Elasticity — Stress, Strain, and Young's Modulus

What Elasticity Really Means

Pull a rubber band and release it — it snaps back. Stretch a piece of chewing gum — it stays deformed. The rubber band is elastic; the gum is plastic. Elasticity is the property of a material to regain its original shape and size after the deforming force is removed.

Every solid material has some elasticity. Steel is highly elastic (it snaps back faithfully). Clay is nearly inelastic (it stays where you put it). Understanding WHY materials behave this way comes down to the intermolecular forces — atoms in a solid are connected like tiny springs. When you stretch or compress the material, you’re stretching or compressing those microscopic springs.

This chapter carries 4-5 marks in CBSE Class 11, and JEE Main tests it at least once per year — usually a numerical on Young’s modulus or a conceptual question on stress-strain curves.

Key Terms and Definitions

Deforming force: Any external force that changes the shape or size of a body.

Restoring force: The internal force developed within the body that opposes deformation and tries to restore the original configuration.

Elasticity: The property by which a body resists deformation and recovers when the deforming force is removed.

Plasticity: The property by which a body undergoes permanent deformation. Lead and putty are plastic materials.

Elastic limit: The maximum stress up to which a material obeys Hooke’s Law. Beyond this, permanent deformation sets in.

Stress — The Internal Fight

When an external force deforms a body, internal restoring forces develop. Stress is defined as the restoring force per unit area:

\sigma = \frac{F}{A}

where $F$ = restoring force (equal in magnitude to applied force), $A$ = cross-sectional area.

Units: $\text{N/m}^2$ or Pascal (Pa)

Dimensional formula: $[ML^{-1}T^{-2}]$

Types of Stress

Tensile stress: Force applied along the length, tending to elongate the body. A wire being pulled at both ends experiences tensile stress.

Compressive stress: Force applied to squeeze the body. Pillars of a building are under compressive stress.

Shear (tangential) stress: Forces applied parallel to the face of the body, causing one layer to slide over another. Cutting with scissors applies shear stress.

Hydraulic (volumetric) stress: Uniform pressure applied from all sides. A submarine hull experiences hydraulic stress from water pressure.

Strain — How Much It Deforms

Strain measures the fractional change in dimension caused by stress. It is dimensionless (no units).

Longitudinal strain: $\varepsilon_L = \dfrac{\Delta L}{L_0}$

Shear strain: $\gamma = \dfrac{x}{h} = \tan\theta \approx \theta$ (for small angles)

Volumetric strain: $\varepsilon_V = \dfrac{\Delta V}{V_0}$

Strain has no units because it is a ratio of two lengths (or volumes). If an examiner asks for the “unit of strain,” the correct answer is “dimensionless” or “no unit.”

Hooke’s Law — The Foundation

Robert Hooke (1678) discovered that within the elastic limit, stress is directly proportional to strain:

\sigma \propto \varepsilon \quad \Rightarrow \quad \sigma = E \cdot \varepsilon

where $E$ is the modulus of elasticity (a material constant).

This is a linear relationship — if you double the stress, strain doubles too. The proportionality constant $E$ tells you how “stiff” the material is. High $E$ = stiff = hard to deform.

Hooke’s Law holds only within the elastic limit. Students often apply it beyond this point. The stress-strain graph is linear only in the initial region — never assume Hooke’s Law applies throughout.

The Three Moduli of Elasticity

Young’s Modulus (Y) — For Stretching/Compressing

Young’s modulus applies when tensile or compressive stress acts along one dimension.

Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L / L_0} = \frac{F \cdot L_0}{A \cdot \Delta L}

Units: $\text{N/m}^2$ or Pa (same as stress, since strain is dimensionless)

Steel: $Y \approx 2 \times 10^{11}$ Pa (very stiff) Rubber: $Y \approx 5 \times 10^6$ Pa (very stretchy)

JEE Main frequently gives a wire problem: find the extension $\Delta L$ given the load, dimensions, and Young’s modulus. Just rearrange: $\Delta L = \frac{FL_0}{AY}$ . This appeared in JEE Main 2023 Session 1.

Shear Modulus (G) — Rigidity Modulus

Shear modulus relates shear stress to shear strain.

G = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\theta}

Liquids have zero shear modulus — they cannot resist shear stress, which is why they flow.

Bulk Modulus (K) — For Volume Changes

K = -\frac{\Delta P}{\Delta V / V_0}

The negative sign ensures $K$ is positive (increase in pressure decreases volume).

Compressibility $= \dfrac{1}{K}$

Liquids have a very high bulk modulus (nearly incompressible). Gases have low bulk modulus (highly compressible).

The Stress-Strain Curve

This graph is one of the most important diagrams in this chapter — CBSE asks for labelled diagrams frequently.

For a ductile metal (like mild steel):

O → A (Proportional Limit): Linear region. Hooke’s Law holds perfectly. $\sigma \propto \varepsilon$ .

