AIRCRAFT LOADING

Overall Consideration

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Requirements

Any flying machine, whether it be a manned aircraft or a guided missile, is required to possess

a structural strength and stiffness which is adequate to enable it to perform its intended role, and continue to perform it throughout its life. The requirements contain much more than conditions to ensure structural strength and stiffness. The scope ranges over performance, handling, detail

design of systems and installations and general operating requirement and procedures.

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Before an aircraft can be operated it is necessary to demonstrate that it meets all the relevant requirements and any special ones which may be necessary.

When the airworthiness authority is satisfied with the general performance and design of the aircraft permission is given to operate it. The permission may be restrictive in some sense and

may not be approved or recognized by other authorities, although it is now normal for design teams to give consideration to all sets of requirements which are likely to be met with.

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FAILURE PROBABILITIES

These probabilities of recurrences which may apply to aircraft structure and systems :-

1. Frequent An event likely to occur often during the life of an aircraft,

numerically more frequent than once in 1000 hours or flights as appropriate.

2. Reasonably probable An event which may occur several times in the life of an aircraft, that is once in 103 to 105 hours or flights 3. Remote Unlikely to occur to any given aircraft in its total life, but

which may occur several times in the total fleet of that aircraft. Once in 105 to 107 hours or flights

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4. Extremely remote Unlikely to occur at all to any aircraft but which nevertheless must be considered. One occurrence in more than 107 hours or flights. 5. Extremely improbable So unlikely that it does not have to be considered as a possibility. Numerically it is relevant to the case where a sequence of extremely improbable failures could, in total, amount to an extremely remote one.

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Loading Action

The aim of loading action analysis is to obtain cases which enable the airframe to be designed and stressed. Many of the methods used for estimating the

loads acting on aircraft are complex and involve a knowledge of

parameters which can only be determined accurately at the later stages of the design.

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Loads and Factors Cause of loads Loads experienced by aircraft fall into two

categories. 1- loads result directly from a) The action of the pilot or the autopilot which are classified as manoeuvring loads since they occur

as the aircraft carries out its intended role.

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Cabin pressure differential and the effect of kinetic heating at higher Mach number can be added to

the above

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The second type of loads arise because of 2- The environment in which the aircraft operates. These may be due to such things as :- a) Atmospheric turbulence, b) pressure differential, c) kinetic heating or d) runway unevenness. Basically this type of loading would not occur in ideal

situation but has to be tolerated in practice.

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Frequency of Loads

Whether the loads be “manoeuvre” or “environmental” they have to be dealt with in two ways for design purposes.

1. There are the so called “limit load” conditions.

The limit load is the actual maximum load of a

particular case which can be experienced in normal conditions. Thus a failure due to the application of limit load must be in the extremely remote

category.

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That is the limit load is one which can be anticipated to be reached, but not exceeded, in 107 hours or flights.

It is in fact the actual maximum load for a particular manoeuvre or

environmental condition and represents the most severe isolated load intensity.

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2. There is a set of loads of varying magnitude which is experienced

by the airframe throughout its life and which arise in any given

category.

Often the majority of these loads are small in comparison with the

limit value, but each may damage the structure and it is necessary

to ensure that the total accumulated damage due to all of them is

within the capability of the structure.

These loads may arise in a specific way and be of known magnitude

and frequency, an example being cabin pressurization.

The majority of these loads cannot be dealt with as simply as this

and some, especially those due to atmospheric turbulence or runway

unevenness are essentially random in nature.

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Load Factors

Some factors are used for the structure design of the aircraft superimposed on the limit load.

They are called a) the proof factor, b) the ultimate factor and c) the load factor.

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Loading factors Load factors are applied to limit load cases. There are two factors which are used :- 1. The first is the Proof Factor which is 1.125 for military and 1.0 for civil aircraft. Under proof loading conditions the airframe must not distort permanently more than a small, specified amount, usually 0.1 – 5% permanent strain depending upon the type of loading. This factor is intended to ensure that the structure will always return to shape after design loading has been applied.

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Secondly there is an Ultimate Factor which in most instances is 1.5 times the limit load in both military and civil requirements.

The Ultimate Factor is a safety factor on the limit load and is intended

to cover such items as variation of material and structural properties

outside the specified limits, deterioration in service, inadequacy of load and stress analysis

assumptions, and possible flight of the aircraft just outside the

stated design limitations.

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During maneuvres or flight through turbulence additional loads are imposed which increase or decrease the net loads on the structure of the aircraft. The amount of this additional load depends upon the severity of the maneuvre or the turbulence, and its magnitude is measured in terms of load factor.

The load factor is a multiplying factor defines load in terms of weight.

The load factors are based on statistical data.

Transport aircraft have values 2 to 3 while fighters aircraft have values 6 to 8.

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life factor

Designers should ensure that the structure is able to continue resisting any design load for a longer period than its design period. The structure should be designed to have an estimated life or safe life three or more times the actually intended in service. This ratio of the demonstrated life to the actually intended is called the life factor.

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The actual numerical value of the life factor is determined by the statistical accuracy of the available design information, especially in relation to actual tests on both components of the structure and the whole airframe.

Structure Life Measures are necessary to safeguard the

integrity of the structure when it is subjected to numerous repetitions of loads over a life which may be more than 40000 flying hours.

Further, flight in transonic and supersonic regimes introduces nonlinearity of load distributions and possibly temperature effects due to kinetic heating.

Each load in the complete spectrum of loads appropriate to a particular case has to be considered, and since the load is usually relatively low the strain resulting from it is elastic, the effect of stress concentration is of vital importance.

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The effect of stress concentrations is of vital importance.

The cumulative damage from all the loads must be evaluated to ascertain the final integrity of the structure at the end of its life.

As accurate specification of the load spectrum is difficult , it is necessary to incorporate safeguard to cover the unknowns.

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Safe Life

Life factor The actual numerical value of the life factor is

determined by the accuracy of the available design information in

relation to actual tests on components of the structure and

to factor each individual load and stresses resulting from it, before evaluating the safe life.

It is a complex process but will enable a more reliable result to be obtained.

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Damage tolerant design

Failsafe It is not unreasonable to accept that the loading

and its effect upon the structure cannot be accurately predicted and that any factors applied to ensure the integrity of the structure are a design penalty.

The structure should be designed so that should a failure occur there is always an alternative load path to enable the aircraft to complete its mission safely.

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The failure would have to be easily detectable on subsequent inspection and repairable. This is the Damage Tolerance approach.

It Should be pointed out that any properly designed structure will possess both damage tolerance and safe life feature.

In practice the degree of damage which may be

tolerated is determined by the rate of crack propagation relative to inspection intervals.

There are some parts of the aircraft where it is impossible to incorporate the damage tolerance idea,

so it important to introduce a reliable means of failure warning which indicate the onset of a crack before it becomes catastrophic.

This is done on mechanical components such as rotor blades.

It is the practice of the designers of aircraft to specify life in the context of repairs and replacements.

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Statistical design Ideally the eventual way to design a structure

would be entirely based upon a statistical approach.

Damage tolerance features would still be incorporated but the emphasis would be on designing the structure to be able to

withstand any possible load it might experience and the total of such possible loads.

This may basically would involve an accurate

knowledge of :-

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1. The total loading spectrum on the most unfortunate aircraft of the type to be produced. 2. The maximum limit load to a stated probability 3. The behaviour of the structure and its individual members under the total loading spectrum and the maximum limit load.

