Stress, strain and Hooke's law

Why it matters

Every structural check on an aircraft part, from a wing spar cap to an engine mount bolt, starts by converting loads into stresses and deformations. Stress tells you whether the material will yield or break; strain and deflection tell you whether the part stays stiff enough to do its job. Hooke's law is the link between the two, and almost every later topic in this subject is built on it.

Key ideas

Stress is the internal force per unit area that one part of a body exerts on the neighbouring part across an imaginary cut. When a bar of area A carries an axial force P, the normal stress on a cross-section is σ = P/A. Tension is taken positive, compression negative. A force acting parallel to the cut produces shear stress τ. Stress is not a vector: its value depends on the plane you cut on (see the stress-transformation topic).

Strain measures deformation and is dimensionless. Normal strain ε = δ/L is the change in length per unit original length. Shear strain γ is the change, in radians, of an angle that was originally 90°.

Hooke's law. For most engineering metals loaded below the proportional limit, stress is directly proportional to strain: σ = E·ε in tension or compression and τ = G·γ in shear. E is Young's modulus and G the shear modulus. Both are material properties, not geometry: a thick and a thin steel bar have the same E, roughly 200–210 GPa, while aluminium alloys are about 70 GPa and titanium alloys about 110 GPa (take exact values from your data book).

The tensile test curve (mild steel) shows, in order: the proportional limit (end of the straight line), the elastic limit (largest stress with no permanent set), the upper and lower yield points, strain hardening up to the ultimate tensile strength, then necking and fracture. Aluminium alloys have no distinct yield point, so a 0.2 % proof stress (offset method) is used instead. Ductile materials show large plastic strain before fracture; brittle materials (cast iron, ceramics, many composites) fracture with little plastic strain.

Engineering versus true stress. Engineering stress uses the original area A₀; true stress uses the current area. In the elastic range the difference is negligible; after necking it is large.

Assumptions behind σ = P/A: the load acts along the centroidal axis, the section is away from load points and abrupt changes of section (Saint-Venant's principle), and the material is homogeneous. Near holes and fillets the local stress is higher, by a stress concentration factor.

Axial deformation. Combining σ = P/A, ε = δ/L and σ = E·ε gives δ = PL/(AE). The product AE is the axial rigidity; L/(AE) is the flexibility of the bar. For bars with several segments, add the segment elongations using the internal force in each segment (draw free-body diagrams to find them).

Formulas

σ = P / A σ = normal stress (Pa), P = axial force (N), A = cross-sectional area (m²). Uniform stress on a section away from load points.

τ = V / A τ = average shear stress (Pa), V = shear force on the plane (N), A = sheared area (m²). Average value only (bolts, pins, rivets).

ε = δ / L ε = normal strain (–), δ = change in length (m), L = original length (m).

σ = E·ε and τ = G·γ E = Young's modulus (Pa), G = shear modulus (Pa), γ = shear strain (rad). Valid up to the proportional limit.

δ = P·L / (A·E) Elongation of a prismatic bar with constant force, area and modulus.

δ = Σ Pᵢ·Lᵢ / (Aᵢ·Eᵢ) Stepped or composite-in-series bar; Pᵢ is the internal force in segment i (tension +).

δ = 4·P·L / (π·E·d₁·d₂) Linearly tapered circular bar with end diameters d₁ and d₂ (m).

δ = ρ·g·L² / (2·E) Elongation of a hanging bar under its own weight; ρ = density (kg/m³), g = 9.81 m/s².

Worked examples

Example 1 (standard). A steel tie rod of diameter 20 mm and length 1.5 m carries an axial tensile load of 50 kN. E = 200 GPa. Find the stress, strain and elongation.

1. Area: A = π·d²/4 = π × 20² / 4 = 314.16 mm².
2. Stress: σ = P/A = 50 000 N / 314.16 mm² = 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 × 1500 mm = 1.194 mm.

Answer: σ ≈ 159 MPa, ε ≈ 7.96 × 10⁻⁴, δ ≈ 1.19 mm.

Example 2 (GATE level). A stepped bar is fixed at end A. Segment AB is aluminium (A = 600 mm², L = 1.0 m, E = 70 GPa); segment BC is steel (A = 300 mm², L = 0.6 m, E = 200 GPa). A force of 20 kN acts at B towards A, and a force of 30 kN acts at the free end C away from A. Find the elongation of C relative to A.

1. Internal force in BC (cut between B and C, keep the part to the right): P_BC = +30 kN (tension).
2. Internal force in AB (cut between A and B, keep the part to the right): P_AB = 30 − 20 = +10 kN (tension).
3. Formula: δ = Σ Pᵢ·Lᵢ / (Aᵢ·Eᵢ).
4. AB: 10 000 × 1000 / (600 × 70 000) = 0.2381 mm.
5. BC: 30 000 × 600 / (300 × 200 000) = 0.3000 mm.
6. Total: 0.2381 + 0.3000 = 0.5381 mm.

Answer: δ_C ≈ 0.538 mm (extension). Note that using the applied loads instead of the internal forces would give a wrong answer.

Common mistakes

• Using the applied loads instead of the internal force in each segment of a stepped bar.
• Mixing units: N with mm² gives MPa directly (1 N/mm² = 1 MPa), but N with m² gives Pa. Keep one consistent set.
• Using diameter instead of radius (or forgetting the 1/4) in the area of a circle.
• Applying Hooke's law beyond the proportional limit, or assuming E changes with the size of the bar.
• Quoting strain in percent in one place and as a pure number in another.
• Forgetting that compression is negative, which flips the sign of δ.

For GATE AE

Expect short numericals on stress and elongation of prismatic, stepped and tapered bars, average shear stress in pins and rivets, and conceptual questions on the stress–strain curve (proportional limit, proof stress, ductile versus brittle behaviour). Practise drawing the axial-force diagram quickly, keeping units in N and mm, and recognising when δ = PL/(AE) has to be summed or integrated.

Quick check

1. A 10 kN load acts on a 100 mm² bar. What is the stress in MPa?
2. Why is a 0.2 % proof stress used for aluminium alloys?
3. Does Young's modulus depend on the diameter of the bar?
4. A bar 2 m long extends by 1 mm. What is the strain?

Answers: 1. 100 MPa. 2. They have no distinct yield point, so an offset of 0.2 % plastic strain defines yield. 3. No, E is a material property. 4. 5 × 10⁻⁴.