3.3 Stress, Strain, Elasticity & Hooke's Law

Stress, Strain, Elasticity & Hooke's Law

Airframe and engine parts carry loads by developing internal forces distributed over cross-sections. Stress quantifies how intense those internal forces are; strain quantifies how much the material deforms. Module 2 requires clear definitions, the SI formulae, Hooke’s law, and the distinction between elastic and plastic behaviour — plus the classic exam trap of mixing up stress and strain.

Types of Loading

Tension

A tensile load pulls on a member along its axis, tending to elongate it. Control cables, bolt shanks in pure tension, and wing-spar lower caps in positive flight load are classic examples. Internal tensile stress is uniform over a straight prismatic bar if the load is axial and centric.

Compression

A compressive load pushes on a member, tending to shorten it. Landing-gear struts, spar upper caps in positive g flight, and engine mount tubes in some load cases carry compression. Slender compression members may buckle (sudden sideways deflection) at loads well below the material’s crushing strength — geometry and end fixity matter as much as material strength.

Shear

Shear loads tend to slide one part of a material over an adjacent part. Rivets and bolts in single or double shear, spar webs carrying beam shear, and adhesive bonds under in-plane load are shear-critical. Shear stress acts parallel to the section surface, not normal to it.

Torsion

Torsion is twisting about the longitudinal axis — a couple applied in a plane perpendicular to the axis. Drive shafts, torque tubes, and some actuator outputs are torsion members. Torsional shear stress varies with radius in a circular shaft (maximum at the outer fibre).

Combined loading (bending plus shear, tension plus torsion) is common in service; Module 2 focuses on recognising pure cases and applying F/A and ΔL/L correctly.

Stress = Force / Area

Normal stress (tension or compression) is:

σ = F / A

where F is the axial force (N) and A is the cross-sectional area (m²). The SI unit is the pascal (Pa) = 1 N/m². Engineering practice often uses megapascals (MPa): 1 MPa = 10⁶ Pa = 1 N/mm², which is convenient when F is in newtons and A in mm².

Shear stress uses the same form:

τ = F / A_shear

with A_shear the area resisting the sliding action (for a bolt in single shear, the shank cross-section; in double shear, twice that area shares the load).

Worked example — tensile stress. A steel tie rod carries 15 000 N. Cross-section diameter is 10 mm, so radius = 5 mm = 0.005 m, A = π(0.005)² ≈ 7.85 × 10⁻⁵ m².

σ = 15 000 / 7.85 × 10⁻⁵ ≈ 1.91 × 10⁸ Pa ≈ 191 MPa

Using mm: A = π(5)² ≈ 78.5 mm²; σ = 15 000 N / 78.5 mm² ≈ 191 N/mm² = 191 MPa. Same result.

Worked example — bolt in double shear. A 12 000 N load is carried by a bolt in double shear (two shear planes). Each plane carries 6 000 N. If shank area is 50 mm², τ = 6 000 / 50 = 120 N/mm² = 120 MPa on each plane.

Strain = Change in Length / Original Length

Direct (normal) strain is:

ε = ΔL / L

where ΔL is the change in length and L is the original length. Strain is a ratio and is therefore dimensionless (sometimes written as m/m or as a percentage: 0.001 = 0.1%). Tensile strain is positive elongation; compressive strain is shortening.

Worked example. A 2.0 m rod elongates by 0.8 mm under load. ΔL = 0.0008 m, L = 2.0 m, ε = 0.0008 / 2.0 = 0.0004 (or 400 microstrain).

Shear strain is the angular distortion (change in a right angle), usually small and expressed in radians; Module 2 emphasises the axial strain definition above.

Stress Versus Strain — The Exam Trap

• Stress = intensity of internal force (Pa).
• Strain = relative deformation (dimensionless).

A very stiff material (high E) can carry high stress with almost invisible strain. A soft elastomer can show large strain at low stress. Questions that ask “which quantity is measured in pascals?” or “which is ΔL/L?” are testing this distinction. Do not say “stress is how much it stretches.”

Elasticity and Hooke’s Law

A material is elastic if it returns to its original shape when the load is removed. Within the elastic range, many engineering metals obey Hooke’s law: stress is proportional to strain.

σ = E ε

or E = σ / ε

E is Young’s modulus (modulus of elasticity), a material property with units of stress (Pa or GPa). Typical order of magnitude: steels ≈ 200 GPa, aluminium alloys ≈ 70 GPa, titanium alloys ≈ 110 GPa. Higher E means stiffer — less strain for a given stress.

Rearrangement for design: if you know allowable stress and E, you can estimate elongation ΔL = (σ / E) × L = (F L) / (A E).

Worked example — elongation. Aluminium rod: L = 1.5 m, A = 100 mm² = 1 × 10⁻⁴ m², F = 10 000 N, E = 70 GPa = 7 × 10¹⁰ Pa.

σ = 10 000 / 1 × 10⁻⁴ = 1 × 10⁸ Pa
ε = σ / E = 1 × 10⁸ / 7 × 10¹⁰ ≈ 0.00143
ΔL = ε L ≈ 0.00143 × 1.5 ≈ 2.14 mm

Elastic Limit, Yield, and Ultimate Strength

On a stress–strain curve for a typical metal:

1. Proportional limit / elastic range — straight line (Hooke’s law); unloading returns to zero strain.
2. Elastic limit / yield region — beyond this, permanent (plastic) deformation remains after unloading. Design usually keeps working stress below yield with a safety factor.
3. Ultimate tensile strength (UTS) — maximum stress on the curve; beyond this the specimen necks and load falls until fracture.
4. Fracture / breaking strength — stress at complete separation (may be lower than UTS after necking).

Aircraft structures are designed with limit loads (maximum expected in service) and ultimate loads (limit × factor of safety, often 1.5 for transport aeroplanes) so that the structure withstands ultimate load without failure for a short duration, and limit load without permanent harmful deformation.

Shear Modulus (Brief)

For pure shear in the linear range, τ = G γ, where G is the shear modulus (modulus of rigidity) and γ is shear strain. G is another material stiffness property, smaller than E for isotropic metals (related by Poisson’s ratio). Recognising that torsion and shear use G rather than E is enough at Module 2 depth.

Maintenance Links

• Torque values and fastener grip lengths protect against tensile overload and shear tear-out.
• Corrosion reduces effective cross-section A, raising stress for the same load — a reason corrosion limits are structural, not cosmetic.
• Over-torquing bolts can yield threads (plastic deformation) even if the joint does not break immediately.
• Hardness and heat-treatment affect yield and ultimate strength; wrong alloy temper changes allowable stress.

Keep formulae tight: σ = F/A, ε = ΔL/L, E = σ/ε in the elastic range; name the load type (tension, compression, shear, torsion); and never swap stress with strain.