Stress, strain and Hooke's law; elastic constants

Why it matters

Every member of a machine or structure is sized by asking two questions: is the stress below what the material can safely carry, and is the deformation small enough for the part to work? Stress, strain and the elastic constants connect loads to both answers, and they are the starting point for every later topic in strength of materials.

Key ideas

Stress. The internal force per unit area that one part of a body exerts on the neighbouring part across an imaginary cut.

• Normal stress σ acts perpendicular to the cut: tensile (positive) or compressive (negative).
• Shear stress τ acts along the cut.
• Average normal stress in an axially loaded bar is σ = P/A, valid away from load points and sudden changes of section (Saint-Venant's principle). Units: Pa = N/m²; in practice MPa = N/mm².

Strain. Deformation per unit length, dimensionless.

• Normal (linear) strain ε = δ/L.
• Shear strain γ is the change, in radians, of an originally right angle.
• Lateral strain is the strain perpendicular to the load; in tension the bar gets thinner.
• Volumetric strain εv = ΔV/V = εx + εy + εz for small strains.

Stress–strain curve of mild steel (tension test). Proportional limit (straight line), elastic limit, upper and lower yield points, strain hardening up to the ultimate tensile strength, then necking and fracture. Brittle materials (cast iron, concrete) show little plastic strain and no clear yield, so a 0.2% proof stress or the ultimate stress is used. Working stress = yield (or ultimate) stress divided by a factor of safety, taken from the relevant code or design data book.

Hooke's law. Up to the proportional limit, stress is proportional to strain: σ = E ε and τ = G γ. Beyond the elastic limit, part of the strain is permanent (plastic).

Poisson's ratio. ν = −(lateral strain)/(longitudinal strain). For steel ν ≈ 0.25–0.3, aluminium ≈ 0.33, rubber ≈ 0.5, cork ≈ 0. For an isotropic material −1 < ν ≤ 0.5; ν = 0.5 means the material is incompressible.

Generalised Hooke's law (isotropic, linear elastic). Each normal stress causes strain in its own direction and Poisson contraction in the other two.

Elastic constants. E (Young's modulus, axial stiffness), G (modulus of rigidity, shear stiffness), K (bulk modulus, resistance to volume change) and ν. For an isotropic material only two are independent; the others follow from the relations below. Typical values (check your data book): steel E ≈ 200–210 GPa, G ≈ 80 GPa; aluminium E ≈ 70 GPa; copper E ≈ 110–120 GPa.

Elongation of bars. For a uniform bar δ = PL/(AE). For stepped bars add the pieces; for tapering bars integrate. A bar hanging under its own weight stretches by ρ g L²/(2E), half of what the same total weight would cause if hung at the end.

Formulas

σ = P / A

• σ = normal stress (Pa), P = axial force (N), A = cross-sectional area (m²).

ε = δ / L

• δ = change in length (m), L = original length (m).

σ = E ε, τ = G γ

• E, G in Pa (GPa); γ in radians.

δ = P L / (A E) (prismatic bar, uniform axial force, within the proportional limit)

ν = − ε_lateral / ε_longitudinal

εx = [σx − ν(σy + σz)] / E (and similarly for εy, εz)

εv = εx + εy + εz = (1 − 2ν)(σx + σy + σz) / E

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

• K = bulk modulus (Pa).

δ = ρ g L² / (2E) (bar hanging under self-weight; ρ = density, kg/m³)

Worked examples

Example 1 (standard). A steel rod 2 m long and 20 mm in diameter carries an axial tensile load of 50 kN. E = 200 GPa, ν = 0.3. Find the stress, the elongation and the change in diameter.

1. Area: A = π/4 × 20² = 314.16 mm².
2. Stress: σ = P/A = 50 000 / 314.16 = 159.15 N/mm² = 159.15 MPa.
3. Strain: ε = σ/E = 159.15 / 200 000 = 7.958 × 10⁻⁴.
4. Elongation: δ = ε L = 7.958 × 10⁻⁴ × 2000 = 1.592 mm.
5. Lateral strain: −ν ε = −0.3 × 7.958 × 10⁻⁴ = −2.387 × 10⁻⁴.
6. Change in diameter: −2.387 × 10⁻⁴ × 20 = −0.00477 mm (a decrease).

Answer: σ = 159.2 MPa, δ = 1.59 mm, diameter decreases by 0.0048 mm.

Example 2 (GATE level). A steel cube of side 100 mm (E = 200 GPa, ν = 0.3) is loaded by σx = 100 MPa (tension) and σy = σz = −40 MPa (compression). Find the strain in x, the volumetric strain, the change in volume, and the values of G and K.

1. εx = [100 − 0.3(−40 − 40)] / 200 000 = (100 + 24)/200 000 = 6.2 × 10⁻⁴.
2. εy = εz = [−40 − 0.3(100 − 40)] / 200 000 = (−40 − 18)/200 000 = −2.9 × 10⁻⁴.
3. εv = 6.2 × 10⁻⁴ − 2 × 2.9 × 10⁻⁴ = 0.4 × 10⁻⁴ = 4 × 10⁻⁵. Check: (1 − 0.6)(100 − 80)/200 000 = 0.4 × 20 / 200 000 = 4 × 10⁻⁵. ✓
4. Change in volume: ΔV = εv V = 4 × 10⁻⁵ × 10⁶ mm³ = 40 mm³.
5. G = E / [2(1 + ν)] = 200 / 2.6 = 76.9 GPa; K = E / [3(1 − 2ν)] = 200 / 1.2 = 166.7 GPa.

Answer: εx = 6.2 × 10⁻⁴, εv = 4 × 10⁻⁵, ΔV = 40 mm³ (increase), G ≈ 76.9 GPa, K ≈ 166.7 GPa.

Common mistakes

• Mixing units: N with mm² gives MPa directly; N with m² gives Pa. Do not combine mm with m in one formula.
• Forgetting the Poisson terms in biaxial or triaxial loading.
• Using ε = σ/E beyond the proportional limit.
• Taking ν > 0.5 or treating G and K as independent of E and ν for an isotropic material.
• Reporting strain with a unit, or as a percentage without saying so.
• Using diameter instead of area, or the radius in πd²/4.

For GATE PI

Expect quick numericals on elongation of uniform, stepped and tapered bars, Poisson's effect on diameter, volumetric strain, and conversions among E, G, K and ν. Conceptual questions test the stress–strain curve (yield, proof stress, necking, ductile vs brittle) and the limits of ν. Practise working in N and mm throughout.

Quick check

1. E = 200 GPa and σ = 100 MPa. What is the strain?
2. If E = 2G, what is ν?
3. What is ν for an incompressible material?
4. A bar of length 1 m stretches by 0.5 mm. What is the strain?

Answers: 1. 5 × 10⁻⁴. 2. Zero. 3. 0.5. 4. 5 × 10⁻⁴.