A → B (Elastic Limit): Still elastic (body recovers), but Hooke’s Law may not hold perfectly.

B → C (Yield Point): Stress barely changes but strain increases rapidly. The material starts to “yield” — this is where permanent deformation begins.

C → D (Plastic Region): Permanent deformation. Stress actually decreases (necking begins in a wire).

D (Breaking Point/Fracture Point): Material breaks.

For CBSE, learn to label the curve with all five points: Proportional limit, Elastic limit, Upper yield point, Lower yield point (sometimes shown), and Fracture point. A neat diagram fetches full marks in 3-mark questions.

Poisson’s Ratio

When a wire is stretched (longitudinal strain), its diameter decreases (lateral strain). The ratio of lateral strain to longitudinal strain is Poisson’s ratio:

\nu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d / d}{\Delta L / L}

Range: $-1 \leq \nu \leq 0.5$ for real materials. For most metals, $\nu \approx 0.3$ .

The negative sign in the definition makes $\nu$ positive (since lateral strain is compressive when longitudinal strain is tensile).

Solved Examples

Example 1 — CBSE Level

A steel wire of length 1.5 m and cross-sectional area $1.5 \times 10^{-4}$ m² stretches by 1.5 mm when a load of 300 N is applied. Find Young’s modulus.

Solution:

Y = \frac{F \cdot L_0}{A \cdot \Delta L} = \frac{300 \times 1.5}{1.5 \times 10^{-4} \times 1.5 \times 10^{-3}}

Y = \frac{450}{2.25 \times 10^{-7}} = 2 \times 10^9 \text{ Pa}

Example 2 — JEE Main Level

A copper wire (Y = $1.2 \times 10^{11}$ Pa) and a steel wire (Y = $2 \times 10^{11}$ Pa) each have length 2 m and area $\pi \times 10^{-6}$ m². They are joined end-to-end and a load of 100 N is applied. Find the total extension.

Solution:

Since both wires are in series, the same force acts on each.

\Delta L_{Cu} = \frac{FL}{AY_{Cu}} = \frac{100 \times 2}{\pi \times 10^{-6} \times 1.2 \times 10^{11}} = \frac{200}{3.77 \times 10^5} \approx 5.3 \times 10^{-4} \text{ m}

\Delta L_{Fe} = \frac{100 \times 2}{\pi \times 10^{-6} \times 2 \times 10^{11}} \approx 3.2 \times 10^{-4} \text{ m}

\Delta L_{total} = 5.3 \times 10^{-4} + 3.2 \times 10^{-4} \approx 8.5 \times 10^{-4} \text{ m}

Example 3 — JEE Advanced Level

A rubber cord of Young’s modulus $5 \times 10^6$ N/m² has length 8 cm and area $1$ cm². How much work is done in stretching it by 2 mm?

Solution:

Work done in stretching an elastic body = $\frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$

W = \frac{1}{2} Y \varepsilon^2 \cdot V

\varepsilon = \frac{2 \times 10^{-3}}{8 \times 10^{-2}} = 2.5 \times 10^{-2}

V = 1 \times 10^{-4} \times 8 \times 10^{-2} = 8 \times 10^{-6} \text{ m}^3

W = \frac{1}{2} \times 5 \times 10^6 \times (2.5 \times 10^{-2})^2 \times 8 \times 10^{-6} = \frac{1}{2} \times 5 \times 10^6 \times 6.25 \times 10^{-4} \times 8 \times 10^{-6}

W = \frac{1}{2} \times 25 \times 10^{-3} = 1.25 \times 10^{-2} \text{ J}

Elastic Potential Energy

When a wire is stretched, work done against restoring forces is stored as elastic potential energy.

Energy stored 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 energy stored:

U = u \times V = \frac{1}{2} \times \frac{F \cdot \Delta L}{1} = \frac{1}{2} F \Delta L

(where $F$ is the stretching force and $\Delta L$ is the extension)

The formula $U = \frac{1}{2} F \Delta L$ looks like the energy stored in a spring ( $\frac{1}{2}kx^2$ ). That’s not a coincidence — a wire under elastic deformation behaves exactly like a spring! This connection is often tested in JEE: “effective spring constant of a wire” = $k = \frac{AY}{L}$ .

Exam-Specific Tips

CBSE Class 11:

Diagrams are essential. The stress-strain curve with all labels is worth 3 marks.
Define all three moduli clearly — definitions alone can fetch 1 mark each.
Derivation of Young’s modulus formula using $F$ , $A$ , $L$ , $\Delta L$ is a standard 2-mark question.

JEE Main:

Wire problems (finding extension) account for ~60% of questions from this chapter.
Spring constant of a wire: $k = AY/L$ — memorise this.
Combined wires (series/parallel) — for series, same force, add extensions; for parallel, same extension, add forces.

