Bulk Properties of Solids

Density, Hooke's Law, stress, strain, and elastic strain energy

SPEC 3.4.2.1 • MATERIALS

What is Density?

Definition

Density is the mass per unit volume of a material.

ρ = \frac{m}{V}

SymbolQuantityUnit

ρ

(rho)Densitykg m⁻³

m

Masskg

V

Volumem³

Key Concept

Density is a property of the material, not the object. All pure copper has the same density regardless of size or shape.

Density in Context

MaterialDensity (kg m⁻³)Context

Air (at STP)1.2Very low density gas

Water1000Reference for liquids

Aluminium2700Light metal

Steel7800Common structural metal

Copper8930Electrical wiring

Gold19300Very dense metal

Unit Conversion

1 g cm⁻³ = 1000 kg m⁻³. Water has density 1 g cm⁻³ = 1000 kg m⁻³.

Worked Example - Density01

A copper wire has length 2.0 m, diameter 0.50 mm, and mass 3.5 g. Calculate the density of copper.

1

Convert units: diameter = 0.50 mm = 0.50 × 10⁻³ m, so radius r = 0.25 × 10⁻³ m

2

Mass: m = 3.5 g = 3.5 × 10⁻³ kg

3

Calculate volume:

V = π r^{2} L = π \times (0.25 \times 1 0^{- 3})^{2} \times 2.0

4

V = π \times 6.25 \times 1 0^{- 8} \times 2.0 = 3.93 \times 1 0^{- 7} m^{3}

5

Calculate density:

ρ = \frac{m}{V} = \frac{3.5 \times 1 0 ^{- 3}}{3.93 \times 1 0 ^{- 7}}

Answer: ρ = 8900 kg m⁻³ (or 8.9 × 10³ kg m⁻³)

Types of Forces on Materials

Forces can change an object's shape. The study of stress and strain examines how forces deform materials.

Tensile Forces

Forces that stretch an object, pulling it apart.

Tensile stress is positive.

Compressive Forces

Forces that squash an object, pushing it together.

Compressive stress is negative.

Sign Convention

In physics, tensile (stretching) forces/stresses are positive, compressive (squashing) forces/stresses are negative.

Hooke's Law

The extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded.

F = k Δ L

SymbolQuantityUnit

F

Applied forceN

k

Spring constant (stiffness)N m⁻¹

Δ L

Extensionm

Spring Constant

The spring constant

k

measures stiffness. A higher

k

means more force is needed for the same extension. It equals the gradient of the linear part of the force-extension graph.

Force-Extension Graph

Important

P and E are NOT the same point! The elastic limit (E) is slightly beyond the limit of proportionality (P). Between P and E, the material is still elastic but doesn't obey Hooke's law.

Key Points Explained

P - Limit of Proportionality

The point beyond which force and extension are no longer proportional. The graph stops being a straight line. Hooke's law no longer applies.

E - Elastic Limit

The point beyond which the material will not return to its original length when the force is removed. Permanent (plastic) deformation begins.

Between P and E

The material is still elastic (returns to original shape) but no longer obeys Hooke's law (F is not proportional to ΔL).

Finding Spring Constant

From a F-x graph, the spring constant

k

equals the gradient of the linear section (O to P):

k = \frac{Δ F}{Δ L}

Worked Example - Hooke's Law02

A spring has natural length 15.0 cm. When a mass of 200 g is hung from it, the spring extends to 18.5 cm. Calculate: (a) the spring constant, (b) the extension when a 350 g mass is hung from it.

1

Extension ΔL = 18.5 - 15.0 = 3.5 cm = 0.035 m

2

Force F = mg = 0.200 × 9.81 = 1.96 N

3

(a) Spring constant:

k = \frac{F}{Δ L} = \frac{1.96}{0.035} = 56 N m^{- 1}

4

(b) New force: F = 0.350 × 9.81 = 3.43 N

5

New extension:

Δ L = \frac{F}{k} = \frac{3.43}{56} = 0.061 m

Answers: (a) k = 56 N m⁻¹, (b) ΔL = 6.1 cm (or 61 mm)

Tensile Stress

Definition

Tensile stress is the force applied per unit cross-sectional area.

σ = \frac{F}{A}

SymbolQuantityUnit

σ

(sigma)Tensile stressPa (or N m⁻²)

F

Force normal to the areaN

A

Cross-sectional aream²

Why Use Stress?

Stress allows us to compare materials fairly. A thick wire and thin wire of the same material have different breaking forces, but the same breaking stress. Stress is a property of the material.

Tensile Strain

Definition

Tensile strain is the fractional change in length (extension per unit original length).

ε = \frac{Δ L}{L}

SymbolQuantityUnit

ε

(epsilon)Tensile strainNo units (ratio)

Δ L

Extensionm

L

Original lengthm

Key Concept

Strain is a dimensionless ratio (no units). It is often expressed as a percentage. A strain of 0.02 means 2% extension.

