Aerospace Design Major Task 2- Dive Brake Calculations

AMIT RAMJI 10241445 A2

AMIT RAMJI – A2 – 10241445 – AEROSPACE DESIGN – MAJOR TASK 2 – DIVE BRAKE

%AMIT RAMJI - 10241445 - A2 - UNIVERSITY OF HERTFORDSHIRE

%Aerospace Design Major Task 2 - Dive Brake Calculations

clear %clear all variables

clc %clear command window

%%%Projected Area Calculations

theta=0:5:45;%setting up row array of operating angles in degrees

L=1.864; %length in meters of dive brake from pivot to tip, for dive brake

area and hinge block

W=1.365; %width of dive brake in meters, constant.

H=L*sind(theta); %working in degrees, vertical distance between tip and

meeting fuselage

A=W*H; %projected area of cross section from 5 to 45 degrees in m^2

%%%Force Calculations

%to calculate drag force, F=Cd*0.5*p*v^2*A where Cd is the drag

coefficient, p is the fluid density, v is velocity of fluid (or object

travelling through it) and A is the projected cross sectional area.

p=1.226; %density of air (assumed at operating altitude-but shouldn't make

much difference) in kg/m^3

Cd=0.7; %drag coefficient

V=180; %max velocity in m/s

F=(1/2)*Cd*p*(V^2)*A; %calculated drag force in Newton’s due to movement

through air at 0 to 45 degrees

%%%Centroid Calculations, Dive brake assumed to be an equivalent

rectangular shape.

Wc=W/2; %position of centroid across width of dive brake

Lc=L/2; %position of centroid along length of dive brake

Hc=H/2; %position of centroid along height

%%%Moment calculations

g=9.81; %ms^-2;

mass=15; %kg

M=(F.*Hc+(mass*g*(1/2)*L*cosd(theta))); %moment about hinge pivot due to

drag force at centroid. Using ".*" for element wise multiplication. With

the sum of force added for the mass of dive brake,15kg

%The mg component's distance can also be found using (1/2)*H/tan(theta),

but since we just calculated the distance for H, we will not use this for

best practice to avoid errors.

Phi=[9.029,19.820,30.732,41.799,53.066,64.608,76.546,89.108,102.809,119.451

]; %angle which jack arm makes with spine rail.

G=2.40289; %GradientBetweenthetaAndPhi

C=9.029; %constant (i.e. value of phi at theta=0)

%Phi=[G*theta()+C] %Can Also use this line instead of above Phi but we have

measured phi so why not use it, so we will not need to go back to change

the theta matrix above.

%Phi is actually a function of theta, when theta changes, so does Phi, by a

predictable amount.

Lj=0.498; %Distance from hinge pivot to jack arm pivot

B=Lj*sind(Phi); % Perpendicular distance from line of force to pivot

%F*Hc=Fj*B, To balance moments about Hinge pivot,

(ForceDueToDrag*PerpendicularDistanceOfForceToHingePivot =

JackForce*PependicularDistanceOfForceToHingePivot)

%Perpendicular as moment calculations are cross product vectors.

%So now Arm Jack force required is; F=M(drag)/B

Fja=M./B;% Arm Jack force, which is made up of x and y coordinates, this

jack will be place in the x direction so the jack force must be made up of

entirely vectors in x direction.

Psi=180-(90-theta)-Phi; %angle the jack arm makes with vertical axis

(yaxis).

Psi=90-(theta-Phi); %equivalent angle to the above Psi

%now Psi can be used along with force of jack arm and work out x component,

%as the y component is provided by the rails and underneath structure.

Fjx=Fja.*sind(Psi); %Jack force made up only of horizontal component


Fjy=Fja.*cosd(90-Psi); %NON ESSENTIAL BUT TO GIVE AN IDEA %Forces in member

in the y direction, provided by brackets and structure

FYstructure=max(Fjy) %%NON ESSENTIAL BUT TO GIVE AN IDEA %MAX Force in

member in the y direction, provided by brackets and structure

% Stress analysis and bending moment of steel rail was carried out, gave

sag of 1.729mm which meant increasing rail height(cross section)and adding

bracket.

