1.9 Stress, strain and the Young modulus
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Unit 1 Mechanics and Materials, AS content, including Core Practical 3. Hooke's law, the stiffness \(k\) of an object, force–extension graphs and elastic strain energy are in 1.8 and are only referred to here.
What this page covers
6 points
- Tensile and compressive stress, \(\sigma = F/A\), including the area of a wire from its diameter
- Tensile and compressive strain, \(\varepsilon = \Delta x/x\)
- The Young modulus, \(E = \sigma/\varepsilon\), and using it to predict extensions
- Stress–strain graphs for a ductile metal; breaking stress (ultimate tensile stress)
- Describing materials as stiff, strong, brittle, ductile, malleable, hard or tough, using graphs as evidence
- Core Practical 3: determining the Young modulus of a material in the form of a wire
In one minute
6 points
- Dividing force by area and extension by original length turns a force–extension graph for one sample into a stress–strain graph for the material, whose gradient is the Young modulus.
- Stress \(\sigma = F/A\) (unit Pa = N m⁻²); strain \(\varepsilon = \Delta x/x\) (no unit). Both can be tensile (stretching) or compressive (squashing).
- Young modulus \(E = \sigma/\varepsilon = Fx/(A\Delta x)\), unit Pa. A thicker or longer wire of the same metal has the same \(E\).
- On a stress–strain graph the gradient of the straight section is \(E\); the highest point is the breaking stress (ultimate tensile stress).
- Stiff = large \(E\); strong = large breaking stress; brittle = breaks with little or no plastic deformation; ductile = large plastic deformation in tension; tough = absorbs a lot of energy before breaking.
- Core Practical 3 uses a long thin wire, a micrometer for the diameter at several places, a fixed scale for the extension, and the gradient of a stress–strain graph.
Contents
12 sections
- What are tensile and compressive stress?
- What is strain?
- What is the Young modulus?
- What does a stress–strain graph show?
- How do you describe a material from its stress–strain graph?
- How do you determine the Young modulus of a wire (Core Practical 3)?
- Exam tips for 1.9
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What are tensile and compressive stress?
A force–extension graph describes one particular wire. Hang the same load on a wire twice as thick and it stretches a quarter as much; use a wire twice as long and it stretches twice as much. To compare materials rather than objects, we need quantities that take out the size of the sample. Stress does that for the force.
- Tensile stress: the forces pull outwards and tend to stretch the sample (a cable, a rope, a guitar string).
- Compressive stress: the forces push inwards and tend to squash it (a column, a table leg, the sole of a shoe).
- The area \(A\) is the cross-section at right angles to the force.
- Stresses in materials are large numbers: 1 MPa = 10⁶ Pa, 1 GPa = 10⁹ Pa. Areas of wires are tiny: 1 mm² = 10⁻⁶ m².
A lamp of mass 42 kg hangs at rest from a single steel wire of diameter 2.4 mm. Calculate the tensile stress in the wire.
The tension equals the weight: \(F = mg = 42\times9.81 = 412\ \text{N}\).
Area from the diameter, in metres: \(A = \pi\times(2.4\times10^{-3})^2\div4 = 4.52\times10^{-6}\ \text{m}^2\).
\(\sigma = 412\div4.52\times10^{-6}\) → σ = 9.1 × 10⁷ Pa (91 MPa).
Putting the diameter into \(\pi r^2\). This makes the area four times too big and the stress four times too small. Halve the diameter first, or use \(\pi d^2/4\), and convert millimetres to metres before squaring.
What is strain?
Strain is a ratio of two lengths, so it has no unit. It may be given as a decimal or as a percentage: a strain of 0.002 is 0.2%. Tensile strain is an extension; compressive strain is a reduction in length.
- Strain compares the change with the size of the sample, so a 1 mm stretch is a big strain for a 10 cm wire but a tiny one for a 10 m cable.
- Metals in their elastic region usually show strains well below 0.01 (1%). This is why the Young modulus practical needs a long wire.
- Use the original length, not the stretched length.
What is the Young modulus?
