QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 8.1NCERT Class 11 · Physics · Chapter 8

Elastic Behaviour of Solids, Stress and Strain

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 10 min read

In kinematics and rigid-body mechanics, bodies were idealized as perfectly non-deformable entities where inter-particle distances remain constant under any external influence. In reality, all solid bodies deform under applied loads. This module examines the microscopic origin of elasticity, the concept of internal restoring force, and the rigorous mathematical definitions of stress and strain underpinning engineering structures.

1. Elasticity and Plasticity: Molecular Foundations

A solid consists of atoms or molecules bonded together by interatomic or intermolecular forces. Under equilibrium conditions at absolute temperature T, the constituent particles occupy positions where the net potential energy is minimized and net interatomic force is zero.

Elasticity Defined

Elasticity is the intrinsic property of a body by virtue of which it tends to recover its original size and shape completely upon the removal of external deforming forces. The restoring force arises from interatomic electrostatic potentials.

Examples: Quartz fibre, phosphor bronze, tempered steel wire.

Plasticity Defined

Plasticity is the property of a body by virtue of which it exhibits permanent deformation and shows no tendency to regain its original dimensions once the deforming load is removed.

Examples: Modeling clay, putty, chewing gum, lead at room temperature.

NCERT Conceptual Benchmark: No real material is perfectly elastic or perfectly plastic. Quartz fibre and phosphor bronze approach near-perfect elasticity, while wet clay or putty approach near-perfect plasticity.
+-----------------------------------------------------------------------------------+ | MICROSCOPIC SPRING-BALL MODEL & POTENTIAL WELL | +-----------------------------------------------------------------------------------+ Tension (Stretching) Equilibrium Compression (r > r0, Attractive Force) (r = r0, F_net = 0) (r < r0, Strong Repulsion) (Atom 1) (Atom 2) (Atom 1) (Atom 2) (Atom 1)(Atom 2) (O)----/\/\/\----(O) (O)--/\/\/\--(O) (O)/\/\(O) <------- r -------> <---- r0 ----> <-- r --> F_restoring inwards U(r) is at Minimum F_restoring outwards Potential Energy Curve U(r): U(r) ^ | | \ Repulsive Branch | 0 +------\------------------------------------> r (Interatomic distance) | \ /------------ Attractive Branch | \ / -U0 | \_______/ ^ r = r0 (Equilibrium separation, F = -dU/dr = 0) +-----------------------------------------------------------------------------------+
Figure 8.1: The spring-ball lattice model and the interatomic potential energy well illustrating restoring forces under tensile and compressive deformation.

2. Concept and Types of Stress

When an external force deforms a solid, internal forces develop across every microscopic cross-section to oppose the deformation. In static equilibrium, the magnitude of this internal restoring force equals the applied deforming force.

Stress (σ) = Internal Restoring Force / Area of Cross-Section = F_int / A

Units & Dimensions:

  • SI Unit: Newton per square metre (N/m²) or Pascal (Pa). (1 Pa = 1 N/m²; 1 MPa = 10⁶ Pa; 1 GPa = 10⁹ Pa)
  • CGS Unit: dyne/cm² (1 N/m² = 10 dyne/cm²)
  • Dimensional Formula: [M L⁻¹ T⁻²] (Identical to pressure, but conceptually distinct because stress can act tangentially and varies with cross-sectional orientation).
Classification of Stress Direction of Force relative to Area Effect on Body Configuration Key Formula
Tensile Stress (Normal) Perpendicular, directed outward from surface Elongation along load axis; cross-section contracts σ_t = F_perpendicular / A
Compressive Stress (Normal) Perpendicular, directed inward toward surface Shortening along load axis; cross-section expands σ_c = F_perpendicular / A
Shearing / Tangential Stress Parallel (tangential) to the surface plane Relative angular displacement of parallel planes (shape change) σ_s = F_parallel / A
Hydraulic / Volumetric Stress Uniformly perpendicular to all surface elements (Fluid pressure) Uniform volume reduction at constant geometric shape σ_v = ΔP = F_perpendicular / A_total

3. Concept and Classification of Strain

Strain measures the relative geometrical deformation produced in a body by an applied stress. It is defined as the ratio of the change in dimension to the original dimension.

