EASA / CAR Part-66 · Module 2 · 2.2 Mechanics · Cat B1/B2

Hooke's Law

Objective: within its elastic limit a spring stretches in direct proportion to the force applied, F = k·x. Load it up and read the straight-line force–extension graph. Push past the elastic limit and it deforms permanently.

Spring & load

Force vs extension

Applied force F 0 N

Spring constant k 20 N/m

Applied force F

Extension x

0cm

State

Unloaded

Energy stored

F = k·x · energy stored = ½·k·x² = ½·F·x (area under the line)

A spring resists being stretched with a force proportional to how far it is stretched, that is Hooke's law, F = k·x, where k (the spring constant or stiffness) is the slope of the force–extension line. A stiffer spring (bigger k) needs more force for the same extension. The energy stored is the area under the line, ½k·x². This proportionality only holds up to the elastic limit; beyond it the material yields and takes a permanent set, so it no longer springs all the way back, the graph bends away from the straight line.

Try this: Increase the load and watch extension grow in step with force, a straight line (Hooke's law).
Try this: Double the spring constant. For the same force the extension halves.
Try this: Push the load past the amber elastic limit, the line bends and the spring is left permanently stretched.
Try this: Reset the spring to remove the permanent set and start again.