Dislocations and Plastic Deformation
Connect crystal slip to line defects, resolved shear, and the dimensional meaning of the Orowan strain-rate relation.
On this page
The question for today
What you’ll understand
- Describe edge and screw dislocations using their line direction and Burgers vector.
- Calculate a slip-system shear strain rate using mobile dislocation density.
- Explain why more total dislocations can strengthen a crystal despite the Orowan rate relation.
A metal paperclip can keep a new shape after you remove the load. Part of its strain was elastic and recovered, but part was plastic and remains. In many crystals, plastic deformation occurs through the movement of dislocations rather than a whole atomic plane sliding at once. A moving defect lets the atomic rearrangement advance locally. Connecting that microscopic motion with a macroscopic strain rate is the purpose of this lesson; a general fracture or alloy-design model would need substantially more information.
A line defect carries a geometrical mismatch
A dislocation is a line defect characterized by a Burgers vector: the lattice-translation mismatch obtained from a circuit around the defect. The vector has units of length. An edge dislocation has Burgers vector perpendicular to its line direction; a screw dislocation has them parallel. A curved dislocation can have mixed character along its length. These descriptions classify lattice geometry, not a stream of extra atoms travelling with the observed macroscopic strain. Continuum elasticity describes the surrounding stress field but cannot resolve the atomic core accurately.
Glide moves a dislocation on a slip plane and produces a permanent relative translation after the defect has passed through the crystal. Climb requires transport of vacancies or atoms and introduces diffusion and temperature sensitivity. Many different microscopic mechanisms can therefore contribute to the same measured plastic strain. A permanent deformation is not automatically a crack, and recoverable elastic distortion still surrounds the moving defects. Separating the elastic and plastic contributions is essential when interpreting unloading: only the elastic part is removed immediately in an ideal quasistatic unload.
External loading must resolve onto an allowed slip system
A crystal does not respond to the magnitude of an applied tensile stress alone. For a chosen slip system, the relevant driving stress is the shear traction resolved along its slip direction. Under uniaxial stress, two directional projections give the Schmid factor. One angle is between the loading axis and the slip-plane normal; the other is between the loading axis and the slip direction. Those directions have a geometric relationship, so their cosines cannot be independently set to arbitrary values for one physical slip system.
Obstacles and lattice resistance oppose motion. Solute atoms, grain boundaries, precipitates, and other dislocations can change the stress needed for appreciable glide. This is why processing and microstructure affect strength even when the basic elastic stiffness changes little. In a polycrystal, neighbouring grains constrain one another, and a single resolved-shear calculation does not predict the whole specimen response. The simplified relation is useful for understanding orientation, provided the uniaxial loading and one-slip-system assumptions are stated rather than silently extended to every stress tensor.
Count the area swept by mobile line defects
A moving line sweeps out slip area. Add the swept areas from the mobile defects in a representative volume and multiply by the Burgers-vector magnitude to obtain plastic shear. Differentiating gives the Orowan relation for a single slip system. Mobile dislocation density is total mobile line length divided by volume, so its units are inverse square metres. Multiply that by Burgers-vector length and average glide speed: the resulting units are inverse seconds, exactly those of a dimensionless shear strain rate.
The relation is kinematic; it does not supply a velocity law or an evolution law for density. Increasing mobile density while holding speed fixed raises the rate, but increasing total stored density can hinder motion and cause work hardening. Total and mobile density are therefore not interchangeable. In the vectors lab, first predict the x projection of a fixed-magnitude vector at 0° and 90°, then test it. This illustrates one directional projection with dimensionless quantities. It does not calculate a complete Schmid factor, a dislocation stress field, defect mobility, or a plastic flow history.
Work it through
From nanometres of slip to a measurable rate
For one active slip system, take mobile dislocation density 10¹² m⁻², Burgers-vector magnitude 0.25 nm, and mean glide speed 4.0 × 10⁻⁶ m/s. Estimate its plastic shear strain after 10 s if those quantities remain constant.
Convert the Burgers-vector magnitude: 0.25 nm = 2.5 × 10⁻¹⁰ m. The mobile density is line length per volume, not a count per metre.
Multiply the Orowan factors: 10¹² × 2.5 × 10⁻¹⁰ × 4.0 × 10⁻⁶ = 1.0 × 10⁻³ s⁻¹.
Integrate the constant rate for 10 s: Δγp = 1.0 × 10⁻³ × 10 = 0.010. This is a dimensionless shear of one percent.
The stated slip-system shear increment is 0.010. This estimate assumes constant mobile density and speed; it does not predict their evolution or directly give the specimen’s axial strain.
The living notebook
Explore the model
Interactive lab · vectors
Inspect one resolved direction
Keep magnitude fixed and predict the x component at 0°, 60°, and 90°. Then use the angle control to check the cosine projection. The displayed vector is dimensionless and models one projection, not the full slip-system stress calculation.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forOrientation controls a resolved component; actual crystal slip also requires the plane normal, the slip direction, and a mobility or resistance law.
Pause & reason
Check your understanding
Your turn
Try it yourself
A slip system needs shear strain rate 0.002 s⁻¹ with b = 0.25 nm and mean speed 10⁻⁵ m/s. What mobile density would the kinematic relation require? Explain why this is not a complete material prediction.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Rearrange the Orowan relation for mobile density and convert nanometres before dividing.
Compare with the solution
The required density is 0.002/(2.5 × 10⁻¹⁰ × 10⁻⁵) = 8 × 10¹¹ m⁻². The calculation supplies a necessary product of mobile density and speed for the chosen slip system. It does not establish that the applied stress can sustain that speed, that enough defects remain mobile, or that neighbouring grains permit the assumed slip.
Keep the idea
Take it with you
- The Burgers vector has units of length and describes the lattice mismatch of a line defect.
- For one slip system, plastic shear rate equals mobile line density times Burgers-vector magnitude times mean glide speed.
- Work hardening concerns resistance and defect evolution; total stored density must not be substituted blindly for mobile density.
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