PRELIMINARY DESIGN OF AN AIRCRAFT DISC BRAKE SYSTEM Roberto Bonfiglio
Department of Aerospace Science and Technologies Course of Aerospace Systems
June 2015
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Preliminary design of an aircraft braking system
Introduction This report has the target to preliminary design and size an aircraft disc brake system, considering the use of different materials (steel and carbon). It is assumed to have the following data.
Aircraft Data ▪ ▪ ▪
Maximum mass at landing Velocity at landing Center of mass height
Environment Data ▪
Landing gear Data ▪ ▪ ▪ ▪ ▪
N° braking wheels Wheel diameter Coefficient friction terrain-tyre Pace % Weight statically on main landing gear
Air Temperature
Brake Disc Data ▪ ▪ ▪
% Energy absorbed by brake disc Disc diameter (max and min) Lining (max thickness, usury coefficient, max pressure, % disc surface)
Aircraft Data Aircraft Data
Disc Materials Data
density, specific heat, maximum temperature, maximum thickness, coefficient friction liningdisc
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Preliminary design of an aircraft braking system
Index Introduction.................................................................................................................................................... 2 Index................................................................................................................................................................. 3 List of symbols ................................................................................................................................................ 4 1.
Problem Description and solution ................................................................................................ 5 1.1
Energy Dissipation ........................................................................................................................ 5 1.1.1
Heat Sizing ............................................................................................................................ 6
1.1.2
N. of discs for wheel ............................................................................................................. 6
1.2
Braking force momentum ............................................................................................................ 7
1.3
Lining usury .................................................................................................................................... 8
2.
Problem data ..................................................................................................................................... 9
3.
Calculus ............................................................................................................................................. 10 3.1
Energy Dissipation ...................................................................................................................... 10
3.2
Braking Momentum .................................................................................................................... 13
3.3
Lining Usury .................................................................................................................................. 15
3.4
Result Resume .............................................................................................................................. 16
Appendix: Matlab ......................................................................................................................................... 17
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Preliminary design of an aircraft braking system
List of symbols symbol
unit
Aircraft mass Landing velocity Centre of mass height Air Temperature
M v H Ta
kg m/s m °C
Fraction of energy converted into heat Energy absorbed by brake system
re Ea
% J
Landing gear pace % weight on braking wheels statically N. braking wheels wheels diameter coefficient friction tyre-land
L rp N Dr Âľt
m % mm -
Disc - maximum diameter Disc - minimum diameter
D1 D2
mm mm
Disc - section area Disc - thickness Disc - max thickness Disc - max temperature N. discs in each wheel Total wheel’s discs thickness Total aircraft’s discs thickness Total aircraft’s discs mass Total Braking Momentum in each wheel
Ad s sd Tmax n sw stot mtot đ??śđ?‘¤_đ?‘Ąđ?‘œđ?‘Ą
mm2 mm mm °C mm mm kg Nm
Total Braking Momentum in aircraft
đ??ś
Nm
Lining - max thickness Lining - % disc area friction Lining - max pressure Lining - usury coefficient
smax ra pp cu
mm % Mpa J/mm3
Material - Specific heat Material - Density Coefficient friction lining-disc
cs Ď Âľp
J/kg*°C kg/m3 -
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Preliminary design of an aircraft braking system
1. Problem Description and solution 1.1 Energy Dissipation The target of a brake system is dissipating energy. During a landing, aircraft’s kinetic energy is dissipated trough many different systems (airbrake, reverse thrust, brakes), each of these gives a different contribute in different instances.
It is common that brake system dissipates nearly 80% of the overall energy, it is one of the most critical system on-board in an aircraft. In this report this percentage of energy dissipation elaborated by brakes is referred as re . In brakes the aircraft’s kinetic energy is transformed into heat. The brake system must absorb all the heat produced by friction without a shock or a break. Commonly this heat isn’t absorbed only by the disc but it diffuses in the whole wheel, tyre and others components. In a preliminary design it is assumed that the brake system can absorb alone without fault all the heat produced in a single landing. Aircraft disc brakes are usually designed with a different approach comparing them to automotive ones. In aeronautics the system’s mission is to absorb an huge amount of heat in few seconds during a landing. In automotive instead, brakes must be always ready to be used.
A ERONAUTICAL BRAKE DISC
For this reason, first ones (aeronautics) have a consistent mass (for absorbing great energy) and have multiple discs for avoiding shocks or break. The have not a complex cooling system,being used only for few seconds in landing.
Second ones (automotive) are usually single disc system, with a complex cooling system ensuring their functioning in every conditions.
