Mechanical properties of matter

What you'll learn

How materials deform under tension and compression, and how to calculate the energy stored in them.
The fundamental definitions of stress, strain, and ultimate tensile strength.
How to measure and calculate the Young modulus of a metal using experimental techniques.
How to distinguish between elastic and plastic deformation, and interpret stress-strain graphs for different material types.

1. Deforming Materials: Force and Extension

When a force is applied to an object, it can change its shape. In tension, forces pull the object, causing extension. In compression, forces push the object, causing it to shorten.

Force-Extension Graphs

If we apply a tensile force $F$ to a material (like a wire or a spring) and measure its extension $x$ , we can plot a force-extension graph.

For many materials, the initial part of this graph is a straight line through the origin. This represents Hooke's law, which states that extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded:

F = k x

where $k$ is the force constant (often called the stiffness of the material or spring constant), measured in newtons per metre ( $m−1\text{N}\,\text{m}^{-1}$ ).

Force-Extension Graph

Definition

Elastic deformation

Elastic deformation is when a material returns to its original shape and length once the deforming forces (load) are removed.

Definition

Plastic deformation

Plastic deformation is when a material is permanently stretched or deformed and does not return to its original shape when the load is removed. This occurs when the load exceeds the elastic limit of the material.

Elastic Potential Energy

Key Idea

Work Done and Energy Stored

The work done in stretching or compressing a material is equal to the area under its force-extension (or force-compression) graph. Within the limit of proportionality, this work is stored entirely as elastic potential energy ( $E$ ).

Mathematically, for a material obeying Hooke's law, the work done (and thus the elastic potential energy stored, $E$ ) is represented by the triangular area under the straight-line graph:

\frac{1}{2}Fx

Since $F = k x$ , we can substitute this into the equation to get:

\frac{1}{2}kx^2

Where:

$E$ is the elastic potential energy in joules ( $J\text{J}$ )
$F$ is the stretching force in newtons ( $N\text{N}$ )
$x$ is the extension in metres ( $m\text{m}$ )
$k$ is the force constant in newtons per metre ( $m−1\text{N}\,\text{m}^{-1}$ )

Example

Calculating energy stored in a spring

A steel spring has an unstretched length of $m0.150\,\text{m}$ and a force constant of $m−1120\,\text{N}\,\text{m}^{-1}$ . Calculate the work required to stretch the spring from a length of $m0.200\,\text{m}$ to a length of $m0.280\,\text{m}$ , assuming it behaves elastically.

Calculate the initial extension ( $x_1$ ) and final extension ( $x_2$ ): The extensions must be calculated from the original unstretched length ( $ml_0 = 0.150\,\text{m}$ ):

x_1 = 0.200\,\text{m} - 0.150\,\text{m} = 0.050\,\text{m}

x_2 = 0.280\,\text{m} - 0.150\,\text{m} = 0.130\,\text{m}

Calculate the initial elastic potential energy ( $E_1$ ): Using the formula $\frac{1}{2}kx^2$ :

E_1 = \frac{1}{2} \times 120\,\text{N}\,\text{m}^{-1} \times (0.050\,\text{m})^2 = 0.15\,\text{J}

Calculate the final elastic potential energy ( $E_2$ ):

E_2 = \frac{1}{2} \times 120\,\text{N}\,\text{m}^{-1} \times (0.130\,\text{m})^2 = 1.014\,\text{J}

Determine the work required to stretch the spring between these two points ( $ΔE\Delta E$ ): The work required is the change in elastic potential energy:

\Delta E = E_2 - E_1 = 1.014\,\text{J} - 0.150\,\text{J} = 0.864\,\text{J}

Rounding to a sensible number of significant figures (2 s.f.):

\Delta E \approx 0.86\,\text{J}

Common Mistake

Incorrect change in extension squared

When calculating the change in elastic potential energy ( $ΔE\Delta E$ ), a very common mistake is to square the difference in extension, i.e., writing $ΔE=12k(x2−x1)2\Delta E = \frac{1}{2}k(x_2 - x_1)^2$ . This is mathematically incorrect! You must calculate the final energy and initial energy separately and subtract them:

\Delta E = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}k(x_2^2 - x_1^2)

2. Tensile Stress and Strain

While force and extension describe a specific specimen, they depend heavily on the dimensions of the sample. To compare the mechanical properties of different materials directly, we must define scale-independent quantities: stress and strain.

