Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 4 doc

Table A5.1 shows the calculations in table form for determining the V, the wing shear in the Z2 direction, the bending moment My or mo- ment about the X axis and My the moment about the

A5.6

BEAMS -~ SHEAR AND MOMENTS The load of 1000 at 45° and applied at point E”

will be referred to point & the centerline of

beam Fig d shows the reaction at E due to the

load at £' The reaction at B should also be

referred to the beam centerline Fig A5.16

shows the beam with the applied loads at points

COE’ F and B’ Figs A5.17, 18 and 19 show

the axial load, vertical shear and bending mo-

ment diagrams under the beam loading of Fig

Bending

3000 pee 500 1965.6"4 Moment Dia

be 207.1 Fig A5 19

The shear diagram is determined in the same manner as explained before The applied exter-

nal couples do not enter into the vertical shear

calculations The bending moment diagram can be

calculated by taking the algebraic sum of all

couples and moments of all forces lying to the

ona side of a particular section If it 1s de~

sired tq use the area of the shear diagram to

obtain the bending moments, it is necessary to

add the couple moments to the shear areas to ob-

tain the true bending moment For example, the

bending moment just to the left of point E will

be equal in magnitude to the area of the shear

diagram between C and E plus the sum of all ap-

plied couple moments between C and E but not in~

cluding that at BE,

To illustrate the calculations are: ~ (- 500 x 5) + (707.8 x 10) = 4878 in Ib (from

area of shear diagram)

(3000 - 4000) = - 1000 In lb (from sum of

couple moments)

Thus bending moment at E

3578 in lb lett * 4578 ~ 1000 =

The bending moment at Eight Will equal that at

Elert Plus the couple moment at & or 3578 + 707

this fact, consider the beam of Fig AS.20,

namely, a simple supported beam with an ex-

ternally applied couple moment of 10 in 13

magnitude at point C the center point of the beam The shear and bending moment diagrams

are as indicated and a maximum bending moment

occurs at C but the shear diagram does not pass

forces Thus if we assume the couple moment has

a dx arm the shear to the right of Cc is one lb and then changes to some unknown negative value and then back to one lb positive as the dist- ance dx is covered in going to the left Thus

the shear goes to zero twice in the region of point Cc

A5.7 Moment Diagrams as Made up of Parts

In calculating the deflection of statically

determinate beams (See Chapter A7) and solving

statically indeterminate structures (See Chapter A8), the area under the bending moment curve is required, thus it is often convenient to treat each load and reaction as a separate acting force and draw the moment diagram for each force The true bending moment at a particular point

will then equal the algebraic summation of the ordinates of all the various moment curves at this particular point or adding the vartous

Separate moment diagrams will give the true bending moment diagram Figs, A5.22 and AS.23 illustrate the drawing of the bending moment diagram in parts In these examples, we start from the left end and proceed to the right end

