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In order to measure the Poisson ratio of a rectangular sample of elastic material I subject it to a vertical load along its major axis using weights of gradually increasing importance and measure the variation in length,$L$. I assume, perhaps wrongly, that volume is conserved and that the area of each face is conserved. As a result I assume that:

  • $W*L=W_0*L_0$
  • $T*L=T_0*L_0$
  • $W*T=W_0*T_0$

Where $L$,$W$ and $T$ are the actual length width and thickness of the rectangular sample of elastic material and $L_0$,$W_0$ and $T_0$ are the original dimensions. This allows me to approximate the Poisson Ratio in the following manner:

$\nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}\approx -\frac{\frac{\frac{W_0*L_0}{L}-W_0}{W_0}}{\Delta{L}/L}=\frac{\Delta{L}/L}{\Delta{L}/L_0}=\frac{L_0}{L}$

Clearly, this results in a very bad approximation that becomes increasingly worse as the Poisson ratio is decreasing in the linear domain when it should be constant. Given that elastomers tend to have constant(and positive) poisson ratio in their linear domain, does this mean that the volume of an elastic material with a prismatic geometry isn't conserved when subject to a uniaxial strain?

After more reflection, I think this can be easily demonstrated:

  1. $\nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}=-\frac{\Delta{T}/T_0}{\Delta{L}/L_0}$ where $-1 \leq \nu \leq 0.5$

  2. Let $\frac{\Delta{L}}{L_0}=\alpha > 0$ then $ \begin{cases} \Delta{W}=\frac{W_0*\nu*\Delta{L}}{L_0}=-W_0*\nu*\alpha \\ \Delta{T}=\frac{T_0*\nu*\Delta{L}}{L_0}=-T_0*\nu*\alpha\\ \end{cases} $

  3. $V'=L*T*W=(L_0+\alpha L_0)*T_0*(1+\nu \alpha)*W_0*(1+\nu \alpha)=L_0*T_0*W_0*(1+\alpha)*(1+\nu \alpha)^2=V*(1+\alpha)*(1+\nu \alpha)^2$

  4. The equation in part 3 actually reduces to:

$\frac{\Delta{V}}{V}=(1+\frac{\Delta{L}}{L_0})*(1+\nu \frac{\Delta{L}}{L_0})^2-1$

In the special case of a cube with sides of length $L_0$ we have,

$\frac{\Delta{V}}{V}=(1+\frac{\Delta{L}}{L_0})*(1-\frac{\Delta{L'}}{L_0})^2-1$

Clearly, for $\nu \leq 0$, $\frac{V'}{V}>1$ so the volume has increased. And for $\nu >0$, the volume is constant only when $\nu = \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$. Since $\nu$ is supposed to be constant and $\alpha$ is allowed to vary we must conclude that the volume is not constant in this case either.

Now, I think my arguments are reasonable but is my conclusion correct?

Note: If I should trust my arguments then for any given non-negative poisson ratio, there is a single value for $\alpha=\Delta{L}/L_0$ such that $V'=V$. The graph of the function $f(\alpha)= \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$ is given below:

enter image description here

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  • $\begingroup$ How does the result of your calculation compare with the experiment you describe at the start of your Question? What is the purpose of your Question? $\endgroup$ – sammy gerbil Jul 6 '16 at 22:13
  • $\begingroup$ I tried to estimate the Poisson Ratio using the assumption of volume conservation and when I realized that this led to a bad approximation I wanted to understand why. Now, I think I might have found an explanation but I'm not sure it's entirely correct. $\endgroup$ – user29305 Jul 7 '16 at 1:20
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Your calculation makes no reference to the properties of any real material. Therefore the conclusion you make is the consequence of your assumptions, not any properties of matter. Those assumptions (constant volume and surface area) can be wrong, as you acknowledge.

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  • $\begingroup$ After more reflection, I try to show in the question details that volume is never conserved for an elastic material with a poisson ratio. Do you think my arguments are correct? $\endgroup$ – user29305 Jul 6 '16 at 7:45
  • $\begingroup$ @AidanRocke : I am sorry, I don't want to check your calculation. I suggest that you compare it with eg en.wikipedia.org/wiki/Poisson%27s_ratio. $\endgroup$ – sammy gerbil Jul 6 '16 at 17:26
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Your analysis for an incompressible material is just done incorrectly. It should read $$\frac{T}{T_0}=\frac{W}{W_0}=\sqrt{\frac{L_0}{L}}$$. So, for small strains, $$\frac{\Delta T}{T_0}=\frac{\Delta W}{W_0}=-\frac{1}{2}\frac{\Delta L}{L_0}$$. So the Poisson ratio of an incompressible material is equal to 1/2.

