aileron presentation gopal

AE 581

DESIGN AND STRESSING OF AIRCRAFT COMPONENTS

ByGOPAL KRISHNA OJHA

10166762

UNDER EXPERT GUIDANCEOF

PROF. JOHN DAVIES

ROYAL MILITARY COLLEGE OF CANADA

AN AILERON IS AN HINGED FLIGHT CONTROL SURFACE USUALLY FORMING PART OF TRAILING EDGE OF EACH OF A FIXED WING AIRCRAFT. AILERONS ARE USED TO CONTROL THE AIRCRAFT IN ROLL.

roll

DESIGN AND STRESSING OF AILERON MECHANISM

OBJECTIVE / REQUIRMENTS OF DESIGN

To design the aileron operating mechanism of a civilian aircraft is mounted aft of the wing spar and operated by a single input rod of 1000mm length. A detailed structural design is required for the following :-

• The operating system that links together the two push-pull rods.

• The bearings and their mountings. Suitable bearings can be selected from applicable catalogues or websites.

• The aileron operating rod. Assume there is a fork end fitting on the aileron structure.

• The aileron deflection is + - 18degrees.


Loads are calculated in accordance with the Canadian aviation regulations CAR 523,

‘ NORMAL UTILITY AEROBATIC AND COMMUTER CATEGORY AEROPLANES’

SUBCHAPTER C – STRUCTURE.

• The aircraft has a dual manual control system which must be tolerant to jamming.

• The critical design case has been derived from CAR 523.397 maximum pilot effort loads

and CAR 523.399 for dual control systems.

• 75% of two pilots acting together corresponds to a limit load of + - 3100N in the input rod.

• In accordance with CAR 523.303, the mechanism and its supporting structure are to be

stressed using proof and ultimate factors of 1.0 and 1.5 respectively.


REQUIREMENTS

• The mechanism shall be designed such that the ratio of movement of the input rod to the

output rod is 1.33:1.

• The mechanism should be easy to maintain and dismantle for the replacement of worn

bearings and damaged parts.

• The input rod must be capable of length adjustment to permit rigging of the Aileron.

• Investigate whether the surrounding aircraft structure has adequate strength to support your

installation.

• Reinforce locally if required.


DESIGN EXERCISE


CHALLENGES OF DESIGN

There were many design challenges faced during design phase:

Keep structural integrity of supporting structure, while working in the space constrain.

To Use of standard parts in attempt to reduce cost and production time

Weight reduction

To protect mechanism from corrosion

To design a simple mechanism to Convert linear motion of input rod into angular motion

of aileron (- + 18O )

To design a mechanism that is easier to operate (follow CAR 523.329 and CAR 523.399,

maintain and assemble.


SELECTION OF MECHANISM

Hydraulic mechanism are much easier to use and can multiply pilots efforts easily

Although hydraulic system uses tandem cylinders and mater cylinder, is still venerable to leakage failures of line or seals or valves, leading to disastrous results.

There is an alternate fail safe mechanism inn hydraulic system, which takes over if one line or one system fails. But this second system allows aileron operation at a reduced hinge movement.

Also hydraulic systems are hard to maintain, and care must be taken as not to allow air bubbles in the fluid tank and a minimum level of hydraulic fluid level is to be maintained in the cylinder, necessitating the need to monitor and topping up of fluid.

Mechanical systems are therefore more reliable requiring only scheduled maintenance.

Mechanical systems may include cables, gears, pulleys, rack and pinion type arrangement etc however uses of push and pull rods along with bell crank seems to be viable choice.


Assembly view


upper assembly


central assembly


lower assembly


Output assembly



Margin of safety

ITEM DESCRIPTION

LOAD CASE MARGIN OF SAFETY

WEIGHT

INPUT ROD BUCKLING 0.008 547.5 gm

OUTPUT ROD BUCKLING HIGH 95 gm

INPUT FORK BENDING >2 38.72 gm

OUTPUT FORK BENDING >2 28 gm

UPPER SUPPORT BENDING 0.93 58.05 gm

LOWER SUPPORT BENDING HIGH 51.3 gm

SLEEVES INPUTOUTPUT

HIGHHIGH

3 X 41gm


ITEM DESCRIPTION

LOAD CASE MARGIN OF SAFETY

WEIGHT

UPPER TENSION FITTING

BENDING OF END PAD

0.72 23 gm

LOWERTENSION FITTING

BENDING OF END PAD

HIGH 23 gm

OUTPUT LUG SHEAR HIGH 31.5 gm

CENTRAL PIVOT TORSION

BENDING

0.45

0.56

250 gm

EYE ENDS TENSION INPUT

TENSION OUTPUT

0.46

0.09

3 X 21 gm

TOTAL WEIGHT 1.450 Kg


STRESS ANALYSIS

• INPUT ROD - BUCKLING

• Material AMS 4081 – AL6061 – Drawn tube

•Euler’s formula for columns WCR = C π2 E I/ L2

• For column’s with both ends hinged C (coefficient of end fixity) = 1

• Moment of inertia (I) = 6991.8 mm4

• E = 9.9 x 103 ksi, L = 1000 mm

• Which gives required crippling load of 4687.73 N, however applied load is 4650 N (1.5 times of 3100 N)

