Dislocation types, motion, and their role in material behavior
Dislocations are line defects in crystalline materials that play a central role in mechanical behavior. Understanding dislocations is essential because they are the primary carriers of plastic deformation in metals and many other crystalline solids. Without dislocations, metals would be either impossibly strong or completely brittle.
In the early 20th century, physicists calculated the theoretical shear strength of a perfect crystal by considering how much stress would be needed to simultaneously break all the bonds across a slip plane and slide one half of the crystal over the other. The result was startling:
Where: $G$ = shear modulus of the material
For a typical metal like copper with $G \approx 45$ GPa, this predicts a theoretical strength of about 4.5 GPa. However, the actual yield strength of annealed copper is only about 10 MPa, roughly 1000 times weaker! This enormous discrepancy between theory and experiment was one of the great puzzles of materials science.
In 1934, Taylor, Orowan, and Polanyi independently proposed that plastic deformation occurs not by simultaneous slip of entire crystal planes, but by the motion of line defects called dislocations. Instead of breaking all bonds at once, a dislocation allows bonds to be broken and reformed sequentially along a line, much like moving a heavy carpet by pushing a wrinkle across it rather than dragging the entire carpet at once.
The actual shear strength of real crystals is therefore:
This dramatic reduction occurs because dislocation motion requires breaking only a small fraction of the bonds at any instant.
Dislocations are fundamental to understanding:
Dislocations can be classified based on the relationship between the dislocation line direction and the slip direction (Burgers vector). The two pure types are edge and screw dislocations, though most real dislocations are mixed.
An edge dislocation can be visualized as an extra half-plane of atoms inserted into the crystal. The dislocation line runs along the edge of this extra half-plane, perpendicular to the direction of slip.
Key characteristics of edge dislocations:
Edge dislocations are often denoted by the symbol $\perp$ (pointing up if the extra half-plane is above the slip plane) or $\top$ (if below).
A screw dislocation is more difficult to visualize. Imagine cutting partway into a crystal, then displacing one surface relative to the other parallel to the cut edge. The atomic planes spiral around the dislocation line like a parking garage ramp.
Key characteristics of screw dislocations:
The ability of screw dislocations to glide on multiple planes is called cross-slip, an important mechanism for dislocation motion around obstacles.
Most real dislocations are neither pure edge nor pure screw but have components of both. A curved dislocation line in a crystal will typically have edge character at some points and screw character at others, with mixed character in between.
For a mixed dislocation, if $\theta$ is the angle between the Burgers vector and the dislocation line:
The edge component is $b\sin\theta$ and the screw component is $b\cos\theta$, where $b$ is the magnitude of the Burgers vector.
Important properties of dislocation lines:
The Burgers vector $\vec{b}$ is the most important characteristic of a dislocation. It defines the magnitude and direction of the lattice distortion caused by the dislocation and determines the slip direction when the dislocation moves.
The Burgers vector is determined using a Burgers circuit:
The Burgers vector represents the "closure failure" of the circuit and physically corresponds to the displacement that occurs as a dislocation passes through a point.
Dislocations with the smallest Burgers vectors are energetically favorable because dislocation energy scales as $b^2$. In close-packed structures, the Burgers vector typically connects nearest-neighbor atoms:
FCC metals: $\vec{b} = \frac{a}{2}\langle110\rangle$, so $|\vec{b}| = \frac{a}{\sqrt{2}}$
BCC metals: $\vec{b} = \frac{a}{2}\langle111\rangle$, so $|\vec{b}| = \frac{a\sqrt{3}}{2}$
HCP metals: $\vec{b} = \frac{a}{3}\langle11\bar{2}0\rangle$, so $|\vec{b}| = a$
Where: $a$ = lattice parameter
The Burgers vector determines:
Problem: Copper (FCC) has a lattice parameter $a = 0.3615$ nm. Calculate the magnitude of the Burgers vector for a perfect dislocation.
