Young's modulus?
5. In Hall–Petch, what happens to the grain-boundary term when grain size is reduced to one quarter?

Answers: 1. 4 atoms, 0.74; 2. Planar (surface) defect; 3. 330 MPa; 4. Young's modulus; 5. It doubles.