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Calculating Yield Strength using a Load vs. Displacement Curve

  1. Apr 7, 2010 #1

    This question came up on my midterm and I had no idea how to answer it without redrawing the entire curve as a stress vs. strain curve (which obviously took too long to do).

    Anyway, I'm just requesting a general procedure, not a numerical answer. If you had a Load (y-axis) vs. Displacement (x-axis) curve, how do you calculate the yield strength? I know for a stress vs. strain curve, you have to have a 0.2% (0.002) offset, but my Professor told me that the offset is not of the same value in a Load vs. Displacement curve (which makes sense).

    However ... what <i>is</i> the actual offset supposed to be? Is there a formula to calculate it?

    Any help would be very much appreciated! :)

    Thank you,
  2. jcsd
  3. Apr 7, 2010 #2


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    The 0.2% (0.002) is a value of strain, which is arbitrary, but it covers somewhat the uncertainty of when a given material actually departs from the purely linear (elastic) relationship between stress and strain.

    Let l = length, then strain ε = (l - lo)/lo, where lo = original length (usually the gauge length). Also, the displacement, d, is given by (l - lo), so strain ε = d/lo.

    Similar the stress, σ, is just the load/force F divided by area A, i.e. σ = F/A, where A is constant until the specimen reaches UTS, which corresponds to the limit of uniform elongation, and necking begins.

    Now back to length/displacement -

    with ε = (l - lo)/lo, one rewrites the equation as ε = l/lo - 1, and reorganizing the terms, l = lo (1+ε),

    so the length equivalent to the strain offset of 0.002 is just l = lo*1.002, or d (0.002) = 0.002 lo.
    Last edited: Apr 11, 2010
  4. Sep 13, 2010 #3
    So, if I have the following data:

    A = 0.36m2
    F = 9800N @ 0.2% offset

    Then [sigma] = 9800/0.36 = 2.7*104 Pa ??

    I just have a problem regarding the constant area. Wouldn't the cross-sectional area be decreasing while the object is strained (stretched)?
  5. Sep 14, 2010 #4


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    That gives you the Engineering Stress (which is always a function of the original cross-sectional area).

    Yes. The stress in this case is called the True Stress and is a function of the instantaneous minimum cross-sectional area of the specimen.

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