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Strength Coefficient and Strain-Hardening Exponent

  1. Aug 28, 2006 #1
    Here's the problem:
    A metal is deformed in a tension test into its plastic region. The starting specimen had a gage length = 2.0 in and an area .50 sq. in. At one point in the tensile test, the gage length = 2.5 in and the corresponding engineering stress = 24000 psi; at another point in the test prior to necking, the gage length = 3.2 in and the corresponding engineering stress = 28000 psi. Determine the strength coefficient and strain-hardening exponent.

    To summarize,
    L0 = 2.0 in
    A0 = .50 sq. in.
    L1 = 2.5 in engineering stress = 24000 psi
    L2 = 3.2 in engineering stress = 28000 psi
    Here's my solution:

    *At the first point given, we can find engineering strain by e = (L1-L0)/L0 = (2.5 - 2.0)/2.0 = .25
    -We can then use the engineering stress and the engineering strain to find the true stress by the relationship:
    true stress = engineering stress (1+engineering strain) = 24000(1+.25) = 30000 psi
    -The true strain is given by ln (L1/L0) = ln (2.5/2.0) = .22

    * For the second point, engineering strain = (L2-L0)/L0 = (3.2-2.0)/2.0
    = .6
    - Hence, true stress = engineering stress (1+engineering strain) =
    28000(1+.6) = 44800 psi
    - True strain is ln (L2/L0) = ln (3.2/2.0) = ln 1.6 = .47

    *So we now have two points on our true stress-strain curve: (.22, 30000) and (.47, 44800).
    The book plots these points using a log scale and then finds the slope to determine the strain hardening exponent. So n = (log 44800 - log 30000)/(log .47 - log .22) = .52
    Does this seem right...I know n for most metals is between 0 and .5, so is .52 a reasonable result???


    *The strength coefficient, K, equals the value of true stress at a true strain value equal to 1.
    Using the point slope equation with the x-coordinate (true strain) equal to 1, I get:
    y - y1 = m(x-x1)
    log y - log 30000 = .52 (log x - log .22)
    log y - log 30000 = .52 (1 - log .22)
    log y - log 30000 = .8619402
    y = 218304 psi
    K = 218304 psi

    Does this seem correct? It seems a little large!

    Thanks for reading! Any help much appreciated!




    I apologize for
     
  2. jcsd
  3. Apr 1, 2011 #2
    Calculations for n look fine.. I'd say the question is the problem ;)

    In fact I'm surprised the answers are even close to reasonable, I'd be worried any time you'd have to do a straight line fit between just two points.

    Most the time I have to calculate a slope from data I use 3 to 5 points to smooth out results due to inevitable noise/errors

    https://www.physicsforums.com/showpost.php?p=1460020&postcount=2

    yeah I think this is the error here.. with log 1 you get something like 123 psi
     
    Last edited: Apr 1, 2011
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