Skip to main content
Physics 3% exam weight

Mechanical Properties

Part of the JEE Main study roadmap. Physics topic phy-008 of Physics.

→ See in Roadmap Build 1-Month Plan

Mechanical Properties

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision.

Mechanical Properties — Key Facts for JEE Main • Core formula: Hooke’s Law — σ = E·ε (stress = Young’s modulus × strain) • Core concept: Within the elastic limit, strain is directly proportional to applied stress; remove the load and the material returns to its original dimensions. • Most common application: Calculating the extension or compression of a rod under an axial load, e.g., a steel column supporting a bridge. • Key numerical value: Young’s modulus of steel ≈ 2.0 × 10¹¹ Pa (≈ 200 GPa); aluminum ≈ 7.0 × 10¹⁰ Pa. • Most tested concept: The σ = E·ε relationship and its derived form ΔL = FL/(AE) for axial elongation. • Common mistake students make: Mixing up stress (force per unit area) with pressure, or forgetting to convert cross-sectional area from cm² to m² before substituting, leading to a wrong numerical answer. ⚡ Exam tip: In any JEE Main problem involving a metal wire under tension, always convert every quantity to SI base units before plugging in — length in m, area in m², force in N. This single habit prevents 90% of dimensional errors.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Mechanical Properties — JEE Main / Advanced Study Guide

When an external force is applied to a body, it deforms. Stress (σ) is the internal resisting force per unit cross-sectional area (σ = F/A), measured in pascals (Pa). Strain (ε) is the dimensionless ratio of change in dimension to the original dimension (for longitudinal strain: ε = ΔL/L).

Hooke’s Law states that within the elastic limit, stress is directly proportional to strain: σ ∝ ε, or σ = E·ε, where E is the Young’s modulus — a material property that quantifies stiffness. Rearranging gives the axial elongation formula: ΔL = FL/(AE).

There are three other elastic moduli: Bulk modulus (K) measures resistance to uniform pressure (K = −V·ΔP/ΔV); Shear modulus (G) governs shape change under shear stress (G = F/Aγ); and Poisson’s ratio (ν) relates lateral to longitudinal strain (ν = −ε_lat/ε_long). These are linked by E = 2G(1 + ν) and K = E/(3 − 6ν).

Solved Example 1: A steel wire (E = 2 × 10¹¹ Pa) of length 2 m and cross-section 1 mm² is pulled by a 1,000 N force. Find extension.

  • σ = F/A = 1,000 / (10⁻⁶) = 10⁹ Pa
  • ε = σ/E = 10⁹ / 2 × 10¹¹ = 5 × 10⁻³
  • ΔL = ε·L = 5 × 10⁻³ × 2 = 10⁻² m = 1 cm

Solved Example 2: A brass rod (E = 1.1 × 10¹¹ Pa), length 50 cm, diameter 2 cm, compressed by 0.1 mm under a force F. Find F.

  • r = 0.01 m → A = π(0.01)² = 3.14 × 10⁻⁴ m²
  • ΔL = 0.1 mm = 10⁻⁴ m; L = 0.5 m → ε = 10⁻⁴/0.5 = 2 × 10⁻⁴
  • σ = E·ε = 1.1 × 10¹¹ × 2 × 10⁻⁴ = 2.2 × 10⁷ Pa
  • F = σ·A = 2.2 × 10⁷ × 3.14 × 10⁻⁴ ≈ 6,908 N

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer timeline.

Mechanical Properties — Comprehensive JEE Notes

Elastic behaviour at the molecular level: Macroscopic deformation corresponds to tiny displacements of atoms from their equilibrium lattice positions. The inter-atomic potential is asymmetric — compressive forces push atoms closer (repulsive regime), while tensile forces pull them apart. Young’s modulus E is essentially the second derivative of this potential at the equilibrium position, which is why it is a material constant independent of geometry.

Stress–strain curve phases: A typical stress–strain curve for a ductile material like mild steel shows: (1) Proportional limit — linear σ–ε (Hooke’s Law holds); (2) Elastic limit — material still returns to origin if unloaded; (3) Yield point — plastic deformation begins with a distinct upper and lower yield stress; (4) Strain hardening region — material strengthens as dislocations multiply; (5) Necking region — localized reduction in cross-section; (6) Fracture point — breaking stress (ultimate tensile strength). The area under the curve up to the elastic limit equals the elastic potential energy per unit volume stored: U = ½ σ ε = ½ E ε².

Relation between elastic constants: Starting from first principles using isotropic linear elasticity theory, we derive the general relationship among E, G, and K: E = 3K(1 − 2ν) = 2G(1 + ν). For an incompressible material (ν ≈ 0.5), K → ∞, meaning volume change requires infinite pressure — consistent with rubber-like materials. Conversely, for cork (ν ≈ 0), K is small, making it easy to compress but hard to shear.

Advanced solved example: A composite bar of length L consisting of two segments — steel (E₁, A₁) and copper (E₂, A₂) — is rigidly fixed at both ends and subjected to a temperature increase ΔT. Find thermal stress induced.

  • Free expansion of each segment: ΔL₁_thermal = α₁ L ΔT; ΔL₂_thermal = α₂ L ΔT
  • After being locked, total internal force F produces opposing elastic compressions: ΔL₁_force = FL/(A₁E₁); ΔL₂_force = FL/(A₂E₂)
  • Compatibility: ΔL₁_force + ΔL₂_force = ΔL₁_thermal + ΔL₂_thermal (net zero change)
  • Solving: F = LΔT(α₁E₁A₁ + α₂E₂A₂) / (A₁E₁ + A₂E₂)² × (A₁E₁ + A₂E₂)? Wait — correctly:
  • F = [α₁E₁A₁ + α₂E₂A₂] · ΔT · (A₁E₁·A₂E₂)/(A₁E₁ + A₂E₂)
  • Thermal stress in steel = F/A₁ = α₁E₁ΔT · (A₂E₂)/(A₁E₁ + A₂E₂)
  • This shows thermal stress can be very large — important in engineered structures like railway tracks (railway expansion joints).

Content adapted based on your selected roadmap duration. Switch tiers using the pill selector above.