M OTORSPORT (F1) BRAKE DISC
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Preliminary design of an aircraft braking system
1.1.1 Heat Sizing All materials are characterised by a maximum limit temperature (đ?‘‡đ?‘šđ?‘Žđ?‘Ľ ) and a specific heat (đ?‘?đ?‘ ). It can be easily calculated the minimum total disc mass, considering đ?‘‡đ?‘Ž as air temperature and đ??¸đ?‘Ž as the energy absorbed by the brake system: ΔTđ?‘šđ?‘Žđ?‘Ľ = đ?‘‡đ?‘šđ?‘Žđ?‘Ľ − đ?‘‡đ?‘Ž 1 đ?‘šđ?‘?đ?‘ Δđ?‘‡đ?‘šđ?‘Žđ?‘Ľ ≼ rđ?‘’ ∗ ( đ?‘€đ?‘Ł 2 ) = đ??¸đ?‘Ž 2
đ?‘šđ?‘Ąđ?‘œđ?‘Ą ≼
BRAKE DISC TEST
đ??¸đ?‘Ž đ?‘?đ?‘ ∗ Δđ?‘‡đ?‘šđ?‘Žđ?‘Ľ
1.1.2 N. of discs for wheel From the previous formula the minimum discs’ total mass has been calculated. Considering the use of annulus disc with a diameter đ??ˇ1 ≤ đ??ˇ ≤ đ??ˇ2 , the total disc thickness in aircraft must be: đ?‘ đ?‘Ąđ?‘œđ?‘Ą =
đ?‘šđ?‘Ąđ?‘œđ?‘Ą đ?œŒđ??´đ?‘‘
đ??ˇ1 2 − đ??ˇ22 ) is 4
, where đ??´đ?‘‘ = đ?œ‹ (
disc’s section area
In each of đ?‘ main landing gear wheels there must be a minimum total disc thickness đ?‘ đ?‘¤ : đ?‘ đ?‘Ąđ?‘œđ?‘Ą đ?‘ đ?‘¤ = N Because each disc has a maximum possible thickness of đ?‘ đ?‘‘ , in each wheel ther must be a minimum of n discs, with n defined as following:
đ?‘ đ?‘¤ đ?‘› = đ?‘?đ?‘’đ?‘–đ?‘™ ( ) đ?‘ đ?‘‘
TOTAL AIRCRAFT DISC THICKNESS
DISC THICKNEES FOR EACH WHEEL
It is so established an available disc thickness đ?‘ ≤ đ?‘ đ?‘‘ so that đ?‘›đ?‘ ≼ đ?‘ đ?‘¤ Having decided the system architecture, the maximum operative temperature can be computed:
đ?‘‡đ?‘œđ?‘?_đ?‘šđ?‘Žđ?‘Ľ = đ?‘‡đ?‘Ž +
đ??¸đ?‘Ž đ?‘?đ?‘ đ?œŒđ??´đ?‘‘ đ?‘ Nđ?‘›
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Preliminary design of an aircraft braking system
1.2 Braking force momentum For each disc can be done the following considerations: In figure aside it is represented the lining coverage of the disc. In particular linings cover ra % of disc area, rđ?‘Ž = đ?‘&#x;1 =
đ??ˇ1 , 2
đ?‘&#x;2 =
đ??ˇ2 , 2
đ?œƒ . 2đ?œ‹
đ??´ = rđ?‘Ž ∗ đ??´đ?‘‘
Can be computed the elementary disc braking momentum as: đ?‘‘đ?‘€đ?‘‘ = đ?œ‡đ?‘? đ?‘ƒđ?‘‘đ??´ ∗ đ?‘&#x; , where P is pressure acting on lining, đ?œ‡đ?‘? the friction coefficient lining-disc, đ?‘‘đ??´ the elementary area considered and đ?‘&#x; the distance to centre. Since linings act on both disc’s surfaces the total disc braking momentum is: đ?‘&#x;1 đ?œƒ
Md = 2 âˆŤ đ?œ‡đ?‘? Pr đ?‘‘đ??´ = 2 âˆŹ đ?œ‡đ?‘? đ?‘ƒđ?‘&#x; ∗ đ?‘&#x; đ?‘‘đ?‘&#x;đ?‘‘đ?œƒ = 2 đ?œ‡đ?‘? đ?‘ƒđ?œƒ ( đ??´