• M.S = (4687.73 / 4650) – 1 = 0.008

……….. Stress analysis report page 40


• Material AMS 4081 – AL6061 – Drawn tube

•Euler’s formula for columns WCR = C π2 E I/ L2

• For column’s with both ends hinged C (coefficient of end fixity) = 1

• Moment of inertia (I) = 6991.8 mm4

• E = 9.9 x 103 ksi, L = 265 mm

• Which gives required crippling load of 71.6 KN, however applied load is 6184.5 N (1.33 times of 4650 N)

• M.S = (4687.73 / 4650) – 1 = 0.008

• OUTPUT ROD - BUCKLING ……….. Stress analysis report page 44


• Material AMS 4126 – AL7075 – T6 die forgings

• Section modulus for H section = ZYY = 588 mm3

• Applied Bending stress = σB = M / ZYY

• Applied Bending moment = M = 176700 N-mm

• σB =300.5MPa

• Using section shape factors [6][8] allowable bending stress = 717 MPa • Ultimate allowable bending moment = MULT = 552570.39 N-mm

• M.S = (MULT / M) – 1 = 2.12 M.S >2

• INPUT FORK - BENDING ……….. Stress analysis report page 51


+4650N

38

-176700 N-mm

- 4650N

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

- 176700 N-mm

38

+4650N

Shear force

Bending moment

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

38

38

176700 N-mm

Shear force

Bending moment

- 4650N



• Section modulus for H section = ZYY = 588 mm3


• Applied Bending moment = M = 158323.2 N-mm

• σB =269.5MPa

• Using section shape factors [6][8]allowable bending stress = 717 MPa • Ultimate allowable bending moment = MULT = 552570.39 N-mm

• M.S = (MULT / M) – 1 = 2.49 M.S >2

• OUTPUT FORK - BENDING ……….. Stress analysis report page 53


Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

-158323.2 N-mm

25.6

+6184.5N

Shear force

Bending moment

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

25.6

25.6

158323.2 N-mm

Shear force

Bending moment

- 6184.5 N25.6

• UPPER SUPPORT - BENDING



• Section modulus for channel section = ZYY = 283.73 mm3


• Applied Bending moment = M = 379320 N-mm

• σB =196.3MPa

• Using section shape factors [6][8]allowable bending stress = 951.4 MPa • Ultimate allowable bending moment = MULT = 733215.43 N-mm

• M.S = (MULT / M) – 1 = 0.93

…….Stress analysis report page 63


Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

-379320 N-mm

97.5

+3890.5N

Shear force

Bending moment

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

97.5

97.5

379320 N-mm

Shear force

Bending moment

- 3890.5 N97.5



• Section modulus for channel section = ZYY = 1931.9 mm3


• Applied Bending moment = M = 56256.165 N-mm

• σB =29.1MPa

• Using section shape factors [6][8]allowable bending stress = 951.4 MPa • Ultimate allowable bending moment = MULT = 733215.43 N-mm

• M.S = (MULT / M) – 1 = 12 M.S HIGH

• UPPER SUPPORT - BENDING …….Stress analysis report page 66


Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

74.07

74.07

56256.1 N-mm

Shear force

Bending moment

- 759.5 N

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

- 56256.1 N-mm

74.07

+759.5N

Shear force

Bending moment

74.07



• Section modulus = ZYY = 792.25 mm3

• Bending stress on section B-B is σb B-B = [F (2d - t) K3] / [t2.a] = 416.3 Mpa

• Allowable bending stress or rupture strength = Fb = fm + fo(k-1) = M/I

• fm = Ftu = 71ksi = 489 Mpa

• fo = 67 ksi = 461.9 Mpa

• k= 1.5

• Using section shape factors [6][8]allowable bending stress = Fb = 104.5 ksi = 717 MPa • M.S = (717 / 416.3) – 1 = 0.72

• UPPER TENSION FITTING – BENDING of END PAD …….Stress analysis report page 72



• Section modulus = ZYY = 792.25 mm3

• Bending stress on section B-B is σb B-B = [F (2d - t) K3] / [t2.a] = 132.2 Mpa

• Allowable bending stress or rupture strength = Fb = fm + fo(k-1) = M/I

• fm = Ftu = 71ksi = 489 Mpa

• fo = 67 ksi = 461.9 Mpa

• k= 1.5

• Using section shape factors [6][8]allowable bending stress = Fb = 104.5 ksi = 717 MPa • M.S = (717 / 132.2) – 1 = 4.4 high M.S

