Stress and strain

Engineering · Senior school and first-year engineering · about 8 min

Pull on a steel rod and it stretches, a tiny bit. Stress describes how hard the material inside is being pulled, and strain describes how much it stretches. Together they tell you whether a part is safe and how much it will deform.

The key idea

\sigma = \frac{F}{A} \qquad \varepsilon = \frac{\Delta L}{L} \qquad E = \frac{\sigma}{\varepsilon}

Stress $\sigma$ is force per area (in pascals), strain $\varepsilon$ is the change in length over the original length (no units), and Young's modulus $E$ is the material's stiffness. Combining them gives the stretch directly: $\Delta L = \dfrac{F L}{A E}$ .

Two horizontal round rods, one above the other, lined up at their left ends. Top rod: unloaded, with a dimension showing its original length. Bottom rod: slightly longer, with a dark red force arrow pulling at each end, and a small dimension showing the extra length at the right end. A thin dashed vertical line at the top rod's right end runs down to the bottom rod. Mark the cross-section area on the bottom rod with a small oval and a leader line. — A rod in tension stretches by a small amount.

Try it yourself

Pull the test bar and follow the curve through the elastic region, yield and necking.

Change the load, area and material, then compare the stress, strain and stretch of the bar.

The stress-strain curve

Test a sample and plot stress against strain:

At first the line is straight: the material is elastic, and returns to its original length when unloaded. The slope of this line is $E$ .
At the yield stress it starts to deform permanently.
It keeps carrying more load up to the ultimate strength, then necks and breaks.

Designs normally keep stresses well below yield, so the straight-line part (Hooke's law) is the one you use most.

A graph with stress on the vertical axis and strain on the horizontal axis. The curve starts at the origin as a straight rising line, bends over at the yield point, rises slowly in a smooth arc to a peak, then falls a little to an end point marked with a small cross. No values on the axes. — A typical stress-strain curve for a ductile metal.

Typical values

Material	Young's modulus $E$
Steel	about $200$ GPa
Aluminium	about $70$ GPa
Concrete	about $30$ GPa
Rubber	about $0.01$ to $0.1$ GPa

A higher $E$ means a stiffer material: the same stress produces less strain.

Worked example: a steel rod in tension

A steel rod $2$ m long and $20$ mm in diameter carries a tensile load of $50$ kN. Take $E = 200$ GPa. Find the stress, the strain and the extension.

A horizontal round steel rod with a dark red force arrow pulling at each end. A dimension along its length, and a small dimension across its diameter at the right end. — The rod in this example.

Show the solution

Area: $A = \pi (0.01)^2 = 3.142 \times 10^{-4}$ m $^2$

Stress:

\sigma = \frac{F}{A} = \frac{50\,000}{3.142 \times 10^{-4}} = 159.2 \text{ MPa}

Strain:

\varepsilon = \frac{\sigma}{E} = \frac{159.2 \times 10^6}{200 \times 10^9} = 7.96 \times 10^{-4}

Extension:

\Delta L = \varepsilon L = 7.96 \times 10^{-4} \times 2 = 1.59 \text{ mm}

Common mistakes

Using the diameter as the radius when finding the area

$A = \pi r^2$ with $r = d/2$ , or use $A = \frac{\pi d^2}{4}$ .

Giving strain units

Strain is a length divided by a length, so it has no units. It's often quoted as a percentage or in microstrain.

Using Hooke's law past the yield point

$E = \frac{\sigma}{\varepsilon}$ only holds in the elastic region.

Check your understanding

1. What are the units of strain?

2. A $10$ kN load acts on a bar with cross-sectional area $100$ mm $^2$ . What is the stress?

3. Two rods carry the same stress. One is steel, one aluminium. Which stretches more?

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