20.5 Poisson’s ratio

Before you read this, I suggest you read post 20.2.

Let’s return to stretching the long, thin wire of post 20.2. In posts 20.2 and 20.3, we neglected that the wire becomes slightly thinner as it gets longer. The reason for thinking about a long, thin wire was that this effect is not easy to see, so the change in shape is negligible – this means that the wire is (almost) subjected to pure tension (post 20.2).

fig1

The picture above exaggerates the simultaneous lengthening and thinning. Originally, the wire has length L and radius r; when a force is applied (in the direction of the red arrows) the length increases to L + ΔL and the radius changes to r + Δr. Since the wire gets thinner, the value of Δr is negative. The wire experiences a strain (post 20.2), in the direction of the force, of

εL = ΔL/L.

Since this strain acts along the length of the wire, it is called a longitudinal strain. But it also experiences a strain in the perpendicular direction, called a radial strain defined by

εr = Δr/r.

Since, in our example, the value of Δr is negative, the value of εr will be negative . Remember that – εr is a compressive strain (post 20.2).

The Poisson’s ratio of the material of the wire is defined as

ν = – εr/ εL.

The minus sign appears in the definition only to ensure that the value of Poisson’s ratio is positive for materials that get thinner as they get longer.

Now let’s suppose that the volume of our wire remains constant when it is stretched. Then

πr2L = (r + Δr)2(L + ΔL) = π[r2 + (Δr)2 + 2rΔr](L + ΔL)

Usually, the value of (Δr)2 will be so small, compared to the dimensions of the wire, that its value is negligible. Then, after dividing both sides of our equation by π, we get

r2L ≈ [r2 + 2rΔr](L + ΔL).

Dividing both sides of this equation by r2L gives

1 ≈ [1 + (2Δr/r)][1 + (ΔL/L)].

Now let’s substitute our definitions of and εL into this result, giving

1 ≈ (1 + 2εr)(1 + εL) = 1 + εL + 2εr + 2εrεL ≈ 1 + εL + 2εr.

The final step arises because, since εL and εr are usually small (post 20.2), 2εrεL will be very small in comparison. Subtracting 1 from both sides of this equation gives, after a bit more manipulation, the result that

εr/εL ≈ 1/2

So, for materials that maintain a constant volume ν ≈ 0.5.

In practice, most metal wires have ν ≈ 0.3. Why is this? It must be because a metal wire changes volume as it stretches. In practice the volume increases because the tensile force pulls the atoms in the metal further apart. How could we measure this increase in the distance between the atoms as the wires stretches? The distance between atoms in a metal is comparable to the wavelength of x-rays (post 19.9). So when x-rays pass through a metal, diffraction occurs; x-ray diffraction can then be used to measure the distance between metal atoms in different directions (post 19.20).

Many textbooks tell us that ν can’t be less than 0. Why not? Because, the authors claim, it is impossible for a wire to get thicker as it becomes longer. This belief arises because most early work on deformation of materials concentrated on metals that consist simply of atoms packed closely together. There is no way such simple materials could increase in width when subjected to tensile stress. But we now need to understand many different materials which may have complicated internal structures.

fig2

Let’s think of the cell, on the left-hand side of the picture above. It is bounded by material that is sufficiently rigid that, when a force is applied in the direction shown (by the red arrows) on the left-hand side, it changes in shape as shown. The force pulls the upper and lower sloping surfaces towards the horizontal (increasing the average height of the cell); as they move towards the horizontal, they push the vertical sides further apart (increasing the width of the cell). The distance L increases to L’ and, as a result, the distance r increases to r’. So both ΔL = L’ – L and Δr = r’ – r are positive. So, for this material, both εL and εr are positive and, by definition, ν is negative.

fig3

Now let’s think of a solid foam that is a material made up of these cells packed closely together. A foam is a three-dimensional structure (solid or liquid) that encloses spaces – so the pictures of the cell and the intact foam are two-dimensional representations of three-dimensional structures. A force acting in the direction shown causes each cell, and therefore the whole sample of the material, to becomes wider as it gets longer. This is because, as the sloping surfaces move towards the horizontal, they push the cells above and below further apart – at the same time as their vertical surfaces are pushed apart.

The foam is an example of an auxetic material – one with a negative Poisson’s ratio. There are also materials with complicated internal structures that have ν > 0.5.

If you read about Poisson’s ratio in most textbooks, you will gain the impression that it is very simple. But the reality is more complicated!

Related posts

20.3 Hooke’s law
20.2 Deformation of objects
16.37 Solids, liquids and gases

Follow-up posts

20.6 Elasticity