Questions: Planar Defects: Grain Boundaries and Interfaces

Question 1

According to the Hall-Petch relationship, if the average grain diameter is reduced from 100 μm to 25 μm (a factor of 4 reduction), how does the grain-boundary strengthening contribution k/√d change?

• A. It doubles, because √(1/25 μm) is twice √(1/100 μm)
• B. It quadruples, because grain boundary area per volume scales as 1/d
• C. It is halved, because smaller grains are softer due to higher boundary fraction
• D. It remains the same, since grain size only affects ductility, not yield strength

Answer: (A)

The Hall-Petch term is k/√d. If d decreases by a factor of 4, √d decreases by a factor of 2, so k/√d increases by a factor of 2. The strengthening doubles. This is why grain refinement is a powerful strengthening strategy — relatively modest reductions in grain size produce significant strength gains.

Question 2

A dislocation moving through Grain A reaches a high-angle grain boundary. Why can it not simply continue into Grain B?

• A. The slip system orientation in Grain B is different, so the dislocation cannot glide on the same plane without a change in Burgers vector or direction
• B. Grain boundaries are lower-density regions, so dislocations lose energy and stop at the boundary due to reduced atomic bonding
• C. The grain boundary absorbs the dislocation permanently by annihilating it with an opposite Burgers vector
• D. Dislocations can cross grain boundaries freely, but the high boundary energy slows them down

Answer: (A)

At a high-angle grain boundary, the crystal lattice orientation changes abruptly. The slip plane and slip direction that carry the dislocation in Grain A are not aligned with any favorable slip system in Grain B. Transmission requires generating a new dislocation with a different Burgers vector, which requires additional applied stress. This is the physical mechanism behind Hall-Petch strengthening.

Question 3

Reducing grain size in a metal usually improves most mechanical properties — strength, ductility, and toughness simultaneously.

• T. True
• F. False

Answer: False

Grain refinement increases yield strength via Hall-Petch strengthening, but it does not universally improve all properties. Finer grains can reduce ductility by limiting dislocation storage capacity and work-hardening, and can make materials more susceptible to grain boundary corrosion or embrittlement in certain environments. Engineering grain size involves trade-offs, not a simple 'finer is always better' rule.

Question 4

High-angle grain boundaries have higher energy than low-angle grain boundaries because the lattice mismatch is too large to be accommodated by a regular array of dislocations.

• T. True
• F. False

Answer: True

Low-angle grain boundaries can be modeled as ordered arrays of edge dislocations whose Burgers vectors account for the small misorientation — the boundary energy scales with misorientation angle. High-angle boundaries (>~15° misorientation) have too large a mismatch for this dislocation model; the interface becomes essentially amorphous over a few atomic spacings, with higher stored energy, higher diffusivity, and greater chemical reactivity than the crystal interior.

Question 5

Why does grain refinement increase yield strength? Explain using the concept of dislocation motion.

Model answer: Grain boundaries are obstacles to dislocation motion because slip system orientations change abruptly across the boundary. A dislocation moving through one grain cannot easily continue into the adjacent grain without additional stress to reorient or transmit. Finer grains mean more grain boundaries per unit volume, so dislocations encounter obstacles more frequently and pile up sooner, requiring higher applied stress to sustain plastic deformation.

The Hall-Petch relationship σ_y = σ_0 + k/√d quantifies this: smaller grain diameter d raises yield strength. The physical mechanism is dislocation pile-up at boundaries. When grain size is reduced, the mean free path for dislocation glide decreases, and each grain boundary must be overcome to continue deformation, directly raising the macroscopic yield stress.