and draw the moment curve for each force as

though the beam was a cantilever with the fixed

ANALYSIS AND DESIGN OF

Final Moment Dia Final Moment Curve

Fig AS 22 Fig AS 23

support at the right end The final bending

moment curve for the true given beam then equals

the sum of these separate diagrams as {llustra-

ted in the figures

STATIC MOMENT CURVES IN SOLVING STATICALLY

INDETERMINATE STRUCTURES

The usual procedure in solving a statically

indeterminate structure is to first make the

structure statically determinate by removing the

necessary redundant or unknown reactions and

then calculating the deflection of this assumed

statically determinate structure as one step in

the overall solution of the problem (See Chapter

A8) In the solution of such structures it is

likewise convenient to treat the bending moment

diagram as made up of parts To illustrate,

Pig AS.24 shows a loaded rectangular frame

fixed at points A and B The reactions at doth

points A and B are unknown in magnitude, di-

rection and location, or each reaction has 3 un-—

known elements or a total of 6 unknowns for the

two reactions With 3 static equilibrium

equations available, the structure is statically

indeterminate to the third degree tg A5.25

illustrates one manner in which the structure

can be made statically determinate, by freeing

the end A to make a bent cantilever beam fixed

FLIGHT VEHICLE STRUCTURES

Tection of the reaction at A

A5.7

at B and thus leaving only 3 unknown elements

of the reaction at 8 Fig AS.25 shows the

bending moment curves for each load acting sep-

arately on this cantilever frame Fig A5.26

Shows the true bending moment as the summation

of the various moment curves of Fig AS.25

As another solution of this fixed ended

frame, one could assume the statically deter~

minate modification as a frame pinned at A and

pinned with rollers at B as illustrated in Fig

A5.27, This assumed stricture is statically

determinate because there P,=10 ars only 3 unknown elements,

namely the magnitude and di- and the magnitude of the re-

action at B For conven- tence the reaction at A is

resolved into two magnitudes

as H and V components The reactions V,, Hy, and Vp can

then be round by statics and Fig A5 27

the results are shown on Fig AS.27 Fig A5.28 shows the bending moment diagram on this frame due to each load or reaction acting separately, starting at A and going clockwise to B Fig

AS5.29 shows the true bending moment diagram as

the summation of the separate diagrams,

30

180 30 in, lb,

Final Bending 2Ð Moment Diagram

of frame is posi- tive moment

Fig AS, 29 45.8 Forces at a Section in Terms of Forces at a Previous Station

STRAIGHT BEAMS

Aircraft structures present many beams

which carry a varying distributed load Mini-

mm structural weight is of paramount importance

in aircraft structural desten thus it 1s de- sirable to have the complete bending moment diagram for the structure so that each portion

of the structure can be proportioned effic- tently To decrease the amount of numerical work required in obtaining the complete shear

A5.8

and bending moment diagrams it usually saves

time to exoress the shear and moment at a given

station in terms of the shear and moment at a

previous station plus the effect of any loads

lying between these two stations To illustrate,

Pig AS.30 shows a cantilever beam carrying a

considerable number of transverse loads F of dif-

ferent magnitudes Fig A5.31 shows a free body

Fie

Bea FF Fy a +,

OMa

te a5 30 Fig A5.31

of the beam portion between stations 1 and 2

The Vertical Shear V, at station 1 equals the

summation of the forces to the left of station 1

and M, the bending moment at station 1 equals

the algebraic sum of the moments of all forces

lying to left of station 1 about station lL

Now considering station 2: - The Vertical

Shear V, = V, + Fite, or stated in words, the

Shear V, equals the Shear at the previous sta-

tion 1 plus the algebraic sum of all forces F

lying between stations 1 and 2 Again consider-

ing Fig A5.21, the bending moment M, at station

2 can be written, Mz = M, + Vid + Pi_.a, or

stated in words, the bending moment M, at sta-

tion 2 {1s equal to the bending moment M, at a

previous station 1, plus the Shear V at the

previous station 1 times the arm d, the dist-

ance between stations 1 and 2 plus the moments

of all forces lying between stations 1 and 2

about station 2

A5.9 Equations for Curved Beams

Many structural beams carry both longitud-

tonal and transverse loads and also the beams

may be made of straight elements to form a frame

or all beam elements may be curved to form a

curved frame or ring For example the airplane

fuselage ring is a curved beam subjected to

forces of varying magnitude and direction along

its boundary due to the action of the fuselage

skin forces on the frame Since the complete

bending moment diagram is usually desirable, it

is desirable to minimize the amount of numerical

work in obtaining the complete shear and bending

moment values Fig A5.32 shows a curved beam

loaded with a number of different vertical loads

F and horizontal loads Q Fig AS.33 shows the

beam portion 1-2 cut out as a free body Hị

represents the resultant horizontal force at

station 1 and equals the algebraic summation of

all the Q forces to the left of station 1 Vy

represents the resultant vertical force at

station 1 and equals the sum of all F forces to

left of station 1, and M, equais the bending

moment about point (1) on station 1 due to the

moments of all forces iying to the left of point 1

BEAMS SHEAR AND MOMENTS

Then from Fig 46.33 we can write for the

resultant forces and moment at point (2) at station 2: '-

Ve = Vi + Pins

Ha =H + Qa

Mẹ EM, + Vid - Hin + Fy.@ - Qi~ab

Having the resultant forces and moments for a

given point on a given station, it is usually necessary in finding beam stresses to resolve

