QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 8.3NCERT Class 11 · Physics · Chapter 8

Moduli of Elasticity and Poisson's Ratio

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 11 min read

Hooke's law establishes that the ratio of stress to strain is constant within the proportional limit. Because deformation can manifest as changes in length, shape, or volume, three distinct elastic moduli exist: Young's Modulus (Y), Shear Modulus (G or η), and Bulk Modulus (B). In addition, lateral contraction accompanying axial elongation is quantified by Poisson's Ratio (ν). This module details their formulations, practical applications, and algebraic interrelations.

1. Young's Modulus of Elasticity (Y)

Young's modulus quantifies the resistance of a solid material to longitudinal (tensile or compressive) deformation along the axis of applied load. It is defined as the ratio of longitudinal stress to longitudinal strain within the proportional limit.

Y = Longitudinal Stress / Longitudinal Strain = (F / A) / (ΔL / L) = (F × L) / (A × ΔL)

If a suspended cylindrical wire of original length L and radius r supports an axial load of mass M:

Y = (M × g × L) / (π × r² × ΔL)  ⇒  ΔL = (M × g × L) / (π × r² × Y)

Key Analytical Consequences of Young's Modulus:

  1. Equivalent Spring Constant of a Stretched Wire:

    Rewriting Hooke's relation as F = (Y A / L) ΔL, and comparing with the ideal spring law F = k x, the wire behaves as a linear spring with effective stiffness:

    k_eff = (Y × A) / L
  2. Elongation of a Heavy Rod Under its Own Weight:

    For a vertical rod of mass M, length L, density ρ, and cross-section A suspended from a rigid ceiling, the tension varies linearly from zero at the bottom to Mg at the top. Integrating the elongation of an infinitesimal element dy yields:

    ΔL_self = (M × g × L) / (2 × A × Y) = (ρ × g × L²) / (2 × Y)
    Crucial Insight: Elongation under self-weight is exactly half the elongation produced by an external suspended load equal to the rod's entire weight Mg placed at the bottom.
  3. Thermal Stress in Rigidly Clamped Rods:

    When the temperature of a rod clamped between immovable supports rises by ΔT, its free expansion would be ΔL = L α ΔT. Because the supports prevent this expansion, an internal compressive strain ε = ΔL / L = α ΔT develops, generating:

    Thermal Stress (σ_th) = Y × α × ΔT
    Thermal Force on Clamps (F_th) = Y × A × α × ΔT

2. Shear Modulus or Modulus of Rigidity (G or η)

Shear modulus quantifies a material's resistance to change in shape without any change in volume. It is defined as the ratio of shearing (tangential) stress to shearing strain within the elastic limit.

G = Shearing Stress / Shearing Strain = (F_parallel / A) / θ = (F × L) / (A × Δx)

For most structural solids, the shear modulus is substantially smaller than Young's modulus, typically G ≈ Y / 3. This implies that solids are inherently easier to twist or shear than to elongate or compress axially.

3. Bulk Modulus (B) and Compressibility (K)

Bulk modulus measures the resistance of a substance to uniform volumetric compression under hydrostatic pressure. It is defined as the ratio of hydraulic stress to volumetric strain.

B = - ΔP / (ΔV / V) = - V × (dP / dV)

The negative sign ensures that Bulk Modulus is always a positive quantity, because an increase in pressure (ΔP > 0) invariably produces a volume contraction (ΔV < 0).

Compressibility (K):

Compressibility is defined as the reciprocal of Bulk Modulus:

Compressibility (K) = 1 / B = - (1 / V) × (dV / dP)

Units: m²/N or Pa⁻¹. Solids have extremely small compressibility (high B ∼ 10¹¹ Pa), liquids are slightly compressible (water B ∼ 2.2 × 10⁹ Pa), while gases are highly compressible (air B ∼ 10⁵ Pa at STP).

4. Poisson's Ratio (ν or σ)

When a solid cylinder or wire is stretched longitudinally by an axial force, it not only elongates along the load axis (longitudinal strain) but simultaneously contracts in its transverse or lateral dimensions (lateral strain).

