Prerequisites
Before studying elastic moduli, make sure you understand:
- Stress and Strain — Foundation concepts of elasticity
What is an Elastic Modulus?
An elastic modulus is a measure of a material’s stiffness or resistance to deformation.
$$\boxed{\text{Elastic Modulus} = \frac{\text{Stress}}{\text{Strain}}}$$Key Points:
- Higher modulus = stiffer material = less deformation for same stress
- Unit: Pa (Pascal) or N/m²
- Dimensions: [ML⁻¹T⁻²] (same as stress and pressure)
There are three types of elastic moduli, corresponding to three types of deformation:
| Modulus | Type of Stress | Type of Strain | Deformation |
|---|---|---|---|
| Young’s Modulus (Y) | Tensile/Compressive | Longitudinal | Change in length |
| Bulk Modulus (K) | Hydraulic | Volume | Change in volume |
| Shear Modulus (η or G) | Shearing | Shearing | Change in shape |
Young’s Modulus (Y)
Young’s Modulus measures resistance to change in length under tensile or compressive stress.
$$\boxed{Y = \frac{\text{Tensile Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L}}$$ $$\boxed{Y = \frac{FL}{A \cdot \Delta L}}$$where:
- F = applied force
- A = cross-sectional area
- L = original length
- ΔL = change in length
Interactive Demo: Visualize Elastic Moduli
Explore how different materials respond to stretching, compression, and shear forces.
Alternative form (from Hooke’s law for wire):
For a wire: $F = \frac{YA}{L} \cdot \Delta L$
This looks like F = kx where spring constant $k = \frac{YA}{L}$
Typical Values of Young’s Modulus
| Material | Young’s Modulus (Pa) | Young’s Modulus (GPa) |
|---|---|---|
| Steel | 2 × 10¹¹ | 200 |
| Iron | 1.9 × 10¹¹ | 190 |
| Copper | 1.2 × 10¹¹ | 120 |
| Aluminum | 0.7 × 10¹¹ | 70 |
| Brass | 1.0 × 10¹¹ | 100 |
| Glass | 0.7 × 10¹¹ | 70 |
| Wood | 1.3 × 10¹⁰ | 13 |
| Bone | 1.5 × 10¹⁰ | 15 |
| Rubber | 5 × 10⁶ | 0.005 |
Stiffness order: Steel > Iron > Copper > Aluminum > Glass ≈ Wood > Rubber
“Superman Is Constantly Avoiding Giant Wooden Rockets”
Important Properties
Temperature dependence: Y generally decreases with increasing temperature
Independent of dimensions: Y depends only on material, not on shape or size
For a wire of length L:
- Extension ∝ L (longer wire stretches more)
- Extension ∝ 1/A (thinner wire stretches more)
Wires in series and parallel:
Series: $\frac{1}{Y_{eff}} = \frac{1}{Y_1} + \frac{1}{Y_2}$ (NO! This is wrong — Y is material property)
Actually, for series: Extensions add up, but Y remains material property.
Parallel: Forces add up: $F_{total} = F_1 + F_2$
A steel wire of length 2 m and cross-section 0.1 cm² is stretched by a force of 200 N. Find the extension. (Y for steel = 2 × 10¹¹ Pa)
Solution:
$$\Delta L = \frac{FL}{AY}$$A = 0.1 cm² = 0.1 × 10⁻⁴ m² = 10⁻⁵ m²
$$\Delta L = \frac{200 \times 2}{10^{-5} \times 2 \times 10^{11}}$$ $$= \frac{400}{2 \times 10^6} = 2 \times 10^{-4} \text{ m} = \boxed{0.2 \text{ mm}}$$Bulk Modulus (K)
Bulk Modulus measures resistance to change in volume under uniform pressure.
