Landing Gear Loads

Landing Gear Loads

Landing Gear Loads

What the tool computes

The landing gear is sized by the energy it must absorb when the airplane meets the ground, and the airframe around it by a short list of prescribed load conditions built from that impact. The tool follows the airworthiness rules in the order a loads engineer uses them: the limit descent velocity and the wing lift the rule allows during the impact; the energy balance of the drop test, which gives the vertical ground reaction of the main gears and the two load factors the rules speak of; the landing conditions (level, three-point, tail-down, one-wheel, side and drift loads, spin-up and spring-back drag) and the ground-handling conditions (braked roll, dynamic braking, turning, nose-wheel loads, towing, jacking, rebound) as vertical, drag and side loads on each gear unit; the loads these put into the trunnion of a cantilever strut; and, as options, the wheel spin-up drag of Appendix D to Part 23, the polytropic air curve of an oleo-pneumatic strut, and the generalized impact solution of Milwitzky and Cook, which estimates the peak load, the stroke and the strut efficiency from the orifice and tire data. The rules are 14 CFR Part 23 as it stood in the 2016 edition (Amendment 23-62, the last with the prescriptive ground-load sections and the Appendix C table of basic landing conditions; Amendment 23-64 of 2017 moved them to accepted means of compliance) and 14 CFR Part 25 as current. Every constant on the page is quoted from the section that prescribes it.

Descent velocity

§ 23.473(d): the limit vertical inertia load factor may not be less than that obtained when landing with a descent velocity V = 4.4 (W/S)^1/4 ft/s, W/S the wing loading in lb/ft², except that V need not be more than 10 ft/s and may not be less than 7 ft/s. § 23.725(a) states the same thing as a free-drop height h = 3.6 (W/S)^1/2 in, not less than 9.2 in nor more than 18.7 in: since V = √(2gh) = √(2 × 32.17 × 0.3) (W/S)^1/4 = 4.39 (W/S)^1/4 ft/s, the two formulas are one, and the 9.2 and 18.7 in bounds are the 7 and 10 ft/s bounds. § 25.473(a) prescribes 10 ft/s at the design landing weight and 6 ft/s at the design take-off weight. The reserve-energy case is 1.2 times the limit velocity for Part 23 (§ 23.723(b); § 23.727 writes it as 1.44 times the drop height) and 12 ft/s for Part 25 (§ 25.723(b)), both with wing lift equal to the weight; the gear may yield but may not fail. An entered velocity replaces the rule value.

Wing lift during the impact

§ 23.473(e) lets wing lift not exceeding two-thirds of the weight be assumed to exist throughout the impact and to act through the centre of gravity, and defines the ground reaction load factor as the inertia load factor minus the ratio of that lift to the weight. § 25.473(b) allows lift up to the whole weight. In the drop test the lift is simulated by dropping a reduced effective weight (§ 23.725(b)), as the next section shows.

The energy balance

With the nose wheel just clear of the ground the main gear units absorb the whole impact. Let W be the landing weight, V the descent velocity, L the wing lift, S_s the vertical travel of the axle from the strut stroke (the stroke times the cosine of the rake angle) and S_t the tire deflection at the maximum load. The airplane arrives with the kinetic energy ½ (W/g) V² and, while the gear compresses, its weight less the lift does the work (W − L)(S_s + S_t). The gear reaction F rises from zero to its maximum over the travel; the work it absorbs is written as the maximum force times the travel times an efficiency, η_s for the strut and η_t for the tire, so that

½ (W/g) V² + (W − L)(S_s + S_t) = F (η_s S_s + η_t S_t)

and F follows at once. The efficiency, also called the planimetric ratio or (in NACA Report 1154) the effectiveness, is the area under the load-stroke curve divided by the enclosing rectangle: 0.80 to 0.90 for an oleo-pneumatic strut with a metering pin, about 0.70 for a plain orifice (Report 1154, Figure 11(h), reproduced by the tool's own integration), 0.50 for a linear spring, 0.45 to 0.47 for a pneumatic tire. Two load factors follow: the ground reaction load factor n_g = F/W, which is what the gear and its attachments feel, and the inertia load factor at the centre of gravity n = n_g + L/W, which is what the airframe and its contents feel and the quantity the rules call n. § 23.473(g) requires n ≥ 2.67 and n_g ≥ 2.0 at the design maximum weight unless taxiing loads are shown to be lower. The reaction is shared equally by the main gear units; the ultimate loads are 1.5 times the limit loads (§ 23.303, § 25.303).

