Poisson’s Ratio and Elastic Potential Energy

So far, you have studied how materials resist three kinds of deformation: stretching (Young’s modulus), sideways distortion (shear modulus), and uniform compression (bulk modulus). But stretching a wire does more than just make it longer. If you pull a rubber band, you will notice it also gets thinner. And all that force you apply to stretch the wire, where does that energy go? This topic answers both questions: first, the neat relationship between lengthwise and sideways deformation captured by Poisson’s ratio, and then the energy that gets locked inside a stretched wire.

Lateral Strain: What Happens Sideways When You Pull Lengthwise

When you stretch a wire along its length, something interesting happens in the perpendicular direction. The wire gets slightly thinner. Its diameter shrinks a little. This sideways contraction, measured as a fraction of the original diameter, is called lateral strain.

If the wire originally has a diameter $d$ and the diameter contracts by $\Delta d$ under the applied tensile force, then:

$\text{Lateral strain} = \frac{\Delta d}{d}$

At the same time, the wire gets longer. If the original length is $L$ and the elongation is $\Delta L$, the longitudinal strain (which you already know from earlier topics) is:

$\text{Longitudinal strain} = \frac{\Delta L}{L}$

The key observation, first pointed out by Simon Poisson, is that these two strains are not independent. Within the elastic limit, the lateral strain is directly proportional to the longitudinal strain. Stretch a wire more, and it also gets proportionally thinner.

Poisson’s Ratio: Connecting the Two Strains

The ratio of lateral strain to longitudinal strain in a stretched wire is called Poisson’s ratio. Putting the two strain expressions together:

$\text{Poisson's ratio} = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{\Delta d / d}{\Delta L / L}$

Dividing by a fraction flips it, so this simplifies to:

$\text{Poisson's ratio} = \frac{\Delta d}{\Delta L} \times \frac{L}{d}$

Key Properties

A few things stand out about this quantity:

• It is dimensionless. Since it is a ratio of two strains (each of which is itself a ratio of two lengths), all length units cancel out. Poisson’s ratio is a pure number with no dimensions and no units.
• It depends only on the material. Just like Young’s modulus or shear modulus, Poisson’s ratio is a material property. Change the material and the ratio changes; change the size of the sample or the magnitude of the force (as long as you stay within the elastic limit), and the ratio stays the same.

Typical Values

Different materials contract sideways by different amounts relative to their lengthwise stretching:

• Steels: Poisson’s ratio falls between 0.28 and 0.30. For every 1% increase in length, the diameter shrinks by roughly 0.28% to 0.30%.
• Aluminium alloys: the value is about 0.33, slightly higher than steel. Aluminium contracts a bit more in the sideways direction for the same lengthwise strain.

These values make physical sense. Atoms in a crystal are connected by bonds in all directions. When you pull atoms apart along one direction, the bonds in the perpendicular direction adjust by pulling the atoms slightly inward. How much they pull inward depends on the bonding geometry and stiffness of the material, which is why Poisson’s ratio varies from one material to another.

Elastic Potential Energy: Where Does the Stretching Energy Go?

When you apply a tensile force to a wire and stretch it, you do work against the inter-atomic forces holding the atoms at their equilibrium positions. Within the elastic limit, this work does not get lost as heat. Instead, it gets stored inside the wire as elastic potential energy (much like a compressed spring stores energy). Release the force, and the wire snaps back to its original length, returning all that stored energy.

Let us derive exactly how much energy is stored.

Setting Up the Problem

Take a wire of original length $L$ and cross-sectional area $A$, made of a material with Young’s modulus $Y$. A deforming force $F$ stretches it. At any instant, suppose the wire has already been elongated by an amount $l$ from its natural length.

From the definition of Young’s modulus (Equation 8.8):

$Y = \frac{F/A}{l/L} = \frac{F \times L}{A \times l}$

Rearranging to express the force $F$ in terms of the current elongation $l$:

$F = \frac{YA \times l}{L}$

Notice that $F$ is not constant. It grows linearly with $l$: the more you stretch the wire, the harder it pulls back. This is Hooke’s law in action.

Integrating to Find the Total Work

To stretch the wire by a tiny extra amount $dl$ when it is already elongated by $l$, the small bit of work done is:

$dW = F \, dl = \frac{YA \, l}{L} \, dl$

The total work $W$ needed to stretch the wire from its natural length ($l = 0$) to a final elongation of $l$ is found by adding up (integrating) all these tiny contributions:

$W = \int_0^l \frac{YA \, l}{L} \, dl$

The constants $Y$, $A$, and $L$ come out of the integral:

$W = \frac{YA}{L} \int_0^l l \, dl$

The integral of $l$ with respect to $l$ is $l^2/2$:

$W = \frac{YA}{L} \times \frac{l^2}{2}$

$W = \frac{YA l^2}{2L}$

Rewriting in Terms of Stress, Strain, and Volume

This result can be rearranged into a more revealing form. The volume of the wire is $V = A \times L$, and the strain is $\varepsilon = l/L$. Rewriting step by step:

$W = \frac{1}{2} \times Y \times \frac{l^2}{L^2} \times (A \times L)$

Recognise $l^2/L^2$ as $\varepsilon^2$ (strain squared) and $A \times L$ as the volume $V$:

$W = \frac{1}{2} \times Y \times \varepsilon^2 \times V$

Since stress $\sigma = Y \times \varepsilon$, you can replace $Y \times \varepsilon$ with $\sigma$:

$W = \frac{1}{2} \times \sigma \times \varepsilon \times V$

In words: the elastic potential energy stored in a stretched wire equals one-half times stress times strain times the volume of the wire.

Why the Factor of One-Half?

The $1/2$ appears because the force is not constant during the stretching. It starts at zero (when the wire is at its natural length) and increases linearly to its maximum value. The average force over the entire stretch is half the final force, so the work done is half of what it would be if the full force had acted over the entire elongation. This is exactly the same reason the area of a triangle (representing force vs. extension) is $\frac{1}{2} \times \text{base} \times \text{height}$.

Energy Density

In many practical situations, you want to know the energy stored per unit volume rather than the total. Dividing the total elastic PE by the volume $V = AL$ of the wire gives the elastic potential energy density $u$:

$u = \frac{1}{2} \sigma \varepsilon \qquad \text{(8.14)}$

This compact formula applies to any linearly elastic material under any type of stress (not just tensile). It tells you that the energy stored per unit volume equals half the product of the stress and the strain at that point.

You can also write the energy density in two other equivalent forms by substituting $\sigma = Y\varepsilon$ or $\varepsilon = \sigma/Y$:

$u = \frac{1}{2} Y \varepsilon^2 \qquad \text{or} \qquad u = \frac{\sigma^2}{2Y}$

All three expressions give the same number; which one you use depends on which quantities you know.