Elastic moduli

What are elastic moduli?

Within the proportional region of the stress-strain curve, the ratio of stress to strain is a constant for a material, called its modulus of elasticity (NCERT Section 8.5, page 170). For the elementary homogeneous, isotropic model, three common moduli describe these loading modes; anisotropic materials need additional directional elastic constants:

In plain words: a modulus says how hard a material is to deform in a particular way. A large Young’s modulus means a wire is hard to stretch; a large shear modulus means a block is hard to shear (its top face pushed sideways relative to the bottom); a large bulk modulus means a body is hard to squeeze into a smaller volume. Two more ideas belong here: Poisson’s ratio, which links stretching to thinning, and the elastic potential energy stored in a stretched wire.

Young’s and shear moduli apply only to solids; the bulk modulus applies to solids, liquids and gases (after NCERT Table 8.4, page 173).

Young’s modulus

The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Young’s modulus and is denoted by the symbol Y.

NCERT Physics Part II (Class 11), Section 8.5.1, page 170

With stress F/A and strain ΔL/L (see stress and strain):

In the small linear range, the model gives equal strain magnitudes for equal tensile and compressive stress magnitudes, so one Y serves for both (NCERT, page 170). The laboratory experiment to measure Young’s modulus of a wire applies this formula, with F = Mg and A = πr^².

Some values from NCERT Table 8.1 (page 170), in units of 10^⁹ N m^⁻² (GPa): aluminium 70, copper 110, iron 190, steel 200, glass 65, concrete 30, wood 13, bone 9.4, polystyrene 3. Metals have large Young’s moduli, so they need a large force for a small change in length; that is why steel, with the largest value here, is preferred in heavy machines and buildings, and why NCERT calls steel “more elastic” than copper, brass and aluminium (page 171). In NCERT’s comparison, “more elastic” means a larger modulus, or greater stiffness: steel is more elastic than rubber, though rubber stretches more, and a perfectly rigid body would have an infinite Y. NCERT’s Points to Ponder (page 177) also says that metals have larger Young’s moduli than alloys; its own Table 8.1 shows steel, an alloy of iron, slightly above iron, so treat that only as a loose remark.

Shear modulus

The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by G. It is also called the modulus of rigidity.

NCERT Physics Part II (Class 11), Section 8.5.2, page 172

Here F acts along a face of area A, Δx is how far that face moves relative to the opposite face, L is the distance between them, and θ is the angle of shear in radians (for small angles, Δx/L = tan θ ≈ θ). Some values from NCERT Table 8.2 (page 172), in GPa: aluminium 25, copper 42, lead 5.6, steel 84, tungsten 150, wood 10.

The shear modulus is generally less than Young’s modulus; NCERT gives G ≈ Y/3 as a rule of thumb (page 172). It is rough: the tables give Y/G of about 2.4 for steel, 2.6 for copper and 2.8 for aluminium.

Bulk modulus and compressibility

The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus. It is denoted by symbol B.

NCERT Physics Part II (Class 11), Section 8.5.3, page 173

Here p is the increase in pressure, Δp. An increase of pressure makes the volume decrease, so ΔV is negative; the minus sign makes B positive. The reciprocal of B is the compressibility (NCERT writes it k), the fractional decrease in volume per unit increase in pressure:

Some values from NCERT Table 8.3 (page 173), in GPa: steel 160, copper 140, aluminium 72, glass 37; water 2.2, mercury 25, glycerine 4.76; air at STP 1.0 × 10^⁻⁴. For the materials and conditions in these tables, solids have larger bulk moduli than liquids and air has the smallest: gases are about a million times more compressible than solids (NCERT, page 174). Atoms in a solid are tightly coupled to their neighbours, molecules in a liquid less so, and molecules in a gas hardly at all.

Poisson’s ratio

A stretched wire gets thinner as well as longer. The strain at right angles to the force is the lateral strain. In the small linear elastic range it is proportional to the longitudinal strain, and their ratio is Poisson’s ratio (NCERT Section 8.5.4, page 174). For a wire of diameter d that thins by Δd while its length L grows by ΔL:

Here Δd denotes the positive decrease in diameter, matching NCERT’s notation. If lateral strain is instead signed as (final diameter − original diameter)/original diameter, the general definition is ν = −(lateral strain)/(longitudinal strain). This sign also permits auxetic materials, which widen on stretching and have negative ν.

