Hooke's law, stress, strain and the Young modulus, elastic and plastic behaviour, stress-strain graphs, and elastic strain energy.

What this dot point is asking

WJEC wants you to use Hooke's law, define and calculate tensile stress, strain and the Young modulus, distinguish elastic from plastic behaviour, interpret stress-strain graphs, and calculate elastic strain energy. Material properties matter to every engineer, and this dot point pairs a tidy specified practical with a set of definitions the examiners expect word-perfect.

The answer

Hooke's law

Stress, strain and the Young modulus

Tensile stress is $\sigma = \dfrac{F}{A}$, the force per unit cross-sectional area, measured in pascals. Tensile strain is $\varepsilon = \dfrac{\Delta L}{L}$, the fractional change in length, with no units.

The Young modulus $E$ is the gradient of the straight (elastic) part of a stress-strain graph. A large Young modulus, such as the $2 \times 10^{11}\,\text{Pa}$ of steel, means the material is stiff and resists stretching.

Elastic and plastic behaviour

Below the elastic limit, a material returns to its original length when the load is removed (elastic behaviour). Beyond it, the material is permanently deformed (plastic behaviour). A stress-strain graph shows the limit of proportionality, the elastic limit, the yield point and the ultimate tensile stress.

Elastic strain energy

Measuring the Young modulus

A specified practical measures $E$ for a wire: hang masses to stretch a long thin wire, measure extension with a marker and reference, and plot stress against strain. Using a long thin wire makes the extension larger and easier to measure, reducing percentage uncertainty.

Examples in context

Example 1. A suspension bridge cable. A steel cable of cross-sectional area $0.020\,\text{m}^2$ carries a tension of $2.0\,\text{MN}$. The stress is $\sigma = F/A = 2.0 \times 10^{6} / 0.020 = 1.0 \times 10^{8}\,\text{Pa}$. Dividing by the Young modulus of steel ($2.0 \times 10^{11}\,\text{Pa}$) gives a strain of $5.0 \times 10^{-4}$, so a $1000\,\text{m}$ span stretches by about $0.5\,\text{m}$. Engineers must keep this stress well below the yield point so the cable always behaves elastically.

Example 2. A climbing rope. Climbing ropes are deliberately designed with a low effective stiffness so they store and release strain energy gently during a fall. A rope stretching $1.5\,\text{m}$ under a $3.0\,\text{kN}$ peak force absorbs roughly $\tfrac{1}{2}(3000)(1.5) = 2.25\,\text{kJ}$, spreading the impact over a longer time and reducing the peak force on the climber, much like a crumple zone.

Try this

Q1. A wire of cross-sectional area $2.0\times10^{-7}\,\text{m}^2$ and length $1.5\,\text{m}$ extends by $1.2\,\text{mm}$ under a $40\,\text{N}$ load. Find the Young modulus. [3 marks]

• Cue. $E = \frac{FL}{A\Delta L} = \frac{40\times1.5}{2.0\times10^{-7}\times1.2\times10^{-3}} = 2.5\times10^{11}\,\text{Pa}$.

Q2. Explain the difference between elastic and plastic deformation. [2 marks]

• Cue. Elastic returns to original length when load removed; plastic leaves a permanent change in length.