Which process leads to higher work-hardening rate of austenitic stainless steel: cross slip or planar slip?
I am not sure if there could be a straight comparison like this, or if the two can be isolated indeed. For the cross slip to take place, that is forcing the dislocations out of their original slip plane, the slip must be saturated at least locally. So the work hardening until that state is provided by the increased dislocation density and their pile-ups due to their slip. And if cross-slip does start it is actually softening mechanism as it provides a way for the slip to continue.
On the other hand, if we consider a hypothetical case where the slip of an individual dislocation (where there is no considerable dislocation density, i.e. the isolated dislocation is relatively far away from the other dislocations) and the cross slip of a pair of partials constituting a stacking fault, the stress required fro the cross-slip action is greater than that for the slip for the same system considered.
Hello Oksana,
Typically a planar slip deformation decreases the probability of dislocation entanglements thereby decreasing the work hardening rate. However, this answer is just based on the comparison of "cross slip" and "planar slip" behavior work hardening can get affected by several other mechanisms.
Ilaksh
I would say cross slip generally plays a more important role in work hardening, as it leads to more dislocation reactions and perhaps formation of Lomer-Cottrell lock.
Cheers,
Ahmad
Dear Ahmad Zafari and Ilaksh Adlakha, could you recommend me some literature on this?
Hi,
I would agree to the first two posts. Maybe this paper helps you:
L. P. Karjalainen, T. Taulavuori, M. Sellman, A. Kyröläinen
Some Strengthening Methods for Austenitic Stainless Steel
Materials Technology, steel research int. 79 No. 6 (2008)
Especially p. 405 "Dislocation Strengthening".
Best regards
I would say that planar slip is much more effective than cross slip. High stacking fault energy metals like pure Al are characterized by easy and abundant cross slip so do not work harden much. Planar slip in low SFE alloys like austenitic steels does not allow dislocations to climb out of obstacles without generating high local stresses and therefore leads to high work hardening rates.
I am not sure if there could be a straight comparison like this, or if the two can be isolated indeed. For the cross slip to take place, that is forcing the dislocations out of their original slip plane, the slip must be saturated at least locally. So the work hardening until that state is provided by the increased dislocation density and their pile-ups due to their slip. And if cross-slip does start it is actually softening mechanism as it provides a way for the slip to continue.
On the other hand, if we consider a hypothetical case where the slip of an individual dislocation (where there is no considerable dislocation density, i.e. the isolated dislocation is relatively far away from the other dislocations) and the cross slip of a pair of partials constituting a stacking fault, the stress required fro the cross-slip action is greater than that for the slip for the same system considered.
Dear Julian H. Driver, could you please recommend me the literature which describes this?
Wenfang Cui, could you please recommend me the literature which describes this?
Not sure I can find a good reference immediately since it refers to an "old" metallurgical problem. Work hardening theories are not always very clear on the question but usually include a term for the recombination stress on split partials or stress to climb out of obstacles (ie SFE dependent). In practice the maximum work hardening coefficient of Al is about 0.2 but the work hardening coefficients of austenitic steels are usually above 0.4.
There is a classic set of articles about dislocations in Metallurgical Transactions A, Dec 1985, vol. 16A, which can help...
- Badges
- Science topic
- Similar topics
- Physical Sciences
- Materials Science