The rules write the same balance as a drop test on one gear unit (§ 23.725). The unit carries its static weight W_M (the whole weight divided among the main units with the nose wheel clear), is dropped from the height h = V²/(2g) with the effective weight W_e = W_M (h + (1 − L) d)/(h + d), d being the tire deflection plus the vertical axle travel, and the recorded load factor n_j = F/W_e is converted to the airplane's inertia factor by n = n_j W_e/W_M + L (§ 23.725(e)). Multiplying out, W_e (h + d) = W_M h + W_M (1 − L) d is exactly the energy of the balance above, so the two forms agree identically; the page prints both so that a drop-test report can be compared with the analysis.

The landing conditions

Part 23 tabulates its basic landing conditions in Appendix C with n the inertia load factor selected under § 23.473, L the lift ratio and K a drag factor (0.25 for airplanes of 3,000 lb or less, 0.33 for 6,000 lb or more, linear between). In the level landing with the nose wheel just clear (§ 23.479(a)(2)(ii)) the main wheels together carry (n − L) W vertically and K n W in drag; in the three-point landing with inclined reactions (§ 23.479(a)(2)(i)) the same loads divide between the mains and the nose by the lever arms a′/d′ and b′/d′ of the appendix's figure, which the tool takes as A/(A + B) and B/(A + B) with the dimensions of the Part 25 appendix; in the tail-down landing (§ 23.481) the reactions are vertical, the wheels being up to speed, and the airplane sits at the stalling angle or the clearance angle, whichever is less; the one-wheel landing (§ 23.483) puts the level-landing loads of one side on one gear; the side-load condition (§ 23.485) is 1.33 W vertical divided equally between the mains with 0.5 W acting inboard on one side and 0.33 W outboard on the other, the struts and tires in their static positions. § 23.479(b) requires the spin-up drag to be combined with the vertical reaction at the same instant and the spring-back (the forward load when the spin-up drag is suddenly relieved) with the vertical at the peak forward load, assuming a tire friction coefficient of 0.8, the drag never less than 25 % of the maximum vertical reaction; the table's K n W is that minimum.

Part 25 prescribes the level landing on the mains (§ 25.479(c)(1)) with an aft drag component of not less than 25 % of the maximum vertical reaction acting at the axle (§ 25.479(d)(1)), the lateral-drift landing of § 25.479(d)(2) (75 % of the maximum vertical with 40 % drag and 25 % side), the nose-and-mains landing if reasonably attainable (§ 25.479(c)(2); the rule gives no distribution, and the tool divides the reaction by the lever arms and says so), the tail-down landing (§ 25.481; the tool applies the 25 % drag to it as well, since the rule asks for the spin-up analysis it does not have), the one-gear landing (§ 25.483), the side loads of § 25.485 (0.8 of the vertical inward on one side and 0.6 outward on the other, combined with one-half of the maximum level-landing vertical reactions) and the rebound load of § 25.487 (20 g on the unsprung parts with the gear extended). Part 25 requires the landing loads to come from a dynamic analysis validated by drop tests (§ 25.473(c), § 25.723); the energy balance is the preliminary-design estimate of that analysis.

