What this chapter is about
This chapter explores how solid materials respond when external forces act on them. When you stretch a rubber band, compress a spring, or twist a metal rod, the material changes shape. The study of these deformations and the forces that cause them forms the core of this chapter.
You will learn about stress (the internal restoring force per unit area within a material) and strain (the fractional change in dimensions). The relationship between these quantities reveals how different materials behave under load. Some materials return to their original shape when the force is removed; others do not. This distinction between elastic and plastic behaviour has enormous practical importance in engineering, construction and manufacturing.
By the end of this chapter, you should be able to calculate stress and strain for different types of deformation, apply Hooke's law within its limits, use the various elastic moduli (Young's modulus, bulk modulus, shear modulus) to solve problems, and understand the energy stored in a deformed elastic body.
Key ideas
- Stress is the internal restoring force per unit area developed inside a material when an external force deforms it. Unit: N/m² or pascal (Pa). Stress = F/A.
- Strain is the fractional change in dimension (length, volume or shape) produced by stress. It is a ratio and has no unit.
- Hooke's law states that within the elastic limit, stress is directly proportional to strain. The constant of proportionality is called the modulus of elasticity.
- Elastic limit is the maximum stress up to which a material returns to its original shape after the deforming force is removed; beyond this, permanent deformation occurs.
- Young's modulus (Y) measures resistance to longitudinal (tensile or compressive) deformation: Y = longitudinal stress / longitudinal strain = (F/A) / (ΔL/L). Unit: Pa.
- Bulk modulus (B) measures resistance to uniform volume change under pressure: B = volume stress / volume strain = −ΔP / (ΔV/V). The negative sign accounts for decrease in volume with increase in pressure.
- Shear modulus or rigidity modulus (G) measures resistance to shape change at constant volume: G = shear stress / shear strain = (F/A) / θ, where θ is the angle of shear in radians.
- Poisson's ratio (σ) is the ratio of lateral strain to longitudinal strain; it lies between 0 and 0.5 for most materials.
Formulas and facts to remember
- Stress = Force / Area = F/A (unit: Pa or N/m²).
- Longitudinal strain = Change in length / Original length = ΔL/L (no unit).
- Volume strain = Change in volume / Original volume = ΔV/V (no unit).
- Shear strain = Angular displacement θ (in radians) ≈ tan θ for small angles.
- Young's modulus: Y = (F × L) / (A × ΔL).
- Bulk modulus: B = −ΔP / (ΔV/V) = −P × V / ΔV.
- Shear modulus: G = (F/A) / θ.
- Elastic potential energy stored per unit volume = (1/2) × stress × strain = (1/2) × Y × (strain)².
Worked examples
Example 1: Calculating Young's modulus
A steel wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² is stretched by a load of 200 N. The wire extends by 0.40 mm. Find Young's modulus of the steel.
Solution
Given: L = 2.0 m, A = 1.0 × 10⁻⁶ m², F = 200 N, ΔL = 0.40 mm = 0.40 × 10⁻³ m.
Stress = F/A = 200 / (1.0 × 10⁻⁶) = 2.0 × 10⁸ Pa.
Strain = ΔL/L = (0.40 × 10⁻³) / 2.0 = 2.0 × 10⁻⁴.
Young's modulus Y = stress / strain = (2.0 × 10⁸) / (2.0 × 10⁻⁴) = 1.0 × 10¹² Pa.
This value (about 10¹¹ Pa) is typical for steel.
Example 2: Finding bulk modulus
A solid copper sphere of volume 0.50 m³ is placed at the bottom of a lake where the excess pressure is 2.0 × 10⁶ Pa. The volume decreases by 1.0 × 10⁻⁴ m³. Calculate the bulk modulus of copper.
Solution
Given: V = 0.50 m³, ΔP = 2.0 × 10⁶ Pa, ΔV = −1.0 × 10⁻⁴ m³ (volume decreases).
Volume strain = ΔV/V = (−1.0 × 10⁻⁴) / 0.50 = −2.0 × 10⁻⁴.
Bulk modulus B = −ΔP / (ΔV/V) = −(2.0 × 10⁶) / (−2.0 × 10⁻⁴) = 1.0 × 10¹⁰ Pa.
Example 3: Energy stored in a stretched wire
An aluminium wire of length 1.5 m, cross-sectional area 2.0 × 10⁻⁶ m² and Young's modulus 7.0 × 10¹⁰ Pa is stretched by 0.30 mm. Find the elastic potential energy stored in the wire.
Solution
Strain = ΔL/L = (0.30 × 10⁻³) / 1.5 = 2.0 × 10⁻⁴.
Stress = Y × strain = 7.0 × 10¹⁰ × 2.0 × 10⁻⁴ = 1.4 × 10⁷ Pa.
Volume of wire = A × L = 2.0 × 10⁻⁶ × 1.5 = 3.0 × 10⁻⁶ m³.
Energy per unit volume = (1/2) × stress × strain = 0.5 × 1.4 × 10⁷ × 2.0 × 10⁻⁴ = 1.4 × 10³ J/m³.
Total energy = 1.4 × 10³ × 3.0 × 10⁻⁶ = 4.2 × 10⁻³ J = 4.2 mJ.
Common mistakes
- Forgetting to convert millimetres to metres before substituting into formulas → always express all lengths in SI units (metres).
- Treating strain as having a unit → strain is a ratio of two lengths (or volumes) and is dimensionless.
- Using positive sign for volume change under compression → volume decreases, so ΔV is negative; the negative sign in the bulk modulus formula keeps B positive.
- Confusing tensile stress with shear stress → tensile stress acts perpendicular to the cross-section; shear stress acts parallel (tangential) to the surface.
- Assuming Hooke's law applies for any size of load → it holds only within the elastic limit; beyond that, the material behaves plastically.
Quick revision
- Stress = F/A; strain = fractional change in dimension; both share the same formula structure but stress has unit Pa, strain has none.
- Hooke's law: stress ∝ strain within elastic limit.
- Y for length change, B for volume change, G for shape change.
- Elastic energy per unit volume = (1/2) × stress × strain.
- Poisson's ratio links lateral and longitudinal strains; it lies between 0 and 0.5.
- Beyond the elastic limit, materials undergo permanent (plastic) deformation.