Definition

Tensile stress

Tensile stress ( $σ\sigma$ ) is defined as the tensile force per unit cross-sectional area of a material.

\sigma = \frac{F}{A}

Where:

$σ\sigma$ is stress in pascals ( $Pa\text{Pa}$ or $m−2\text{N}\,\text{m}^{-2}$ )
$F$ is the applied force in newtons ( $N\text{N}$ )
$A$ is the cross-sectional area in square metres ( $m2\text{m}^2$ )

Definition

Tensile strain

Tensile strain ( $ε\varepsilon$ ) is defined as the extension per unit original length of a material. It is a ratio of two lengths and therefore has no units (it is dimensionless).

\varepsilon = \frac{\Delta L}{L}

Where:

$ε\varepsilon$ is strain
$ΔL\Delta L$ is the extension in metres ( $m\text{m}$ )
$L$ is the original length of the material in metres ( $m\text{m}$ )

Definition

Ultimate Tensile Strength (UTS)

The Ultimate Tensile Strength (UTS) is the maximum tensile stress a material can withstand before it breaks or fractures. Beyond this stress level, the material will undergo "necking" (narrowing at its weakest point) and quickly snap.

3. The Young Modulus

Definition

Young Modulus

The Young modulus ( $E$ ) is the ratio of tensile stress to tensile strain for a material within its limit of proportionality. It is a measure of the stiffness of a material.

\frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon}

Because strain is dimensionless, the unit of the Young modulus is the same as stress: pascals ( $Pa\text{Pa}$ ) or newtons per square metre ( $m−2\text{N}\,\text{m}^{-2}$ ).

We can combine the expressions for stress and strain to form a single equation for the Young modulus:

\frac{F / A}{\Delta L / L} = \frac{FL}{A \Delta L}

Example

Calculating the extension of a loaded wire

A brass wire of original length $m2.5\,\text{m}$ and diameter $mm0.60\,\text{mm}$ is suspended vertically. A mass of $kg4.5\,\text{kg}$ is hung from the free end. The Young modulus of brass is $1.0 \times 10^{11}\,\text{Pa}$ . Calculate the extension of the wire. Use $9.81\,\text{m}\,\text{s}^{-2}$ .

Calculate the tensile force (tension $F$ ) acting on the wire: The tension is equal to the weight of the hung mass:

4.5\,\text{kg} \times 9.81\,\text{m}\,\text{s}^{-2} = 44.145\,\text{N}

Calculate the cross-sectional area ( $A$ ) of the wire: First, find the radius $r$ from the diameter $0.60\,\text{mm} = 6.0 \times 10^{-4}\,\text{m}$ :

\frac{d}{2} = 3.0 \times 10^{-4}\,\text{m}

Now, calculate the circular area:

\pi r^2 = \pi \times (3.0 \times 10^{-4}\,\text{m})^2 = 2.8274 \times 10^{-7}\,\text{m}^2

Rearrange the Young modulus equation to solve for extension ( $ΔL\Delta L$ ):

\frac{FL}{A\Delta L} \implies \Delta L = \frac{FL}{AE}

Substitute the known values and solve:

\Delta L = \frac{44.145\,\text{N} \times 2.5\,\text{m}}{2.8274 \times 10^{-7}\,\text{m}^2 \times 1.0 \times 10^{11}\,\text{Pa}}

\Delta L \approx 3.9 \times 10^{-3}\,\text{m}

The extension of the wire is $mm3.9\,\text{mm}$ (expressed to 2 significant figures, matched to the input data).

4. Experimental Determination of the Young Modulus

In the laboratory, you can determine the Young modulus of a metal wire using a bench-top experimental setup (aligned with Practical Activity Group 2 — PAG2).

Young Modulus Experimental Setup

Method and Measurements

Measure the original length $L$ : Measure the distance from the wooden clamp blocks to the paper tape marker using a metre ruler.
Measure the wire diameter $d$ : Use a micrometer screw gauge to measure the diameter of the wire at several points along its length and at different angles. Calculate the average diameter and find the area $\frac{\pi d^2}{4}$ .
Add load and measure extension: Add slotted masses to the hanger, recording the force $F = m g$ applied. For each mass, measure the new position of the paper marker on the metre ruler. Subtract the initial position to find the extension $ΔL\Delta L$ .

