Aircraft loads: flight, ground and inertial

Why it matters

Every spar, skin panel, frame and fitting in an aircraft is sized for a specific list of load cases: manoeuvres, gusts, landing impact, taxiing, pressurisation. If a load case is missed or its magnitude underestimated, the structure is either unsafe or carries weight it does not need, and weight is what the whole aircraft pays for. Load analysis is therefore the first step of structural design and the input to everything else in this subject.

Key ideas

Classes of load.

• Flight (aerodynamic) loads: lift, drag and pitching moment distributed over the wing, tail and fuselage, plus control-surface loads. They vary with speed, altitude, angle of attack and configuration.
• Ground loads: landing impact, braking, taxiing over rough ground, turning on the ground, towing and jacking. They enter the structure through the landing gear and its attachment points.
• Inertial loads: when the aircraft accelerates, every item of mass (structure, fuel, engines, payload) needs a force to accelerate it. Seen from the aircraft, each mass m carries a combined weight-plus-inertia load n·m·g acting opposite to the acceleration (downward in a pull-up). Inertia loads relieve the wing bending caused by lift: fuel and engines spread along the span reduce the root bending moment.
• Other loads: cabin pressurisation (hoop and longitudinal stress in the fuselage), engine thrust and torque, thermal loads, and loads from systems.

Load factor. The load factor is n = L/W (strictly, the aerodynamic force normal to the flight path divided by weight). In steady level flight n = 1. In a manoeuvre every part of the aircraft "feels" n times its weight, so structure is designed in terms of n. Load factor is dimensionless and is often written in "g".

• Level coordinated turn at bank angle φ: n = 1/cos φ.
• Symmetric pull-up at the bottom of a vertical circle of radius R: n = 1 + V²/(g·R).

Limit and ultimate loads. The limit load is the largest load expected in service. The structure must carry it without permanent deformation. The ultimate load = factor of safety × limit load, and the structure must carry it without failure for a short time. The airworthiness codes (FAR/CS 23 and 25) use a factor of safety of 1.5. Limit manoeuvre load factors are set by the code for each category: for a transport aircraft typically +2.5 and −1.0. Take exact values for a given category from the code.

V–n diagram (flight envelope). A plot of load factor against equivalent airspeed. At low speed the curve is limited by stall: n = ρ·V²·S·C_Lmax/(2W), a parabola in V. At the speed where this parabola meets the positive limit load factor, the design manoeuvring speed V_A = V_s·√n_max, the aircraft can just reach limit load before stalling. Above V_A the limit load factor caps the envelope, up to the design dive speed V_D. The structure must be strong enough at every point on the boundary. Gust lines are drawn on the same diagram.

Gust loads. A sharp-edged vertical gust of velocity U raises the angle of attack by about U/V, which adds lift ΔL = ½·ρ·V²·S·a·(U/V). Dividing by W gives an increment Δn that rises linearly with speed and with lift-curve slope and falls with wing loading W/S. Light aircraft with high aspect ratio wings, flying fast in turbulence, see the largest gust load factors. Real gusts build up gradually and the aircraft starts to respond, so codes apply a gust alleviation factor K (less than 1) whose expression is given in the code.

Ground loads. In a landing the gear shock absorber must remove the vertical kinetic energy of the aircraft over its stroke. A longer stroke gives a smaller deceleration and smaller gear reaction. It is common to assume the wing still carries lift equal to the weight at touchdown.

Formulas

n = L / W

• L: lift (N), W: weight (N), n: load factor (dimensionless). Valid for any flight condition; n = 1 in steady level flight.

F_inertia = n·m·g

• m: mass of an item (kg), g = 9.81 m/s², F_inertia: inertia load (N) acting opposite to the acceleration. Use for engines, fuel, payload and the structure itself.

n = 1 / cos φ

• φ: bank angle. Level coordinated turn only.

n = 1 + V² / (g·R)

• V: speed (m/s), R: radius of the flight path (m). Bottom of a pull-up.

n_stall = ρ·V²·S·C_Lmax / (2·W) and V_A = V_s·√n_max

• ρ: air density (kg/m³), S: wing area (m²), C_Lmax: maximum lift coefficient, V_s: 1-g stall speed (m/s), n_max: positive limit load factor.

Δn = ρ₀·U_e·V_e·a·K / (2·W/S)

• ρ₀ = 1.225 kg/m³ (sea-level density), U_e: equivalent gust velocity (m/s), V_e: equivalent airspeed (m/s), a: lift-curve slope (per rad), K: gust alleviation factor (K = 1 for a sharp-edged gust), W/S: wing loading (N/m²). Gust load factor n = 1 ± Δn.