the forces into components normal and parallel

to the beam cross-section and also transfer their location to a point on the neutral axis

of the beam cross-section

For example Fig 45.34 shows the resultant

Gob EE Mg=M, -Ne

Fig AS 34 Fig AS 35 Fig AS 36

forces and moment at point 1 of a beam cross-

section They can be resolved into a normal foree N and a shear for S plus a moment M, as shown in Fig A5.35 where,

N=Hcosa+V sina S=Vcosa-H sing later on when the beam section is being de-

Sigmed it may be found that the neutral axis lies at point 0 instead of point l Fig

45.36 shows the forces and moments referred to

point 0, with M, being equal to M, - Ne

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 45.10 Torsional Moments,

The loads which cause only bending of a

beam are located so that their line of action

passes through the flexural axis of the beam

Quite often, the loading on a beam does not act

through the flexural axis of the beam and thus

the beam undergoes both bending and twisting

The moments which cause the twisting action are

usually referred to as torsional moments The

airplane wing 1s an excellent example of a beam

structure that is subjected to combined bending

and torsion Since the center of pressure of

the airfoil forces changes with angle of attack,

and since there are many flight conditions it is

impossible to eliminate torsional moments under

all conditions of flight and landing For the

fuselage, the Vertical tail surfaces is norm-

ally located above the fuselage and thus a load

on this tail unit causes combined bending and

twisting of the fuselage

Fig AS.&4 illustrates a cantilever tube

being subjected to a load P acting at point A on

a fitting attached to the tube end The flex-

Fig A5 34 Fig, A5, 35

ural axis coincides with the tube centerline, or

axis 1-1 Fig AS.25 shows the load P being

moved to the point (0) on the tube axis 1-1,

however the original force P had a moment about

{O) equal to Pr, thus the moment Pr must be

added to the load P acting at (0) if the force

system at point (0) is to be equivalent to the

original force P at point A The force P acting

through (0) causes bending without twist and the

moment Pr causes twisting only

For the resolution of moments into various

Trasultant planes of action, the student should

refer to any textbook on statics

45.11 Shears.and Moments on Wing

Arts A4.5 and A4.6 of Chapter A4 discusses

the airloads on the wing and the equilibrium of

the airplane as a whole in flight As explain-

ed, it ig customary to replace the distributed

air forces on an airfoil by two resultant

forces, namely, lift and drag forces acting

through the aerodynamic center of the airfoil

plus a wing moment The airflow around a wing

is not uniform in the spanwise direction, thus

the airfoil force coefficients Cy, Cp and Cy

vary spanwise along the wing Fig AS.26 shows

a typical spanwise variation of the Cy and Cp

force coefficients in terms of a uniform span-

wise variation C, and Cp

Any particular type of airplane is destgned

to carry out a certain job or duty and to do

that job requires a certain maximm airplane

LIFT AND DRAG COEFFICIENTS

(In terms of uniform distribution)

velocity with the maneuvering limited to certain Maximum accelerations These limiting acceler-

ations are usually specified with reference to

the X Y Z axes of the airplane Since the di- rections of the lift and drag forces change with angle of attack it is simpler and convenient in stress analysis to resolve all forces with ref-

erence to the X Y Z axes which remain fixed in direction relative to the airplane