Minimising Uncertainties

Long Wire: Using a very long wire (usually $2\,\text{m}$ ) ensures that the extension $ΔL\Delta L$ is large enough to keep the percentage uncertainty in the extension reading low.
Diameter Checks: Measuring the diameter at multiple points and angles accounts for any non-uniformity in the wire's cross-section.
Unloading: Remove the weights in stages to ensure the marker returns to the starting position. This confirms that the wire has not undergone permanent plastic deformation.

Graphical Analysis

Since $\left( \frac{EA}{L} \right) \Delta L$ , we can plot a graph of Force $F$ (y-axis) against Extension $ΔL\Delta L$ (x-axis).

The gradient of the linear region is equal to $EAL\frac{EA}{L}$ .
Therefore, the Young modulus can be determined from the gradient:

\text{gradient} \times \frac{L}{A}

Alternatively, plotting Stress $σ\sigma$ (y-axis) against Strain $ε\varepsilon$ (x-axis) produces a straight line through the origin where the gradient is directly equal to the Young modulus $E$ .

Example

Determining Young modulus from experimental parameters

An experiment is conducted using a copper wire of original length $m2.85\,\text{m}$ and diameter $mm0.46\,\text{mm}$ . A plot of Force $F$ against Extension $ΔL\Delta L$ yields a straight line with a gradient of $1.11 \times 10^4\,\text{N}\,\text{m}^{-1}$ . Calculate the Young modulus of copper.

Calculate the cross-sectional area of the wire: The radius is $\frac{0.46 \times 10^{-3}\,\text{m}}{2} = 2.3 \times 10^{-4}\,\text{m}$ .

\pi r^2 = \pi \times (2.3 \times 10^{-4}\,\text{m})^2 = 1.6619 \times 10^{-7}\,\text{m}^2

Relate the gradient of the graph to the Young modulus: Since $\left( \frac{EA}{L} \right) \Delta L$ , the gradient of the force-extension graph is:

\text{gradient} = \frac{EA}{L}

Rearrange and solve for $E$ :

\text{gradient} \times \frac{L}{A}

\times 10^4\,\text{N}\,\text{m}^{-1}) \times \frac{2.85\,\text{m}}{1.6619 \times 10^{-7}\,\text{m}^2}

\times 10^{11}\,\text{Pa}

The Young modulus of copper is $1.9 \times 10^{11}\,\text{Pa}$ (or $GPa190\,\text{GPa}$ ), expressed to 2 significant figures.

5. Stress-Strain Graphs for Different Materials

The shape of a stress-strain graph tells us how different classes of materials behave under load.

Stress-Strain Graphs

Material Classes

Definition

Ductile material

A ductile material can be drawn into wires and shows a large amount of plastic deformation before breaking (e.g., copper, mild steel).

Definition

Brittle material

A brittle material shows little or no plastic deformation. It obeys Hooke's law up to its breaking point, where it fractures suddenly (e.g., glass, cast iron).

Definition

Polymeric material

A polymeric material is made of long-chain molecules (polymers). These materials can undergo massive extensions (often elastic) and generally do not obey Hooke's law (e.g., rubber, polythene).

Rubber and Hysteresis

Rubber is a polymeric material. It can undergo massive elastic strain because its long, tangled molecular chains can easily unfold and stretch when loaded.

When rubber is loaded and then unloaded, the unloading curve lies below the loading curve. This forms a loop known as a hysteresis loop.

The area under the loading curve represents the work done per unit volume to stretch the rubber.
The area under the unloading curve represents the useful elastic energy returned per unit volume as it contracts.
The area enclosed within the loop represents the mechanical energy per unit volume that is lost as thermal energy (heating up the rubber).

Example

Calculating thermal energy dissipated in rubber hysteresis

A rubber band of unstretched length $m0.12\,\text{m}$ is loaded and unloaded. The area under the loading force-extension curve is $J0.48\,\text{J}$ . The area under the unloading curve is $J0.32\,\text{J}$ . Calculate: a) The work done in stretching the rubber band. b) The useful elastic energy recovered during unloading. c) The thermal energy transferred to the rubber band during this cycle.

