⚗️ Materials Science · Crystal Physics
📅 Березень 2026⏱ 11 min🟡 Середній

Dislocations: How Crystals Deform Plastically

In the 1920s, theoretical calculations predicted that metals should require shear stresses ~1000× higher to deform than what experiments measured. The resolution — proposed independently by Taylor, Orowan, and Polanyi in 1934 — was the dislocation: a line defect that lets crystals deform by moving the boundary between slipped and unslipped regions, one atomic row at a time.

1. The Theoretical Strength Problem

The theoretical shear strength of a perfect crystal can be estimated by considering how much force it takes to slide one atomic plane over the next. Frenkel's (1926) calculation:

Frenkel theoretical shear strength: τ_th ≈ G / (2π) ≈ G/6 (varies by model) G = shear modulus (Fe: ~80 GPa) τ_th ≈ 80/6 ≈ 13 GPa for iron Experimental shear yield stress for pure iron: τ_exp ≈ 25–30 MPa (annealed polycrystal, room temperature) Discrepancy factor: ~500–1000× This ~3 orders of magnitude discrepancy demanded a mechanism that allows deformation at much lower stress. Resolution (1934): Dislocations — line imperfections in crystal lattice. A dislocation moves like an inchworm: you don't need to slide all atoms simultaneously — just move the boundary between slipped/unslipped, one atomic spacing at a time. Energy required ∝ line length, not area.

2. Edge and Screw Dislocations

Edge dislocation: An extra half-plane of atoms inserted into the lattice. Dislocation line: bottom edge of the extra half-plane. Burgers vector b: ⊥ perpendicular to dislocation line. Stress field: compressive above slip plane, tensile below. Mobile under resolved shear stress τ on slip plane. Movement: "glide" — parallel to Burgers vector in slip plane. Screw dislocation: Lattice planes form a helical ramp (like a spiral staircase). Burgers vector b: ∥ parallel to dislocation line. No extra half-plane — instead, lattice "screws" around the line. Can cross-slip (change slip plane) more easily than edge dislocations — important for plastic deformation at high temperatures. Mixed dislocation: Real dislocations are usually a combination. Decompose into edge + screw components at angle θ: b_edge = b·sin(θ) b_screw = b·cos(θ) Strain energy per unit length: E_L = G·b² / (4π) × ln(r_outer/r_core) → proportional to b²: dislocations prefer smallest Burgers vector → Frank's rule: a dislocation b→b₁+b₂ is energetically favoured if |b|²>|b₁|²+|b₂|²

3. Burgers Vector

The Burgers vector b is the fundamental descriptor of a dislocation. It is defined via the Burgers circuit — a closed loop around the dislocation in a perfect crystal compared with the same circuit in the real crystal:

Burgers circuit: 1. In perfect crystal: make a right-hand closed loop (MNOPQ→M) counting N steps each direction. 2. In real crystal around dislocation: same sequence of steps does NOT close → closure failure vector = b. Burgers vector magnitude in common metals (FCC): FCC: b = (a/2)⟨110⟩, |b| = a/√2 (e.g., Cu: b = 0.256 nm) BCC: b = (a/2)⟨111⟩, |b| = a√3/2 (e.g., Fe: b = 0.248 nm) HCP: b = (a/3)⟨11̄20⟩ (basal) or (a/3)⟨11̄23⟩ (pyramidal) Peierls-Nabarro stress: minimum stress to move a dislocation through lattice τ_PN ≈ (2G/(1-ν)) · exp(-2πw/b) w = dislocation core width ∝ d (interplanar spacing) Wide planes, close-packed: low Peierls stress → easy slip (FCC {111}) Narrow planes, less packed: high Peierls stress → harder (BCC, ceramics)

4. Slip Systems

Slip system = slip plane {hkl} + slip direction ⟨uvw⟩ Conditions for a slip system to be active: 1. Must be the most densely packed plane → widest interplanar spacing → lowest Peierls stress 2. Slip direction must have the shortest Burgers vector (smallest b) Number of independent slip systems: FCC: {111}⟨110⟩ → 4 planes × 3 directions = 12 systems (high ductility) BCC: {110}⟨111⟩ (principle) + {112} + {123} → up to 48 systems (many systems, but narrower planes → needs higher stress) HCP: {0001}⟨11̄20⟩ (basal) = 3 systems → low ductility at RT Additional pyramidal systems at elevated T → improved formability Schmid's Law (resolved shear stress): τ = σ · cos φ · cos λ σ = applied uniaxial stress φ = angle between load axis and slip plane normal λ = angle between load axis and slip direction m = cos φ · cos λ = Schmid factor (max = 0.5 for 45°/45°) Active slip: τ ≥ τ_CRSS (critical resolved shear stress, ~10 MPa for Al)

5. Dislocation Density and Work Hardening

Dislocation density ρ: ρ = total dislocation line length / unit volume (m⁻²) Typical values: Annealed pure metal: ρ ≈ 10¹⁰–10¹² m⁻² Heavily cold-worked: ρ ≈ 10¹⁴–10¹⁶ m⁻² Grain boundary region: ρ ≈ 10¹⁵ m⁻² Taylor hardening (obstacle hardening): τ_c = τ₀ + α·G·b·√ρ τ_c = critical resolved shear stress (flow stress) α = Taylor factor ≈ 0.2–0.5 G = shear modulus b = Burgers vector magnitude √ρ = square root of dislocation density Work hardening: as material deforms, ρ increases → τ_c increases → material becomes harder and stronger (work hardening, strain hardening) Conversion to macroscopic yield stress: σ_y = M · τ_c (M = Taylor factor for polycrystal ≈ 3.06 for FCC)
Cold working and annealing: Cold rolling steel sheet increases ρ from ~10¹² to 10¹⁶ m⁻², doubling or tripling its strength — but severely reducing ductility. Annealing at ~600°C allows recovery (dislocations rearrange into lower-energy configurations) and recrystallisation (new, defect-free grains nucleate and grow). The final microstructure shows fine equiaxed grains: high strength from grain boundary hardening, good ductility restored. This cycle — cold work + anneal — is the basis of sheet metal manufacturing.

6. Frank-Read Sources and Dislocation Multiplication

A key question: how do crystals develop dislocation densities of 10¹⁶ m⁻² during deformation when they start with only 10¹⁰ m⁻²? The answer is dislocation multiplication via Frank-Read sources:

Frank-Read source mechanism: 1. A dislocation segment is pinned at two points (by precipitates, junctions, or grain boundaries) separated by distance L. 2. Applied shear stress τ bows the segment outward. 3. Segment continues to bow... sweeps around the pin points... 4. The two ends meet behind the source and annihilate. 5. A complete dislocation loop is emitted + source reforms. 6. Process repeats → one source emits thousands of loops/second. Critical stress to operate a Frank-Read source: τ_FR = α·G·b / L Example: L = 1 μm, G = 80 GPa, b = 0.25 nm τ_FR = 0.5 × 80×10⁹ × 0.25×10⁻⁹ / 10⁻⁶ ≈ 10 MPa ✓ (reasonable) This mechanism explains: • Why plastics strain occurs at near-constant stress in Stage I (easy glide) • Why work hardening accelerates in Stage II (loops pile up, mutual blocking)

7. Engineering Controlled Dislocations