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Loading Cases

All airplanes are designed to possess symmetryabout a vertical plane passing through the centrelineof the body. However manned aircraft are asymmetricwith reference to a horizontal plane passing throughthe body centreline.Arrangement possessing single symmetry can besubjected to both symmetric and asymmetric loadingcases.

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Categories of load cases 1. Symmetric flight case :- This arises either due to pilot initiated manoeuvre in the longitudinal or pitching plane or from asymmetric vertical air turbulence or gust. 2. Asymmetric flight case :- This arises when a pilot operates his rudder to initiate a yawing motion or his ailerons to initiate rolling. Also engine cut has the same effect. Horizontal air turbulence, a cross wind, or a non symmetric vertical gust will also give rise to asymmetric loads.

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Additional loads

a) grounds loads b) Longitudinal loads c) Local loads

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3. Ground case :- From the point of view of the undercarriage design the normal take off and landing loads are basically symmetric,

but landing with sideways drift introduces an asymmetric component, and

there is also a special asymmetric one wheel landing case. These landing loads are the result of the vertical deceleration which occurs when the aircraft lands or encounters a runway irregularity and they are associated with fore and aft deceleration.

Ground maneouvering is to be considered as well.

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4. Longitudinal load case :- Fore and aft loading usually quite small on aircraft except for arrester landings and crash cases.

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Design Speeds

The structural loading requirements areexpressed in terms of a set of design speeds,complemented where appropriate by thecorresponding Mach numbers.Equivalent airspeed is used in most cases and itis that speed related to the sea level air densityused to define the dynamic pressure.

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Stalling speed, VS

It is defined as the minimum steady flight speed at whichcontrol can be maintained VS = (2ngm/ρ0SCNMAX)½ (1)

Where ρ0 is the sea level density

VS is an equivalent air speed

S is the wing area g is gravitational acceleration m is the mass n is the nominal acceleration factor, one in level flight CNMAX is the maximum normal force coefficient at a Mach number

corresponding to a true speed at a given altitude . It is normally equal to the lift coefficient

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There are some definitions of stalling speed

a) VS1 is the stalling speed in a specified condition, typically with high-lift devices, and undercarriage retracted and with engines idling.

b) VS0 is the stalling speed in the landing configuration

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Manoeuvre speed VA

It is the lowest speed at which the aircraft can attain the prescribed maximum limit normal manoeuvre factor n1

VA = VS1/(n1)½ (2)

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Design cruising speed VC

This speed is intended to cover the maximum normal operating condition speed VNO.

JAR-25 (large aircraft) VC must be sufficiently

greater than the gust design speed VB.

Thus VC =VB+1.32Uref (Uref reference gust velocity )

providing that VC does not exceed the maximum

speed in level flight for the corresponding altitude36

for light civil aircraft (JAR-23) VC (knots) is equal

to 33(W/S)½ for W/S below 20 lb/ft2 and to 28.6(W/S)½ for W/S of 100lb/ft2 except for aerobatic aircraft where the value is 36(W/S)½

W is weight in lbs (mg) and S is the reference wing area in ft2.

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Maximum horizontal speed VH

VH is defined as the maximum speed attainable

in level flight with powerplant set at the maximum

continuous cruise condition.

For military aircraft designed for dive bombing or ground attack duties VH is assumed to be equal to

VD

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Design (diving) speed, VD

It is not the maximum speed the aircraft could attain in a dive of specified steepness. In case of aircraft designed to JAR-25 the definition depends on whether the aircraft is designed to operate into the transonic range or not. For an aircraft flying at low speed VD can be set

at 1.25VC or MD at 1.25MC

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Gust speed VB

It is the design speed for maximum gust intensity. JAR-25-335 state that VB may not be less than

VBMIN = VS1(nG+1)½

Where nG is the incremental load factor resulting from the aircraft encountering a gust of magnitude Uref when flying at a speed VC

VBMIN need not exceed VC.

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To determine VBMIN the air gust is assumed to

be 20 m/s (66ft/s) EAS between sea level and

6097 m (20000 ft) then falling linearly to 11.6 m/s

(38ft/s) EAS at 15240 m (50000 ft)41

Flap and high-lift devices design speeds

JAR-25 defined VF for each flap or high-lift device setting as not less than

a) 1.6 times the stalling speed at the maximum take off mass with the high lift device in the take-off position, or b) 1.8 times the stalling speed at the design landing mass with the high-lift devices in the approach position, or c) 1.8 times the stalling speed at the landing mass with the high-lift devices in landing position.

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Flight Loading

Due to induced manoeuvre by the pilot or auto pilot the aircraft will change from steady level flight condition to a climbing or diving path. The greatest loads arise when the aircraft pulls out of a dive or when it is in a banked turn. Symmetric manoeuvre is a general case

covering a number of specific loading conditions

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For an aircraft entering a dive from steady level flight ( see fig) the pilot moves the pitch motivator ( elevator or horizontal stabilizer) to induce a nose down pitching acceleration. The aircraft gains a nose-down pitching velocity ‘q’ causing it to fly on a circular path with a corresponding centrifugal load which has to be balanced by the lift which is in this case is

negative.44

When the aircraft approaches the required angle of dive the pilot uses opposite motivator deflection to arrest the pitching velocity. At the end of the dive reverse procedure is done.

In general the aircraft must be designed to withstand lifting loads greater or less than the weight, and with or without nose-up or nose-down pitching accelerations.

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To ensure safety, governmental agencies have put some requirements for structural integrity.

Load factor (n) expresses structure-strength requirements which associated with the n-V diagram.

Velocity-Load Diagram (V-n )or

Flight envelope (n-V )diagram

The symmetric flight loading can be described by a flight envelope of forward velocity and acceleration perpendicular, or normal to the flight path. The greatest air loads usually come from the generation of lift during high-g manoeuvre. The factor ‘n’ represents the manoeuvreing of an aircraft as a multiple of the standard acceleration due to gravity. In operational circumstances usual manoeuvre loads lie within the boundaries of the envelope.

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Newton’s Law states that change from an equilibrium state requires an additional applied force which is associated with some form of acceleration, ‘a’. When applied in the pitch Plane, the force appears as an increment in lift, ∆L, and would overcome the weight to an

increased altitude initiated by rotation of the aircraft

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ΔL = centrifugal acceleration X mass = a * W/g The resultant force is going to be L + ΔL = W + (a * W/g) = W(1 + a/g) L is the lift equalling weight W thus load factor

‘n’ is defined as n = (1 + a/g) = (L/W) + (ΔL/W) = 1 + ΔL/w The load factor ‘n’ indicates the increase in

force contributed by the centrifugal acceleration, ‘a’.

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n = 2 indicates a twofold increase in weight i.e a 100 kg person would experience a 200 kg weight. The load factor ‘n ‘ is also the g-load

When an aircraft experiences a force more than its normal weight it is called Positive load and vice versa. A high g-load damages the human body with the human limits of the instantaneous g-load higher than for continuous g-load.

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g-load is associated with pitch plane manoeuvres, pitch changes are related to change in the angle of attack ‘α’, and the velocity V.

There is a variation in CL up to its limit of Clmax in both positive and negative sides of the wing incidence to airflow.

This is the relation represented in the n-V diagram.

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Limits- Load and Speeds Limit load is defined as the maximum load that an aircraft can be subjected to in its life cycle. Under the limit load, any deformation recovers to its original shape and would affect structural integrity. Structural performance is defined in terms of stiffness and strength. Stiffness is related to flexibility and deformations and has implications for aeroelasticity and flutter.

