Figure 7.16. Bizjet undercarriage positions

In all cases, the nose wheel attachment point remains unchanged but the wheel base and wheel track change with new attachment points for the main wheels on the wing or elsewhere.

The industry conducts a trade-off study to examine which options offer the maximum cost benefit to operators. This book uses the second option – that is, start with the biggest variant shown in Figure 7.9. Iteration is likely to occur after accurate sizing (see Chapter 11).

7.14 Worked-Out Examples

The worked-out example continues with the aircraft configuration developed in Chapter 6 (both civil and military). The heaviest and the longest in the family is the most critical from the undercarriage design perspective (see Figures 7.17 and 7.18 for undercarriage positioning).

7.14.1 Civil Aircraft: Bizjet

The largest 14/16 passenger variant design is the most critical to maintain for undercarriage commonality. For the largest variant, the wheel base is 28.54 ft (8.66 m), β = 15, and γ = 14 deg. The undercarriage and tire sizing for the family of civil aircraft variants are more involved procedures than combat aircraft because the fuselage length and weight changes are relatively high, affecting both the wheel base and the strut load.

The undercarriage is the tricycle type and retractable. The aftmost CG position is considered first, placed at 35% (40% is a limiting situation) of the wing MAC. It will be revised when the CG is determined from actual weight computation. The nose wheel load is based on the forwardmost CG position.

The growth version of the MTOM = 11,000 kg (24,250 lb) (at this time, it is estimated from the baseline MTOM of 9,500 kg [21,000 lb]). It will be subsequently sized to a more accurate mass. The wing and empennage sizes are the same as for the baseline aircraft. The main-wheel attachment point is at a reinforced location on the rear-wing spar and is articulated with the ability to fold inward for retraction. It is assumed that there is no problem for stowage space in the fuselage with adequate wheel clearance. The following dimensions are of interest; the large growth variant is shown in Figure 7.16. At this time, the CG position is estimated and is refined in Chapter 8 when the undercarriage position is iterated.

Figure 7.17. AJT undercarriage positions

1.Fuselage length = 17.74 m (58.2 ft). The wing is positioned as shown in Figure 7.17.

2.The aftmost CG position is estimated at 10 m (33 ft) from the zero reference plane and 1.83 m (6 ft) above the ground, just below the centerline.

3.The forwardmost CG position is estimated at 8.8 m (28.87 ft) from the zero reference plane.

4.Place the main wheel at 55% of the wing MAC and check that the fuselage clearance angles, γ = 14 deg and β = 15 deg, are sufficient for rotation. This measures 10.4 m (34.1 ft) from the zero reference plane with a height of 1 m from the wing attachment point to the ground in a normal loading condition. The ground clearance from the bottom of the fuselage is 0.914 m (3 ft). The main wheel extends 1 ft in a free unloaded situation (i.e., it clears the fuselage aft end at rotation).

5.The nose wheel is kept at the front bulkhead at 1.7 m (5.6 ft) from the zero reference plane. It folds forward into the nose-cone bay and extends 22.86 cm (9 in) in the free unloaded situation.

6.The wheel base is computed as lBASE = 10.4 − 1.7 = 8.7 m (26.2 ft).

7.The wheel track = 2.9 m (9.5 ft). Use Figure 7.3 to compute the turn-over angle θ . It results in a low angle of 40 deg in a very stable aircraft.

8.The CG angle β can be worked out as tan−1[(10.4–9.4)/1.83]. This results in β = 28.6 deg, a safe angle well over 15 deg.

Wheel-loading, LCN, and tire sizing (use Figure 7.17 to compute wheelloading):

The main-wheel load is computed at the aftmost CG, which gives lREAR = 9.4– 1.7 = 7.7 m (25.26 ft).

Equation 7.2 gives RMAI N = (lREAR × MTOW)/ lBASE = (7.7 × 11,000)/8.7 = 9,736 kg (21,463 lb).

The load per strut is 4,868 kg (10,732 lb). It is better to keep the wheel load below 10,000 lb in order to have a smaller wheel and tire.

Then, make the twin-wheel arrangement. For this arrangement, Equation 7.5 gives the ESWL = 4,868/1.5 = 3,245 kg (7,155 lb).

A typical airfield LCN for this aircraft class is low, anywhere from 10 to 20. Table 7.6 provides several options. From the tire catalogs (see Appendix E), a suitable match is the New Tire Type (i.e., Type VII, with inch code) with a designation of 22 × 6.6–10 (18-ply), an inflation pressure of 260 psi, and a maximum wheel load of 10,700 lb. The maximum speed capability is 200 knots (this aircraft class does not exceed 130 knots during an approach). Although this is about the smallest size for wheel-loading, it has some redundancy with a twin wheel.

