Elasticity

Every solid changes shape a little when you push or pull it. A bridge deck sags under a loaded truck on the Prithvi Highway, a steel cable in a Trishuli suspension bridge stretches, a rubber band grows longer. Elasticity asks two questions: how much does a body deform under a given force, and does it come back when the force is removed? The whole unit is built on two quantities — stress (force per unit area) and strain (fractional change in size) — and on the constants that link them.

8.1 Hooke's law: force constant

When an external force acts on a body, its molecules are pulled apart from (or pushed towards) their equilibrium separations. The inter-molecular forces then act to restore the original spacing. The external force is called the deforming force; the internal force that opposes it is the restoring force. As long as the deformation is small, the two are equal in magnitude.

Robert Hooke stated the experimental law in 1678: for small deformations the extension is proportional to the load.

The constant k is the force constant (also called stiffness, or spring constant for a spring). It is the force needed to produce unit extension:

The force constant is not a property of the material alone. For a wire of length L and area of cross-section A made of a material of Young modulus Y (section 8.3),

so a short thick wire is stiffer than a long thin one of the same metal. When springs of constants k_1 and k_2 are joined in series the combination is softer, 1/k = 1/k_1 + 1/k_2; joined side by side in parallel it is stiffer, k = k_1 + k_2.

8.2 Stress, Strain, Elasticity and plasticity

Stress has the same dimensions as pressure, but the two are not the same thing: pressure in a fluid always acts normal to a surface, while stress may act along a surface (shear). There are three kinds of stress and a matching strain for each.

Stress	How the force acts	Strain produced	Modulus
Tensile / compressive (longitudinal)	normal to the end faces, along the length	\text{longitudinal strain} = e/L	Young modulus Y
Hydraulic (volume)	normal and equal on every face	\text{volume strain} = \Delta V/V	Bulk modulus B
Tangential (shear)	parallel to one face, the opposite face fixed	\text{shear strain} = \theta = x/L	Shear modulus \eta

Elasticity is the property by which a body regains its original shape and size after the deforming force is removed. Plasticity is the opposite property: the body keeps the new shape. No real material is perfectly elastic or perfectly plastic; a quartz fibre is the nearest to perfectly elastic and wet clay or putty the nearest to perfectly plastic.

Reading the curve from the origin:

• O to P — proportional limit. Stress is proportional to strain; Hooke's law holds and the slope of this line is the Young modulus.
• P to E — elastic limit. The graph bends, so Hooke's law fails, but the wire still returns exactly to its original length when unloaded.
• Beyond E — permanent set. Unloading now leaves the wire longer than it started. At the yield point Y the wire begins to extend with almost no extra load.
• U — ultimate tensile stress, the largest stress the material can take. After U the wire develops a local neck, so the true area falls and the curve drops until the wire snaps at the breaking point B.

A material that stretches a great deal between Y and B (copper, mild steel, gold) is ductile; one that breaks very soon after the elastic limit (glass, cast iron, ceramic tile) is brittle.

Two related effects are worth naming. Elastic after-effect is the delay some materials (glass, rubber) show in returning to their original size. Elastic fatigue is the loss of elastic strength when a body is subjected to repeated cycles of stress — the reason a wire finally breaks after being bent back and forth many times, and why bridges and aircraft parts are replaced long before they look worn.

8.3 Elastic modulus: Young modulus, bulk modulus, shear modulus

Young modulus

where e is the extension of a wire of original length L and area A under a load F. Rearranging gives F = (YA/L)e, which compared with F = kx proves the result k = YA/L quoted in section 8.1. Only solids have a Young modulus, because only a solid can be pulled into a wire.

Bulk modulus

If a uniform increase of pressure p on all sides reduces a volume V by \Delta V, then

The minus sign is there because \Delta V is negative when p increases, so that B comes out positive. Solids, liquids and gases all have a bulk modulus. Its reciprocal is the compressibility, K = 1/B, measured in Pa⁻¹.

