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Aluminium alloys.

  1. Sep 3, 2010 #1

    vanesch

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    Hello,

    Does anybody know whether the alloy elements (Cr, Mg, Mn, Fe) in the aluminium alloy Al - 5083 are mainly interstitial or substitutional. The reason I want to know this is that this has an effect on the thermal neutron diffraction of this material...

    thanks!
     
  2. jcsd
  3. Sep 3, 2010 #2

    Astronuc

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    http://www.keytometals.com/article55.htm

    Aluminum and aluminum alloys
    Physical Metallurgy
    See Figure 1 and Table 1.

    Mg has significant solubility, and Mn is sparingly soluble, and Cr is less soluble than Mn, and Fe less so. Solubility is strongly a function of temperature also.
     
  4. Sep 5, 2010 #3

    vanesch

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    Thanks!

    I guess that "solubility" means "substitutional" in the crystal lattice, although that doesn't have to be the case, right ? It is just that thermodynamically, it doesn't imply a phase separation, but it could just as well be interstitial as substitutional ?

    I'll explain you why it makes a difference. The scattering cross section of a (thermal) neutron on an atom has a coherent and an incoherent part. The coherent part is the one that makes diffraction patterns, the incoherent one is essentially 4 pi uniform diffusion without any interference pattern.

    Now, I'm supposed to find out (compare measurements and model) the behavior of relatively thin aluminium alloy plates on a thermal neutron beam.

    Point is, for pure aluminium, you have a well-defined diffraction pattern (the coherent cross section and the crystal lattice), and a well-defined incoherent diffusion pattern (and also some absorption but that's easy).

    But if you have an alloy, if the alloy atoms are interstitial, their "coherent" part of the diffusion cross section will behave incoherently because of the "random" positions (there is no coordinated interference of the waves), so you can consider the entire diffusion cross section as incoherent (you add the "coherent" and "incoherent" together, as the coherent part will also be "randomized" and hence behave incoherently).

    However, if it is substitutional, the situation is more complicated: part of the coherent diffusion on the substitutional atom will behave as if it were an aluminium atom, and part of it will be act as if it is incoherent.

    If you do the calculations, this makes a substantial difference in what happens to the beam.

    I know that in nuclear power applications, people usually don't make any difference between coherent and incoherent diffusion, and treat everything as incoherent, as an approximation, but in my case, it makes a difference.
     
  5. Sep 5, 2010 #4

    Gokul43201

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    Correct.

    In this case, however, the listed elements are almost certainly substitutional (all of them) or more specifically, they are not interstitially substituted. Al-5083 is about 95% Al. There are two likely scenarios for the microstructure:

    1. It is a single phase FCC alloy (like the [itex]\alpha[/itex]-phase of the Al-Cu binary alloy),
    2. It is some eutectic-like multi-phase alloy, composed of a mixture of primarily the alpha-phase along with small amounts of different intermetallic compounds formed with/by the less soluble elements (like Fe). You can estimate approximate upper bounds for the fraction of these intermetallic phases in the alloy if you know the exact composition.

    Here's the typical composition of generic Al-5083:
    Code (Text):
    Aluminum      Balance
    Chromium    0.05 - 0.25
    Copper          0.1 max
    Iron            0.4 max
    Magnesium   4 - 4.9
    Manganese   0.4 - 1
    Remainder Each  0.05 max
    Remainder Tot   0.15 max
    Silicon     0.4 max
    Titanium    0.15 max
    Zinc            0.25 max
    If we approximate this as a 2-component (Al-Mg) system it is almost certainly a single phase FCC, which is a close-packed structure and therefore has pretty small interstitials. There are two kinds of interstitial sites in an FCC lattice: the octahedral site (radius about 40% of Al atomic radius), and the tetrahedral site (radius about 20% of the Al radius).

    200px-Sites_interstitiels_cubique_a_faces_centrees.svg.png

    Al is already a pretty small atom with an atomic radius of about 120pm. It can really only accept the very small elements (H, O, N, C) interstitially. Approximate atomic radii of Cr, Mg, Mn and Fe are respectively 170, 140, 160 and 160pm (the radii in a crystal vary with co-ordination number, but the atomic radius is a good enough approximation for now). Mg is certain to be substitutional in this alloy.

