The effect of Al alloying on the deformation behavior of hcp Ti is summarized. It is found that the hardening rate (i.e., the slope of CRSS vs. Al concentration curve) of prismatic slip is higher than that of basal slip. We attributed this phenomenon to the preferred Al atomic substitution on prismatic plane of hcp Ti (supported by ab initio calculation). Now we receive a comment as follows:
No matter the Al atoms are randomly distributed or form short-range order, both basal and prismatic slips have to overcome larger CRSS due to the distortion of the lattice. If Al atoms substitute regularly on a prismatic plane, then all basal slips and 2/3 of prismatic slips have to cut though the Al plane and 1/3 of prismatic slips do not. So the situation is indeed against the main assumption of the manuscript, i.e. substitution on prismatic planes affect prismatic slip more.
Now I want to know (i) the origin of 2/3 and 1/3 percentage, and (ii) whether slip on basal or prismatic plane of hcp Ti only depends on the atomic shear or not. Also, can we interpret this phenomenon using the theory of solid solution strengthening (solute atoms impede the dislocation motion)?
Interesting question.
My instinct is the reviewer is wrong, but I could be completely wrong this needs some thought. I think he is implying that slip on one of the 3 prismatic planes will not intersect itself but will intersect the other two prismatic planes. Whereas the basal plane always intersects all 3 prismatic ones.
But I think he misses the point as a dislocation is more likely to meet one of these Al atoms as it glides on the Prismatic plane that the Al atoms are also on, than by gliding on an arbitrary plane that crosses the prismatic plane with the Al atoms on. Will be dependent on how many of the Al atoms there are etc and I don't think this is as easily solved as the reviewer thinks.
Apologies I couldn't help, I think if you are to suggest what you say about sitting on prismatic planes though you should be explicit in how this effects the dislocation.
Many thanks. At least you provide a better understanding of the origin of 1/3 or 2/3 pencentage. Actually for hcp metals, prismatic Al atoms may be regarded as basal atoms if the Al atoms are randomly distributed. From this viewpoint, we think the reviewer's comments may be not convincing.
hcp consists of four slip system, basal slip {00,1}/, prismatic slip {10,0}/, pyramid 1 {10,1}/ and pyramid 2 {11,2}/. The pyramid 2 may be not cross the Al atom on a prismatic plane.
Hi Hao, I look forward to your publication. It sounds like a nice study. I'd really appreciate it if you could you inform me when it is published so I can read it. Good luck Tom.
Dear Hao Wu,
I think your result is interesting. Being retired I do not have access to research facilities like a library. So my questions arising from curiosity are based on common sense.
1. Am I correct to assume that all Al atoms are in solid solution? Then the amount of Al atoms in solid solution at room temperature is rather small: only about 6 at% at max I guess. Due to the entropy of mixing I would suggest that these Al atoms are randomly distributed on the Ti lattice points. For a non-random distribution supplementary evidence would be helfpful, e.g. via electron microscopy? This independent route would give a confirmation of your ab initio calculation.
2. Could twinning be invoked to explain the observed difference in hardening rate between prismatic and basal slip?
Dear P. Van Mourik,
For the first question, the solid solubility of Al in Ti is 6 wt.%, rather than at.%. The ordeing of solute atoms is energetically driven or favored, and the non-random distribution of Al atoms is hard to be observed by STEM because of similar atom numbers of Ti and Al. Therefore an initio calculation is strongly recommended.
Secondly, we only focus on the slip bahavior, i.e., the effects of solute atoms on the motion of dislocations. In the annealled Ti(Al) solid solution, twinning is not occurred, because there is no tensile stress applied in this work. Also thanks for your suggestions.
Best,
Hao
Dear Hao Wu,
Thank you for your prompt reply. For the moment I accept that you don't invoke twinning as an explanation for the observed differences. You did not mention the actual composition of your alloy under study. Supposing that the Al amount present is about 6 wt % Al another question remains: can you exclude the presence of Ti3Al phase? See also attached link.
http://www.calphad.com/titanium-aluminum.html
Dear P. Van Mourik,
Sorry. The attached website may be shielded by our network.
The Al contents are 6.5at.%, 7.5at.%, and 10.2at.% (no more than 6.6wt.%), respectively, for three kinds of Ti-Al alloys in our work. Actually, till now, the critical solubility for the precipitation of Ti3Al at room temperature is not strictly clear. The reviewer also suggested us to examine the microstructure of Ti-10.2%Al alloys by TEM, which is the aim of next work.
Best,
Hao
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