Elasticity, stress and strain
A rubber band springs back after a gentle pull, but a wet clay ball keeps its new shape. Link to earlier classes : the idea of a rigid body is now relaxed : real bodies stretch, compress and bend a little when forces act.
Elasticity is the property of a body to regain its original size and shape when the deforming force is removed. Stress is the restoring force per unit area. Strain is the fractional change in size.
Rule 1: Stress — stress = F ÷ A (F = force acting perpendicular or along the area, A = area). Unit : N m⁻² = pascal (Pa). Dimensions : M L⁻¹ T⁻².
- Easy : F = 100 N on A = 2 × 10⁻⁴ m² ⇒ stress = 5 × 10⁵ Pa.
- Medium : The same force on a wire of half the area gives twice the stress.
- Exam : A 600 N load hangs from a wire of radius 1.0 mm. A = πr² = 3.14 × 10⁻⁶ m² ⇒ stress = 1.9 × 10⁸ Pa.
Rule 2: Strain — strain = change in size ÷ original size. It has no unit. Longitudinal strain = ΔL ÷ L; shear strain = Δx ÷ h; volume strain = ΔV ÷ V.
- Easy : A 2.0 m wire stretches by 1 mm ⇒ strain = 0.001 ÷ 2.0 = 5 × 10⁻⁴.
- Medium : A 20 cm bar compresses by 0.1 mm ⇒ strain = 5 × 10⁻⁴ (compressive).
- Exam : Two wires of different lengths, both stretched by 1 mm, have different strains because strain = ΔL ÷ L.
Stress = F ÷ A (in Pa); strain = fractional change (no unit).
Dividing the force by the wrong area (use the cross-section of the wire, not its side area).
Memory clue : "Stress is Squeeze per area; Strain is Stretch per length."
Spot the difference
| Quantity | Meaning | Unit |
|---|---|---|
| Stress | force per unit area | pascal |
| Strain | fractional change | none |
Hooke's law and the stress–strain curve
A spring balance has an evenly spaced scale because the stretch grows in step with the load. Link to earlier classes : spring force F = −kx of chapter 4 is now stated for stress and strain.
Hooke's law : within the elastic limit, stress is directly proportional to strain. The ratio is a constant called the modulus of elasticity.
Rule 1: Hooke's law — stress = E × strain (E = modulus of elasticity). The unit of E is the same as that of stress (Pa).
- Easy : Stress 2 × 10⁸ Pa gives strain 1 × 10⁻³ ⇒ E = 2 × 10¹¹ Pa.
- Medium : If the stress is doubled within the elastic limit, the strain doubles.
- Exam : A material has E = 1 × 10¹¹ Pa. A strain of 2 × 10⁻³ needs stress = 2 × 10⁸ Pa.
Rule 2: Reading the curve — Along the curve there are the proportional limit, the elastic limit, the yield point, the plastic region and the fracture (breaking) point. Up to the elastic limit the body returns to its shape. Beyond the yield point it is permanently deformed.
In the figure : The straight part is Hooke's law; the peak is the ultimate strength; the end of the curve is the breaking point.
- Easy : A steel wire is pulled until it returns to its original length when released : it is still within the elastic limit.
- Medium : A material that breaks soon after the elastic limit is brittle (glass); one that stretches a long way is ductile (copper).
- Exam : A bridge cable has a breaking stress of 4 × 10⁸ Pa and a safety factor of 4. Its safe working stress = 1 × 10⁸ Pa; on a cross-section of 5 × 10⁻⁴ m² it can carry 5 × 10⁴ N.
Hooke's law holds only up to the proportional/elastic limit.
Applying Hooke's law beyond the yield point.
Memory clue : "Proportional, Elastic, Yield, Plastic, Fracture : PEYPF — go along the curve in this order."
Spot the difference
| Region | Behaviour on removing load |
|---|---|
| Elastic | returns fully |
| Plastic | permanent change remains |
Young's modulus
A thin steel wire and a rubber cord of the same size stretch very differently under the same load. Link to earlier classes : Hooke's law of the previous section is now used for stretching a wire.
Young's modulus Y is the ratio of longitudinal stress to longitudinal strain. It measures how hard it is to stretch or compress a material.
Rule 1: Young's modulus — Y = (F ÷ A) ÷ (ΔL ÷ L) = F L ÷ (A ΔL). Steel : about 2.0 × 10¹¹ Pa; copper : about 1.1 × 10¹¹ Pa; rubber is far smaller.
