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NCERT · Class XI · Physics

Mechanical Properties
of Solids

Chapter 8  ·  Unit IV — Properties of Bulk Matter

Explore how solids resist deformation — from the microscopic bonding forces to the macroscopic elastic response described by stress, strain, and the elastic moduli.

📐 Stress & Strain
🔩 Young's Modulus
🧲 Elastic Moduli
📊 Hooke's Law
⚙️ JEE / NEET Relevant
Chapter
8
Class
XI — Physics
Unit
IV — Bulk Matter
Key Moduli
Y, B, η (G)
Exams
JEE · NEET · CBSE
Difficulty
Intermediate
§ 01

Introduction to Mechanical Properties

Every solid object around us — from a steel bridge to a rubber band — responds in a characteristic way when external forces are applied. The study of Mechanical Properties of Solids examines how solids deform under applied loads and how they recover once those loads are removed.

At the atomic level, atoms in a solid are held together by interatomic bonds that act like tiny springs. When a deforming force is applied, atoms are displaced from their equilibrium positions, generating restoring forces. It is this restoring mechanism that gives rise to the concept of elasticity.

Elasticity
The property of a solid by virtue of which it tends to recover its original shape and size after the deforming force is removed.
Plasticity
The property of a solid by virtue of which it does not recover its original shape and size after the removal of the deforming force — the deformation is permanent.

Real materials lie on a spectrum: elastic solids (steel, glass), plastic solids (clay, putty), and those with mixed behaviour (rubber, human tissue).

§ 02

Elastic Behaviour of Solids

In a solid, atoms are arranged in a regular lattice. The interatomic potential energy curve shows a minimum at the equilibrium separation $r_0$. For small displacements from $r_0$, the restoring force varies linearly with displacement — analogous to a spring:

Interatomic Restoring Force (Linear Approximation)
$$F = -k\,(r - r_0)$$ where $k$ is the effective interatomic spring constant.

This microscopic spring-like behaviour gives rise to macroscopic elasticity. The elastic limit is the maximum stress up to which a material behaves elastically. Beyond it, permanent deformation sets in.

💡 Key Insight
For most engineering materials, the elastic limit is well within the region where the interatomic force–displacement relation is linear, justifying the use of Hooke's Law in practical calculations.
§ 03

Stress & Strain

Stress

When a deforming force $F$ is applied to a body of cross-sectional area $A$, the restoring force per unit area set up inside the body is called stress.

Stress
$$\sigma = \frac{F}{A} \qquad \text{SI Unit: N m}^{-2} = \text{Pa (Pascal)}$$

Types of Stress:

  • Longitudinal (Tensile/Compressive) Stress: Force along the axis of the body; causes change in length.
  • Shearing (Tangential) Stress: Force acts tangentially to the surface; causes change in shape (shear).
  • Hydraulic (Volume) Stress: Uniform pressure applied on all sides; causes change in volume.
Strain

Strain is the fractional change in dimension produced by stress. It is a dimensionless quantity.

Type of Strain Definition Formula
Longitudinal Strain Change in length ÷ Original length $\varepsilon = \Delta L / L$
Shearing Strain Transverse displacement per unit length $\gamma = x / h = \tan\theta \approx \theta$
Volumetric Strain Change in volume ÷ Original volume $\varepsilon_V = \Delta V / V$
§ 04

Hooke's Law

Robert Hooke (1678) observed that within the elastic limit, the stress developed in a body is directly proportional to the strain produced.

Hooke's Law
$$\text{Stress} \propto \text{Strain}$$ $$\Rightarrow \quad \frac{\text{Stress}}{\text{Strain}} = \text{Modulus of Elasticity (constant)}$$

The proportionality constant is the Modulus of Elasticity (or Elastic Modulus), which depends on the nature of the material, not on the dimensions of the body.

📌 Validity Condition
Hooke's Law holds only within the proportionality limit — a subset of the elastic limit. Beyond the proportionality limit, the stress–strain relationship becomes non-linear even though the material may still be elastic.
§ 05

Stress–Strain Curve

The stress–strain curve for a ductile material (e.g., mild steel) reveals several critical regions:

📊 Stress–Strain Curve (Ductile Metal)
Strain → Stress ↑ A B C D E F Elastic Region Plastic Region O A: Proportionality Limit B: Elastic Limit E: Fracture Point
  • O → A (Proportionality Limit): Linear stress–strain; Hooke's Law valid. Slope = Young's Modulus.
  • A → B (Elastic Limit): Non-linear but still elastic — body returns to original shape on unloading.
  • B → C (Yield Point / Upper & Lower): Sudden extension at almost constant stress. The material "yields".
  • C → D (Strain Hardening): Stress must be increased for further deformation; the material hardens.
  • D (Ultimate Tensile Strength): Maximum stress the material can withstand before necking begins.
  • D → E (Necking & Fracture): Cross-sectional area reduces rapidly (necking); fracture occurs at E/F.
§ 06

Young's Modulus (Y)

Within the proportionality limit, for a wire (or rod) under longitudinal stress, the ratio of longitudinal stress to longitudinal strain is a constant called Young's Modulus, denoted $Y$.