%%% Efficiency Calculations.

X0=535.499; %distance of trolley from datum (forward end of rod)

x=[X0-X0,X0-524.847,X0-505.547,X0-477.394,X0-440.122,X0-393.308,X0-

336.191,X0-267.209,X0-182.552,X0-68.363]; %distance travveled by jack, it

is x0-value due to direction chosen as positive.

FJack=Fjx/1000; %to get jack force in KN (kilo Newtons)

%when plotting for efficiency it does not

matter which units are used as a ratio of

areas are to be used, but for convenience

let us use mm for

%extension and KN for jack force.

figure, plot(x,FJack,

'color','k','LineStyle','-')

title('A graph showing the force required

by the jack against jack extension');

xlabel('Jack Extension (mm)'); ylabel('Jack

Force (KN)');

legend('Efficiency = Area under Curve /

Entire Area'); grid on; hold on;

%cftool; %OPEN CURVE FIT TOOL

%THEN SELECT DATA THEN CURVE FIT TO THE

DATA, THEN USE INTEGRATE FUNCTION

%function createFit(x,FJack)

%%%Selecting the right Jack

FMax=max(FJack) % Max Jack force in KN, This function filters through the

values of jack force and outputs the highest value.

FMaxSafetyOf2=2*FMax % Safety factor of 2, therefore we will select a jack

of bore=50mm, Max Ext=43.6KN, Max Retrct=39.2KN

%Reason for this is that the bore diameter of 40mm provides a

%MaxExtn=27.9KN, and MaxRetrct=25.1KN. The MaxExtn of 40mm bore is very

%close to out factor of safety (twice the force), so we will select the

40mm bore in this case, however further testing may be required.

%NewSafetyFactor=43.6/FMax DEPENDING ON THE EXTRA MASS OF THE 50mm BORE,

THE 40mm WOULD GO UNDER MORE SCRUTINY, SO THE 50mm WOULD BE USED IN

INDUSTRY FOR ADDITIONAL SAFETY AS THIS COMPONENT IS AN IMPORTANT COMPONENT

DURING FLIGHT. But by possibly compensating in additional mass.

%%%FROM CFTOOL INTEGRATE AT XI=467.136 INTEGRAL FROM XI=0 TO XI=467.136 IS

%4917.33

EfficiencyPercentage=(4917.33./(max(x)*FMax))*100 %where max(x)=467.136 is

the range in x (extension range) and Fmax is the range in y (range of

force). Therefore the area under curve using 4th degree polynomial curve

fit tool divided by total are of graph is efficiency as a decimal, multiply

by 100 to get as percentage.

%%

TotalStrokeLengthIfMaxAt60Percent=max(x)/6*10 %Therefore we shall require a

jack bore length in excess of 780mm

OUTPUT >>

FYstructure =1.2917e+004

FMax =12.9170

FMaxSafetyOf2 =25.8340

EfficiencyPercentage = 81.4939

TotalStrokeLengthIfMaxAt60Percent = 778.5600


θ

φ φ

180-(90-θ)-φ

90-θ ψ

Lj.Sin(φ)

Lj

Fja

Fja.sin(ψ)

Fja.cos(ψ)

Hc/2

Fair

O

AH BG

DE CF BG AH

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This drawing is our property; it can't be reproduced or communicated without our written agreement.