$$\boxed{K = \frac{\text{Hydraulic Stress}}{\text{Volume Strain}} = \frac{-\Delta P}{\Delta V/V}}$$ $$\boxed{K = -V \frac{\Delta P}{\Delta V}}$$where:
- ΔP = change in pressure = P - P₀
- V = original volume
- ΔV = change in volume = V - V₀
Negative sign: Increase in pressure causes decrease in volume (ΔP and ΔV have opposite signs)
Compressibility (κ) is the reciprocal of bulk modulus:
$$\boxed{\kappa = \frac{1}{K} = -\frac{1}{V}\frac{\Delta V}{\Delta P}}$$- High K = low κ = incompressible material
- Low K = high κ = compressible material
Typical Values of Bulk Modulus
| Material | Bulk Modulus (Pa) | Bulk Modulus (GPa) |
|---|---|---|
| Steel | 1.6 × 10¹¹ | 160 |
| Iron | 1.0 × 10¹¹ | 100 |
| Copper | 1.4 × 10¹¹ | 140 |
| Aluminum | 0.7 × 10¹¹ | 70 |
| Glass | 0.4 × 10¹¹ | 40 |
| Water | 2.2 × 10⁹ | 2.2 |
| Air (at STP) | 1.0 × 10⁵ | 0.0001 |
Solids have higher K than liquids, liquids higher than gases.
This is why:
- Solids are nearly incompressible
- Liquids are slightly compressible
- Gases are highly compressible
Air has K about 1 million times smaller than steel!
Special Case: Gases
For an ideal gas undergoing:
Isothermal process (constant temperature):
$$\boxed{K_{isothermal} = P}$$Adiabatic process (no heat exchange):
$$\boxed{K_{adiabatic} = \gamma P}$$where γ = Cp/Cv (ratio of specific heats)
For air: γ ≈ 1.4
Calculate the decrease in volume of 100 liters of water when subjected to pressure of 10⁵ Pa (approximately 1 atm). Bulk modulus of water = 2.2 × 10⁹ Pa.
Solution:
$$K = -V \frac{\Delta P}{\Delta V}$$ $$\Delta V = -\frac{V \cdot \Delta P}{K} = -\frac{100 \times 10^{-3} \times 10^5}{2.2 \times 10^9}$$ $$= -4.55 \times 10^{-6} \text{ m}^3 = -4.55 \text{ mL}$$Decrease = 4.55 mL (only 0.00455% of original volume!)
Water is nearly incompressible!
Shear Modulus (η or G)
Shear Modulus (also called Modulus of Rigidity) measures resistance to change in shape under shearing stress.
$$\boxed{\eta = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\theta}}$$ $$\boxed{\eta = \frac{F/A}{\Delta x/L}}$$where:
- F = tangential force
- A = area parallel to force
- θ = shearing strain (angle of deformation in radians)
- Δx = lateral displacement
- L = perpendicular height
Typical Values of Shear Modulus
| Material | Shear Modulus (Pa) | Shear Modulus (GPa) |
|---|---|---|
| Steel | 0.8 × 10¹¹ | 80 |
| Iron | 0.7 × 10¹¹ | 70 |
| Copper | 0.4 × 10¹¹ | 40 |
| Aluminum | 0.25 × 10¹¹ | 25 |
| Brass | 0.4 × 10¹¹ | 40 |
| Lead | 0.055 × 10¹¹ | 5.5 |
Liquids and gases cannot sustain shearing stress.
η = 0 for all fluids!
This is why liquids flow — they have no rigidity. Any shearing stress causes continuous deformation (flow).
Only solids have a shear modulus.
A cube of rubber of side 5 cm has its lower face fixed. A tangential force of 100 N is applied to the upper face, causing a displacement of 0.2 cm. Find the shear modulus.
Solution:
A = (5 × 10⁻²)² = 25 × 10⁻⁴ m²
Shearing stress = F/A = 100/(25 × 10⁻⁴) = 4 × 10⁴ Pa
Shearing strain = Δx/L = (0.2 × 10⁻²)/(5 × 10⁻²) = 0.04 rad
$$\eta = \frac{\text{Stress}}{\text{Strain}} = \frac{4 \times 10^4}{0.04} = \boxed{10^6 \text{ Pa} = 1 \text{ MPa}}$$(Rubber has very low shear modulus compared to metals)
Poisson’s Ratio (σ)
When a material is stretched in one direction, it contracts in perpendicular directions.
Poisson’s Ratio is the ratio of lateral strain to longitudinal strain.
$$\boxed{\sigma = -\frac{\text{Lateral strain}}{\text{Longitudinal strain}} = -\frac{\Delta d/d}{\Delta L/L}}$$where:
- Δd = change in diameter (or width)
- d = original diameter
- ΔL = change in length
- L = original length
Negative sign: Lateral and longitudinal strains have opposite signs (one expands, other contracts)
Range of values: 0 < σ < 0.5
- σ = 0: No lateral contraction (cork, ideal material for bottle stoppers!)
