What an elastic modulus is
Hooke's law states that for small deformations, stress is proportional to strain: $\text{stress} = k \times \text{strain}$. The constant of proportionality $k$ is the modulus of elasticity. Because strain is dimensionless, every modulus carries the units of stress, namely $\text{N m}^{-2}$ or pascal (Pa), with dimensional formula $[\text{ML}^{-1}\text{T}^{-2}]$.
There is not one modulus but three, because a solid can be deformed in three distinct ways — stretched, sheared, or squeezed uniformly. Each mode pairs a particular kind of stress with a particular kind of strain, and the ratio of the two defines a separate elastic constant. The whole subtopic is best held in mind as a single parallel structure, which the comparison table below makes explicit.
Young's modulus
When a wire or rod is stretched by a force $F$ applied normal to its cross-section of area $A$, it experiences a tensile (longitudinal) stress $F/A$ and elongates by $\Delta L$ on an original length $L$. The longitudinal strain is $\Delta L / L$. Young's modulus $Y$ is their ratio:
$$ Y = \frac{\text{tensile stress}}{\text{longitudinal strain}} = \frac{F/A}{\Delta L / L} = \frac{F\,L}{A\,\Delta L} $$
NCERT records that experimentally the strain is the same whether the stress is tensile or compressive, so a single $Y$ serves both. Rearranged, the same relation gives the Hooke's-law working form and the elongation:
$$ F = \frac{Y\,A\,\Delta L}{L}, \qquad \Delta L = \frac{F\,L}{Y\,A} $$
Metals have large Young's moduli, so they need a large force for a small change in length. NCERT illustrates this with steel: a force of 2000 N is needed to stretch a thin steel wire of cross-section $0.1~\text{cm}^2$ by 0.1%, whereas aluminium, brass and copper of the same area need only 690 N, 900 N and 1100 N respectively for the same strain. Steel therefore has the highest Young's modulus of these four — it is "more elastic" — which is why it is preferred in heavy machines and structural designs. Wood, bone, concrete and glass have comparatively small moduli.
Shear modulus (modulus of rigidity)
When two equal and opposite forces are applied parallel to the faces of a body, opposite faces slide relative to one another. The tangential force per unit area is the shearing stress $\sigma_s = F/A$, and the body twists through a small angle $\theta$. The relative displacement $\Delta x$ of one face over the height $L$ gives the shearing strain $\Delta x / L = \tan\theta \approx \theta$ for small angles. The shear modulus $G$, also called the modulus of rigidity, is their ratio:
$$ G = \frac{\text{shearing stress}}{\text{shearing strain}} = \frac{F/A}{\Delta x / L} = \frac{F/A}{\theta} = \frac{F}{A\,\theta} $$
From this, the shearing stress can be written compactly as $\sigma_s = G\,\theta$. NCERT notes that for most materials $G \approx Y/3$, so the modulus of rigidity is generally smaller than Young's modulus. Crucially, shear changes the shape of the body while leaving its volume unchanged — which is why the stretching of a coil spring is governed by the shear modulus, not Young's modulus.
Bulk modulus & compressibility
When a body is immersed in a fluid under pressure, the fluid presses normally and uniformly at every point of its surface. The body shrinks in volume by $\Delta V$ on an original volume $V$ with no change of shape. The hydraulic stress equals the applied pressure $p$, and the volume strain is $\Delta V / V$. The bulk modulus $B$ is:
$$ B = -\frac{p}{\Delta V / V} $$
The negative sign is essential: an increase in pressure produces a decrease in volume, so if $p$ is positive then $\Delta V$ is negative, and the minus sign keeps $B$ positive for any system in equilibrium. The reciprocal of the bulk modulus is the compressibility:
$$ k = \frac{1}{B} = -\frac{1}{\Delta p}\left(\frac{\Delta V}{V}\right) $$
From the NCERT data, bulk moduli for solids are far larger than for liquids, which are again far larger than for gases. Solids are therefore the least compressible and gases the most — gases are about a million times more compressible than solids. Unlike $Y$ and $G$, the bulk modulus applies to all three states of matter, because volume change is possible in solids, liquids and gases alike.
The three moduli compared
The three moduli share one definition — stress over strain within the elastic limit — but differ in the stress applied, the strain produced, and the state of matter to which they apply. The table consolidates NCERT Table 8.4.
| Modulus | Definition (stress / strain) | Formula | Deforming stress | Changes | Applies to |
|---|---|---|---|---|---|
| Young's, $Y$ | Longitudinal (tensile/compressive) stress / longitudinal strain | Y = FL / (A·ΔL) |
Two equal, opposite forces perpendicular to opposite faces | Size (length); not volume | Solids only |
| Shear, $G$ | Shearing (tangential) stress / shearing strain $\theta$ | G = F / (A·θ) |
Two equal, opposite forces parallel to opposite faces | Shape; not volume | Solids only |
| Bulk, $B$ | Hydraulic stress (pressure) / volume strain | B = −p / (ΔV/V) |
Forces normal everywhere; pressure uniform on all surfaces | Volume; not shape | Solids, liquids, gases |
For most metals the three are ordered $Y > B > G$, with $G \approx Y/3$ as the handy NCERT estimate. Representative NCERT table values (in GPa, i.e. $10^{9}~\text{N m}^{-2}$) help calibrate the scale: among liquids, water has $B = 2.2$, glycerine $4.76$, ethanol $0.9$ and carbon disulphide $1.56$; air at STP has the tiny bulk modulus $1.0 \times 10^{-4}~\text{GPa}$, confirming how readily gases compress. Young's modulus of structural steel is $2.0 \times 10^{11}~\text{N m}^{-2}$; copper is $1.1 \times 10^{11}~\text{N m}^{-2}$; bone is about $9.4 \times 10^{9}~\text{N m}^{-2}$.
