Bulk properties of solids

Welcome to the bulk properties of solids! In mechanics so far, we've often treated objects as perfect, rigid blocks that don't bend or break. In reality, everything stretches, squashes, or snaps when you push it hard enough.

What you'll learn in this topic:

How to define and calculate density.
How springs behave under tension (Hooke's Law) and how much energy they store.
The difference between a whole object's stiffness and a material's inherent properties (Stress and Strain).
How to read force-extension and stress-strain graphs to identify brittle and plastic materials.
How engineers use these concepts to design safe, ethical transport.

Let's build this up step by step.

1. Density

Before we stretch anything, we need a basic property of solids: density. You will have met this at GCSE, but it remains a crucial foundation at A-Level.

Definition

Density

Density ( $ρ\rho$ ) is the mass per unit volume of a material.

\rho = \frac{m}{V}

Where $m$ is mass in kg, $V$ is volume in $m3\text{m}^3$ , and $ρ\rho$ is density in $m−3\text{kg m}^{-3}$ .

When dealing with regular shapes, you can calculate the volume geometrically (e.g. $\times w \times h$ for a block, or $\pi r^2 l$ for a wire). For irregular shapes, you might estimate the volume by submerging the object in water and measuring the displaced volume.

2. Hooke's Law and Springs

If you hang a weight from a spring, it extends. For many objects, the extension is directly proportional to the force applied, provided you don't stretch it too far.

Definition

Hooke's Law

Hooke's Law states that the extension of an elastic object is directly proportional to the force applied to it, up to the limit of proportionality.

k\Delta L

Where $F$ is the force in N, $ΔL\Delta L$ is the extension in m, and $k$ is the spring constant (or stiffness) in $m−1\text{N m}^{-1}$ .

The greater the value of $k$ , the stiffer the spring. You need more force to get the same extension.

What happens if you pull too hard?

Eventually, Hooke's Law breaks down. If you plot a graph of Force against Extension, it starts as a straight line through the origin, but eventually begins to curve.

Force-Extension Graph

There are two critical points on this graph you must know:

Limit of proportionality: The exact point where the graph stops being a straight line. Hooke's law is no longer obeyed beyond this point.
Elastic limit: The maximum force you can apply without causing permanent deformation. If you stretch an object beyond its elastic limit, it will not return to its original length when the force is removed. This permanent stretch is called plastic deformation.

Common Mistake

Confusing the two limits

Students often use "limit of proportionality" and "elastic limit" interchangeably. They are technically different! The limit of proportionality always happens slightly before the elastic limit. A material can stop obeying Hooke's Law (it curves) but still return to its original length.

3. Elastic Strain Energy

When you stretch a spring, you are doing work. This work is stored in the spring as elastic strain energy.

Key Idea

Area under the graph

The work done (and therefore the elastic strain energy stored) is equal to the area under the force–extension graph.

For the straight-line portion of the graph (where Hooke's Law is obeyed), the area is a triangle. The base is $ΔL\Delta L$ and the height is $F$ .

\frac{1}{2} F \Delta L

Because $k\Delta L$ , we can substitute $F$ into the energy equation to get a second, incredibly useful version:

\frac{1}{2} k (\Delta L)^2

Energy Conservation in Action

Springs are fantastic for storing energy and releasing it as Kinetic Energy (KE) or Gravitational Potential Energy (GPE). If a spring is used to launch a projectile vertically, the elastic strain energy converts first to kinetic energy, and then to GPE as the projectile climbs.

Example

Vertical spring launch

A spring with a stiffness of $\text{ N m}^{-1}$ is compressed by $\text{ m}$ . A ball of mass $\text{ kg}$ is placed on top. The spring is released. Calculate the maximum height the ball reaches above its starting position. Assume air resistance is negligible.

First, calculate the elastic strain energy initially stored in the spring using $\frac{1}{2}k(\Delta L)^2$ .

\frac{1}{2} \times 400 \times (0.05)^2

\times 0.0025 = 0.5 \text{ J}

State the energy conversion. By the principle of conservation of energy, all the elastic strain energy converts into gravitational potential energy at the maximum height (since the ball stops momentarily, KE is zero).

\text{Elastic Strain Energy} = \text{Gravitational Potential Energy}

\text{ J} = mg\Delta h

Rearrange for $Δh\Delta h$ and substitute the values (using $\text{ m s}^{-2}$ ).

\Delta h = \frac{0.5}{mg}

\Delta h = \frac{0.5}{0.15 \times 9.81}

\Delta h = \frac{0.5}{1.4715} \approx 0.34 \text{ m}

4. Stress and Strain

The spring constant $k$ is great for a specific object (like "that one specific spring on your desk"). But what if we want to compare the stiffness of copper vs steel?

