Why a steel bridge cable and a rubber band respond to load so differently — putting exact numbers on how solids stretch, twist, and compress before they break.
No solid is perfectly rigid — every real material deforms at least slightly under load, and most solids spring back to their original shape once that load is removed. This tendency to recover is called elasticity. Push the load past a certain point, though, and the deformation becomes permanent — the material has been stretched into plastic deformation.
Plotting stress against strain as a material is loaded reveals distinct regions, all worth knowing by name:
Relates longitudinal stress to longitudinal strain — the standard measure of how stiff a material is when stretched or compressed along its length (wires, rods, columns).
Relates shearing stress to shearing strain — how much a material resists having its layers slide past each other, rather than stretching or compressing.
Relates volume (hydraulic) stress — uniform pressure from every direction — to the resulting fractional change in volume.
Stretching (or compressing) a material within its elastic limit stores energy, exactly like a spring (Chapter 5) — because, within Hooke's Law, a wire genuinely behaves like a spring.
Real engineering depends on knowing these numbers precisely, not just qualitatively: bridge cables and building beams are sized so that everyday loads stay well within the elastic limit, with a safety margin. Choosing a beam shape — a thick, hollow cylinder rather than a solid rod of the same weight, for instance — increases resistance to bending far more efficiently than adding solid material, since bending stress is greatest at the surface and weakest near the centre.
A wire of length 1 m and cross-sectional area 2×10⁻⁶ m² stretches by 1 mm under a force of 200 N. Find Young's modulus of the material.
Solution: Y = FL/(AΔL) = (200 × 1) / (2×10⁻⁶ × 1×10⁻³) = 200 / (2×10⁻⁹) = 1×10¹¹ N/m².
A wire is stretched by 2 mm under a force that rises steadily from 0 to 50 N, within its elastic limit. Find the elastic potential energy stored.
Solution: U = ½FΔL = ½ × 50 × 0.002 = 0.05 J.