Shear modulus (modulus of rigidity)

A tangential force F on the top face of area A of a block of height L slides that face sideways by x, turning the vertical edges through a small angle \theta = x/L radians. Then

Only solids have a shear modulus: a liquid or a gas cannot sustain a tangential stress — it simply flows. For most metals \eta is roughly Y/3.

Approximate room-temperature values (1 GPa = 10^{9} Pa):

Material	Y (GPa)	B (GPa)	\eta (GPa)	Poisson ratio \sigma
Steel	200	160	84	0.29
Copper	110	140	44	0.34
Brass	91	61	35	0.34
Aluminium	70	70	25	0.33
Glass	55	37	23	0.25
Water	—	2.2	0	—
Rubber	0.0005 – 0.05	≈ 2	≈ 0.0007	≈ 0.50

8.4 Poisson's ratio

A stretched wire does not only get longer — it also gets thinner. If a wire of length L and diameter d stretches by e and its diameter shrinks by \Delta d, then

Theory allows \sigma to lie between -1 and +0.5. For all common materials it is positive and lies between about 0.2 and 0.4; rubber is the extreme case at nearly 0.5. The upper bound has a simple physical meaning. For a wire of volume V = \pi r^{2}L, small changes give

If \sigma = 0.5 the volume does not change at all when the wire is stretched; \sigma larger than 0.5 would mean the volume shrinks on stretching, which no material does.

8.5 Elastic potential energy

Stretching a wire takes work, and that work is stored in the wire as elastic potential energy. It is recovered when the wire is released — this is how a catapult, a bow and a clock spring work.

Chapter summary

• Hooke's law: within the elastic limit F = kx, where the force constant k = F/x (N m⁻¹). For a wire k = YA/L, so k depends on shape as well as material.
• Stress = F/A (Pa, [ML^{-1}T^{-2}]); strain = change ÷ original value (no unit). Modulus = stress ÷ strain.
• The stress–strain curve runs: proportional limit P, elastic limit E, yield point Y, ultimate stress U, breaking point B. Ductile materials have a long Y–B stretch; brittle ones break just after E.
• Young modulus Y = FL/Ae; bulk modulus B = -pV/\Delta V with compressibility K = 1/B; shear modulus \eta = F/(A\theta) with \theta = x/L. Only B applies to liquids and gases.
• Poisson's ratio \sigma = (\Delta d/d)\div(e/L); no unit, theoretical range -1 to 0.5, real materials 0.2–0.4. Also \Delta V/V = (1-2\sigma)e/L.
• Elastic potential energy U = \frac{1}{2}Fe = \frac{1}{2}\dfrac{YAe^{2}}{L}, and energy per unit volume u = \frac{1}{2}\times stress \times strain = \frac{1}{2}Y\varepsilon^{2}.
• Breaking force depends on the area of cross-section, never on the length.

Practice questions

Group A — Multiple choice (1 mark each)

1. The SI unit of stress is [1]aN mbN m⁻²cN m⁻¹dit has no unit
2. Strain is [1]ameasured in N m⁻²bmeasured in mca pure numberdmeasured in N
3. For a perfectly rigid body the Young modulus is [1]azerobinfinitecunitydnegative
4. The theoretical limits of Poisson's ratio are [1]a0 to 1b-1 to 0.5c-0.5 to 1d0 to 0.25
5. The energy stored per unit volume of a stretched wire equals [1]astress × strainb½ × stress × straincstress ÷ straind½ × stress ÷ strain

Answers to Group A

1. (b) — stress is force per unit area, N m⁻² = Pa.

2. (c) — it is a ratio of two lengths (or two volumes), so it has no unit.

3. (b) — a rigid body has zero strain, and Y = \text{stress}/\text{strain} \to \infty.

4. (b) — \sigma larger than 0.5 would make the volume fall on stretching.

5. (b) — the area under the stress–strain line is \frac{1}{2}\times stress \times strain.

