Category AIRCRAF DESIGN

This chapter discussing undercarriage design is a relatively large, complex, and standalone chapter without which an aircraft design cannot be completed. Only the preliminary information – what is needed by aircraft designers to conduct a con­ceptual study – is presented here. Details of the undercarriage design are imple­mented by specialists after the go-ahead on a project is obtained. Aircraft designers and undercarriage designers maintain communication to integrate the undercar­riage with the aircraft, doing it right the first time.

There is a tendency to minimize undercarriage design work in coursework exer­cises, possibly because of time constraints. As now understood, this is an involved procedure; if time is a constraint, then the undercarriage should be addressed in a second term, using CAD and including work on retraction kinematics. A good spreadsheet must be prepared for the calculations because they are required for subsequent iterations.

In summary, the chosen undercarriage should be the tricycle type with retrac­tion. The runway LCN and ESWL decide tire pressure (the higher level of inflation pressure may be necessary), which in turn decides the number of wheels and struts required. Tire manufacturers’ catalogs list the correct sizes of the tires.

The methodologies for civil and military aircraft undercarriages and tire sizing are nearly the same. The differences are in operational requirements. In general, civil aircraft design poses more difficulty in maintaining component commonality

within the variants. The cost options for component commonality for variant designs must be decided early during the conceptual design phase. Trade-off studies on cost versus weight must be conducted.

7.13 Undercarriage and Tire Data

Table 7.8 gives some production aircraft undercarriage and tire data.

The worked-out example of an AJT aircraft, as developed in Chapter 6 (see Fig­ure 6.14), continues with sizing of the undercarriage and tires. The combat air­craft tire sizing for the variant designs is considerably simpler because the affecting geometries are not altered, only the weight changes. This military aircraft exam­ple has only two variants, the AJT and CAS versions. The CAS role of the variant aircraft has the same geometric size but is heavier.

Because the affecting geometries (i. e., the wheel base and wheel track) do not change, the cost option logic is less stringent for maintaining undercarriage and tire component commonality – especially for the worked-out example. In this case, it is designed for the heaviest variant with shaved-off metal for the baseline trainer version. Therefore, only the CAS version design is discussed; it is assumed that the AJT baseline will have lighter struts but the same undercarriage and tires.

The reference lines are constructed using Figure 7.17. The aircraft centerline is taken conveniently through the center of the engine exhaust duct. Although at this stage, when the exact CG is not known and placement of the undercarriage is based on a designer’s experience, the example of the AJT given here has been sized to avoid repetition. The exercise for readers is to start with their own layout and then iterate to size.

Even as a tandem seat arrangement, the aircraft CG travel in this aircraft class would be less than that of the civil aircraft example – for example, from 20 to 35% (aftmost) of the wing MAC. The CAS version of the CG variation is from 25 to 38% (fully loaded).

The nose wheel load is based on the forwardmost CG position. An armament payload is placed around the aircraft CG, and the CAS aircraft CG movement is insignificant between a clean configuration and a loaded configuration. Following is the relevant information for the two variants. Only the heaviest aircraft needs to be considered in this case, as explained previously.

The most critical situation for both the main wheel and the nose wheel and the tire is the CAS version with a fully loaded MTOM equal to 9,000 kg. The aftmost CG position from the reference plane is 6.8 m (22.32 ft). The AJT has an MTOM equal to 6,500 kg.

The following information is used to determine wheel loading, LCN, and tire sizing (use Figure 7.17 to compute wheel loading):

• Place the main-wheel ground contact point at 45% of the MAC. 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 known by actual compu­tation.

• The fuselage clearance angle, y > 15 deg, is adequate for both variants.

• Then, the most critical CG angle в = tan-1(0.874/1.9); i. e., в = 24.7 deg.

• The main-wheel load is computed at the aftmost CG, which gives IReaR =

6.8 – 2.2 = 4.6 m (15.1 ft).

• Equation 7.2 gives Rmain = (Irear x MTOW)/Ibase

• Or Rmain = (4.6 x 9,000)/5.33 = 7,767 kg (17,127 lb).

The load per strut is 3,883.5 kg (8,563 lb). It is better to keep the wheel load below 10,000 lb in order to have a smaller wheel and tire. Appendix E provides the tire data from which to choose; there are many options.

