Crystal defects and plastic deformation

Why it matters

A perfect metal crystal would need a shear stress of several gigapascals to yield, yet real copper yields at a few tens of megapascals. The difference is defects, above all dislocations. Every strengthening method an engineer uses, from cold rolling and alloying to grain refinement, works by making dislocations harder to move, so this topic is the bridge between atomic structure and the stress-strain curve.

Key ideas

Point defects (zero-dimensional).

• Vacancy: a missing atom. Vacancies exist in equilibrium; their fraction rises exponentially with temperature and they make substitutional diffusion possible.
• Self-interstitial: an extra host atom squeezed into an interstitial site; high energy, rare in metals.
• Impurity atoms: substitutional (similar size, e.g. Ni in Cu) or interstitial (small atoms such as C, N, H in iron). They distort the lattice and give solid-solution strengthening.
• In ionic solids, a Schottky defect is a cation–anion vacancy pair and a Frenkel defect is an ion displaced to an interstitial site; both keep the crystal neutral.

Line defects: dislocations (one-dimensional).

• Edge dislocation: the edge of an extra half-plane of atoms. The Burgers vector b is perpendicular to the dislocation line.
• Screw dislocation: atoms are displaced so the planes form a helical ramp around the line. b is parallel to the line.
• Mixed dislocation: any curved dislocation has edge and screw character along its length. The Burgers vector is the same everywhere on one dislocation.
• The Burgers vector is found by a closed circuit around the dislocation compared with the same circuit in a perfect crystal; its magnitude is usually the shortest lattice translation (a/2⟨110⟩ in FCC, a/2⟨111⟩ in BCC).
• Dislocation density ρ is total line length per unit volume (m/m³ = m⁻²): about 10¹⁰ m⁻² in annealed metals and 10¹⁴–10¹⁶ m⁻² in heavily cold-worked metals.

Planar defects (two-dimensional). External surfaces, grain boundaries (high-angle and low-angle), twin boundaries (mirror planes), stacking faults (an error in the ABC stacking of FCC) and phase boundaries.

Volume defects (three-dimensional). Pores, cracks, inclusions and second-phase particles.

Plastic deformation by slip. A dislocation moving through the crystal shifts the part above the slip plane by one Burgers vector, breaking only one row of bonds at a time. That is why the real yield stress is far below the theoretical shear strength (about G/2π to G/30). Slip occurs on a slip system — the densest plane and densest direction in it: FCC {111}⟨110⟩ (12 systems), BCC {110}⟨111⟩ (plus {211} and {321}), HCP basal (0001)⟨112̄0⟩ (3 systems).

Schmid's law. Slip starts in a single crystal when the shear stress resolved on a slip system reaches the critical resolved shear stress (CRSS), a material property. The Schmid factor cos φ·cos λ is at most 0.5 (φ = λ = 45°).

Twinning. A region of the crystal shears into a mirror image of the parent across a twin plane. It is important in HCP metals and in BCC metals at low temperature or high strain rate, where slip is difficult. Twinning gives small strains but reorients grains so slip can continue.

Strengthening mechanisms (all hinder dislocations).

• Strain (work) hardening: plastic deformation multiplies dislocations; they tangle and pin each other.
• Grain-size reduction: grain boundaries block slip; finer grains mean higher yield strength (Hall–Petch) and, unusually, better toughness too.
• Solid-solution strengthening: solute atoms' strain fields pin dislocations.
• Precipitation / dispersion hardening: fine particles must be cut or bypassed (Orowan looping).

Cold work is removed by annealing: recovery (dislocations rearrange, internal stresses fall), recrystallisation (new strain-free grains form, typically at 0.3–0.5 of the melting temperature in kelvin) and grain growth.

Formulas

τR = σ·cos φ·cos λ τR = resolved shear stress on the slip system (Pa); σ = applied uniaxial stress (Pa); φ = angle between the stress axis and the slip-plane normal; λ = angle between the stress axis and the slip direction. Valid for a single crystal under uniaxial load.

σy = τCRSS / (cos φ·cos λ)max Yield stress of a single crystal: slip begins on the system with the largest Schmid factor.