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Low speed limit

The low speed limit in a V-n diagram is established at the velocity at which the aircraft stalls in an acceleration flight load of ‘n’ until it reaches the limit load factor.

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The main design of an aircraft can be given by stating the limiting values of the acceleration and speed in addition to maximum value of the applied gust velocity. The design loading requirements for a certain aircraft could be stated as follows “”The proposed aircraft shall be designed for applied positive and negative accelerations of + 6.0 g and – 3.5 g respectively at all speeds from that corresponding to Clmax up to 1.4 times the maximum level speed. Furthermore the aircraft shall withstand any applied loads due to a 30 ft/sec gust acting in

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any direction up to the restricted speed of 1.4 times the maximum level speed. A design factor of safety of 1.5 shall be used on these applied loads.”” The design requirements can be represented by

plotting load factor and velocity to obtain a diagram generally called “ Velocity-acceleration diagram”. The results of this specification would be similar

to the diagram shown in the following slide

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The left-hand corner of the diagram are determined by the stalling characteristics of the aircraft in both upright and inverted flight. The upper left-hand corner is the intersection of the stall line and the maximum normal acceleration factor ‘n’ at the manoeuvre speed VA. The right-hand corners are determined by the maximum speed conditions of the aircraft. The extreme right-hand side being the design speed VD

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There are four definitive values of the normal acceleration factor in any given case. They are

n1 the maximum positive, upright, value

n2 the lower normal acceleration factor at VD

n3 the maximum negative, inverted, value

n4 the maximum positive value at VD when it differs from n1

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The n-V diagram describes all points corresponding to every specified symmetric manoeuvre the aircraft is permitted to perform.

The corners correspond to the maximum manoeuvres permitted on the assumed probability of occurrence and it is sufficient to examine these corners when considering loading cases.

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As can be seen from the fig above the maximum lift load factor equals 1.0 at level flight stall speed. The aircraft can be stalled at a higher speed by trying to exceed the available load factor, such as in a steep turn. The point at the extreme top left corner, (can be

called high angle of attack A.O.A ) is the lowest speed at which the maximum load

factor can be reached without stalling.

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At this part of the envelope the load on the wing is approximately perpendicular to the flight direction, not

the body-axis vertical direction. At high angle of attack the load direction may

actually be forward of the aircraft body-axis vertical direction, thus causing

a forward load component on the wing structure

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The point at the top right of the diagram (maximum speed or dive speed) represent the maximum dynamic pressure and maximum load factor.

At this condition the aircraft is at low angle of attack due to

high dynamic pressure and the load is approximately vertical in the body-axis.

For subsonic aircraft max. or dive speeds is about 50% higher than level-flight cruise speed and for supersonic about Mach 0.2 faster

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Gust loads Transport aircraft encountering air turbulence

may experience load factors due to gust ranging from a negative 1.5 to a positive 3.5 g or more.

When aircraft experience a gust the effect is increase or decrease in angle of attack.

The fig below show the effect of an upward gust of velocity U.

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The change in angle of attack is approximately U/V and the change in aircraft lift is proportional to the gust velocity

Δα = tan-1 (U/V) ≡ U/V ΔL = 1/2ρV2S(CLαΔα)

= 1/2ρVSCLαU

Resulting change in load factor is Δn = ΔL/W = ρUVCLα/(2W/S)

Increment in the load factor in this case assumed that the gust instantly affected the aircraft which unrealistic, as the gust follow a cosine-like intensity increase as the aircraft flies through, allowing it time to react to the gust.

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To account for this effect a gust alleviation factor K was devised to measure gust data.

Thus the gust velocity U in the equation aboveRegarding increment in load factor can be defined

as follows U = KUde

For subsonic K = 0.88μ/(5.3+μ)Supersonic K = μ1.03/(6.95+μ1.03) Mass ratio μ = 2(W/S)/ρgcCLα

Where Ude is the standard vertical gust speed

The loading of an aircraft in flight can be placed in

the following ;-

a) The forces and moments present while the aircraft is in steady flight, referred to as the trim condition.

In case of a conventional aircraft having a vertical plane of symmetry the loading in the trim condition is limited to that plane, that is, to longitudinal forces and moments.

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b) The forces and moments consequent upon the departure of the aircraft from the trim conditions as a result of control inputs or atmospheric disturbances, referred to as TRANSIENT LOADING

Static stability and the dynamic stability characteristics have an influence upon the loading when the aircraft performs manoeuvres or encounters atmospheric turbulence.

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Forces and moments in symmetric flight

Aircraft in symmetric flight has three degrees offreedom, they are, translation in flight direction, translation perpendicular to the flight direction, androtation in pitch about centre of gravity.Forces and moments experienced by the aircraftarise from aerodynamic, powerplant, inertial andatmospheric effects which are illustrated in fig below :

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The forward velocity along the flight path is defined by the wind axis OX’ is V0 and

it is inclined at an a climb angle y relative to the earth axis OX”.

The body axis OX is inclined at an angle of attack αB relative to the wind axis.

The aircraft is pitching relative to the earth axis with an angular velocity dθ/dt. The acceleration along the flight path is ‘mg’ and perpendicular to it the acceleration is ‘ng’ . (m and n are acceleration factors )

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lWB lift on aircraft

lT horizontal stabilizer lift

D total dragT net engine thrustm aircraft mass ( weight/gravitational

acceleration)C MAChc distance of centre of gravity behind leading edge

Hoc distance of aerodynamic centre behind leading edge

L’T distance along body from cg to the point of

application of lT

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ky radius of gyration of aircraft in pitch about cg

ZD perpendicular distance of drag axis below cg

ZT perpendicular distance of thrust axis below cg

ßT angle of the thrust axis to the body axis

M0 pitch moment at zero lift

MT pitching moment due to the deflection of the

elevator relative to the aerodynamic centre of

the horizontal stabilizer Mθ damping pitch moment

V0 velocity of aircraft along the flight direction

ρ local density of the air77

For an aircraft in level flight ‘ϒ’ is zero and if the thrust is equal to the drag, m is unity.

The drag force usually passes close to the cg so ZD may be neglected as well as Mθ, MT.

For tailless aircraft Mθ and MT arte significant and should considered.

Resolving forces along the line of flight OX’ axis

T- D – mgm = 0 lWB + lT - ngm = 0

M0 - lWB (H0 – h) c - lT L’T + TZT - mky2θ = 0

horizontal stabilizer load is

lT = [ M0 + ngm(h – H)c + TZT – ky2θ]/ L’T

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Wing-body lift lWB

Dynamic pressure q = (ρV0EAS2)/2

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Aeroelastic considerations and

Stiffness requirements

In addition to ensuring that a flying vehicle is structurally strong enough to withstand the loads imposed upon it, it is essential that it should

possess adequate stiffness. Reason for this is that the structure must be capable

of retaining its design shape so that any deformation which may occur is not so large as

to seriously impair the aerodynamic performance of the vehicle.

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Airframe distortion introduces both structural and aerodynamic effects, thus both the static and dynamic properties of the aircraft be considered and this requiresconsideration of the aerodynamic and structural inertias and damping as well as stiffness.

Examples

Examples to understand how various effects occur.

a) problem of control reversal Consider an aileron which is deflected

downwards to cause the aircraft to roll by increasing the local lift.

This lift increment due to control deflection acts well aft on the chord, almost aft of the local section centre of twist and introduces a tendency for the section to pitch nose down relative to wing root. 82

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If the wing is insufficiently stiff in torsion thisdownward pitching will introduce a downward lift increment of the same order of magnitude as that produced by the aileron. At a certain speed, known as reversal speed, the two forces are equal and the control lose its effectiveness. At higher speeds a roll in a direction opposite to that intended will occur.

