Stress, Strain and the Tensile Test

Pull two steel bars with an identical 10 kN. The 20 mm² one fails; the 200 mm² one is fine. Force alone cannot predict failure, because strength is a property of the *material* and force is a fact about the *loading*.

Stress σ = P/A normalises by the area carrying the load, and immediately a single number explains both results: mild steel yields at about 250 MPa, so 500 MPa fails and 50 MPa does not.

Strain ε = δL/L does the same for deformation. An extension of 1 mm is meaningless until you know whether the bar was 10 mm or 10 m long. Strain is dimensionless, which is what makes it comparable across every size of part.

Load a specimen gradually and at first stress and strain rise in exact proportion — Hooke's law, σ = E·ε. Remove the load and the specimen returns to its original length exactly.

The constant E is Young's modulus, the slope of that line: about 200 GPa for steel, 70 GPa for aluminium, 30 GPa for concrete.

Steel is not three times *stronger* than aluminium — it is three times stiffer. Stiffness is resistance to deformation and is E; strength is resistance to failure and is the yield or ultimate stress. High-strength steel and mild steel have virtually the same E and very different yield points, which is exactly why the distinction matters.

At the yield point proportionality ends and, more importantly, the deformation stops being recoverable. Dislocations in the crystal begin sliding past each other permanently, so releasing the load leaves a permanent set.

Mild steel shows this dramatically, extending at almost constant stress between an upper and lower yield point. Aluminium, copper and high-strength steels have no sharp yield at all, so a 0.2% proof stress — the stress leaving 0.2% permanent strain — is used by agreement in its place.

This is the number most designs are checked against. A component that has yielded is no longer the shape it was designed to be, which is usually a failure even though it has not broken.

1. 1Strain hardening. Past yield, tangled dislocations obstruct each other and more stress is needed to continue deforming. Cold working exploits this deliberately, trading ductility for strength.
2. 2Ultimate tensile strength. The curve reaches a maximum — about 400 MPa for mild steel. This is the highest *engineering* stress the specimen ever carries.
3. 3Necking. Deformation localises into one narrow region whose area shrinks rapidly.
4. 4Fracture. The neck separates, typically in a cup-and-cone shape for a ductile metal.

The falling tail after the ultimate point confuses almost everyone, and the explanation is a definition rather than the material. Engineering stress divides by the original area A₀, so as the neck shrinks it reports a falling stress. True stress divides by the actual current area and keeps rising until fracture. Nothing got weaker.

The factor of safety is σ_yield / σ_working, and it is what a design code actually specifies: typically 1.5–2 for ductile metals in a controlled environment, and considerably more where loads are uncertain or the material is brittle.

A ductile material — mild steel, copper, aluminium — deforms visibly before it breaks, giving warning and absorbing a great deal of energy. A brittle one — cast iron, concrete, glass, ceramic — has almost no curve past the elastic line and fails suddenly.

• Ductile materials are designed against yield; brittle ones against ultimate stress, because there is nothing in between.
• Brittle materials are far weaker in tension than compression, which is precisely why concrete is reinforced with steel on its tension side.
• Ductility is not fixed: steel becomes brittle at low temperature, and cold working reduces it.
• The area under the curve is the energy absorbed before fracture, which is why a brittle failure is so much more dangerous even at the same strength.

Young's modulus E

Direct stress over direct strain.

Poisson's ratio ν

Lateral strain over longitudinal strain, about 0.3 for most metals. Stretch a bar and it gets thinner — that is why a uniaxial load produces a triaxial strain state.

Modulus of rigidity G

Shear stress over shear strain. E = 2G(1 + ν).

Bulk modulus K

Volumetric stress over volumetric strain. E = 3K(1 − 2ν).

Only two of the four are independent. Note also what E = 3K(1 − 2ν) implies: at ν = 0.5 the bulk modulus becomes infinite, so a material with ν = 0.5 is perfectly incompressible. Rubber is close, which is why a rubber block confined on all sides behaves almost like a liquid.