Close packed planes have highest atomic density and their interplanar distance is large hence they become the favorable plane for slip to occur. There are favorable directions also. The slip plane and slip direction both form a slip system. When the critical stress is maximum for a particular slip system the slip occurs

Close packed planes have high atomic density that is the atoms are very close to each other. Thus, effort required to move dislocations on the application of the external stress, which acts as the resolved shear stress, on a close packed plane is comparatively small and hence the dislocation can move relatively easily as compared to that on the atomic planes where the atoms are widely spaced and hence have low atomic density.

In case of FCC {111} family of planes are most favourable planes for slipping to take place because it is the most closely packed plane. As said by Arshad, combination of a slip plane and slip direction forms a slip system. (111) is not a slip system. It is a plane.

I completely agree with Abhinav. I suppose you know stress is required for movement of dislocations. This stress (to cause dislocation motion) increases exponentially with every step that the dislocation move (read Burgers vector). For easy movement of dislocation, you therefore require small repeat distances between atoms (in other words high atomic density). This condition is best satisfied in closed pack planes since more atoms are stuffed within this plane compared to others. Therefore slip is preferred in closed pack planes. Abhinav has already given the closed pack configurations in some crystal structures.

The energy for slip of dislocation system to move are need be the lowest (related the b vector burgers^2), ref. Mechanicall Metallurgy-Dieter.

What happens when u push 1 bicycle in a packed cycle stand

Dear Sathosh, I fully agree with Anil;s simple example.No doubt a close packed

plane has the highest APF, a small shear stress slightly more than the CRS will

definitely move the atoms quickly there by causing quick slip of the plane.

A.V.Sethuraman

Hindustan University

FCC (111) plane have 2 effective atoms and area about 0.886a^2, hence planer density about 2.31/a^2. It is the closest pack one. Therefore the resolved stress for dislocation movement is lower than other set of planes. Further if you consider the Peierls-nabarro criteria for slip it also sites that stress for dislocation movement reduces exponentially with the distance of atoms on a particular plane (Courtesy: George E Dieter). As (111) is the closest one this distance is minimum and dislocations move very easily.

Koushik. : one half of your answer is about the Pierls nabarro function as many others have described..

But the first half of resolved stress in your explanation:- are you referring to schmid law ? if so then the justification is incomplete. Schmid's law says that dislocation moves when the resolved stress reaches a critical value.. Ofcourse, if the specimen is loaded along tensile axis, the resolved stress in the slip plane along the slip direction will be lesser than the applied normal stress along the loading direction. but that doesnt justify that slip occurs. It only tells that for the same applied load, each plane and direction has different resolved stress. According to schmid law, the resolved stress is same for BCC and FCC as both are cubic structures.

Hariharan, thank you for your kind suggestion. But if you kindly notice my ans. In first part I have mentioned planer density factor and in the second part I mentioned the energetics of the precess in support of my previous statement. I am unable to found schemid's law in my ans..

Buddy, you are from MGIT. Ask Viplav Kumar sir! He will be more than glad to answer you.

By the way, the cycle stand example is the closest that I can think of too.

From thermodynamically analysis, (which is suitable for large scale but however often be used for micro-scale as well), the process will happen on the direction of lowest system energy . The preferable slip plain is following the same law. All the above mentioned reasons are the sub-reasons bellow the thermodynamic law.

Dear Santhosh,

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

A.V.Sethuraman

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.

Ease of deformation can be interpreted in many ways: 

1.) Directional parameter of dislocation in a lattice.( More slip systems more is the      ease or the cases hence FCC > BCC > SCC) [ with BCC>FCC>SCC At some threshold elevated temperatures cause addition of more slip systems in BCC]. 

2.) Energy associated with the deformation to be produced.( For the same element in these three forms we can say Accordance to Frank's rule Energy of dislocation as FCC=BCC > SCC) each proportional to square of radii of element.

Due to minimum peierl's-nabarro stress along close-packed planes

I am not very good at this subject. If I may, I would like to take part in this discussion.In my opinion, the stress required to cause motion of dislocation get increase exponentially with the length of the Burgers vector. As a result of it, the slip direction should have a small repeat distance or high linear density. The close-packed direction in metals and alloys satisfy this criterion.

Dear Santosh,

The atoms in such planes are bonded more strongly to each other than the atoms in one such plane are bonded to the atoms in a neighbouring plane. Slip systems are vital for deformation in a metal to occur. The application of shear stress along the length of an object causes crystal lattices to glide along each other and form slip systems. Slip systems are unique to the lattice where they are present. Slip planes are the planes with the highest density of atoms.

For easy movement of dislocation, you therefore require small repeat distances between atoms (in other words high atomic density). This condition is best satisfied inclosed pack planes since more atoms are stuffed within this plane compared to others. Therefore slip is preferred in closed pack planes.

Ashish

Dear Santosh,

The atoms in such planes are bonded more strongly to each other than the atoms in one such plane are bonded to the atoms in a neighbouring plane. Slip systems are vital for deformation in a metal to occur. The application of shear stress along the length of an object causes crystal lattices to glide along each other and form slip systems. Slip systems are unique to the lattice where they are present. Slip planes are the planes with the highest density of atoms.

For easy movement of dislocation, you therefore require small repeat distances between atoms (in other words high atomic density). This condition is best satisfied inclosed pack planes since more atoms are stuffed within this plane compared to others. Therefore slip is preferred in closed pack planes.

Ashish

Hi

Closed packed plane is the plane of greatest atomic density should be obvious plane for slip,because ON THIS PLANE only there will be proper co-ordination implexation impetus to carry forward deformation stimuli for Effective Slip Materialisation.

thanks & regards

g.sudhakar

phd(material engineering)

HCU