Wing torsional divergence Consider the wing in fig.The aerodynamic lift loadinglocated along span wise axis at aparticular chord-wise station at a distance ec forward of thestructural flexural axis. The effect of this is to twist thelocal section through an angle Ѳ.

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The aerodynamic load increment due to angle Ѳ is :-

{(ρ V2c/2).dy} a1Ѳ where

V is the airspeed c is the local chord a1 is the local lift curve slope

the corresponding moment about the flexural axis is :-

MA = (ρ/2)V2a1Ѳec2 dy

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This moment is restricted by the torsional stiffness of the structure. If the stiffness is mѲ then the torque due to deflection Ѳ is

MS = mѲ. Ѳ

When MA is less than MS the strip is stable but if the opposite it is of static instability.

That is the application of aerodynamic forces will cause a continuously increasing angle, Ѳ, until the wing break.

The critical case is when MA =MS that is when

mѲ = (ρ/2)V2a1ec2dy

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The torsional divergence speed for the strip is :-

VT = {2mѲ/ρa1ec2dy}1/2

The divergence speed for the wing as a whole requires an integration of both the aerodynamic and structural moments over the span and it depends upon the planform shape and structural section properties.

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Inertia and Mass distribution

The structural density of a wing is defined as :- δW = (W1+kW2)/S1c1 where

S1 is the wing area outside the fuselage

c1 is the corresponding mean chord

W1 is the corresponding wing weight

W2 is the weight of non-structural loads in the wing

k is a factor which is usually taken as 0.588

When the wing is of low structural density the location of the inertia loads tend to be more critical than if the reverse is true, but it must be clear that this is really only a relative effect.

If heavy concentrated loads or fuel tanks are housed within or below a wing they should be located so that their individual centres

of gravity are as far forward as possible of the flexural axis, should it be not possible, they should be located as near to the root as can be achieved.

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In case of wing fuel tanks the system should be arranged to keep the fuel centre of gravity as far forward as is feasible in all tank conditions.

Where external and wing tips fuel tanks are used it is essential to ensure that the

centre of gravity of the fuel does not move too far forward or far backward.

Very large wing inertias, such as lift engine pods, present a difficult problem.

The feasible solution is the provision of rear fins to increase the aerodynamic damping. This was been used on some large external fuel tanks.

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Stiffness Criteria and considerations

Wing stiffness One of the most important stiffness

requirements is that concerned with wing torsional stiffness. Three main considerations that decide the

minimum value required are flutter, aileron reversal and divergence.

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Flutter To ensure that what is called classical flutter did not

occur earliest attempts resulted in the use of a wing torsional stiffness criteria of the form :-

k = (1/VD){mѲ/ρdcm2}1/2

Where k is a non dimensional criteria dependent upon wing density

VD is design dividing speed

d is 90% of distance from root to tip of wing cm is wing mean chord

mѲ is torsional stiffness relative to the wing root at a section 70% of the

distance from the root to tip ρ is air density

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The value of k arranged to ensure the flutter speed would be 1.5 to 2 times the

design diving speed. This formula makes no allowance for sweep,

Compressibility, aspect ratio, taper ratio and certain other important criteria.

The value of mѲ is defined as that concentrated torsion couple located at 0.7 semispan which would give one radian rotation relative to the root.

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As subsonic compressibility effects should be considered in accordance to civil aircraft requirements an appropriate correction is done :-

k = (1/VD)(mѲ/dc2)1/2(1-M2)1/4

where M is the Mach number appropriate to VD

mѲ is asymmetrical stiffness measured at 0.75 semispan

k = 0.024{1+(δW/2)} or 0.36 whichever smaller or

If the wing engines forward of the leading edge are fitted k = 0.028 for 0<δW<2

where δW is wing density.94

To allow for other important factors the following formula can be used to calculate the flutter speed which could be at least 1.25VD

VF ={mѲ/ρdcm2}1/2{(0.9-0.33k)(1-0.1r)/[0.9(g-0.1)(1.3-h)]}

*sec3/2(Λ-π/16)f(M) ##

where

VF is critical flutter speed

mѲ is antisymmetric torsional stiffness at 0.7 semispan

g is location of inertia axis subject to 0.35<g<0.55 h is location of flexural axis aft of the leading edge k is taper ratio provided 0.25<k<1.0 r is stiffness ratio f(M) = (1-M2)1/4 Mach No. correction where 0<M<0.8 = 0.775 where 0.8<M<0.95 Λ is sweep back of a line 5% behind line of maximum thickness rad.

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Note :- This formula does not apply to swept

wing aircraft with a fuselage of low pitching inertia or for

wings of low aspect ratio, wings carrying twin booms, those with large concentrated masses or those flying at transonic or supersonic speeds.

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Aileron Reversal For ensuring sufficient wing torsional stiffness to prevent aileron reversal is given by :- VR = {(2mѲ/ρsCm

2)(a2/a1m)}1/2 where :-

VR is aileron reversal speed

s is semispan mѲ is anti-symmetric stiffness at the mid aileron section

cm is wing mean chord

a1 is the wing curve slope

a2 is lift curve slope due to aileron angle

m is slope curve of nose down moment VR must be at least 1.15VD

This applies only to conventional wings97

Tailplane Stiffness

Stiffness requirements for a tailplane to which is attached an elevator .

k = (1/VD){Tt/(btct2/2)}1/2f(M)

Tt is tailplane torsional stiffness measured at 0.8

semiplane outboard of centreline and perpendicular

to the main spar, or relative to fuselage at root if

tailplane is moveable

bt is span of tailplane measured along spar

ct is geometric mean chord of tailplane elevator

k is stiffness, ≥ 0.036 for tailplane without fins and

rudder and 0.045 if these are fitted

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Elevator stiffness and flutter

For reasons of stability and control, stiffness criteria for elevators

k = (1/VD){Te/sece2}1/2f(M) where

Te is elevator torsional stiffness measured between the body side and 0.1be from the extreme tip, perpendicular to the hinge line.

be is elevator span

se is elevator semispan

ce is mean chord behind the hinge

k = ≥0.024 for plain elevators or ≥0.028 for balanced elevators

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Control Surface Mass Balance

Mass balances can be either concentrated weights or distributed. Mass balance is not required on aircraft whose VD is less than 130 knots.

The mass balance requirements for all controls state that at the neutral position and at plus or minus 100

deflection the product of inertia for the complete control shall be zero, and is defined as

Σm.x.y100

m is an element of control surface mass x is distance of m from hinge line measured normal to hinge line. y is the perpendicular distance of m from a fore and aft axis. This axis is normally the wing root for ailerons and the body torsion axis for rudders and elevators

101

Ground Load Cases The vertical energy of a landing aircraft is

absorbed in the tyre, shock strut system and the airframe. It is dissipated largely by the shock strut, which

may include the hop damper in a bogie layout, during the initial impact and subsequent rebounds of the aircraft.

102

These rebounds do not normally imply that the wheels leave the ground after the initial impact, but that

the aircraft centre of gravity oscillates vertically on

the shock strut system.

103

The performance of shock strut unit is expressed in terms of a load-deflection diagram as shown in fig. The form of diagram depends upon the type of shock strut used and the rate of load application. The majority of aircraft use either an air or oil spring to absorb the energy and an oil dashpot todissipate it.

dynamic static

c Deflection stroke

R is maximum reaction R = λP, where P is proportion of weight carried by shock strut λ is maximum reaction factor

load

R

104

The efficiency of a shock absorber is defined as the ratio of the energy absorbed to that equivalent to the product of the maximum load and stroke.