Solution:
For FCC metals, the Burgers vector is $\frac{a}{2}\langle110\rangle$.
The magnitude is:
$$|\vec{b}| = \frac{a}{2}\sqrt{1^2 + 1^2 + 0^2} = \frac{a}{2}\sqrt{2} = \frac{a}{\sqrt{2}}$$
$$|\vec{b}| = \frac{0.3615}{\sqrt{2}} = 0.256 \text{ nm}$$
This is equal to the nearest-neighbor distance in FCC, confirming that slip occurs along close-packed directions.
Plastic deformation occurs through dislocation motion. Understanding how dislocations move is key to understanding and controlling mechanical properties. There are two fundamental modes of dislocation motion: glide (conservative) and climb (non-conservative).
Glide is the motion of a dislocation along its slip plane under the influence of shear stress. During glide, no atoms are created or destroyed; they simply rearrange along the slip plane. This is called conservative motion.
The force per unit length on a dislocation due to an applied shear stress is given by the Peach-Koehler equation. For the simple case of a shear stress $\tau$ acting on the slip plane in the slip direction:
Where: $F$ = force per unit length on dislocation, $\tau$ = resolved shear stress, $b$ = Burgers vector magnitude
The more general Peach-Koehler formula for arbitrary stress states is:
Where: $\vec{\sigma}$ = stress tensor, $\vec{b}$ = Burgers vector, $\vec{\xi}$ = unit vector along dislocation line
An edge dislocation can only glide on the unique plane containing both the dislocation line and the Burgers vector. A screw dislocation, however, can glide on any plane containing its line (since $\vec{b}$ is parallel to the line, every plane containing the line also contains $\vec{b}$).
This difference allows screw dislocations to cross-slip: they can move from one slip plane to another, enabling them to bypass obstacles.
Climb is the motion of an edge dislocation perpendicular to its slip plane. This requires adding or removing atoms from the extra half-plane, which means climb requires diffusion of vacancies or interstitials. Climb is therefore:
Climb allows dislocations to bypass obstacles that would block glide, making it important for high-temperature creep deformation.
Even in a perfect crystal, dislocations experience resistance to motion because they must overcome energy barriers between equilibrium positions. The stress required to move a dislocation through the lattice is called the Peierls stress or Peierls-Nabarro stress:
Where: $w$ = dislocation core width, $\nu$ = Poisson's ratio, $b$ = Burgers vector
The Peierls stress is low for close-packed metals (where $w$ is large relative to $b$) and high for covalent materials like silicon (where $w$ is small). This explains why metals are generally ductile while covalent ceramics are brittle at room temperature.
Plastic deformation by dislocation glide occurs on specific crystallographic planes in specific directions. A slip system consists of a slip plane and a slip direction. The number and geometry of available slip systems profoundly affects the ductility of crystalline materials.
Slip occurs most easily on close-packed planes (highest atomic density, largest interplanar spacing) in close-packed directions (shortest Burgers vector, lowest dislocation energy).
| Structure | Slip Plane | Slip Direction | Number of Systems |
|---|---|---|---|
| FCC | {111} | $\langle110\rangle$ | 12 (4 planes × 3 directions) |
| BCC | {110}, {112}, {123} | $\langle111\rangle$ | 48 (many planes, same direction) |
| HCP | {0001} (basal) | $\langle11\bar{2}0\rangle$ | 3 (1 plane × 3 directions) |
The von Mises criterion states that for a polycrystalline material to undergo arbitrary plastic deformation, each grain must have at least 5 independent slip systems. This ensures that neighboring grains can accommodate each other's shape changes.
Slip begins when the shear stress resolved onto a slip system reaches a critical value called the critical resolved shear stress (CRSS), denoted $\tau_{CRSS}$ or $\tau_0$.