A thick steel wire is harder to stretch than a thin steel wire. A long steel wire stretches more than a short steel wire under the same force. To compare materials, we need to remove the dimensions (length and thickness) from the equation. We do this by calculating Stress and Strain.

Definition

Tensile Stress

Tensile stress ( $σ\sigma$ ) is the force applied per unit cross-sectional area.

\sigma = \frac{F}{A}

Where $F$ is the tension force in N, and $A$ is the cross-sectional area in $m2\text{m}^2$ . The unit of stress is the Pascal ( $Pa\text{Pa}$ ), which is equivalent to $m−2\text{N m}^{-2}$ .

Definition

Tensile Strain

Tensile strain ( $ε\varepsilon$ ) is the extension per unit original length.

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

Where $ΔL\Delta L$ is the extension in m, and $L$ is the original length in m. Strain is a ratio of two lengths, so it has no units.

Tip

Area of a wire

In exam questions, wires are almost always cylindrical. To find the cross-sectional area $A$ , you will usually be given the diameter $d$ . Remember to halve it to get the radius $r$ , and then use $\pi r^2$ .

5. Material Behaviours and Stress-Strain Curves

Just as we plotted force against extension for a specific object, we can plot Stress against Strain for a specific material. The shape of this curve tells us how the material behaves under extreme tension.

Stress-Strain Graphs

Let's define some key behaviours shown on these graphs:

Elastic behaviour: The material returns to its original length when the stress is removed.
Plastic behaviour: The material permanently deforms. On a molecular level, layers of atoms are sliding past each other.
Ductile: A ductile material (like copper) has a large plastic region. It can be easily drawn into wires because it stretches a lot before breaking.
Brittle: A brittle material (like glass or cast iron) snaps without any noticeable plastic deformation. Its graph is a steep straight line that just abruptly ends. It doesn't "give" before it shatters.
Fracture: The point where the material physically snaps.
Breaking stress (Ultimate Tensile Stress): The maximum stress a material can withstand before it breaks. It is the peak of the stress-strain curve.

Example

Calculating Breaking Stress

A steel wire of original length $\text{ m}$ and diameter $\text{ mm}$ hangs vertically. It breaks when a mass of $\text{ kg}$ is attached to the end. Calculate the breaking stress of the steel.

First, calculate the cross-sectional area in $m2\text{m}^2$ . The diameter is $\text{ mm}$ , so the radius is $\text{ mm} = 0.4 \times 10^{-3} \text{ m}$ .

\pi r^2 = \pi \times (0.4 \times 10^{-3})^2

\pi \times 1.6 \times 10^{-7} \approx 5.03 \times 10^{-7} \text{ m}^2

Calculate the force (weight) that caused the wire to break.

\times 9.81 = 1177.2 \text{ N}

Calculate the breaking stress using $σ=FA\sigma = \frac{F}{A}$ .

\sigma = \frac{1177.2}{5.03 \times 10^{-7}}

\sigma \approx 2.34 \times 10^9 \text{ Pa}

6. Ethical Transport Design

Physics isn't just about abstract wires; it saves lives. AQA specifically wants you to appreciate how energy conservation and material properties apply to ethical transport design.

When a car crashes, its massive kinetic energy has to go somewhere.

If the car is completely rigid, the kinetic energy is transferred rapidly to the passengers, causing huge forces and lethal injuries.
Modern cars are designed with crumple zones. These are parts of the car chassis engineered to undergo plastic deformation during a crash.

By deforming plastically, the crumple zone does two things:

It absorbs a huge amount of the kinetic energy (converting it into the work done to permanently deform the metal, and heat).
It increases the time taken for the car to come to a stop, which reduces the deceleration and therefore reduces the force on the passengers.

The Ethical Balance: Engineers face ethical decisions. They must balance passenger safety with the mass of the vehicle (heavier, highly-reinforced cars burn more fuel and release more emissions) and cost (making a perfectly safe car might make it completely unaffordable for the average family).

Exam technique

In the exam

Watch your prefixes: Strain and extension questions are notorious for mixing units. You will see lengths in metres, extensions in millimetres, diameters in micrometres, and stresses in MegaPascals ( $MPa\text{MPa}$ ). Convert absolutely everything to standard SI units ( $m\text{m}$ , $N\text{N}$ , $Pa\text{Pa}$ ) before pressing a button on your calculator.
Squaring errors: When using $\pi r^2$ , don't forget to square the $×10−3\times 10^{-3}$ part of the radius if you are converting from millimetres! Writing $0.5 \times 10^{-3})^2$ is safer than doing it in your head.
Read the graph axes: If you are asked to find energy from a graph, check the $x$ -axis. Is it in $m\text{m}$ or $mm\text{mm}$ ? If it's in $mm\text{mm}$ , your area calculation will be off by a factor of 1000 unless you convert it.