Identify the meaning of the area under the loading curve: The area under the loading curve represents the work done on the rubber band to stretch it:

\text{Work done} = 0.48\,\text{J}

Identify the meaning of the area under the unloading curve: The area under the unloading curve represents the work done by the rubber band as it contracts (the useful elastic potential energy recovered):

\text{Recovered energy} = 0.32\,\text{J}

Calculate the thermal energy dissipated using the difference between the two areas: The energy that was put in but not recovered as mechanical work is dissipated as thermal energy (heating up the rubber band):

\text{Thermal energy} = \text{Work done} - \text{Recovered energy}

\text{Thermal energy} = 0.48\,\text{J} - 0.32\,\text{J} = 0.16\,\text{J}

Tip

Units check for Stress and Young Modulus

Always check the prefix of the stress or Young modulus values in exam questions! They are usually given in megapascals ( $Pa\text{MPa} = 10^6\,\text{Pa}$ ) or gigapascals ( $Pa\text{GPa} = 10^9\,\text{Pa}$ ). Always convert to the base unit ( $Pa\text{Pa}$ ) before carrying out your calculations.

Common Mistake

Area and radius conversions

One of the most frequent mathematical traps is converting wire diameters in millimetres ( $mm\text{mm}$ ) to cross-sectional areas in square metres ( $m2\text{m}^2$ ). Remember:

Divide diameter by 2 to get radius $r$ .
Convert $r$ from $mm\text{mm}$ to $m\text{m}$ (multiply by $10^{-3}$ ).
Square the radius (which squares the $10^{-3}$ conversion factor, turning it into $10^{-6}$ ).
Multiply by $π\pi$ .

\pi \left( \frac{d}{2} \times 10^{-3} \right)^2

Exam technique

In the exam

Always read the ruler scale direction: In Young modulus experimental questions, check if the extension measurements are cumulative or absolute positions. If positions are given, subtract the initial reading from each subsequent reading to get the true extensions.
Distinguish force-extension vs. stress-strain: Force-extension graphs depend on the physical dimensions of the specimen. Stress-strain graphs describe the bulk material properties, meaning their gradient (Young modulus) is constant for a given material regardless of its thickness or length.
Show all intermediate conversions: In multi-step calculation questions (like calculating extension), write down the values of cross-sectional area, force, and conversions in standard form. This secures partial marks even if you make a calculation error at the final stage.
Learn the experimental details: Be prepared to describe how to measure the diameter of a wire (using a micrometer in multiple directions) and how to ensure the elastic limit is not exceeded (by checking if the wire returns to its original length after unloading).

Self review

Check yourself

Can you explain the difference between the limit of proportionality and the elastic limit on a force-extension graph?
Why is a thin, long wire preferred over a thick, short wire when setting up an experiment to determine the Young modulus of a metal?
How is the energy dissipated per unit volume represented on a stress-strain graph for a rubber band during a loading and unloading cycle?

What you'll learn

How materials deform under tension and compression, and how to calculate the energy stored in them.
The fundamental definitions of stress, strain, and ultimate tensile strength.
How to measure and calculate the Young modulus of a metal using experimental techniques.
How to distinguish between elastic and plastic deformation, and interpret stress-strain graphs for different material types.

1. Deforming Materials: Force and Extension

When a force is applied to an object, it can change its shape. In tension, forces pull the object, causing extension. In compression, forces push the object, causing it to shorten.

Force-Extension Graphs

If we apply a tensile force $F$ to a material (like a wire or a spring) and measure its extension $x$ , we can plot a force-extension graph.

F = k x

where $k$ is the force constant (often called the stiffness of the material or spring constant), measured in newtons per metre ( $m−1\text{N}\,\text{m}^{-1}$ ).

Force-Extension Graph

Definition

Elastic deformation

Elastic deformation is when a material returns to its original shape and length once the deforming forces (load) are removed.