As a time saving element in wing stress

analysis, it is customary to make unit load an- alysis for wing shears and moments The wing

shears and moments for any design condition

then follows as a matter of simple proportion and addition For example it is customary: - (1) To assume a total arbitrary unit load act- ing on the wing in the Z direction through the aerodynamic section of the airfoil section and distributed spanwise according

to that of the Cy; or lift coefficient

(2) A similar total load as in (1) but acting

in the X direction

{3} To assume a total unit wing load acting in

the Z direction through the aerodynamic center and distributed spanwise according

to that of the Cp or drag coefficient

(4) Same as (3) but acting in the X direction

(5) To assume a unit total wing moment and

distributed spanwise according to that of the ony „9 moment coefficient

The above unit load conditions are for con- ditions of acceleration in translation of the

A5.10

airplane as a rigid body Unit load analyses

are also made for angular accelerations of the

airplane which can also occur in flight and

landing maneuvers

The subject of the calculation of loads on the airplane is far too large to cover ina

structures book, This subject is usually cover-

ed in a separate course in most aeronautical

curricula after a student has had initial

courses in aerodynamics and structures To il-

lustrate the type of problem that is encountered

in the calculation of the applied loads on the

airplane, simplified problems concerning the

wing and fuselage will be given

A5.12 Example Problem of Calculating Wing Shears and

Moments for One Unit Load Condition

Pig AS.37 shows the half wing planform of

a cantilever wing Fig AS5.38 shows a wing

section at station 0 The reference Y axis has

been taken as the 40 percent chord line which

happens to be a straight line in this particular

The total wing area is 17760 sq in For convenience a total unit distributed load of

17760 lbs will be assumed acting on the half

wing and acting upward in the 2 direction and

through the airfoil aerodynamic center The

spanwise distribution of this load will be ac-

cording to the (cy) lift coefficient spanwise

distribution For simplicity in this example

it will be assumed constant

Table A5.1 shows the calculations in table

form for determining the (V,) the wing shear in

the Z2 direction, the bending moment My or mo-

ment about the X axis and My the moment about

the Y axis for a number of stations between the

wing tip station 240 and the centerline station

9

Column 1 of the table shows the number of Stations selected Column 2 shows the Tư

BEAMS SHEAR AND MOMENTS

ratio or the spanwise variation of the lift coefficient C, in terms or a uniform distribu-

tion ổt, In this example we have taken this

ratio as unity since we nave no wind tunnel or aerodynamic calculations for this wing relative

to the spanwise distribution of the lift force

coefficient In an actual problem involving an

airplane a curve such as that given in Fig

A5.36 would be available and the values to

place in Column 3 of Table AS.1 would be read from such a curve, Column (2) gives the wing

chord length at each station Column (4)

gives the wing running ioad per inch of span

at each station point Since a total unit load of 17760 lb was assumed acting on the half wing and since the wing area is 17760 sq

in., the ruming load per inch at any station

equals the wing chord length at that station

In order to find shears and moments at the various station points, the distributed load is now broken down into concentrated loads which are equal to the distributed load on a strip and this concentrated strip load is taken as acting through the center of gravity of this

distributed strip load Columns 5, 6, and 7 show the calculations for determining the

(APz) strip loads Colum 8 shows the lo-

cation of the AP, load which is at the centroid

of a tropizcidal distributed load whose end values are given in Colum (4) In determin- ing these centroid locations it is convenient

to use Table 43.4 of Chapter As

The values of the shear V, and the mo-

ment My at each station are calculated by the method explained in Art A5.8, Columns 9, 10,

11 and 12 of Table AS.1 give the calculations

For example, the value of My > 9884 in Col-

wm (12) for station 220 equals 2436, the My moment at the previous station in Column (12) plus 4908 in Colum (10) which is the shear

at the previous station (230) times the dist-

ance 10 inches plus the moment 2540 in Colum

(9) due to the strip load between stations

230 and 220, which gives a total of 9884 the

value in Column (12)