+-----------------------------------------------------------------------------------+ | LATERAL CONTRACTION & POISSON'S RATIO GEOMETRY | +-----------------------------------------------------------------------------------+ Axial Tensile Load F <--- ---> Axial Load F +---------------------------------------------------------+ | d | +---------------------------------------------------------+ <--------------------------- L ---------------------------> After stretching: +---------------+-----------------------------------------+---------------+ | (d - Δd) | | | +---------------+-----------------------------------------+---------------+ <-- ΔL/2 --><-------------------- L --------------------><-- ΔL/2 --> Longitudinal Strain (ε_long) = + ΔL / L (Elongation) Lateral Strain (ε_lat) = - Δd / d (Contraction) Poisson's Ratio (ν) = - (ε_lat / ε_long) = - (- Δd / d) / (ΔL / L) = (Δd / d) / (ΔL / L) +-----------------------------------------------------------------------------------+
Figure 8.5: Axial stretching producing longitudinal elongation accompanied by transverse lateral contraction.
Permissible Limits of Poisson's Ratio:
  • Theoretical Limits: Classical theory of isotropic elastic solids dictates: -1.0 ≤ ν ≤ +0.5.
  • Practical Limits: For all known engineering solids, longitudinal extension produces lateral thinning, ensuring ν > 0. Thus practically: 0.0 ≤ ν ≤ +0.5.
  • Special Cases: Cork has ν ≈ 0 (when corked into a wine bottle, its diameter does not swell). Vulcanized rubber has ν ≈ 0.5 (perfectly incompressible, volume remains strictly constant upon stretching: ΔV = 0).

5. Fundamental Relations Connecting Elastic Constants

For a homogeneous, isotropic elastic medium, the four elastic parameters (Y, G, B, ν) are interconnected by rigorous mathematical relations. Specifying any two constants completely determines the remaining two:

Fundamental Relations:

  1. Y = 3 B (1 − 2 ν)
  2. Y = 2 G (1 + ν)
  3. 9 / Y = 1 / B + 3 / G
  4. ν = (3 B − 2 G) / (6 B + 2 G)

NCERT Derivation Insight:

From Relation 1: Since Y and B must be positive physical quantities, 1 − 2ν > 0 ⇒ ν < 0.5.

From Relation 2: Since Y and G must be positive, 1 + ν > 0 ⇒ ν > −1.0.

Combining both inequalities rigorously yields the universal theoretical domain: −1.0 < ν < +0.5.

6. Solved Examples & Numerical Applications

Example 1: A uniform steel wire of length 3.0 m, cross-sectional area 4.0 mm², and density 7.8 × 10³ kg/m³ is suspended vertically from a ceiling. Young's modulus of steel is 2.0 × 10¹¹ N/m². (a) Find the elongation of the wire solely due to its own weight. (b) Find the additional elongation if a mass of 20 kg is suspended at its bottom end. (Take g = 9.8 m/s²).
Solution:

Step 1: Compute wire mass and self-weight:

Area A = 4.0 × 10⁻⁶ m², Length L = 3.0 m.

Volume V = A × L = 4.0 × 10⁻⁶ × 3.0 = 1.2 × 10⁻⁵ m³.

Mass of wire M = ρ × V = (7.8 × 10³) × (1.2 × 10⁻⁵) = 0.0936 kg.

Step 2: Calculate elongation due to self-weight (ΔL_self):

ΔL_self = (M g L) / (2 A Y) = (0.0936 × 9.8 × 3.0) / (2 × 4.0 × 10⁻⁶ × 2.0 × 10¹¹)

ΔL_self = 2.752 / (1.6 × 10⁶) = 1.72 × 10⁻⁶ m (1.72 μm).

Step 3: Calculate additional elongation due to 20 kg suspended load (ΔL_load):

ΔL_load = (m g L) / (A Y) = (20 × 9.8 × 3.0) / (4.0 × 10⁻⁶ × 2.0 × 10¹¹)

ΔL_load = 588 / (8.0 × 10⁵) = 7.35 × 10⁻⁴ m = 0.735 mm.

Example 2: A railway line laid with steel rails of length 20 m each at 15 °C is rigidly clamped at both ends without any gap. What compressive thermal stress and thermal thrust force are developed when the ambient temperature rises to 45 °C? (Given for steel: α = 1.2 × 10⁻⁵ K⁻¹, Y = 2.0 × 10¹¹ N/m², cross-sectional area of each rail A = 50 cm²).
Solution:

Step 1: Compute temperature change and thermal strain:

ΔT = 45 °C − 15 °C = 30 °C (or 30 K).