Definition

Plastic deformation

Elastic Potential Energy

Key Idea

Work Done and Energy Stored

Mathematically, for a material obeying Hooke's law, the work done (and thus the elastic potential energy stored, $E$ ) is represented by the triangular area under the straight-line graph:

\frac{1}{2}Fx

Since $F = k x$ , we can substitute this into the equation to get:

\frac{1}{2}kx^2

Where:

$E$ is the elastic potential energy in joules ( $J\text{J}$ )
$F$ is the stretching force in newtons ( $N\text{N}$ )
$x$ is the extension in metres ( $m\text{m}$ )
$k$ is the force constant in newtons per metre ( $m−1\text{N}\,\text{m}^{-1}$ )

Example

Calculating energy stored in a spring

Calculate the initial extension ( $x_1$ ) and final extension ( $x_2$ ): The extensions must be calculated from the original unstretched length ( $ml_0 = 0.150\,\text{m}$ ):

x_1 = 0.200\,\text{m} - 0.150\,\text{m} = 0.050\,\text{m}

x_2 = 0.280\,\text{m} - 0.150\,\text{m} = 0.130\,\text{m}

Calculate the initial elastic potential energy ( $E_1$ ): Using the formula $\frac{1}{2}kx^2$ :

E_1 = \frac{1}{2} \times 120\,\text{N}\,\text{m}^{-1} \times (0.050\,\text{m})^2 = 0.15\,\text{J}

Calculate the final elastic potential energy ( $E_2$ ):

E_2 = \frac{1}{2} \times 120\,\text{N}\,\text{m}^{-1} \times (0.130\,\text{m})^2 = 1.014\,\text{J}

Determine the work required to stretch the spring between these two points ( $ΔE\Delta E$ ): The work required is the change in elastic potential energy:

\Delta E = E_2 - E_1 = 1.014\,\text{J} - 0.150\,\text{J} = 0.864\,\text{J}

Rounding to a sensible number of significant figures (2 s.f.):

\Delta E \approx 0.86\,\text{J}

Common Mistake

Incorrect change in extension squared

\Delta E = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}k(x_2^2 - x_1^2)

2. Tensile Stress and Strain

Definition

Tensile stress

Tensile stress ( $σ\sigma$ ) is defined as the tensile force per unit cross-sectional area of a material.

\sigma = \frac{F}{A}

Where:

$σ\sigma$ is stress in pascals ( $Pa\text{Pa}$ or $m−2\text{N}\,\text{m}^{-2}$ )
$F$ is the applied force in newtons ( $N\text{N}$ )
$A$ is the cross-sectional area in square metres ( $m2\text{m}^2$ )

Definition

Tensile strain

Tensile strain ( $ε\varepsilon$ ) is defined as the extension per unit original length of a material. It is a ratio of two lengths and therefore has no units (it is dimensionless).

\varepsilon = \frac{\Delta L}{L}

Where:

$ε\varepsilon$ is strain
$ΔL\Delta L$ is the extension in metres ( $m\text{m}$ )
$L$ is the original length of the material in metres ( $m\text{m}$ )

Definition

Ultimate Tensile Strength (UTS)

3. The Young Modulus

Definition

Young Modulus

The Young modulus ( $E$ ) is the ratio of tensile stress to tensile strain for a material within its limit of proportionality. It is a measure of the stiffness of a material.

\frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon}

Because strain is dimensionless, the unit of the Young modulus is the same as stress: pascals ( $Pa\text{Pa}$ ) or newtons per square metre ( $m−2\text{N}\,\text{m}^{-2}$ ).

We can combine the expressions for stress and strain to form a single equation for the Young modulus:

\frac{F / A}{\Delta L / L} = \frac{FL}{A \Delta L}

Example

Calculating the extension of a loaded wire

Calculate the tensile force (tension $F$ ) acting on the wire: The tension is equal to the weight of the hung mass:

4.5\,\text{kg} \times 9.81\,\text{m}\,\text{s}^{-2} = 44.145\,\text{N}

Calculate the cross-sectional area ( $A$ ) of the wire: First, find the radius $r$ from the diameter $0.60\,\text{mm} = 6.0 \times 10^{-4}\,\text{m}$ :

\frac{d}{2} = 3.0 \times 10^{-4}\,\text{m}

Now, calculate the circular area:

\pi r^2 = \pi \times (3.0 \times 10^{-4}\,\text{m})^2 = 2.8274 \times 10^{-7}\,\text{m}^2

Rearrange the Young modulus equation to solve for extension ( $ΔL\Delta L$ ):

\frac{FL}{A\Delta L} \implies \Delta L = \frac{FL}{AE}

Substitute the known values and solve:

\Delta L = \frac{44.145\,\text{N} \times 2.5\,\text{m}}{2.8274 \times 10^{-7}\,\text{m}^2 \times 1.0 \times 10^{11}\,\text{Pa}}

\Delta L \approx 3.9 \times 10^{-3}\,\text{m}

The extension of the wire is $mm3.9\,\text{mm}$ (expressed to 2 significant figures, matched to the input data).

