Solids Under Stress

What you'll learn

How to use Hooke’s law, stress, strain and Young modulus quantitatively.
How strain energy is found from the area under a force-extension graph.
How metals, brittle materials and rubber behave differently when stretched.
How to determine the Young modulus of a metal wire and investigate rubber experimentally.

1. Stretching a solid: force and extension

When a solid is pulled, it may extend. For small loads, many materials behave in a simple, predictable way: doubling the force doubles the extension.

Definition

Extension

The extension, $Δl\Delta l$ , is the increase in length of an object compared with its original length. If the original length is $l$ and the new length is $\Delta l$ , then the extension is $Δl\Delta l$ .

Hooke’s law

Hooke’s law states that the force applied to an object is directly proportional to its extension, as long as the limit of proportionality has not been exceeded.

F = k x

Here, $F$ is force in newtons (N), $x$ is extension in metres (m), and $k$ is the spring constant in newtons per metre (N m⁻¹). The spring constant is the force per unit extension.

Key Idea

Hooke’s law graph

On a force-extension graph, Hooke’s law gives a straight line through the origin. The gradient of this line is the spring constant $k$ .

Example

Finding a spring constant

A spring extends by 24 mm when a force of 3.6 N is applied. Find its spring constant.

Convert the extension into metres:
$24\text{ mm} = 2.4 \times 10^{-2}\text{ m}$
Rearrange Hooke’s law and substitute:
$\frac{F}{x} = \frac{3.6\text{ N}}{2.4 \times 10^{-2}\text{ m}}$
Calculate the gradient value:
$\times 10^2\text{ N m}^{-1}$

So the spring constant is 150 N m⁻¹.

Common Mistake

Using millimetres directly

Always convert extension into metres before using $F = k x$ . If you use millimetres, your spring constant will be wrong by a factor of 1000.

2. Stress, strain and Young modulus

A thin wire and a thick rod can be made of the same material, but they will not extend by the same amount under the same force. To describe the material itself, rather than the size of the sample, we use stress and strain.

Definition

Tensile stress

Tensile stress, $σ\sigma$ , is the force applied per unit cross-sectional area:

σ=FA\sigma = \frac{F}{A}

It is measured in pascals (Pa), where one pascal is one newton per square metre.

Definition

Tensile strain

Tensile strain, $ε\varepsilon$ , is the extension per unit original length:

ε=Δll\varepsilon = \frac{\Delta l}{l}

Strain has no unit because it is a ratio of two lengths.

The Young modulus, $E$ , measures the stiffness of a material in the Hooke’s law region.

\frac{\sigma}{\varepsilon}

A large Young modulus means a material is stiff: it needs a large stress to produce a small strain.

Key Idea

Young modulus from a graph

On a stress-strain graph, the gradient of the straight-line elastic region is the Young modulus $E$ .

Example

Calculating the Young modulus of a wire

A metal wire has original length 1.50 m and diameter 0.400 mm. A force of 20.0 N produces an extension of 1.20 mm. Find $E$ .

Calculate the cross-sectional area using $\frac{\pi d^2}{4}$ , with $\times 10^{-4}\text{ m}$ :
$\frac{\pi (4.00 \times 10^{-4}\text{ m})^2}{4} = 1.26 \times 10^{-7}\text{ m}^2$
Calculate the stress and strain:
$Pa\sigma = \frac{20.0\text{ N}}{1.26 \times 10^{-7}\text{ m}^2} = 1.59 \times 10^8\text{ Pa}$ $m=8.00×10−4\varepsilon = \frac{1.20 \times 10^{-3}\text{ m}}{1.50\text{ m}} = 8.00 \times 10^{-4}$
Divide stress by strain:
$\frac{1.59 \times 10^8\text{ Pa}}{8.00 \times 10^{-4}} = 1.99 \times 10^{11}\text{ Pa}$

So the Young modulus is about $2.0 \times 10^{11}\text{ Pa}$ .

3. Strain energy and work done

When you stretch a solid, you do work on it. This work is stored as strain energy if the deformation is elastic.

Definition

Strain energy

Strain energy is the energy stored in a deformed object due to stretching, compression or bending.

The work done in deforming a solid is equal to the area under the force-extension graph.