The typical airfield LCN for this class of CAS aircraft is low. From Figure 7.13, the LCN is below 15 for the CAS variant and the AJT is still lower. This means there is good flotation and the aircraft can operate from semiprepared airfields. Several options are listed in Table 7.6. From the tire catalogs (see Appendix E), a suitable match is the New Tire Type (Type VII; although the equivalent inch code is available, the metric code is used to familiarize readers) with a designation of 450 x 190 – 5 (22-ply) and an inflation pressure of 15.5 bar (225 psi) that takes 4,030 kg (8,886 lb). The maximum speed capability is 190 mph (165 knots), which is a sufficient margin because this aircraft class does not exceed 130 knots during an approach. The AJT is much lighter with a MTOM of 6,500 kg and, therefore, the same tire at a reduced pressure can be used (or a suitable smaller tire can be used, if changing the hub is required).

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 6.07 m (19.92 ft) from the zero reference plane gives IFORwARd = 7.53 – 6.07 = 1.46 m (4.8 ft).

Equation 7.3 gives Rnose = (Iforward x MTOW)/Ibase, or Rnose = (1.46 x 9,000)/5.33 = 2,465.3 kg (5,436 lb). A single nose wheel of a smaller size is chosen.

From the tire catalogs (see Appendix E), a suitable match is the New Tire Type (Type VII, inch code) with a designation of 17.5 x 4.4 – 8 (14-ply) and an infla­tion pressure of 220 psi (14.47 bar) that takes 6,000 lb (2,721 lb). The maximum speed capability is 210 mph (182 knots). The AJT is much lighter with the MTOM of 6,500 kg; therefore, the same tire with a reduced pressure can be used.

Deflection is estimated as in the civil aircraft case and therefore is not shown here. The high-wing configuration also shows sufficient clearance. The author sug­gests that readers undertake the computation.

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), в = 15, and y = 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.

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, y = 14 deg and в = 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 Ibase = 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 в. It results in a low angle of 40 deg in a very stable aircraft.

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

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

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 x 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 Iforward = 10.4-8.8 = 1.6 m (5.25 ft).

Equation 7.3 gives Rnose = (Iforward x MTOW)/Ibase = (1-6 x 11,000)/

8.7 = 2,203 kg (4, 460lb).

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 x 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 x Vvert2 = 0-5 x 0.95 x 21, 463 x 122 = 1, 468, 070 lb ft2

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

(1/2 x Vvert2)/g = X x (n x kstrut x Sstrut + m x ktire x Stire)

In Equation 7.18:

Stire = D/2 – Rload = 22/2 – 9.6 = 1.4in = 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 Stire = 0.1167 ft

Then, Equation 7.14 becomes:

(1/2 x 122)/32.2 = 2 x (2 x 0.7 x Sstrut + 4 x 0.47 x 0.1167)

or 1.12 = 1.4 x Sstrat + 0.22

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

The total deflection is Stire + Sstrut = 0.1167 + 0.72 = 0.837 ft = 10 in (25.4 cm).

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 y = 15 deg, sufficient for rotation, and в = 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:

Irear = 8.75 – 1.8 = 6.95 m (22.8 ft)

Rmain = (6.95 x 9,500)/7.925 = 8,331 kg (18,3671 lb). The load per strut = 4,166kg (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 x 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 x 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 x 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). [11]

• The CG from the zero reference plane = 7.7 m (25.26 ft), which gives Irear =

7.7 – 1.8 = 5.9 m (19.35 ft)

• Fuselage clearance angle y =16 deg, sufficient for rotation, and в = 16 deg

• Distance of main wheel from nose wheel reference = 8.29 m (27.2 ft) (at about 55% of the wing MAC)

• Wheel base = 7.1 m (23 ft)

• Wheel track = 2.9 m (9.5 ft)

• Rmain = /7.1 = 5,817kg (12,824lb). Load per strut = 6,412 lb.

In this case, one wheel with the same tire designation of 22 x 5.75 – 12 (12-ply) would suffice (see Figure 7.8). The nose wheel tire also could remain the same.

It would be beneficial to shave off metal from the strut shank, if possible, and to reduce the height for both the nose and main wheels. Readers can proceed with the remaining calculations in the same manner; the spreadsheet method is helpful.

After obtaining the necessary information available on the undercarriage as a sys­tem, the next step is to systematically lay down the methodology to configure the undercarriage arrangement in order to integrate it with the aircraft conceived in Chapter 6. All aircraft designers benefit from existing designs by having guidelines – this is what is meant by “experience”: past designs provide a good databank.

