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Homework Help: Modulus of Elasticity

  1. Aug 23, 2009 #1
    Hello, having some problems with calculating young's modulus, please help.

    1. A Strain gauge records the following strain when a block of material is pulled:

    Force(N) Strain(%)
    100 0.01
    1000 0.1
    2000 0.179
    3100 0.9

    Area of block = 10cmx10cm

    Work out the modulus of elasticity and when the material fails?

    3. The attempt at a solution

    Area over which force acts = 3.14x5x5
    = 78.5cm2

    In inches = 78.5/2.54
    = 30.9 inch2

    Stress = force/area
    = 3100/30.9
    =100.3 psi

    E = Stress/Strain
    = 100.3/0.9
    = 111.4psi
  2. jcsd
  3. Aug 23, 2009 #2


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    Perhaps first study how to compute the area of a square. If the material cross-sectional area is not a square, please clarify the problem statement. Also, try to avoid converting to a nondecimal, nonstandard, incoherent measurement system. Just convert cm to mm, and all stresses will be in MPa. Also study how to compute the slope of a straight line.
  4. Aug 26, 2009 #3
    Thanks for the advice. I have made another attempt.

    Area = 10x10
    = 100cm2 or 10000mm2

    E = Force applied x original length/area of cross section x change in length
    = 1700 x 100/10000 x 0.17
    = 100 Mpa

    Not sure where I should use slope of straight line calculation.
  5. Aug 26, 2009 #4


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    boyblair: Yes, use 10 000 mm^2 for the area. You could compute the stress (force divided by area) at point 1 or 2. Then, to obtain E, in your particular case, you could divide stress by strain at point 1 or 2. I don't quite understand from where you got 1700 N and 0.17 %, but you somehow got the correct answer, nonetheless (except the unit symbol for megapascal is spelled MPa).
  6. Aug 26, 2009 #5


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    You are going to have to clarify the problem. Is the length of the bar given? Strain is a dimensionless quantity (change in length divided by original length). You indicate it as a percent; if that's the case, under the 100N load, for example, the strain is 0.0001, and the change in length (elongation) is 0.01 mm, if the bar is 100 mm long. The stress strain curve is linear for the first 2 load cases. I'd use one of those values for determining E. Then the strain goes way up (non linear) under the 3100N load. Is that the failure load? This value of strain under that load condition should not be used for determining E. Then be consistent in determining E. You can use either of your formulas
    (E = stress/strain or E = FL/A(elongation)), but watch your units and values to use.
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