SCALE

1:1WEIGHT (kg) DRAWING NUMBER

Aft-End Jack Bracket

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A3 Amit Ramji - 10241445

CHECKED BY:

Amit RamjiDATE:

10th Jan 2012

DESIGNED BY:

Amit RamjiDATE:

10th Jan 2012 Aerospace Design Major Task 2

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Notes:DO NOT SCALE OFF DRAWING- All Dimensions in mm- General Machine Tolerance 0.01 mm (Unless Otherwise Stated)- Deburr all edges- Material: Aluminium Alloy 2024-T3

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DESIGNED BY:

Amit RamjiDATE:


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Notes:DO NOT SCALE OFF DRAWING- All Dimensions in mm- General Machine Tolerance 0.01 mm (Unless Otherwise Stated)- Deburr all edges- Finish & Storage: Coat in oil- Material: Steel S96

1812

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Amit RamjiDATE:


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I _Notes:DO NOT SCALE OFF DRAWING- All Dimensions in mm- General Machine Tolerance 0.01 mm (Unless Otherwise Stated)- Deburr all edges- Finish & Storage: Keep in oil- Material: Steel S96


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DESIGNED BY:

Amit RamjiDATE:


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Vertical Y-Arm

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Amit RamjiDATE:


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Dive Brake Assembly

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CHECKED BY:

Amit RamjiDATE:

10th Jan 2012

DESIGNED BY:

Amit RamjiDATE:


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Notes:DO NOT SCALE OFF DRAWING- Finish & Storage: Keep Roller Wheels and Rail Track Oiled Using B-373-SAE-90 Oil.

Isometric view Stringers and Skin OmmitedScale: 1:8

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Dive Brake Assembly 2

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CHECKED BY:

Amit RamjiDATE:

10th Jan 2012

DESIGNED BY:

Amit RamjiDATE:


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Notes:DO NOT SCALE OFF DRAWING- Finish & Storage: Keep Roller Wheels and Rail Track Oiled Using B-373-SAE-90 Oil.


To be riveted around FS6445To be riveted around FS5545

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CHECKED BY:

Amit RamjiDATE:

10th Jan 2012

DESIGNED BY:

Amit RamjiDATE:


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Notes:DO NOT SCALE OFF DRAWING

Exploded View: Stringers, FS4645, FS5545, FS5995, FS6445, Hinge Block & Skin Ommitted Scale: 1:12

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Quantity Part Name Balloon Number1 HingeBlock -1 SpineRailPortSide 101 SpineRailStarboardSide 115 DoubleCChannelFrame 126 Stringer -1 FrontRailBracket 32 FrameBracketInner 82 FrameBracketOuter 92 Y-Arm 72 Rail 21 RailJackBracket 48 RollerWheel 151 Aft-JackBracket 11 JackPistonRod 141 JackOuterCasing 138 EN 22341 CLEVIS PIN 8X45 STEEL WITH

HEAD TYPE B WITH SPLIT PIN HOLE 24

6 EN 22341 CLEVIS PIN 20X60 STEEL WITHHEAD TYPE B WITH SPLIT PIN HOLE

18

1 EN 22341 CLEVIS PIN 22X160 STEELWITH HEAD TYPE B WITH SPLIT PIN HOLE

16

8 EN ISO 1234 SPLIT PIN 2X12 STEEL FORBOLTS AND SCREWS

24

8 EN ISO 1234 SPLIT PIN 5X25 STEEL FORBOLTS AND SCREWS

25

4 EN ISO 4016 BOLT M12X55 STEEL GRADEC HEXAGON HEAD

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4 EN ISO 4016 BOLT M10X70 STEEL GRADEA HEXAGON HEAD

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8 EN ISO 4016 BOLT M10X45 STEEL GRADEA HEXAGON HEAD

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1 EN 22341 CLEVIS PIN 20X50 STEEL WITHHEAD TYPE B WITH SPLIT PIN HOLE

17

2 EN ISO 4016 BOLT M10X100 STEEL GRADEC HEXAGON HEAD

21

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SCALE


Dive Brake Assembly

SHEET

1/1

SIZE


CHECKED BY:

Amit RamjiDATE:

10th Jan 2012

DESIGNED BY:

Amit RamjiDATE:


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Notes:DO NOT SCALE OFF DRAWING- All Dimensions in mm (Unless Otherwise Stated)

Dive brake at 45Scale: 1:14



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Aerospace Design Major Task 2- Dive Brake Calculations