Interactive Lecture
Understanding Crystal Defects and Material Strength
by Dr Arnaldo Delli Carri
Dislocations are line defects in crystalline materials where atoms are misaligned from their perfect lattice positions. They are the primary mechanism by which metals and alloys undergo plastic (permanent) deformation. Dislocation behaviour governs material strength, ductility, and formability.
Real metals deform plastically at stresses far below the theoretical shear strength of a perfect crystal, and dislocations are the reason:
The theoretical shear strength of perfect crystals predicts that materials should be ~1000× stronger than observed. In 1934, the dislocation theory was proposed to explain this discrepancy - dislocations allow plastic deformation to occur at much lower stresses than perfect crystal shear.
An extra half-plane of atoms inserted into the crystal structure. The dislocation line is perpendicular to the Burgers vector.
Atoms are displaced in a helical pattern around the dislocation line. The dislocation line is parallel to the Burgers vector.
Most real dislocations have both edge and screw components. A dislocation loop typically has edge character at some points and screw character at others.
The Burgers vector (b) defines the magnitude and direction of lattice distortion caused by a dislocation.
It represents:
For a perfect dislocation: \(\vec{b} = \frac{a}{2}\langle 110 \rangle\) (FCC)
For a perfect dislocation: \(\vec{b} = \frac{a}{2}\langle 111 \rangle\) (BCC)
Where \(a\) is the lattice parameter.
Plastic deformation occurs by dislocation glide on specific crystallographic planes and directions:
FCC metals have 12 slip systems: {111}⟨110⟩
BCC metals have 48 slip systems: {110}, {112}, {123}⟨111⟩
Dislocations begin to move when the resolved shear stress on the slip system reaches the CRSS:
\(\tau_R = \sigma \cos\phi \cos\lambda\)
Where:
Schmid's Law: Slip begins when \(\tau_R = \tau_{CRSS}\)
\(\sigma_y = \frac{\tau_{CRSS}}{m}\)
Strengthening materials involves impeding dislocation motion:
Where:
Watch how a dislocation glides through a crystal lattice under resolved shear stress. Adjust the Schmid factor angles and stress to explore when slip occurs.
Compare different strengthening mechanisms and see how they affect the stress-strain response. Use Snapshot to compare configurations.
The Hall-Petch relationship shows that reducing grain size increases strength without sacrificing ductility — a rare win-win in materials design. Thermomechanical processing to refine grain size is a key manufacturing strategy for exactly this reason.
| Concept | Equation | Description |
|---|---|---|
| Schmid's Law | \(\tau_R = \sigma \cos\phi \cos\lambda\) | Resolved shear stress on slip system |
| Yield Criterion | \(\sigma_y = \frac{\tau_{CRSS}}{m}\) | Yield stress from critical resolved shear stress |
| Hall-Petch | \(\sigma_y = \sigma_0 + \frac{k}{\sqrt{d}}\) | Grain size strengthening |
| Work Hardening | \(\sigma = K\epsilon^n\) | Strain hardening relationship |
| Dislocation Density | \(\tau \propto \sqrt{\rho}\) | Strength from dislocation interactions |
A single crystal of aluminium (FCC) is loaded in tension. The slip plane normal makes an angle of 60° with the tensile axis, and the slip direction makes an angle of 35° with the tensile axis. The critical resolved shear stress is 0.5 MPa. Calculate:
Show Solution
Given: \(\phi = 60°\), \(\lambda = 35°\), \(\tau_{CRSS} = 0.5\) MPa
a) Schmid factor:
\(m = \cos\phi \cos\lambda\)
\(m = \cos(60°) \cos(35°)\)
\(m = 0.5 \times 0.819 = 0.410\)
b) Required tensile stress:
\(\sigma_y = \displaystyle \frac{\tau_{CRSS}}{m} = \frac{0.5}{0.410}\)
\(\sigma_y = 1.22\) MPa
The single crystal will begin to deform plastically at a tensile stress of 1.22 MPa.
A steel has a yield strength of 250 MPa with an average grain diameter of 100 μm. After grain refinement processing, the grain size is reduced to 25 μm. The Hall-Petch parameters are \(\sigma_0 = 50\) MPa and \(k = 2.0\,\text{MPa}\cdot\text{m}^{1/2}\). Calculate:
Show Solution
Given: \(d_1 = 100\) μm = \(100 \times 10^{-6}\) m, \(d_2 = 25\) μm = \(25 \times 10^{-6}\) m
\(\sigma_0 = 50\) MPa, \(k = 2.0\,\text{MPa}\cdot\text{m}^{1/2}\)
a) Initial yield strength:
\(\sigma_{y1} = \sigma_0 + \displaystyle \frac{k}{\sqrt{d_1}}\)
\(\sigma_{y1} = 50 + \displaystyle \frac{2.0}{\sqrt{100 \times 10^{-6}}}\)
\(\sigma_{y1} = 50 + \displaystyle \frac{2.0}{0.01} = 50 + 200 = 250\) MPa ✓
b) Yield strength after refinement:
\(\sigma_{y2} = 50 + \displaystyle \frac{2.0}{\sqrt{25 \times 10^{-6}}}\)
\(\sigma_{y2} = 50 + \displaystyle \frac{2.0}{0.005} = 50 + 400 = 450\) MPa
c) Percentage increase:
\(\displaystyle \frac{\sigma_{y2} - \sigma_{y1}}{\sigma_{y1}} \times 100\% = \frac{450 - 250}{250} \times 100\% = 80\%\)
Reducing grain size by 4× increases yield strength by 80%.
A metal follows the work hardening relationship σ = 400ε^0.3 MPa. Calculate:
Show Solution
Given: \(\sigma = 400\epsilon^{0.3}\) MPa
a) Stress at ε = 0.1:
\(\sigma_1 = 400 \times (0.1)^{0.3}\)
\(\sigma_1 = 400 \times 0.50119 = 200.48\) MPa
b) Stress at ε = 0.3:
\(\sigma_2 = 400 \times (0.3)^{0.3}\)
\(\sigma_2 = 400 \times 0.69685 = 278.74\) MPa
c) Strength increase:
\(\Delta\sigma = \sigma_2 - \sigma_1 = 278.74 - 200.48 = 78.26\) MPa
Percentage increase: \(\displaystyle \frac{78.26}{200.48} \times 100\% = 39.04\%\)
Increasing strain from 0.1 to 0.3 increases the flow stress by 39% due to work hardening.