What is elasticity?
Every solid is made of atoms held by spring-like forces. When you pull a solid, the atoms move apart a tiny bit. The springs pull back. This pull-back force is called the restoring force.
Elasticity is the property of a body to come back to its original size and shape when the force is removed. Steel and glass are highly elastic. Plasticity is the opposite: clay or putty stays deformed. No real material is perfectly elastic or perfectly plastic.
Surprise: steel is more elastic than rubber. It is harder to stretch steel, so it needs a larger force for the same strain. That is what a larger modulus means.
Stress and strain
Stress = restoring force ÷ area = F/A. Unit: N/m² = pascal (Pa). Three kinds:
- Tensile or compressive (longitudinal) stress: force along the length, pulling or pushing.
- Shearing stress: force along the surface (tangential), making layers slide.
- Hydraulic stress: equal pressure from all sides, like a stone deep under water.
Strain = change in size ÷ original size. It has no unit.
- Longitudinal strain = ΔL/L
- Shearing strain = Δx/L = tan θ ≈ θ (sideways shift ÷ height)
- Volume strain = ΔV/V
Hooke's law and the stress–strain curve
Hooke's law: for small deformations, stress is proportional to strain. So stress = k × strain, where k is called the modulus of elasticity.
Walk along the stress–strain curve of a metal wire (step 3 of the 3D):
- O to A (straight line): Hooke's law holds. A is the proportional limit.
- A to B: not straight any more, but the wire still returns fully. B is the elastic limit or yield point; stress there is the yield strength.
- B to D: the wire takes a permanent set. Remove the load and it does not come back. D shows the ultimate tensile strength.
- D to E: the wire thins at one spot (necking) and breaks at E, the fracture point.
If D and E are far apart, the material is ductile (copper, easily drawn into wires). If they are close, it is brittle (glass, cast iron).
Elastomers like rubber and the aorta in your body stretch a lot but do not follow Hooke's law; their curve is not a straight line.
Young's modulus
For stretching or compressing a wire or rod: Y = (F/A) ÷ (ΔL/L) = FL / (A ΔL). Unit: Pa.
Typical values: steel ≈ 2 × 10¹¹ Pa, copper ≈ 1.1 × 10¹¹ Pa, aluminium ≈ 0.7 × 10¹¹ Pa. Y depends on the material only, not on the length or thickness of the wire. A longer or thinner wire stretches more, but Y stays the same.
Lab idea (Searle's method): two identical wires hang side by side. One carries a fixed load (reference), the other gets extra loads. A spirit level and micrometer screw measure the tiny ΔL. The slope of the load–extension graph gives Y.
Bulk modulus and shear modulus
Bulk modulus B: when pressure rises by p on all sides, volume falls by ΔV. B = −p ÷ (ΔV/V). The minus sign makes B positive because ΔV is negative. Compressibility k = 1/B. Solids have the biggest B, liquids smaller, gases the smallest; so gases are the easiest to squeeze.
Shear modulus (modulus of rigidity) G: a tangential force F on the top face of area A makes the block tilt by angle θ. G = (F/A) ÷ θ. For most metals, G is about Y/3. Liquids and gases have no shear modulus because they cannot hold a fixed shape; their layers just flow.
Poisson's ratio
Pull a rubber band and it becomes thinner. The change in width is the lateral strain Δd/d. The change in length is the longitudinal strain ΔL/L.
Poisson's ratio σ = lateral strain ÷ longitudinal strain (sign ignored). It has no unit. For steel it is about 0.28–0.30; for aluminium about 0.33. It can never be more than 0.5 for ordinary materials.
Elastic potential energy in a stretched wire
While you stretch a wire, the force grows from 0 to F. So the average force is F/2. Work done = average force × stretch:
U = ½ F ΔL
Put F = YAΔL/L. Then U = ½ × Y × (ΔL/L)² × AL. Since AL is the volume:
Energy per unit volume u = ½ × stress × strain = ½ × Y × strain²
This energy is stored, not lost. That is why a stretched catapult throws a stone.
Uses of elastic behaviour
- Crane ropes: the rope is made of many thin wires twisted together (easier to make, flexible, strong). Its thickness is chosen so that the stress stays well below the yield strength with a safety margin.
- Beams in buildings and bridges: the sag of a beam of length l, breadth b, depth d, with load W at the middle is δ = Wl³/(4bd³Y). Depth comes as d³, so beams are laid on their narrow edge. An I-shaped cross-section saves material but keeps strength.
- Height of mountains: rock at the base must not flow under the weight above. This limits mountains on Earth to about 10 km.
- Springs, shock absorbers, musical strings and bridges are designed so they always stay in the elastic region.
