How Much Does a Steel Wire Stretch When It Holds a Load?

A rubber band springs back, a steel cable holds a lift cage, and a bridge flexes slightly as trucks cross it. Elasticity describes how solids deform under force and recover — and how much force they can take before they stop recovering.

This lesson covers stress, strain and Hooke's law, the three elastic moduli, and the energy stored in a stretched wire.

Stress is the restoring force per unit area inside a deformed body, strain is the fractional change in its size or shape, and Hooke's law states that within the elastic limit stress is proportional to strain.

Stress, $\frac{F}{A}$, in N m${}^{-2}$ or pascals:

- Longitudinal — tensile or compressive, along the length
- Shearing — parallel to a surface
- Volume — equal pressure from all sides

Strain has no units: longitudinal strain $\frac{\Delta L}{L}$, volume strain $\frac{\Delta V}{V}$, and shearing strain $\theta$, the angle of deformation.

Hooke's law. Stress is proportional to strain, so $\frac{\text{stress}}{\text{strain}}$ is a constant for the material, called its modulus of elasticity.

Stress-strain curve of a metal wire:

- Proportional limit — Hooke's law holds up to here
- Elastic limit — beyond it, the wire no longer returns fully to its length
- Yield point — strain grows rapidly with little extra stress
- Breaking point — the wire snaps at its ultimate tensile strength

Worked example. A 2.0 m wire of cross-section $1.0\times 10^{-6}$ m${}^2$ carries a 100 N load and stretches by 1.0 mm:

$$\text{stress}=\frac{100}{1.0\times 10^{-6}}=1.0\times 10^8\text{ Pa}\text{strain}=\frac{1.0\times 10^{-3}}{2.0}=5.0\times 10^{-4}$$

An everyday example. The steel ropes of a lift in a Mumbai tower are designed so their stress stays far below the elastic limit, letting them stretch and recover on every trip.

The substance. Hooke's law holds only for small strains — beyond the proportional limit, stress and strain are no longer proportional.

Young's modulus is longitudinal stress over longitudinal strain, bulk modulus is volume stress over volume strain, and the modulus of rigidity is shearing stress over shearing strain.

$$Y=\frac{FL}{A\Delta L}B=-\frac{\Delta p}{\Delta V/V}G=\frac{F/A}{\theta }$$

- The minus sign keeps B positive, since volume falls as pressure rises; compressibility is $\frac{1}{B}$
- Only solids have a modulus of rigidity, because fluids cannot resist shear

Worked example 1 — Young's modulus. For the wire above, $Y=\frac{1.0\times 10^8}{5.0\times 10^{-4}}=2.0\times 10^{11}$ Pa, typical of steel.

Worked example 2 — extension of a steel wire. A 3.0 m steel wire of radius 1.0 mm, with $Y=2.0\times 10^{11}$ Pa, holds a 20 kg mass, with $g=9.8$ m s${}^{-2}$:

$$\Delta L=\frac{FL}{AY}=\frac{196\times 3.0}{\pi \times (1.0\times 10^{-3}{)}^2\times 2.0\times 10^{11}}=9.4\times 10^{-4}\text{ m}$$

Worked example 3 — bulk modulus. Water, with $B=2.2\times 10^9$ Pa, under an extra pressure of $1.0\times 10^7$ Pa shrinks by $\frac{\Delta V}{V}=\frac{1.0\times 10^7}{2.2\times 10^9}=4.5\times 10^{-3}$.

Worked example 4 — rigidity. A rubber block 10 cm tall with a top face of 0.010 m${}^2$ is sheared by 50 N, and its top shifts 5.0 mm, so $\theta =0.050$ and

$$G=\frac{50/0.010}{0.050}=1.0\times 10^5\text{ Pa}$$

An everyday example. Railway tracks are made of steel rather than aluminium partly because steel's much larger Young's modulus means far less bending under heavy trains.

The substance. Young's modulus belongs to the material, not the wire — a longer or thicker wire of the same steel stretches differently but has the same Y.

**The work done in stretching a wire is stored as elastic potential energy, $U=\frac{1}{2}F\Delta L$, so the energy stored per unit volume is $\frac{1}{2}\times \text{stress}\times \text{strain}$.

Derivation.** At extension x, the restoring force is $F=\frac{YAx}{L}$, so the work done stretching the wire by $\Delta L$ is

$$W={\int }_0^{\Delta L}\frac{YA}{L}xdx=\frac{1}{2}\frac{YA}{L}(\Delta L{)}^2=\frac{1}{2}F\Delta L$$

Dividing by the volume AL gives the energy density:

$$u=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{{\text{stress}}^2}{2Y}$$

Worked example. For the steel wire above, with $F=196$ N and $\Delta L=9.4\times 10^{-4}$ m:

$$U=\frac{1}{2}\times 196\times 9.4\times 10^{-4}=0.092\text{ J}$$

Its stress is $\frac{196}{3.14\times 10^{-6}}=6.24\times 10^7$ Pa, so its energy density is $\frac{(6.24\times 10^7{)}^2}{2\times 2.0\times 10^{11}}=9.7\times 10^3$ J m${}^{-3}$.

An everyday example. A bow at a village archery contest stores elastic potential energy in its bent limbs and string, then releases it into the arrow.

The substance. The stored energy is half of force times extension — the force grows from zero as the wire stretches, so its average value is only half the final force.