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Homework Help: Finding Yield Strength, E, and TS

  1. Jan 18, 2008 #1
    Heres my given problem:
    Gauge Length = 2 in.
    Area = 0.5 in^2

    The specimen yields under a load of 32,000 lb
    The corresponding gauge length is 2.0083 (the 0.2 yield point)

    Max load = 60,000 lb @ gage length = 2.60 in

    I need to determine the yield strength, modulus of elasticity, and Tensile strength

    This problem has no graph, and this books has no examples

    I tried to answer it by having ys= 32,000 lb/ 0.5 in^2 (6.895x10^-3 Mpa / 1 lb/in^2) = 441.28 MPa

    e= (2.0083 - 2)/2 =.00415

    E = 441.28/.00415 = 106 x 10^3 MPa

    TS = 60,000/0.5 in^2 (6.895x10^-3 Mpa / 1 lb/in^2) = 827.4 MPa

    Im no too sure about my answers because I cant find where it states explicitly how to get the yield strength or what the it is in terms of a formula
  2. jcsd
  3. Jan 18, 2008 #2


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    Gold Member

    Your problem is given in units of pounds and inches, which are quite valid units in the USA and one or 2 other countries, so there is no need to convert to MPa. Pounds per square inch (psi) will suffice nicely.
    The problem seems to have some inconsistencies and missing data and the missing graph. You note the the gauge length at the 0.2 yield point (that is e=.002) corresponds to its yield stress at that point. But the gauge length of 2.0083 corresponds to a strain of 0.004 (as you noted), which is inconsistent with a yield stress definition of stress at 0.2%strain. Then there is the problem of definition of tensile strength; and to calculate E, you realy need to know the proportional limit stress prior to yield(i.e, separation between the elastic and plastic state). Now if the problem were to be simplified to an approximate curve with a straight line stress strain curve in the elastic region, with a slope of E up to its yield point at e=.002, then a horizontal line beyond yield to a certain strain value, then with an increasingly sloped lineto the max tensile stress, then the gage length at max load is not necessary, and you have for the yield strengh, ys=P/A = 32000/0.5 = 64000psi, which you have correrctly calculated (before your conversion to the ''forbidden' pascals!). And for the E modulus, that'd be yield stress over yield strain =64000/.002 = 32,000,000 psi; and your tensile strength is the given max load of 60000 pounds (120,000psi max stress per your correct calc). But I'm just approximating...a graph and more data would help. (the material looks a lot like 60ksi yield steel).
    Last edited: Jan 18, 2008
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