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Normal Strain

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    ε = xe-x2
    L = L

    2. Relevant equations

    ε = ΔL/L


    3. The attempt at a solution

    xe-x2 = (ΔL - L)/L

    ∴ ΔL = Lxe-x2 + L


    I really don't understand how the solution got "dx" on the first step, also the delta L. I want to understand the first step, the rest of the solution I understand. Thanks!
     

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    Last edited: Apr 10, 2014
  2. jcsd
  3. Apr 10, 2014 #2
    I've been around stress and strain all my professional life, and I can tell you that this is an unbelievably poorly worded problem. I think what they are trying to say is that the axial strain ε is a function of the distance x from the left end of the bar. If u is the axial displacement of a material cross section from its initial location before the bar was strained, then the displacement is related to the strain by du/dx = ε. You integrate this equation from x = 0 to x = L to get the axial displacement of the right end of the bar.

    Chet
     
  4. Apr 11, 2014 #3
    Thank you for your response,
    I'm sorry, I still don't understand how you are getting du/dx = ε, I understand up to the point where you say, "is related to the strain". The rest I understand, thanks!

    Is it just that du is the extension and dx is the original length of a segment of length?

    But then what is δL? Is that du?
     
    Last edited: Apr 11, 2014
  5. Apr 11, 2014 #4
    What this problem is trying to do is to show you what to do in cases where the local strain in the bar is non-uniform.

    Let x represent the axial location of a material cross section before the non-uniform strain is imposed, and let x + u(x) represent the location of the same material cross section after the strain is imposed. The parameter u is called the local displacement. Suppose we focus on the short segment of material that was originally between x and x + dx in the unstrained bar. The original length of this short segment of material was dx. If we now focus on this same short segment of material after the non-uniform strain is imposed, the new length of this same short segment of material is now (dx + du). The change in length of the short segment of material is du, and its original length was dx, so the local strain of the material is du/dx (the change in length divided by the original length). So,

    ε(x) = du/dx

    This equation is called the "strain-displacement" relationship for the non-uniform deformation.

    Chet
     
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