It is thus the ratio of the area under the load-deflection diagram to the area of the rectangle circumscribing it.

A simple spring with a linear characteristic has an efficiency of 50% and under static conditions many shock absorbers approach this condition.

105

Under dynamic conditions appropriate to landing the efficiency can be much higher.

Typical values are 80% of a liquid (oil) spring and 85% for an oleo-pneumatic system.

These characteristics are obtained by introducing valves into the units so that the initial stiffness is relatively high but at a given dynamic pressure they open to reduce the stiffness.

106

Tyres have a spring characteristic such that the efficiency is 47% which implies that the load-deflection curve is slightly hollowed out from a linear characteristic.

When designing an undercarriage unit the tyre size is chosen on a load capacity basis. The number of tyres and pressure are determined

by such matters as runway loading and stowage when retracted.

Tyre size is a function either of the maximum static load or the limiting dynamic load which is about three times the static criterion.

107

The vertical energy equation of a landing aircraft is written in the form :-

Wv2/2g = W.n.z where (1) v is the vertical velocity of descent at the moment of impact W is the aircraft weight n is the mean normal deceleration factor z is the vertical movement of the centre of gravity required to bring it to rest.

108

The energy absorbed in the system must be equivalent to the potential energy on the right hand side of this expression.

Under normal circumstances it is required that this energy should be entirely absorbed in the main tyre and shock absorber units.

In most cases the main shock absorber can be given a characteristic which is defined in terms of the vertical travel of the axle, so that the reference load is also vertical.

109

In these circumstances a total load of “W” on all the shock absorber units being used is equivalent to

the static condition and if there are “N” units and the maximum load on

each unit during landing is “λW/N” then “λ’ represents the maximum vertical acceleration

of the centre of gravity. The total energy absorbed in each shock

absorber and tyre unit is (λW/N)(ηsS+ηtSt) where (2)

110

λW/N is the maximum vertical load S is the vertical axle travel appropriate to λW/N St is the tyre deflection “ “ “ “ “ “ ηs, ηt are the shock absorber and tyre energy absorption efficiencies appropriate to vertical motion and λW/N λ is known as the vertical reaction factor

111

Equation (2) can be written as energy absorbed = P(ηsS+ηtSt)

combining (1) and (2) Wv2/2g = N(λW/N)(ηsS+ηtSt)

or v2/2g = λ(ηsS+ηtSt) (3)

112

A short undercarriage is the lightest and hence within the limitation of ground geometry the stroke should be made as small as possible.

Short stroke may imply a high value of the reaction factor, λ, and since this implies also a high load, any weight saved by the use of a short undercarriage may be more than offset by the heavier construction and in particular the fatigue effects on the airframe.

For the majority of aircraft “λ” varies between 1.5 and 2.5 with a definite tendency to lower values on aircraft designed on low normal manoeuvre factors and long life. 113

A fighter type aircraft has a high strength airframe and can tolerate a value of “λ” of three or more without undue penalty.

Naval aircraft have increased energy absorption requirements to meet coupled with severe size restrictions so reaction factors of 4 are common.

Strokes are of the order of 10 to 18 inches and may exceed to 2 feet.

114

Energy absorption requirements

The requirements are specified in terms of design vertical descent velocities.

The loading cases are directly derived from velocities, thus proof and ultimate values are quoted.

The ultimate values are 1.18 times the proof value for military aircraft and 1.2 for civil aircraft, so that the ultimate energy requirements are 40% to 44% greater than the proof condition.

115

Retraction of undercarriage

For military aircraft it is possible to operate the retraction and lowering mechanism at speeds in the range 1.2 -1.5 times the stalling speed at appropriate aircraft weight.

The retraction and lowering times required vary between 5-25 seconds according to the type of the aircraft.

The civil aircraft are similar except that the speed range is from 1.0-1.6 the stalling speed.

116

Brakes The energy absorption per brake can be estimated

as :- energy absorption per brake = (k-/Nb)(WV2/2g)

where Nb is the number of brakes on the main wheel

W is the design weight V is the speed from which a stop must be made K- is a factor to allow for aerodynamic drag contribution

117

K usually taken as 0.7 for a tail wheel and 1.0 for a nose wheel aircraft.

It may be possible to use the drag effect of a nose wheel aircraft by maintaining a high incidence until late in the landing run, by delaying the dropping of the nose.

In this case K may be as low as 0.8.

118

Landing dynamics

119

In some cases it is necessary to carry out a full dynamic analysis of the landing behaviour in order to deduce the correct loading system.

The general configuration of an aircraft of weight W during landing is shown in the previous fig. where

L is the resultant trimmed lift acting at the c.g LT is the damping force on the tail due to

landing motion

120

D is the aerodynamic drag Tr Is the thrust

Dn, Dm are the drags due to contact of the wheels

with the ground μ is the coefficient of friction lm, ln, lt are respective distances of the main

undercarriage, nose u/c and tailplane from c.g KB is radius of gyration in pitch about c.g

Km, Kn are the stiffness of the main and nose shock

strut units Bm, Bn are the damping coefficients of the main and

nose units and are proportional to velocity

121

Energy absorption The nose, tail loads are here specified in terms of a

parameter δ which represents the proportion of the load that they are considered to react.

δn = (lm+0.4h)/(lm+ln)

δt = lm/(lm+lt)

h is the vertical distance of the centre of gravity above a line joining the nose and main wheel In order to ensure there is no change in pitch during a

three point landing , the travels of the nose and main legs must be equal and relative energies absorbed as follows :-

122

Main wheels = (WLv2/2Ng)[(ln-0.4h)/(lm+ln)]

Nose wheel = (WLv2/2g)[(lm+0.4h)/(lm+ln)]

The design vertical velocities, proof, are specified as follows ;-

Naval aircraft v = 14-20 ft/sec civil aircraft v = (5+0.06vso) ft/sec

where vso is the stall speed at WL, flaps down, but

7 ≤ v ≤ 10 ft/sec

123

Loading Cases

landing with drag and side load The process of spinning up the wheels during landing,

and the resulting drag, is assumed to continue until the maximum reaction is developed.

The case is appropriate to a two point landing for the main undercarriage and a three point landing for the nose or tail u/c.

The vertical load on the leg is the product of the maximum reaction factor and the proportion of the static load reacted by it.

124

The lift is assumed to equal the weight at the moment of impact.

The nose and main wheel reaction factors are chosen so that no overall pitching occurs in the three point landing. Thus ….

N1 = 2M1 {(lm+0.4h)2/[(lm+ln)(ln-0.4h)]}1/2

Where N1 and M1 are the nose and main wheel reactions.

Drag loads are assumed to act at the wheel axle unless they are due to brake torque, they are assumed to act at the ground line.

125

Dynamic braking The vertical load on the nose wheel is N3 = [ WT/(lm+ln)][lm+(2cHD/WT)] where c is the coefficient expressing the increase in nose wheel reaction due to pitching, app. 3.0 for military and 2.0 civil aircraft. D is the drag about 0.8 times the vertical reaction for civil aircraft.

