According to scmid’s law, resolved shear stress will be zero for a situation in which the slip plane is parallel to tensile stress.
In this case, despite the presence of shear stress on slip plane, even if the applied tensile stress is enormous, slip in single crystal will not commence. Why does applied shear stress do not cause a slip on this plane?
Single crystal is not like poly crystalline materials. In single crystal, there are preferred plane where slip moves
Slip will commence - just not in that orientation. The crystal will rotate into a different orientation, where the Schmid factor is higher. Schmid law only gives the relative ease of slip for a given orientation; it doesn't give the ease of crystal rotation (which is dependent on the testing geometry).
Thanks
But the question is regardless of the other slip systems, why such a slip system will not be active despite the presence of shear stress on this plane?
For example consider a single crystal of FCC metal oriented such that a tensile stress is applied along a [1 -1 0] direction . For this situation, schmid's factor will be zero for (111) [110] slip system ( The tensile stress is parallel to (111) plane ). So the resolved shear stress for this system will be equal to zero and even if the applied tensile stress is enormous, slip in this plane will not occur wherease there is shear stress on this plane ( 111 ) equal to applied tensile stress. why?
(1) If the Schmid factor is 0, there is no shear stress on that particular slip system. The resolved shear stress is exactly that - the shear stress on a given system due to an applied normal(tensile) stress in some direction.
(2) Why will there be no shear stress parallel to the normal stress? If you look into the way shear stress is defined in mechanics (or easier, look at a Mohr's circle), you will find that upon application of normal stress, shear stress is zero only in two directions-parallel and perpendicular to the normal stress (and maximum at 45degrees). Therefore, you don't have a shear stress in the same direction. Extending this to 3 dimensions, now shear stress is 0 in the slip system if either slip plane normal or slip direction is parallel/perpendicular to the normal stress direction.
(3) (111)[110] is not a slip system. (111)[-110], (111)[-101] and (111)[0-11] are the slip systems on (111).
You are assuming that there is a shear stress (equal to tensile stress) on a plane parallel to the tensile axis. This is simply not correct.
On cylindrical sample, you are not applying tensile stress from one point or part of cross-section, but across the complete cross-section. In this case there cannot be any shear stress. However, if you apply tensile stress on part of the cross-section, there will be shear stress parallel to the tensile axis.
Please see the following figure for more details.
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