Elastic Moduli and Stress-Strain
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Direct answer
Hooke's law states that within the proportional limit, stress is proportional to strain: the ratio defines the modulus. Young's modulus Y = (F/A)/(delta-L/L) for stretching a wire, bulk modulus K = −P/(delta-V/V) for uniform squeezing, shear modulus G = (F/A)/theta for tangential distortion. Poisson's ratio sigma, the ratio of lateral to longitudinal strain, lies between −1 and 0.5 in principle and near 0.3 for most metals, and the three moduli are linked: Y = 2G(1 + sigma) = 3K(1 − 2 sigma). The elastic energy stored per unit volume is (1/2) stress × strain, and steel's Y of about 2 × 10^11 Pa — a hundred times wood's — is the standard reference value JEE papers quote.
What you must remember
- Stress and strain: stress = F/A in Pa; strain is dimensionless (delta-L/L, delta-V/V or the shear angle), so every modulus shares the dimensions of pressure.
- Young's modulus: Y = F L/(A delta-L); the elongation of a hanging wire of own weight is delta-L = rho g L^2/(2Y), doubling length quadruples self-extension.
- Bulk modulus: K = −P/(delta-V/V); the minus sign keeps K positive since pressure increase shrinks volume; the reciprocal 1/K is compressibility.
- Shear modulus: G = (F/A)/theta; fluids have no shear rigidity (G = 0), which is the mechanical definition of a fluid.
- Poisson's ratio and interconnection: sigma = −(lateral strain)/(longitudinal strain), restricted to −1 to 0.5 (sigma = 0.5 means incompressible), with Y = 2G(1 + sigma) = 3K(1 − 2 sigma).
- Elastic energy: U = (1/2) F × delta-L for a wire loaded within the limit; energy density = (1/2) stress × strain = (1/2) Y strain^2.
- Thermal stress: a clamped rod prevented from expanding carries stress Y alpha delta-T, force = Y A alpha delta-T — numbers rails and pipelines obey.
- Pattern note: Main tests Y = FL/(A delta-L) rearrangements and energy density; Advanced adds self-weight elongation, thermal stress and breaking-load comparisons.
Two wires, one comparison
A steel wire (Y = 2 × 10^11 Pa) and a copper wire (Y = 1.1 × 10^11 Pa) of identical length and cross-section hang from a ceiling, each carrying the same load. Steel elongates less — delta-L = F L/(A Y) halves when Y doubles — but stores more modestly: U = (1/2) F delta-L, so the softer copper stores more elastic energy for the same force. Deciding which quantity is held fixed — force or extension — before substituting is the whole skill in moduli questions.
Self-weight adds the second standard layer. A wire of density rho and length L hanging under its own weight stretches delta-L = rho g L^2/(2Y): each element carries only the weight below it, so integrating a linearly falling tension gives the square in L and the 2 in the denominator. Thermal stress closes the triangle: a steel rail welded between rigid supports on a day 30 degrees hotter carries stress Y alpha delta-T = 2 × 10^11 × 1.2 × 10^-5 × 30, about 7 × 10^7 Pa, independent of the rail's cross-section — the independence being the examined insight.
Where students slip
Strain is dimensionless and stress is not; candidates who slip a length into strain carry wrong dimensions through the whole numerical. The area in F/A is the original cross-section, and for a wire of diameter d it is pi d^2/4 — forgetting the factor 4 is worth one wrong digit. The stress-strain graph deserves respect: Hooke's law dies at the proportional limit, slightly before the elastic limit, and the yield point and ultimate strength beyond it are distinct landmarks; a wire unloaded beyond the elastic limit keeps a permanent set.
Frequently asked questions
How are Young's, bulk and shear moduli defined?
Y = (F/A)/(delta-L/L) for length change, K = −P/(delta-V/V) for volume change, G = (F/A)/theta for a shearing angle — each the ratio of one stress type to its matching strain.
What is Poisson's ratio and its allowed range?
The negative ratio of lateral to longitudinal strain; thermodynamic stability restricts it to −1 to 0.5, real materials sit near 0.2-0.4, and 0.5 implies incompressibility.
How much does a wire stretch under its own weight?
delta-L = rho g L^2/(2Y), growing as the square of length, because each cross-section carries only the weight hanging below it.
Why does a clamped rod develop stress on heating even without applied force?
Being prevented from expanding by delta-L = L alpha delta-T, the rod must generate the opposing elastic strain Y alpha delta-T all by itself, giving force = Y A alpha delta-T.
Where does the elastic energy of a stretched wire come from and how is it stored?
External work during loading, stored as (1/2) F delta-L in the deformed lattice, recoverable on release only within the elastic limit.