Elastic moduli and Hooke's law

When a force deforms a solid, the material develops an internal restoring force that opposes the deformation. Stress is this restoring force per unit cross-sectional area:

where: $F$ is the applied (restoring) force in newtons and $A$ is the cross-sectional area in square metres, so $\sigma$ carries units of $\text{N/m}^2$, or pascal.

Strain is the fractional deformation — always a dimensionless ratio, never carrying units. There are three matched pairs of stress and strain, one for each way a solid can be pushed out of shape.

• Longitudinal (tensile/compressive) stress and strain — a force along the length of a rod or wire produces strain $\varepsilon = \Delta L / L$, the fractional change in length.
• Bulk (hydrostatic) stress and strain — uniform pressure from every side produces volume strain $\Delta V / V$, the fractional change in volume.
• Shearing stress and strain — a tangential force produces shear strain $\theta \approx \tan\theta = \Delta x / L$, the angle through which one face tilts relative to the opposite face.

The figure below shows all three side by side, so the geometry of each deformation is easy to keep straight.

Hooke's law states that, within the elastic limit, stress is directly proportional to strain:

where: $E$ is a constant called the modulus of elasticity, with a different value and a different name for each type of deformation. Beyond the elastic limit the proportionality breaks down, and the material undergoes permanent (plastic) deformation — it no longer returns fully to its original shape when the force is removed. The three moduli, each stress divided by its matching strain, are:

where: $Y$ is Young's modulus (stretching/compression along a length), $K$ is the bulk modulus (the minus sign shows volume shrinks as pressure rises; $1/K$ is the compressibility), and $\eta$ is the shear modulus, or modulus of rigidity (twisting). All three share the unit $\text{N/m}^2$, because strain is dimensionless. A larger modulus means a stiffer material for that type of deformation — steel's Young's modulus is roughly $10^5$ times rubber's, so a steel wire barely stretches where a rubber cord of the same size and load would stretch enormously. One more constant completes the picture: Poisson's ratio $\sigma_P$ compares how much a stretched wire thins sideways to how much it lengthens,

It is dimensionless and typically falls between about $0.2$ and $0.4$ for common solids.

A steel wire of length $2.0\ \text{m}$ and cross-sectional area $1.0\times10^{-6}\ \text{m}^2$ is stretched by a force of $100\ \text{N}$. Young's modulus of steel is $2.0\times10^{11}\ \text{N/m}^2$. Find (a) the extension, and (b) the extension if a second, identical wire were used instead but with double the cross-sectional area.

Step 1. Use the extension formula derived above, $\Delta L = FL/(AY)$, with the given numbers: $\Delta L = \dfrac{(100)(2.0)}{(1.0\times10^{-6})(2.0\times10^{11})} = 1.0\times10^{-3}\ \text{m} = 1.0\ \text{mm}$. This matches the two-step stress-then-strain calculation, as it must, since it is the same physics rearranged.

Step 2. For part (b), double the area to $A' = 2.0\times10^{-6}\ \text{m}^2$, keeping $F$, $L$, and $Y$ unchanged. Since $\Delta L \propto 1/A$ (everything else fixed), doubling $A$ halves the extension: $\Delta L' = \dfrac{(100)(2.0)}{(2.0\times10^{-6})(2.0\times10^{11})} = 0.5\times10^{-3}\ \text{m} = 0.5\ \text{mm}$.

Step 3. Interpret the comparison: a wire twice as thick (in cross-sectional area) stretches only half as much under the same load, because the same force is now spread over twice the area, halving the stress, and strain follows stress in direct proportion. Answer: the original wire stretches $1.0\ \text{mm}$; the wire with double the area stretches only $0.5\ \text{mm}$.

Competitive stretch

A solid rubber cube of side $10\ \text{cm}$ is subjected to a uniform pressure increase of $2.0\times10^{6}\ \text{Pa}$, and its volume decreases by $6\ \text{cm}^3$. Find its bulk modulus, and compare it with a typical metal. The original volume is $V = (0.10)^3 = 1.0\times10^{-3}\ \text{m}^3 = 1000\ \text{cm}^3$, so the volume strain is $\Delta V/V = 6/1000 = 6.0\times10^{-3}$. The bulk modulus is then $K = \Delta p / (\Delta V/V) = (2.0\times10^{6})/(6.0\times10^{-3}) \approx 3.3\times10^{8}\ \text{N/m}^2$ — roughly a thousand times smaller than a typical metal's bulk modulus of order $10^{11}\ \text{N/m}^2$, confirming that rubber is far more compressible than steel or aluminium.

• Confusing stress with pressure: pressure is one special case of stress (hydrostatic/bulk), but stress can equally be longitudinal or shearing.
• Forgetting Hooke's law holds only up to the elastic limit — beyond it, stress and strain stop being proportional, and permanent deformation can set in.
• Mixing up which stress pairs with which strain — each modulus matches a specific deformation type: longitudinal-longitudinal for $Y$, volume-volume for $K$, shear-shear for $\eta$.
• Assigning strain a unit. Strain is always a ratio of two lengths or two volumes, so it is dimensionless; only stress and the moduli carry units of $\text{N/m}^2$.

Ask the tutor: Want a walkthrough of how to combine two wires in series or parallel using these moduli, or more practice numericals on stress, strain, and extension?