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Working out strain and stress of an elastic band (youngs modulus)

  1. Oct 29, 2012 #1
    This is regarding youngs elastic modulus
    I need help as I am trying to figure out the stress and strain of an elastic band when a force has been applied - in youngs modulus.

    Does anyone know how to go about this?
     
  2. jcsd
  3. Oct 29, 2012 #2
    What do you mean by "in youngs modulus"?
    I suppose you mean "in Young's modulus" but still not clear.
    To calculate the stress you need to know the dimensions of the band (the area of the cross section) and the force.
    Then you can find the strain from Hooke's law (this has Young's modulus in it).
     
  4. Oct 29, 2012 #3
    Hi, thank you for your reply, I think the dimensions of the elastic band are (height: 7cm) (width 4cm) and thickness (2mm)

    So I think the area is (5.6nm)

    I am still not sure how to work out the stress, strain and force
     
  5. Oct 29, 2012 #4
    An elastomeric band is a different kind of animal from a metal bar, the latter of which is characterized in the region of elastic behavior (in extension) by Young's modulus.

    An elastomer behaves differently. Its region of elastic behavior is much larger than that of a metal (in terms of the magnitude of the elastic deformations that can be sustained, followed by full recovery to the initial length). In addition, for an elastomer, the stress is not a linear function of the strain. Finally, because of the larger extensions that an elasomeric band suffers, its cross sectional area decreases significantly.

    The real issue is how do you calculate the strain and the stress for large deformations. For uniaxial extensions like you are interested in, there are several strain measures that can be used and two different measures of stress:

    ε =( l/l0 - 1 )

    or ε =(( l/l0)2 -1)/2

    or ε = (1-(l0/l)2)/2

    where l is the stretched length of the band and l0 is the initial unstretched length.

    The stress can be calculated in terms of the force per unit cross sectional area (the so-called true stress) or in terms of the force per unit initial cross sectional area (the so-called engineering stress):

    σtrue = F/A = (F/A0) (l / l0)

    σengineering = F/A0=(F/A) (l0/l)

    If doesn't matter which measure of strain you choose to work with or which measure of stress you choose to work with. Whatever the case, the stress will be a unique non-linear function of the strain which you can measure experimentally (once and for all for the material in question).

    σ = σ (ε)

    The functionalities between all the various stress and strain measures are known, so you can convert from one strain measure to another and/or from one stress measure to another.

    Elastomeric materials are typically cross linked polymers, and the fundamental deformational behavior of these types of materials was studied in the 1930's and 1940's using statistical thermodynamics. Various mathematical descriptions of the stress-strain behavior of elastomers are presented in the literature, based on these fundamental studies. The most well-known of these is the Mooney-Rivlin model.
     
  6. Oct 30, 2012 #5
    No, the volume is 56 cm^2. The area of cross section will be 4mmx2mm = 8mm^2. This is, assuming that you stretch the band along the 7 cm side.
    You can calculate the initial stress as F/A. This will be the initial or "engineering" stress. However, as Chestermiller have explained in detail, the area changes significantly when you stretch the rubber band.
     
  7. Oct 30, 2012 #6
    Watch those units. Actually the area is 80 mm2 = 0.8 cm2. This is the initial cross sectional area of the sample.

    Do you have data on the measured force versus the measured length? If you do, then calculate the engineering stress by the equation I gave you, and calculate the strain by the first equation I gave you. Then plot a graph of the stress as a function of the strain, and see what it looks like. Then you can decide what kind of parametric relationship would be best for fitting to the data.
     
  8. Oct 30, 2012 #7
    Right. I did not see that the width is in cm. Thank you for correction.
     
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