If Hooke’s law is obeyed, the graph is a straight line through the origin, so the area is a triangle:

\frac{1}{2}Fx

where $W$ is the work done in joules (J).

Example

Finding strain energy from a force-extension graph

A wire obeys Hooke’s law up to a force of 8.0 N. At this force, its extension is 16 mm. Find the strain energy stored.

Convert the extension into metres:
$16\text{ mm} = 1.6 \times 10^{-2}\text{ m}$
Use the triangular area under the graph:
$\frac{1}{2}Fx$
Substitute the force and extension:
$\frac{1}{2}(8.0\text{ N})(1.6 \times 10^{-2}\text{ m}) = 6.4 \times 10^{-2}\text{ J}$

The strain energy stored is 0.064 J.

Tip

Area means energy

If the force-extension graph is curved, do not use $12Fx\frac{1}{2}Fx$ unless Hooke’s law applies. Estimate the area under the curve instead.

4. Elastic and plastic behaviour in metals

A material shows elastic deformation if it returns to its original shape when the force is removed. It shows plastic deformation if it is permanently deformed.

The graph below shows the main regions for a ductile metal such as copper.

$Force-extension graph for a ductile metal showing elastic region, plastic region, necking and fracture$

Key regions on the metal graph

Limit of proportionality: the point beyond which force is no longer directly proportional to extension.
Elastic limit: beyond this, the material will not return fully to its original length.
Yield region: a small increase in force causes a large increase in extension.
Plastic region: permanent deformation occurs.
Ultimate tensile stress: maximum stress reached by the material.
Necking: the material becomes narrower in one region.
Ductile fracture: the material finally breaks after significant plastic deformation.

Dislocations and strengthening metals

A dislocation is a line defect in a crystal structure. Plastic deformation in metals happens largely because dislocations move through the crystal lattice.

Metals can be strengthened by making it harder for dislocations to move. Barriers include:

foreign atoms, as in alloys
other dislocations, caused by work hardening
grain boundaries, made more numerous by having smaller grains

Key Idea

Why barriers strengthen metals

If dislocations cannot move easily, a larger stress is needed to produce plastic deformation.

5. Classes of solid

Solids can be classified by their internal structure.

Crystalline solids

A crystalline solid has atoms arranged in a repeating lattice. Metals are crystalline, often made of many small crystals called grains.

Amorphous solids

An amorphous solid has no long-range repeating structure. Glass is the classic example, and ceramics are treated with this brittle group in this topic.

Polymeric solids

A polymeric solid is made from long chain molecules. Rubber is a polymer with chains that can uncoil and straighten when stretched.

6. Brittle materials and rubber

Different materials have very different force-extension and stress-strain graphs.

Stress-strain graphs comparing brittle glass, ductile metal and rubber hysteresis

Brittle materials such as glass

A brittle material fractures with little or no plastic deformation. Glass is approximately elastic and obeys Hooke’s law up to fracture.

Brittle fracture often happens by crack propagation. A tiny surface crack concentrates stress at its tip, so the crack grows rapidly.

Breaking stress can be increased by:

reducing surface imperfections, as in thin glass fibres
putting the surface under compression, as in toughened glass
pre-stressing concrete, so tensile cracks are less likely to open

Common Mistake

Thinking thick glass is always stronger

Surface flaws often control glass failure. Thin fibres can have a higher breaking stress because they are less likely to contain serious surface cracks.

Rubber

Rubber only approximately obeys Hooke’s law. It has a low Young modulus, so it stretches a lot under relatively small stress.

In rubber, stretching mainly straightens long chain molecules. Thermal motion tends to keep the chains randomly coiled, so the stretch happens against this thermal opposition.

Rubber also shows hysteresis: the unloading curve is different from the loading curve. The area inside the loop on a force-extension graph is energy transferred to the surroundings, mostly as thermal energy.

Key Idea

Rubber hysteresis

For rubber, not all the work done during stretching is recovered during unloading. The missing energy is dissipated, which is why repeatedly flexed rubber can warm up.

7. Practical: Young modulus of a metal wire

In the specified practical, you determine $E$ for a metal wire by measuring force, extension, length and diameter.

