Questions: Mechanisms of Plastic Deformation and Slip

Question 1

The theoretical shear strength of a perfect copper crystal is roughly G/10 (where G is the shear modulus). In practice, pure copper yields at roughly G/10,000 — a factor of 1000 lower. What is the correct explanation for this enormous discrepancy?

• A. Copper always contains impurity atoms that reduce its strength far below the theoretical ideal
• B. Dislocations allow slip to propagate one atomic row at a time rather than simultaneously across an entire plane, requiring far less stress — analogous to moving a wrinkle across a carpet rather than lifting the whole carpet
• C. Thermal vibrations at room temperature provide sufficient energy to overcome the theoretical barrier
• D. The theoretical calculation assumes simple cubic geometry; the FCC structure of copper is inherently weaker by a factor of 1000

Answer: (B)

The wrinkle analogy is the key insight. Sliding one entire plane over another requires every bond across that plane to stretch and break simultaneously, demanding very high stress. Dislocation motion advances the slip by one Burgers vector at a time — only a small number of bonds are stressed at once. The dislocation sweeps across the entire slip plane incrementally, producing the same macroscopic strain at a tiny fraction of the theoretical stress. This explains why real metals are ductile rather than brittle, and why dislocation theory was such a transformative insight in materials science.

Question 2

FCC metals like aluminum are generally much more ductile than HCP metals like magnesium at room temperature. Which explanation is most consistent with slip system theory?

• A. FCC metals have weaker interatomic bonds, so dislocations require less energy to move
• B. FCC metals have 12 equivalent {111}⟨110⟩ slip systems, providing many orientations for dislocation motion; HCP metals have very few independent slip systems, severely restricting how the crystal can accommodate shape change
• C. FCC metals work-harden more slowly, so they retain ductility after initial deformation
• D. HCP planes are more closely packed than FCC planes, making dislocation motion on HCP slip planes inherently more difficult

Answer: (B)

The number of independent slip systems directly controls ductility. With 12 equivalent slip systems, FCC metals can accommodate deformation in almost any direction — there is nearly always a well-oriented slip system to activate. HCP metals have only 3 easily activated basal slip systems at room temperature, all nearly parallel, so they cannot accommodate arbitrary shape changes without cracking. This is why magnesium alloys require elevated temperature (which activates additional prismatic and pyramidal slip systems) for significant ductility.

Question 3

Work hardening increases a metal's yield strength because plastic deformation increases dislocation density, and a higher dislocation density makes it harder for further dislocations to move.

• T. True
• F. False

Answer: True

True. As dislocations multiply during plastic deformation, they intersect and tangle with each other, forming jogs and sessile segments that act as pinning points. The stress required to push additional dislocations through this 'forest' of obstacles increases with dislocation density. Macroscopically, the material requires greater applied stress to continue deforming — it has work-hardened. The trade-off is that suppressing dislocation motion also reduces the capacity for further deformation, decreasing ductility.

Question 4

Plastic deformation in metals occurs when the applied stress is large enough to simultaneously rupture most atomic bonds across a slip plane.

• T. True
• F. False

Answer: False

False — this describes the theoretical strength, which is orders of magnitude higher than actual yield stresses. Real plastic deformation occurs through dislocation motion: a dislocation sweeps across the slip plane one Burgers vector at a time, with only a small region of bonds stressed at any moment. The carpet wrinkle analogy captures this: moving a wrinkle takes a fraction of the force needed to lift the entire carpet. If plastic deformation required simultaneous bond rupture, metals would be brittle rather than ductile.

Question 5

What does Schmid's law predict, and why does it mean that differently oriented grains in a polycrystal will begin to yield at different applied stresses?

Model answer: Schmid's law states that the resolved shear stress on a slip system is τ = σ cos φ cos λ, where σ is the applied tensile stress, φ is the angle between the loading axis and the slip plane normal, and λ is the angle between the loading axis and the slip direction. Slip initiates when τ reaches the critical resolved shear stress (CRSS), a material constant. The Schmid factor cos φ cos λ varies between 0 and 0.5 depending on grain orientation relative to the loading axis. A grain with a favorably oriented slip system (Schmid factor near 0.5) will activate slip at a much lower applied stress than a grain oriented with all slip systems nearly parallel or perpendicular to the loading direction. This is why polycrystals yield progressively — grain by grain — rather than all at once.

Schmid's law is the crystallographic equivalent of resolving a force onto a plane: only the component of stress acting in the slip direction on the slip plane can drive dislocation motion. Grains most favorably aligned (so-called 'soft orientations') yield first and transfer stress to neighboring grains, which then yield in turn. This grain-by-grain yielding underlies the gradual onset of plastic flow seen in the stress-strain curves of polycrystalline metals.