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Yield strength

  1. Feb 13, 2005 #1
    Can someone help me explain how a plate of metal's yield strength changes as it is 1) Rolled (flattened) then 2) Heated for some hours and cooled, and then finally 3) Cooled and rolled additionally.

    Does the yield strength increase at first, then drop, and then increase again? That's what I think, but I'm not sure. And if so, can someone please explain to me why this happens? Thank you. :shy:
  2. jcsd
  3. Feb 13, 2005 #2


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    You're correct.

    If the "metal" you are referring to is being "rolled" at a temperature less than or equal to 0.4 times its absolute melting temperature, then this qualifies as "cold working". One of the key points of cold working is that the grains which make up the microscopic structure (but on a larger scale than the crystal, atomic structure) of the metal are flatened and elongated in the direction of deformation. That changes the area of the grain boundaries in some way (I can't remember exactly how) thus changing the strain energy. This deformation, combined with the increase in the number of dislocations (errors in the crystal lattice) causes the material to become stronger and harder. It also becomes more brittle. But that seems counterintuitive. Why would more defects cause the material to become stronger? Because the mechanism behind plastic deformation is the *motion* of dislocations. Many dislocations confine each others movement, preventing this motion, or slip. The overall strengthening effect of cold working is known as strain hardening. Those two terms can be regarded as synonymous, I think. I'm not going to explain to you how annealing the metal i.e. subjecting it to a heat treatment, has the reverse effect. I encourage you to look it up. All I'm going to say is that by annealing the metal, you can change its structure, and therefore its mechanical properties, back to the way they were before cold working. The mechanism for this transition involves three processes: recovery, recrystallization, and grain growth. For future reference, there is a dedicated materials engineering forum, and the guys there probably know a lot more than I do. But if this is hw, you should try to actively seek the answers to these conceptual questions.
  4. Feb 13, 2005 #3
    Wow, thanks Cepheid! Another question for you; how should I alter the process I described above in order to achieve the best possible yield strength? Is the best thing to do to abort the process after "rolling" it the first time, or is there another better way? And while I'm writing, how does the grain-size affect the yield strength in the end? I think the larger the grains, the lower the yield strength gets. Or? :uhh:
  5. Feb 13, 2005 #4


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    As Cepheid pointed out, cold working increases yield strength (increases the elastic range). The objective in cold working is to change dimension in one direction while increasing in the others - as in rolling a billet to a slab or a slab to plate or plate to strip.

    At some point, a level is reached at which the material may crack. Before this occurs, the material can be annealed (heated) to the 'recrystallization' temperature, where by the dislocation density decreases - by nucleating new grain boundaries.

    For large quantities of material, it is desirable to 'hot work', e.g. hot forge, the material. This avoids the contraints imposed on cold-working, and saves energy by not having to 'reheat'.

    Annealing between cold-work steps is critical for hcp materials like Ti and Zr alloys. During cold work, hcp materials develop elongated grain structures. Heating to recrystallization causes the elongated grains to form into equiaxed grains (dislocations form new grain boundaries) and the material is 'softer' and easier to deform without cracking.
  6. Feb 14, 2005 #5
    Thanks alot!
    A project yields these results; Time for 50% recrystallization t = 1 min at 210 degrees celsius. At 144 degrees it is 100 min. The question is then; what will the time be at 84 degrees?
    I have this formula; [tex]\frac{1}{t}=Ae^{-\frac{\Delta G}{RT}}[/tex] where R is given as 8.3 J/K and A and delta G are said to be constant at all temperatures, but I have no idea what they are! Anyone know how I can solve this problem? I don't even know where to start... :confused:
  7. Feb 15, 2005 #6


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    You have 2 unknowns, but are given sufficient data to write down 2 equations. Solve for these 2 unknowns from the equations. Having thus determined A and \Delta G, you can find the recrystallization time at any temperature.

    PS : Notice that the equation you have is just that for the half-life of first order reaction.
    Last edited: Feb 15, 2005
  8. Feb 15, 2005 #7


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    One way to tackle the equation is take the ln of both sides and make it linear.

    [tex]\frac{1}{t}=Ae^{-\frac{\Delta G}{RT}}[/tex]


    [tex]ln \frac{1}{t}= ln (Ae^{-\frac{\Delta G}{RT}})[/tex]

    [tex]-\ln{t}=ln A -\frac{\Delta G}{RT}[/tex]
  9. Feb 15, 2005 #8
    Again, thanks for all the help. I'm really stuggekling with this! I'm having some trouble determining how the grainsize, and the grainshape changes along the process I wrote at the top of this thread. I didn't think the size and shape of the grains changed at all during cold deformation?
  10. Feb 15, 2005 #9


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    Ferritic, martensitic and austenitic stainless steels exhibit the so-called Hall-Petch relationship between yield strength and grain size.

    Generally - [itex]\sigma[/itex]y = [itex]\sigma[/itex]o + kyd-1/2, where[itex]\sigma[/itex]y is the yield strength, [itex]\sigma[/itex]o is some constant (reference strength), ky is a proportionality constant, and 'd' is the grain diameter.

    Google on "Hall-Petch","steel"

    Hall–Petch behaviour of 316L austenitic stainless steel at room temperature
    http://www.ingentaconnect.com/content/maney/mst/2002/00000018/00000002/art00007 (registration required, but article free).
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