126

Braked taxying For this case the aircraft is assumed to be in a

non-airborne condition with the nose wheel centralized.

a) main wheel loads for nose wheel aircraft M4 = [FWT/2(lm+ln)][ln-2HDM/FWT]

for tail wheel aircraft M4 = (F/2)WT where

F = 2.33 for tyre below 30 inch diameter and 1.8 for that above 60 inch diameter with linear variation between them and = 1 for supersonic airliner DM = 1.33D for a supersonic airliner

= D for civil aircraft127

The associated drag load for subsonic aircraft is D and for military aircraft 0.33D at the hub plus 1.0D at the point of tyre contact.

b) Nose wheel load The vertical nose wheel load with nose wheel in

any castored position is N4 = [FNWT/(lm+ln)][lm+HDM/FNWT] where

FN = 1.0 for supersonic airliner, but not to

exceed 0.8M4

The drag load in range 0-0.4N4 and side load not less than 025N4

128

c) Tail wheel load

With the wheel in all positions the vertical load is R3 = 1.75WT[lm/(lm+lt)]

The drag load 0-0.4R3 and the side load is 0.25R3 for civil aircraft.

For military aircraft fitted with hydraulic dampers the side load must be not less than 0.9WT[lm/(lm+lt)] and not less than half the value for other types

129

Turning and swinging Main undercarriage load is given by Nose wheel aircraft M5 = (WT/2){[lm/(lm+ln)]-+R/t} where

M5 is vertical load

t is main wheel track There is no drag load but a side load of 0.5M5

for tail wheel aircraft M5 = (WT/2)[1-+1.2H/t]

And the side load is 0.6M5130

Rolling back

Here the vertical main wheel reaction is M6 = (WT/2)[ln/(lm+ln)] and

the drag load is 0.7M6

Nose wheel load N5 = WT[lm/(lm+ln)] with

forward drag load 0.7N5

131

Take off

Main wheel vertical load is M7 = FWT/2 where

F is as defined before

the side load is 0-0.2M7

the drag load is 0-0.4M7

132

General Loads on Aircraft Before the structural design of an aircraft can be made, the external loads acting on the aircraft in flight, on landing and take off conditions must be known. There is a wide range of wing configurations, such as the straight tapered wing, the swept wing and the delta wing, and many of these wings include leading and trailing

edge devices for promoting better lift or control characteristics.

133

The presence of powerplant nacelle units, external fuel tanks, etc…, are units that

affect the air flow around the wing and thus affect the magnitude and distribution of the air

forces on the wing. Also the fuselage itself influences the air flow

over the wing. All the loads can be calculated aerodynamically.

134

The Design loads

As aircraft is designed to carry out a definite job, there result many types of aircraft relative to size, configuration and performance.

The magnitude of the air forces ( loads) on an aircraft depend on the

velocity of the aircraft and the rate at which the velocity changed in magnitude and direction. The magnitude of the flight acceleration factor is governed by the capacity of the human body to

withstand without injury as for fighters and for civil aircraft determined by what is necessary to safely transport passengers from one place to another.

135

To insure safety in the air transportation, along with uniformity and efficiency of design governments aeronautical agencies (civil and military) have definite requirements for various types of aircraft relative to the magnitude of loads to be used in the structural design of aircraft. These loads are :-

a) limit or Applied loads limit loads are the maximum loads anticipated on the aircraft during its lifetime of service.

136

The aircraft structure shall be capable of supporting the limit loads without suffering detrimental permanent deformations.

At all loads up to the limit loads the deformation of the structure shall be such as not to interfere with the safe operation of the aircraft.

b) Ultimate or Design loads Ultimate or design loads are equal to the Limit loads multiplied by a factor of safety (F.S). In general the over-all F.S is 1.5

137

The broad general category of external loads on conventional aircraft can be classified as follows :-

a) Air loads due to aircraft maneouvre under the control of the pilot and due to air gust which is not under the pilot control. b) Landing loads due to landing on land, on water, arresting etc… c) Power plant loads due to thrust and torque. d) Take-off loads e) Special loads due to hoisting, towing, fuselage pressurization etc…

138

Weight and Inertia forces Weight is that constant force proportional to the mass, which tends to draw every body towards the centre of

the earth. An aircraft in steady flight ( uniform velocity) is acted

upon by a system of forces in equilibrium, the weight of the aircraft, the air forces on the whole

aircraft and engine forces. This balance can be changed by the pilot by changing

the engine power or by operating the control surfaces to change the direction or altitude.

139

Inertia forces

If the unbalanced forces acting on a rigid body cause only a change in the magnitude of the velocity of the body and not its direction,

the motion is called translation. From basic physics the accelerating force F = Ma.

M is the mass of the body or W/g. F W motion The unbalanced force F caused the body to move to the right. The effect of this unbalanced force produces a force on each mass particle of W the body.

140

effective Force = Σma = Ma m1a m2a m3a

The effective force Σma = Ma. If the effective forces are reversed they are referred to as inertia forces. The external forces and the inertia forces form a

force system in equilibrium. For a motion of translation and constant acceleration V –V0 = at

S = V0t+(1/2)at2

V2 – V02 =2aS

141

When an aircraft pull up from a steady flight or pull up from a dive it will follow a curved path as shown in fig.

142

If at point A the velocity is increasing along it path, the aircraft is being subjected to

two accelerations “at” tangential to the curve at point A and equal in

magnitude to at =ŕa and

an = ŕω2 an acceleration normal to the flight path at A and directed toward the centre of rotation O.

143

144

From Newton’s law the effective forces due to these accelerations Fn = Mŕω2 =Mv2/r

Ft =Mŕa where

ω is angular velocity at A a angular acceleration ŕ is radius of curvature

The inertia forces are equal and opposite to the effective forces as shown in fig. These forces are considered to be part of the

total force system on the aircraft which is in equilibrium.

145

If the velocity of the aircraft along the path is constant, then

at = zero and thus the inertia force

Ft =zero, leaving only the normal inertia force Fn.

If the angular acceleration is constant, the following relationship hold ω - ω0 = at

ϴ = ω0t + ½ at2

ω2 – ω02 = 2aϴ

146

147

Where ω is angular velocity after time t ω0 is initial angular velocity in rad/sec

ϴ is angle of rotation in time t

148

The moment To of the inertia forces about the

centre of rotation “o” is Mŕa(ŕ) = Mŕ2a. The term Mŕ2 is the mass moment of inertia of the aircraft about point “o”. As the aircraft has considerable pitching moment of inertia about its c.g axis thus by parallel axis To =Ioa+Ic.ga

where Io = Mŕ2 and

Ic.g = moment of inertia of aircraft about Y

axis through c.g of the aircraft.

In flight an air gust may strike the horizontal tail producing a tail force which has a moment

about the aircraft c.g. In some landing conditions the ground or

water loads do not pass through the aircraft c.g, thus

producing a moment about the aircraft c.g which causes the aircraft to

rotate about the Y axis through the c.g.

149

150

In such case and for the above mentioned case the centre of rotation is going to be c.g instead of o. thus ŕ =0 and Fn, Ft are zero and

the only inertia force for the pure rotation is Ic.ga and thus the moment of this inertia about the c.g = Tc.g = Ic.ga

Air loads on wing The wing of an aircraft carries the major portion of the air load.In level steady flight the verticalupward force of the air on the wing equals the weight of the aircraft. Figs. Illustrate the air pressure intensity diagram due to an air stream flowing round an airfoil for both positive and

negative angle of attack.

151

Wings are attached to fuselage, they may support power plants, wing tip tanks etc… The wings are tapered in planform and thickness and some possess leading and trailing slats and flaps to produce high lift or control effect. The airflow around the wing is affected by such factors and thus wind tunnel tests are usually necessary to obtain a true picture of the air loads on a wing relative to their chordwise and spanwise distribution.