Thermal strain ε_th = α ΔT = (1.2 × 10⁻⁵ K⁻¹) × 30 K = 3.6 × 10⁻⁴.

Step 2: Calculate compressive thermal stress:

σ_th = Y × ε_th = (2.0 × 10¹¹ N/m²) × (3.6 × 10⁻⁴) = 7.2 × 10⁷ N/m² (72 MPa).

Step 3: Calculate total thermal thrust force on rail supports:

Area A = 50 cm² = 50 × 10⁻⁴ m² = 5.0 × 10⁻³ m².

F_th = σ_th × A = (7.2 × 10⁷ N/m²) × (5.0 × 10⁻³ m²) = 3.6 × 10⁵ N (360 kN).

(This immense force can easily buckle track lines, which is why expansion gaps and fishplates are left during rail laying).

Example 3: For an isotropic structural alloy, Young's modulus is measured as 1.2 × 10¹¹ N/m² and its shear modulus is 4.8 × 10¹⁰ N/m². Calculate: (a) Poisson's ratio (ν), and (b) Bulk modulus (B) of this material.
Solution:

Step 1: Determine Poisson's ratio using Y = 2G(1 + ν):

1 + ν = Y / (2 G) = (1.2 × 10¹¹) / (2 × 4.8 × 10¹⁰) = 1.2 / 0.96 = 1.25.

ν = 1.25 − 1 = 0.25.

Step 2: Determine Bulk Modulus using Y = 3B(1 − 2ν):

B = Y / [3 (1 − 2 ν)] = (1.2 × 10¹¹) / [3 × (1 − 2 × 0.25)]

B = (1.2 × 10¹¹) / [3 × 0.50] = (1.2 × 10¹¹) / 1.5 = 8.0 × 10¹⁰ N/m² (80 GPa).

Example 4: A cylindrical rubber cord of original length L and radius r is stretched longitudinally by an axial force such that its elongation is ΔL. If Poisson's ratio of the rubber is exactly 0.5, prove that the fractional change in its volume is identically zero (ΔV / V = 0).
Solution:

Step 1: Express cylinder volume:

V = π r² L.

Taking the natural logarithm of both sides: ln V = ln π + 2 ln r + ln L.

Differentiating logarithmically for small fractional variations:

ΔV / V = 2 (Δr / r) + (ΔL / L).

Step 2: Substitute definition of Poisson's ratio:

By definition, ν = − (Δr / r) / (ΔL / L) ⇒ Δr / r = − ν (ΔL / L).

Substituting this into the fractional volume equation:

ΔV / V = 2 (− ν ΔL / L) + (ΔL / L) = (1 − 2 ν) (ΔL / L).

Step 3: Evaluate for ν = 0.5:

ΔV / V = (1 − 2 × 0.5) (ΔL / L) = (1 − 1) (ΔL / L) = 0.

Hence, an elastic material with Poisson's ratio 0.5 preserves its total volume perfectly during deformation.

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Why are bridge structures and tall building frames fabricated from steel rather than copper or aluminum?
Young's modulus of steel (Y ≈ 2.0 × 10¹¹ N/m²) is significantly higher than that of copper (1.1 × 10¹¹ N/m²) and aluminum (0.7 × 10¹¹ N/m²). Under heavy traffic loads or wind shear, steel undergoes minimal elastic strain, providing structural stiffness, high yield strength, and minimal sag.
Q2. What is the isothermal vs adiabatic bulk modulus of an ideal gas?
For an ideal gas, isothermal process obeys PV = constant, differentiating gives dP/dV = -P/V, so isothermal bulk modulus B_iso = P. For an adiabatic process, PV^γ = constant, giving adiabatic bulk modulus B_ad = γ P. Hence, B_ad / B_iso = γ = C_p / C_v.
Q3. Can Poisson's ratio be negative for real materials?
Yes, specially engineered micro-architectured materials termed auxetic materials (such as certain expanded PTFE polymers and re-entrant honeycomb foams) possess negative Poisson's ratio (-1 < ν < 0). When stretched longitudinally, their microstructure unfolds causing them to expand laterally rather than contract.
Q4. What is the difference between elastic fatigue and elastic after-effect?
Elastic after-effect is the temporary delay in recovering the exact original configuration after removing deforming forces. Elastic fatigue is the progressive loss in mechanical strength and eventual fracture of a material subjected to repeated alternating or cyclic stresses over extended periods.
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