đ?‘&#x;2 0
đ?‘&#x;13 − đ?‘&#x;23 ) 3
Considering that in every wheel there are n braking discs, the braking torque in each wheel is: đ??śđ?‘¤ = n ∗ Md = 2đ?œ‡đ?‘? đ?‘ƒđ?‘›đ?œƒ (
đ?‘&#x;13 − đ?‘&#x;23 ) 3
The total braking torque in the aircraft, given by different wheels is: đ??śđ?‘Ąđ?‘œđ?‘Ą = N ∗ đ??śđ?‘¤ The aircraft can be analysed as a mechanical system as follows:
Where: đ?‘Š = đ?‘€đ?‘”, aircraft weight in baricenter đ??ś = N ∗ đ??śđ?‘¤ , total wheels braking momentum , đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ N = đ?‘›Â° đ?‘?đ?‘&#x;đ?‘Žđ?‘˜đ?‘–đ?‘›đ?‘” đ?‘¤â„Žđ?‘’đ?‘’đ?‘™
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Preliminary design of an aircraft braking system
â–Ş
Applying horizontal force equilibrium : đ??š=đ?‘‡
â–Ş
Applying momentum equilibrium in front wheel - terrain contact point: đ?‘Š ∗ đ?‘&#x;đ?‘? đ??ż = Ď• ∗ đ??ż + đ??š ∗ đ??ť
â–Ş
Applying rear wheel momentum equilibrium: đ??ś =đ?‘‡âˆ—
đ??ˇđ?‘&#x; 2
Considering wheel blocking limit conditions: đ?‘‡ = đ?œ‡đ?‘Ą Ď• ➨ đ??š = đ?‘‡ = đ?œ‡đ?‘Ą Ď• đ?‘Š ∗ đ?‘&#x;đ?‘? đ??ż = Ď• ∗ đ??ż + đ?œ‡đ?‘Ą Ď• ∗ đ??ť ➨ đ?‘Š ∗ đ?‘&#x;đ?‘? đ??ż = Ď• ∗ ( đ??ż + đ?œ‡đ?‘Ą đ??ť ) It is so determined vertical reaction in limit conditions as: Ď•=
đ?‘Šđ?‘&#x;đ?‘? đ??ż đ??ż + đ?œ‡đ?‘Ą đ??ť
It is so determined max braking torque đ??śđ?‘šđ?‘Žđ?‘Ľ in wheels blocking conditions â–Ş
đ??ˇ
đ??śđ?‘šđ?‘Žđ?‘Ľ = đ?‘‡ ( 2đ?‘&#x; ) = (đ?œ‡đ?‘Ą Ď•)
đ??ˇđ?‘&#x; 2
Braking Momentum is obtained by lining action as function of pressure as said before â–Ş
đ??ś = 2đ?‘ đ?‘› ∗ đ?œ‡đ?‘? đ?‘ƒđ?œƒ (
đ?‘&#x;13 −đ?‘&#x;23 3
)
For having đ??śđ?‘šđ?‘Žđ?‘Ľ a pressure đ?‘ƒ must acts: đ??ˇđ?‘&#x; đ??śđ?‘šđ?‘Žđ?‘Ľ 3 đ?œ‡đ?‘Ą Ď• 2 3 đ?‘ƒ= ( 3 )= ( 3 ) 3 2đ?‘ đ?‘› ∗ đ?œ‡đ?‘? đ?œƒ đ?‘&#x;1 − đ?‘&#x;2 2đ?œ‡đ?‘? đ?‘ đ?‘›đ?œƒ đ?‘&#x;1 − đ?‘&#x;23 đ?‘šđ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘œđ?‘› đ?‘™đ?‘–đ?‘›đ?‘–đ?‘›đ?‘” đ?‘“đ?‘œđ?‘&#x; đ?‘Žđ?‘Łđ?‘œđ?‘–đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘¤â„Žđ?‘’đ?‘’đ?‘™ đ?‘?đ?‘™đ?‘œđ?‘?đ?‘˜đ?‘–đ?‘›đ?‘”
1.3
Lining usury
In each landing the brake system absorbs energy: 1 đ??¸đ?‘Ž = rđ?‘’ ∗ ( đ?‘€đ?‘Ł 2 ) 2 In each landing it is so spent a certain amount of lining volume in order to absorb energy. In every landing it’s consumed
đ??¸đ?‘Ž đ?‘?đ?‘˘
[đ?‘šđ?‘š3 ] of lining volume, where đ?‘?đ?‘˘ is the lining usury coefficient.
