What is the Dilatation of a Cube Under Extreme Pressure?

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In summary, the question is asking what the dilatation of a cube would be if it is subjected to a large pressure that causes its pores to close completely. With an initial volume of 1, if 60% of the cube is porous, the new volume would be 0.4. Using the formula ΔV/Vo, the dilatation would be -0.6. However, if the deformation is small, the dilatation could also be calculated by adding the strains, which in this case would equal 0.2.
  • #1
Nexttime35
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Homework Statement


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?


Homework 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)

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.
 
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  • #2
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?
 
  • #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.
 
  • #4
OK - so if 60% of the initial volume is holes, and you remove all the holes, what percentage of the original volume is left?
 
  • #5
40% of the original volume.
 
  • #6
So you are saying that ##V=0.4V_0## ?
 
  • #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?
 
  • #8
Yes, if the original volume is 1, and you shrink it by 60%, then the new volume should be 0.4, correct?
 
  • #9
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?
 
  • #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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