Properties of Materials

This lesson covers AQA Spec 3.4.2.1–3.4.2.2 in detail: elastic and plastic behaviour, brittle/ductile/polymeric materials, loading–unloading curves, and hysteresis.

Spec mapping (AQA 7408 §3.4.2.1–§3.4.2.2): This lesson develops Hooke's law as F = kx, springs in series and parallel, elastic versus plastic deformation, the elastic energy ½kx² = ½Fx stored in a stretched spring, force-extension graphs for ductile, brittle and polymeric materials, hysteresis (loading and unloading curves diverging), and the loading-unloading cycle for metal wires beyond the elastic limit. Strength, stiffness, toughness and hardness are defined precisely. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Material properties cross-connect several A-Level topics. (i) Young's modulus (Lesson 7, §3.4.2.2) is the continuum-mechanics analogue of the spring constant: a wire of length L and cross-section A obeys F = (EA/L) × x, so k_effective = EA/L. The spring framework here generalises to wires through this relation. (ii) Elastic potential energy in mechanics (Lesson 5, §3.4.1.7) is the same ½kx² result, now applied to bulk materials and seen as the area under a force-extension graph. (iii) Simple harmonic motion of mass-spring systems (Year 13, §3.6.1.2) uses Hooke's law as the linear restoring force that produces SHM; the period T = 2π√(m/k) is the dynamical consequence of the elastic relation derived here. (iv) Hysteresis in magnetic materials (Year 13) reuses the same conceptual structure — loading and unloading curves diverge, and the enclosed area is energy dissipated per cycle. Material hysteresis here is the introduction to a deep cross-subject theme.

---

1 Hooke's Law

Hooke's Law: The extension of a spring (or wire) is directly proportional to the applied force, provided the limit of proportionality is not exceeded.

F = kx

where k is the spring constant (or stiffness) in N m⁻¹, and x is the extension (or compression).

Worked Example 1 — Spring Constant

A spring extends by 40 mm when a force of 5.0 N is applied. Find (a) the spring constant and (b) the extension when a 12 N force is applied (assuming Hooke's law still applies).

Solution:

(a) k = F/x = 5.0 / (40 × 10⁻³) = 125 N m⁻¹

(b) x = F/k = 12 / 125 = 0.096 m = 96 mm

---

2 Springs in Series and Parallel

Springs in Series

For two springs with constants k₁ and k₂ in series:

1/k_total = 1/k₁ + 1/k₂

The total spring is softer (lower k).

Springs in Parallel

For two springs in parallel:

k_total = k₁ + k₂

The total spring is stiffer (higher k).

Worked Example 2

Two springs, k₁ = 100 N m⁻¹ and k₂ = 200 N m⁻¹, support a 6.0 N load. Find the extension when arranged (a) in parallel and (b) in series.

Solution:

(a) Parallel: k_total = 100 + 200 = 300 N m⁻¹

x = F/k = 6.0/300 = 0.020 m = 20 mm

(b) Series: 1/k_total = 1/100 + 1/200 = 2/200 + 1/200 = 3/200

k_total = 200/3 = 66.7 N m⁻¹

x = F/k = 6.0/66.7 = 0.090 m = 90 mm

---

3 Elastic and Plastic Deformation

Elastic deformation: The material returns to its original shape when the load is removed. No permanent deformation. The atoms are displaced from their equilibrium positions but return when the force is removed.

Plastic deformation: The material is permanently deformed when the load is removed. The atoms have moved to new equilibrium positions — planes of atoms have slipped past each other.

The elastic limit is the maximum stress (or load) that can be applied without causing permanent deformation. Beyond this, some deformation is plastic and the material does not fully recover.

---

4 Elastic Potential Energy

The energy stored in a stretched spring (or wire) that obeys Hooke's law:

E = ½Fx = ½kx²

This is the area under the force–extension graph (a triangle for a linear spring).

Worked Example 3

A spring with k = 80 N m⁻¹ is compressed by 0.15 m. Find the elastic potential energy stored.