The strip loads AP, act through the

aerodynamic center (a.c.) of each airfoil strip Colum (13) and (14) give the x arms which is the distance from the a.c to the reference Y

axis (See Fig 45.38) Colum 15 gives the

My moment for each strip load and Colum 16

the My moment at the various stations which

equals the summation of the strip moments as one progresses from station 240 to zero

Fig A5.3S shows the results at station (0} as taken from Table 45.1

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AG.11

TABLE A5.1 CALCULATION OF WING SHEAR V, AND WING MOMENTS DUE TO TOTAL UNIT DISTRIBUTED HALF WING LOAD OF 17 AND My

Wnen the time comes to design the structural

make-up of a cross-section to withstand these

applied shears and moments, the structural de-

signer may wish to refer the forces to another

Y axis as for example one that passes through

the shear center of the given section This

transfer of a force system with reference to an-

other set of axes presents no difficulty

SHEARS AND MOMENTS CN AIRPLANE BODY

A5,13 Introduction

The body of an airplane acts essentially as

& beam and in some conditions of flight or land-

ing as a beam column which may be also subjected

to twisting or torsional forces Thus to design

an airplane body requires a complete picture of

the shearing, bending, twisting and axial forces

which may be encountered in flight or landing

In the load analysis for wings, the direct air

Checks Total Limit Load Assumed on Haif

forces are the major forces For the body load

analysis the direct air pressures are secondary, the major forces being of a concentrated nature

in the form of loads or reactions from units

attached to the body, as the power plant, wing, landing gear, tail, etc Im addition, since the

body usually serves as the load carrying medium,

important forces are produced on the body in re-

sisting the inertia forces of the weight of the

interior equipment, installations, pay load etc,

AS in the case of the wing, a large part

of the load analysis can be made without much

consideration as to the structural analysis of

the body The load analysis of an airplane body involves a large amount of calculation, and

thus the treatment in this chapter must be of 2

simplified nature, and is presented chiefly fcr the purpose of showing the student in general how the problem of load analysis for an air- plane body is approached

AS, 12

AS.14 Design Conditions and Design Weights

The airplane body must be designed to witn~

stand ali loads from specified flight conditions

for both maneuver and gust conditions Since

accelerations due to air gusts vary inversely

as the airplane weight, it customary to analyze

or check the body for a light load condition for

flight conditions in general, the design

weights are specified by the government agen-

cites For landing conditions, however, the

normal gross weight is used since it would be

more critical than a lightly loaded condition

The general design conditions which are

usually investigated in the design of the body

are as follows:

Flight Conditions:

HAA (High angle of attack)

LALA (Low angle of attack)

I.L-A.A {Inverted low angle of attack)

I.H.A.A {Inverted high angle of attack}

The above conditions generally assume only

translational acceleration In addition, it is

sometimes specified that the forces due to 4a

certain angular acceleration of the airplane

about the airplane c.g must be considered

The body is usually required to witastand

special tail loads both symmetrical and unsym-

metrical which may be produced by air gusts,

engine forces, etc Also, the body should be

checked for forces due to unsymmetrical air

loads on the wing

Landing Conditions:

In general, the body 1s investigated for the

following landing conditions The detailed re-

quirements for each condition are given in the

government specifications for both military and

commercial airplanes

Landplanes: Level landing

Level landing with side load

Three point landing

Three point landing with ground loop

Nose over or turn over

condition

arresting (Usually for only

Navy Carrier based air- planes)

