Stress, Strain & Young Modulus in Tension and Compression

Hang 200 N from a steel wire 1.00 m long with a cross-section of 1.00 mm² and it stretches 1.00 mm. Put four such wires side by side and each carries 50 N, so the bundle stretches only 0.25 mm; join two end to end and each stretches 1.00 mm, so the pair stretches 2.00 mm. Force and extension describe a sample, not the steel. Divide the force by the area carrying it and the extension by the length that produced it, and all three collapse onto one statement: 200 MPa with strain 1.00 × 10⁻³ for the single and the doubled wire, 50 MPa with 2.5 × 10⁻⁴ for the bundle, and σ/ε = 200 GPa every time. That is the whole reason for the definitions. Stress σ = F/A is in pascals, N m⁻², usually quoted in MPa; strain ε = ΔL/L₀ is a ratio of lengths with no unit; and their ratio has shed every trace of the sample's size.

Stress is an internal quantity, so find it the way statics finds any internal force: cut the member at right angles to its axis and draw a free-body diagram of one piece. In a straight member loaded only along its axis, the cut face carries the full axial force, spread over the face's area A and directed normal to it — hence normal stress. Tension pulls the two faces apart and is counted positive; compression pushes them together and is counted negative; σ = F/A serves both. Use the original area: a steel bar at ε = 1.0 × 10⁻³ narrows in diameter by only about 0.03%, so its area changes by 0.06%, far inside the uncertainty of any tabulated E. Two cautions. F/A is the true local stress only where the load has spread across the section, a few widths away from grips, bolt holes and sudden changes of size, where the local stress runs higher. And for a round bar A = πd²/4; writing πd², or using d where r belongs, is a factor-of-four error.

Load a specimen in small steps and plot σ against ε. For metals, glass, bone and timber loaded along the grain, the first part of the graph is a straight line through the origin, and its gradient is the Young modulus E = σ/ε, in pascals. The straightness has a microscopic reason: close to their equilibrium spacing, the force between neighbouring atoms grows in proportion to their displacement, as it does near any smooth minimum of potential energy, so a solid bar behaves as a vast network of tiny springs. Typical values: steel ≈ 200 GPa, copper ≈ 120 GPa, aluminium and glass ≈ 70 GPa, bone ≈ 17 GPa, oak along the grain ≈ 11 GPa. For most metals one E describes tension and compression alike, because pushing atoms slightly together is the reverse of pulling them apart. The line does not continue forever: for mild steel it ends at the proportional limit, near 250 MPa and a strain of about 1.25 × 10⁻³.

Chain the definitions and the deformation of a uniform member follows: ΔL = εL₀ = σL₀/E = FL₀/(EA). That is Hooke's law F = kΔL with k = EA/L₀, and unlike E, k belongs to the member: it grows with area and falls with length. A 2.00 m aluminium rod 20.0 mm in diameter has A = 3.14 × 10⁻⁴ m² and k = (70 × 10⁹ Pa)(3.14 × 10⁻⁴ m²)/(2.00 m) = 1.10 × 10⁷ N m⁻¹, so a 22.0 kN load stretches it 2.00 mm; a steel rod of the same size is 2.9 times stiffer. The spring rules carry over. Segments in series carry the same force and their changes in length add, so a stepped bar gives most in its thinnest, softest, longest part. Members in parallel are forced to share one ΔL and their forces add, so load divides in proportion to EA/L₀. That rule settles supports that equilibrium could not: when there are more unknown forces than equations, the missing equations come from requiring the deformations to fit together.

ΔL = FL₀/(EA) assumes the same force and area at every cross-section. When they vary, apply σ = Eε to a thin slice of length dx, whose extension is ε dx = F(x) dx/(EA(x)), and add up the slices: ΔL = ∫₀^L₀ F(x)/(EA(x)) dx. The cleanest case is a uniform rod hanging under its own weight. A cut at height x above the free end holds up only the rod below it, so F(x) = ρgAx and σ(x) = ρgx, rising from zero at the bottom to ρgL₀ at the top. The integral gives ΔL = ρgL₀²/(2E), half the stretch the same weight would cause hung from the end, because the average slice carries half the weight. The area has cancelled from the stress, and that has a hard consequence: a mild-steel rod (ρ = 7850 kg m⁻³, E = 200 GPa) reaches 250 MPa at its top when L₀ = σ/(ρg) = 3.2 km, however thick it is, and by then it has stretched 2.0 m under its own weight.

Working strains are usually below 10⁻³. A 5.00 m steel column at 100 MPa shortens by 2.50 mm, a 0.05% change in geometry, which is why the equilibrium topics could draw beams and cables undeformed: no moment arm or angle moves enough to matter. Deformation becomes the point when it is the design requirement, as for a lift that must stop level with a floor, when supports are statically indeterminate, and when parts of different stiffness are clamped together. Before trusting ΔL = FL₀/(EA), run three checks. Is |σ| below the proportional limit? Beyond it, strain grows faster than stress and the model underestimates the deformation. Is a compressed member stocky? A slender strut bows sideways at a load set by its length, shape and E, which can be far below the load that would crush the material, so a correct shortening calculation can describe a column that has already buckled. Is the material linear at all? Rubber and soft tissue curve from the origin, and for them E is at best a local gradient.