Worked Example - Stress and Strain03

A steel cable has diameter 8.0 mm and original length 25 m. Under a load of 15 kN, it extends by 4.5 mm. Calculate: (a) the tensile stress, (b) the tensile strain.

1

Radius r = 4.0 mm = 4.0 × 10⁻³ m

2

Area:

A = π r^{2} = π \times (4.0 \times 1 0^{- 3})^{2} = 5.03 \times 1 0^{- 5} m^{2}

3

(a) Stress:

σ = \frac{F}{A} = \frac{15000}{5.03 \times 1 0 ^{- 5}} = 2.98 \times 1 0^{8} Pa

4

(b) Strain:

ε = \frac{Δ L}{L} = \frac{4.5 \times 1 0 ^{- 3}}{25} = 1.8 \times 1 0^{- 4}

Answers: (a) σ = 298 MPa (or 3.0 × 10⁸ Pa), (b) ε = 1.8 × 10⁻⁴ (or 0.018%)

Force-Extension vs Stress-Strain

Force-Extension Graph

Describes the behaviour of a specific object.

Depends on dimensions of sample

Different wires give different graphs

Useful for springs and specific components

Stress-Strain Graph

Describes the behaviour of a material.

Independent of sample dimensions

Same material = same graph

Used to compare materials

Ultimate Tensile Stress (UTS)

The maximum stress a material can withstand. On a stress-strain graph, this is the highest point before the material begins to neck and weaken.

Exam Distinction

Force-extension graphs are object-specific. Stress-strain graphs describe the material itself and allow fair comparison between different materials.

Elastic Strain Energy

Definition

Elastic strain energy is the energy stored in a material when it is stretched or compressed elastically.

E = \frac{1}{2} F Δ L = \frac{1}{2} k (Δ L)^{2}

Graphical Method

The elastic strain energy equals the area under the force-extension graph.

Energy Stored in a Spring

For a spring obeying Hooke's law, the energy stored can be written in three equivalent forms:

Form 1

E = \frac{1}{2} F Δ L

When F and ΔL known

Form 2

E = \frac{1}{2} k (Δ L)^{2}

When k and ΔL known

Form 3

E = \frac{F ^{2}}{2 k}

When F and k known

Derivation

Work done = average force × distance. Since force increases from 0 to F linearly, average force = ½F.

Work done = ½F × ΔL = ½FΔL

Substituting F = kΔL gives: E = ½k(ΔL)²

Important

These formulas only apply in the elastic region where Hooke's law is obeyed. Beyond the limit of proportionality, use the area under the graph method.

Worked Example - Elastic Strain Energy04

A spring with spring constant 80 N m⁻¹ is compressed by 12 cm. Calculate the elastic potential energy stored. If this energy is used to launch a 50 g ball vertically, calculate the maximum height reached.

1

Convert: ΔL = 12 cm = 0.12 m, m = 50 g = 0.050 kg

2

Elastic PE:

E = \frac{1}{2} k (Δ L)^{2} = \frac{1}{2} \times 80 \times (0.12)^{2}

3

E = \frac{1}{2} \times 80 \times 0.0144 = 0.576 J

4

At max height: Elastic PE = Gravitational PE, so

0.576 = m g h

5

h = \frac{0.576}{0.050 \times 9.81} = 1.17 m

Answers: E = 0.58 J, h = 1.2 m

Types of Material Behaviour

Elastic

Returns to original shape when force removed. Energy is recoverable. Occurs before the elastic limit.

Plastic

Permanently deformed when force removed. Energy is dissipated as heat. Occurs after the elastic limit.

Brittle

Fractures suddenly with little or no plastic deformation. Breaks at or near the elastic limit. Examples: glass, ceramics, hard sweets.

Ductile

Can be drawn into wires. Shows significant plastic deformation before breaking. Examples: copper, gold, steel.

Key Distinction

Brittle materials break with very little extension. Ductile materials undergo extensive plastic deformation (large extension) before fracture.

Stress-Strain Curves

Reading Stress-Strain Curves

Brittle: Straight line to fracture, small strain, no plastic region

Ductile: Linear region, then curves, shows UTS peak, large strain before breaking

Polymeric: Low stress, very large strain, non-linear from start

Breaking Stress

Definition

Breaking stress is the stress at which the material fractures.

Ultimate tensile stress (UTS) is the maximum stress a material can withstand before it begins to neck and weaken.

Material Property

Breaking stress is a property of the material, not the sample. A thicker wire has a higher breaking force but the same breaking stress as a thinner wire of the same material.

MaterialBreaking Stress (MPa)

Steel400 - 800

Copper200 - 250

Glass30 - 90

Rubber15 - 25

Worked Example - Breaking Stress05

A steel cable has diameter 6.0 mm and breaking stress 500 MPa. Calculate the maximum load the cable can support before breaking.

