A metal component operates reliably at room temperature but creeps under sustained load at 0.5 Tm. The mechanism responsible for this high-temperature plasticity is:
AIncreased dislocation glide velocity because thermal energy lowers the Peierls barrier
BDislocation climb, which allows dislocations to bypass obstacles by absorbing or emitting vacancies through thermally-activated diffusion
CGrain boundary melting that allows slip between adjacent grains at elevated temperature
DMultiplication of dislocations by Frank-Read sources, which becomes active only above a threshold temperature
Glide is conservative — it requires only bond rearrangement at the dislocation core and can occur at any temperature. At room temperature, dislocations glide until they pile up against obstacles (grain boundaries, precipitates, other dislocations) and are stuck. Climb requires vacancy diffusion to move the dislocation perpendicular to its slip plane, allowing it to step over the obstacle. Vacancy diffusion is thermally activated and becomes significant only above roughly 0.4 Tm. This is why creep is predominantly a high-temperature phenomenon: glide alone cannot bypass obstacles, but climb can.
Question 2 Multiple Choice
Both an edge dislocation and a screw dislocation in an FCC metal are subjected to an applied shear stress on their primary slip plane. Which dislocation can move to an entirely different crystallographic plane to avoid an obstacle?
AThe edge dislocation, because its extra half-plane can tilt to intersect other slip planes
BBoth equally, because any dislocation under sufficient stress can switch slip planes
CThe screw dislocation, because its Burgers vector is parallel to its line, so it has no unique slip plane and can cross-slip
DNeither; both are confined to their original slip plane unless they climb
A screw dislocation's defining geometric feature — its Burgers vector being parallel to the dislocation line — means it has no unique slip plane. Any plane containing the Burgers vector direction is a valid slip plane for a screw dislocation. This enables cross-slip: the dislocation moves from one slip plane to another when stress geometry favors it, allowing it to circumvent obstacles that pure glide could not bypass. An edge dislocation's Burgers vector is perpendicular to its line, fixing it to a unique slip plane (glide plane). It cannot cross-slip; it can only leave its plane by climb.
Question 3 True / False
For an edge dislocation, the Burgers vector is perpendicular to the dislocation line; for a screw dislocation, the Burgers vector is parallel to the dislocation line.
TTrue
FFalse
Answer: True
This geometric relationship is the defining characteristic of each dislocation type and has direct physical consequences. For edge dislocations, the perpendicular Burgers vector means there is an extra half-plane of atoms pointing toward the dislocation line — the dislocation is the terminus of this half-plane. For screw dislocations, the parallel Burgers vector creates the helical ramp geometry. Most real dislocations are mixed (the Burgers vector makes an intermediate angle with the line), but decomposing them into edge and screw components remains useful for analyzing glide, climb, and cross-slip behavior.
Question 4 True / False
Dislocation climb is essentially a faster or thermally-assisted version of dislocation glide, driven by the same bond-rearrangement mechanism.
TTrue
FFalse
Answer: False
Glide and climb are fundamentally different mechanisms, not different speeds of the same process. Glide is conservative: the dislocation moves within its slip plane by sequential bond rearrangement at the core, requiring no net mass transport and no diffusion. It can occur at cryogenic temperatures. Climb is non-conservative: the dislocation moves perpendicular to its slip plane by absorbing or emitting vacancies — atoms must diffuse away from or to the dislocation core, which requires thermal activation. This is why climb is negligible at room temperature and becomes important only above roughly 0.4 Tm.
Question 5 Short Answer
Why can dislocations enable plastic deformation at stresses orders of magnitude below the theoretical strength of a perfect crystal, and what is the mechanistic difference between dislocation glide and dislocation climb?
Think about your answer, then reveal below.
Model answer: In a perfect crystal, plastic deformation would require simultaneously breaking all bonds across an entire slip plane — an enormous stress (~G/10, where G is the shear modulus). Dislocations allow deformation one atomic step at a time: only the bonds at the dislocation core need to break and reform at any instant, requiring far less stress. Glide moves the dislocation within its slip plane through sequential bond rearrangement with no diffusion required. Climb moves the dislocation perpendicular to its slip plane by exchanging atoms with the surrounding lattice (absorbing or emitting vacancies), which requires thermally-activated vacancy diffusion and is therefore significant only at elevated temperatures.
The analogy is moving a rug by pushing a wrinkle across it versus sliding the whole rug at once. The dislocation is the wrinkle — a localized defect that requires only local atomic rearrangement to advance. This is why metals yield at stresses 10–1000× lower than theoretical predictions for defect-free crystals. Glide vs. climb distinguishes two regimes of deformation: glide dominates at low temperatures and produces work hardening as dislocations pile up; climb allows dislocations to recover and anneal at high temperatures, enabling creep.