Stress, strain and Hooke's law; elastic constants

Why it matters

Every part of a vehicle, from a suspension link to a connecting-rod bolt, is sized by asking two questions: is the stress below what the material can take, and is the deformation small enough for the part to work? Stress, strain and the elastic constants are the language for both. Almost every later topic (thermal stress, bending, torsion, pressure vessels) uses the relations set up here.

Key ideas

Stress. The intensity of internal force on a cut through a loaded body.

• Normal stress σ acts perpendicular to the cut: tensile (positive) or compressive (negative).
• Shear stress τ acts parallel to the cut. For a straight bar under an axial load through the centroid, σ = P/A is uniform away from the ends and from holes or shoulders. Near those, stress concentrations raise the local stress (Saint-Venant's principle says the disturbance dies out within about one bar width).

Strain. A dimensionless measure of deformation.

• Normal strain ε = change in length ÷ original length.
• Shear strain γ = change in a right angle (radians). Engineering strain uses the original dimensions and is accurate for the small strains of elastic design.

Stress–strain curve of mild steel in tension. Proportional limit (end of the straight line), elastic limit (beyond which some set remains), upper and lower yield points, strain hardening up to the ultimate tensile strength, then necking and fracture. Materials with no clear yield point (aluminium alloys, high-strength steels) use the 0.2 % offset proof stress. Ductile materials (steel, aluminium, copper) show large plastic strain before fracture; brittle materials (cast iron, ceramics, glass) fracture with little plastic strain and are much stronger in compression than in tension.

Hooke's law. Within the proportional limit, stress is proportional to strain: σ = E·ε and τ = G·γ. E (Young's modulus) measures stiffness, not strength: a high-strength steel and a mild steel have nearly the same E.

Poisson's ratio. A bar stretched axially contracts laterally. ν = −(lateral strain) ÷ (axial strain) under uniaxial stress. For isotropic materials 0 ≤ ν ≤ 0.5; metals are about 0.25–0.35, rubber is close to 0.5 (nearly incompressible), cork is near 0.

Generalised Hooke's law (isotropic, linear elastic). Each normal strain has a direct part and Poisson parts from the other two stresses, e.g. ε_x = [σ_x − ν·(σ_y + σ_z)] / E. Shear strains depend only on their own shear stress.

Volumetric strain and bulk modulus. ε_v = ΔV/V = ε_x + ε_y + ε_z. Under equal all-round pressure p, ε_v = −p/K, where K is the bulk modulus.

Elastic constants. E, G, K and ν are linked; for an isotropic material only two are independent. Values depend on the material and its condition, so take them from a data book unless the problem gives them. Typical round figures: steel E ≈ 200–210 GPa, G ≈ 80 GPa; aluminium alloys E ≈ 70 GPa; copper E ≈ 110–120 GPa; grey cast iron about 100–120 GPa.

Axial deformation. Elongation of a prismatic bar under an axial load P is δ = P·L/(A·E). For stepped bars, add the pieces; for a bar of varying section, integrate.

Allowable stress. Design keeps the working stress below the yield (ductile) or ultimate (brittle) strength divided by a factor of safety, which comes from the applicable code or company practice.

Formulas

σ = P / A, τ = V / A (average)

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

ε = δ / L, γ = change in right angle (rad)

• δ: change in length (m); L: original length (m).

σ = E·ε, τ = G·γ

• E: Young's modulus (Pa); G: modulus of rigidity (Pa). Linear elastic range only.

δ = P·L / (A·E), stepped bar δ = Σ Pᵢ·Lᵢ / (Aᵢ·Eᵢ)

• Bars with uniform force and section in each segment.

δ = 4·P·L / (π·E·d₁·d₂)

• Bar tapering linearly from diameter d₁ to d₂.

δ = ρ·g·L² / (2·E)

• Elongation of a bar hanging under its own weight; ρ: density (kg/m³).

ν = −ε_lateral / ε_axial

• Uniaxial stress.