Initially, with new linings: đ??żđ?‘–đ?‘›đ?‘–đ?‘›đ?‘” đ?‘‘đ?‘–đ?‘ đ?‘? đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž = đ?‘&#x;đ?‘Ž ∗ đ??´đ?‘‘ đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘™đ?‘–đ?‘›đ?‘–đ?‘›đ?‘” đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ = (đ??żđ?‘–đ?‘›đ?‘–đ?‘›đ?‘” đ?‘‘đ?‘–đ?‘ đ?‘? đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž) ∗ đ?‘ đ?‘šđ?‘Žđ?‘Ľ ∗ 2đ?‘›đ?‘ It is so possible to determine the n. of possible landings with a set of linings. đ?‘ ° đ?‘œđ?‘“ đ?‘?đ?‘œđ?‘ đ?‘ đ?‘–đ?‘?đ?‘™đ?‘’ đ?‘™đ?‘Žđ?‘›đ?‘‘đ?‘–đ?‘›đ?‘”đ?‘ =
đ?‘?đ?‘˘ ∗ đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘™đ?‘–đ?‘›đ?‘–đ?‘›đ?‘” đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ??¸đ?‘Ž
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Preliminary design of an aircraft braking system
2. Problem data Aircraft mass
M
52000 kg
Landing velocity
v
50 m/s
Centre of mass height
H
3m
Air temperature
Ta
40 °C
Fraction of energy converted into heat
re
75 %
Landing gear pace
L
20 m
% weight on braking wheels statically
rp
90 %
N. braking wheels
N
4
wheels diameter
Dr
1100 mm
coefficient friction tyre-land
µt
0.9
Disc - maximum diameter
D1
650 mm
Disc - minimum diameter
D2
250 mm
Lining - max thickness
smax
10 mm
Lining - % disc area friction
ra
80 %
Lining - max pressure
pp
1.3 Mpa
Lining - usury coefficient
cu
2000 J/mm3
Disc - max temperature
Tmax
500 °C
Specific heat
cs
500 J/kg*°C
Density
ρ
7800 kg/m3
Disc - max thickness
sd
10 mm
Coefficient friction lining-disc
µp
0.25
Disc - max temperature
Tmax
800 °C
Specific heat
cs
1250 J/kg*°C
Density
ρ
1700 kg/m3
Disc - max thickness
sd
15 mm
Coefficient friction lining-disc
µp
0.30
STEEL
CARBON
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Preliminary design of an aircraft braking system
3. Calculus 3.1 â–Ş
Energy Dissipation
Amount of energy absorbed by brake system: 1
1
đ??¸đ?‘Ž = rđ?‘’ ∗ (2 đ?‘€đ?‘Ł 2 ) = 0.75 ∗ 2 ∗ 52000 đ?‘˜đ?‘” ∗ (50 đ?‘šâ „đ?‘ )2 = 48 750 000 đ??˝ â–Ş
Aircraft Weight: đ?‘
đ?‘Š = đ?‘€ ∗ đ?‘” = 52000 đ?‘˜đ?‘” ∗ 9.81 đ?‘˜đ?‘” = 510 120 đ?‘ â–Ş
Disc area: đ?œ‹(6502 − 2502 ) đ??´đ?‘‘ = đ?‘šđ?‘š2 = 282 700 đ?‘šđ?‘š2 4
3.1.1 Steel Disc â–Ş
â–Ş
Minimum Total Disc Mass: đ??¸đ?‘Ž đ?‘šđ?‘Ąđ?‘œđ?‘Ą ≼ = đ?‘?đ?‘ ∗ Δđ?‘‡đ?‘šđ?‘Žđ?‘Ľ 500
48 750 000 đ??˝ = đ?&#x;?đ?&#x;?đ?&#x;? đ?’Œđ?’ˆ = đ?’Žđ?’Žđ?’Šđ?’? đ??˝ ∗ (500 − 40)°đ??ś đ?‘˜đ?‘” °đ??ś
Minimum Total Disc Thickness đ?‘šđ?‘šđ?‘–đ?‘› 212 đ?‘˜đ?‘” đ?‘ đ?‘Ąđ?‘œđ?‘Ą = = = 9.6 10−2 đ?‘š = 96 đ?‘šđ?‘š đ?‘˜đ?‘” đ?œŒ ∗ đ??´đ?‘‘ 7800 3 ∗ 0.2827 đ?‘š2 đ?‘š
â–Ş
Minimum Total Disc Thickness for each wheel đ?‘ đ?‘Ąđ?‘œđ?‘Ą 96 đ?‘šđ?‘š đ?‘ đ?‘¤ = = = 24 đ?‘šđ?‘š đ?‘ 4
â–Ş
Number of discs: đ?‘ đ?‘¤ 24 đ?‘› = đ?‘?đ?‘’đ?‘–đ?‘™ ( ) = đ?‘?đ?‘’đ?‘–đ?‘™ ( ) = đ?‘?đ?‘’đ?‘–đ?‘™(2.4) = 3 đ?‘›Â° đ?‘‘đ?‘–đ?‘ đ?‘?đ?‘ đ?‘“đ?‘œđ?‘&#x; đ?‘’đ?‘Žđ?‘?â„Ž đ?‘¤â„Žđ?‘’đ?‘’đ?‘™ đ?‘ đ?‘‘ 10
In total there must be đ?‘ ∗ đ?‘› = 4 ∗ 3 = đ?&#x;?đ?&#x;? brake discs with a thickness 8 đ?‘šđ?‘š ≤ đ?‘ ≤ 10 đ?‘šđ?‘š â–Ş
Maximum operative temperature as function of disc thickness (s) đ?‘‡đ?‘œđ?‘?_đ?‘šđ?‘Žđ?‘Ľ (đ?‘ ) = đ?‘‡đ?‘Ž +
đ??¸đ?‘Ž = 40°đ??ś + đ?‘?đ?‘ ∗ đ?œŒđ??´đ?‘‘ đ?‘ ∗ đ?‘ đ?‘› 500
48 750 000 đ??˝ 1 ( ) đ?‘˜đ?‘” đ??˝ ∗ 7800 3 ∗ 0.1728 đ?‘š2 ∗ 12 đ?‘ đ?‘˜đ?‘” °đ??ś đ?‘š
1 = 40°đ??ś + 6.028 °đ??ś đ?‘š ( ) đ?‘ Disc thickness can assume different value, it is necessary to analyze its impact on total weight and maximum operative temperature.