4. Experimental Determination of the Young Modulus

In the laboratory, you can determine the Young modulus of a metal wire using a bench-top experimental setup (aligned with Practical Activity Group 2 — PAG2).

Young Modulus Experimental Setup

Method and Measurements

Measure the original length $L$ : Measure the distance from the wooden clamp blocks to the paper tape marker using a metre ruler.
Measure the wire diameter $d$ : Use a micrometer screw gauge to measure the diameter of the wire at several points along its length and at different angles. Calculate the average diameter and find the area $\frac{\pi d^2}{4}$ .
Add load and measure extension: Add slotted masses to the hanger, recording the force $F = m g$ applied. For each mass, measure the new position of the paper marker on the metre ruler. Subtract the initial position to find the extension $ΔL\Delta L$ .

Minimising Uncertainties

Long Wire: Using a very long wire (usually $2\,\text{m}$ ) ensures that the extension $ΔL\Delta L$ is large enough to keep the percentage uncertainty in the extension reading low.
Diameter Checks: Measuring the diameter at multiple points and angles accounts for any non-uniformity in the wire's cross-section.
Unloading: Remove the weights in stages to ensure the marker returns to the starting position. This confirms that the wire has not undergone permanent plastic deformation.

Graphical Analysis

Since $\left( \frac{EA}{L} \right) \Delta L$ , we can plot a graph of Force $F$ (y-axis) against Extension $ΔL\Delta L$ (x-axis).

The gradient of the linear region is equal to $EAL\frac{EA}{L}$ .
Therefore, the Young modulus can be determined from the gradient:

\text{gradient} \times \frac{L}{A}

Example

Determining Young modulus from experimental parameters

Calculate the cross-sectional area of the wire: The radius is $\frac{0.46 \times 10^{-3}\,\text{m}}{2} = 2.3 \times 10^{-4}\,\text{m}$ .

\pi r^2 = \pi \times (2.3 \times 10^{-4}\,\text{m})^2 = 1.6619 \times 10^{-7}\,\text{m}^2

Relate the gradient of the graph to the Young modulus: Since $\left( \frac{EA}{L} \right) \Delta L$ , the gradient of the force-extension graph is:

\text{gradient} = \frac{EA}{L}

Rearrange and solve for $E$ :

\text{gradient} \times \frac{L}{A}

\times 10^4\,\text{N}\,\text{m}^{-1}) \times \frac{2.85\,\text{m}}{1.6619 \times 10^{-7}\,\text{m}^2}

\times 10^{11}\,\text{Pa}

The Young modulus of copper is $1.9 \times 10^{11}\,\text{Pa}$ (or $GPa190\,\text{GPa}$ ), expressed to 2 significant figures.

5. Stress-Strain Graphs for Different Materials

The shape of a stress-strain graph tells us how different classes of materials behave under load.

Stress-Strain Graphs

Material Classes

Definition

Ductile material

A ductile material can be drawn into wires and shows a large amount of plastic deformation before breaking (e.g., copper, mild steel).

Definition

Brittle material

A brittle material shows little or no plastic deformation. It obeys Hooke's law up to its breaking point, where it fractures suddenly (e.g., glass, cast iron).

Definition

Polymeric material

A polymeric material is made of long-chain molecules (polymers). These materials can undergo massive extensions (often elastic) and generally do not obey Hooke's law (e.g., rubber, polythene).

Rubber and Hysteresis

Rubber is a polymeric material. It can undergo massive elastic strain because its long, tangled molecular chains can easily unfold and stretch when loaded.

When rubber is loaded and then unloaded, the unloading curve lies below the loading curve. This forms a loop known as a hysteresis loop.

The area under the loading curve represents the work done per unit volume to stretch the rubber.
The area under the unloading curve represents the useful elastic energy returned per unit volume as it contracts.
The area enclosed within the loop represents the mechanical energy per unit volume that is lost as thermal energy (heating up the rubber).

