Stress, strain and Hooke's law; elastic constants

Why it matters

Every load cell, torque sensor and strain-gauge transducer in a mechatronic system works by measuring a tiny elastic strain and converting it to force through Hooke's law. The same relations size tie rods, actuator pins and machine frames so that they stay elastic and stiff enough for accurate positioning. If you know stress, strain and the elastic constants well, you can both design a part and design the sensor that measures its load.

Key ideas

Normal stress. The internal force per unit area acting perpendicular to a cut: σ = P/A for an axial load P on area A, tensile positive. It assumes the load acts through the centroid and the section is far from holes, shoulders and load points (Saint-Venant's principle); near such features stress concentrations raise the local value.

Shear stress. Internal force per unit area acting along the cut: τ = V/A on average, for example in a pin in single shear (A = one cross-section) or double shear (A = two cross-sections).

Normal strain and shear strain. Normal strain ε = δ/L is the change in length per unit length (dimensionless, often quoted in microstrain, με = 10⁻⁶). Shear strain γ is the change in a right angle, in radians.

Stress-strain curve (mild steel, tension test). In order: proportional limit (end of the straight line), elastic limit, upper and lower yield points, strain hardening up to the ultimate tensile strength, then necking and fracture. Engineering stress uses the original area, so it falls after the ultimate point; true stress uses the current area and keeps rising. Ductile materials (steel, aluminium) show large plastic strain; brittle ones (cast iron, ceramics) fracture with little. Aluminium and many alloys have no sharp yield point, so a 0.2 % proof stress is used.

Hooke's law and elastic constants. In the linear-elastic range σ = E·ε and τ = G·γ. Under axial stress a bar also contracts laterally: Poisson's ratio ν = −ε_lateral/ε_axial, about 0.25 to 0.33 for metals, 0.5 for rubber (incompressible), and theoretically between −1 and 0.5. Under equal pressure on all faces, volume changes according to the bulk modulus K. For an isotropic material only two constants are independent; E, G, K and ν are linked by the relations below.

Generalised Hooke's law. With stresses in three directions, the strain in each direction includes the Poisson effects of the other two: ε_x = [σ_x − ν·(σ_y + σ_z)]/E. Volumetric strain is ε_v = ε_x + ε_y + ε_z.

Axial deformation. For a uniform bar δ = P·L/(A·E). For a stepped bar, add the segments. For a bar hanging under its own weight, δ = ρ·g·L²/(2E), half of what the full weight would produce if applied at the end. For a conical (linearly tapered) circular bar, δ = 4·P·L/(π·E·d₁·d₂).

Working stress and factor of safety. Allowable stress = yield (ductile) or ultimate (brittle) strength divided by a factor of safety. Take strengths from your materials data book or the relevant code.

Formulas

σ = P / A, τ = V / A

• P: axial force (N); V: shear force (N); A: area resisting it (m²); σ, τ in Pa (N/m²; 1 MPa = 1 N/mm²).

ε = δ / L, σ = E·ε, τ = G·γ

• δ: change in length (m); L: original length (m); E: Young's modulus (Pa); G: shear modulus (Pa); γ: shear strain (rad). Linear-elastic range only.

δ = P·L / (A·E), δ = Σ P_i·L_i / (A_i·E_i)

• Uniform bar, and stepped or composite-in-series bar.

ν = −ε_lateral / ε_axial

• Poisson's ratio (dimensionless).

ε_x = [σ_x − ν·(σ_y + σ_z)] / E

• Generalised Hooke's law for isotropic material; similarly for y and z.

ε_v = ΔV / V = ε_x + ε_y + ε_z, for uniaxial stress ε_v = ε·(1 − 2ν)

• Volumetric strain (dimensionless).

E = 2G·(1 + ν) = 3K·(1 − 2ν), E = 9K·G / (3K + G)

• K: bulk modulus (Pa). Relations for an isotropic linear-elastic material.