Apparatus for determining the Young modulus of a metal wire

Method outline

Measure the original length $l$ of the wire using a metre rule.
Measure the diameter $d$ at several points using a micrometer, checking for zero error.
Calculate the mean diameter, then the area using $\frac{\pi d^2}{4}$ .
Add masses gradually. The load force is $F = m g$ .
Measure the extension $Δl\Delta l$ each time, using a scale, pointer or travelling microscope.
Plot force against extension, or stress against strain.

For a force-extension graph in the Hooke’s law region:

\frac{AE}{l}\Delta l

So if the gradient is $G$ :

\frac{Gl}{A}

Tip

Improving the measurement

Use a long, thin wire because it gives a larger extension, reducing percentage uncertainty. Measure diameter carefully: because $A$ depends on $d^2$ , the percentage uncertainty in area is about twice the percentage uncertainty in diameter.

Common Mistake

Safety

Wear eye protection and do not overload the wire. A stretched wire can snap suddenly, and falling masses can damage the bench or injure feet.

8. Practical: force-extension relationship for rubber

For rubber, add masses in small steps and record the extension. Then remove the masses step by step and record the unloading extensions too.

Plot force against extension. You should expect:

a curved loading graph, not a perfect straight line
a different unloading curve
a hysteresis loop, showing energy dissipated as thermal energy

This investigation is a good place to comment on uncertainty: rubber may continue stretching slowly after a load is added, so take readings after a consistent waiting time.

Exam technique

In the exam

Check whether the question is about a sample or a material: force and extension depend on dimensions, but stress, strain and Young modulus describe the material.
For graph questions, use the correct feature: gradient gives $k$ on a force-extension graph, but $E$ on a stress-strain graph.
For energy, look for area under the force-extension graph; use $12Fx\frac{1}{2}Fx$ only for a straight Hooke’s law region.

Self review

Check yourself

Why does measuring the diameter of a wire carefully matter so much in a Young modulus experiment?
What is the difference between elastic deformation and plastic deformation?
Why can toughened glass have a higher breaking stress than ordinary glass?

What you'll learn

How to use Hooke’s law, stress, strain and Young modulus quantitatively.
How strain energy is found from the area under a force-extension graph.
How metals, brittle materials and rubber behave differently when stretched.
How to determine the Young modulus of a metal wire and investigate rubber experimentally.

1. Stretching a solid: force and extension

When a solid is pulled, it may extend. For small loads, many materials behave in a simple, predictable way: doubling the force doubles the extension.

Definition

Extension

Hooke’s law

Hooke’s law states that the force applied to an object is directly proportional to its extension, as long as the limit of proportionality has not been exceeded.

F = k x

Here, $F$ is force in newtons (N), $x$ is extension in metres (m), and $k$ is the spring constant in newtons per metre (N m⁻¹). The spring constant is the force per unit extension.

Key Idea

Hooke’s law graph

On a force-extension graph, Hooke’s law gives a straight line through the origin. The gradient of this line is the spring constant $k$ .

Example

Finding a spring constant

A spring extends by 24 mm when a force of 3.6 N is applied. Find its spring constant.

Convert the extension into metres:
$24\text{ mm} = 2.4 \times 10^{-2}\text{ m}$
Rearrange Hooke’s law and substitute:
$\frac{F}{x} = \frac{3.6\text{ N}}{2.4 \times 10^{-2}\text{ m}}$
Calculate the gradient value:
$\times 10^2\text{ N m}^{-1}$

So the spring constant is 150 N m⁻¹.

Common Mistake

Using millimetres directly

Always convert extension into metres before using $F = k x$ . If you use millimetres, your spring constant will be wrong by a factor of 1000.

2. Stress, strain and Young modulus

Definition

Tensile stress

Tensile stress, $σ\sigma$ , is the force applied per unit cross-sectional area:

σ=FA\sigma = \frac{F}{A}

It is measured in pascals (Pa), where one pascal is one newton per square metre.

Definition

Tensile strain

Tensile strain, $ε\varepsilon$ , is the extension per unit original length:

ε=Δll\varepsilon = \frac{\Delta l}{l}

Strain has no unit because it is a ratio of two lengths.

The Young modulus, $E$ , measures the stiffness of a material in the Hooke’s law region.

\frac{\sigma}{\varepsilon}

A large Young modulus means a material is stiff: it needs a large stress to produce a small strain.

Key Idea

Young modulus from a graph

On a stress-strain graph, the gradient of the straight-line elastic region is the Young modulus $E$ .

