Stress, Strain and the Young Modulus

Spec mapping: OCR H556 Module 3.4 — Materials (tensile stress $\sigma = F/A$σ=F/A; tensile strain $\varepsilon = \Delta\ell/\ell_0$ε=Δℓ/ℓ0​; Young modulus $E = \sigma/\varepsilon$E=σ/ε; experimental determination via the long-wire (Searle's) method; elastic energy density). This is a core anchor lesson for OCR PAG 2 (Materials). Refer to the official OCR H556 specification document for exact wording.

Hooke's law (previous lesson) tells you how a particular spring behaves — but its stiffness $k$k depends on the spring's geometry as well as its material. To say something universal about, say, steel or copper, you must normalise force by cross-sectional area (giving stress) and extension by original length (giving strain). Their ratio in the elastic regime is the Young modulus $E$E, an intrinsic property of the material that bridges atomic-bond stiffness and bulk engineering behaviour. This lesson develops stress, strain and the Young modulus, derives the universal connection between them and the spring constant of a uniform wire, and lays out the OCR PAG 2 long-wire apparatus for experimental determination.

Key Definition: Tensile stress $\sigma = F/A$σ=F/A is the force per unit cross-sectional area transmitted at right angles through a sample under tension. Tensile strain $\varepsilon = \Delta\ell/\ell_0$ε=Δℓ/ℓ0​ is the dimensionless fractional change in length. The Young modulus $E = \sigma/\varepsilon$E=σ/ε, an intrinsic property of the material in the elastic regime, has SI unit pascal (Pa), with engineering values typically quoted in GPa.

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Tensile Stress

For a sample of cross-sectional area $A$A under an axial force $F$F acting perpendicular to that area, the tensile stress is:

$\sigma = \frac{F}{A}$σ=AF​

• $F$F in newtons (N).
• $A$A in square metres (m$^2$2) — the area perpendicular to the force.
• $\sigma$σ in pascals (Pa), equal to N m$^{-2}$−2 (same unit as pressure — the analogy is exact, since pressure is just compressive stress).

Engineering stresses are typically large. A steel piano wire of diameter $1$1 mm under a tension of $200$200 N carries:

$\sigma = \frac{200}{\pi \times (0.5 \times 10^{-3})^2} = \frac{200}{7.85 \times 10^{-7}} = 2.55 \times 10^8\;\text{Pa} = 255\;\text{MPa}$σ=π×(0.5×10−3)2200​=7.85×10−7200​=2.55×108Pa=255MPa

Typical maximum-allowable working stresses for mild structural steel are $\sim 200$∼200 MPa; the failure (fracture) stress is around $400$400 MPa for mild steel and rises to $\sim 2$∼2 GPa for cold-drawn high-tensile piano wire and to $\sim 3.5$∼3.5 GPa for ultra-high-tensile aerospace alloys.

Common Exam Mistake: Using the diameter $d$d in place of the radius $r$r when computing $A = \pi r^2$A=πr2. The area is $\pi r^2 = \pi (d/2)^2 = \pi d^2 / 4$πr2=π(d/2)2=πd2/4 — be explicit which form you are using and convert the diameter to radius before squaring.

Worked Example 1 — Stress in a Wire

A wire of diameter $0.50$0.50 mm supports a $4.0$4.0 kg static load. Find the tensile stress.

Cross-sectional area: $A = \pi r^2 = \pi \times (0.25 \times 10^{-3})^2 = \pi \times 6.25 \times 10^{-8} = 1.963 \times 10^{-7}$A=πr2=π×(0.25×10−3)2=π×6.25×10−8=1.963×10−7 m$^2$2.

Force: $F = m g = 4.0 \times 9.81 = 39.24$F=mg=4.0×9.81=39.24 N.

Stress: $\sigma = F/A = 39.24 / 1.963 \times 10^{-7} = 2.00 \times 10^8$σ=F/A=39.24/1.963×10−7=2.00×108 Pa $= 200$=200 MPa.