δ = ρ·g·L² / (2E), δ = 4·P·L / (π·E·d₁·d₂)

• Self-weight extension of a hanging uniform bar (ρ: density, kg/m³); extension of a conical bar with end diameters d₁ and d₂ (m).

Worked examples

Example 1 (standard): actuator tie rod. Given: a steel tie rod of diameter 12 mm and length 1.5 m carries 20 kN tension; E = 200 GPa, ν = 0.3. Find stress, strain, elongation and the change in diameter.

1. Area: A = π/4 × 12² = 113.1 mm².
2. σ = P/A = 20 000/113.1 = 176.8 MPa (N/mm²).
3. ε = σ/E = 176.8/200 000 = 8.84 × 10⁻⁴ (884 με).
4. δ = ε·L = 8.84 × 10⁻⁴ × 1500 = 1.33 mm.
5. Lateral strain = −ν·ε = −2.65 × 10⁻⁴, so Δd = −2.65 × 10⁻⁴ × 12 = −0.0032 mm. Answer: σ ≈ 177 MPa, ε ≈ 884 με, δ ≈ 1.33 mm, diameter decreases by about 3.2 μm. Check the stress against the allowable for the grade of steel used (data book value divided by the factor of safety).

Example 2 (GATE level): column load cell. Given: a hollow aluminium load-cell column, outside diameter 30 mm, inside 24 mm, E = 70 GPa, ν = 0.33. An axial strain gauge reads −400 με under a compressive load. Find the load, the reading of a circumferential gauge, the volumetric strain, and G and K for the alloy.

1. Area: A = π/4 × (30² − 24²) = π/4 × 324 = 254.5 mm².
2. Stress: σ = E·ε = 70 000 × 400 × 10⁻⁶ = 28.0 MPa (compression).
3. Load: P = σ·A = 28.0 × 254.5 = 7125 N.
4. Circumferential (lateral) strain = −ν·ε_axial = −0.33 × (−400) = +132 με (the column bulges).
5. Volumetric strain: ε_v = ε·(1 − 2ν) = −400 × (1 − 0.66) = −136 με (a decrease in volume).
6. G = E/[2(1 + ν)] = 70/2.66 = 26.3 GPa; K = E/[3(1 − 2ν)] = 70/1.02 = 68.6 GPa. Answer: P ≈ 7.13 kN, circumferential gauge +132 με, ε_v = −136 με, G ≈ 26.3 GPa, K ≈ 68.6 GPa. In a real load cell the axial and circumferential gauges are wired in a Wheatstone bridge, so the circumferential gauge adds sensitivity and compensates for temperature.

Common mistakes

• Mixing units: with N and mm, stress comes out in MPa; with N and m, in Pa. Keep E in the same system (200 GPa = 200 000 N/mm²).
• Using the gross area when a hole or thread reduces the net area that carries the load.
• Applying σ = E·ε beyond the proportional limit.
• Forgetting that Poisson contraction has the opposite sign to the axial strain.
• Using the uniaxial ε = σ/E when stresses act in two or three directions.
• Using single-shear area for a pin that is in double shear (or the reverse).

For GATE ME

Expect elongation of uniform, stepped, tapered and self-weight-loaded bars, relations between E, G, K and ν, volumetric strain of bars and cubes, generalised Hooke's law with two- or three-dimensional stresses, and reading of the stress-strain diagram (proof stress, ductility, toughness as area under the curve). Practise unit handling and the sign of lateral strain.

Quick check

1. E = 200 GPa and ν = 0.3. Find G.
2. A 2 m wire stretches by 2 mm. What is the strain?
3. For which value of ν is a material incompressible?
4. What is the stress when 1000 N acts on 0.02 m²?
5. A steel bar (E = 200 GPa) is stressed to 50 MPa. What is the strain in microstrain?

Answers: 1. 76.9 GPa. 2. 0.001. 3. ν = 0.5. 4. 50 000 Pa (50 kPa). 5. 250 με.