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B Poisson's ratio, steel rod for example

  1. Jun 24, 2017 #1
    Steel has a "Poisson ratio" μ of 0.29 (in my book of problems at least). Am I understanding the meaning of this ratio correctly: if a steel cylinder is stretched along its symmetry axis by a factor p (something like 1.01 or 0.99) then the radius will change by a factor 1-μ(p-1)?
    [So then the volume will go like p(1-μ(p-1))2 and the density inversely?]
     
  2. jcsd
  3. Jun 24, 2017 #2
    I think, your understanding, calculation and inference is correct.
     
  4. Jun 25, 2017 #3
    Is μ really constant for strains in the elastic regime, or is it just the lowest order coefficient for small strain?

    And presumably this holds for general (not just circular) cross sections?
     
  5. Jun 25, 2017 #4
    So I stretch/compress a rod by a factor p(≈1) then I will have V=Vop(1-μ((p-1))2 ... If we let dV/dp be zero then we'll get 2pμ = 1. I take this to mean a rod of material μ≈0.5 will approximately maintain it's density for small strains? (Taken a bit farther, a rod of material μ will approximately maintain it's density for small additional strains about the strain p=1/(2μ), assuming of course p is not so large that theory (or the rod!) breaks down.)

    How can I understand the theory of all this? If I stretch a rod I do work, presumably to separate the molecules a bit? The density is maintained in the above case but the average intermolecular distance changes? How do you generally analyze the energy in a strained object? [etc...]

    If anyone has insight or ideally a book where I can study these theories, it would be appreciated.
    (The book I mentioned in the OP is only problems, of many kinds.)
     
  6. Jun 26, 2017 #5
    Ok then...

    What if I stretch a rectangular-bar in all three directions normal to it's faces. Then am I correct in thinking to find the change in length along one direction we would have to consider the 'poisson-effect' from the other two directions?

    This effect becomes particularly simple if you assume uniform pressure on all faces, so let me pose the above question in that context:

    Suppose a rectangular-bar is subjected to a uniform pressure P.
    (The material of the bar has Young's modulus E and Poisson's ratio μ.)

    Then what I am asking is if the change in length will look like this:
    ΔLx=(P/E)(-1+2μ)Lx for each (principle) dimension?
     
    Last edited: Jun 26, 2017
  7. Jun 26, 2017 #6
    Is ## \rho ## in the formula density of the material? Do you have a reference to this definition of Poisson ratio?
    The one I know is just the ratio of the strains, like here.
    https://en.wikipedia.org/wiki/Poisson's_ratio
     
  8. Jun 26, 2017 #7
    It was an arbitrary p not rho:
    Anyway the OP is equivalent to 'the ratio of strains' definition, so you can ignore the first post.

    What I would like to know now is if we can superimpose the strains as described in post #5; that is, if we compress the bar along the x direction, it will expand in the y and z directions, but now if we compress the bar in all three directions, do I just superimpose the effect of each compression? (Yielding ΔLx=(P/E)(-1+2μ)Lx?)

    Also I would like to understand the general theory better, such as the question of analyzing the energy in post #4. I suppose this falls under continuum mechanics? I'm just having trouble finding relevant theoretical material.
     
  9. Jun 26, 2017 #8
    Yes, the effects are additive in the linear regime. A document with the general theory is for example this:
    http://web.mit.edu/16.20/homepage/3_Constitutive/Constitutive_files/module_3_with_solutions.pdf
    If this is too general and you are interested just in the the isotropic case, this is treated on page 62.
    You can see that the strain along one direction ( ## \epsilon_{11} ## for example) is given by contributions from the three stress components (## \sigma _{11}, \sigma _{22}, \sigma _{33}## ).
    If you just have the same pressure applied on the sample (like in so called isostatic pressing) , then the three stress components are the same and equal to the pressure.
     
  10. Jun 27, 2017 #9

    Orodruin

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    I would say what the OP really wants in this case is on page 66 and note that it is true not only in the case of hydrostatic pressure. Taking the trace of the relation between the stiffness and stress tensors, you will always find (assuming a homogeneous isotropic material) that the volumetric strain is related to the stress tensor as
    $$
    e = \epsilon_{ii} = \frac{1}{E}[(1+\nu)\sigma_{ii} - \nu \sigma_{kk} \delta_{ii}] = \frac{1-2\nu}{E} \sigma_{kk} = \frac{3(1-2\nu)}{E} \frac{\sigma_{kk}}{3}
    = \frac{\sigma_{kk}}{3K},
    $$
    where ##K## is the bulk modulus. Subjecting the material to hydrostatic pressure would lead to ##\sigma_{kk} = 3p## and therefore ##e = p/K## as stated in the text, but the linearity is also clear since the trace ##\sigma_{kk}## is just the sum of the principal stresses.
     
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