This is at the upper end of safe working stresses for mild steel. A modest extra load would push it into the plastic regime, and a brief overload (e.g. a sudden weight increase) could fracture it. Engineers design with safety factors of $2$2–$4$4 on the failure stress to avoid this.

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Tensile Strain

For a sample of natural length $\ell_0$ℓ0​ that extends by $\Delta\ell$Δℓ under tension, the tensile strain is:

$\varepsilon = \frac{\Delta\ell}{\ell_0}$ε=ℓ0​Δℓ​

• $\Delta\ell$Δℓ in metres (m).
• $\ell_0$ℓ0​ in metres (m).
• $\varepsilon$ε is dimensionless — a pure ratio. Often expressed as a percentage or in microstrain ($1$1 microstrain $= 10^{-6}$=10−6).

A strain of $0.001$0.001 ($0.1\%$0.1% or $1000$1000 microstrain) means the sample has extended by one-thousandth of its original length. A $1$1 m wire with strain $0.001$0.001 has extended by $1$1 mm.

Strains in real engineering materials are tiny. Steel and copper yield at strains of about $0.001$0.001–$0.005$0.005 ($0.1$0.1–$0.5\%$0.5%); concrete cracks at $\sim 10^{-4}$∼10−4. Polymers and rubber, by contrast, can sustain strains of $1$1 (i.e. $100\%$100%, doubling the natural length) and elastomers can stretch to $\varepsilon \sim 5$ε∼5 or more before fracture.

Worked Example 2 — Strain in a Loaded Rod

A $2.40$2.40 m steel rod extends by $3.0$3.0 mm when loaded. Find the strain.

$\varepsilon = \frac{\Delta\ell}{\ell_0} = \frac{3.0 \times 10^{-3}}{2.40} = 1.25 \times 10^{-3} = 0.125\%$ε=ℓ0​Δℓ​=2.403.0×10−3​=1.25×10−3=0.125%

This is well below the typical yield strain of mild steel ($\sim 0.2\%$∼0.2%), so the rod is still firmly in the elastic regime. On unloading, it would return exactly to its $2.40$2.40 m natural length.

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The Young Modulus

For an isotropic material in its elastic regime (i.e. below the limit of proportionality), stress is directly proportional to strain:

$\sigma \propto \varepsilon$σ∝ε

The proportionality constant is the Young modulus $E$E (sometimes written $Y$Y):

$E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta\ell/\ell_0} = \frac{F\,\ell_0}{A\,\Delta\ell}$E=εσ​=Δℓ/ℓ0​F/A​=AΔℓFℓ0​​

• SI unit: pascal (Pa), since strain is dimensionless and stress has units Pa.
• Engineering values are typically very large, so gigapascals ($1$1 GPa $= 10^9$=109 Pa) are standard.

The Young modulus is sometimes called stiffness modulus to distinguish it from related elastic constants (shear modulus $G$G, bulk modulus $K$K, Poisson ratio $\nu$ν). For an isotropic material these are related by $E = 2G(1 + \nu) = 3K(1 - 2\nu)$E=2G(1+ν)=3K(1−2ν). At A-Level only the Young modulus matters; the others appear in undergraduate solid mechanics.

Typical Values

Material	$E$E (GPa)	Notes
Rubber	$0.01$0.01–$0.1$0.1	strongly strain-dependent (rubber is non-Hookean)
Polyethylene	$0.3$0.3	thermoplastic
Wood (along grain)	$10$10–$15$15	highly anisotropic
Bone (cortical)	$14$14	biocomposite
Concrete	$20$20	brittle in tension
Aluminium	$69$69	aerospace standard
Glass	$70$70	brittle
Brass	$100$100	machinable
Copper	$117$117	electrical conductor
Steel (mild)	$\sim 200$∼200	construction standard
Tungsten	$\sim 400$∼400	very stiff
Diamond	$\sim 1050$∼1050	stiffest natural material

A larger Young modulus means a stiffer material — one that resists elastic deformation. Steel ($E \approx 200$E≈200 GPa) is twice as stiff as copper ($117$117 GPa) and several thousand times as stiff as rubber.