The ground-handling conditions

Ground-handling conditions are taken with no wing lift and the struts and tires static (§ 25.489). Braked roll (§ 23.493 with n = 1.33, § 25.493 with n = 1.2 at the landing weight) applies a drag of 0.8 times the vertical reaction at each braked wheel in the level-landing attitudes. In the three-point attitude with zero pitching acceleration the drag at the ground and the inertia force at the centre-of-gravity height E transfer load to the nose; moments about the nose contact give V_m = n W A/(A + B + μE) on the mains and V_n = n W − V_m on the nose. § 25.493(d)–(e) adds the dynamic pitching of sudden braking, V_N = W_T/(A + B) [B + f μ A E/(A + B + μE)] with f = 2.0 unless a lower factor is substantiated; setting f = 1 recovers the static balance, so f multiplies only the braking increment. § 25.495 (turning): 1.0 g vertical and 0.5 g lateral at the centre of gravity with the side reaction of each wheel 0.5 of its vertical; the rolling moment 0.5 W E transfers 0.5 W E/T between the outer and inner mains. § 25.499(a) (nose-wheel yaw): side load 0.8 of the static nose reaction. § 25.507 (reversed braking): forward loads of 0.55 of the vertical at each braked wheel. § 23.499 (nose-wheel supplementary conditions): 2.25 times the static nose load with 0.8 of it aft, 0.4 of it forward or 0.7 of it sideways. § 23.507 (jacking): 1.35 times the static reactions with 0.4 of them horizontally. Towing: 0.225 W at the main gear and 0.3 W at the auxiliary gear for Part 23 (§ 23.509); for Part 25 the tow load F_TOW is 0.3 W_T below 30,000 lb, (6 W_T + 450,000)/70 between 30,000 and 100,000 lb and 0.15 W_T above, applied as 0.75 F_TOW per main gear unit and 1.0 F_TOW at the auxiliary gear (§ 25.509). The geometry symbols A, B, E and T are those of Figure 1 of Appendix A to Part 25.

Trunnion loads

Each condition's ground loads are carried to the trunnion of a cantilever strut: the strut axis is inclined by the rake φ from the vertical (positive with the axle forward of the trunnion), the axle lies l along the axis below the trunnion and t aft of it (the trail), and the ground contact is r below the axle. The ground loads are first expressed in airplane axes at the pitch attitude θ of the condition, F_x = D cos θ − V sin θ aft and F_z = D sin θ + V cos θ up (so that in a tail-down landing a vertical reaction has a forward component on the strut), then resolved along the strut: axial P = F_x sin φ + F_z cos φ, shear Q = F_x cos φ − F_z sin φ, side shear S. The bending moment at the trunnion in the fore-and-aft plane is the moment of (F_x, F_z) acting at the contact point, positive when it pushes the wheel aft; the side load gives a bending moment S (l + r cos φ) in the side plane and a torsion S (t + r sin φ) about the strut axis. The largest axial load and bending moment are named with their conditions, and an entered ultimate allowable gives the axial margin of safety.

Wheel spin-up (Appendix D to Part 23)

The drag that spins the wheel up to the ground speed is, by the NACA TN 863 method the appendix adopts, F_Hmax = (1/r_e) √(2 I_w (V_H − V_c) μ F_Vmax/t_s): the drag impulse ∫μF dt over the spin-up time must equal the wheel's change of angular momentum I_w (V_H − V_c)/r_e, and with the vertical load rising linearly to F_Vmax in t_s the drag at the end of spin-up is as written. It need not exceed 0.8 F_Vmax; the spring-back load is the spin-up load reversed. The tool combines each with the vertical reaction at the instant of spin-up and adds them to the condition table when the option is on.