First, the undercarriage commonality for variant designs is considered. In gen­eral, in civil aircraft design, the baseline aircraft is the middle of the three main sizes (other variants are possible). Therefore, the largest version is more critical to the undercarriage layout for carrying the heaviest load. The methodology described herein should be applied to the largest of the variants and then all other variants should be checked for commonality. At the conceptual design stage, all versions have an identical undercarriage layout except for the wheel base and wheel track. A production version has the scope to shave off metal, making it lighter for smaller variants; this requires only minor changes in the manufacturing setup. Following is the stepwise approach for the undercarriage layout geometry, load estimation, and tire sizing:

1. Determine the type of undercarriage: nose wheel or tail wheel. For the reason explained in Section 7.3, it is a nose wheel type. Low-speed smaller aircraft sen­sitive to weight and drag could have a tail wheel type.

2. If the aircraft operating speed exceeds 150 knots, the undercarriage should be retractable.

3. Estimate from statistics and experience the CG position; it is suggested 20 to 40% of the wing MAC at about the fuselage centerline, depending on the wing position. Ensure that the CG angle в with the vertical is about 15 deg.

4. The main-wheel strut length should allow full rotation that clears the fuselage aft end, solves oleo-collapse problems, allows for full flap-deflection clearance, and makes a trade-off with the wheel track to prevent the aircraft from turning over. The fuselage clearance angle y is between 12 and 16 deg, which should clear the fuselage aft end at the rotation when the oleo is fully extended at liftoff.

5. The nose wheel strut length should be relative to the main-wheel strut length in order to keep civil aircraft payload floorboards level (for a propeller-driven aircraft, ensure that there is propeller clearance at the nose-oleo collapse). Avoid making the strut length too long. The nose wheel attachment point ideally should be located ahead of the cockpit.

6. Determine that the wheel-track turn-over angle в is less than the recommended value and satisfies the desired turn radius. A trade-off may be required but, in general, runway widths are wide enough. A wider wheel track is better but it should not be too wide, which would create turning and structural problems on the wing attachment.

7. Determine the stowage space; less articulation is simpler and requires less main­tenance and lower cost. In general, civil aircraft articulation is simpler than the military aircraft type. Ensure that the storage space has adequate tire clearance. At this point, it is assumed that stowage space is available.

8. Compute the loads on each point of support and determine the number of wheels (see Section 7.9). Establish the operational airfield LCN. If an aircraft is to use a Type 1 airfield, it is better for the wheel load to be less than 10,000 lb; then, the LCN also will be low. Use Table 7.6 to determine tire pressure; there are options: the higher the pressure for the LCN, the smaller is the tire size.

9. Use tire manufacturers’ catalogs to select a tire (see Appendix E and www. airmichelin. com and www. goodyearaviation. com).

Two worked-out examples follow. As discussed previously, the civil and military aircraft design methodologies are similar but differ considerably in their operational mission profile. There are various levels of options available to maintain component commonality, including the following:

1. Low-Cost Option. Maintain the same undercarriage for all variants even when there are performance penalties. In this situation, design the undercarriage for the biggest aircraft and then use it for other variants. The biggest aircraft may have tighter design criteria to sacrifice some margin in order to benefit smaller designs.

2. Medium-Cost Option. In this situation, design the undercarriage for the biggest aircraft and then make minor modifications to suit the smaller variants and to retrieve some of the performance loss associated with the low-cost option. These modifications maintain the external geometry but shave off metal to lighten the structure, to the extent possible. The smallest aircraft may require a shortening of the strut length without affecting the shock-absorber geome­try, which may require a spring change. Wheel, brake, and tire size are kept the same. The smallest variant is nearly half the weight of the largest but may be unchanged; it may be possible to change a dual wheel to a single wheel.

3. High-Cost Option. Make major modifications to the undercarriage. In this situation, design three strut lengths for each variant, maintaining the maxi­mum manufacturing-process commonality – this would reduce costs when NC machines are used. Maintain the other items with the maximum component commonality. The spring of the smallest variant may change without affecting the external geometry. Performance gain could be maximized by this option.

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).

Ground movement would experience friction between the tire and the ground. Dur­ing the takeoff run, this friction is considered drag that consumes engine power. Figure 7.15 is a representation of the ground-rolling friction coefficient, л, versus aircraft speed for various types of runways. Conceptual studies use the value for the friction coefficient, л.