Poisson’s ratio is a ratio of two strains; it is a pure number and has no dimensions or units.

NCERT Physics Part II (Class 11), Section 8.5.4, page 174

Its value depends only on the material: between 0.28 and 0.30 for steels and about 0.33 for aluminium alloys (NCERT, page 174). Beyond the NCERT text: for ordinary materials it lies between 0 and 0.5 (for stable isotropic linear elastic materials with finite positive bulk and shear moduli, −1 < ν < ½); rubber is close to 0.5, which means its volume hardly changes when it is stretched.

Elastic potential energy in a stretched wire

When a wire is put under a tensile stress, work is done against the inter-atomic forces. This work is stored in the wire in the form of elastic potential energy.

NCERT Physics Part II (Class 11), Section 8.5.5, page 174

Let a wire of length L and cross-section A be stretched by an amount l (NCERT uses l here for the extension, ΔL above). From the definition of Y, the force at an extension x is YAx/L, which grows in proportion to x. The work done in stretching it from 0 to l, as for any variable force, is the area under the force-extension line:

with F = YAl/L the final force: that is, ½ × stress × strain × volume. This is the same form as the ½kx^² of a spring, with k = YA/L. Per unit volume:

Since σ = Yε, this can also be written u = σ^²/(2Y) = ½Yε^².

Solved examples

The first five are NCERT Examples 8.1 to 8.5 (pages 171–174). Examples 1–3 use Young’s modulus, Example 4 the shear modulus, Example 5 the bulk modulus and Example 6 the elastic energy; Poisson’s ratio is practised in Q3. Take g = 9.8 m s^⁻² unless stated.

Printing slips in NCERT Section 8.5. Several cross-references in NCERT’s text still carry numbers from an older edition: Examples 8.2 and 8.3 cite “Table 9.1” for Table 8.1, Example 8.3 cites “Eq. (9.8)” for Eq. (8.8), and Section 8.5.2 cites “Eq. (9.4)”, “Table 9.2” and “Table 9.1” for Eq. (8.4), Table 8.2 and Table 8.1. Equation number (8.12) is used twice, for σ_s = Gθ and for the bulk modulus. Section 8.5.3 refers to hydraulic stress as seen “in Section (8.3)”; it was Section 8.2. In Example 8.4 the working uses 9.4 × 10^⁴ N, though the question gives 9.0 × 10^⁴ N; the printed result, 1.80 × 10^⁶ N m^⁻², matches 9.0 × 10^⁴ N, as used above. The same answer refers to “Fig. 8.6” for the slab, which is printed as Fig. 8.5, and the equation numbers jump from (8.8) to (8.10).

Common mistakes to avoid

Quick revision summary

• A modulus of elasticity is stress/strain in the proportional region; unit Pa, dimensions [ML^⁻¹T^⁻²].
• Young’s modulus Y = FL/(AΔL); metals have large Y, and steel has the largest value in NCERT Table 8.1.
• For small shear angles, G ≈ F/(Aθ); for common metals NCERT’s rough rule is G ≈ Y/3 (its steel, copper and aluminium values give Y/G ≈ 2.4–2.8).
• Bulk modulus B = −p/(ΔV/V); compressibility = 1/B; solids are least compressible, gases most.
• Poisson’s ratio is minus signed lateral strain divided by longitudinal strain; NCERT instead uses a positive diameter contraction. It is dimensionless, about 0.3 for many metals.
• Elastic energy stored in a stretched wire = ½ × stress × strain × volume; per unit volume u = ½σε.

Questions students ask

How are Young’s modulus, shear modulus, bulk modulus and Poisson’s ratio related?

