I initially thought that it had something to do with the number of slip systems in FCC vs. BCC, but they're both the same.
You are mistaken to imply that FCC metals do not undergo a transition from brittle to ductile behavior. FCC metals can fracture at sufficiently high stress - at some temperature. Please provide more information in your question if you believe otherwise.
The poster above is incorrect: FCC metals do not have a ductile to brittle transition temperature and instead remain ductile at low temperatures. This is because the stress required to move dislocations is not strongly temperature-dependent in FCC metals, and thus failure occurs by plastic flow instead of crack propagation.
In BCC metals, the stress required significantly increases at low temperatures while the cracks propagation stress is not strongly temperature-dependent. Thus BCC metals fail by crack propagation at low temperatures. More info here: http://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php
The only time that FCC metals experience brittle fracture is in the final stage of fatigue failure.
The presence of a ductile to brittle transition temperature implies there are insufficient (ductile) deformation modes at low temperatures to support plastic deformation and therefore fracture occurs to release energy/load.
In FCC materials, dislocation slip of both edge and screw dislocations is relatively athermal and due to the number of active slip systems it is relatively homogeneous. Furthermore, there are more than 5 independent slip systems that can activate to accommodate arbitrary plastic strain (a requirement of von Mises' deformation criteria).
In BCC materials, screw dislocation motion is not athermal. In particular, the cores of the screw dislocations tend to 'unzip' into a sessile configuration when the material is not under load. Under load, the dislocations cores, with some thermal activation, may reconfigure into a glissile dislocation core. These dislocations can then move and enable plastic deformation. Deformation in BCC materials without screw dislocations is not sufficiently independent to accommodate arbitary shape changes and so the material often deforms through fracture. The critical temperature for screw dislocation mobility (the thermal contribution to the rearrangement of the BCC screw dislocation core structure) is the root cause of the DBTT in the BCC materials.
For what it's worth (http://materialiaindica.ning.com/forum/topics/ductile-brittle-transition ):
"Let us compare BCC and FCC here. At high temperatures, both of these have mobile dislocations, and thus they can sustain large plastic deformations without undergoing fracture.
At low temperatures however, while dislocations in BCC are no longer mobile, dislocations in FCC can still move very quickly. This lack of dislocation movement makes BCC brittle, while FCC stays ductile...
Now the key question is thus why do dislocations in FCC stay mobile at low temperatures while dislocations in BCC find it increasingly difficult to move as the temperature is lowered.
This is easy to understand: in FCC there are closest-packed planes belonging to each slip system and slip means a corner atom being moved to centre of the face.
In BCC though there are as many total number of slip systems as in FCC(12), the movement of dislocations happen only as a line of atom jumps from one potential energy valley to another (so-called Peierls valleys) - a process that could be enhanced by application of stress, or by thermal activation. The planes are not as close-packed and any slip means a corner atom being moved to centre of the cube. Thus one would see kink nucleation and propagation in BCC dislocation movement, since cost of moving an entire row of atoms from corner to centre at the same time is too high.
Thus to summarize, BCC dislocations' movement is thermally activated while (relatively) FCC dislocations' movement needs significantly smaller activation. This leads to BCC materials becoming brittle at low temperatures while FCC staying ductile irrespective of temperature - but probably not at 0 Kelvin since an atom still has to be moved. I think however that there is an activation energy for FCC slip, only it is much less than kbT."
protected by Community♦ Jul 7 '16 at 20:16
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?