Strain (ε) = Change in Configuration / Original Configuration = Δξ / ξ

Because strain is a ratio of two identical physical dimensions (length/length, volume/volume), it is a dimensionless and unitless quantity with dimensional formula [M⁰ L⁰ T⁰].

Three Fundamental Types of Strain:

  1. Longitudinal Strain: Produced by normal (tensile or compressive) stress. It is the fractional change in length:
    Longitudinal Strain = ΔL / L
    Tensile strain is positive (ΔL > 0), compressive strain is negative (ΔL < 0).
  2. Shearing Strain (θ): Produced by tangential stress. When a tangential force F acts on the top surface of a cube of height L while the opposite bottom surface is held rigidly fixed, the top surface is displaced laterally by Δx:
    Shearing Strain = tan θ ≈ θ = Δx / L (in radians)
    Here θ is the angle of shear through which a vertical line perpendicular to the fixed plane tilts.
  3. Volume Strain: Produced by hydraulic stress. It is the fractional change in volume of a body subjected to uniform pressure:
    Volume Strain = ΔV / V
+-----------------------------------------------------------------------------------+ | DIAGNOSTIC GEOMETRY OF THREE STRAIN MODES | +-----------------------------------------------------------------------------------+ (a) Longitudinal Strain: +=====================+ F ---> | Original Length L |====================> +------+ +=====================+ | ΔL | Strain = ΔL / L <------------------ L ---------------------><--ΔL-> (b) Shearing Strain (Tangential): F_tangential ---> +----------------+ /----------------/ Top plane shifts Δx /| /| / /| / | / | / / | +--+-------------+ | ==> +----------------+ | | | | | |θ\ | | tan θ ≈ θ = Δx / L L | +-------------+--+ | \-------------+--+ | / | / L | / | / |/ |/ | / |/ +----------------+ +-/--------------+ Bottom plane fixed <------- L ------> <-- Δx --> (c) Volumetric Strain (Hydraulic): | | | (P_fluid) +-v-v-v-+ ---> | --- | <--- | | V | | Volume contracts uniformly: V ---> V - ΔV ---> | --- | <--- Volume Strain = ΔV / V +-^-^-^-+ +-----------------------------------------------------------------------------------+
Figure 8.2: Visual geometry of longitudinal extension, tangential shear distortion, and uniform volumetric compression under fluid pressure.

4. Solved Examples & Numerical Applications

Example 1: A structural steel cable of length 4.0 m and cross-sectional diameter 2.0 cm supports an elevator cabin of mass 2000 kg. If g = 9.8 m/s², compute: (a) the tensile stress developed in the cable, and (b) the elongation of the cable given Young's modulus of steel Y = 2.0 × 10¹¹ N/m².
Solution:

Step 1: Calculate tensile load and cross-sectional area:

Load F = M × g = 2000 kg × 9.8 m/s² = 19,600 N.

Radius r = d / 2 = 1.0 cm = 1.0 × 10⁻² m.

Area A = π r² = 3.1416 × (1.0 × 10⁻²)² = 3.1416 × 10⁻⁴ m².

Step 2: Calculate tensile stress:

σ = F / A = 19,600 / (3.1416 × 10⁻⁴) = 6.24 × 10⁷ N/m² (62.4 MPa).

Step 3: Calculate elongation ΔL:

From definition of Young's modulus, Y = σ / (ΔL / L) ⇒ ΔL = (σ × L) / Y.

ΔL = (6.24 × 10⁷ × 4.0) / (2.0 × 10¹¹) = 1.248 × 10⁻³ m = 1.25 mm.

Example 2: A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force of 9.0 × 10⁴ N applied tangentially to its upper narrow face (50 cm × 10 cm) while the opposite narrow face is riveted to the floor. If the shear modulus of lead is 5.6 × 10⁹ N/m², find the lateral displacement of the upper edge.
Solution:

Step 1: Identify dimensions and shear area:

Area of the narrow face experiencing the tangential force: A = length × width = 0.50 m × 0.10 m = 0.05 m².