Try it: rubber band Hooke test
Hang a rubber band on a nail and tie a small bag to it. Add equal coins one at a time and measure the length with a ruler. Write stretch vs number of coins. First few coins: equal steps (Hooke's law). Many coins: the steps change, because rubber is an elastomer. In the 3D, step 6, set 10 kg and read ΔL, then predict ΔL for 20 kg before moving the slider.
Key formulas and definitions
- Stress = F / A (Pa)
- Longitudinal strain = ΔL / L
- Young's modulus Y = FL / (A ΔL)
- Bulk modulus B = −p / (ΔV/V); compressibility = 1/B
- Shear modulus G = (F/A) / θ
- Poisson's ratio σ = (Δd/d) / (ΔL/L)
- Elastic energy U = ½ F ΔL; energy per volume = ½ × stress × strain
- Beam sag δ = W l³ / (4 b d³ Y)
Worked examples
1. A force of 100 N acts on a wire of cross-section 2 × 10⁻⁶ m². Find the stress.
Step 1: Stress = F/A. Step 2: = 100 / (2 × 10⁻⁶). Answer: 5 × 10⁷ Pa.
2. A 2 m wire stretches by 1 mm. Find the strain.
Step 1: ΔL = 1 mm = 1 × 10⁻³ m. Step 2: strain = ΔL/L = 10⁻³ / 2. Answer: 5 × 10⁻⁴ (no unit).
3. A steel wire (Y = 2 × 10¹¹ Pa) is 2 m long with radius 0.5 mm. A 10 kg load hangs on it. Find the extension. (g = 9.8 m/s²)
Step 1: F = 10 × 9.8 = 98 N. Step 2: A = πr² = 3.14 × (5 × 10⁻⁴)² = 7.85 × 10⁻⁷ m². Step 3: ΔL = FL/(AY) = 98 × 2 / (7.85 × 10⁻⁷ × 2 × 10¹¹). Step 4: = 196 / 1.57 × 10⁵. Answer: ΔL ≈ 1.25 × 10⁻³ m = 1.25 mm.
4. A copper wire and a steel wire of the same length and thickness carry the same load. Y(steel) = 2 × 10¹¹ Pa, Y(copper) = 1.1 × 10¹¹ Pa. Find the ratio of their extensions.
Step 1: ΔL = FL/(AY), and F, L, A are the same. So ΔL ∝ 1/Y. Step 2: ΔL(copper)/ΔL(steel) = Y(steel)/Y(copper) = 2/1.1. Answer: about 1.8. Copper stretches 1.8 times more.
5. Water is taken 1 km deep in the sea, where extra pressure is 1 × 10⁷ Pa. Bulk modulus of water is 2.2 × 10⁹ Pa. By what fraction does its volume fall?
Step 1: ΔV/V = −p/B. Step 2: = −(1 × 10⁷)/(2.2 × 10⁹). Answer: ΔV/V ≈ −4.5 × 10⁻³, a fall of about 0.45%. Water is hard to compress.
6. A metal block has a top face 0.5 m × 0.5 m. A tangential force of 9 × 10⁴ N shifts the top face by 0.2 mm relative to the bottom. The block is 0.5 m high. Find the shear modulus.
Step 1: shear stress = F/A = 9 × 10⁴ / 0.25 = 3.6 × 10⁵ Pa. Step 2: shear strain θ = Δx/L = 0.2 × 10⁻³ / 0.5 = 4 × 10⁻⁴. Step 3: G = stress/strain = 3.6 × 10⁵ / 4 × 10⁻⁴. Answer: G = 9 × 10⁸ Pa.
7. A wire 1 m long, area 1 × 10⁻⁶ m², Y = 2 × 10¹¹ Pa, is stretched by 1 mm. Find the elastic energy stored, and the lateral strain if Poisson's ratio is 0.3.
Step 1: F = YAΔL/L = 2 × 10¹¹ × 10⁻⁶ × 10⁻³ / 1 = 200 N. Step 2: U = ½ F ΔL = ½ × 200 × 10⁻³ = 0.1 J. Step 3: longitudinal strain = 10⁻³. Step 4: lateral strain = σ × 10⁻³ = 3 × 10⁻⁴. Answer: U = 0.1 J, lateral strain = 3 × 10⁻⁴ (the wire gets 0.03% thinner).
Common mistakes
- Thinking rubber is more elastic than steel. Steel has a much bigger Young's modulus, so it resists deformation more and returns better.
- Forgetting to convert mm to m and mm² to m² before putting values in Y = FL/(AΔL).
- Thinking Y changes when the wire is made longer or thinner. Only the extension changes; Y depends on the material alone.
- Using U = F ΔL instead of U = ½ F ΔL. The force grows from zero, so only the average force F/2 counts.