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Homework Help: Calculating yield stress?

  1. Jul 4, 2014 #1
    I have a cylindrical steel sample tested in tension. It's cross-sectional area is 500 mm2. I am also told that:

    • Force at yield = 205 kN
      Maximum force = 258 kN
      Force at fracture = 200 kN

    I am asked to calculate the yield stress, ultimate tensile stress and fracture stress if the reduction of area at fracture is 40%.


    Yield Stress:

    Yield Stress = 205 x103 / 5 x10-4 = 4.1 x108 Nm-2

    Ultimate Tensile Strength

    Ultimate Tensile Stress = 258 x103 / 5 x10-4 = 5.16 x108 Nm-2

    Fracture Stress

    For this question I reduced the 5 x10-4 by 40% to get 3 x10-4. I then did 200 x 103 / 3 x10-4 = 6.7 x108 Nm-2.

    Are my answers correct? Thanks.
  2. jcsd
  3. Jul 4, 2014 #2


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    The method looks OK.

    Stress values are usually given in MPa, not Nm-2.
  4. Jul 4, 2014 #3
    Hi AlephZero,

    Isn't fracture stress usually reported based on the original cross section (so called engineering stress), rather than the actual cross section (so-called true stress)? That would be consistent with how yield stress and ultimate stress were calculated by the OP. What is your experience in this regard?

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