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Volumetric Change Problem

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Given a cube that is 60% porous, and you subject it to a very large pressure such that the pores close completely, what is the dilatation of the cube?


    2. Relevant equations

    Dilatation (Δ) = ΔV/Vo where Vo is the original volume.
    Vnew equals = (∂x-εxx∂x)(∂y - εyy∂y)(∂z - εzz∂z)

    εxx= change in side parallel to x-axis/ (original side)

    3. The attempt at a solution

    Say the original side, x=1, and since we're working with a cube, then (∂x-εxx∂x)3 = V .

    So, Solving for εxx = (0.4x-x)/x= -0.6.
    Then Dilatation (Δ) = ΔV/V = (1 - 0.6)^3 - 1 / 1 = -0.936

    Is this logic correct? I am trying to solve for dilatation of the cube if it shrinks, say, by 60%.

    Thanks for your help.
     
  2. jcsd
  3. Sep 18, 2014 #2

    Simon Bridge

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    What does it mean to say that 60% of the cube is porous?
    Does it mean that 60% of the volume is holes or something else?
     
  4. Sep 18, 2014 #3
    Yes, say it's sand or something. After applying stress to the cube on all sides equally such that those holes are filled, what is the dilatation? Thus, each new side would be (x- 0.6x) for some side x.
     
  5. Sep 18, 2014 #4

    Simon Bridge

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    OK - so if 60% of the initial volume is holes, and you remove all the holes, what percentage of the original volume is left?
     
  6. Sep 18, 2014 #5
    40% of the original volume.
     
  7. Sep 18, 2014 #6

    Simon Bridge

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    So you are saying that ##V=0.4V_0## ?
     
  8. Sep 18, 2014 #7
    I understand that part of the question, and I have read that if the strains are small, then we can assume that exx+eyy+ezz is equal to the dilatation. So if exx=eyy=ezz and exx = 1-(0.6), then would the dilatation be 0.4*3=1.2?
     
  9. Sep 18, 2014 #8
    Yes, if the original volume is 1, and you shrink it by 60%, then the new volume should be 0.4, correct?
     
  10. Sep 19, 2014 #9

    Simon Bridge

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    That is what you are telling me... so what was that formula for the volume dilatation again?
    $$(\Delta)=\frac{V-V_0}{V_0}$$... is that right? Or is ##\Delta V## not change in volume?
     
  11. Sep 19, 2014 #10
    ΔV is the change in volume.

    Thus, my logic is that since Vnew equals = (∂x-εxx∂x)(∂y - εyy∂y)(∂z - εzz∂z) = (∂x-εxx∂x)3 (since we are working with a cube,

    then Dilatation = Vnew-Vo/Vo, plug in for Vnew.

    However, my text says that if the deformation is so small, then dilatation could just equal εxx+εyy +ezz.
     
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