- σ = 0.5: Incompressible material (rubber ≈ 0.5)
- Most materials: σ ≈ 0.2 to 0.4
| Material | Poisson’s Ratio |
|---|---|
| Cork | ~0.0 |
| Concrete | 0.1 - 0.2 |
| Steel | 0.28 - 0.30 |
| Aluminum | 0.33 |
| Copper | 0.34 |
| Gold | 0.42 |
| Rubber | ~0.5 |
Relationships Between Elastic Constants
For isotropic materials (same properties in all directions), the three moduli are related:
$$\boxed{Y = 2\eta(1 + \sigma)}$$ $$\boxed{Y = 3K(1 - 2\sigma)}$$ $$\boxed{K = \frac{Y}{3(1 - 2\sigma)}}$$ $$\boxed{\eta = \frac{Y}{2(1 + \sigma)}}$$ $$\boxed{K = \frac{\eta(2 + 6\sigma)}{3(1 - 2\sigma)}}$$Most important relationship:
$$\boxed{Y = \frac{9K\eta}{3K + \eta}}$$Theoretical limits:
For real materials:
- If σ = 0: Y = 2η, K = Y/3
- If σ = 0.5: Y = 3η, K → ∞ (incompressible)
For most metals, σ ≈ 0.3, so:
Y ≈ 2.6η (roughly Y ≈ 2.5 to 3 times η)
Y ≈ 1.2K (roughly Y ≈ K to 1.5K)
For a material with Y = 12 × 10¹⁰ Pa and σ = 0.25, find bulk modulus K and shear modulus η.
Solution:
Shear modulus:
$$\eta = \frac{Y}{2(1 + \sigma)} = \frac{12 \times 10^{10}}{2(1 + 0.25)}$$ $$= \frac{12 \times 10^{10}}{2.5} = \boxed{4.8 \times 10^{10} \text{ Pa}}$$Bulk modulus:
$$K = \frac{Y}{3(1 - 2\sigma)} = \frac{12 \times 10^{10}}{3(1 - 0.5)}$$ $$= \frac{12 \times 10^{10}}{1.5} = \boxed{8 \times 10^{10} \text{ Pa}}$$Summary Table
| Property | Formula | Applies To | Unit |
|---|---|---|---|
| Young’s Modulus (Y) | Y = (F/A)/(ΔL/L) | Solids | Pa |
| Bulk Modulus (K) | K = -V(ΔP/ΔV) | Solids, Liquids, Gases | Pa |
| Shear Modulus (η) | η = (F/A)/θ | Solids only | Pa |
| Poisson’s Ratio (σ) | σ = -(Δd/d)/(ΔL/L) | Solids | Dimensionless |
| Compressibility (κ) | κ = 1/K | All materials | Pa⁻¹ |
Practice Problems
Level 1: JEE Main Basics
A copper wire of length 3 m and area 2 mm² is stretched by 1.5 mm under a force F. If Y for copper = 1.2 × 10¹¹ Pa, find F.
Solution:
$$Y = \frac{FL}{A \cdot \Delta L}$$ $$F = \frac{Y \cdot A \cdot \Delta L}{L}$$A = 2 mm² = 2 × 10⁻⁶ m²
$$F = \frac{1.2 \times 10^{11} \times 2 \times 10^{-6} \times 1.5 \times 10^{-3}}{3}$$ $$= \frac{1.2 \times 2 \times 1.5}{3} \times 10^2 = 1.2 \times 10^2 = \boxed{120 \text{ N}}$$A rubber ball of volume 500 cm³ is subjected to a pressure of 10⁶ Pa. If bulk modulus of rubber is 5 × 10⁸ Pa, find the decrease in volume.
Solution:
$$K = -V \frac{\Delta P}{\Delta V}$$ $$\Delta V = -\frac{V \cdot \Delta P}{K} = -\frac{500 \times 10^{-6} \times 10^6}{5 \times 10^8}$$ $$= -10^{-3} \text{ m}^3 = -1000 \text{ cm}^3$$Wait, this is twice the original volume — impossible!
Error check: Actually should be:
$$= -\frac{500 \times 10^{-6} \times 10^6}{5 \times 10^8} = -10^{-3} \text{ m}^3$$Converting: -10⁻³ m³ = -1000 cm³ is wrong!