• A Type 3 runway (concrete pavement) = 0.02 to 0.025 (0.025 is recommended for coursework)

• A Type 2 runway = 0.025 to 0.04 (0.03 is recommended for coursework)

• A Type 1 runway = 0.04 to 0.3, depending on the surface type, as follows:

hard turf = 0.04

grass field = 0.04 to 0.1 (0.05 is recommended for a maintained airfield) soft ground = 0.1 to 0.3 (not addressed in this book)

The braking friction coefficient, ль would be much higher depending on the run­way surface condition (e. g., dry, wet, slush, or snow – or ice-covered) (Table 7.7). A typical value is лЬ « 0.5. Locked wheels skid that wear out a tire to the point of a possible blowout. Most high-performance aircraft that touch down above 80 knots

have an antiskid device when the p, b value could be as high as 0.7. Slipping wheels are not considered during the conceptual study phase. Tire-tread selection should be compatible with the runway surface condition (e. g., to avoid hydroplaning). The braking friction coefficient, p, b = 0.45 to 0.5, is the average value used in this book. The tire load is based on a brake coefficient of 0.8.

The pavement-loading (i. e., flotation) limit is one of the drivers for tire design. This section presents relevant information for preliminary tire sizing to establish the sec­tion width (Wg), height (H), and diameter (D), as shown in Figure 7.14. The rim diameter of the hub is designated d. Under load, the lower half deflects with the radius, Rload. The number of wheels and tire size is related to its load-bearing capac­ity for inflation pressure and the airfield LCN for an unrestricted operation. For heavy aircraft, the load is distributed over the number of wheel and tires. The FAA regulates tire standards.

Tires are rated based on (1) unloaded inflation pressure, (2) ply ratings for holding shape under pressure, (3) maximum static load for the MTOW (i. e., flota­tion consideration), and (4) maximum aircraft speed on the ground. Basically, there are mainly three types of tires from nine categories, as described herein. (See the Michelin and Goodyear data sourcebooks listed in the references.)

Types I and II: These types are becoming obsolete and are no longer produced. Type I is intended for a fixed undercarriage.

Type III: This type includes low-pressure tires that provide a larger footprint or flotation effect. They have a relatively small rim diameter (d) compared to overall tire diameter. Speed is limited to less than 160 mph. The tire designa­tion is expressed by its section width, WG, and rim diameter, d (Figure 7.14). All dimensions are in inches. For example, a typical small-aircraft tire desig­nation of 6.00-6 means that it has a width of 6.00 inches (in hundredths) and a rim diameter hub of 6 inches.

Types IV, V, and VI: These types no longer exist.

Type VII: These are high-pressure tires that are relatively narrower than other types. They are widely used in aircraft with pressure levels from 100 to more than 250 psi that operate on Type 2 and Type 3 airfields. Military- aircraft tire pressure can reach as high as 400 psi. Tire designation is expressed by the overall section diameter (D) and the nominal section width, WG, with the multiplication sign (x) in between. All dimensions are in inches. For example, 22 x 5.5 has an overall section diameter of 22 inches and a section width of 5.5 inches.

New Design Tire (three-part nomenclature): Except for Type III tires, all newly designed tires are in this classification. A Type VIII tire also has this designa­tion. This type uses a three-part designation shown as (outside diameter, D) x (section width, WG) – (rim diameter, d). These are also known as biased tires, which are intended for high-speed aircraft with high tire-inflation pres­sures. Dimensions in FPS are in inches and dimensions in SI are in milli­meters but the rim diameter is always in inches. For example, a B747 tire has the designation, 49 x 19.0-20, meaning that it has an outside diameter of 49 inches, a section width of 19 inches, and a rim diameter of 20 inches. New

tires also have radial types; the three-part designation has an “R” instead of a hyphen. An example of a radial tire in SI is 1400 x 530 R 23. There is a spe­cial designation that precedes the three-part nomenclature tires with a B, C, or H. The description of these construction details is beyond the scope of this book.

There are small tires not approved by the FAA that are used in the homebuilt aircraft category. This book addresses only Types III and VII and the New Design Tire.

Several tire manufacturers are available from which to choose, as in the case of the automobile industry. Tire manufacturers (e. g., Goodyear, Goodrich, Dunlop, and Michelin) publish tire catalogs, which provide important tire data (e. g., dimen­sions and characteristics) in extensive detail. Appendix E lists data from the manu­facturers’ catalogs needed for the coursework in this book. Aircraft designers have the full range of tire catalogs and contact tire manufacturers to stay informed and benefit mutually from new tire designs.

