Stress, Strain and Elastic Constants
Normal and shear stress, strain, Hooke's law, the stress-strain curve, elastic constants E, G, K and Poisson's ratio, bars of varying section, composite bars and thermal stress.
Strength of materials starts with two ideas: stress (internal force per unit area) and strain (deformation per unit length). Hooke's law links them while the material stays elastic.
Stress and strain
- Normal stress σ = P/A (tensile positive). Shear stress τ = V/A.
- Normal strain ε = δL/L (no units). Shear strain γ = change in a right angle, in radians.
- Hooke's law: σ = Eε and τ = Gγ, where E is Young's modulus and G the shear (rigidity) modulus.
- Elongation of a bar: δ = PL/(AE).
The stress–strain curve (mild steel)
In a tension test, the curve shows the proportional limit, the elastic limit, the upper and lower yield points, strain hardening up to the ultimate tensile strength, then necking and fracture. Ductile materials show large plastic strain before breaking; brittle ones (cast iron, glass) break with little strain.
Elastic constants
- Poisson's ratio ν = −(lateral strain)/(longitudinal strain); for metals about 0.25–0.33, and at most 0.5.
- Bulk modulus K = (hydrostatic stress)/(volumetric strain).
- E = 2G(1 + ν) = 3K(1 − 2ν), and E = 9KG/(3K + G).
- Volumetric strain of a bar under axial stress: ε_v = ε(1 − 2ν). Of a body under equal stress σ on all sides: ε_v = 3σ(1 − 2ν)/E.
Bars in series and parallel
- Stepped bar (series): same load in each part; δ = Σ PLᵢ/(AᵢEᵢ).
- Tapered bar (circular, diameters d₁ to d₂): δ = 4PL/(πEd₁d₂).
- Bar under its own weight: δ = ρgL²/(2E) = WL/(2AE), half that of the same load at the end.
- Composite bar (parallel): both parts stretch equally; P = P₁ + P₂ and P₁L/(A₁E₁) = P₂L/(A₂E₂).
Thermal stress
A bar free to expand grows by δ = αLΔT with no stress. If it is fully restrained, the stress is
σ = EαΔT (compressive on heating).
If the support yields by a gap Δ, then σ = E(αLΔT − Δ)/L.
Worked example
A steel bar (E = 200 GPa) has two parts: 1 m with area 400 mm² and 0.5 m with area 200 mm². It carries 40 kN. Find the total elongation.
- Part 1: 40000 × 1000 / (400 × 200000) = 0.5 mm.
- Part 2: 40000 × 500 / (200 × 200000) = 0.5 mm.
- Total δ = 1.0 mm.
Practice
A material has E = 210 GPa and G = 80 GPa. What is its Poisson's ratio? (Give 4 decimal places.)
Show answer and explanation
Answer: 0.3125ν = E/(2G) − 1 = 210/160 − 1 = 0.3125.
A steel rail (E = 200 GPa, α = 11 × 10⁻⁶ /°C) is fully restrained and heated by 30 °C. What is the thermal stress in MPa?
Show answer and explanation
Answer: 66σ = EαΔT = 200000 × 11 × 10⁻⁶ × 30 = 66 MPa (compression).
A bar 2 m long, area 100 mm², E = 100 GPa, carries 10 kN. What is its elongation in mm?
Show answer and explanation
Answer: 2δ = PL/(AE) = 10⁴ × 2000/(100 × 10⁵) = 2 mm.
What is the maximum possible value of Poisson's ratio for an isotropic elastic material?
Show answer and explanation
Answer: 0.5K = E/(3(1 − 2ν)) must stay positive, so ν < 0.5. At ν = 0.5 the material is incompressible.
A bar hanging under its own weight W stretches by how much compared with the same bar carrying a load W at its free end?
Show answer and explanation
Answer: BSelf-weight gives δ = WL/(2AE), half of WL/(AE).