- \(E\) measures the stiffness of a material: a large \(E\) means a large stress is needed for a small strain.
- It does not depend on the shape or size of the sample. Compare the stiffness of an object (1.8): \(k = F/\Delta x = EA/x\), which does depend on \(A\) and \(x\).
- Typical values: steel about 200 GPa, copper about 120 GPa, aluminium about 70 GPa, rubber only a few MPa.
A granite column of square cross-section, 0.30 m × 0.30 m, supports a compressive load of 1.8 × 10⁵ N. The column is 4.5 m tall and the load shortens it by 0.21 mm. Calculate (a) the compressive stress, (b) the compressive strain and (c) the Young modulus of the granite.
(a) \(\sigma = 1.8\times10^{5}\div(0.30\times0.30) = 1.8\times10^{5}\div0.090\) → σ = 2.0 × 10⁶ Pa.
(b) Both lengths in metres: \(\varepsilon = 2.1\times10^{-4}\div4.5\) → ε = 4.67 × 10⁻⁵ (no unit).
(c) \(E = 2.0\times10^{6}\div4.67\times10^{-5}\) → E = 4.3 × 10¹⁰ Pa (43 GPa).
A copper wire is 1.6 m long and 0.50 mm in diameter. A load of 25 N is hung from it. The Young modulus of copper is 1.2 × 10¹¹ Pa; assume the wire obeys Hooke's law. (a) Calculate the extension. (b) A second wire of the same copper has the same length but twice the diameter. State its extension under the same load.
(a) \(A = \pi\times(0.50\times10^{-3})^2\div4 = 1.96\times10^{-7}\ \text{m}^2\), so \(\sigma = 25\div1.96\times10^{-7} = 1.27\times10^{8}\ \text{Pa}\).
\(\varepsilon = \sigma/E = 1.27\times10^{8}\div1.2\times10^{11} = 1.06\times10^{-3}\), so \(\Delta x = \varepsilon x = 1.06\times10^{-3}\times1.6\) → Δx = 1.7 mm.
(b) Doubling \(d\) multiplies \(A\) by 4, so the stress and the strain are quartered: \(1.70\div4\) → 0.42 mm. The Young modulus is unchanged because it is the same material; only the object's stiffness changes.
What does a stress–strain graph show?
A stress–strain graph has the same shape as the force–extension graph of the same sample, but it describes the material: any sample of it gives the same curve. The points on it are the ones you met in 1.8.
- Gradient of the straight section = Young modulus \(E\).
- P, limit of proportionality: the end of the straight line.
- E, elastic limit: beyond this the sample does not return to its original length when the load is removed.
- Y, yield point: the material starts to deform plastically with little or no increase in stress.
- Breaking stress (ultimate tensile stress, UTS): the greatest stress the material can withstand, the highest point of the curve.
- For a ductile metal, beyond the UTS the wire forms a narrow "neck"; the force it can support falls and it breaks soon after.
Use graphs A and B for this metal. (a) Determine the Young modulus. (b) A wire of this metal has diameter 0.80 mm. Calculate the greatest load it can support. (c) A 1.2 m length of the wire is stretched to a strain of 2.0 × 10⁻³ and the load is removed. Estimate its permanent extension.
(a) Gradient from the origin to P: \(E = 90\times10^{6}\div0.75\times10^{-3}\) → E = 1.2 × 10¹¹ Pa.
(b) Greatest load = breaking stress × area: \(A = \pi\times(0.80\times10^{-3})^2\div4 = 5.03\times10^{-7}\ \text{m}^2\), \(F = 220\times10^{6}\times5.03\times10^{-7} = 111\ \text{N}\) → 1.1 × 10² N.
(c) At \(\varepsilon = 2.0\times10^{-3}\) the stress is about 116 MPa. On unloading the elastic part is recovered along a line of gradient \(E\): elastic strain \(= 116\times10^{6}\div1.2\times10^{11} = 9.7\times10^{-4}\). Permanent strain \(\approx 2.0\times10^{-3} - 0.97\times10^{-3} = 1.0\times10^{-3}\), so permanent extension \(\approx 1.0\times10^{-3}\times1.2\) → ≈ 1.2 mm.