1

Radius r = 3.0 mm = 3.0 × 10⁻³ m

2

Cross-sectional area:

A = π r^{2} = π \times (3.0 \times 1 0^{- 3})^{2} = 2.83 \times 1 0^{- 5} m^{2}

3

Breaking stress = 500 MPa = 500 × 10⁶ Pa

4

Rearrange σ = F/A: F = σA = 500 × 10⁶ × 2.83 × 10⁻⁵

5

F = 14150 N ≈ 14 kN

Answer: Maximum load = 14 kN (or 1.4 × 10⁴ N)

Loading and Unloading

When a material is stretched beyond its elastic limit then unloaded:

Key Features

The unloading line is parallel to the original loading line (same gradient = same k)

The material does not return to zero extension - there is permanent deformation

The area between the curves = energy dissipated as heat during plastic deformation

Springs in Series and Parallel

Each spring feels the full force F.

Extensions add up:

Δ L_{t o t a l} = Δ L_{1} + Δ L_{2}

\frac{1}{k _{e ff}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}

Each spring has the same extension.

Forces add up:

F_{t o t a l} = F_{1} + F_{2}

k_{e ff} = k_{1} + k_{2}

Remember

Series: softer (smaller k), Parallel: stiffer (larger k). This is the opposite of resistors!

Worked Example - Springs06

Two springs have spring constants k₁ = 60 N m⁻¹ and k₂ = 40 N m⁻¹. Calculate the effective spring constant when they are connected: (a) in series, (b) in parallel.

1

(a) Series:

\frac{1}{k _{e ff}} = \frac{1}{k _{1}} + \frac{1}{k _{2}} = \frac{1}{60} + \frac{1}{40}

2

\frac{1}{k _{e ff}} = \frac{2}{120} + \frac{3}{120} = \frac{5}{120} = \frac{1}{24}

3

Therefore k_eff = 24 N m⁻¹ (softer than either spring alone)

4

(b) Parallel: k_eff = k₁ + k₂ = 60 + 40 = 100 N m⁻¹ (stiffer)

Answers: (a) k = 24 N m⁻¹, (b) k = 100 N m⁻¹

Energy Conservation

Elastic strain energy can be converted to other forms:

Spring to Kinetic

Compressed spring releases and accelerates an object:

\frac{1}{2} k (Δ L)^{2} = \frac{1}{2} m v^{2}

Spring to Gravitational PE

Stretched spring lifts an object:

\frac{1}{2} k (Δ L)^{2} = m g h

Problem-Solving Strategy

Apply conservation of energy: Initial elastic PE = Final KE + Final GPE + Energy dissipated

Energy Conservation in Transport

Safety features of vehicles rely on material properties to protect passengers.

The Problem

In a collision, a large amount of kinetic energy must be dissipated. If this energy is transferred to passengers, it causes injury.

Crumple Zones

Parts of vehicle that permanently deform

Show plastic behaviour

Convert kinetic energy into work done deforming metal

Energy dissipated as heat in material

Passengers experience less force

Seatbelts

Stretch slightly to increase stopping time

Convert passenger's KE to elastic strain energy

Reduce force on passenger (F = Δp/Δt)

Distribute force over larger area

Practice

Question 1: A uniform steel rod has length 1.5 m and cross-sectional area 2.0 × 10⁻⁴ m². The density of steel is 7800 kg m⁻³. Calculate the mass of the rod.

1

Volume: V = A × L = 2.0 × 10⁻⁴ × 1.5 = 3.0 × 10⁻⁴ m³

2

Mass: m = ρV = 7800 × 3.0 × 10⁻⁴ = 2.34 kg

Answer: m = 2.3 kg

Practice

Question 2: Two identical springs, each with spring constant 40 N m⁻¹, are connected in series and stretched by a force of 8.0 N. Calculate the total extension.

1

For springs in series, each spring experiences the full 8.0 N force

2

Extension of each spring: ΔL = F/k = 8.0/40 = 0.20 m

3

Total extension = 0.20 + 0.20 = 0.40 m

Answer: Total extension = 0.40 m (or 40 cm)

Practice

Question 3: A wire of original length 2.5 m is stretched by a force of 200 N. The wire extends by 1.2 mm. Calculate the elastic strain energy stored in the wire.

1

Convert extension: ΔL = 1.2 mm = 1.2 × 10⁻³ m

2

Using E = ½FΔL:

E = \frac{1}{2} \times 200 \times 1.2 \times 1 0^{- 3}

3

E = 0.12 J

Answer: E = 0.12 J

Summary

Key Formulas

QuantityFormulaUnits

Density

ρ = \frac{m}{V}

kg m⁻³

Hooke's Law

F = k Δ L

N, N m⁻¹, m

Tensile Stress

σ = \frac{F}{A}

Pa

Tensile Strain

ε = \frac{Δ L}{L}

No units

Elastic Strain Energy

E = \frac{1}{2} F Δ L = \frac{1}{2} k (Δ L)^{2}

J

Key Exam Skills

Identify P (limit of proportionality) and E (elastic limit) on F-x graphs

Distinguish force-extension (object) vs stress-strain (material) graphs

Describe elastic, plastic, brittle, and ductile behaviour

Apply energy conservation with elastic strain energy

Explain crumple zones and seatbelts using material properties