- Easy : F = 200 N, L = 1 m, A = 1 × 10⁻⁶ m², ΔL = 1 × 10⁻³ m ⇒ Y = 2 × 10¹¹ Pa.
- Medium : A steel wire (Y = 2 × 10¹¹ Pa), L = 2.0 m, A = 1.0 × 10⁻⁶ m² carries 100 N. ΔL = F L ÷ (A Y) = 100 × 2 ÷ (1 × 10⁻⁶ × 2 × 10¹¹) = 1.0 × 10⁻³ m.
- Exam : Same load and size on copper (Y = 1.1 × 10¹¹ Pa). The extension ratio copper : steel = 2.0 ÷ 1.1 = 1.8, so copper stretches 1.8 times more.
Rule 2: Elastic potential energy — Energy stored in a stretched wire = work done = ½ F ΔL = ½ × stress × strain × volume.
- Easy : F = 100 N and ΔL = 1 × 10⁻³ m ⇒ energy = ½ × 100 × 10⁻³ = 0.05 J.
- Medium : Energy per unit volume = ½ × stress × strain = ½ × 1 × 10⁸ × 5 × 10⁻⁴ = 2.5 × 10⁴ J m⁻³.
- Exam : Doubling the extension of a wire makes the stored energy four times larger (energy ∝ ΔL²).
Y = F L ÷ (A ΔL); a larger Y means a stiffer material; steel is stiffer than copper and rubber.
Calling "more stretch" a sign of a higher Young's modulus : it is the opposite.
Memory clue : "Y = F L ÷ (A ΔL) : Force and Length on top, Area and Elongation below."
Spot the difference
| Material | Young's modulus | Stretching |
|---|---|---|
| Steel | about 2.0 × 10¹¹ Pa | very hard |
| Copper | about 1.1 × 10¹¹ Pa | less hard |
| Rubber | much smaller | very easy |
Shear modulus
Pushing the top cover of a thick book sideways while the bottom stays fixed slides the pages over each other. Link to earlier classes : a force along a surface (tangential) is the kind of force that causes friction; here it changes the shape.
Shear stress is the tangential force per unit area. Shear strain is the angle (or the ratio Δx ÷ h) by which the layers slide. The shear modulus G is the ratio of the two.
Rule 1: Shear modulus — G = (F ÷ A) ÷ (Δx ÷ h) = F h ÷ (A Δx) (h = height of the body, Δx = sideways shift). Liquids and gases have no shear modulus because they flow.
- Easy : A shear force of 1 × 10⁴ N acts on a face of 0.01 m² ⇒ shear stress = 1 × 10⁶ Pa.
- Medium : The top face of a 10 cm block shifts by 1 mm ⇒ shear strain = 0.001 ÷ 0.1 = 0.01.
- Exam : A 10 cm steel cube (G = 8.4 × 10¹⁰ Pa) has a tangential force of 8.4 × 10⁴ N on its top face. Stress = 8.4 × 10⁴ ÷ 0.01 = 8.4 × 10⁶ Pa ⇒ strain = 1 × 10⁻⁴ ⇒ Δx = 1 × 10⁻⁵ m.
Shear changes the shape, not the volume.
Taking the force perpendicular to the face in shear : it must be along the face.
Memory clue : "Shear = Slide : the layers slide, nothing is squeezed."
Spot the difference
| Stress | Force direction | Change in |
|---|---|---|
| Tensile / compressive | perpendicular to the face | length |
| Shear | along the face | shape |
Bulk modulus
A diver feels pressure rising all around the body, and the whole body is squeezed from every side. Link to earlier classes : pressure from class 9 is now used to measure a change in volume.
Bulk modulus B is the ratio of hydraulic stress (pressure change) to volume strain. Its inverse (1 ÷ B) is called compressibility.
Rule 1: Bulk modulus — B = −Δp ÷ (ΔV ÷ V). The minus sign shows that an increase in pressure reduces the volume. Water : about 2.2 × 10⁹ Pa; steel : about 1.6 × 10¹¹ Pa; gases are very compressible.
- Easy : Δp = 2 × 10⁷ Pa and ΔV ÷ V = 1 × 10⁻³ (decrease) ⇒ B = 2 × 10¹⁰ Pa.
- Medium : For water (B = 2.2 × 10⁹ Pa) an increase of 2.2 × 10⁷ Pa gives ΔV ÷ V = 0.01, i.e. a 1% decrease in volume.