    Look at the well-studied Al-Cu alloy system for reference (Cu has an atomic radius of about 150pm) - where Cu dissolves substitutionally in the Al lattice up to about 0.3 atomic % (much more at higher temperatures - see here). Beyond that, you get into a 2-phase region with small inclusions of an Al-Cu intermetallic compound.

    If the composition of your plates are close to that in the table above, my guess is that you will almost certainly have a dominant alpha-phase with all the Mg dissolved substitutionally in the Al (as well as some of the Cr, Fe, Mn, etc). But in addition, there will likely be some tiny inclusions of Al3Fe and/or Al12Mg2Cr (and probably some others as well).

    In your case, the inclusions will affect the neutron scattering quite differently from the alpha-phase, though they will have a pretty small cross-section. But it is highly unlikely, no matter what the actual composition, that any of the elements will incorporate interstitially in the Al-lattice.
     
    Last edited: Sep 5, 2010
  6. Sep 5, 2010 #5

    Astronuc

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    As Gokul indicated, I believe the alloying elements are largely substitutional, but it's more complicated than that. There are other secondary phases or intermetallics formed.

    When one refers to a pure Al alloy, is one also referring to a fully annealed state as opposed to cold worked. And is the material single crystal or polycrystalline.

    Looking at Al -

    •Space group: Fm-3m (Space group number: 225)
    •Structure: ccp (cubic close-packed)
    •Cell parameters:
    ◦a: 404.95 pm
    ◦b: 404.95 pm
    ◦c: 404.95 pm
    ◦α: 90.000°
    ◦β: 90.000°
    ◦γ: 90.000°

    Mg

    •Space group: P63/mmc (Space group number: 194)
    •Structure: hcp (hexagonal close-packed)
    •Cell parameters:
    ◦a: 320.94 pm
    ◦b: 320.94 pm
    ◦c: 521.08 pm
    ◦α: 90.000°
    ◦β: 90.000°
    ◦γ: 120.000°

    •Space group: Fd-3m (Space group number: 227)
    •Structure: diamond [??]
    •Cell parameters:
    ◦a: 543.09 pm
    ◦b: 543.09 pm
    ◦c: 543.09 pm
    ◦α: 90.000°
    ◦β: 90.000°
    ◦γ: 90.000°

    In the Zr-Sn binary alloy, Sn is a solution solution element.

    Zr is hcp, but Sn is tetragonal.
    Sn
    •Space group: I41/amd (Space group number: 141)
    •Structure: tetragonal
    •Cell parameters:
    ◦a: 583.18 pm
    ◦b: 583.18 pm
    ◦c: 318.19 pm
    ◦α: 90.000°
    ◦β: 90.000°
    ◦γ: 90.000°

    I believe elements like H, C, N, O and some others are predominantly interstitial, while others are predominantly substitutional, to varying degrees.

    In Zircaloys, Zr-Sn-(Fe, Cr, Ni, O, Si), the elements Fe, Cr, Ni are sparingly soluble, while Sn is quite soluble. There are two predominant secondary phases, Zr(Fe,Cr)2 and Zr2(Fe,Ni), as well as Zr3Si as a very small population on which the other two predominant phases precipitate, and then there is ZrO2 dispersed, and perhaps some Zr3O, or ZrxOy. ZrO2 on the surface can be tetragonal or monoclinic depending on conditions. Zr3O can be found at the metal oxide interface. If hydrogen is present then Zr hydrides (ZrH2 or more likely ZrH1.6 - ZrH1.7 may form.

    With irradiation, one will find amorphous Zr(Fe,Cr) and Zr(Fe,Ni) with Fe being the more predominant element in the amorphous region, but that may also depend on the temperature.
     
  7. Sep 5, 2010 #6
    If I understand correctly, to a first approximation, an annealed metal has its alloy elements as substitutional if their diameter is similar to the principal metal, and interstitial if they are somewhat smaller.
     
  8. Sep 5, 2010 #7

    Gokul43201

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    As a general rule of thumb, yes. Some crystal structures have bigger interstitial voids than others: in a simple cubic lattice you can squeeze in an atom in the central void that is nearly as big as the solvent atom. Close-packed structures, like the fcc, on the other hand, have the smallest voids.
     
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