10⁻³ m³ = 1000 cm³
Actually: 500 × 10⁻⁶ m³ = 500 cm³ ✓
$$\Delta V = -\frac{500 \times 10^{-6} \times 10^6}{5 \times 10^8} = -\frac{0.5}{500} = \boxed{-0.001 \text{ m}^3 = -1 \text{ cm}^3}$$Very small compression!
Level 2: JEE Main/Advanced
Two wires of same material and length are in the ratio of radii 1:2. They are stretched by the same force. Find ratio of: (a) Stress (b) Strain (c) Extension
Solution:
Let radii be r and 2r, so areas are πr² and 4πr².
(a) Stress ratio:
$$\frac{\sigma_1}{\sigma_2} = \frac{F/(\pi r^2)}{F/(4\pi r^2)} = \frac{4\pi r^2}{\pi r^2} = \boxed{4:1}$$(b) Strain ratio:
Same material → same Y
Strain = Stress/Y
$$\frac{\epsilon_1}{\epsilon_2} = \frac{\sigma_1}{\sigma_2} = \boxed{4:1}$$(c) Extension ratio:
Extension = Strain × Length = (ΔL/L) × L = ΔL
Same length:
$$\frac{\Delta L_1}{\Delta L_2} = \frac{\epsilon_1}{\epsilon_2} = \boxed{4:1}$$A material has Y = 10¹¹ Pa and η = 4 × 10¹⁰ Pa. Find Poisson’s ratio and bulk modulus.
Solution:
From $Y = 2\eta(1 + \sigma)$:
$$\sigma = \frac{Y}{2\eta} - 1 = \frac{10^{11}}{2 \times 4 \times 10^{10}} - 1$$ $$= \frac{10}{8} - 1 = 1.25 - 1 = \boxed{0.25}$$From $K = \frac{Y}{3(1 - 2\sigma)}$:
$$K = \frac{10^{11}}{3(1 - 0.5)} = \frac{10^{11}}{1.5} = \boxed{6.67 \times 10^{10} \text{ Pa}}$$Level 3: JEE Advanced
A wire of length L and radius r is stretched by a force F. Another wire of same material has length 2L and radius 2r. What force will produce: (a) Same stress? (b) Same strain? (c) Same extension?
Solution:
(a) Same stress:
Stress = F/A
For stress₁ = stress₂:
$$\frac{F_1}{\pi r^2} = \frac{F_2}{\pi(2r)^2}$$ $$F_2 = F_1 \times 4 = \boxed{4F}$$(b) Same strain:
Same material → same Y → same stress gives same strain
$$\boxed{F_2 = 4F}$$(same as part a)
(c) Same extension:
$$\Delta L = \frac{FL}{AY}$$For ΔL₁ = ΔL₂:
$$\frac{F_1 \cdot L}{\pi r^2 \cdot Y} = \frac{F_2 \cdot 2L}{\pi(2r)^2 \cdot Y}$$ $$F_1 \cdot L \cdot 4 = F_2 \cdot 2L$$ $$F_2 = 2F_1 = \boxed{2F}$$A wire suspended vertically from a fixed point is stretched by its own weight. If density is ρ, length is L, and Young’s modulus is Y, prove that total extension is:
$$\Delta L = \frac{\rho g L^2}{2Y}$$Solution:
Consider element at distance x from bottom, of thickness dx.
Weight below this element = (mass below) × g = (ρ × A × x) × g
Tension at x: T(x) = ρgAx
Stress at x: σ(x) = T(x)/A = ρgx
Strain at x: ε(x) = σ(x)/Y = ρgx/Y
Extension of element dx:
$$d(\Delta L) = \epsilon(x) \cdot dx = \frac{\rho gx}{Y} dx$$Total extension:
$$\Delta L = \int_0^L \frac{\rho gx}{Y} dx = \frac{\rho g}{Y} \int_0^L x \, dx$$ $$= \frac{\rho g}{Y} \cdot \frac{L^2}{2} = \boxed{\frac{\rho g L^2}{2Y}}$$Proved!
Note: This is half the extension if the same total weight (ρgAL) were hung at the bottom!
Related Topics
Within Properties of Matter
- Stress and Strain — Foundation of elasticity
- Fluid Pressure — Bulk modulus applications
- Viscosity — Fluid resistance to shear
- Surface Tension — Surface elasticity
Connected Chapters
- Simple Harmonic Motion — Spring constant and Y
- Waves — Wave speed depends on elastic moduli
- Gravitation — Tidal forces and material strength