Elasticity

Elasticity is a material's ability to spring back. Learn Hooke's law, Young's, shear and bulk moduli, axial deformation δ = PL/AE, stiffness, thermal stress and composite bars.

Below the yield point, stress and strain are proportional and the material returns to its original shape when unloaded. The constant of proportionality — Young's modulus E — is one of the most important numbers in engineering: it sets how stiff a part is.

Strength decides whether a part breaks; stiffness (E and geometry) decides whether it deflects too much. Many designs are stiffness-limited, not strength-limited.

An analogy

A spring: pull twice as hard and it stretches twice as far, then snaps back. Every solid is a very stiff spring.

Hooke's law and Young's modulus

E is the slope of the elastic part of the stress–strain curve. Steel ≈ 200 GPa, aluminium ≈ 70 GPa, titanium ≈ 110 GPa, concrete ≈ 30 GPa, wood ≈ 10 GPa, nylon ≈ 2–3 GPa. Heat treatment and alloying change strength a lot but hardly change E.

Stiffness and axial deformation

A bar behaves like a spring of stiffness k = AE/L. Bars in series add flexibilities; in parallel (composite bars) they add stiffness and share load in proportion to AE.

Thermal stress

Free expansion is δ = αLΔT. If ends are fixed, that expansion is prevented and stress σ = EαΔT appears regardless of length. Rails, pipelines and bridges need expansion joints or loops.

Anisotropy and specific stiffness

Composites and wood have different E in different directions. Designers compare specific stiffness E/ρ: steel, aluminium and titanium are all ≈ 25 MJ/kg, while carbon fibre composites reach 100+ — the reason aircraft and race cars use them.

Formulas

• Hooke's law: σ = E · ε, τ = G · γ
• Axial deformation: δ = P L / (A E)
• Moduli relation: E = 2G(1 + ν) = 3K(1 − 2ν)
• Thermal stress (fully restrained): σ = E · α · ΔT

Worked example: Elongation of a tie rod

Given: Ø20 mm steel, L = 2 m; P = 50 kN, E = 200 GPa

Find: Elongation.

• A = 314.2 mm²
• δ = 50 000 × 2000/(314.2 × 200 000) = 1.59 mm

Answer: δ ≈ 1.6 mm

Worked example: Thermal stress in a rail

Given: Continuously welded rail, fully restrained; ΔT = +30 °C, α = 12 × 10⁻⁶/°C

Find: Stress.

• σ = 200 000 × 12 × 10⁻⁶ × 30 = 72 MPa (compressive)

Answer: 72 MPa compression

Common mistakes

• Wrong: Choosing a stronger steel to reduce deflection. — Right: All steels have E ≈ 200 GPa; increase section size or change geometry instead.
• Wrong: Forgetting unit conversion of E (GPa → MPa = N/mm²). — Right: 200 GPa = 200 000 N/mm².

Applications

• Machine-tool frames: Stiffness, not strength, controls machining accuracy — cast iron and polymer concrete bases.
• Bolt preload: Bolts are stretched elastic springs; tightening elongation sets clamp force.
• Expansion joints: Pipelines, bridges and exhaust systems accommodate αLΔT growth.

Key takeaways

• σ = Eε in the elastic range.
• δ = PL/AE; bar stiffness = AE/L.
• Restrained thermal stress σ = EαΔT.
• Stiffness is geometry + E; strength is a separate check.

Going further

• Stiffness matrices: k = AE/L for a bar becomes the building block of the finite element stiffness matrix [K]{u} = {F}.
• Viscoelasticity: Polymers creep and relax over time; E depends on temperature and load duration.
• Energy methods: Strain energy U = P²L/2AE leads to Castigliano's theorem for deflections of complex structures.