A theoretical calculation predicts that a pure aluminum crystal should require ~1 GPa to yield plastically, but real aluminum yields at roughly 10 MPa. What is the primary explanation for this discrepancy?
AAluminum bonds are weaker than models assume, so less stress is needed to break them
BStress concentrates at grain boundaries, locally exceeding the theoretical threshold
CDislocations allow slip to propagate sequentially — one bond breaks and reforms at a time — rather than simultaneously across the entire slip plane
DReal aluminum crystals are not truly periodic at the atomic scale, so the theoretical model is inapplicable
The theoretical yield stress assumes all atomic bonds along a slip plane break simultaneously (like lifting the entire rug), requiring ~G/10. Dislocations act like a wrinkle in the rug: only the bonds near the dislocation core break at any instant, propagating slip one atomic spacing at a time at a tiny fraction of the theoretical stress. The macroscopic slip is identical, but achieved incrementally.
Question 2 Multiple Choice
As a metal is cold-worked (repeatedly deformed), it becomes progressively harder to deform further. What mechanism is primarily responsible for this work hardening?
AGrain boundaries fill with precipitates during deformation, blocking dislocation motion
BDislocations multiply and tangle, creating a network that impedes further dislocation motion
CThe Burgers vector grows with each deformation cycle, requiring more energy to move each dislocation
DScrew dislocations convert to edge dislocations during cold working, and edge dislocations move more slowly
Cold working dramatically increases dislocation density (from ~10¹² to ~10¹⁶ m⁻²). At high density, dislocations interact and tangle, generating stress fields that obstruct each other's glide. The material strengthens because there are too many tangled dislocations to move easily — every strengthening mechanism in metals ultimately works by making dislocation motion more difficult.
Question 3 True / False
For an edge dislocation, the Burgers vector is perpendicular to the dislocation line direction.
TTrue
FFalse
Answer: True
An edge dislocation is defined by having its Burgers vector (the lattice distortion direction) perpendicular to the dislocation line. This contrasts with a screw dislocation, where the Burgers vector is parallel to the dislocation line. Mixed dislocations have both edge and screw character.
Question 4 True / False
A heavily cold-worked metal is typically weaker than an annealed metal of the same composition, because cold working creates many defects that disrupt the lattice.
TTrue
FFalse
Answer: False
Cold working makes a metal stronger, not weaker. The high dislocation density produced by cold working causes dislocations to tangle and impede each other, raising the yield stress. Annealed metals have low dislocation density and are relatively soft. This is precisely why work hardening is a useful industrial process — repeated deformation progressively increases strength.
Question 5 Short Answer
Using the 'rug wrinkle' analogy, explain why pushing a wrinkle across a rug requires less force than dragging the whole rug — and how this maps onto why real metals yield at stresses far below the theoretical prediction.
Think about your answer, then reveal below.
Model answer: Dragging the entire rug simultaneously means overcoming all friction at once. Pushing a wrinkle moves only a small region at a time, requiring much less local force. In a crystal, the theoretical yield stress requires all bonds along the slip plane to break simultaneously. A dislocation allows slip to propagate by breaking and reforming bonds one at a time near the dislocation core, just like advancing the wrinkle. The full lattice displacement (one Burgers vector) still occurs, but the stress required is orders of magnitude lower because only a tiny fraction of bonds are strained at any instant.
The analogy captures the key insight: sequential bond breaking vs. simultaneous bond breaking. The dislocation mechanism means the required stress scales with the local distortion energy near the core, not with the energy needed to shear an entire perfect plane — hence the factor-of-1000 reduction from theoretical to observed yield strength.