Common Exam Mistake: Confusing Young modulus $E$E (intrinsic material property) with spring constant $k$k (geometry-dependent stiffness of a particular object). $E$E is independent of the specimen's shape; $k = EA/\ell_0$k=EA/ℓ0​ varies with $A$A and $\ell_0$ℓ0​.

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Worked Example 3 — Young Modulus from Experimental Data

A $1.80$1.80 m copper wire of diameter $0.40$0.40 mm is loaded with $4.0$4.0 kg and extends by $2.25$2.25 mm. Calculate the Young modulus.

Force: $F = m g = 4.0 \times 9.81 = 39.24$F=mg=4.0×9.81=39.24 N.

Cross-sectional area: $A = \pi r^2 = \pi \times (0.20 \times 10^{-3})^2 = 1.257 \times 10^{-7}$A=πr2=π×(0.20×10−3)2=1.257×10−7 m$^2$2.

Stress: $\sigma = F/A = 39.24 / 1.257 \times 10^{-7} = 3.12 \times 10^8$σ=F/A=39.24/1.257×10−7=3.12×108 Pa.

Strain: $\varepsilon = \Delta\ell/\ell_0 = 2.25 \times 10^{-3} / 1.80 = 1.25 \times 10^{-3}$ε=Δℓ/ℓ0​=2.25×10−3/1.80=1.25×10−3.

Young modulus: $E = \sigma/\varepsilon = 3.12 \times 10^8 / 1.25 \times 10^{-3} = 2.50 \times 10^{11}$E=σ/ε=3.12×108/1.25×10−3=2.50×1011 Pa $= 250$=250 GPa.

The literature value for copper is $\sim 117$∼117 GPa. Our measured value is about a factor of $2$2 too high. Plausible sources of systematic error: (i) the wire was perhaps already slightly extended before loading ("zero-extension" datum off); (ii) the diameter was over-stated by $\sim 20\%$∼20% (since $A$A enters squared); (iii) there was some plastic-deformation contamination from prior loading. A real PAG 2 result close to $\sim 120$∼120 GPa would be considered satisfactory.

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Young Modulus and Spring Constant — The Universal Link

For a uniform wire of length $\ell_0$ℓ0​ and area $A$A made from a material of Young modulus $E$E, the spring constant $k$k (relating tension to extension via $F = kx$F=kx) is:

$k = \frac{E A}{\ell_0}$k=ℓ0​EA​

Derivation:

$F = \sigma A = E \varepsilon A = E\,A\,\frac{\Delta\ell}{\ell_0}$F=σA=EεA=EAℓ0​Δℓ​

Comparing with the spring form $F = k\,\Delta\ell$F=kΔℓ:

$k = \frac{E A}{\ell_0}$k=ℓ0​EA​

This decomposition is enormously useful. It says that a wire's stiffness is the product of:

• The material factor $E$E (Young modulus, intrinsic).
• The geometric factor $A / \ell_0$A/ℓ0​ (cross-sectional area divided by length).

A shorter, fatter wire is stiffer (larger $k$k); a longer, thinner wire is floppier (smaller $k$k). This is why piano wires are kept thin and long (to give modest stiffness over the full string length, supporting controlled vibration in the audible range); why suspension bridge cables are bundled from thousands of thin steel wires (each thin enough to handle the bending around the saddles, yet collectively thick enough to be stiff under the bridge load); why aircraft wings use long, slender spars (which provide flexure for aerodynamic load relief while remaining stiff under twisting).

Worked Example 4 — Effective Stiffness of a Steel Wire

A steel wire ($E = 200$E=200 GPa) has length $1.0$1.0 m and cross-sectional area $1.0$1.0 mm$^2 = 1.0 \times 10^{-6}$2=1.0×10−6 m$^2$2. Find its spring constant.