Height perpendicular to fixed surface: L = side = 0.50 m.

Step 2: Calculate shearing stress:

σ_s = F_tangential / A = (9.0 × 10⁴ N) / 0.05 m² = 1.8 × 10⁶ N/m².

Step 3: Compute shear strain θ and lateral displacement Δx:

θ = σ_s / G = (1.8 × 10⁶) / (5.6 × 10⁹) = 3.214 × 10⁻⁴ radians.

Lateral displacement Δx = L × θ = 0.50 m × 3.214 × 10⁻⁴ = 1.607 × 10⁻⁴ m = 0.16 mm.

Example 3: A solid copper sphere of volume 0.5 m³ is lowered to the bottom of the Mariana Trench where the gauge water pressure is 1.1 × 10⁸ Pa. If the Bulk Modulus of copper is 1.4 × 10¹¹ N/m², determine the fractional change in volume and the absolute volume reduction.
Solution:

Step 1: Relate hydraulic stress to bulk modulus:

Hydraulic stress ΔP = 1.1 × 10⁸ Pa.

Bulk Modulus B = - ΔP / (ΔV / V) ⇒ Fractional volume reduction |ΔV / V| = ΔP / B.

Step 2: Calculate fractional change in volume:

|ΔV / V| = (1.1 × 10⁸) / (1.4 × 10¹¹) = 7.86 × 10⁻⁴ (or 0.0786%).

Step 3: Compute absolute volume reduction ΔV:

ΔV = 7.86 × 10⁻⁴ × 0.5 m³ = 3.93 × 10⁻⁴ m³ (393 cm³).

Example 4: A uniform heavy steel rod of length L, mass M, and cross-sectional area A is suspended vertically from a rigid ceiling. Derive the tensile stress as a function of distance y measured from its free bottom end, and find where the stress is maximum.
Solution:

Step 1: Formulate the weight supported by cross-section at height y:

At a cross-section located at distance y from the bottom, the tensile force F(y) is solely due to the weight of the rod segment below this section.

Mass of segment of length y: m(y) = (M / L) × y.

Tension at section y: T(y) = m(y) × g = (M g / L) × y.

Step 2: Compute tensile stress σ(y):

σ(y) = T(y) / A = (M g y) / (A L).

Step 3: Analyze boundary conditions:

  • At the bottom free end (y = 0): σ(0) = 0.
  • At the midpoint (y = L/2): σ(L/2) = M g / (2 A L / L) = M g / (2 A).
  • At the top fixed support (y = L): σ_max = M g / A.

Thus, stress is linearly distributed along the length, attaining its maximum at the point of suspension.

Frequently Asked Questions (Class 11 & JEE/NEET)

Q1. Why does a solid possess shearing elasticity while liquids and gases do not?
Solids have fixed lattice positions with directional interatomic chemical bonds that exert restoring forces when neighboring atomic planes slide past each other. Fluids lack fixed lattice bonds; any applied tangential force causes continuous flow (shear rate) rather than a static shearing strain.
Q2. Is stress a scalar or a vector quantity?
Stress is neither a pure scalar nor a simple vector; it is a second-rank tensor because specifying stress fully requires two directional indicators: the orientation of the surface normal and the direction of the restoring force vector acting across that surface. At an elementary level, it is treated as a physical scalar magnitude for a given plane.
Q3. What is the breaking stress (ultimate strength) of a wire? Does it depend on length or area?
Breaking stress is the maximum stress a material can withstand before fracture. It is an intrinsic intensive material property depending solely on the substance, temperature, and heat treatment. It does NOT depend on the length or cross-sectional area of the wire. Breaking force, however, is extensive and scales directly with cross-sectional area: F_break = σ_break × A.
Q4. How does temperature affect the elastic properties of materials?
In general, increasing temperature increases the average amplitude of atomic vibrations and interatomic separation, weakening interatomic bonding forces. Consequently, the elastic moduli (Young's, Bulk, Shear) decrease with rising temperature. (An exception is Invar steel, whose thermal expansion is exceptionally small).
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