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Preliminary design of an aircraft braking system
Thickness [mm]
Total mass [kg]
đ?‘‡đ?‘œđ?‘?_đ?‘šđ?‘Žđ?‘Ľ ≤ 500 °đ??ś
S=8 S=9 S = 10
M = 212 M = 238 M = 264
500 449 408
9 mm disc’s thickness seems to be a good compromise between weight and max temperature
đ?‘ = 9 đ?‘šđ?‘š x 3 đ?‘€ = 238 đ?‘˜đ?‘” đ?‘‡_đ?‘œđ?‘?_ max = 408 °đ??ś
CONFIGURATION – STEEL DISCS
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Preliminary design of an aircraft braking system
3.1.2 Carbon Disc â–Ş
Minimum Total Disc Mass: đ??¸đ?‘Ž 48′ 750′ 000 đ??˝ đ?‘šđ?‘Ąđ?‘œđ?‘Ą ≼ = = đ?&#x;“đ?&#x;?. đ?&#x;‘ đ?’Œđ?’ˆ = đ?’Žđ?’Žđ?’Šđ?’? đ?‘?đ?‘ ∗ Δđ?‘‡đ?‘šđ?‘Žđ?‘Ľ 1250 đ??˝ ∗ (800 − 40)°đ??ś đ?‘˜đ?‘” °đ??ś
â–Ş
Minimum Total Disc Thickness: đ?‘šđ?‘šđ?‘–đ?‘› 51.3 đ?‘˜đ?‘” đ?‘ đ?‘Ąđ?‘œđ?‘Ą = = = 1.06 10−1 đ?‘š = 106 đ?‘šđ?‘š đ?œŒ ∗ đ??´đ?‘‘ 1700 đ?‘˜đ?‘”3 ∗ 10−6 đ?‘š2 đ?‘š
â–Ş
Minimum Total Disc Thickness per wheel: đ?‘ đ?‘Ąđ?‘œđ?‘Ą 106 đ?‘šđ?‘š đ?‘ đ?‘¤ = = = 26.5 đ?‘šđ?‘š đ?‘ 4
â–Ş
Number of discs for wheel: đ?‘ đ?‘¤ 26.5 đ?‘› = đ?‘?đ?‘’đ?‘–đ?‘™ ( ) = đ?‘?đ?‘’đ?‘–đ?‘™ ( ) = đ?‘?đ?‘’đ?‘–đ?‘™(1.77) = 2 đ?‘›Â° đ?‘‘đ?‘–đ?‘ đ?‘?đ?‘ đ?‘“đ?‘œđ?‘&#x; đ?‘’đ?‘Žđ?‘?â„Ž đ?‘¤â„Žđ?‘’đ?‘’đ?‘™ đ?‘ đ?‘‘ 15
In total there must be đ?‘ ∗ đ?‘› = 4 ∗ 2 = đ?&#x;– brake discs with a thickness (s) 13.25 đ?‘šđ?‘š ≤ đ?‘ ≤ 15 đ?‘šđ?‘š â–Ş
Maximum operative temperature as function of disc thickness đ?‘‡đ?‘œđ?‘?_đ?‘šđ?‘Žđ?‘Ľ = đ?‘‡đ?‘Ž +
đ??¸đ?‘Ž = 40°đ??ś + đ?‘?đ?‘ ∗ đ?œŒđ??´đ?‘‘ đ?‘ ∗ đ?‘ đ?‘›
48′ 750′ 000 đ??˝ 1 ( ) đ?‘˜đ?‘” đ??˝ đ?‘ 1250 ∗ 1700 3 ∗ 0.1728 đ?‘š2 ∗ 8 đ?‘˜đ?‘” °đ??ś đ?‘š
1 = 40°đ??ś + 16.6 °đ??ś đ?‘š ( ) đ?‘ It is possible to analyze thickness impact on weight and maximum temperature.
