Stress, strain and the Young modulusEdexcel International A Level Physics: Revision notes
Section 1
Stress and strain
Tensile stress is the force per unit cross-sectional area:
σ = F / A, in pascals (Pa = N m⁻²)
Tensile strain is the extension per unit original length:
ε = ΔL / L, with no unit (a ratio, sometimes given as a percentage)
For a compression the same definitions apply, using the compressive force and the decrease in length.
For a wire of diameter d, the area is A = πd²/4. Convert mm to m before calculating: 0.80 mm = 0.80 × 10⁻³ m.
Use the area of the cross-section (πr² or πd²/4), not the length of the wire, and convert the diameter from mm to m first.
Section 2
The Young modulus
The Young modulus is the ratio of stress to strain for a material, within its limit of proportionality:
E = σ / ε = (F/A) / (ΔL/L), in Pa
Unlike stiffness k, E depends only on the material, not on the size or shape of the sample. A large E means a stiff material (steel is about 2 × 10¹¹ Pa).
Worked example: a wire of length 2.50 m and diameter 0.80 mm carries 120 N and extends by 3.0 mm. A = π(0.40 × 10⁻³)² = 5.0 × 10⁻⁷ m²; σ = 120 / 5.0 × 10⁻⁷ = 2.4 × 10⁸ Pa; ε = 3.0 × 10⁻³ / 2.50 = 1.2 × 10⁻³; E = 2.4 × 10⁸ / 1.2 × 10⁻³ = 2.0 × 10¹¹ Pa.
Section 3
Stress-strain graphs
A stress-strain graph has stress on the vertical axis and strain on the horizontal axis.
- In the straight section the gradient is the Young modulus.
- The graph curves after the limit of proportionality. A ductile metal then shows elastic limit, yield and a long region of plastic deformation.
- The breaking stress is the stress at which the material fractures.
- A brittle material, such as a ceramic or glass, gives a straight line up to fracture with no plastic region, so it breaks suddenly.
Because stress and strain remove the effect of size, the stress-strain graph is the same for any sample of the same material, whereas the force-extension graph depends on the dimensions of the sample.
Section 4
Core practical 3: the Young modulus of a wire
Use a long, thin wire (large extension for a given load, small area so large stress), clamped at one end and passing over a pulley with masses on the free end.
- Measure the original length L from the clamp to a marker with a metre rule.
- Measure the diameter with a micrometer at several points and in perpendicular directions; take the mean and find A = πd²/4.
- Add masses in equal steps, measuring the extension from the marker against a scale each time.
- Plot force against extension: a straight line within the limit of proportionality, gradient = EA/L.
- E = gradient × L / A.
Safety: wear goggles and use a box of sand under the masses in case the wire snaps. Do not exceed the elastic limit.
Taking the diameter at several places reduces the effect of a non-uniform wire, and a long wire reduces the percentage uncertainty in the extension.
Must Know
- σ = F/A; ε = ΔL/L; E = σ/ε
- E depends on the material only; k depends on the object
- Stress-strain gradient (straight section) = E
- Breaking stress = stress at fracture
- Brittle: no plastic region; ductile: long plastic region
- E = gradient × L / A from a force-extension graph
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Stress, strain and the Young modulus
- A steel wire of original length 2.50 m and diameter 0.80 mm hangs vertically from a rigid support. A load of 120 N is attached to the lower end of the wire and the wire extends by 3.0 mm. The wire obeys Hooke's law.Calculate the Young modulus of the steel.2 marks
- A ductile metal wire of cross-sectional area 2.0 × 10⁻⁷ m² is stretched until it breaks. The stress is directly proportional to the strain up to a stress of 2.4 × 10⁸ Pa, and the wire breaks when the stress reaches 3.6 × 10⁸ Pa.A second wire is made of the same metal but has twice the diameter. State and explain how its breaking stress and the force needed to break it compare with the first wire.2 marks
- A student determines the Young modulus of a copper wire. One end of the wire is clamped to a bench and the wire passes over a pulley, with masses added to the free end. The original length of the wire, measured from the clamp to a marker with a metre rule, is 2.00 m. The mean diameter of the wire, measured with a micrometer screw gauge, is 0.50 mm. The student plots a graph of force against extension, which is a straight line through the origin of gradient 1.2 × 10⁴ N m⁻¹.Explain why the student uses a long, thin wire.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).