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  • $\begingroup$ This sounds reasonable. But, where does the first formula come from? Can you please add references? $\endgroup$ – user29305 Jul 6 '16 at 12:55
  • $\begingroup$ Does this mean that my analysis for all other poisson ratios other than $\nu = 0.5$ is correct? And, can you please point out where I made a mistake in my reasoning for the case $\nu = 0.5$? $\endgroup$ – user29305 Jul 6 '16 at 13:16
  • $\begingroup$ For the deformation you described, the stresses in the transverse directions are both zero. So the strains in these directions must be equal. Understand that the concept of Poisson ratio only applies to small strains. For an incompressible material, the sum of the three strains must be equal to zero, since this is the volumetric strain. This leads to the relationship I presented. If the material is not incompressible, then the volumetric strain is $\epsilon_L+\epsilon_T+\epsilon _W=(1-2\nu )\epsilon_L$ $\endgroup$ – Chet Miller Jul 6 '16 at 16:38
  • $\begingroup$ Ok. That's clear. But, what about my conclusion that $V'=L*T*W=V*(1+\alpha)*(1-\nu \alpha)^2$ for any compressive material with the geometry of a rectangular prism? $\endgroup$ – user29305 Jul 7 '16 at 0:40
  • $\begingroup$ What do you get if you linearize your equation with respect to $\alpha$? $\endgroup$ – Chet Miller Jul 7 '16 at 0:44
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After more reflection, I realized that this fact can be easily demonstrated for Poisson Ratios in the range $ -1 \leq \nu < 0.5$:

  1. We are given $ \nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}=-\frac{\Delta{T}/T_0}{\Delta{L}/L_0}$ where $ -1 \leq \nu < 0.5$

  2. Let $ \frac{\Delta{L}}{L_0}=\alpha > 0$ then $ \begin{cases} \Delta{W}=\frac{W_0*\nu*\Delta{L}}{L_0}=-W_0*\nu*\alpha \\ \Delta{T}=\frac{T_0*\nu*\Delta{L}}{L_0}=-T_0*\nu*\alpha\\ \end{cases} $

  3. $V'=L*T*W=(L_0+\alpha L_0)*T_0*(1+\nu \alpha)*W_0*(1+\nu \alpha)=L_0*T_0*W_0*(1+\alpha)*(1+\nu \alpha)^2=V*(1+\alpha)*(1+\nu \alpha)^2$

  4. The equation in part 3 actually reduces to:

$ \frac{\Delta{V}}{V}=(1+\alpha)*(1+\nu \alpha)^2-1$

In the special case of a cube with sides of length $ L_0$ we have,

$ \frac{\Delta{V}}{V}=(1+\alpha)*(1-\alpha)^2-1$

Clearly, for $ \nu \leq 0$, $ \frac{V'}{V}>1$ so the volume has increased. And for $ \nu >0$, the volume is constant only when $ \nu = \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$. Since $ \nu$ is supposed to be constant and $latex \alpha$ is allowed to vary we must conclude that the volume is not constant in this case either.

If I should trust my arguments then for any given non-negative poisson ratio, there is a single value non-trivial value for $ \alpha=\Delta{L}/L_0$ such that $ V'=V$. The graph of the function $ f(\alpha)= \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$ is given below:

enter image description here

Now, we must deal with the case $ \nu = \frac{1}{2}$ separately:

Let’s suppose, without loss of generality, that our material has the geometry of a cylinder. Then $ V=\pi r^2 L$. If its volume is constant(i.e. it’s incompressible), then:$ dV= 2\pi r L dr +\pi r^2 dL = 0$ $ \nu \equiv \frac{dr/r}{dL/L} = \frac{1}{2}$

In fact, for incompressible materials we have:

$$\frac{\Delta{T}}{T_0}=\frac{\Delta{W}}{W_0}=\sqrt{\frac{L_0}{L}}-1$$

So the moral of the story is that for most materials that we interact with in everyday life, their volume isn't conserved when stretched or compressed unless they happen to have a poisson ratio of $ \nu = \frac{1}{2}$.

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