Example

Calculating thermal energy dissipated in rubber hysteresis

Identify the meaning of the area under the loading curve: The area under the loading curve represents the work done on the rubber band to stretch it:

\text{Work done} = 0.48\,\text{J}

Identify the meaning of the area under the unloading curve: The area under the unloading curve represents the work done by the rubber band as it contracts (the useful elastic potential energy recovered):

\text{Recovered energy} = 0.32\,\text{J}

Calculate the thermal energy dissipated using the difference between the two areas: The energy that was put in but not recovered as mechanical work is dissipated as thermal energy (heating up the rubber band):

\text{Thermal energy} = \text{Work done} - \text{Recovered energy}

\text{Thermal energy} = 0.48\,\text{J} - 0.32\,\text{J} = 0.16\,\text{J}

Tip

Units check for Stress and Young Modulus

Common Mistake

Area and radius conversions

One of the most frequent mathematical traps is converting wire diameters in millimetres ( $mm\text{mm}$ ) to cross-sectional areas in square metres ( $m2\text{m}^2$ ). Remember:

Divide diameter by 2 to get radius $r$ .
Convert $r$ from $mm\text{mm}$ to $m\text{m}$ (multiply by $10^{-3}$ ).
Square the radius (which squares the $10^{-3}$ conversion factor, turning it into $10^{-6}$ ).
Multiply by $π\pi$ .

\pi \left( \frac{d}{2} \times 10^{-3} \right)^2

Exam technique

In the exam

Always read the ruler scale direction: In Young modulus experimental questions, check if the extension measurements are cumulative or absolute positions. If positions are given, subtract the initial reading from each subsequent reading to get the true extensions.
Distinguish force-extension vs. stress-strain: Force-extension graphs depend on the physical dimensions of the specimen. Stress-strain graphs describe the bulk material properties, meaning their gradient (Young modulus) is constant for a given material regardless of its thickness or length.
Show all intermediate conversions: In multi-step calculation questions (like calculating extension), write down the values of cross-sectional area, force, and conversions in standard form. This secures partial marks even if you make a calculation error at the final stage.
Learn the experimental details: Be prepared to describe how to measure the diameter of a wire (using a micrometer in multiple directions) and how to ensure the elastic limit is not exceeded (by checking if the wire returns to its original length after unloading).

Self review

Check yourself

Can you explain the difference between the limit of proportionality and the elastic limit on a force-extension graph?
Why is a thin, long wire preferred over a thick, short wire when setting up an experiment to determine the Young modulus of a metal?
How is the energy dissipated per unit volume represented on a stress-strain graph for a rubber band during a loading and unloading cycle?

What you'll learn

1. Deforming Materials: Force and Extension

Force-Extension Graphs

Elastic Potential Energy

2. Tensile Stress and Strain

3. The Young Modulus

4. Experimental Determination of the Young Modulus

Method and Measurements

Minimising Uncertainties

Graphical Analysis

5. Stress-Strain Graphs for Different Materials

Material Classes

Rubber and Hysteresis

Review Mechanical properties of matter by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mechanical properties of matter

What you'll learn

1. Deforming Materials: Force and Extension

Force-Extension Graphs

Elastic Potential Energy

2. Tensile Stress and Strain

3. The Young Modulus

4. Experimental Determination of the Young Modulus

Method and Measurements

Minimising Uncertainties

Graphical Analysis

5. Stress-Strain Graphs for Different Materials

Material Classes

Rubber and Hysteresis

Review Mechanical properties of matter by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mechanical properties of matter

What you'll learn

1. Deforming Materials: Force and Extension

Force-Extension Graphs

Elastic Potential Energy

2. Tensile Stress and Strain

3. The Young Modulus

4. Experimental Determination of the Young Modulus

Method and Measurements

Minimising Uncertainties

Graphical Analysis

5. Stress-Strain Graphs for Different Materials

Material Classes

Rubber and Hysteresis

Review Mechanical properties of matter by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mechanical properties of matter Revision Guide

Mechanical properties of matter

What you'll learn

1. Deforming Materials: Force and Extension

Force-Extension Graphs

Elastic Potential Energy

2. Tensile Stress and Strain

3. The Young Modulus

4. Experimental Determination of the Young Modulus

Method and Measurements

Minimising Uncertainties

Graphical Analysis

5. Stress-Strain Graphs for Different Materials

Material Classes

Rubber and Hysteresis

Review Mechanical properties of matter by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Mechanical properties of matter Revision Guide