Self review

Check yourself

Can you write down the equations for density, stress, and strain from memory, including their SI units?
What is the difference between the limit of proportionality and the elastic limit?
How would you spot a brittle material from its stress-strain curve compared to a ductile one?
Why do crumple zones in cars rely on plastic deformation rather than elastic deformation?

What you'll learn in this topic:

How to define and calculate density.
How springs behave under tension (Hooke's Law) and how much energy they store.
The difference between a whole object's stiffness and a material's inherent properties (Stress and Strain).
How to read force-extension and stress-strain graphs to identify brittle and plastic materials.
How engineers use these concepts to design safe, ethical transport.

Let's build this up step by step.

1. Density

Before we stretch anything, we need a basic property of solids: density. You will have met this at GCSE, but it remains a crucial foundation at A-Level.

Definition

Density

Density ( $ρ\rho$ ) is the mass per unit volume of a material.

\rho = \frac{m}{V}

Where $m$ is mass in kg, $V$ is volume in $m3\text{m}^3$ , and $ρ\rho$ is density in $m−3\text{kg m}^{-3}$ .

2. Hooke's Law and Springs

If you hang a weight from a spring, it extends. For many objects, the extension is directly proportional to the force applied, provided you don't stretch it too far.

Definition

Hooke's Law

Hooke's Law states that the extension of an elastic object is directly proportional to the force applied to it, up to the limit of proportionality.

k\Delta L

Where $F$ is the force in N, $ΔL\Delta L$ is the extension in m, and $k$ is the spring constant (or stiffness) in $m−1\text{N m}^{-1}$ .

The greater the value of $k$ , the stiffer the spring. You need more force to get the same extension.

What happens if you pull too hard?

Eventually, Hooke's Law breaks down. If you plot a graph of Force against Extension, it starts as a straight line through the origin, but eventually begins to curve.

Force-Extension Graph

There are two critical points on this graph you must know:

Limit of proportionality: The exact point where the graph stops being a straight line. Hooke's law is no longer obeyed beyond this point.
Elastic limit: The maximum force you can apply without causing permanent deformation. If you stretch an object beyond its elastic limit, it will not return to its original length when the force is removed. This permanent stretch is called plastic deformation.

Common Mistake

Confusing the two limits

3. Elastic Strain Energy

When you stretch a spring, you are doing work. This work is stored in the spring as elastic strain energy.

Key Idea

Area under the graph

The work done (and therefore the elastic strain energy stored) is equal to the area under the force–extension graph.

For the straight-line portion of the graph (where Hooke's Law is obeyed), the area is a triangle. The base is $ΔL\Delta L$ and the height is $F$ .

\frac{1}{2} F \Delta L

Because $k\Delta L$ , we can substitute $F$ into the energy equation to get a second, incredibly useful version:

\frac{1}{2} k (\Delta L)^2

Energy Conservation in Action

Example

Vertical spring launch

First, calculate the elastic strain energy initially stored in the spring using $\frac{1}{2}k(\Delta L)^2$ .

\frac{1}{2} \times 400 \times (0.05)^2

\times 0.0025 = 0.5 \text{ J}

State the energy conversion. By the principle of conservation of energy, all the elastic strain energy converts into gravitational potential energy at the maximum height (since the ball stops momentarily, KE is zero).

\text{Elastic Strain Energy} = \text{Gravitational Potential Energy}

\text{ J} = mg\Delta h

Rearrange for $Δh\Delta h$ and substitute the values (using $\text{ m s}^{-2}$ ).

\Delta h = \frac{0.5}{mg}

\Delta h = \frac{0.5}{0.15 \times 9.81}

\Delta h = \frac{0.5}{1.4715} \approx 0.34 \text{ m}

4. Stress and Strain

The spring constant $k$ is great for a specific object (like "that one specific spring on your desk"). But what if we want to compare the stiffness of copper vs steel?

Definition

Tensile Stress

Tensile stress ( $σ\sigma$ ) is the force applied per unit cross-sectional area.

\sigma = \frac{F}{A}

Where $F$ is the tension force in N, and $A$ is the cross-sectional area in $m2\text{m}^2$ . The unit of stress is the Pascal ( $Pa\text{Pa}$ ), which is equivalent to $m−2\text{N m}^{-2}$ .

Definition

Tensile Strain

Tensile strain ( $ε\varepsilon$ ) is the extension per unit original length.

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

Where $ΔL\Delta L$ is the extension in m, and $L$ is the original length in m. Strain is a ratio of two lengths, so it has no units.