ε_x = [σ_x − ν·(σ_y + σ_z)] / E (and cyclic for y, z)

• Generalised Hooke's law, isotropic material.

ε_v = ε_x + ε_y + ε_z; uniaxial ε_v = ε·(1 − 2ν); all-round pressure K = p / ε_v

• K: bulk modulus (Pa).

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

• Relations between elastic constants (isotropic).

Worked examples

Example 1 (standard): stepped bar. Given: a steel bar (E = 200 GPa) has a 300 mm length of 20 mm diameter and a 500 mm length of 30 mm diameter, and carries an axial pull P = 50 kN. Find the stresses and the total elongation.

1. Areas: A₁ = π/4 × 20² = 314.16 mm²; A₂ = π/4 × 30² = 706.86 mm².
2. σ = P/A: σ₁ = 50 000 / 314.16 = 159.2 MPa; σ₂ = 50 000 / 706.86 = 70.7 MPa.
3. δ = (P/E)·(L₁/A₁ + L₂/A₂) = (50 000 / 200 000) × (300/314.16 + 500/706.86) = 0.25 × (0.9549 + 0.7074) = 0.4156 mm. Answer: σ₁ = 159.2 MPa, σ₂ = 70.7 MPa, δ ≈ 0.416 mm. (Using N, mm and N/mm² keeps the units consistent.)

Example 2 (GATE level): dimensional and volume change. Given: a steel bar 2 m long with a 40 mm × 20 mm section carries an axial tension of 160 kN; E = 200 GPa, ν = 0.3. Find the changes in length, width, thickness and volume, and the values of G and K.

1. σ = 160 000 / (40 × 20) = 200 MPa (below typical yield for structural steel; check the actual grade).
2. Axial strain ε = σ/E = 200 / 200 000 = 0.001, so ΔL = 0.001 × 2000 = 2.0 mm.
3. Lateral strain = −ν·ε = −0.0003: Δb = −0.0003 × 40 = −0.012 mm; Δt = −0.0003 × 20 = −0.006 mm.
4. ε_v = ε·(1 − 2ν) = 0.001 × 0.4 = 0.0004. V = 2000 × 40 × 20 = 1.6 × 10⁶ mm³, so ΔV = 640 mm³ (an increase).
5. G = E / (2(1 + ν)) = 200 / 2.6 = 76.9 GPa; K = E / (3(1 − 2ν)) = 200 / 1.2 = 166.7 GPa. Answer: ΔL = +2.0 mm, Δb = −0.012 mm, Δt = −0.006 mm, ΔV = +640 mm³, G ≈ 76.9 GPa, K ≈ 166.7 GPa.

Common mistakes

• Unit slips: mixing N with kN, mm² with m², MPa with GPa. 1 MPa = 1 N/mm² is the safest working set.
• Using Hooke's law beyond the proportional limit.
• Treating E as strength. Stiffness and strength are different properties.
• Ignoring Poisson effects when stresses act in two or three directions.
• Forgetting the factor (1 − 2ν) in volumetric strain, or thinking a bar in tension loses volume (for ν < 0.5 it gains volume).
• Using an outside diameter where the net area at a hole or thread root should be used.

For GATE ME

Expect elongation of stepped, tapered and self-weight-loaded bars, relations between E, G, K and ν (including the limit ν = 0.5), volumetric strain under uniaxial or triaxial stress, and reading properties from a stress–strain curve. Practise working in N, mm and MPa and check whether ν values given are physically possible.

Quick check

1. E = 210 GPa and G = 80 GPa. Find ν.
2. A bar under uniaxial tension has ν = 0.5. What is its volumetric strain?
3. What stress corresponds to a strain of 0.0005 in aluminium with E = 70 GPa?
4. A 1 m steel rod of 10 mm diameter carries 10 kN. Find its elongation (E = 200 GPa).

Answers: 1. ν = E/(2G) − 1 = 0.3125. 2. Zero. 3. 35 MPa. 4. δ = 10 000 × 1000 / (78.54 × 200 000) = 0.64 mm.