152

Aerodynamic Centre

Since aircraft flies at many different angles of attack, it means that the centre of pressure changes for the many flight design considerations. There is one point on the airfoil that the moment due to the lift and drag forces is constant for any angle of attack. This is the aerodynamic centre (a.c) and its approximate location is at 25% of chord. Thus the resultant force can be replaced lift and drag at that point plus a wing moment Ma.c

153

The fig illustrate in general the main forces on the aircraft in an accelerated flight condition.

T engine thrust L total wing lift plus fuselage lift D total drag Ma moment of L and D

with reference to wing a.c

154

W weight of aircraft IL inertia force normal to

the flight path ID inertia force parallel to

the flight path Im rotation inertia moment

E tail load normal to flight path

155

156

Equations of equilibrium for steady flight

ΣFx = 0, D + W sinѳ-T cos β=0

ΣFg = 0, L – Wcosѳ + T sinβ

-E = 0ΣMy = 0, -Ma-La-Db

+ Tc.cos β +Ee = 0

157

Equations of equilibrium in accelerated Flight

ΣFx = 0, D+W sinѳ – T cosβ – ID = 0

ΣFg = 0, L – W cosѳ +T sinβ – IL – E = 0

ΣMy = 0, -Ma –La –Db +Tc cosβ + Ee +Im = 0

158

Load Factor

Load factor is defined as thenumerical multiplying factor bywhich the forces on the aircraftin steady flight are multiplied toobtain a static system of forcesequivalent to the dynamic forcesystem acting during theacceleration of the aircraft.L is the total lift and W = L

159

Now assume the aircraft isaccelerated upward along the Z axis.The fig shows the additional inertiaforce Wag/g acting downwards, or

opposite to the direction ofacceleration. The total lift for the un-accelerated condition must be

multiplied by the load factor ng to

produce static equilibrium in theZ direction.

160

Thus, ngL – W – Wag/g = 0

Since L = W then ng = 1 + ag/g

If the aircraft acceleratealong the X axis as well asthe Z axis thrust will begreater than drag and theinertia force will be nxW = Wax/g

ΣFx = 0, T-D-nxW = 0

nx = (T-D)/W

161

The applied or limit load factors are the maximum load factors that might occur during the service of the particular aircraft. These loads must be taken by the aircraft structure without appreciable permanent deformation.

The design load factors are equal to the limit load factors multiplied by the factor of safety, and these design loads must be carried by the structure without rupture or collapse, or complete failure.

162

Any aircraft flight altitude can be defined by stating the existing values of load factors (acceleration) and the airspeed ( more properly the dynamic pressure. The accelerations are produced by maneuvers and air gusts.

The acceleration due to gust is not under the control of the pilot as it depends upon the direction and velocity of the air gust. According to the available data it has been found a gust of 30ft/s appeared sufficient

163

The speed of the aircraft affects the loads on the aircraft. The higher the speed the higher the aerodynamic wing moment. The gust acceleration increase with aircraft speed. Thus it is customary to limit the particular aircraft to a definite maximum flight speed.

164

Gust load Factors

When a gust strikes an aircraft in a direction normal to the thrust line (X axis), a sudden change takes place in the wing angle of attack with no sudden change in aircraft speed.

The normal force coefficient (CZA) can be assumed to vary linearly

with the angle of attack. CZA

C KU

B V Δα Δα

α

165

Let point B represents the normalaircraft force coefficient CZA necessary

to maintain level flight with a velocity Vand point C the value of CZA after the

gust KU caused a sudden change Δα inangle of attack without change in V. Thetotal increase in aircraft load in the Zdirection can be expressed by the ratioCZA at B. From (b) for small angles

Δα = KU/V and from (a)ΔCZA = m Δα

where m is the slope of the aircraftnormal curve (CZA per radian )

Δα (a)

KU

Δα

V (b)

CZAC

ΔCZAB

166

The load factor increment due to the gust KU can then be expressed as Δn = ΔCZA/CZA = (KU*m/V)(ρ*V2*S/2W)

= (KU*V*Sm/575W) -----(a)

Where U is the gust velocity K is gust correction factor depending on wing loading V is indicated air speed S is the wing area w is gross weight of aircraft

167

If U is taken as 30 ft/sec, and m as the change in CZA with respect to angle of attack in absolute units per degree, Equation (a) reduces to

Δn = 3KmV/(w/S) Therefore the gust load factor n when the

aircraft is flying in horizontal attitude equals n = 1± 3KmV/(W/S) And when aircraft is in vertical attitude n = ± 3KmV/(w/S)

168

Velocity-Load Factor Diagram

The main design flight conditions for an aircraft is given by stating the limiting values of the acceleration and speed in addition to the maximum value of the applied gust velocity.The design loading requirement can be stated as follows: The proposed aircraft shall be designed for applied positive and negative acceleration of +6.0g and -3.5g at all speeds from that corresponding to Clmax up to 1.4 times the maximum level flight speed.

169

170

The lines AB and CD represent the restricted positive andNegative maneuvre load factorswhich are limited to speedsinside line BD which is taken as1.4 times the maximum levelFlight speed in this illustration.These restricted maneuvrelines are terminated at pointsA and C by their intersectionwith the maximum Cl values ofthe aircraft.

171

At speeds between A and Bthe aircraft must not exceedthe maneuver acceleration.At speeds below A and C theloads and maneuverproducing Clmax would give

an accelerating less than thelimited values given by linesAB and CD.

172

The positive and negativegust acceleration due to a30ft/sec gust normal toflight path as shown, apositive gust is not criticalwithin the restricted velocityof the aircraft since the gustlines intersect the line BDbelow the line AB.

173

For a negative gust thegust load factor becomescritical at velocitiesbetween F and D with amaximum acceleration asgiven by point E. Foraircraft having relativelylow maneuver factor thegust acceleration may becritical for both positiveand negative accelerations.

174

on the diagram, points A and B correspond in general to high angle of attack and low angle of attack respectively and points C and D the inverted high and low angle of attack conditions respectively.

175

For stress analysis purposes all speeds are expressed as indicated airspeeds. The indicated airspeed is defined as the speed which would be indicated by a perfect air-speed indicator, that is, one that would indicate true air speed at sea level under standard atmospheric conditions. The relationship between actual airspeed Va and indicated airspeed Vi is

Vi = Vax√(ρa/ρo)

Where Vi indicated airspeed

Va actual airspeed

ρo standard air density at sea level

ρa density of air in which Va is attained

176

Special flight Design Conditions

There are types of flights which may be critical for certain potions of the wing or fuselage structure. Aircraft are equipped with flaps to decrease the landing speed and be lowered at speeds at least twice the minimum landing speed. Since the flapped aerofoil has different values for the magnitude and location of the airfoil characteristics, the wing structure must be checked for all possible flap conditions within the specified requirement relative to maximum speed at which the flap may be operated.

177

The flap conditions will effect the wing portion inboard of the flap and only critical for the rear beam web and for the top and bottom walls of the torsion box. This due to the movement of the centre of pressure backwards due to the deflection of the flaps. This will produce more shear load on the rear wall as well as torsional moment on the conventional cantilever box metal beam. Also operation of ailerons produce a different air load on each side of the aircraft wing which produces an angular rolling acceleration of the aircraft.

178

Accelerated Motion of aircraft

It is general practice to place the aircraft under accelerated conditions of motion into a condition of static equilibrium by adding the inertia forces to the applied force system acting on it.