Under load, a tire deflects and creates a footprint on the ground. Therefore:

load on tire = (footprint x tire pressure) (7.17)

For tire static deflection:

5tire = (maximum radius at no load) – (minimum radius under static load)

= D/2 – Rload, (7.18)

where Rload equals the radius of the depressed tire under load. It can be expressed as a percentage of the maximum radius.

Table 7.6 lists the typical tire pressures for the range of aircraft weights.

Under a typical static load, tire deflection is kept at a maximum of a third of the maximum height (H). As aircraft speed increases, the load also increases on tires as dynamic loading. During landing impact, the deflection would be higher and would recover sooner, with the tire acting as a shock absorber. Bottom-out occurs at maximum deflection (i. e., three times the load); therefore, shock absorbers take the

Figure 7.15. Ground friction coefficient

impact deflection to prevent a tire from bottoming-out. Section 7.9 discusses tire – deflection calculations; corresponding typical tire pressures for the sizes are given in Table 7.6.

Tire sizing is a complex process and depends on the static and dynamic loads it must sustain. This book addresses tire sizing for Type 2 and Type 3 runways. One of the largest tires used by the B747-200F has a main and nose gear tire size of 49 x 19-20 with an unloaded inflation pressure of 195 psi. Sizes used by existing designs of a class are a good guideline for selecting tire size.

Use of an unprepared runway (i. e., Type 1) demands a low-pressure tire; higher pressure tires are for a metal runway (i. e., Types 2 and 3). The higher the pressure, the smaller is the tire size. Civil aircraft examples in this book use a Type 3 airfield; military aircraft examples use Types 2 and 3 airfields. Small aircraft use a Type 1 airfield for club usage.

The LCN is airfield-specific and aircraft must comply with it. Subsequently, ICAO introduced a new classification system, known as the aircraft classification number (ACN), which represents the tire-loading limit, and another system that represents the airfield pavement-strength limit, known as the pavement classification number (PCN). Both numbers must be the same to operate at an airport without any restric­tions. However, the LCN method is still in use and conversion is needed to use the ACN/PCN method. This book uses Figure 7.13 to obtain the LCN.

The ACN/PCN method is described in [9]. According to the design manual, the ACN/PCN method is intended only for publication of pavement-strength data in the Aeronautical Information Publication (AIP). It is not intended for design or

evaluation of pavements, nor does it contemplate the use of a specific method by the airport authority for either the design or evaluation of pavements. The ACN/PCN method is more elaborate and involved. Parameters like the California Bearing Ratio (CBR) for subgrade-strength soil tests are required to determine tire pres­sure. The LCN method is still in use and can be converted to the ACN and PCN. According to the AIP, “The ACN of an aircraft is numerically defined as two times the derived wheel load, where the derived single wheel load is expressed in thou­sands of kilograms.”

The author was able to locate the Airbus publication for the largest passenger­carrying aircraft, the A380-800F model; pertinent data are listed in Table 7.4. The weight per wheel is distributed relative to the wheel arrangement (see Figure 7.11). The braking deceleration is 10 ft/s2. The horizontal ground load is calculated at a brake coefficient of 0.8. The main landing gears can take as much as 95.5% of the weight.

The ICAO, as an international agency, established ground rules to match aircraft and runway performance requirements. The ICAO developed the strength classifi­cations of Type 2 and Type 3 runways by designating a load classification number (LCN) that represents the extent of load that a runway can accommodate based on construction characteristics. All Type 2 and Type 3 runways must have a LCN and the aircraft undercarriage design must comply with it. The LCN range of the airfield’s type is grouped under the load classification group (LCG). For example, an aircraft with the LCN 62 can operate on any airfield with an LCG of I to III. Table 7.2 provides the LCN range for the types of runways.

The relationship among the LCN, tire pressure, and ESWL is presented in Figure 7.13. The procedure is to first obtain the LCN of the airfield in question. Then, compute the ESWL of the undercarriage (see Section 7.9). Finally, find the tire pressure required using Table 7.6 (see Section 7.11); this provides a guideline to choose tire size. Section 7.13 outlines the methodology followed by worked-out

examples (see the references for more details on other types). Typical examples of aircraft complying with the LCN and the corresponding MTOM and tire pressures are given in Table 7.3.

The B757, which is twice as heavy as the B737, maintains nearly the same LCN by having more wheels to distribute load per tire.