Young's Modulus
$$Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}$$
SI Unit: $\text{N m}^{-2}$ (Pascal). Typical values: Steel $\approx 2\times10^{11}$ Pa, Rubber $\approx 10^6$ Pa.
Experimental Determination (Searle's Method)

Two identical wires — one reference wire and one experimental wire — hang from a rigid support. A vernier scale measures the differential elongation $\Delta L$ when a load $W = Mg$ is applied to the experimental wire.

From Searle's Experiment
$$Y = \frac{4MgL}{\pi d^2 \,\Delta L}$$ where $d$ = diameter of wire, $L$ = original length.
MaterialYoung's Modulus (GPa)
Steel190 – 210
Aluminium70
Copper120
Glass50 – 90
Rubber0.001 – 0.01
Bone9 – 21
§ 07

Shear Modulus (Modulus of Rigidity, G or η)

When a tangential (shear) force deforms a body without changing its volume, the ratio of shear stress to shear strain is the Shear Modulus (also called Modulus of Rigidity).

Shear Modulus / Modulus of Rigidity
$$G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\theta} = \frac{F\,h}{A\,x}$$ Typical values: Steel $\approx 80$ GPa, Aluminium $\approx 26$ GPa.
🔗 Relation to Young's Modulus
For an isotropic material: $\;G = \dfrac{Y}{2(1+\nu)}\;$ where $\nu$ is Poisson's ratio.
§ 08

Bulk Modulus (B or K)

When a uniform pressure $P$ is applied over the entire surface of a body (hydraulic stress), the volume changes by $\Delta V$. The Bulk Modulus is defined as the ratio of hydraulic stress to volumetric strain.

Bulk Modulus
$$B = -\frac{\Delta P}{\Delta V / V} = -V\,\frac{\Delta P}{\Delta V}$$ The negative sign ensures $B$ is positive (pressure increase → volume decrease).

Compressibility $K = \dfrac{1}{B}$ — a measure of how easily a material can be compressed. Gases have very high compressibility; solids and liquids have very low compressibility.

MaterialBulk Modulus (GPa)
Steel160
Copper140
Aluminium72
Water2.2
Air$1.0 \times 10^{-4}$
§ 09

Poisson's Ratio (ν)

When a wire is stretched longitudinally, it contracts laterally. The ratio of lateral strain to longitudinal strain (within the elastic limit) is a constant for a given material called Poisson's Ratio.

Poisson's Ratio
$$\nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d/d}{\Delta L/L}$$ Typical range: $0 < \nu < 0.5$. For most metals: $\nu \approx 0.25$–$0.35$. For rubber: $\nu \approx 0.5$ (incompressible).
🔗 Inter-modulus Relations
$$Y = 3B(1-2\nu) \qquad Y = 2G(1+\nu) \qquad \frac{1}{Y} = \frac{1}{3G}+\frac{1}{9B}$$
§ 10

Elastic Potential Energy

When a body is deformed elastically, work is done by the external forces. This work is stored as elastic potential energy (strain energy) in the body, which is fully recovered upon unloading.

Elastic PE per unit volume (Energy Density)
$$U = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{\sigma^2}{2Y} = \frac{1}{2}Y\varepsilon^2$$
Total Elastic PE of a stretched wire (length L, area A)
$$U = \frac{1}{2} \times F \times \Delta L = \frac{F^2 L}{2YA}$$

This is analogous to the spring PE: $U = \tfrac{1}{2}kx^2$, where the "spring constant" of the wire is $k_{wire} = \dfrac{YA}{L}$.

§ 11

Real-World Applications

🏗️
Structural Engineering
Steel I-beams are designed using Young's modulus to minimise deflection under load while saving material.
🌉
Bridge Design
Suspension cables must sustain tensile stress well below the elastic limit to prevent permanent deformation.
⚙️
Spring Design
Helical springs exploit shear modulus; coil spring stiffness $k = \frac{Gd^4}{8nD^3}$.
🔩
Bolts & Fasteners
Bolts are preloaded to a tensile stress that keeps joints tight while below the elastic limit.
🏊
Submarine Design
Bulk modulus determines how hull material responds to enormous hydraulic stress at ocean depths.
🦷
Dental & Bone Implants
Implant materials must match the elastic modulus of bone to avoid stress-shielding effects.
§ 12

Quick Revision Formulae

Stress
$\sigma = F/A \;\; \text{(Pa)}$
Longitudinal Strain
$\varepsilon = \Delta L / L$
Young's Modulus
$Y = \sigma/\varepsilon = FL/(A\Delta L)$
Shear Modulus
$G = F/(A\theta)$
Bulk Modulus
$B = -\Delta P / (\Delta V/V)$
Compressibility
$K = 1/B$
Poisson's Ratio
$\nu = -(\Delta d/d)/(\Delta L/L)$
Elastic PE Density
$u = \tfrac{1}{2}\sigma\varepsilon$
Wire Spring Constant
$k_{wire} = YA/L$
Relation Y, G, ν
$Y = 2G(1+\nu)$
Relation Y, B, ν
$Y = 3B(1-2\nu)$
Relation Y, G, B
$\frac{9}{Y}=\frac{3}{G}+\frac{1}{B}$
§ 13

Chapter Resources