A-Level Physics

Materials

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The mechanical response of a material depends on both the material itself and the geometry of the sample. Tensile stress is force divided by original cross-sectional area, σ = \(\dfrac{F}{A}\), while tensile strain is extension divided by original length, ε = \(\dfrac{\Delta L}{L}\). Strain is dimensionless. In the initial linear elastic region, stress is proportional to strain and the Young modulus is E = \(\dfrac{\sigma }{\varepsilon }\). A large Young modulus means a material is stiff: a relatively large stress is required to produce a given strain.

Elastic deformation is reversible when the load is removed, whereas plastic deformation leaves a permanent change. The limit of proportionality marks the end of linear stress-strain behaviour. Yielding, elastic limit and fracture are distinct ideas and should not be collapsed into one point unless a particular simplified graph justifies it. Ductile materials can undergo substantial plastic strain before fracture; brittle materials fracture with little plastic deformation. Strength and stiffness are also different: strength concerns the stress a material can withstand, whereas stiffness is related to Young modulus.

The work done in stretching a sample is represented by the area under a force-extension graph. For a linear spring-like response this reduces to \(\dfrac{1}{2}F\Delta L\). Stress-strain graphs allow different sample geometries to be compared because force and extension have been normalised by area and original length.

Stress, strain and Young modulus

Tensile stress is force per original cross-sectional area, σ = \(\dfrac{F}{A}\). Tensile strain is extension divided by original length, ε = \(\dfrac{\Delta L}{L}\), and has no unit. In the linear elastic region, Young modulus E = \(\dfrac{\sigma }{\varepsilon }\). Young modulus measures stiffness of the material, not the strength of a particular sample.

Because stress and strain normalise for geometry, they allow different samples and dimensions to be compared more meaningfully than force and extension alone.

Force-extension and stress-strain behaviour

For an ideal Hookean region, force is proportional to extension and stress is proportional to strain. The limit of proportionality marks the end of this linear relation. Elastic deformation is recovered when the load is removed; plastic deformation leaves permanent strain.

On a stress–strain graph, the yield point is where large plastic strain begins, the ultimate tensile stress is the maximum stress, and the fracture point is where the sample breaks. A material can be stiff but brittle, or less stiff but able to absorb much more energy before fracture.

Experimental determination

Young modulus can be determined by measuring force, original length, extension and diameter. Diameter is particularly important because area depends on diameter squared. A micrometer should be used at several positions and orientations if the wire may not be perfectly uniform.

A long, thin wire gives a larger extension for a given stress, improving percentage resolution, provided the deformation remains elastic and safe.

Material comparison

Stiffness, strength, ductility, brittleness and toughness describe different properties. Do not use them interchangeably. Young modulus compares stiffness; ultimate tensile stress concerns maximum stress; toughness relates to energy absorption before fracture.

Exam technique

Use original dimensions for engineering stress and strain unless told otherwise. Convert diameter to radius before calculating area. When comparing materials, name the property supported by the graph rather than using generic words such as 'stronger'.

Worked example

Worked example 1: A wire carries 120 N and has area 2.0×10⁻⁶ m², so stress = 6.0×10⁷ Pa. If strain is 3.0×10⁻⁴, Young modulus = \(\dfrac{6.0\times 10^{7}}{3.0\times 10^{-4}}\) = 2.0×10¹¹ Pa.

Worked example 2: A 2.0 m wire of area 1.5 × 10⁻⁶ m² extends by 1.2 mm under 120 N. Stress = 8.0 × 10⁷ Pa, strain = 6.0 × 10⁻⁴, so Young modulus = 1.33 × 10¹¹ Pa.

For stress 1.5 × 10⁸ Pa and strain 7.5 × 10⁻⁴ in the linear region, E = 2.0 × 10¹¹ Pa.

Worked example 3: For F = 90 N on area 1.2 × 10⁻⁶ m², stress = 7.5 × 10⁷ Pa. If strain is 5.0 × 10⁻⁴, E = 1.5 × 10¹¹ Pa.

Misconception check

A strong material is not necessarily stiff. Strength concerns failure stress, while stiffness is measured by Young modulus in the elastic region.

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Define tensile stress.

σ = \(\dfrac{F}{A}\), using the original cross-sectional area.

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