By the end of this chapter you'll be able to…

  • 1Define stress and strain and their types
  • 2State Hooke's law and interpret the stress-strain curve
  • 3Distinguish Young's, bulk, and shear moduli
  • 4Calculate elastic potential energy stored in a stretched solid
  • 5Define and apply Poisson's ratio
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Why this chapter matters
Why does a steel bridge hold while a rubber band stretches? Stress, strain, Hooke's law, and the elastic moduli describe how solids deform under load -- essential knowledge for engineering, construction, and material selection.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Mechanical Properties of Solids

'The important thing in science is not so much to obtain new facts as to discover new ways of thinking about them.' — William Lawrence Bragg

1. Chapter Overview

Why does a rubber band stretch? Why does a steel bridge not collapse under its own weight? This chapter explores how SOLIDS respond to external forces. You will learn about STRESS, STRAIN, HOOKE'S LAW, and the three ELASTIC MODULI that characterise a material's stiffness. The chapter also covers ELASTIC POTENTIAL ENERGY and some practical applications.


2. Stress and Strain

Stress (σ)

  • Definition: Restoring force per unit area SET UP INSIDE the body
  • σ = F/A
  • SI unit: N/m² or pascal (Pa)
  • Tensile Stress: When forces tend to STRETCH the body
  • Compressive Stress: When forces tend to SQUEEZE the body
  • Shear Stress: When forces act PARALLEL to the surface

Strain (ε)

  • Definition: Ratio of CHANGE in dimension to ORIGINAL dimension
  • ε = ΔL/L (longitudinal strain), ΔV/V (volume strain), Δθ (shear strain)
  • Strain is DIMENSIONLESS (no unit)

Types of Stress and Strain

TypeStressStrainExample
LongitudinalTensile/CompressiveΔL/LStretching a wire
Volume (Hydraulic)Pressure (ΔP)ΔV/VCompression under water
ShearTangential force/AreaAngle of shear θRivet in shear loading

3. Hooke's Law and Elastic Limit

  • Hooke's Law: Within the ELASTIC LIMIT, stress ∝ strain
  • σ = E × ε (where E is the elastic modulus)
  • Elastic Limit: The maximum stress after which the material does NOT return to its original shape
  • Beyond the elastic limit → PERMANENT deformation (plastic behaviour)

Stress-Strain Curve

  • Region OA: Proportional limit (Hooke's law holds)
  • Region AB: Elastic limit (material still recovers fully)
  • Region BC: Yielding (large strain for small stress increase)
  • Point C: Yield point
  • Region CD: Plastic flow (material deforms permanently)
  • Point D: Ultimate tensile strength (maximum stress)
  • Region DE: Necking (cross-section reduces)
  • Point E: Fracture point

4. Elastic Moduli

Young's Modulus (Y)

  • Definition: Ratio of longitudinal stress to longitudinal strain
  • Y = (F/A) / (ΔL/L) = FL/AΔL
  • Measures a material's RESISTANCE to stretching/compression
  • Typical values: Steel ≈ 2×10¹¹ Pa, Aluminium ≈ 7×10¹⁰ Pa

Bulk Modulus (B)

  • Definition: Ratio of volumetric stress to volumetric strain
  • B = -ΔP/(ΔV/V) (negative sign because volume decreases with pressure increase)
  • Compressibility k = 1/B
  • Solids have HIGHER bulk modulus than liquids and gases

Shear Modulus (G or η)

  • Definition: Ratio of shear stress to shear strain
  • G = (F/A)/θ = (F/A)/(Δx/L)
  • Also called MODULUS OF RIGIDITY

Comparison

ModulusSymbolWhat it resistsExpression
Young'sYLinear deformationFL/AΔL
BulkBVolume change-VΔP/ΔV
ShearGShape change(F/A)/θ

5. Elastic Potential Energy

  • When a wire is stretched, work done is stored as ELASTIC POTENTIAL ENERGY
  • U = ½ × Stress × Strain × Volume
  • U = ½ × (F/A) × (ΔL/L) × AL = ½ × F × ΔL = ½ × (YA/L) × (ΔL)²
  • SI unit: J (joule)

Worked Problem

Q: A steel wire of length 2 m and cross-section area 1 mm² is stretched by 1 mm. Find energy stored. (Y = 2×10¹¹ Pa) A: U = ½ × (YA/L) × (ΔL)² = ½ × (2×10¹¹×10⁻⁶/2) × (0.001)² = ½ × 10⁵ × 10⁻⁶ = 0.05 J.


