A steel rod is loaded past its yield point and then completely unloaded. What happens to the deformation, and what atomic-scale event explains this outcome?
AThe deformation fully recovers, because steel bonds are strong enough to return all atoms to their original positions
BThe deformation is permanent, because dislocations have moved along slip planes — atoms have broken bonds with old neighbors and formed bonds with new ones, leaving no restoring force
CThe rod slowly returns to its original shape over hours as residual stresses relax elastically
DThe rod fractures during unloading because yield-point deformation always causes immediate failure
Plastic deformation is fundamentally different from elastic deformation. In elastic deformation, atoms are displaced from equilibrium but remain bonded to the same neighbors — the stretched bonds pull them back. In plastic deformation, dislocations glide along slip planes: as a dislocation moves through, atoms sequentially break bonds with one set of neighbors and form bonds with the next, advancing the crystal one Burgers vector. There is no restoring force after dislocation passage, so the shape change is permanent.
Question 2 Multiple Choice
A materials engineer proposes cold-rolling a steel sheet to strengthen it before use in a structural application. A colleague objects: 'Introducing more defects into the crystal will only weaken it.' Who is correct?
AThe colleague is correct — any lattice defect reduces mechanical strength
BThe engineer is correct — cold-rolling causes work-hardening: the increased dislocation density causes dislocations to tangle and impede each other's motion, raising the stress required for further deformation
CNeither is correct — cold-rolling has no effect on yield strength, only on surface finish
DThe colleague is correct that strength decreases, but the reduced ductility makes the material more useful in structures
This targets the key counterintuitive insight of work-hardening. Cold-rolling plastically deforms the material, multiplying dislocation density. But higher dislocation density means more mutual obstruction — dislocations tangle, pin each other, and create local stress fields that impede further dislocation motion. The material becomes harder to deform further: the yield strength increases. This is work-hardening (strain-hardening), and it is the basis of many industrial strengthening processes including cold drawing, shot peening, and wire-drawing.
Question 3 True / False
A material with a large gap between its yield strength and ultimate tensile strength is more likely to fail suddenly without warning than a material whose yield strength and UTS are nearly identical.
TTrue
FFalse
Answer: False
A large gap between yield strength and UTS indicates high work-hardening capacity and ductility. The material distributes plastic deformation broadly before failing locally, giving visible warning (necking, elongation) before fracture. A small gap — yield strength close to UTS — indicates limited work-hardening: once yielding begins, fracture follows quickly with little additional deformation and little warning. High-strength brittle materials or overaged alloys can exhibit this dangerous behavior.
Question 4 True / False
Work-hardening increases the stress required for further plastic deformation because the growing dislocation density causes dislocations to tangle and impede each other's motion.
TTrue
FFalse
Answer: True
This is the mechanism of strain-hardening. As dislocations multiply and move during plastic deformation, they increasingly encounter other dislocations and interact — forming tangles, jogs, and local stress fields that obstruct further glide. The material becomes progressively harder to deform (higher flow stress required), which is exactly what the rising slope of the stress–strain curve above the yield point represents. The quantitative relationship is captured by the power-law σ = K·εⁿ, where n is the work-hardening exponent.
Question 5 Short Answer
Explain why plastic deformation is permanent while elastic deformation is not, using atomic-scale reasoning.
Think about your answer, then reveal below.
Model answer: In elastic deformation, atoms are displaced from their equilibrium positions but remain bonded to the same neighbors. The stretched or compressed interatomic bonds act like springs and pull atoms back to equilibrium when the load is removed — like stretching a rubber band. In plastic deformation, dislocations glide along crystallographic slip planes. As a dislocation moves through the lattice, atoms at the dislocation core sequentially break bonds with one set of neighbors and reform bonds with the next set, shifting one region of the crystal one Burgers vector relative to the other. After the dislocation passes, the atoms are in new bonding arrangements with new neighbors. There is no potential energy gradient pulling them back to their original positions, so the shape change is irreversible.
The dislocation mechanism is why metals can be ductile: rather than requiring every atomic bond across a plane to break simultaneously (which would require enormous stress), a dislocation allows a wave of sequential bond-switching to propagate through the crystal at much lower stress. This is analogous to moving a large rug by forming a wrinkle and sliding it across rather than dragging the whole rug at once.