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Solid Mechanics - Stress-Strain diagram - absorbed energy

  1. Jun 9, 2015 #1
    1. The problem statement, all variables and given/known data
    graph.png

    1.The maximum energy (per unit volume) that can be absorbed by the steel alloy without
    sustaining permanent deformation is _______ lb/in^2

    1.4. The maximum energy (per unit volume) that can be absorbed bythe steel prior to fracturing is _______ kip/in^2.

    .

    2. Relevant equations


    3. The attempt at a solution
    If I remember correctly, stored energy is the area under the graph.
    Since I don't know the equation I will work with estimates.

    First, I used the 0.2 % offset method to find the Yield strength of the material and I found the yield strength to be about 65 ksi.
    The area under the straight line is (1/2)(.002)(60E3)=60
    I estimate the rest of the area as a rectangle: (0.002)(60)=120
    which gives a grand total of 180 lb/in^2.

    Yet the solution, which is not worked out, claims that the energy is only the first part, namely 60 lb/in^2. I thought a material wouldn't be deformed before it gets to the point of yield strength, and in the previous step I found the yield strength correctly. I'm thinking they did something wrong because the solution even goes on to say "Modulus of resilience"=60PSI.

    1.4 should just be the same thing, total area under curve?
     
  2. jcsd
  3. Jun 9, 2015 #2

    SteamKing

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    IDK why you are using the 0.2% offset method to determine yield strength, IIRC, 0.2% offset is used for various non-ferrous metals and alloys (like aluminum) which don't exhibit a clear linear stress-strain relationship in the stress-strain curve.

    For the test results shown in the graph, there is clearly a linear stress-strain relationship up to a stress of approx. 60 ksi.

    http://en.wikipedia.org/wiki/Yield_(engineering)

    Once the material is stressed beyond yield, there will be a permanent set (it's kinda the definition of what yield is.)

    The modulus of resilience is defined in this article:

    http://en.wikipedia.org/wiki/Resilience_(materials_science)
     
  4. Jun 9, 2015 #3
    So now I'm thoroughly confused as to what the difference between yield strength and elastic limit is.... And if I don't use the .2 % method, then how am I supposed to know where the yield point is? It's not given to me in the problem.

    Nevermind, the information was in the wikipedia article, thanks.
     
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