"Breaking stress" means the maximum stress on the curve, not the stress at the point where the line stops. For a ductile metal the line ends lower than the maximum because the wire has necked.
A hoist must hold a load of mass 2500 kg on a single steel cable. The breaking stress of the steel is 1.6 × 10⁹ Pa and, for safety, the stress in the cable must not exceed one fifth of this. The Young modulus of the steel is 2.0 × 10¹¹ Pa. (a) Calculate the minimum diameter of the cable. (b) Calculate the extension of a 32 m length of this cable when it carries the load.
Greatest allowed stress \(\sigma = 1.6\times10^{9}\div5 = 3.2\times10^{8}\ \text{Pa}\); load \(F = 2500\times9.81 = 2.45\times10^{4}\ \text{N}\).
(a) \(A = F/\sigma = 2.45\times10^{4}\div3.2\times10^{8} = 7.66\times10^{-5}\ \text{m}^2\), so \(d = \sqrt{4A/\pi}\) → d = 9.9 mm.
(b) \(\varepsilon = \sigma/E = 3.2\times10^{8}\div2.0\times10^{11} = 1.6\times10^{-3}\), so \(\Delta x = 1.6\times10^{-3}\times32\) → Δx = 51 mm.
The cable diameter must be rounded up to a size that is available, so the real stress is a little lower.
How do you describe a material from its stress–strain graph?
Everyday words like "strong" and "tough" have precise meanings in physics, and each one is linked to a different feature of the stress–strain graph.
| Property | Meaning | Evidence on a stress–strain graph |
|---|---|---|
| Stiff | large stress needed for a small strain | steep straight section (large \(E\)) |
| Strong | large stress needed to break it | high maximum (large breaking stress) |
| Brittle | breaks with little or no plastic deformation | line ends at or just after the straight section |
| Ductile | can be drawn out into a wire; large plastic deformation under tension | long plastic region before breaking |
| Malleable | can be hammered or pressed permanently into a new shape | large plastic deformation (shown on a compressive graph) |
| Hard | resists scratching and denting of its surface | not read from the graph; hard materials usually yield at high stress |
| Tough | absorbs a lot of energy before breaking; resists cracks | large area under the curve up to breaking |
The properties are independent: glass is stiff and strong but brittle; copper is ductile but not very strong; a material can be strong without being tough. The opposite of brittle is tough or ductile, not "strong"; the opposite of stiff is flexible, not "weak".
Use the graph of materials P, Q and R. Identify, with a reason from the graph, the material that is (a) the stiffest, (b) the strongest, (c) brittle, (d) the most ductile and (e) the toughest.
(a) Gradients of the straight sections: P: 480 MPa ÷ 0.008 = 6.0 × 10¹⁰ Pa; Q: 280 MPa ÷ 0.007 = 4.0 × 10¹⁰ Pa; R: 80 MPa ÷ 0.010 = 8.0 × 10⁹ Pa → P is the stiffest.
(b), (c) P reaches 480 MPa, above Q (400 MPa) and R (120 MPa) → P is the strongest; its line ends where it stops being straight → P is brittle.
(d) R breaks at a strain of 0.090, Q at 0.080, P at 0.008 → R is the most ductile.
(e) Area under P = \(\tfrac12\times480\times10^{6}\times0.008 \approx 1.9\times10^{6}\ \text{J m}^{-3}\); Q's area is about 15 times this and R's about 5 times → Q is the toughest. P is the strongest material but the least tough: strength and toughness are different properties.
Using "strong" for everything. If a question asks which material is stiffer, the answer must be about the gradient; if it asks which is tougher, it must be about the area. Name the feature of the graph each time.
How do you determine the Young modulus of a wire (Core Practical 3)?
Idea: load a wire, measure its extension, and plot stress against strain; the gradient of the straight section is \(E\).
- Why long and thin: metal strains are tiny. A long wire (2–3 m) gives a larger extension for the same strain; a thin wire gives a large stress for a modest load. Both make the extension big enough to measure with a small percentage uncertainty.