Ultimate load = 1.5 × limit load

• Factor of safety per FAR/CS 23 and 25.

a = v² / (2·s) and R_gear = m·a (lift = weight at touchdown)

• v: vertical sink speed (m/s), s: effective shock-absorber stroke (m), a: mean deceleration (m/s²), R_gear: total vertical gear reaction (N).

Worked examples

Example 1 (standard): turn and engine inertia load. Given: a transport of weight W = 600 kN in a level coordinated turn at φ = 60°. Each wing engine has mass 3000 kg.

1. Load factor: n = 1 / cos φ = 1 / cos 60° = 1 / 0.5 = 2.0.
2. Lift required: L = n·W = 2.0 × 600 kN = 1200 kN.
3. Inertia load of one engine on its pylon: F = n·m·g = 2.0 × 3000 kg × 9.81 m/s² = 58 860 N.
4. This load acts downward relative to the wing, opposite to the lift, and so relieves wing root bending. Answer: n = 2.0, L = 1200 kN, engine inertia load = 58.9 kN.

Example 2 (GATE level): gust load factor and envelope. Given: wing loading W/S = 4000 N/m², lift-curve slope a = 5.0 /rad, C_Lmax = 1.5, the aircraft flies at V_e = 150 m/s EAS and meets a sharp-edged vertical gust U_e = 10 m/s EAS (K = 1). Positive limit manoeuvre load factor n_max = 2.5.

1. Gust increment: Δn = ρ₀·U_e·V_e·a / (2·W/S) = 1.225 × 10 × 150 × 5.0 / (2 × 4000) = 9187.5 / 8000 = 1.148.
2. Gust load factors: n = 1 ± 1.148, i.e. +2.15 (upgust) and −0.15 (downgust). The upgust case is below the manoeuvre limit of 2.5, so the manoeuvre case still governs at this speed.
3. 1-g stall speed at sea level: V_s = √(2·(W/S) / (ρ₀·C_Lmax)) = √(2 × 4000 / (1.225 × 1.5)) = √4353.7 = 65.98 m/s.
4. Design manoeuvring speed: V_A = V_s·√n_max = 65.98 × √2.5 = 104.3 m/s. Answer: n_gust = +2.15 / −0.15; V_s = 66.0 m/s; V_A = 104.3 m/s.

Example 3 (landing): gear reaction. Given: m = 50 000 kg, sink speed v = 3.0 m/s, effective stroke s = 0.40 m, lift = weight at touchdown.

1. Mean deceleration: a = v² / (2·s) = 3.0² / (2 × 0.40) = 11.25 m/s².
2. Gear reaction: R = m·a = 50 000 × 11.25 = 562 500 N.
3. As a multiple of weight: R/W = 11.25 / 9.81 = 1.15. Answer: R ≈ 563 kN (1.15 W).

Common mistakes

• Treating load factor as a force. n is dimensionless; the load is n·W.
• Applying inertia loads in the direction of the acceleration. In the aircraft's frame they act opposite to it, which is why wing-mounted masses relieve wing bending.
• Using n = 1/cos φ for a turn that is not level and coordinated, or forgetting the "1 +" in the pull-up formula.
• Mixing true and equivalent airspeed in the gust formula: with ρ₀ both U and V must be EAS.
• Designing to ultimate load for "no permanent deformation". Limit load is the yield criterion; ultimate (1.5 × limit) is the failure criterion.
• Forgetting the negative side of the envelope: downgusts and push-overs load the structure in reverse, and compression panels change sides.

For GATE AE

Expect short numericals: load factor in a turn or pull-up, lift and inertia loads from n, stall speed and manoeuvring speed from the V–n diagram, and the sharp-edged gust increment. Conceptual questions test limit versus ultimate load, the factor of safety of 1.5, and why inertia relief reduces wing root bending. Practise reading a V–n diagram and doing the gust formula with consistent units.

Quick check

1. An aircraft is in a level turn at 45° bank. What is n?
2. Why do wing-mounted engines reduce wing root bending moment in flight?
3. Doubling wing loading, with everything else fixed, changes the gust increment Δn how?
4. What must the structure do at limit load and at ultimate load? Answers: 1. n = 1/cos 45° = 1.41; 2. their inertia load n·m·g acts opposite to lift; 3. Δn halves; 4. no permanent deformation at limit load, no failure at ultimate load (1.5 × limit).