• LOWER TENSION FITTING – BENDING of END PAD …….Stress analysis report page 76



• Kbru = shear bearing efficiency = 1.5

• Area = A = π (122 - 62) / 4 = 84.8mm2

• F / (Kbru x A) = σ = 48.6 MPa

• Allowable stress = 489 MPa • M.S = (489 / 48.6) – 1 = 9.06 high M.S

• OUTPUT LUG – SHEAR …….Stress analysis report page 78


• Material AISI 4130 alloy steel

• Section modulus = Z = 394 mm3

• ME = equivalent bending moment = (1/2) { M + [ M2 + T2]1/2 } = 217422.9 N-mm

• M = net bending moment = 155267.25 N-mm

• T = net torque = 232500 N-mm

•Bending stress σ B = ME / Z = 551.8 Mpa

• Allowable bending stress = 861.8 Mpa • M.S = (861.8 /551.8) – 1 = 0.56

• CENTRAL PIVOT – BENDING …….Stress analysis report page 63


Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

150

150

Shear force

Bending moment

- 759.5 N

Y(N)

X(mm)

X(mm)

Y(N-mm)

0

0

95317.25 N-mm

150

+3890.5NShear force

Bending moment

759.5N

15024.5 126.5

+ 969N

5215.5N

24.5

126.5

122566.48 N-mm


• Material AISI 4130 alloy steel

• TE = equivalent torsion = [ M2 + T2 ]1/2 = 279578.5 N-mm

• M = net bending moment = 155267.25 N-mm

• T = net torque = 232500 N-mm

•

• Allowable shear stress = 517.1 Mpa

• TE = {π τ (dO4 – dI

4)} / (16dO), gives shear stress applied τ = 354.6 Mpa • M.S = (517.1 /354.6) – 1 = 0.45

• CENTRAL PIVOT – TORSION …….Stress analysis report page 63

FEATURES OF DESIGN

Features of this design are as follows:

Uses H and channel section for forks and supports for reduced weight and higher strength

Use of standard parts already available in market. E.g. use of AMS 4081 AL6061 drawn tubes OD 7/8” and ID 0.5” from ALCOBRA METALS INC [1] is machined.

Other examples are those of maintenance free SKF spherical bearings and eye ends [1], splined rod , standard size nuts and bolts.

Complete weight of whole mechanism with supports is 1..450 Kg

Design uses a simple bell crank mechanism that is easier to mount and un mount i.e maintain



Parts can be easily replaced if damaged

As per requirements uses ultimate factor of 1.5

Only components needed to be manufactured are supports, output lug, tension fittings

and forks.

Design provides a alignment hole for neutral position.

Mechanism can be dismantled by displacing the input or output rod and then

un mounting the supports.

• Although AL is naturally protected to corrosion as it forms a thin layer of Al2O3 , but still parts are susceptible to corrosion form environment including from oceanic/ salty environment, galvanic corrosion, pitting, chemical reactions from fuels.

• Protection from corrosion can be done using paints and primer coatings, electroplating, anodizing, grease.

• The choice over here in this design is use of ALODINE EC2 Electro Ceramic Coating of HENKEL Surface Technologies. [7]

• ALODINE EC2 is non hazardous under 29 CFR 1910.1200 (hazard communication)

• Is a 2-15 micron ceramic coating which is a chromium free product specially formulated for coating AL, titanium and their alloys.


Corrosion protection


• Does not add weight or change dimensions as does the paints and other coating does.

• Easy to apply, environmentally safe

• Extends life of coated components

• Secondary coatings may be used, pH 2.4 to 3.0

• Protects from corrosion even when scratched, E.g. Alodined AL2024 withstand salt spray for 150 – 160 hrs, whereas untreated AL2024 withstand less than 24 hrs before forming white corrosion.

• Simple hex nuts can be replaced by lock nuts, or drilled jam nuts

• Countersunk rivets could be used instead of blind rivets, wherever they expose to outer surface

• Ribs can be used for strengthening supports

• Suitable tolerances must be used while machining

• Surface finish of parts machined is good, eliminating need of bushes and bearings at many places.

• Slotted shoulder bolts could be used.

• Corrosion protection must be done as per the procedure specified in catalogue by HENKEL technologies[3].

• It is preferred to machine (mill) parts from solid block of material in order to retain grain direction.

• Regular schedule maintenance must be done of mechanism, and must be checked for area of crack propagation.

RECOMMENDATION


1) http://www.skf.com/binary/96-122020/6116_1-EN.pdf

2) ‘ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES’ BY E.F.BRUHN, B.S, M.S, C.E, Dr. Eng.

3) http://na.henkel-adhesives.com/us/content_data/232666_Alodine_EC2EN.pdf

4) LOCKHEED STRESS MEMO NO. 88a, SEPTEMBER 15, 1955, REVISED 1, 1968


References

http://www.skf.com/binary/96-122020/6116_1-EN.pdf

http://na.henkel-adhesives.com/us/content_data/232666_Alodine_EC2EN.pdf

http://na.henkel-adhesives.com/us/content_data/232666_Alodine_EC2EN.pdf

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