The strut air curve

An oleo-pneumatic strut carries the static load on its air spring and the dynamic load mostly on the hydraulic resistance of its orifice. The air is compressed polytropically from the extended volume V₀ at the gauge pressure p₀: F_a(s) = A_a [(p₀ + p_atm)(V₀/(V₀ − A_a s))ⁿ − p_atm] (NACA TN 2755, eq. 3, with the ambient pressure acting on the outside of the piston), n = 1.0 for a slow isothermal compression, about 1.12 in drop tests, 1.3 for a fast near-adiabatic one. The tool draws the curve, locates the static stroke, gives the compression ratio and the pressure at full stroke, and integrates the energy the air alone stores (closed form of the polytropic integral) with its planimetric ratio. The curve should pass below the energy-balance load at full stroke: if it reaches it earlier the strut carries that load on air alone and the hydraulic force must fall away near the end of the stroke for the entered efficiency to be realised.

The generalized impact of Milwitzky and Cook

NACA TN 2755 (1952, reissued as Report 1154) derives the equations of motion of a two-mass landing gear with an orifice strut, a polytropic air spring, bearing friction and a nonlinear tire, compares them with drop tests, and shows that for many purposes the air spring, the lower mass and friction may be neglected and the tire taken as linear. The simplified system then reduces to one dimensionless form: with u = z A g/W₁, θ = t √(a g/W₁), A = ρ A_h³/(2 (C_d A_o)²) the hydraulic constant (strut force A ṡ²) and a the tire stiffness,

(u₁′ − u₂′)² + u₁″ = 0,   (u₁′ − u₂′)² − u₂ = 0,   u₁″ + u₂ = 0

whose only parameter is the initial velocity u₀′ = V √(A² g/(W₁ a)) (eq. 21 and Figure 11 of the report). The tool integrates these equations for the entered orifice, fluid and tire data with the sprung weight per unit as W₁, and scales back to the peak strut force, the stroke, the tire deflection, the time to the peak and the strut and landing-gear effectiveness of the report's Figure 11(h) (about 0.70 to 0.73 and 0.70 to 0.60 over the range). Lift equal to the weight is built into the model, as in a drop test with lift simulation. The report's comparison with the exact two-mass solution for its test gear (2.50 g against 2.73 g at 8.86 ft/s, Figure 12) is reproduced in the tool's test script, which also checks the integration against an independent scipy solution. The estimate tells whether the efficiency entered in the energy balance is plausible for the strut at hand: a plain orifice cannot exceed about 0.73, and a higher value presumes a metering pin.

Default example

The default is a 2,400 lb Part 23 airplane with 160 ft² of wing (W/S = 15 lb/ft²), two main gears with a 6 in vertical stroke at 80 % efficiency and 2 in of tire deflection at 47 %, lift of two-thirds of the weight, a nose wheel 57 in ahead of the centre of gravity and the mains 8 in behind it, a 48 in c.g. height and a 100 in track. The rule gives V = 4.4 × 15^1/4 = 8.66 ft/s (h = 13.9 in); the kinetic energy is 2,797 ft-lb and the potential term 533 ft-lb, so F = 3,330/(0.80 × 0.5 + 0.47 × 0.1667) = 6,961 lb on both mains, 3,481 lb per gear, n_g = 2.90 and n = 3.57, with 84 % of the energy in the struts; the drop-test form gives W_e = 909 lb and n_j = 3.83; the reserve-energy case at 10.39 ft/s gives 4,209 lb per gear. The level landing puts 3,481 lb vertical and 1,070 lb drag (K = 0.25) on each main; the braked roll with the mains only gives the largest trunnion bending moment, 35,100 lb-in. Every number is checked in the test script against the rule text, closed forms and an independent numpy computation.

Limits

The energy balance is a rigid-airplane, single-degree-of-freedom estimate: it does not know the load-stroke history, tire bottoming, strut friction or the forward-speed dynamics, and Part 25 requires the full dynamic analysis of § 25.473(c) validated by the tests of § 25.723. The three-point impact is treated by the lever-arm distribution of the appendix table, not by a separate energy balance for the nose gear. Multiple-wheel and bogie units (§ 25.511), pivoting (§ 25.503), tail-wheel airplanes, skis and floats, flexible-airframe response and the nose gear's own attachment loads are not covered. The conditions are limit loads; the structure is designed to the ultimate factor entered.