Copper's theoretical shear strength is approximately 3 GPa (about G/30), yet copper actually yields around 50 MPa. What accounts for this 60-fold discrepancy?
ACopper is a soft metal with unusually weak metallic bonds compared to harder metals
BImpurities in commercial copper lower its strength significantly below the theoretical pure-crystal value
CDislocations allow sequential atomic-scale slip — like pushing a wrinkle across a rug — which requires far less stress than sliding an entire plane simultaneously
DThe theoretical value assumes single-crystal copper; polycrystalline copper is weaker because grain boundaries act as stress concentrators
The rug analogy captures the key insight: sliding a whole rug simultaneously is nearly impossible, but pushing a wrinkle across it is trivial, and the net result is that the rug moved. Dislocations are that wrinkle — a line defect that allows one atomic spacing of slip to propagate sequentially through the crystal at a fraction of the theoretical stress for simultaneous sliding. This is why real metals yield at stresses far below G/30 and why controlling dislocation motion (through alloying, work hardening, grain refinement) is the central strategy of strengthening.
Question 2 Multiple Choice
A single FCC crystal is loaded in tension with the tensile axis oriented 45° to a {111} slip plane normal and 45° to a ⟨110⟩ slip direction. According to Schmid's law, what is the Schmid factor and what does this imply?
ASchmid factor = 1.0 — this orientation activates slip at the lowest possible applied stress
BSchmid factor = 0.5 — this orientation maximizes the resolved shear stress on this slip system, so slip initiates at the lowest applied stress
CSchmid factor = 0.5 — but this is the worst orientation for slip; ⟨100⟩ loading gives a factor of 1.0
DSchmid factor = 0.25 — the factor is cos²(45°) because both angles are equivalent
Schmid factor = cos(φ)·cos(λ). Both angles are 45°, so the factor = (1/√2)(1/√2) = 0.5. This is the maximum possible Schmid factor — no orientation can exceed 0.5 — so this orientation initiates slip at the lowest applied tensile stress. Grains with ⟨100⟩ or ⟨111⟩ axes parallel to loading have low Schmid factors on all slip systems and require much higher applied stress to yield, which is exploited in texture strengthening.
Question 3 True / False
Plastic deformation by dislocation motion requires most atoms in the slip plane to move simultaneously.
TTrue
FFalse
Answer: False
The entire point of dislocation theory is that atoms move sequentially, not simultaneously. A dislocation is the boundary between a slipped and unslipped region; as it glides through the crystal, each successive row of atoms slips by one spacing while the rest remain stationary. This sequential mechanism is why real yield stresses are ~60× below theoretical values. Requiring simultaneous movement of an entire plane (as in the theoretical calculation) demands an enormous stress; moving a dislocation through the lattice requires only a tiny local distortion.
Question 4 True / False
FCC metals have more active slip systems than HCP metals, which contributes to their greater ductility.
TTrue
FFalse
Answer: True
True. FCC metals have 12 slip systems ({111}⟨110⟩ — 4 planes × 3 directions), so there is almost always a favorably oriented system available no matter how the crystal is loaded. HCP metals typically have only 3 easy slip systems on the basal (0001) plane, which are easily exhausted. When all available slip systems are blocked, the material cannot accommodate further deformation and fractures instead of deforming plastically. This is why magnesium and zinc (HCP) are brittle at room temperature without the additional slip and twinning mechanisms that activate at higher temperatures.
Question 5 Short Answer
Why do FCC metals deform plastically much more easily than HCP metals under comparable stress conditions, even when their bond strengths are similar?
Think about your answer, then reveal below.
Model answer: The key difference is the number of available slip systems. FCC metals have 12 slip systems (4 close-packed {111} planes × 3 ⟨110⟩ directions), so a favorably oriented slip system is almost always available for any loading direction — plastic deformation can proceed without exhausting all options. HCP metals have only 3 easy slip systems on the basal plane, and once these are loaded unfavorably or exhausted, no alternative system is available. The crystal cannot accommodate the imposed strain and fractures instead. More slip systems = more pathways for strain accommodation = greater ductility.
This tests whether students can connect the crystallography (number and orientation of slip systems) to macroscopic mechanical behavior. The insight is that ductility is not just about bond strength — it's about geometric flexibility in accommodating deformation.