Example

Calculating the Young modulus of a wire

A metal wire has original length 1.50 m and diameter 0.400 mm. A force of 20.0 N produces an extension of 1.20 mm. Find $E$ .

Calculate the cross-sectional area using $\frac{\pi d^2}{4}$ , with $\times 10^{-4}\text{ m}$ :
$\frac{\pi (4.00 \times 10^{-4}\text{ m})^2}{4} = 1.26 \times 10^{-7}\text{ m}^2$
Calculate the stress and strain:
$Pa\sigma = \frac{20.0\text{ N}}{1.26 \times 10^{-7}\text{ m}^2} = 1.59 \times 10^8\text{ Pa}$ $m=8.00×10−4\varepsilon = \frac{1.20 \times 10^{-3}\text{ m}}{1.50\text{ m}} = 8.00 \times 10^{-4}$
Divide stress by strain:
$\frac{1.59 \times 10^8\text{ Pa}}{8.00 \times 10^{-4}} = 1.99 \times 10^{11}\text{ Pa}$

So the Young modulus is about $2.0 \times 10^{11}\text{ Pa}$ .

3. Strain energy and work done

When you stretch a solid, you do work on it. This work is stored as strain energy if the deformation is elastic.

Definition

Strain energy

Strain energy is the energy stored in a deformed object due to stretching, compression or bending.

The work done in deforming a solid is equal to the area under the force-extension graph.

If Hooke’s law is obeyed, the graph is a straight line through the origin, so the area is a triangle:

\frac{1}{2}Fx

where $W$ is the work done in joules (J).

Example

Finding strain energy from a force-extension graph

A wire obeys Hooke’s law up to a force of 8.0 N. At this force, its extension is 16 mm. Find the strain energy stored.

Convert the extension into metres:
$16\text{ mm} = 1.6 \times 10^{-2}\text{ m}$
Use the triangular area under the graph:
$\frac{1}{2}Fx$
Substitute the force and extension:
$\frac{1}{2}(8.0\text{ N})(1.6 \times 10^{-2}\text{ m}) = 6.4 \times 10^{-2}\text{ J}$

The strain energy stored is 0.064 J.

Tip

Area means energy

If the force-extension graph is curved, do not use $12Fx\frac{1}{2}Fx$ unless Hooke’s law applies. Estimate the area under the curve instead.

4. Elastic and plastic behaviour in metals

A material shows elastic deformation if it returns to its original shape when the force is removed. It shows plastic deformation if it is permanently deformed.

The graph below shows the main regions for a ductile metal such as copper.

$Force-extension graph for a ductile metal showing elastic region, plastic region, necking and fracture$

Key regions on the metal graph

Limit of proportionality: the point beyond which force is no longer directly proportional to extension.
Elastic limit: beyond this, the material will not return fully to its original length.
Yield region: a small increase in force causes a large increase in extension.
Plastic region: permanent deformation occurs.
Ultimate tensile stress: maximum stress reached by the material.
Necking: the material becomes narrower in one region.
Ductile fracture: the material finally breaks after significant plastic deformation.

Dislocations and strengthening metals

A dislocation is a line defect in a crystal structure. Plastic deformation in metals happens largely because dislocations move through the crystal lattice.

Metals can be strengthened by making it harder for dislocations to move. Barriers include:

foreign atoms, as in alloys
other dislocations, caused by work hardening
grain boundaries, made more numerous by having smaller grains

Key Idea

Why barriers strengthen metals

If dislocations cannot move easily, a larger stress is needed to produce plastic deformation.

5. Classes of solid

Solids can be classified by their internal structure.

Crystalline solids

A crystalline solid has atoms arranged in a repeating lattice. Metals are crystalline, often made of many small crystals called grains.

Amorphous solids

An amorphous solid has no long-range repeating structure. Glass is the classic example, and ceramics are treated with this brittle group in this topic.

Polymeric solids

A polymeric solid is made from long chain molecules. Rubber is a polymer with chains that can uncoil and straighten when stretched.

6. Brittle materials and rubber

Different materials have very different force-extension and stress-strain graphs.

Stress-strain graphs comparing brittle glass, ductile metal and rubber hysteresis

Brittle materials such as glass

A brittle material fractures with little or no plastic deformation. Glass is approximately elastic and obeys Hooke’s law up to fracture.