- Set-up: clamp one end firmly; run the wire along the bench over a pulley; attach a small pointer (a tape flag) near the pulley, against a fixed scale; start with a small load to straighten the wire.
- Original length \(x\): from the clamp to the pointer, with a metre rule or tape measure.
- Diameter \(d\): a micrometer at several points along the wire and in two directions at right angles at each point; take the mean. This finds the average size and allows for a wire that is not perfectly round.
- Extension: add masses in equal steps; record the pointer position each time and subtract the starting reading. A vernier scale (or a travelling microscope) gives a finer reading than a millimetre scale.
- Check: remove the masses one at a time and read again; returning readings show the wire stayed elastic.
- Safety: eye protection in case the wire snaps; a box of padding under the masses; keep feet clear.
The micrometer readings of the wire's diameter were 0.30, 0.29, 0.31, 0.30 and 0.30 mm. The original length was 2.40 m. (a) Calculate the mean diameter and its percentage uncertainty. (b) Show how the point for a 2.0 kg load was calculated: its extension was 3.4 mm. (c) Use the graph to determine the Young modulus of the steel.
(a) Mean, with half the range as the uncertainty: d = 0.300 mm ± 0.01 mm, percentage uncertainty = \(0.01\div0.30\times100\) → 3.3%.
(b) \(A = \pi\times(0.300\times10^{-3})^2\div4 = 7.07\times10^{-8}\ \text{m}^2\); \(\sigma = 19.62\div7.07\times10^{-8} = 2.78\times10^{8}\ \text{Pa}\); \(\varepsilon = 3.4\times10^{-3}\ \text{m}\div2.40 = 1.42\times10^{-3}\).
(c) Large triangle on the best-fit line, from \((0.50\times10^{-3},\ 0.99\times10^{8}\ \text{Pa})\) to \((2.50\times10^{-3},\ 5.01\times10^{8}\ \text{Pa})\): \(E = 4.02\times10^{8}\div2.0\times10^{-3} = 2.01\times10^{11}\ \text{Pa}\) → E = 2.0 × 10¹¹ Pa.
For comparison: the largest extension, 6.6 mm, read to ± 0.05 mm, has a percentage uncertainty of 0.76%, and the length (2.40 m ± 0.5 mm) only 0.02%. The diameter dominates, and it is squared in the area, so measure it with the most care.
When asked to improve this experiment, aim at the largest percentage uncertainty: take more diameter readings along the wire, or use a longer wire to increase the extension. Repeating extension readings while unloading also shows whether the elastic limit was passed.
Exam tips for 1.9
Change millimetres to metres before calculating an area: 0.50 mm = 0.50 × 10⁻³ m, and (0.50 × 10⁻³)² = 2.5 × 10⁻⁷ m². An area left in mm² gives a stress a million times too large.
Stress axes are often labelled in MPa or 10⁸ Pa and strain axes in 10⁻³ or %. Write the powers of ten into the gradient calculation; a Young modulus for a metal should come out at around 10¹⁰–10¹¹ Pa.
The Young modulus, breaking stress and properties such as "ductile" belong to the material. Stiffness \(k\), extension and breaking force belong to a particular sample and change with its length and area.
When you describe a material, back each word with a feature: "stiff: the straight section is steep", "brittle: it breaks at the end of the straight section with no plastic region", "tough: large area under the curve".
For each quantity give the instrument and how it is used: micrometer on the diameter at several places, metre rule for the original length from the clamp to the pointer, fixed scale or vernier for the extension, equal load steps, then a graph of stress against strain.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A thicker wire of the same metal has a larger Young modulus. | A larger area reduces the stress and the strain in the same proportion. | \(E\) is a property of the material; thickness changes the object's stiffness \(k\). |
| Strain is measured in metres. | It is a length divided by a length. | Strain has no unit; it may be written as a percentage. |
| Strong and stiff mean the same thing. | Stiffness is about strain under a stress; strength is about the stress at breaking. | Stiff = steep gradient; strong = high breaking stress. |
| Brittle materials are weak. | Brittleness is about how a material breaks, not the stress needed. | Glass and ceramics can be very strong but still brittle. |
| The breaking stress is where the graph ends. | A ductile wire necks after the maximum, so the line ends lower. | Breaking stress (UTS) is the highest stress on the curve. |
| Hard means tough. | Hardness concerns the surface; toughness concerns energy absorbed before breaking. | A hard material such as glass can be brittle and not tough. |
How this topic connects
1.8 Hooke's law and elastic strain energy
The force–extension graph of a sample becomes a stress–strain graph when you divide by \(A\) and \(x\); the object's stiffness is \(k = EA/x\), and the area under the force–extension graph is the elastic strain energy.