6. Poisson's Ratio (σ or μ)

  • When a wire is stretched longitudinally, it CONTRACTS laterally
  • Poisson's ratio σ = (Lateral strain) / (Longitudinal strain) = -(ΔD/D)/(ΔL/L)
  • Range: -1 < σ < 0.5 (typically 0.2 — 0.4 for most materials)
  • Cork: σ ≈ 0 (no lateral contraction)
  • Rubber: σ ≈ 0.5 (nearly incompressible)

7. Common Mistakes

  1. Confusing stress and pressure: Stress is RESTORING force/area INSIDE material; pressure is EXTERNAL force/area
  2. Hooke's law applies ONLY within elastic limit: Beyond yield point, a small force causes large deformation
  3. Young's modulus is NOT the same for all wires: It's a MATERIAL property, not a wire property
  4. Bulk modulus sign stays positive: The negative sign in B = -ΔP/(ΔV/V) accounts for volume decreasing
  5. Stress is NOT force: Stress = Force/Area; doubling area halves stress for same force

8. CBSE Exam Focus

  1. Stress-strain curve interpretation (3-mark)
  2. Young's modulus numericals (wire stretching problems)
  3. Bulk and shear modulus numerical problems
  4. Elastic potential energy derivation (3-mark)
  5. Poisson's ratio definition and applications

9. Key Formulas

  • σ = F/A (Stress)
  • ε = ΔL/L (Longitudinal strain)
  • Y = FL/AΔL
  • B = -ΔP/(ΔV/V)
  • G = (F/A)/θ
  • U = ½ × Stress × Strain × Volume = ½ × F × ΔL
  • σ (Poisson's) = -Lateral strain/Longitudinal strain

10. Self-Test (5+ Q&A)

Q1: A 3 m long steel wire (Y = 2×10¹¹ Pa) of area 2 mm² is stretched by a 100 kg load. Find elongation (g = 10 m/s²). A: F = mg = 1000 N. ΔL = FL/AY = 1000×3/(2×10⁻⁶×2×10¹¹) = 3000/(4×10⁵) = 7.5×10⁻³ m = 7.5 mm.

Q2: What is the elastic limit? A: The MAXIMUM stress a material can withstand without permanent deformation. Beyond this, Hooke's law fails.

Q3: Which has higher Young's modulus — rubber or steel? A: Steel (≈ 2×10¹¹ Pa). Rubber has much lower Y (≈ 10⁶ Pa), meaning it stretches more for the same stress.

Q4: Water in a lake is compressed by 1%. Find the pressure increase. (B = 2.2×10⁹ Pa) A: B = -ΔP/(ΔV/V) → ΔP = B × |ΔV/V| = 2.2×10⁹ × 0.01 = 2.2×10⁷ Pa = 22 MPa.

Q5: Why does a wire become thinner when stretched? A: Due to POISSON'S EFFECT: longitudinal strain causes a lateral contraction. The ratio of lateral to longitudinal strain is Poisson's ratio (σ).


