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3.7.6Forces and Motion

Young’s Modulus

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3.7.6 Young's Modulus — OCR A Level Physics A notes: Define and calculate Young's modulus and describe how to measure it for a wire.For the region where stress is proportional to strain, the ratio is a constant of the material called the Young modulus: E = stress / strain = (F/A)…

  • Worked example 1
  • Worked example 2
  • Worked example 3
  • Micrometer
  • Hookean region
  • Energy density

Section 1 of 8Definition

For the region where stress is proportional to strain, the ratio is a constant of the material called the Young modulus: E = stress / strain = (F/A) / (Δl/l) = Fl / AΔl.

Definition
Its unit is the pascal (Pa), the same as stress because strain has no unit; typical values for metals are of the order 1011 Pa (steel about 2.0 × 1011 Pa, copper about 1.2 × 1011 Pa),
  • and for rubber near 107 Pa.
A high value means a stiff material that undergoes little strain for a given stress.

Unlike the force constant k, E is independent of the sample's length and area, so it is a property of the material. The two are linked by k = EA / l: a wire is stiffer when it is thicker (larger A), shorter (smaller l) or made of stiffer material.

Section 2 of 8On the stress–strain graph

The gradient of the straight (Hookean) part of a stress–strain graph equals the Young modulus.

On the stress–strain graph
If the graph is a force–extension graph, the gradient is k = F/Δl, and the Young modulus is found from E = (gradient × l) / A.
The area under a stress–strain graph is energy per unit volume, so the energy stored in a stretched wire is Eel = ½ × stress × strain × volume = ½FΔl.

Section 3 of 8Experiment: Young modulus of a wire

A long thin wire (about 2 m of copper or constantan) is clamped at one end, passes over a pulley, and carries a hanging mass holder. A reference marker (a strip of tape) is attached near the load and a metre rule fixed beside it, or a travelling microscope or vernier is used to read the extension. Method:

1
Measure the original length l from the fixed clamp to the marker with a metre rule (uncertainty about ±1 mm on 2 m is small).
2
Measure the diameter of the wire at several points along the length and in perpendicular directions with a micrometer, and average.
  • The area A = πd2/4.
3
Add a small initial load to straighten the wire, then add masses in equal steps, recording the extension each time.
  • Also unload, checking that the wire returns to its original length.
4
Plot force F (or stress) against extension (or strain). The gradient is F/Δl, so E = gradient × l / A.

The diameter is the biggest source of uncertainty since A depends on d2: a 2% error in d gives a 4% error in A. Safety: wear goggles and keep feet clear because a breaking wire can whip, and cushion the floor beneath the masses.

Section 4 of 8Why E is a useful quantity

Engineers use E to predict deflection of beams, stretch of cables and to choose materials for stiffness. At atomic level E is related to the slope of the force–separation curve at equilibrium spacing: stronger bonds produce a larger E. Composite materials such as carbon fibre combine high E with low density, so are used where stiffness-to-weight is key.

Section 5 of 8Worked strategy and links

When asked for the Young modulus, first find the area from the diameter, then the stress, then the strain, then divide, keeping units in SI throughout.

1
If a graph is supplied, find the gradient using a large triangle.
2
Link to energy: the energy stored per unit volume in a stretched wire equals ½ × stress × strain,
  • so a material with a high E can store energy only with a high stress.
3
Link also to 3.7.1: k = EA/l shows how a wire's stiffness depends on both the material and its geometry.
4
In evaluation answers, note that using a longer wire gives a larger extension and smaller percentage uncertainty in the extension.

Section 6 of 8Worked examples

  1. A steel wire 2.5 m long with diameter 0.60 mm carries a load of 80 N and extends 4.2 mm. Find the Young modulus.

  2. A wire has k = 600 N m−1, length 1.8 m and area 1.0 × 10−7 m2. Find its Young modulus.

  3. A copper wire (E = 1.2 × 1011 Pa) of length 3.0 m and cross-section 1.5 × 10−6 m2 carries a force of 90 N. Find the extension and the energy stored.

Section 7 of 8Key terms

TermMeaning
Young modulusRatio of stress to strain in the linear region, in Pa.
StiffnessResistance to deformation, high E means stiff.
MicrometerInstrument to measure wire diameter to ±0.01 mm.
GradientSlope of a graph; here gives F/Δl or E.
Hookean regionRegion where stress is proportional to strain.
Energy densityEnergy stored per unit volume, ½ × stress × strain.

Section 8 of 8Exam technique

1
Quoting the unit of E as N or N m. It is Pa (N m⁻²).
2
Using the diameter in place of the area, or radius in place of the diameter, in the area formula.
3
Using the extended length in strain instead of the original length.
4
Taking the gradient of a graph with extension on the y-axis and force on the x-axis as k.
5
Using non-linear region data to determine E.
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