1.2 Forces and Newton's laws
The force in stress = \(F/A\) is usually a tension or a weight found from equilibrium: the same free-body diagrams as for ropes and hanging loads.
2.8 Resistivity and 3.3 The AS core practicals
Finding the resistivity of a wire uses the same micrometer readings and the same area \(\pi d^2/4\); both experiments are most sensitive to the diameter. Mean and half-range uncertainties and gradients from a large triangle are practised here too.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\sigma = F/A\) | tensile or compressive stress (Pa) | yes |
| \(\varepsilon = \Delta x/x\) | tensile or compressive strain (no unit) | yes |
| \(E = \sigma/\varepsilon\) | Young modulus (Pa) | yes |
| \(A = \pi d^2/4\) | area of a wire from its diameter | no — learn it |
| \(E = Fx/(A\Delta x)\) | Young modulus from raw measurements | no — derive it from the three above |
| \(\Delta x = Fx/(EA)\) | predicting an extension | no — derive it |
| \(k = EA/x\) | stiffness of a wire or rod (link to 1.8) | no — derive it |
Check your understanding
1. Express the pascal in SI base units.
1 Pa = 1 N m⁻² = 1 kg m s⁻² ÷ m² = 1 kg m⁻¹ s⁻².
2. A nylon cord 2.50 m long stretches by 45 mm. Calculate the strain.
\(\varepsilon = 0.045\div2.50 = 0.018\) (1.8%), with no unit.
3. Wire B is made of the same metal as wire A but is twice as long and has twice the diameter. Both carry the same load. Compare their extensions.
\(\Delta x = Fx/(EA)\). Doubling \(x\) doubles \(\Delta x\); doubling \(d\) quadruples \(A\). B's extension is \(2\div4 = 0.5\) times A's.
4. Explain the difference between a ductile and a malleable material.
Both undergo large plastic deformation without breaking. Ductile describes this under tension (the material can be drawn into a wire); malleable describes it under compression (it can be hammered or rolled into sheets).
5. Explain why a long wire is used in the Young modulus core practical.
For a given stress the strain is fixed, so the extension is proportional to the original length. A long wire gives a larger extension, which reduces its percentage uncertainty.
6. A steel wire (E = 2.0 × 10¹¹ Pa) of length 3.0 m and cross-sectional area 1.5 × 10⁻⁷ m² carries 30 N. Calculate its extension.
\(\Delta x = Fx/(EA) = 30\times3.0\div(2.0\times10^{11}\times1.5\times10^{-7}) = 3.0\times10^{-3}\ \text{m} = 3.0\ \text{mm}\).
Specification checklist
- (28) I can calculate tensile and compressive stress from a force and a cross-sectional area, including the area of a wire from its diameter.
- (28) I can calculate tensile and compressive strain and know that strain has no unit.
- (28) I can define the Young modulus as stress ÷ strain and use it to find stresses, strains and extensions.
- (30) I can sketch and interpret stress–strain graphs, find the Young modulus from the gradient and identify the breaking stress (ultimate tensile stress).
- (30) I can describe materials as stiff, strong, brittle, ductile, malleable, hard or tough, using features of their stress–strain graphs as evidence.
- (31) I can describe how to determine the Young modulus of a wire, naming the instruments and explaining the choice of a long, thin wire.
- (31) I can process the results: mean diameter and its uncertainty, stress and strain for each load, and the Young modulus from the gradient.