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Prove Poisson's Ratio is .5 for Small Strains

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Poisson's ratio v is the negative ratio of the transverse strain to the axial strain. For an incompressible (density doesn't change), homogeneous (everything is the same molecule), isotropic (it doesn't matter which direction you pull or push on it, it will act the same way) linear elastic (Hookean) material, v = .5 for small strains. Using a first order approximation (sort of like a Taylor Series approximation, prove v = .5.

    2. Relevant equations
    strain = ε = ΔL/L0

    Poisson's ratio = - εT / εA= - (ΔLT/LT) / (ΔLA/LA) = -(ΔLTLA)/(ΔLALT)

    ΔVolume = 0

    3. The attempt at a solution

    I drew a picture of a cube with original dimensions lwh = LA3. Then I applied a transverse strain to the box and drew a new box inside and extending from the original box. The new box's dimensions are (ΔLA + LA)(LA - ΔLT)(LA - ΔLT) assuming that ΔLT and ΔLA are positive values. Since the change in volume is zero, the amount the box "shrinks" and the amount the box "extends" should be equal, so I tried summing the new volumes and setting them equal to zero:

    0 = -(Volume that the cube shrunk) + (volume that the cube elongated)

    0 = - (2(ΔLT)(LA) + (ΔLT)2LA) + ΔLA(LT - ΔLT)

    2(ΔLT)(LA) + (ΔLT)2LA = ΔLA(LT - ΔLT)

    I tried to factor out the terms that I wanted to be the numerator and denominator:

    (ΔLTLA)(2 + LT) = ΔLALT - ΔLAΔLT

    (ΔLTLA)(2 + LT) + ΔLAΔLT = ΔLALT

    Divide both sides by ΔLALT

    v(2 + LT + ΔLT/LT = 1

    v = (1 - ΔLT/LT)/(2 + LT)

    Simplifying...

    v = (LT - ΔLT) / (LT(2 + LT)

    What should I be doing differently?
     
  2. jcsd
  3. Nov 8, 2014 #2

    OldEngr63

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    This problem makes an interesting assertion about Poisson's ratio. It is particularly interesting since, for common metals like steel, Poisson's ratio is about 0.3.

    Poisson's ratio = 0.5 applies to a material such as rubber, but not to most metals.
     
  4. Nov 8, 2014 #3

    haruspex

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    That last equation is dimensionally wrong. It mixes quadratic and cubic powers of length.
    Once you have the correct equation, you need to make an approximation for small deformations. If L >> ΔL, what can you say about L ΔL compared with ΔL2?
     
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