Theoretical calculations predict that a perfect metal crystal should require ~10 GPa of shear stress to deform plastically. Real metals yield at 10–100 MPa — a factor of 100–1000 lower. What explains this enormous discrepancy?
AReal metals contain many grain boundaries that act as weak planes, reducing the required stress
BDislocations allow plastic deformation to occur by propagating local bond-breaking sequentially rather than sliding entire planes simultaneously
CThe theoretical calculation assumes room temperature; real metals are softer because thermal vibrations assist deformation
DReal metals have impurities that lubricate the slip planes, dramatically lowering friction between atomic layers
The theoretical shear strength assumes all atoms in a plane slide simultaneously — a synchronized collective motion requiring enormous force. Dislocations short-circuit this: a dislocation line represents the boundary between slipped and unslipped regions, and it glides forward by breaking only a few atomic bonds at a time (like the ripple in a rug). The energy required is orders of magnitude smaller because only local, sequential bond-breaking is needed. Grain boundaries (option A) actually impede dislocation motion and increase yield stress, not decrease it.
Question 2 Multiple Choice
FCC metals like copper are much more ductile and formable than HCP metals like magnesium at room temperature. The primary structural reason is...
AFCC metals have lower melting points, making them softer under applied stress
BFCC metals have 12 independent slip systems compared to only 3 primary slip systems in HCP, making it easier to find active slip systems for any loading direction
CHCP metals have stronger covalent bonds that resist dislocation motion more effectively
DFCC metals have larger Burgers vectors, which means more plastic displacement per dislocation passage
Ductility requires that slip systems be activated in multiple directions as a material deforms. FCC has 4 {111} close-packed planes × 3 ⟨110⟩ directions = 12 slip systems, so whatever direction stress is applied, several systems are favorably oriented (high Schmid factor). HCP has only 3 primary basal slip systems — loading at most orientations will find no system with a high Schmid factor, so the material fractures instead of yielding plastically. Option D is backwards: smaller Burgers vectors (not larger) are energetically favored since dislocation energy scales as |b|².
Question 3 True / False
The elastic strain energy stored in a dislocation is proportional to |b|², which is why dislocations in crystals preferentially form with the shortest possible Burgers vectors.
TTrue
FFalse
Answer: True
Dislocation energy per unit length scales approximately as μb²/2, where μ is the shear modulus. The shortest lattice translation vectors — which lie along close-packed directions — minimize |b| and therefore minimize the energy cost of creating and maintaining the dislocation. This energy criterion explains why slip preferentially occurs on close-packed planes in close-packed directions: not just because the planes are widely spaced, but because these directions provide the smallest Burgers vectors.
Question 4 True / False
Dislocations prefer to glide on close-packed planes because those planes have the highest atomic density, making them the strongest and most resistant planes in the crystal.
TTrue
FFalse
Answer: False
The reasoning is opposite. Close-packed planes are favored because they are the most widely *spaced* planes in the crystal — there is more distance between them, so it costs less energy to create the planar separation needed for slip. High atomic density on the plane itself means the atoms are tightly packed laterally, which relates to the Burgers vector length, but the key reason for preferring close-packed planes for slip is their large interplanar spacing (low energy to shear apart), not their in-plane strength.
Question 5 Short Answer
Why does dislocation glide require so much less applied stress than sliding one complete atomic plane over another in a perfect crystal, even though both ultimately move atoms the same net distance?
Think about your answer, then reveal below.
Model answer: Sliding an entire plane simultaneously requires every atom to pass over the energy barrier of its neighbor at once — all bonds must be stretched and broken in concert, requiring enormous collective stress (~10 GPa). Dislocation glide breaks only the few bonds at the dislocation core at any moment; the rest of the crystal remains undisturbed. As the dislocation advances one atomic spacing, only a small number of local bonds break and reform. The stress needed is only sufficient to drive this local, sequential process, not the global simultaneous process — hence 100–1000× lower yield stress.
An analogy: dragging a large rug across the floor requires enormous friction force on the entire area simultaneously. Propagating a ripple across the rug requires only the force to lift and slide a small buckle at a time. The net displacement is the same; the local force required is dramatically different. This analogy directly maps to the dislocation mechanism.