$k = \frac{EA}{\ell_0} = \frac{200 \times 10^9 \times 1.0 \times 10^{-6}}{1.0} = 2.0 \times 10^5\;\text{N m}^{-1}$k=ℓ0​EA​=1.0200×109×1.0×10−6​=2.0×105N m−1

A $1$1 kN load (about a $100$100 kg mass) extends it by $F/k = 1000 / 2.0 \times 10^5 = 5$F/k=1000/2.0×105=5 mm — a strain of $0.5\%$0.5%, right at the yield-strain limit for mild steel. Metals barely deform elastically; this is why measuring Young modulus with a wire requires a long wire and a fine extensometer.

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Elastic Energy Density

The elastic PE stored in a stretched wire is $\tfrac{1}{2}F\Delta\ell$21​FΔℓ. Substituting $F = EA\varepsilon$F=EAε and $\Delta\ell = \varepsilon\ell_0$Δℓ=εℓ0​:

$E_{\text{elastic}} = \tfrac{1}{2}(EA\varepsilon)(\varepsilon\ell_0) = \tfrac{1}{2}E\varepsilon^2 \times (A\ell_0) = \tfrac{1}{2}E\varepsilon^2 \times V$Eelastic​=21​(EAε)(εℓ0​)=21​Eε2×(Aℓ0​)=21​Eε2×V

where $V = A\ell_0$V=Aℓ0​ is the volume of the wire. Dividing by volume gives the energy density (elastic PE per unit volume):

$u = \frac{E_{\text{elastic}}}{V} = \tfrac{1}{2}\sigma\varepsilon = \tfrac{1}{2}E\varepsilon^2 = \frac{\sigma^2}{2E}$u=VEelastic​​=21​σε=21​Eε2=2Eσ2​

Units of $u$u: J m$^{-3}$−3 (equivalent to Pa).

Energy density quantifies the toughness of a material — how much energy it can absorb per unit volume before fracture. For mild steel at yield ($\sigma_y \approx 250$σy​≈250 MPa, $\varepsilon_y \approx 0.001$εy​≈0.001), $u \approx \tfrac{1}{2} \times 2.5 \times 10^8 \times 10^{-3} = 1.25 \times 10^5$u≈21​×2.5×108×10−3=1.25×105 J m$^{-3}$−3 — modest. For a soft polymer like rubber that can sustain $\varepsilon = 5$ε=5 at $\sigma = 10$σ=10 MPa, $u \approx 2.5 \times 10^7$u≈2.5×107 J m$^{-3}$−3 — two hundred times larger, despite a thousand-fold smaller Young modulus. This is why rubber dampers and bumpers are excellent shock absorbers: they store a lot of energy per kilogram.

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Measuring the Young Modulus — OCR PAG 2 Apparatus

OCR PAG 2 specifies a long-wire (or Searle's apparatus) method:

1. Mount a long ($2$2–$3$3 m) thin wire horizontally between a rigid clamp at one end and a pulley over the bench-edge at the other. Suspend a hanger from the free end of the wire over the pulley.
2. Datum mark. Attach a reference marker to the wire near the pulley end and align it with a fixed scale (Vernier or travelling microscope).
3. Diameter measurement. Using a micrometer with $\pm 0.01$±0.01 mm precision, measure the wire diameter at at least three points along its length and average — wires are rarely uniform.
4. Length measurement. With a metre rule (or steel tape), measure the length $\ell_0$ℓ0​ from clamp to reference marker.
5. Loading. Add masses to the hanger in steps (typically $\sim 1$∼1 N or $\sim 100$∼100 g per step). After each addition, wait for the wire to settle and read the new position of the reference marker. Calculate the extension $\Delta\ell$Δℓ.
6. Unloading. Reverse the procedure, removing masses one at a time and recording the marker position. The loading and unloading curves should coincide (within scatter); systematic divergence indicates plastic deformation.
7. Plot $\sigma = F/A$σ=F/A on the $y$y-axis against $\varepsilon = \Delta\ell/\ell_0$ε=Δℓ/ℓ0​ on the $x$x-axis. In the linear region, gradient = Young modulus $E$E.

Why a long, thin wire?

