Why use stress and strain rather than force and extension?

Force and extension depend on the sample’s size. Stress and strain normalise for area and length, so the resulting graph and Young modulus describe the material itself.

Why does strain have no units?

It is extension divided by original length — metres over metres — so the units cancel.

What is a hysteresis loop?

For materials like rubber, loading and unloading follow different curves. The area between them is energy per unit volume transferred to heat in each cycle.

Why use a long, thin wire in the practical?

Length gives a measurable extension; thinness gives large stress for a small load. Both reduce percentage uncertainty in the key readings.

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54–58MechanicsStretch

Stress, Strain and the Young Modulus

A spring constant describes one particular spring — a thicker or shorter wire of the same metal would have a different one. $\textbf{Stress}$ and $\textbf{strain}$ remove the dependence on size, and their ratio, the $\textbf{Young modulus}$ , is a property of the material itself.

40 min Topic 4 · Materials

What you'll be able to do

Define and calculate tensile stress and tensile strain
Define and use the Young modulus $E = \sigma/\varepsilon$
Describe the experiment to determine the Young modulus of a wire
Interpret stress–strain graphs for ductile, brittle and polymeric materials

Stress and strain

$\textbf{Tensile stress}$ is force per unit cross-sectional area, $\sigma = \dfrac{F}{A}$ , measured in pascals. It captures how intensely the material is loaded, regardless of how thick the sample is.

$\textbf{Tensile strain}$ is extension per unit original length, $\varepsilon = \dfrac{\Delta x}{x}$ . It is a ratio of two lengths and so has no units.

Normalising in this way means a thick wire and a thin wire of the same material, loaded to the same stress, show the same strain.

\sigma = \dfrac{F}{A}

Stress, in Pa.

\varepsilon = \dfrac{\Delta x}{x}

Strain — no units.

Example 1A wire of diameter 0.60 mm carries a load of 45 N. Find the stress.

1Radius

3.0\times10^{-4}

m, so

A = \pi(3.0\times10^{-4})^{2} = 2.83\times10^{-7}\ \mathrm{m^{2}}

\sigma = \dfrac{45}{2.83\times10^{-7}}

Answer

1.6\times10^{8}\ \mathrm{Pa}

Tip — Stresses in metals are enormous in pascals — $10^{8}$ is typical. A small answer usually means the area was left in square millimetres.

The Young modulus

Within the limit of proportionality, stress is proportional to strain, and the constant of proportionality is the Young modulus $E = \dfrac{\sigma}{\varepsilon}$ , in pascals.

Substituting the definitions gives $E = \dfrac{Fx}{A\Delta x}$ , which is the form used with experimental data.

A high Young modulus means a stiff material. Steel has about $2\times10^{11}$ Pa, rubber around $10^{7}$ Pa — so steel needs roughly ten thousand times the stress for the same strain.

E = \dfrac{\sigma}{\varepsilon}

Young modulus.

E = \dfrac{Fx}{A\Delta x}

In measurable quantities.

Example 1A 2.0 m steel wire of cross-sectional area

1.0\times10^{-7}\ \mathrm{m^{2}}

extends by 1.0 mm under 10 N. Find the Young modulus.

\sigma = \dfrac{10}{1.0\times10^{-7}} = 1.0\times10^{8}

Pa.

\varepsilon = \dfrac{1.0\times10^{-3}}{2.0} = 5.0\times10^{-4}

E = \dfrac{1.0\times10^{8}}{5.0\times10^{-4}}

Answer

E = 2.0\times10^{11}\ \mathrm{Pa}

Measuring the Young modulus

The standard practical uses a long, thin wire. Length makes the extension large enough to measure; thinness makes the stress large for a modest load.

Measure the original length with a metre rule, the diameter at several points with a micrometer (averaging to reduce random error), then add masses in steps and record the extension each time using a vernier scale or a marker against a fixed ruler.

Plot stress against strain and take the gradient of the straight section. Using a gradient from many readings is more reliable than a single calculation, and it automatically discards any zero error in the extension readings.

Tip — Wear eye protection and keep the load below the elastic limit — a wire that snaps under tension recoils violently.

Reading stress–strain graphs

A $\textbf{ductile}$ material such as copper shows a straight region (Hooke’s law), then a $\textbf{yield point}$ where strain increases rapidly with little extra stress, a long region of plastic deformation, and finally fracture. The peak stress is its $\textbf{ultimate tensile strength}$ .

A $\textbf{brittle}$ material such as glass or cast iron stays almost straight right up to fracture, with no plastic region. It breaks suddenly without warning.

A $\textbf{polymer}$ such as rubber has a curved graph and a large strain, and unloading follows a different path. The area enclosed between loading and unloading curves — the $\textbf{hysteresis loop}$ — is energy transferred to heat per unit volume, which is why a repeatedly stretched rubber band warms up.

The area under a stress–strain graph is energy stored per unit volume, the material equivalent of $\tfrac12 F\Delta x$ .

\text{energy per unit volume} = \tfrac{1}{2}\sigma\varepsilon

Area under the linear region.

Example 1A material has

E = 1.2\times10^{11}

Pa and is loaded to a stress of

6.0\times10^{7}

Pa within its elastic region. Find the energy stored per unit volume.

1Strain:

\varepsilon = \dfrac{6.0\times10^{7}}{1.2\times10^{11}} = 5.0\times10^{-4}

2Energy per unit volume:

\tfrac12 \times 6.0\times10^{7} \times 5.0\times10^{-4}

Answer

1.5\times10^{4}\ \mathrm{J\,m^{-3}}

Tip — Strength and stiffness are different properties. Stiffness is the gradient (Young modulus); strength is the stress at fracture. Glass is stiff but not tough.

Equation recap

\sigma = \tfrac{F}{A}

Stress.

\varepsilon = \tfrac{\Delta x}{x}

Strain.

E = \tfrac{\sigma}{\varepsilon} = \tfrac{Fx}{A\Delta x}

Young modulus.

\tfrac12\sigma\varepsilon

Energy per unit volume.

Common mistakes to avoid

Using diameter instead of radius in

A = \pi r^{2}

Halve the diameter first; using it directly quadruples the area.

Using total length instead of original length for strain.

Strain is extension divided by the

\textit{original}

length.

Confusing stiff with strong.

Stiffness is the Young modulus; strength is the breaking stress.

Giving strain a unit of metres.

Strain is a ratio of lengths and is dimensionless.

Key takeaways

Stress $\sigma = F/A$ is in pascals; strain $\varepsilon = \Delta x/x$ has no units.
The Young modulus $E = \sigma/\varepsilon$ is a material property and the gradient of the linear region.
Measure it with a long thin wire, repeated diameter readings and a stress–strain gradient.
Ductile materials yield and deform plastically; brittle ones fracture without warning.
Area under a stress–strain graph is energy per unit volume.

Test yourself

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