A value of thickness đ?‘ = 14 đ?‘šđ?‘š seems to be a good choice 1 đ?‘‡đ?‘œđ?‘?_đ?‘šđ?‘Žđ?‘Ľ |đ?‘ =0.014 đ?‘š = 40°đ??ś + 16.6 °đ??ś đ?‘š ( ) = 764 °đ??ś 0.014 đ?‘š
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Preliminary design of an aircraft braking system
With this thickness total discs’ mass is: kg
đ?‘€ = đ?‘›đ?‘ ∗ đ?œŒ ∗ đ??´đ?‘‘ đ?‘ = 8 ∗ 1700 m3 ∗ 282′ 700 đ?‘šđ?‘š2 ∗ 14 đ?‘šđ?‘š = 53.8 đ?‘˜đ?‘”
đ?‘ = 14 đ?‘šđ?‘š x 2
CONFIGURATION CARBON DISCS
đ?‘€ = 53.8 đ?‘˜đ?‘” đ?‘‡_đ?‘œđ?‘?_ max = 764 °đ??ś
3.2 â–Ş
â–Ş
Braking Momentum
Brake Disc size: đ??ˇ1 đ?‘&#x;1 = = 325 đ?‘šđ?‘š , 2
đ?‘&#x;2 =
đ??ˇ2 = 125 đ?‘šđ?‘š 2
Lining disc coverage: đ?œƒ = 2đ?œ‹ rđ?‘Ž = 2đ?œ‹ 0.8 = 5.03 đ?‘&#x;đ?‘Žđ?‘‘
Mechanical System Analysis : â–Ş
Vertical reaction during braking in blocking limit conditions: đ?‘ đ?‘Šđ?‘&#x;đ?‘? đ??ż đ?‘€đ?‘” ∗ đ?‘&#x;đ?‘? đ??ż 52000 đ?‘˜đ?‘” ∗ 9.81đ?‘˜đ?‘” ∗ 0.9 ∗ 20đ?‘š Ď•= = = = 404 500 đ?‘ đ??ż + đ?œ‡đ?‘Ą đ??ť đ??ż + đ?œ‡đ?‘Ą đ??ť 20 đ?‘š + 0.9 ∗ 3 đ?‘š
â–Ş
In this conditions on braking wheels there is nearly 80% of aircraft’s weight instead of 90% in static conditions: ϕ 404 500 = = 0.79 � 510 120
â–Ş
Maximum Braking Momentum: on aircraft: đ??ś ≤ đ??śđ?‘šđ?‘Žđ?‘Ľ = đ?œ‡đ?‘Ą Ď• ∗ (đ??ˇ2đ?‘&#x;) = 0.9 ∗ 404500 đ?‘ ∗ 550 10−3 đ?‘š = 200 228 đ?‘ đ?‘š on a single wheel: đ??śđ?‘¤_đ?‘Ąđ?‘œđ?‘Ą ≤ đ??śđ?‘¤_đ?‘šđ?‘Žđ?‘Ľ =
đ??śđ?‘šđ?‘Žđ?‘Ľ đ?‘
=
200228 đ?‘ đ?‘š 4
= 50 057 đ?‘ đ?‘š
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Preliminary design of an aircraft braking system
3.2.1 Steel disc Maximum Pressure on lining: đ?‘ƒ=
đ??śđ?‘¤_đ?‘šđ?‘Žđ?‘Ľ 3 50 057 đ?‘ đ?‘š 3 ∗ 109 = 0.615 đ?‘€đ?‘ƒđ?‘Ž ( 3 )= 3 3 2đ?‘›đ?œ‡đ?‘? đ?œƒ đ?‘&#x;1 − đ?‘&#x;2 2 ∗ 3 ∗ 0.25 ∗ 5.03 (325 − 1253 )đ?‘š3
3.2.2 Carbon Disc Maximum Pressure on lining: đ?‘ƒ=
đ??śđ?‘¤_đ?‘šđ?‘Žđ?‘Ľ 3 50 057 đ?‘ đ?‘š 3 ∗ 109 = 0.769 đ?‘€đ?‘ƒđ?‘Ž ( 3 )= 3 3 2đ?‘›đ?œ‡đ?‘? đ?œƒ đ?‘&#x;1 − đ?‘&#x;2 2 ∗ 2 ∗ 0.30 ∗ 5.03 (325 − 1253 )đ?‘š3
pressure available < 1.3 MPa maximum pressure In both cases pressure requested < pressure available â&#x17E;˘ steel configuration â&#x20AC;&#x201C; less pressure is requested since there are more discs (12) â&#x17E;˘ carbon configuration â&#x20AC;&#x201C; more pressure is requested since there are less discs (8)
X â&#x20AC;&#x201C; axis: pressure on discs Y â&#x20AC;&#x201C; axis: aircraft braking torque
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Preliminary design of an aircraft braking system
3.3
Lining Usury
Absorbed energy in each landing: 𝐸𝑎 = 48 750 000 𝐽 𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑛𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 = (2𝑟𝑎 𝐴𝑑 )𝑠𝑚𝑎𝑥 𝑁 ∗ 𝑛 = (2 ∗ 0.8 ∗ 282700 𝑚𝑚2 ∗ 10 𝑚𝑚 ∗ 4) ∗ 𝑛 = 18 092 800 ∗ 𝑛 𝑚𝑚3
3.3.1 Steel disc 𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑛𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 = (2𝑟𝑎 𝐴𝑑 )𝑠𝑚𝑎𝑥 𝑁 ∗ 𝑛 = 18 092 800 𝑚𝑚3 ∗ 3 = 5.42 107 𝑚𝑚3 𝐽 7 3 𝑐𝑢 ∗ 𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑛𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 2000 𝑚𝑚3 ∗ 5.42 10 𝑚𝑚 𝑁° 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑙𝑎𝑛𝑑𝑖𝑛𝑔𝑠 = = = 2227 𝐸𝑎 48 750 000 𝐽
3.3.2 Carbon disc 𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑛𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 = (2𝑟𝑎 𝐴𝑑 )𝑠𝑚𝑎𝑥 𝑛𝑁 = 18 092 800 𝑚𝑚3 ∗ 2 = 3.62 107 𝑚𝑚3 𝐽 7 3 𝑐𝑢 ∗ 𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑛𝑖𝑛𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 2000 𝑚𝑚3 ∗ 3.62 10 𝑚𝑚 𝑁° 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑙𝑎𝑛𝑑𝑖𝑛𝑔𝑠 = = = 1484 𝐸𝑎 48 750 000 𝐽
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16
Preliminary design of an aircraft braking system