Nose-wheel and tire sizing are based on the forwardmost CG position; therefore, the aft CG nose-wheel load is not computed. The nose-wheel load at the forward CG at 8.8 m (28.87 ft) from the zero reference plane gives lFORWARD = 10.4–8.8 = 1.6 m (5.25 ft).

Equation 7.3 gives RNOSE = (lF ORWARD 8.7 = 2,203 kg (4, 460 lb).

To maintain a smaller nose wheel, a twin-wheel tire arrangement is chosen. For the twin-wheel arrangement, the ESWL = 2,203/1.5 = 1,469 kg (3,238 lb).

From the tire catalogs, a suitable match is Type VII, with the designation of 18 × 4.4 (10-ply) and an inflation pressure of 185 psi that can take 3,550 lb (1,610 kg). The maximum speed capability is 210 mph, the same as for the main wheel.

Landing is most critical for deflection. Typically, the maximum landing weight is 95% of the MTOM. To calculate deflection on landing (in Table 7.1, FAR25 VVert = 12 fps), the energy to be absorbed is given in Equation 7.10 (it is computed in correlate with FPS to FAR data) as:

Eab = 1/2ML × VVert2 = 0.5 × 0.95 × 21, 463 × 122 = 1, 468, 070 lb ft2

In Equation 7.14, total deflection by tire and strut is:

(1/2 × VVert2)/g = x × (n × kstrut × δstrut + m × ktire × δtire)

In Equation 7.18:

δtire = D/2 − Rload = 22/2 − 9.6 = 1.4 in = 0.1167 ft

This can be expressed as a percentage of the maximum radius. Tire and inflation pressures result in the tire footprint (see Equation 7.17) from which the tire deflection can be computed. Use the following values:

x = 2(maximum civil aircraft g-load at landing)

m = 4(number of tires), n = 2(number of struts)

kstrut = 0.7 ktire = 0.47 δtire = 0.1167 ft

Then, Equation 7.14 becomes:

(1/2 × 122)/32.2 = 2 × (2 × 0.7 × δstrut + 4 × 0.47 × 0.1167)

or 1.12 = 1.4 × δstrut + 0.22

δstrut = 0.643 ft plus 0.077 as the margin, totaling 0.72 ft (220 mm)

Baseline Aircraft with 10 Passengers at a 33-Inch Pitch

To maintain component commonality, the same undercarriage and tire size are used. The tire-ply rating can be reduced to 10 to make it less expensive. It is left to readers to repeat the calculation as given previously, using the following data:

Baseline aircraft MTOM = 9,500 kg (≈ 21,000 lb) (refined in Chapter 8)

Fuselage length = 15.24 m (50 ft): The CG from the zero reference plane = 8.75 m (28.7 ft)

Fuselage clearance angle γ = 15 deg, sufficient for rotation, and β = 15 deg

Distance of the main wheel from the nose wheel reference = 9.2 m (30.2 ft) (at about 55% of the wing MAC)

Wheel base = 7.925 m (26 ft)

Wheel track = 2.9 m (9.5 ft) (no change)

To size the main tire, the aftmost CG position is considered:

lREAR = 8.75 − 1.8 = 6.95 m (22.8 ft)

RMAI N = (6.95 × 9,500)/7.925 = 8,331 kg (18,3671 lb). The load per strut

= 4,166 kg (9,184 lb).

It will require a twin-wheel arrangement. From Equation 7.5, the ESWL = 4,166/ 1.5 = 2,777 kg (6,123 lb).

From the tire catalogs, a lower-width tire designation of 22 × 5.75 − 12 (12-ply) with a speed rating of 230 mph and a tire inflation pressure of 220 psi can be used.

A nose wheel tire is sized based on the forwardmost CG position at 8.14 m (26.7 ft) from the zero reference plane. This gives lREAR = 8.14 − 1.8 = 6.34 m (20.8 ft).

RNOSE = (6.34 × 9,500)/7.925 = 7,600 kg (16,755 lb), giving a nose wheel load RNOSE = 1,900 kg (4,400 lb). The ESWL = 1,900/1.5 = 1,267 kg (2,792 lb).

To maintain commonality, the same tire designation of 18 × 4.4 (10-ply) that takes 3,550 lb is used.

Readers can proceed in the same way with deflection calculations. The deflection will be lower than the previous case and the spring could change without changing the geometric size.

Shrunk Aircraft (Smallest in the Family Variant) with 6 Passengers at a 33-Inch Pitch

The last task is to check the smallest aircraft size to maintain as much component commonality of the undercarriage and tire size as possible (which may not be easy to do).

Small variant aircraft MTOM = 7,000 kg (15,400 lb) (refined in Chapter 8)

Fuselage length = 13.56 m (44.5 ft)