- Exam : At the bottom of a 4 km deep sea, p = ρgh = 1000 × 9.8 × 4000 = 3.9 × 10⁷ Pa ⇒ ΔV ÷ V = 3.9 × 10⁷ ÷ 2.2 × 10⁹ = 0.018. Water is compressed by about 1.8%, so its density rises by about 1.8%.
Solids and liquids have large B (hard to compress); gases have a small B.
Forgetting that B uses the change in pressure, not the total pressure.
Memory clue : "Bulk = Body squeezed from all sides : pressure in, volume out."
Spot the difference
| Modulus | Strain type | Applies to |
|---|---|---|
| Young's Y | length change | solids |
| Shear G | shape change | solids only |
| Bulk B | volume change | solids, liquids, gases |
Poisson's ratio and elastic energy
When you stretch a rubber band lengthwise, it becomes thinner. Link to earlier classes : longitudinal strain from section [1] is now related to the sideways (lateral) strain.
Poisson's ratio σ is the ratio of lateral strain to longitudinal strain in a stretched wire. It has no unit.
Rule 1: Poisson's ratio — σ = lateral strain ÷ longitudinal strain. For most materials σ lies between 0 and 0.5.
- Easy : Longitudinal strain 2 × 10⁻³ and lateral strain 5 × 10⁻⁴ ⇒ σ = 0.25.
- Medium : If σ = 0.3 and a rod is stretched by a strain of 1 × 10⁻³, its width shrinks by a strain of 3 × 10⁻⁴.
- Exam : A wire has σ = 0.25. For a longitudinal strain of 4 × 10⁻³ the fractional decrease in diameter = 0.25 × 4 × 10⁻³ = 1 × 10⁻³.
Rule 2: Elastic energy density — u = ½ × stress × strain (energy per unit volume, J m⁻³).
- Easy : stress = 1 × 10⁸ Pa and strain = 5 × 10⁻⁴ ⇒ u = 2.5 × 10⁴ J m⁻³.
- Medium : Using stress = Y × strain, u = ½ Y (strain)².
- Exam : For Y = 2 × 10¹¹ Pa and strain 1 × 10⁻³, u = ½ × 2 × 10¹¹ × 10⁻⁶ = 1 × 10⁵ J m⁻³.
σ = lateral strain ÷ longitudinal strain; energy per volume u = ½ × stress × strain.
Treating the lateral strain as having the same sign as the longitudinal strain; a stretched rod becomes thinner.
Memory clue : "Pull long, get thin : Poisson is the thinning-to-lengthening ratio."
Spot the difference
| Ratio | Formula | Unit |
|---|---|---|
| Poisson's ratio | lateral ÷ longitudinal strain | none |
| Young's modulus | stress ÷ strain | pascal |
Applications of elastic behaviour
Cranes use thick steel ropes made of many thin wires, and the bridge girders are shaped like the letter I. Link to earlier classes : the moduli of the earlier sections now guide the choice of material and shape.
Engineers use the elastic constants to choose a safe material, cross-section and shape so that the stress stays below the elastic limit.
Rule 1: Safe design — Allowed stress = breaking stress ÷ safety factor. Required area = load ÷ allowed stress. A rope made of many thin strands is more flexible than a single solid rod of the same area.
- Easy : Breaking stress 4 × 10⁸ Pa, safety factor 4 ⇒ allowed stress = 1 × 10⁸ Pa.
- Medium : Load 2 × 10⁴ N, allowed stress 1 × 10⁸ Pa ⇒ area = 2 × 10⁻⁴ m².
- Exam : For the same load, a larger Young's modulus makes the sag smaller (ΔL ∝ 1 ÷ Y), so steel girders sag less than copper ones.
Rule 2: Beams — A beam loaded at the middle bends by a small amount; the bending (depression) δ is proportional to W l³ ÷ (4 b d³ Y) for a rectangular beam of length l, breadth b and depth d. Making the beam deep is far more effective than making it broad.
- Easy : Doubling the depth d reduces the bending to 1 ÷ 8 of its value (because of d³).
- Medium : Doubling the breadth b only halves the bending.
- Exam : A beam of length l loaded W bends by δ. For the same load a beam twice as long bends 8 times as much (δ ∝ l³).
Deep, hollow, I-shaped sections save material and resist bending; columns use a wider top to spread the load.
Making a beam broad instead of deep.
Memory clue : "Depth is the cube : d³ in the denominator."
Spot the difference
| Change | Effect on bending |
|---|---|
| Depth doubled | bending ÷ 8 |
| Breadth doubled | bending ÷ 2 |
| Length doubled | bending × 8 |