σy = σ0 + ky·d^(−1/2) Hall–Petch: σ0 = friction stress (Pa); ky = locking parameter (Pa·m^½), a material constant from data; d = mean grain diameter (m). Valid for grain sizes from about 1 µm upward.

Nv / N = exp(−Qv / (k·T)) Nv = vacancies per m³; N = atomic sites per m³; Qv = vacancy formation energy (eV); k = 8.617 × 10⁻⁵ eV/K; T = absolute temperature (K).

Δτ ≈ α·G·b·√ρ Taylor hardening: α ≈ 0.2–0.5 (empirical); G = shear modulus (Pa); b = Burgers vector magnitude (m); ρ = dislocation density (m⁻²).

%CW = (A0 − Ad) / A0 × 100 Percent cold work from original area A0 and deformed area Ad.

Worked examples

Example 1 (standard): resolved shear stress. Given: σ = 150 MPa, φ = 45°, λ = 30°.

1. τR = σ·cos φ·cos λ.
2. cos 45° = 0.7071, cos 30° = 0.8660, Schmid factor = 0.6124.
3. τR = 150 × 0.6124 = 91.9 MPa.

Example 2 (GATE level): single-crystal yield and Hall–Petch. (a) A single crystal has CRSS = 1.0 MPa. It is pulled along an axis that makes φ = 60° with the slip-plane normal and λ = 35° with the slip direction. Find the tensile stress at which slip starts.

1. Schmid factor = cos 60° × cos 35° = 0.5 × 0.8192 = 0.4096.
2. σy = 1.0 / 0.4096 = 2.44 MPa.

(b) A steel has σ0 = 70 MPa and ky = 0.74 MPa·m^½ (data). Find σy for d = 25 µm, and the grain size needed for σy = 300 MPa.

1. d^(−1/2) = 1/√(25 × 10⁻⁶ m) = 200 m^(−1/2).
2. σy = 70 + 0.74 × 200 = 218 MPa.
3. For 300 MPa: d^(−1/2) = (300 − 70)/0.74 = 310.8 m^(−1/2), so d = 1/310.8² = 1.035 × 10⁻⁵ m = 10.4 µm.

Example 3: vacancies in copper near its melting point. Given: Qv = 0.9 eV, T = 1357 K, N = 8.49 × 10²⁸ atoms/m³.

1. Qv/(kT) = 0.9 / (8.617 × 10⁻⁵ × 1357) = 7.70.
2. Nv/N = e^(−7.70) = 4.54 × 10⁻⁴.
3. Nv = 8.49 × 10²⁸ × 4.54 × 10⁻⁴ = 3.86 × 10²⁵ vacancies/m³.

Common mistakes

• Writing Schmid's law with a division or a factor of 2; it is τR = σ cos φ cos λ.
• Measuring φ from the slip plane itself instead of from its normal.
• Assuming the Schmid factor can exceed 0.5, or that φ + λ must equal 90° (only if the slip direction, normal and load axis are coplanar).
• Using T in °C in the Arrhenius vacancy expression.
• Forgetting to convert grain size to metres in Hall–Petch.
• Saying the Burgers vector of an edge dislocation is parallel to its line (that is the screw case).

For GATE PI

One-mark questions test defect classification (point, line, planar), the relation of b to the line for edge and screw dislocations, slip systems of FCC/BCC/HCP, and which mechanisms strengthen a metal. Two-mark numericals use Schmid's law, Hall–Petch, percent cold work and vacancy concentration. Practise angle bookkeeping in Schmid's law and unit conversions in Hall–Petch.

Quick check

1. What is the maximum possible Schmid factor?
2. For a screw dislocation, how is the Burgers vector oriented relative to the dislocation line?
3. What happens to yield strength when grain size is reduced by a factor of 4, if σ0 is negligible?
4. Name the slip system of FCC metals.
5. A bar is cold-drawn from 100 mm² to 70 mm² cross-section. What is the percent cold work?

Answers: 1. 0.5; 2. Parallel; 3. It doubles; 4. {111}⟨110⟩; 5. 30 %.