I thought it had to do with the amount of slip systems...because BCC doesn't have as many slip planes, it cracks at low temperatures, while FCC has enough slip planes that it is not dependent upon temperature. However, I don't understand this for HCP metals...They don't have a lot of slip planes either? How is it that they don't have a ductile-to-brittle transition temperature?

edit referring to crystalline unit structures here: Body-Centered Cubic (BCC), Face-Centered Cubic (FCC), Hexagonal Close Packed (HCP)

  • $\begingroup$ I have no idea, what HCP, BCC, FCC are you should really write them out. $\endgroup$ – Gonenc Jul 21 '15 at 21:39
  • $\begingroup$ Lots of factors. The fcc and hcp slip systems are similar in theory, since they are both close packed planes. Hcp just has fewer than the higher-symmetry fcc, and for a given grain may not be optimally oriented to slip. Bcc doesn't have any close packed planes, and the dislocations that might lead to slip often decompose into partials lying on different planes, so their movement is very difficult without thermal activation to reconfigure onto one slip plane. $\endgroup$ – Jon Custer Jul 21 '15 at 22:01

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