Group B — Short answer (5 marks each)

1. Define stress and strain and state their units. State Hooke's law and hence define modulus of elasticity. [5]
2. Draw a stress–strain curve for a ductile material and mark the proportional limit, elastic limit, yield point and breaking point. Distinguish between elasticity and plasticity. [5]
3. Define Young modulus. A steel wire 3.0\ \text{m} long with cross-sectional area 2.0\ \text{mm}^{2} stretches by 1.5\ \text{mm} under a load W. Find W, given Y = 2.0\times10^{11}\ \text{Pa}. [5]
4. Define bulk modulus and compressibility. The pressure on a sample of water is increased by 4.4\times10^{7}\ \text{Pa}. Find the percentage decrease in its volume if B = 2.2\times10^{9}\ \text{Pa}. [5]
5. Show that the energy stored per unit volume of a stretched wire is \frac{1}{2}\times stress \times strain. A wire of volume 1.0\times10^{-6}\ \text{m}^{3} is stretched so that its strain is 1.0\times10^{-3}. Find the energy stored if Y = 2.0\times10^{11}\ \text{Pa}. [5]

Answers to Group B

3. F = \dfrac{YAe}{L} = \dfrac{2.0\times10^{11}\times2.0\times10^{-6}\times1.5\times10^{-3}}{3.0} = 200\ \text{N}.

4. \dfrac{\Delta V}{V} = \dfrac{p}{B} = \dfrac{4.4\times10^{7}}{2.2\times10^{9}} = 0.020, i.e. a decrease of 2.0\ \%.

5. Outline: work done = \int_0^e (YA/L)x\,dx = \frac{1}{2}Fe; divide by volume AL to get u = \frac{1}{2}(F/A)(e/L). Numerically U = \frac{1}{2}Y\varepsilon^{2}V = \frac{1}{2}\times2.0\times10^{11}\times(1.0\times10^{-3})^{2}\times1.0\times10^{-6} = 0.10\ \text{J}.

Group C — Long answer (8 marks each)

1. (a) Define the three elastic moduli and write the expression for each. [3](b) Derive an expression for the elastic potential energy stored in a stretched wire and hence for the energy per unit volume. [3](c) A wire of length 2.0\ \text{m} and area 1.0\ \text{mm}^{2} is stretched by 1.0\ \text{mm} with Y = 1.1\times10^{11}\ \text{Pa}. Find the energy stored. [2]
2. A copper wire and a steel wire, each 2.0\ \text{m} long and 1.0\ \text{mm} in diameter, are joined end to end and a force of 100\ \text{N} is applied to the free ends. Given Y_{Cu} = 1.1\times10^{11}\ \text{Pa} and Y_{St} = 2.0\times10^{11}\ \text{Pa}, find the stress in each wire, the extension of each wire and the total extension. [8]

Answers to Group C

1. (c) F = YAe/L = (1.1\times10^{11}\times1.0\times10^{-6}\times1.0\times10^{-3})/2.0 = 55\ \text{N}, so U = \frac{1}{2}Fe = \frac{1}{2}\times55\times1.0\times10^{-3} = 2.75\times10^{-2}\ \text{J}.

2. The two wires are in series, so each carries the same force 100\ \text{N} and, having the same diameter, the same stress.

A = \pi d^{2}/4 = 7.854\times10^{-7}\ \text{m}^{2}, hence stress = 100/7.854\times10^{-7} = 1.273\times10^{8}\ \text{Pa} in both.

Copper: e = \text{stress}\times L / Y = (1.273\times10^{8}\times2.0)/1.1\times10^{11} = 2.32\times10^{-3}\ \text{m} = 2.32\ \text{mm}.

Steel: e = (1.273\times10^{8}\times2.0)/2.0\times10^{11} = 1.27\times10^{-3}\ \text{m} = 1.27\ \text{mm}.

Total extension = 2.32 + 1.27 = 3.59\ \text{mm}. The copper wire stretches nearly twice as much because its Young modulus is about half that of steel.