Solution:

E = ½kx² = ½ × 80 × 0.15² = ½ × 80 × 0.0225 = 0.90 J

---

5 Force–Extension Graphs for Different Materials

Ductile Material (e.g., Copper)

Described diagram — Force–extension graph for copper wire:

• Linear region (Hooke's law) up to the limit of proportionality.
• The curve bends over as plastic deformation begins.
• Large plastic extension before fracture — the material "necks" (cross-section reduces locally).
• On unloading from the plastic region, the wire follows a line parallel to the original loading line, but offset — showing a permanent extension.

Brittle Material (e.g., Glass)

Described diagram — Force–extension graph for glass:

• Straight line (linear, elastic) right up to the fracture point.
• No plastic deformation — the material snaps suddenly.
• Glass obeys Hooke's law until it breaks.

Polymeric Material (e.g., Rubber)

Described diagram — Force–extension graph for a rubber band (loading and unloading):

• Non-linear curve — rubber can stretch to several times its original length.
• The unloading curve lies below the loading curve — this is called hysteresis.
• The area between the loading and unloading curves represents the energy dissipated as heat within the rubber.
• Rubber returns to its original length (elastic), but energy is lost.

---

6 Hysteresis

Key Definition: Hysteresis is the phenomenon where the loading and unloading curves on a force–extension (or stress–strain) graph do not coincide. The area enclosed between the curves represents the energy dissipated per cycle.

Applications of Hysteresis

Application	Requirement	Material
Car tyres	Low hysteresis (less heating)	Synthetic rubber with low hysteresis
Shock absorbers	High hysteresis (dissipate energy)	Rubber with high hysteresis
Squash balls	Moderate hysteresis	The ball warms up during play

Worked Example 4 — Hysteresis Energy

A rubber band is stretched and released. The loading curve has an area under it of 3.5 J and the unloading curve has an area of 2.8 J. Find the energy dissipated as heat.

Solution:

Energy dissipated = area between curves = 3.5 − 2.8 = 0.7 J

---

7 Comparing Material Behaviour

Property	Ductile	Brittle	Polymeric
Plastic deformation	Large	None	Little/none
Breaking strain	Large	Very small	Very large
Stress–strain shape	Curves over, necks	Straight line to fracture	Non-linear, elastic
Energy absorption	Large (area under curve)	Small	Moderate (with hysteresis loss)
Examples	Copper, mild steel, gold	Glass, ceramic, cast iron	Rubber, polythene, nylon

Ductile vs Brittle — Key Distinction

A ductile material can be drawn into wires. It shows significant plastic deformation before fracture.

A brittle material fractures with little or no plastic deformation. Cracks propagate rapidly through the material. Brittle materials are often strong in compression but weak in tension.

Polymeric Materials

Polymers consist of long-chain molecules. When stretched:

• Initially, the tangled chains straighten and uncoil — easy to stretch
• Once straightened, further stretching requires extending the bonds between atoms — much stiffer
• This produces the characteristic S-shaped (non-linear) curve

---

8 The Loading–Unloading Cycle for a Metal Wire

When a metal wire is loaded beyond its elastic limit and then unloaded:

1. Loading: Follows the stress–strain curve up through elastic and into plastic region.
2. Unloading: Returns along a straight line parallel to the original elastic loading line.
3. Permanent extension: The wire does not return to zero strain. The intercept on the strain axis is the permanent (plastic) strain.

The area between the loading and unloading lines represents the energy dissipated during plastic deformation.

Worked Example 5

A wire is loaded to a stress of 300 MPa (strain = 0.005) then unloaded. The permanent strain is 0.002. Young's modulus is 200 GPa. Find the elastic strain recovered and the energy density dissipated.

Solution:

Elastic strain recovered = total strain − permanent strain = 0.005 − 0.002 = 0.003

Check: elastic strain = σ/E = 300 × 10⁶ / (200 × 10⁹) = 0.0015

Hmm — this discrepancy means the 0.005 strain includes both elastic and plastic. Let's recalculate.

At 300 MPa, elastic strain = 300 × 10⁶ / (200 × 10⁹) = 0.0015

So upon unloading, the wire recovers 0.0015 of strain.