The undercarriage design depends on how the wheels interact with the airfield sur­face. An airport runway surface must be designed to withstand an aircraft’s weight not only at the static condition but also at dynamic loading (e. g., for a heavy land­ing). Runway pavement loading is known as flotation. Among airports, the runway pavement strength varies. There are three main types of surfaces, as follows:

2. Type 2: Prepared Macadam Surface. These are asphalt – or tar-topped runways with strength built in by the thick macadam filler; these are designated as a Type 2 surface. These surfaces are less expensive to prepare by using a heav­ily rolled macadam filler. However, local depressions can cause the surface to undulate, and it requires frequent maintenance with longer downtime. This type of runway can accommodate heavy aircraft such as the B747.

3. Type 3: Prepared Concrete Surface. This is a rigid concrete runway designated as a Type 3 surface. These runways are built with pavement-quality concrete (i. e., about a half-meter thick) and are covered by asphalt (e. g., 150 mm thick). All major international airports have Type 3 runways, which can take a load similar to a Type 2 surface and do not have to be as thick. This type is expensive to prepare and maintenance downtime is minimal. Cracks are the typical type of failure that occurs. A Type 3 surface can accommodate heavy aircraft such as the B747 and the A380.

Aircraft designers must design aircraft to be compatible with existing airfields in order to operate. If the market demand necessitates larger and heavier aircraft, then designers must make the aircraft comply with the pavement strength of exist­ing airfields or the airfield must be reinforced to accept the heavy aircraft. Runway reinforcement depends on new designs; therefore, airport authorities communicate with aircraft manufacturers to remain current with market demand. When the B747 began operating, almost all international airfields needed reinforcement to accept them – some were not operational for several years.

Both the tire and the shock absorber absorb the energy to cushion the impact of an aircraft’s vertical descent rate at landing in order to maintain structural integrity and avoid the tire bottoming out. FAA safety requirements limit the vertical descent velocity, VVert> for civil aircraft applications; military specifications limit military applications. Table 7.1 lists limits for various types of aircraft. In turn, VVert pro­duces g-load at the sudden termination of VVert at landing – it can be expressed as load factor n (see Section 5.5). Equation 5.4 gives n = (1 + a/g); it is loosely termed as the number of the g-load; for an undercarriage design, it is designated ni (see Table 7.1). During landing, nl takes a positive value; that is, it would experience heavier weight. For example,

nl = x (a number) means that the weight has changed by x times. (7.10)

These are extreme values for safety; in practice, 4 fps is a hard landing in a civil aircraft operation. The maximum landing aircraft mass ML is taken as 0.95 MTOM for aircraft with a high wing-loading.

The vertical velocity kinetic energy to be absorbed is

Eab = 1/2Ml X VVert2 (7.11)

This is the energy to be absorbed by all the main wheels (m wheels) and struts (n struts) at touchdown during landing. The nose wheel touches the ground much later, after the main wheels have already absorbed the impact of landing.

Eab = Eabstrut + EabJire (7.12)

energy absorption by strut (Let n be the number of struts.)

Assume that a landing is even and all struts have equal deflection of 5strut.

Then, energy absorbed by all the struts is

Eabstrut = n X nl X gML X kstrut X ^strut, (7.13)

where kstrut is an efficiency factor representing the stiffness of the spring and

has values between 0.5 and 0.8, depending on the type of shock absorber used.

In this book, 0.7 is used for modern aircraft and 0.5 is used for small club and homebuilt categories.

energy absorption by tire (Let m be the number of tires.)

Assume that a landing is even and all tires have equal deflection of 5tire. Then, energy absorbed by all the tires is

Eab-tire = m X Hi X gML X ktire X Stire (7.14)

where ktire = 0.47 is an efficiency factor representing the stiffness of all types of tires.

The following can be written by equating Equation 7.11 with Equation 7.12 and then substituting Equations 7.13 and 7.14 in Equation 7.12 and replacing ni by Equa­tion 7.10. Here, the load factor ni is replaced by x:

Eab = [5] [6] [7] [8]/2Ml X VVet = H X x X gML X kstrut X 5strut + m X x X gML X ktire X Stire

Simplifying as follows:

(1/2 X VVert)/g = xx(h X kstrut X Sstrut + m X ktire X Stire) (7.15)

7.9.1 Deflection under Load

The total vertical deflection of the strut and tire during landing can be computed by using Equation 7.15. Other types of lateral strut deflection during turning and other maneuvers are not addressed in this book.

Total deflection is

S = [9]strut + Stire (7.16)

It is recommended that a cushion be kept in the strut deflection (compression) so that ends do not hit each other. In general, 1 inch (2.54 cm) is the margin.