What is the Dilatation of a Cube Under Extreme Pressure?

[Nexttime35]

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/V_o where V_o is the original volume.
V_new equals = (∂x-ε_xx∂x)(∂y - ε_yy∂y)(∂z - ε_zz∂z)

ε_xx= 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-ε_xx∂x)³ = V .

So, Solving for ε_xx = (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.

 

[Simon Bridge]

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?

 

[Nexttime35]

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.

 

[Simon Bridge]

OK - so if 60% of the initial volume is holes, and you remove all the holes, what percentage of the original volume is left?

 

[Nexttime35]

40% of the original volume.

 

[Simon Bridge]

So you are saying that $V=0.4V_0$ ?

 

[Nexttime35]

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?

 

[Nexttime35]

Yes, if the original volume is 1, and you shrink it by 60%, then the new volume should be 0.4, correct?

 

[Simon Bridge]

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?

 

[Nexttime35]

ΔV is the change in volume.

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

then Dilatation = V_new-V_o/V_o, plug in for V_new.

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