Permanent strain = 0.005 − 0.0015 = 0.0035

Energy density of loading (total area ≈ trapezium) is approximately ½ × 300 × 10⁶ × 0.005 = 750 000 J m⁻³

Energy density recovered (elastic triangle) = ½ × 300 × 10⁶ × 0.0015 = 225 000 J m⁻³

Energy density dissipated = 750 000 − 225 000 = 525 000 J m⁻³ = 525 kJ m⁻³

Exam Tip: In loading–unloading problems, the elastic strain recovered always equals σ/E. The permanent strain is the total strain minus the elastic strain.

---

9 Strength, Stiffness, and Toughness

These terms have precise meanings in physics:

Term	Definition	Related to...
Stiffness	Resistance to elastic deformation	High Young's modulus
Strength	Maximum stress before fracture (UTS)	High breaking stress
Toughness	Energy absorbed before fracture	Large area under stress–strain curve
Hardness	Resistance to surface indentation	Not directly from stress–strain graph

Common Misconception: Strong ≠ stiff ≠ tough. A material can be strong but brittle (glass), or tough but not very stiff (rubber). Steel is strong, stiff, AND tough.

Worked Example 6 — Combined Spring Systems With Energy

Three springs each with spring constant k = 50 N m⁻¹ are arranged: spring 1 supports a 0.40 kg mass directly; springs 2 and 3 are connected in parallel below spring 1 to support the same mass. Find the total extension when the system is in equilibrium.

Solution:

Springs 2 and 3 in parallel: k_23 = k₂ + k₃ = 50 + 50 = 100 N m⁻¹.

Springs 1 and (2-and-3) in series: 1/k_total = 1/50 + 1/100 = 2/100 + 1/100 = 3/100 k_total = 100/3 = 33.3 N m⁻¹.

Load weight: W = mg = 0.40 × 9.81 = 3.924 N.

Total extension: x = W/k_total = 3.924/33.3 = 0.118 m = 11.8 cm.

The same load on spring 1 alone would extend by W/k₁ = 3.924/50 = 0.0785 m = 7.85 cm. So adding the parallel pair below adds 11.8 − 7.85 = 3.95 cm of extra extension, which is W/k_23 = 3.924/100 = 0.0392 m = 3.92 cm ✓ (small rounding difference). The series arrangement of "stiffer below" still softens the total system.

Worked Example 7 — Loading Beyond the Elastic Limit

A copper wire is loaded to a stress of 350 MPa (well into its plastic region). Young's modulus is 130 GPa. The wire's natural strain at the elastic limit is 0.0019 (= 190 MPa / 130 GPa). At 350 MPa, the total strain is measured as 0.030 (a strain 16× the elastic-limit strain). The load is then removed.

Find (a) the elastic strain recovered on unloading, (b) the permanent (plastic) strain remaining, and (c) the energy density dissipated as plastic work.

Solution:

(a) Elastic strain recovered = σ/E = 350 × 10⁶ / 130 × 10⁹ = 2.69 × 10⁻³.

(b) Permanent strain = total − elastic = 0.030 − 0.00269 = 0.0273 ≈ 0.0273 (about 2.7%).

(c) Energy density during loading is approximately (taking the loading curve as roughly linear from σ = 0 to σ = 350 MPa over strain 0 to 0.030):

u_load ≈ ½ × 350 × 10⁶ × 0.030 = 5.25 × 10⁶ J m⁻³ (this is an over-estimate because the plastic region is at high stress for most of the strain; a more accurate estimate uses the actual loading curve).

Energy density recovered (elastic triangle on unloading):

u_recovered = ½ × 350 × 10⁶ × 2.69 × 10⁻³ = 4.71 × 10⁵ J m⁻³.

Energy density dissipated as plastic work:

u_dissipated = u_load − u_recovered ≈ 5.25 × 10⁶ − 4.71 × 10⁵ = 4.78 × 10⁶ J m⁻³.

This dissipated energy heats the wire and contributes to the permanent reorganisation of atomic positions in the metal — the microscopic source of plastic deformation.