NEET:

Mostly conceptual — “which material has the highest Young’s modulus” or “rubber vs steel — which is more elastic?”
Remember: Steel is more elastic than rubber (counterintuitive but correct — steel returns to its original shape more faithfully for the same stress).

Common Mistakes to Avoid

Mistake 1: Thinking rubber is more elastic than steel because it stretches more. Correction: Elasticity is about how well a material RETURNS to its original shape, not how much it stretches. Steel does this better.

Mistake 2: Using the area at the deformed cross-section instead of the original cross-section in stress calculations. We always use the original (undeformed) area.

Mistake 3: Forgetting units. Stress is in Pa (or N/m²), strain has no unit, modulus of elasticity is in Pa. Young’s modulus values for metals are in the range $10^{10}$ to $10^{11}$ Pa — if your answer is in a completely different range, recheck.

Mistake 4: Applying Hooke’s Law beyond the elastic limit. Always check if the question specifies the material is within elastic limits.

Mistake 5: Confusing longitudinal strain and shear strain. Longitudinal strain = $\Delta L / L$ (change in length). Shear strain = $\theta$ (angle of shear).

Practice Questions

Q1. A wire of length 2 m, radius 1 mm, and Y = $2 \times 10^{11}$ Pa is stretched by 1 mm. Find the force applied.

$F = \frac{Y A \Delta L}{L} = \frac{2 \times 10^{11} \times \pi \times (10^{-3})^2 \times 10^{-3}}{2} = \frac{2 \times 10^{11} \times \pi \times 10^{-6} \times 10^{-3}}{2} \approx 314$ N

Q2. A rubber ball of bulk modulus $9.8 \times 10^8$ Pa is taken to a depth where the pressure increases by $10^5$ Pa. Find the fractional decrease in volume.

$\frac{\Delta V}{V} = \frac{\Delta P}{K} = \frac{10^5}{9.8 \times 10^8} \approx 1.02 \times 10^{-4}$

Q3. Two wires of same material and same length but areas $A$ and $2A$ are connected in parallel and a load $F$ is applied. Find the extension.

For parallel connection: both have same extension $\Delta L$ . Forces add: $F = F_1 + F_2$ . Effective area = $3A$ . So $\Delta L = \frac{FL}{3AY}$ .

Q4. The Young’s modulus of a material is $2 \times 10^{11}$ Pa and its breaking stress is $10^9$ Pa. Find the maximum length of a wire of this material that can hang under its own weight. (Density = $8000$ kg/m³, $g = 10$ m/s²)

Maximum stress = $\frac{\text{Weight}}{A} = \frac{\rho A L g}{A} = \rho g L$ . Setting equal to breaking stress: $L = \frac{10^9}{8000 \times 10} = 12500$ m = 12.5 km.

Q5. What is the ratio of energy stored per unit volume in two wires of same material with same length but radii $r$ and $2r$ , under the same load $F$ ?

Energy density $= \frac{\sigma^2}{2Y}$ . Stress in wire 1: $\sigma_1 = F/\pi r^2$ . Stress in wire 2: $\sigma_2 = F/\pi(2r)^2 = F/4\pi r^2$ . Ratio: $u_1/u_2 = \sigma_1^2/\sigma_2^2 = 16$ .

FAQs

Why is steel more elastic than rubber? Elasticity measures how faithfully a material returns to its original shape. Steel has a higher Young’s modulus (stiffer), meaning it resists deformation better and always returns precisely to its original shape. Rubber stretches a lot but is also elastic in everyday use — however, for the same stress, steel undergoes much less strain and returns more precisely.

What happens beyond the elastic limit? Beyond the elastic limit, permanent (plastic) deformation occurs. The material does not return to its original shape. If stress continues to increase, the material eventually fractures at the breaking point.

Can liquids have Young’s modulus? No. Liquids and gases cannot sustain tensile or shear stress — they flow. Only solids have Young’s modulus and shear modulus. All three states have bulk modulus.

What is meant by “neck formation” in a stress-strain experiment? Beyond the yield point, the wire starts narrowing at one region more than others — this region is called the “neck.” Further stretching concentrates here until it breaks. Ductile materials show clear necking; brittle materials break without necking.

How is elasticity different from stiffness? Stiffness measures resistance to deformation (high Young’s modulus = stiff). Elasticity refers to the ability to recover. A material can be stiff (resists deformation) and elastic (recovers perfectly), stiff and brittle (resists then snaps), or flexible and elastic (like rubber).

What is the significance of the area under the stress-strain curve? The area under the curve up to any point represents the energy stored per unit volume. The total area under the fracture point is the modulus of toughness — a measure of how much energy the material can absorb before breaking.

Why is Poisson’s ratio always positive for most materials? Because when you stretch a material along one axis (positive longitudinal strain), it contracts laterally (negative lateral strain). The ratio of these strains (with a negative sign in the definition) gives a positive value. Materials with negative Poisson’s ratios (auxetics) do exist but are special engineered materials.