Below are some examples

179

The fig shows an aircraft landing on a Navy aircraft carrier and being arrested by a cable pull T on the aircraft arresting hock. Aircraft weight 12000 lb and the aircraft given a constant acceleration of 3.5g(112.7

ft/sec2). Find the hock pull T, the wheel reaction R, and the distance “d” if the landing velocity is 60 MPH. What is the stopping distance.

180

SolutionOn contact of the aircraftwith the arresting cablethe aircraft decelerate,and the motion istranslation horizontally.The inertia force is :- Ma = (w/g)a = (12000/g)x3.5g = 42000 lb

181

By using the staticequation of equilibriumT and R can be found.ΣFx = -42000 + Tcos10o= 0

T = 42700 lbΣFg = - 12000 + R

-42700xsin 10o =0 R = 19420 lbTo find “d” Take moments

about c.g of the aircraft

182

ΣMc.g = 19420x24 – 42700d =0

d = 10.9 inchesLanding velocity V0

V0 = 60 MPH

V2- V02 = 2as

02 – 882 = 2(-112.7)s s = 34.4 ft. Hence the stoppingdistance is 34.4 ft

183

As shown in fig an aircraftequipped with float is catapultedinto the air from a navy cruiser. ThePulling force P gives the aircraft aconstant horizontal acceleration of3g (96.6 ft/sec2). The weight of theaircraft is 9000 lbs and the pullingtrack is 35 feet long. Find the forceP and reactions R1 and R2 ofCatapult car. The engine thrust is900 lb. What is the velocity ofaircraft at end of the run.

184

SolutionThe forces will be determinedjust after the beginning of thecatapult run where the carvelocity is small and thus thelift and drag can be neglected.The inertia force acting towardsthe tail is Ma = (9000/g)x3.0g = 27000 lbΣFx = - 900-P+27000 = 0 P = 26100 lb

185

To find R2 take moments about point A

ΣMA = 900x55+27000x78 – 900x83- 85R2 = 0 R2 = 29800 lb

ΣFg = 29800- 9000+R1 =0 R1 = - 20800 lb

To find velocityV2- V0

2 = 2as

V2 – 0 = 2x96.6x35 V = 82 ft/sec = 56 MPH

186

Example 3

A transport aircraft touched down in landing and a braking force of 35000 lb on wheels was applied to bring it to rest. The landing horizontal velocity is 85 MPH (125 ft/sec). Neglecting air forces on the aircraft and assuming engine power is zero find the ground reactions R1 and R2. What is the landing run distance with the constant braking force.

187

The aircraft is beingDecelerated Horizontally andinertia force through the c.gacts toward the front of theaircraft. Using equilibriumequation solve fo decelerationfactorΣFx = 35000 – Max = 0

Max = 35000 lb Or

(W/g)ax = 35000 lb hence

ax =35000x32.2/100000

ax = 11.27 ft/sec2

188

To find the landing run V2- V0

2 = 2axs

0 – 1252 = 2x(- 11.27)s s = 695 feetTo find R2 take moments about point A

ΣMA = 100000x21- 35000x9 + 38R2 = 0

R2 = 47000 lb

ΣFg + 47000 – 100000 + R1 = 0

R1 = 53000 lb189

ExampleAir craft in fig weighs 14000 lbflying horizontally with velocity of500 MPH (733 ft/sec) when thepilot pulls it upward into a curvedpath of radius of curvature of 2500ft. The thrust and drag are equal.Find a) acceleration of aircraft in Z

direction b) wing lift and tail forces c) aircraft load factor

190

Acceleration ag = V2/r = 7332/2500 = 215 ft/sec2

Or 214.5/32.2 = 6.68g The inertia force normal tothe flight path actingdownwardsMaz = (14000/g)6.68g

= 93520 lbPutting this force on theAircraft Through the c.gpromotes static equilibrium.

191

To find tail load T take moments about C.PΣmaz = - (14000 + 93520)x8 + 210T = 0

T = 4100 lb (down)To find wing lift (L) useΣFz = - 4100 – 14000 – 93520 + L = 0

L = 111600 lb

Aircraft Load Factor = Aircraft lift/W = (111600 – 4100)/14000 = 7.7

9

192

ExampleAssume the aircraft of the previous example is in thesame attitude and the pilot suddenly furthermaneuvered it by pushing the control stick forward togive the aircraft a pitching acceleration of 4 rad/sec2.Finda) The inertia forces and the tail load T, assuming the lift

force on the wing does not change.b) The forces on the jet engine which weighs 1500 lb

and its centre of gravity as shown in fig. Assume moment of inertia Iy equals 300000 lb.sec2 in

193

The figure shows the freebody of the aircraft with liftand inertia forces as foundbefore. The additional inertiaForce due to the angularacceleration α = 4rad/sec2

Iy = 300000x4 = 1200000 in.lb

Which acts clockwise orcounter to the direction of the

angular acceleration.

194

The aircraft is now in static equilibrium. To find the tail load T take moments about aircraft c.g

ΣMc.g = 1200000 – 111600x8

-218T = 0 T = 1409 lbTo find Mag take

ΣFg = 111600 – 14000 + 140 – Mag = 0

Mag = 99009 lb

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hence ag = (99009/14000)g = 7.1g ft/sec2

The engine c.g is 50 inches aft the aircraft c.g. The force on the engine will be its own weight of 1500 lb and the inertia forces due to ag and α.

Inertia force due to ag is Mag = (1500/g)7.1g = 10630 lb

Inertia force due to angular velocity α is Mr α = [1500/(32.2x12)]x50x4 = 776 lbTherefore the resultant force on the engine is 1500 + 10630 + 776 = 12906 lb

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ExampleA large transport aircraft of gross weight 100000 lbwith pitching mass moment of inertia Iy=40000000 lb.sec2.in is

making a level landing with nose wheel slightly off ground. Thereaction on the rear wheels is 319000 lb inclined at such anangle to give a drag component of 100000 lb and a verticalcomponent of 300000 lb.

Find a) The inertia forces on the aircraft b) The resultant load on the pilot whose weight is 180 lb and

whose location is as shown in fig.

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SolutionThe inertia forces onthe aircraft are forcesMax and Mag and

couple Ic.gα.

To find Max Take

ΣFx = 100000 – Max = 0

Max = 100000 lb

Hence ax = 100000/M

= (100000/100000)g ax = 1g

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To find Mag take

ΣFg = 300000 – 100000

- Mag = 0

Mag = 200000 lb

hence ag = 200000/M

= (200000/100000)g ag = 2g

To find the inertia coupleIc.gα take moments about

aircraft c.g

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ΣMc.g = - 100000x120

– 300000x84 + Ic.gα = 0

Ic.gα = 37200000 lb

Hence angular acceleration α equals

α = 37200000/40000000 α = 0.93 rad/sec2

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The forces on the pilotconsist of the pilot weight(180 lb) and the variousinertia forces as shown in fig. Max = (180/g)g = 180 lb

Mag = (180/g)2g = 360 lb

The inertia force due to αacts normal to the radiusarm between the aircraftc.g and the pilot. This forcewill be replaced by its g andx components

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Fx = Mzα

= (180/32.2x12) x40x093 = 17 lbFg = Mxα

=(180/32.2x12)x372x0.93 = 161 lbTotal force in x direction onpilot is 180-17 = 163 lbTotal force in g direction is 360 + 180 – 161 = 379 lbhence resultant force is√(3792 + 1632) = 410 lb

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