Stress and strain

Why it matters

Statics tells you the force in a member. Strength of materials asks what that force does to the material inside it. A 50 kN pull is nothing for a thick steel bar and catastrophic for a thin wire. To compare members of different sizes, engineers divide by area to get stress, and divide stretch by length to get strain. These two quantities, and the link between them, underpin every structural and machine calculation that follows.

Stress

Normal stress is the internal force per unit area acting perpendicular to a cut through the member:

σ = F / A

Tension (pulling) is positive and compression negative. The SI unit is the pascal, 1 Pa = 1 N/m², which is tiny, so engineers use megapascals: 1 MPa = 1 N/mm² = 10⁶ Pa. Working in newtons and millimetres gives MPa directly, which is why many engineers prefer N and mm for stress problems.

Shear stress τ = V / A acts parallel to the cut, as in a bolt that a lap joint tries to slice. This lesson focuses on axial (normal) stress.

Strain

Normal strain is the change in length divided by the original length:

ε = δ / L

It has no units (mm/mm). Strains in structural metals are small: steel at a typical working stress stretches by well under a tenth of a percent. Engineers sometimes write strain in microstrain (με, 10⁻⁶).

Hooke's law and Young's modulus

For most materials, at small strains, stress is proportional to strain:

σ = E · ε

E is Young's modulus (the modulus of elasticity), a measure of stiffness. Rough typical values:

Combining the three equations gives the elongation of an axially loaded bar:

δ = F · L / (A · E)

This only holds in the elastic range, below the yield point, where unloading returns the bar to its original length. Beyond it the material deforms permanently, which is covered in Materials Science Fundamentals.

Worked example: a steel tie rod

A steel rod of diameter 20 mm and length 2.0 m carries a tensile force of 50 kN. E = 200 GPa = 200,000 MPa.

A = π · d² / 4 = π · 20² / 4 = 314.2 mm²
σ = F / A = 50,000 N / 314.2 mm² = 159.2 MPa
ε = σ / E = 159.2 / 200,000 = 7.96 × 10⁻⁴   (about 0.08 %)
δ = ε · L = 7.96 × 10⁻⁴ · 2000 mm = 1.59 mm

A 2 m steel rod stretches less than 2 mm under five tonnes of pull. If the rod were aluminium (E = 70 GPa) with the same diameter, the stress would be the same 159.2 MPa (stress does not depend on the material), but it would stretch 200/70 times as much: about 4.5 mm.

Poisson's effect

A bar that is stretched also gets slightly thinner. The ratio of sideways strain to lengthwise strain is Poisson's ratio ν, about 0.3 for steel and aluminium:

ε_lateral = −ν · ε_axial

For the rod above, its diameter shrinks by about 0.3 × 7.96 × 10⁻⁴ × 20 mm ≈ 0.005 mm.

Thermal strain

Temperature changes also cause strain: ε_thermal = α · ΔT, where α is the coefficient of thermal expansion (about 12 × 10⁻⁶ per °C for steel). If a member is prevented from expanding, that strain turns into stress: a fully restrained steel bar heated by 30 °C develops σ = E·α·ΔT = 200,000 × 12 × 10⁻⁶ × 30 = 72 MPa of compression. This is why bridges have expansion joints.

Common mistakes

• Mixing GPa and MPa. 200 GPa = 200,000 MPa. Dividing 159 MPa by 200 (GPa) gives a strain a thousand times too large.
• Using diameter instead of radius (or forgetting the /4) in A = π·d²/4.
• Applying δ = FL/AE beyond yield. It is only valid in the linear elastic range.
• Thinking stress depends on the material. For a statically determinate member, σ = F/A depends only on load and area; the material decides whether that stress is safe and how much the member stretches.

Recap

• σ = F/A (MPa = N/mm²), ε = δ/L (dimensionless).
• In the elastic range σ = E·ε, so δ = F·L/(A·E).
• Steel is about three times stiffer than aluminium.
• Restrained thermal expansion creates stress: σ = E·α·ΔT.

Try it

On the Do tab, finish an elongation function and check the steel tie rod from this lesson.