The resolved shear stress on a slip system is:
Where: $\sigma$ = applied tensile stress, $\phi$ = angle between stress axis and slip plane normal, $\lambda$ = angle between stress axis and slip direction
The term $\cos\phi \cos\lambda$ is called the Schmid factor, $m$. Slip occurs first on the slip system with the highest Schmid factor when:
This is known as Schmid's Law. The maximum possible Schmid factor is 0.5 (when $\phi = \lambda = 45°$).
Problem: A single crystal of aluminum (FCC) is loaded in tension along the [100] direction. The CRSS for slip on {111}$\langle110\rangle$ systems is 1.0 MPa. Calculate the tensile yield stress.
Solution:
We need to find the slip system with the maximum Schmid factor. For loading along [100], consider the (111) plane with slip direction $[1\bar{1}0]$:
Angle between [100] and (111) normal [111]:
$$\cos\phi = \frac{[100] \cdot [111]}{|[100]||[111]|} = \frac{1}{\sqrt{3}}$$
Angle between [100] and $[1\bar{1}0]$:
$$\cos\lambda = \frac{[100] \cdot [1\bar{1}0]}{|[100]||[1\bar{1}0]|} = \frac{1}{\sqrt{2}}$$
Schmid factor:
$$m = \cos\phi \cos\lambda = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{6}} = 0.408$$
Yield stress:
$$\sigma_y = \frac{\tau_{CRSS}}{m} = \frac{1.0 \text{ MPa}}{0.408} = 2.45 \text{ MPa}$$
Real materials contain many dislocations that interact with each other and with other features of the microstructure. These interactions are responsible for work hardening, the increase in strength that occurs during plastic deformation.
Dislocation density $\rho$ is defined as the total length of dislocation line per unit volume (units: m/m³ = m⁻² or equivalently lines per m²).
| Material Condition | Typical $\rho$ (m⁻²) |
|---|---|
| Annealed metal | $10^{10}$ to $10^{12}$ |
| Heavily cold-worked | $10^{15}$ to $10^{16}$ |
| Carefully grown single crystal | $10^{6}$ to $10^{8}$ |
The macroscopic plastic strain rate is related to dislocation motion by the Orowan equation:
Where: $\dot{\gamma}$ = shear strain rate, $\rho_m$ = mobile dislocation density, $b$ = Burgers vector magnitude, $v$ = average dislocation velocity
This equation connects microscopic dislocation behavior to macroscopic deformation. It shows that plastic strain rate depends on how many dislocations are moving and how fast they move.
During plastic deformation, dislocation density increases dramatically. The primary mechanism for dislocation multiplication is the Frank-Read source: a dislocation segment pinned at both ends.
Under stress, the pinned segment bows out. When the stress reaches a critical value, the segment bows into a complete loop, regenerating the original segment and creating a new dislocation loop. This process can repeat, generating many dislocations from a single source.
The stress required to operate a Frank-Read source is:
Where: $G$ = shear modulus, $b$ = Burgers vector, $L$ = length of the pinned segment
Shorter segments require higher stresses to operate, which is one reason why fine microstructures are stronger.
As dislocation density increases, dislocations increasingly interact with each other:
The flow stress due to dislocation interactions follows approximately:
Where: $\tau_0$ = friction stress, $\alpha$ = constant (~0.3-0.6), $\rho$ = dislocation density
This square-root dependence explains why the work hardening rate decreases as strain increases.
Dislocations have energy associated with the elastic strain field around them. The energy per unit length is approximately:
This is per unit length. The factor of 1/2 is approximate; exact values depend on dislocation character and core energy.
This line energy means dislocations act like elastic strings under tension. They tend to be as short as possible (straight lines) and will respond to forces that reduce their total energy.
Problem: An aluminum sample ($G = 26$ GPa, $b = 0.286$ nm) has an initial dislocation density of $10^{12}$ m⁻². After cold rolling, the density increases to $10^{15}$ m⁻². Using $\alpha = 0.4$, estimate the increase in flow stress.