If the stress and strain definitions feel shaky, revisit stress and strain and Hooke's law before going further.
Poisson's ratio
A wire that stretches lengthwise also thins sideways. The strain perpendicular to the applied force is the lateral strain, and within the elastic limit it is proportional to the longitudinal strain. Their ratio is Poisson's ratio $\sigma$. If the original diameter is $d$ and contracts by $\Delta d$ while the length $L$ elongates by $\Delta L$:
$$ \sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{\Delta d / d}{\Delta L / L} = \frac{\Delta d}{\Delta L}\cdot\frac{L}{d} $$
Poisson's ratio is a ratio of two strains, so it is a pure number with no units or dimensions, and its value depends only on the nature of the material. NCERT gives it as 0.28 to 0.30 for steels and about 0.33 for aluminium alloys; common materials lie in the 0.2 to 0.4 band. Because a deforming force in one direction generally produces strains in other directions too, a single modulus cannot fully describe such behaviour — Poisson's ratio is the extra constant that captures it.
Elastic potential energy in a stretched wire
Stretching a wire does work against the inter-atomic forces, and that work is stored as elastic potential energy. For a wire of length $L$ and area $A$, the force at elongation $l$ is $F = YA\,(l/L)$ — it grows from zero as the wire extends. Integrating the work $dW = F\,dl$ from $0$ to the final elongation gives:
$$ W = \int_0^{l} \frac{YA\,l'}{L}\,dl' = \frac{1}{2}\,\frac{YA}{L}\,l^{2} = \frac{1}{2}\times\text{stress}\times\text{strain}\times\text{volume} $$
The energy stored per unit volume is therefore $u = \tfrac{1}{2}\,\sigma\varepsilon$, and the total stored energy can be written equivalently as $U = \tfrac{1}{2}\,\text{stress}\times\text{strain}\times\text{volume} = \tfrac{1}{2}\,F\,\Delta L$. The factor of one-half is the entire point: the force is not constant during stretching but rises linearly from zero, so the average force is half the final force.
Worked examples
A structural steel rod of radius 10 mm and length 1.0 m is stretched by a 100 kN force along its length. Find (a) the stress, (b) the elongation, and (c) the strain. Take $Y = 2.0\times10^{11}~\text{N m}^{-2}$.
(a) Stress. $A = \pi r^{2} = 3.14\times(10^{-2})^{2} = 3.14\times10^{-4}~\text{m}^{2}$, so $\sigma = F/A = (10^{5})/(3.14\times10^{-4}) = 3.18\times10^{8}~\text{N m}^{-2}$.
(b) Elongation. $\Delta L = \dfrac{(F/A)\,L}{Y} = \dfrac{3.18\times10^{8}\times 1}{2.0\times10^{11}} = 1.59\times10^{-3}~\text{m} = 1.59~\text{mm}$.
(c) Strain. $\varepsilon = \Delta L/L = 1.59\times10^{-3} = 0.16\%$.
The average depth of the Indian Ocean is about 3000 m. Find the fractional compression $\Delta V/V$ of water at the bottom, given the bulk modulus of water is $2.2\times10^{9}~\text{N m}^{-2}$ and $g = 10~\text{m s}^{-2}$.
Pressure at depth. $p = h\rho g = 3000\times1000\times10 = 3\times10^{7}~\text{N m}^{-2}$.
Fractional compression. $\dfrac{\Delta V}{V} = \dfrac{p}{B} = \dfrac{3\times10^{7}}{2.2\times10^{9}} = 1.36\times10^{-2}$, that is about $1.36\%$.
Elastic moduli in one breath
- Modulus = stress / strain within the elastic limit; units Pa, dimensions $[\text{ML}^{-1}\text{T}^{-2}]$.
- Young's $Y = FL/(A\,\Delta L)$ — change of length; solids only; steel highest among common metals.
- Shear $G = F/(A\theta)$ — change of shape, not volume; $G \approx Y/3$; governs coil-spring stretching.
- Bulk $B = -p/(\Delta V/V)$ — change of volume; keep the minus sign; all states of matter; solids $>$ liquids $>$ gases.
- Compressibility $k = 1/B$, the reciprocal of bulk modulus.
- Poisson's ratio $\sigma = $ lateral / longitudinal strain; dimensionless; 0.2–0.4 typically.
- Elastic PE per unit volume $u = \tfrac{1}{2}\,\sigma\varepsilon$; total $U = \tfrac{1}{2}F\,\Delta L$ — never forget the one-half.