11. Conclusion

The mechanical properties of solids determine how materials behave under load — critical knowledge for engineering and construction. Hooke's law provides a LINEAR model of elasticity. The three moduli (Young's, Bulk, Shear) quantify a material's stiffness in different deformation modes. Elastic potential energy shows that stretched materials STORE energy, explaining everything from catapults to seismic waves.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Stress and strain
stress = F/A (Pa); strain = delta-L/L (dimensionless)
Stress is the internal restoring force per unit area.
Young's modulus
Y = (F/A)/(delta-L/L) = FL/(A delta-L)
Measures resistance to stretching; a material property.
Bulk and shear moduli
B = -delta-P/(delta-V/V); G = (F/A)/theta
B resists volume change, G (rigidity) resists shape change.
Elastic potential energy
U = (1/2) x stress x strain x volume = (1/2) F delta-L
Energy stored in a stretched wire.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Confusing stress with pressure
Stress is the internal restoring force per unit area within the material; pressure is an external force per unit area.
WATCH OUT
Applying Hooke's law beyond the elastic limit
Hooke's law (stress proportional to strain) holds only within the elastic limit; beyond the yield point deformation is plastic.
WATCH OUT
Thinking Young's modulus depends on the wire's dimensions
Young's modulus is a property of the material, not the particular wire's length or area.
WATCH OUT
Dropping the negative sign in bulk modulus
The minus sign in B = -delta-P/(delta-V/V) keeps B positive because volume decreases as pressure increases.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1MEDIUM· Young's Modulus
A 3 m steel wire (Y = 2e11 Pa, area 2 mm^2) carries a 100 kg load. Find the elongation (g = 10 m/s^2).
Show solution
F = mg = 1000 N. delta-L = FL/(AY) = (1000 x 3)/(2e-6 x 2e11) = 3000/4e5 = 7.5e-3 m = 7.5 mm.
Q2MEDIUM· Bulk Modulus
Water is compressed by 1%. Find the pressure increase (B = 2.2e9 Pa).
Show solution
B = -delta-P/(delta-V/V), so delta-P = B x |delta-V/V| = 2.2e9 x 0.01 = 2.2e7 Pa = 22 MPa.
Q3MEDIUM· Energy
A steel wire (L = 2 m, area 1 mm^2, Y = 2e11 Pa) is stretched by 1 mm. Find the energy stored.
Show solution
U = (1/2)(YA/L)(delta-L)^2 = (1/2)(2e11 x 1e-6 / 2)(0.001)^2 = (1/2)(1e5)(1e-6) = 0.05 J.
Q4EASY· Concept
Which has the higher Young's modulus, rubber or steel?
Show solution
Steel (about 2e11 Pa). Rubber's Young's modulus is far lower (about 1e6 Pa), so it stretches much more for the same stress.
Q5EASY· Poisson
Why does a wire become thinner when stretched?
Show solution
Because of the Poisson effect: longitudinal extension is accompanied by lateral contraction. The ratio of lateral to longitudinal strain is Poisson's ratio.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Stress = F/A (Pa); strain = change in dimension / original dimension (dimensionless).
  • Hooke's law: stress proportional to strain within the elastic limit.
  • Stress-strain curve: proportional limit, elastic limit, yield point, ultimate strength, fracture.
  • Young's modulus Y = FL/(A delta-L) for stretching.
  • Bulk modulus B = -delta-P/(delta-V/V) for volume change; shear modulus G for shape change.
  • Elastic PE = (1/2) x stress x strain x volume.
  • Poisson's ratio = -lateral strain / longitudinal strain (typically 0.2-0.4).

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 4-6 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Elastic moduli numericals31Young's, bulk, shear modulus calculations
Stress-strain curve2-31Interpreting the curve and elastic limit
Energy / Poisson's ratio2-31Elastic PE and lateral strain
Prep strategy
  • Memorise definitions and formulae for the three moduli
  • Practise wire-stretching numericals
  • Learn the labelled stress-strain curve
  • Understand Poisson's ratio and its range

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Structural engineering

Elastic moduli decide which materials can safely bear the loads in bridges, buildings, and cranes.

Material selection

Designers pick materials by stiffness and strength for springs, cables, and machine parts.

Seismology

Elastic properties of rock govern how seismic waves travel through the Earth.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Write the correct modulus formula before substituting
  2. Keep track of units (mm^2 to m^2, mm to m)
  3. Label the stress-strain curve regions
  4. Use U = (1/2)F delta-L for energy questions

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Relate the three elastic moduli and Poisson's ratio for isotropic materials.
  • Analyse thermal stress produced when a clamped rod is heated.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examMedium
JEE Main and Advanced (Elasticity)Medium
NEET PhysicsMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The curve shows how a material deforms under increasing stress. The initial straight portion obeys Hooke's law (proportional limit). Up to the elastic limit the material recovers fully when unloaded. Beyond the yield point it deforms plastically and does not return to its original shape. The highest stress is the ultimate tensile strength, after which necking occurs until the fracture point. The shape of the curve tells us whether a material is ductile, brittle, strong, or stiff.

Steel has a much higher Young's modulus (about 2e11 Pa) than copper, meaning it deforms very little under large loads -- it is stiffer. It also has a high elastic limit and tensile strength, so it can bear heavy stresses without permanent deformation or fracture. These properties make steel ideal for load-bearing structures like bridges and buildings.
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Last reviewed on 29 May 2026. Written and reviewed by subject-matter experts — read about our process.
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