Q1. Many papers report that mainly appear (111) in very thin foils of stainless steel, nickel or aluminum.
Does it promote the biaxial (111) orientation of crystals in metallic thin foils of FCC under rolling stress conditions especially tensile stress, or stress relax tender to (111) reconstruction during high temperature annealing?
Q2. If the idea in Q1 is correct, why does to get (100) orientation at high temperature annealing of the FCC copper foil?
Many researcher reported that (111) is more stable than (100) orientation in FCC structure thermodynamically.
However very thin copper foil becomes to all (100) after annealing, which previously had coexisted (100) and (111) in copper foil.
I don't know the reason exactly.
What am I confused about somewhere?
Dear Angelo Oñate ,
Thank you for your kind advice.
I Have deleted BCC in Q1, as following your advice Q1.
For Q2, the orientation is after applying during high temperature annealing above 800℃, the orientation is (100) dominantly for copper foil.
However sometimes copper foil has (111) orientation dominantly by very weak tensile stress for thin foil .
The tensile stress is just as self weight load by hanging the foil in furnace.
The annealing condition is almost same, except applying weak tensile stress.
Dear Angelo Oñate,
Annealing conditions are almost same, vacuum, temperature and time.
Just foil insert conditions are different.
The foil is inserted bent into a round quartz tube, and the tube is charged horizontally in a bigger quartz tube furnace.
Another method is to fold the copper foil and hang vertically it on the upper edge of quartz plate.
And the quartz plate is inserted into a round tube. Heating is same as the tube furnace.
The difference of two methods is that the copper foil is usually inserted bent into the round tube, and the other copper foil is folded and hung vertically on the upper edge of the quartz plate.
I guess that just very small difference of stress level between the round bending and the hanging vertically.
Kind regards,
Kang-Hyung Kim
Dear Angelo Oñate,
Thank you for your kind advice.
I agree that the temperature must be high enough for kinematics to work.
So I modified in detail Q1.
In addition to thermal stress, the copper foil has a rolling stress that is expected to require high temperature and time until the stress is relaxed.
What I don't understand is that the bent copper foil becomes (100) dominant orientation during annealing, which is a different behavior from other FCC metal foils, and this is common sense in most copper thin foil annealing.
However, the same copper thin foil is annealed vertically down, which is why it becomes (111) dominant. This is not usual phenomenon.
Best regards,
dear dr. Kim,
I will answer here regarding your question/answer of another topic, since it seems to be better answered here.
Yes, inner energy U is driving force for those things. U=TS-pV+mueN is a function of extensive variables S(entropy), V(volume) and N(amount of material). T(temperature), -p(p: pressure, can be negative for solid material), and mue(chemical potential) are 1st partial derivatives of U. Note: U is defined for some material having some phase structure. Other phase structure, other function U. Now look for stability of phases and according transformation. Here we need to change coordinates first, since simply looking for lowest U does not work. We can transform to enthalpy H(T,p,N)=U+pV (dU=TdS-pdV+muedN =>dH=TdS+Vdp+muedN), called Legendre-Transformation and find H=TS+mueN. We transform a 2nd time towards Gibbs enthalpy G(T,p,N)=H-TS (now dG=SdT+Vdp-muedN). Here we can compare G(T,p,N) for different phases or different orientation for some fixed given T and p and can state that the material always tends to go to that structure having the lower value of G(note: N is just a multiplier, simply divide by N first, because comparison always needs to be made by same amount N. It is allowed to do comparison only for same T and p, never compare e. g. for different T to gather driving forces.). More clear:
Compute G1(T,p,1) for structure 100
Compute G2(T,p,1) for structure 111
compare for some given p and T, and then you know what state the material wants to be. 100 or 111.
The material always wants to take the lower one. It cannot be explained easily why this works. Much more theoretical background is needed to understand that.
The difference works as driving force for dynamic behaviour also (if compared for same T and p). For thin layers, you need to consider surface energy, which depends on orientation and temperature. For tempering, it can be needed that some nucleids are necessary to start transformation, i. e. some minimum difference depending on number of defect places is needed.
This is a very compressed answer for some very complex physics, and according theory is not easy (or in other words: It is very easy to make errors, and then results are wrong). You may read some text books to understand more details. I understood after working through Falk: Theoretische Physik II Thermodynamik (the only book I know that explains correctly why G is working).
It is a rather old book in german, possibly also available in english.
It may happen that not everything happening for copper foils can be explained by thermodynamic theory, or it is too complicated for practical approaches. On the other hand, I was really impressed which effects can be explained e. g. for steel by using thermocalc and dictra, just applying those thermodynamics.
With kind regards
Klaus Weinzierl
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