For a homogeneous isotropic material in the linear elastic range, with the same properties in every direction, Young’s modulus Y, the shear modulus G, the bulk modulus B and Poisson’s ratio ν are linked by Y = 2G(1 + ν) and Y = 3B(1 − 2ν), so only two of the four are independent (a result beyond the NCERT text). With ν ≈ 0.3, typical of metals, these give G ≈ Y/2.6, within the 2.4–2.8 of NCERT’s tables, and B ≈ Y/1.2; for steel, Y = 200 GPa gives B ≈ 170 GPa, close to NCERT’s 160 GPa. Since G and B must be positive, ν lies between −1 and ½, and a material with ν close to ½, such as rubber, is very hard to compress compared with how easily it stretches.

What is the bulk modulus of a gas?

The bulk modulus of an ideal gas equals its pressure p for an isothermal compression and γp for a reversible adiabatic compression, where γ = C_p/C_v. Slow compression with good thermal contact can approximate the first process; negligible heat exchange and near-equilibrium compression approximate the second. Speed alone does not specify the thermodynamic path. So air at atmospheric pressure has an isothermal bulk modulus of about 1.0 × 10^⁵ Pa, the value in NCERT Table 8.3, and an adiabatic one of about 1.4 × 10^⁵ Pa, about a millionth of the bulk moduli of metals. The adiabatic value is the one that sets the speed of sound in a gas.

How much does a rod or wire stretch under its own weight?

A uniform rod or wire hanging from one end stretches under its own weight by ΔL = ρgL^²/(2Y) = MgL/(2AY), half as much as if its whole weight Mg hung from its lower end. This is because the tension at each point is the weight of the part below it, so the stress grows steadily from zero at the bottom to ρgL at the top, and the average stress is half the greatest. For example, a steel wire 50 m long, with ρ = 7.8 × 10^³ kg m^⁻³ and Y = 2.0 × 10^¹¹ N m^⁻², stretches by 7.8 × 10^³ × 9.8 × 50^²/(4.0 × 10^¹¹) ≈ 0.48 mm under its own weight, whatever its thickness.

What is the equivalent Young’s modulus of two wires joined end to end or side by side?

For two wires of equal length and equal cross-section, the equivalent Young’s modulus is (Y₁ + Y₂)/2 when they hang side by side and share the load, and 2Y₁Y₂/(Y₁ + Y₂) when they are joined end to end. Side by side they stretch equally, like springs in parallel, so their spring constants add; end to end they carry the same tension, like springs in series, so their extensions add. For copper (1.1 × 10^¹¹ Pa) and steel (2.0 × 10^¹¹ Pa) these give about 1.55 × 10^¹¹ Pa and 1.42 × 10^¹¹ Pa.

Are Poisson’s ratio and elastic energy in the JEE Main and NEET syllabus?

Poisson’s ratio and the elastic energy of a stretched wire are not named in the official JEE Main 2027 or NEET (UG) 2026 syllabus, whose Unit 7 lists elastic behaviour, the stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus and modulus of rigidity. Both are in NCERT Sections 8.5.4 and 8.5.5, and the stored energy, ½ × stress × strain per unit volume, follows directly from Young’s modulus and the work done by a variable force, so learn them with the rest of Section 8.5.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Two wires of the same material carry equal loads. One has length L and radius r; the other has length 2L and radius 2r. Compare their extensions.

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   ΔL = FL/(AY) ∝ L/r^². The first gives L/r^²; the second 2L/(4r^²) = ½ × L/r^². The first stretches twice as much as the second.

2. Q2. By how much does the volume of 1.0 litre of water decrease when the pressure on it rises by 1.0 × 10^⁷ Pa? Take B = 2.2 × 10^⁹ Pa.

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   ΔV = −Vp/B = −1.0 × 10^⁻³ × 1.0 × 10^⁷/(2.2 × 10^⁹), about −4.5 × 10^⁻⁶ m^³: a decrease of about 4.5 cm^³.

3. Q3. A steel wire 2.0 m long and 1.0 mm in diameter is stretched by 1.0 mm. Taking Poisson’s ratio as 0.30, by how much does its diameter decrease?

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   Longitudinal strain = 1.0 × 10^⁻³/2.0 = 5.0 × 10^⁻⁴, so lateral strain = 0.30 × 5.0 × 10^⁻⁴ = 1.5 × 10^⁻⁴. Δd = 1.5 × 10^⁻⁴ × 1.0 mm = 1.5 × 10^⁻⁴ mm, about 0.15 µm.