Slip plane with highest atomic density is preferred for
deformation because the distance between atoms are so small
that dislocation movement due to lower applied stress is easier and higher
atomic packing factor is an indication of easiness of deformation
What needs to be understood here is that the slip is between the planes, planes as integral units; not be overlooked for movement of individual atoms.
During shear, it must be ensured that the crystal structure remains unchanged due to slip. And this process must occur with minimum distortion in the lattice.
Closest packed planes have most no. of atoms per unit area. This means that they are structurally most strong and stiff. Another implication of this is that a direction orthogonal to the closest packed plane would be the weak link. A direction perpendicular to which shear is the easiest. Since the closed packed direction on the closest packed plane is the most strongest of all, a direction perpendicular to it would be the weakest.
This explains why slip occurs in this direction.
Another way to look at this is through the concept of burgers vector. If you move planes relative to one another, the burgers vector is the shortest in the direction of closed packing; meaning the strain distortion produced is the least. I guess this is where thermodynamics would also pop in.
Burgers vector is smallest for the slip direction in that slip plane, so the energy of dislocation would be small. That's why it is the slip plane in fcc.
Indian Institute of Technology Kanpur
A. K. Kavala
Indian Institute of Technology Patna
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National Metallurgical Laboratory
National Institute of Foundry & Forge Technology
Indian Institute of Science
Indian Institute of Technology Gandhinagar