Tip

Area of a wire

5. Material Behaviours and Stress-Strain Curves

Stress-Strain Graphs

Let's define some key behaviours shown on these graphs:

Elastic behaviour: The material returns to its original length when the stress is removed.
Plastic behaviour: The material permanently deforms. On a molecular level, layers of atoms are sliding past each other.
Ductile: A ductile material (like copper) has a large plastic region. It can be easily drawn into wires because it stretches a lot before breaking.
Brittle: A brittle material (like glass or cast iron) snaps without any noticeable plastic deformation. Its graph is a steep straight line that just abruptly ends. It doesn't "give" before it shatters.
Fracture: The point where the material physically snaps.
Breaking stress (Ultimate Tensile Stress): The maximum stress a material can withstand before it breaks. It is the peak of the stress-strain curve.

Example

Calculating Breaking Stress

A steel wire of original length $\text{ m}$ and diameter $\text{ mm}$ hangs vertically. It breaks when a mass of $\text{ kg}$ is attached to the end. Calculate the breaking stress of the steel.

First, calculate the cross-sectional area in $m2\text{m}^2$ . The diameter is $\text{ mm}$ , so the radius is $\text{ mm} = 0.4 \times 10^{-3} \text{ m}$ .

\pi r^2 = \pi \times (0.4 \times 10^{-3})^2

\pi \times 1.6 \times 10^{-7} \approx 5.03 \times 10^{-7} \text{ m}^2

Calculate the force (weight) that caused the wire to break.

\times 9.81 = 1177.2 \text{ N}

Calculate the breaking stress using $σ=FA\sigma = \frac{F}{A}$ .

\sigma = \frac{1177.2}{5.03 \times 10^{-7}}

\sigma \approx 2.34 \times 10^9 \text{ Pa}

6. Ethical Transport Design

Physics isn't just about abstract wires; it saves lives. AQA specifically wants you to appreciate how energy conservation and material properties apply to ethical transport design.

When a car crashes, its massive kinetic energy has to go somewhere.

If the car is completely rigid, the kinetic energy is transferred rapidly to the passengers, causing huge forces and lethal injuries.
Modern cars are designed with crumple zones. These are parts of the car chassis engineered to undergo plastic deformation during a crash.

By deforming plastically, the crumple zone does two things:

It absorbs a huge amount of the kinetic energy (converting it into the work done to permanently deform the metal, and heat).
It increases the time taken for the car to come to a stop, which reduces the deceleration and therefore reduces the force on the passengers.

Exam technique

In the exam

Watch your prefixes: Strain and extension questions are notorious for mixing units. You will see lengths in metres, extensions in millimetres, diameters in micrometres, and stresses in MegaPascals ( $MPa\text{MPa}$ ). Convert absolutely everything to standard SI units ( $m\text{m}$ , $N\text{N}$ , $Pa\text{Pa}$ ) before pressing a button on your calculator.
Squaring errors: When using $\pi r^2$ , don't forget to square the $×10−3\times 10^{-3}$ part of the radius if you are converting from millimetres! Writing $0.5 \times 10^{-3})^2$ is safer than doing it in your head.
Read the graph axes: If you are asked to find energy from a graph, check the $x$ -axis. Is it in $m\text{m}$ or $mm\text{mm}$ ? If it's in $mm\text{mm}$ , your area calculation will be off by a factor of 1000 unless you convert it.

Self review

Check yourself

Can you write down the equations for density, stress, and strain from memory, including their SI units?
What is the difference between the limit of proportionality and the elastic limit?
How would you spot a brittle material from its stress-strain curve compared to a ductile one?
Why do crumple zones in cars rely on plastic deformation rather than elastic deformation?

1. Density

2. Hooke's Law and Springs

What happens if you pull too hard?

3. Elastic Strain Energy

Energy Conservation in Action

4. Stress and Strain

5. Material Behaviours and Stress-Strain Curves

6. Ethical Transport Design

Review Bulk properties of solids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Bulk properties of solids

1. Density

2. Hooke's Law and Springs

What happens if you pull too hard?

3. Elastic Strain Energy

Energy Conservation in Action

4. Stress and Strain

5. Material Behaviours and Stress-Strain Curves

6. Ethical Transport Design

Review Bulk properties of solids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Bulk properties of solids

1. Density

2. Hooke's Law and Springs

What happens if you pull too hard?

3. Elastic Strain Energy

Energy Conservation in Action

4. Stress and Strain

5. Material Behaviours and Stress-Strain Curves

6. Ethical Transport Design

Review Bulk properties of solids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Bulk properties of solids Revision Guide

Bulk properties of solids

1. Density

2. Hooke's Law and Springs

What happens if you pull too hard?

3. Elastic Strain Energy

Energy Conservation in Action

4. Stress and Strain

5. Material Behaviours and Stress-Strain Curves

6. Ethical Transport Design

Review Bulk properties of solids by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Bulk properties of solids Revision Guide