Glasses and amorphous materials
Which statement about plastic deformation of metallic glasses at room temperature is correct?
- AThe yield strength is high, but deformation concentrates in thin shear bands, so in tension they tend to fracture with almost no elongation
- BWith no grain boundaries they are highly ductile and elongate uniformly by several tens of percent in a room-temperature tensile test before fracturing
- CDislocations glide in them just as in crystals, so they work-harden strongly, and they neck extensively before fracturing in a ductile manner
Show answer and explanation
Answer: A. The yield strength is high, but deformation concentrates in thin shear bands, so in tension they tend to fracture with almost no elongation
In metallic glasses, shear stress makes small groups of atoms move cooperatively, locally increasing free volume and softening the material. Because this softening invites further deformation, deformation concentrates in shear bands about 10 to a few tens of nm thick, and with no work hardening, a single shear band runs through the specimen in tension and it fractures with almost no elongation. In compression and bending, many shear bands form and some plastic deformation is seen. Work hardening and necking by dislocation glide are features of crystalline metals, and the absence of grain boundaries does not guarantee ductility. As a countermeasure, composites with dispersed crystalline phases that hinder shear band propagation are being studied. In the supercooled-liquid range, on the other hand, metallic glasses can deform greatly by viscous flow, which is used for precision forming.