Brittle fracture often happens by crack propagation. A tiny surface crack concentrates stress at its tip, so the crack grows rapidly.

Breaking stress can be increased by:

reducing surface imperfections, as in thin glass fibres
putting the surface under compression, as in toughened glass
pre-stressing concrete, so tensile cracks are less likely to open

Common Mistake

Thinking thick glass is always stronger

Surface flaws often control glass failure. Thin fibres can have a higher breaking stress because they are less likely to contain serious surface cracks.

Rubber

Rubber only approximately obeys Hooke’s law. It has a low Young modulus, so it stretches a lot under relatively small stress.

In rubber, stretching mainly straightens long chain molecules. Thermal motion tends to keep the chains randomly coiled, so the stretch happens against this thermal opposition.

Key Idea

Rubber hysteresis

For rubber, not all the work done during stretching is recovered during unloading. The missing energy is dissipated, which is why repeatedly flexed rubber can warm up.

7. Practical: Young modulus of a metal wire

In the specified practical, you determine $E$ for a metal wire by measuring force, extension, length and diameter.

Apparatus for determining the Young modulus of a metal wire

Method outline

Measure the original length $l$ of the wire using a metre rule.
Measure the diameter $d$ at several points using a micrometer, checking for zero error.
Calculate the mean diameter, then the area using $\frac{\pi d^2}{4}$ .
Add masses gradually. The load force is $F = m g$ .
Measure the extension $Δl\Delta l$ each time, using a scale, pointer or travelling microscope.
Plot force against extension, or stress against strain.

For a force-extension graph in the Hooke’s law region:

\frac{AE}{l}\Delta l

So if the gradient is $G$ :

\frac{Gl}{A}

Tip

Improving the measurement

Common Mistake

Safety

Wear eye protection and do not overload the wire. A stretched wire can snap suddenly, and falling masses can damage the bench or injure feet.

8. Practical: force-extension relationship for rubber

For rubber, add masses in small steps and record the extension. Then remove the masses step by step and record the unloading extensions too.

Plot force against extension. You should expect:

a curved loading graph, not a perfect straight line
a different unloading curve
a hysteresis loop, showing energy dissipated as thermal energy

This investigation is a good place to comment on uncertainty: rubber may continue stretching slowly after a load is added, so take readings after a consistent waiting time.

Exam technique

In the exam

Check whether the question is about a sample or a material: force and extension depend on dimensions, but stress, strain and Young modulus describe the material.
For graph questions, use the correct feature: gradient gives $k$ on a force-extension graph, but $E$ on a stress-strain graph.
For energy, look for area under the force-extension graph; use $12Fx\frac{1}{2}Fx$ only for a straight Hooke’s law region.

Self review

Check yourself

Why does measuring the diameter of a wire carefully matter so much in a Young modulus experiment?
What is the difference between elastic deformation and plastic deformation?
Why can toughened glass have a higher breaking stress than ordinary glass?

What you'll learn

1. Stretching a solid: force and extension

Hooke’s law

2. Stress, strain and Young modulus

3. Strain energy and work done

4. Elastic and plastic behaviour in metals

Key regions on the metal graph

Dislocations and strengthening metals

5. Classes of solid

Crystalline solids

Amorphous solids

Polymeric solids

6. Brittle materials and rubber

Brittle materials such as glass

Rubber

7. Practical: Young modulus of a metal wire

Method outline

8. Practical: force-extension relationship for rubber

Review Solids Under Stress by teaching Genie

Remember key concepts with flashcards

Solids Under Stress Revision Guide

Solids Under Stress

What you'll learn

1. Stretching a solid: force and extension

Hooke’s law

2. Stress, strain and Young modulus

3. Strain energy and work done

4. Elastic and plastic behaviour in metals

Key regions on the metal graph

Dislocations and strengthening metals

5. Classes of solid

Crystalline solids

Amorphous solids

Polymeric solids

6. Brittle materials and rubber

Brittle materials such as glass

Rubber

7. Practical: Young modulus of a metal wire

Method outline

8. Practical: force-extension relationship for rubber

Review Solids Under Stress by teaching Genie

Remember key concepts with flashcards

Solids Under Stress Revision Guide