• Long wire: extension $\Delta\ell = \varepsilon\,\ell_0$Δℓ=εℓ0​ scales with $\ell_0$ℓ0​. A $2$2 m wire with strain $0.001$0.001 extends $2$2 mm — measurable with a Vernier; a $20$20 cm wire would extend only $0.2$0.2 mm, near the Vernier resolution.
• Thin wire: stress $\sigma = F/A$σ=F/A scales as $1/A$1/A. A diameter of $\sim 0.5$∼0.5 mm gives reasonable stresses for $\sim 1$∼1 kg loads, avoiding the need for heavy weights that risk yielding the wire.

Sources of uncertainty:

Source	Typical magnitude	Contribution to $E$E
Diameter (micrometer)	$\pm 0.01$±0.01 mm at $d = 0.40$d=0.40 mm $\Rightarrow \pm 2.5\%$⇒±2.5%	$\pm 5\%$±5% in $E$E (since $A \propto d^2$A∝d2)
Length (metre rule)	$\pm 1$±1 mm over $2$2 m $\Rightarrow \pm 0.05\%$⇒±0.05%	$\pm 0.05\%$±0.05% in $E$E
Extension (Vernier/microscope)	$\pm 0.01$±0.01 mm	depends on extension; typically $\pm 1$±1–$3\%$3%
Load (standard masses)	$\pm 0.1\%$±0.1%	$\pm 0.1\%$±0.1%

Diameter is almost always the dominant uncertainty because it enters squared in $A$A. Best practice is to take 5–10 diameter readings around the wire's circumference at different points along its length and use the mean.

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Worked Example 5 — Combined Wire-Spring Energy

A copper wire ($E = 117$E=117 GPa) of length $1.5$1.5 m and cross-sectional area $0.50$0.50 mm$^2$2 is stretched by a force of $50$50 N. Find (a) the extension, (b) the elastic PE stored, (c) the elastic energy density.

(a) From $k = EA/\ell_0 = 117 \times 10^9 \times 0.50 \times 10^{-6} / 1.5 = 3.9 \times 10^4$k=EA/ℓ0​=117×109×0.50×10−6/1.5=3.9×104 N m$^{-1}$−1. Extension $\Delta\ell = F/k = 50 / 3.9 \times 10^4 = 1.28 \times 10^{-3}$Δℓ=F/k=50/3.9×104=1.28×10−3 m $= 1.28$=1.28 mm.

(b) Elastic PE: $\tfrac{1}{2}F\Delta\ell = \tfrac{1}{2} \times 50 \times 1.28 \times 10^{-3} = 3.2 \times 10^{-2}$21​FΔℓ=21​×50×1.28×10−3=3.2×10−2 J.

(c) Energy density: divide by volume $V = A\ell_0 = 0.50 \times 10^{-6} \times 1.5 = 7.5 \times 10^{-7}$V=Aℓ0​=0.50×10−6×1.5=7.5×10−7 m$^3$3. $u = 0.032 / 7.5 \times 10^{-7} = 4.27 \times 10^4$u=0.032/7.5×10−7=4.27×104 J m$^{-3}$−3. Or equivalently $u = \sigma^2/(2E) = (50/0.5 \times 10^{-6})^2 / (2 \times 117 \times 10^9) = (10^8)^2 / (2.34 \times 10^{11}) = 4.27 \times 10^4$u=σ2/(2E)=(50/0.5×10−6)2/(2×117×109)=(108)2/(2.34×1011)=4.27×104 J m$^{-3}$−3. ✓

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Stress-Strain Graph — A First Look

Plotting stress on the $y$y-axis and strain on the $x$x-axis gives a curve characteristic of the material (independent of sample shape). For a typical ductile metal like mild steel:

flowchart LR
  A[Origin] --> B[Linear region: Hooke obeyed, gradient = E]
  B --> C[P: limit of proportionality]
  C --> D[E: elastic limit]
  D --> F[Plastic region: large extension at near-constant stress]
  F --> G[UTS: ultimate tensile stress]
  G --> H[Necking begins]
  H --> I[Fracture / breaking]