3.4
Result Resume
STEEL ▪ ▪ ▪ ▪ ▪
CARBON 238 kg (112%) 12 (4x3) discs with s = 14mm 449 °C (90%) Pmax = 0.61 MPa 2240 landings
▪ ▪ ▪ ▪ ▪
54 kg (105%) 8 (4x2) discs with S = 9mm 764 °C (95%) Pmax = 0.77 MPa 1480 landings
Concluding brake discs can be realized in steel or in carbo-ceramic materials, each has pro and contros, which are resumed in the following scheme STEEL
CARBON
pro + n. landings without linings substitution + pressure (low)
pro + weight
contro - weight
contro - n. landings without linings substitution - pressure (high)
It is necessary an aircraft’s cost and mission analysis in order to decide which configuration best fits.
▪
N. landings without lining substitution Brake linings need to be changed after a certain number of landings. It is not a difficult operation, being brakes easily reachable by maintainers, but surely it affects on general aircraft’s costs and it keeps aircraft unavailable for a certain time. ▪ Pressure Brake calipers are supplied by an hydraulic system, it is clear that if less pressure is requested the whole system will be easier, lighter and cheaper ▪ Weight In Aeronautics weight cleraly plays an important role. Less weight less lift less drag less fuel
16
Preliminary design of an aircraft braking system
Appendix: Matlab script MATLAB® implemented and used in this report clear all; close all; %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------%% DATA %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------M = 52000; % aircraft mass [kg] v = 50; % landing velocity [m/s] re = 0.75; % energy absorbed by discs [%] E = re*(1/2)*M*(v^2); % energy absorbed by discs [J] Ta = 40; % air temperature [°C] H = 3; % center mass height [m] L = 20; % landing gear pace [m] rp = 0.9; % weight on main landing gear statically [%] N = 4; % n° braking wheels Dr = 1100 * 10^-3; % wheel diameter [m] D1 = 650 * 10^-3; % min diameter disc [m] D2 = 250 * 10^-3; % min diameter disc [m] Ad = pi*( (D1^2 - D2^2) / 4) ; % disc area [m^2] ra = 0.8; % lining area/disc area [%] P_max = 1.3 * 10^6; % max lining pressure [Pa] s_max = 10 * 10^-3; % max lining thickness [m] mu_t = 0.9; % friction tyre-land [ / ] cu = 2000 ; % usury lining coefficient [J/(mm^3)] % steel data Tmax_s = 500; cs_s = 500; rho_s = 7800; sd_s = 10 * 10^-3; mu_p_s = 0.25 ;
% max temperature [°C] % specific heat [ J / (kg °C)] % density [kg / m^3] % max disc thickness [m] % friction lining-disc [ / ]
% carbon data Tmax_c = 800; cs_c = 1250; rho_c = 1700; sd_c = 15 * 10^-3; mu_p_c = 0.3 ;
% max temperature [°C] % specific heat [ J / (kg °C)] % density [kg / m^3] % max disc thickness [m] % friction lining-disc [ / ]
%----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------%% SIZING %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------% STEEL m_tot_s = (E) / ( (Tmax_s - Ta)*cs_s ); % minimum total discs mass s_tot_s = m_tot_s / ( rho_s * Ad); % min total disc thickness sw_s = s_tot_s / N; % min wheel total disc thickness n_s = ceil (sw_s / sd_s); % min n° discs for wheel a_s = m_tot_s / (rho_s * Ad * N * n_s ) ; % min disc thickness b_s = sd_s; % max disc thickness sgrid_s = linspace( a_s, b_s, 100 ); % grid disc thickness t_s = @(s) 40 + ( (E)./(rho_s * Ad * s * N* n_s * cs_s)); % disc temperature function(s) mass = @(s) rho_s * Ad * s * n_s * N; % total disc mass function(s) % graphic temperature/mass function of thickness figure('name','Steel sizing','NumberTitle','off'); subplot(1,2,1); plot( t_s(sgrid_s), sgrid_s,'k', [Tmax_s, Tmax_s], [a_s, b_s],'--r'); grid on; title ('temperature - thickness'); xlabel('temperature [°C]'); ylabel('thickness [m]'); subplot(1,2,2); plot( mass(sgrid_s), sgrid_s,'k', [m_tot_s, m_tot_s], [a_s, b_s],'--r'); grid on; title ('mass - thickness'); xlabel('mass [kg]'); ylabel('thickness [m]'); %____________ % CARBON m_tot_c = (E) / ((Tmax_c - Ta)*cs_c); % minimum total discs mass s_tot_c = m_tot_c / ( rho_c * Ad); % minimum total disc thickness sw_c = s_tot_c / N; % minimum wheel total disc thickness n_c = ceil (sw_c / sd_c); % minimum n° discs for wheel a_c = m_tot_c / (rho_c * Ad * N * n_c ) ; % min disc thickness b_c = sd_c; % max disc thickness sgrid_c = linspace( a_c, b_c, 100 ); % grid disc thickness t_c = @(s) 40 + ( (E)./(rho_c * Ad * s * N * n_c * cs_c)); % disc temperature function(s) mass = @(s) rho_c * Ad * s * n_c * N; % total disc mass function(s) % graphic temperature/mass function of thickness figure('name','Carbon sizing','NumberTitle','off'); subplot(1,2,1); plot( t_c(sgrid_c), sgrid_c,'k', [Tmax_c, Tmax_c], [a_c, b_c],'--r'); grid on; title ('temperature - thickness'); xlabel('temperature [°C]'); ylabel('thickness [m]'); subplot(1,2,2); plot( mass(sgrid_c), sgrid_c, 'k',[m_tot_c, m_tot_c], [a_c, b_c],'--r'); grid on; title ('mass - thickness'); xlabel('mass [kg]'); ylabel('thickness [m]'); %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------%% PRESSURE - LINING %----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------theta = 2 * pi * ra; W = ( M * 9.81 ); No = ( W * rp * L ) / ( L + mu_t * H); pgrid = linspace(0, P_max, 100);
% equivalent lining angle % aircraft weight % vertical reaction on braking wheel in limit condition % pressure grid possible
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Preliminary design of an aircraft braking system
C_max = (mu_t * No * (Dr/2)); % braking momentum in limit condition figure('name','lining pressure','NumberTitle','off'); %_______________________ % STEEL C_ws = @(p) 2*N*mu_p_s * p * n_s * theta * ( ( (D1/2)^3 - (D2/2)^3 ) / 3); % braking wheel momentum function(p) ' p = pressure ' p_s = ( mu_t * No *(Dr/2) * 3 ) / (2* mu_p_s * N*n_s * theta * ( (D1/2)^3 - (D2/2)^3) ); % maximum pressure subplot(2,1,1); plot ( pgrid, C_ws(pgrid),'k', [ p_s, p_s ], [0, C_max],'--r', [0, p_s], [C_max, C_max],'--r'); grid on; title('Steel'), xlabel('pressure [Pa] '); ylabel('momentum [Nm]');ylim([0 4*10^5]);xlim([0 12*10^5]); %_______________________ % CARBON C_wc = @(p) 2*N*mu_p_c * p * n_c * theta * ( ( (D1/2)^3 - (D2/2)^3 ) / 3); % braking wheel momentum function(p) - pressure p_c = ( mu_t * No *(Dr/2) * 3 ) / (2* mu_p_c *N* n_c * theta * ( (D1/2)^3 - (D2/2)^3) ); % maximum pressure subplot(2,1,2); plot ( pgrid, C_wc(pgrid),'k', [ p_c, p_c ], [0, C_max],'--r', [0, p_c], [C_max, C_max],'--r'); grid on; title('Carbon'); xlabel('pressure [Pa]'); ylabel('momentum [Nm]');ylim([0 4*10^5]);xlim([0 12*10^5]); %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------%% N. LANDING %-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------%____________ % STEEL V_p_s = 2*(N * n_s) * (s_max*10^3) * (Ad*10^6) * ra; % total lining volume n_landing_s = ceil ( (cu * V_p_s) / E ) - 1; % n° landings %____________ % CARBON V_p_c = 2*(N * n_c) * (s_max *10^3)* (Ad*10^6) * ra; % total lining volume n_landing_c = ceil ( (cu * V_p_c) / E ) - 1; % n° landings % available lining volume in function of n. landings V_s = @(k) V_p_s - (E/cu) * k; V_c = @(k) V_p_c - (E/cu) * k; figure('name','landings','NumberTitle','off'); plot( 1:n_landing_s , V_s(1:n_landing_s),'r', 1:n_landing_c , V_c(1:n_landing_c),'b', [0,n_landing_s],[0,0],'k'); grid on; legend('steel','carbon'); xlabel('n. landings'); ylabel('available lining volume [m^3]');
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