Solution:
Using $\Delta\tau = \alpha Gb(\sqrt{\rho_f} - \sqrt{\rho_i})$:
$$\sqrt{\rho_i} = \sqrt{10^{12}} = 10^6 \text{ m}^{-1}$$
$$\sqrt{\rho_f} = \sqrt{10^{15}} = 3.16 \times 10^7 \text{ m}^{-1}$$
$$\Delta\tau = 0.4 \times (26 \times 10^9) \times (0.286 \times 10^{-9}) \times (3.16 \times 10^7 - 10^6)$$
$$\Delta\tau = 0.4 \times 26 \times 0.286 \times 30.6 = 91 \text{ MPa}$$
This substantial increase in flow stress demonstrates why cold working is an effective strengthening mechanism.
Since plastic deformation requires dislocation motion, anything that impedes dislocation motion increases strength. This principle underlies all major strengthening mechanisms in metals. Each mechanism creates obstacles that dislocations must overcome, requiring higher applied stresses.
Dislocations can be impeded by:
Solute atoms create local stress fields that interact with moving dislocations. Atoms larger than the matrix create compressive regions; smaller atoms create tensile regions. Dislocations are attracted to or repelled from these regions, requiring extra stress to break free.
The strengthening increment is approximately:
Where: $c$ = solute concentration. The square-root dependence arises from statistical averaging of randomly distributed solute atoms.
Second-phase particles can block dislocation motion by two mechanisms:
Particle cutting: If particles are small and coherent, dislocations can cut through them. The strength increases with particle size because larger particles require more energy to cut.
Orowan bowing (bypass): If particles are large and/or incoherent, dislocations cannot cut through and must bow around them. The stress required for Orowan bypass is:
Where: $\lambda$ = spacing between particles (mean free path)
For cutting, strength increases with particle size. For Orowan bypass, strength decreases with increasing particle size (at constant volume fraction) because spacing increases. Maximum strength occurs at the transition between cutting and bowing, which corresponds to the peak-aged condition in precipitation hardening.
Grain boundaries are strong barriers to dislocation motion because slip directions change across the boundary. Dislocations pile up at grain boundaries, creating stress concentrations that eventually activate slip in the neighboring grain.
The Hall-Petch equation describes the relationship between yield strength and grain size:
Where: $\sigma_y$ = yield strength, $\sigma_0$ = friction stress (lattice resistance), $k_y$ = Hall-Petch coefficient (material-dependent), $d$ = average grain diameter
Smaller grains mean more boundaries, shorter slip distances, and higher strength. This is why grain refinement is such an effective strengthening method.
Multiple strengthening mechanisms can operate simultaneously. Their contributions often add according to:
Where subscripts indicate: SS = solid solution, ppt = precipitation, HP = Hall-Petch, WH = work hardening
In practice, the superposition is sometimes linear (as shown) and sometimes root-sum-square, depending on whether the mechanisms operate on the same or different length scales.
Problem: A low-carbon steel has $\sigma_0 = 70$ MPa and $k_y = 0.74$ MPa·m$^{1/2}$. Calculate the yield strength for grain sizes of (a) 100 µm and (b) 10 µm.
Solution:
(a) For $d = 100$ µm $= 100 \times 10^{-6}$ m:
$$\sigma_y = 70 + 0.74 \times (100 \times 10^{-6})^{-1/2}$$
$$\sigma_y = 70 + 0.74 \times 100 = 144 \text{ MPa}$$
(b) For $d = 10$ µm $= 10 \times 10^{-6}$ m:
$$\sigma_y = 70 + 0.74 \times (10 \times 10^{-6})^{-1/2}$$
$$\sigma_y = 70 + 0.74 \times 316 = 304 \text{ MPa}$$
Reducing grain size by a factor of 10 more than doubled the yield strength, demonstrating the powerful effect of grain refinement.
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