BEAMS SHEAR AND MOMENTS

AS.15 Body Weight and Balance Distribution

The resisting inertia forces due to the dead wetght of the body and its contents plays

an important part in the load analysis for the

airplane body ‘hen the initial aerodynamic and general layout and arrangement of the air-

plane is made, it is necegsary that a complete

weight and balance estimate of the airplane be made This estimate is usually made by an en~

gineer from the weight control section of the

engineering department who has lad experience

in estimatinz she weight end distribution of airplane units This astimate wnicn 1s pre-

sented in report form gives the weights and

{c.g.} locations of all major airplane units

or installations as well as for many of tne

minor units which make up these major airplane assemblies or installations This weignt and balance report forms the oasis for the dead wetent inertia load analysis which forms an

important part in the load analysis of the air-

plane body The use of this weizht and balance estimate will be illustrated in the example

problem to follow later

A&.16 Load Analysis Unit Analysis

Due to the many design conditions such as those listed in Art 45.14, the zensral pro+

cedure in the load analysis of an airplane body

is to dase it on a series of unit analyses

The loads for any particular design condition

then 2cllows as a certain combination of the

unit results with the proper multiplying fac- tors A simplified example problem follows which illustrates this unit metnod of appreacn

45.17 Example Problem Milustrating the Calculation of Shears and Moments on Fuselage [ e to Unit

Load Conditions

Pig A5.40 and AS.41 shows a layout of the

airplane body tobe used in this example pr«b-

lem It happens to be the body of an actuas airplane and the wing used in the previous ex~

ample problem was the wing that went with the

are — nh

ANALYSIS AND DESIGN OF

Table AS.2 gives the Weight and Balance

estimate for the total airplane This table is

usually formulated by the Weight and Balance

Section of the engineering department and it is

necessary to nave this tnformation before the

airplane load analysis can be made

Vert (Z) arms measured from thrust line

Reference ) (+ ia up)

Axis forward prop ¢ Horiz (x) arms measured from 2 axis 5” (+ is aft}

tem we | Bortz, | Bortz, | Vert | Vert

No Name và, „|, Arm | Moment | Arm | Mom

Grose wt x= 376500/4300 + 98,5” aft of Ref Axis

| z + ~40280/4300 = 9.4”' below thrust Line

SOLUTION:

WEIGHT AND BALANCE OF BODY ITEMS

WEIGHT DISTRIBUTION

Table AS.3 gives the weight and balance

calculations for all items attached to fuselage

or carried in t*e fuselage, except the wing and

items attached to the wing as tha front landing

gear and the fuel

In order to obtain 4 close approximation to

the true shears and moments on the fuselage due

to the dead weight inertia loads, it is neces-

sary to distribute the weights of the various

items as given in Table A5.3 Fig AS.42 shows

a side view of the airplane with the center of

gravity locations of the weight items of Table

A5.3 indicated by the (+) signs In the various

design conditions, the direction of the weight

inertia forces changes, thus it is convenient

and customary to resolve the inertia forces into

X and Z components Thus, in Fig AS.43, the

weights as given in Table A5.3 are assumed act-

ing in the Z direction through their (c.g ) lo-

cations The loads as shown would not give a

true picture as to the shears and moments along

the fuselage, thus these loads should ve dis-

tributed in a manner which should simulate the

actual weight distribution In most weight and

valance reports, the weight items are broken

down into considerable more detail than that

shown in Table AS.3, which makes the weight dis-

tribution more evident The person making the

WEIGHT AND BALANCE OF AIRPLANE LESS WING GROUP AND INSTALLATIONS IN AND ON WING

% equal arm from thrust line + la up, Ret Axes {x is distance from z Ref Axis 5" forward af

propeller + la aft

Boriz | Boriz | Vert | Vert

em Name Wetght | ‘arm | Mom | Arm | Mom,

& | Tlectricai aygtea| 140 | 1 7830| + 520

8 | Tail Wheel group 3 306 10700 ~L0 - 350

overall structural arrangement as to its possi-

ple influence on fuselage weight distribution

The whole process involves considerable common

sense if a good approximation to the welght dis-

tribution is to be obtained Fortunately the

large dead weight loads, such as the power

plant, tail, etc are definitely located, thus

small errors in the distribution of the minor

distributed weights does not change the over- all shears and moments an appreciable amount

In order to obtain reasonable accuracy, the fuselage or body is divided into a series of stations or sections In Fig A5.42, the sec- tions selected are designed as stations which represent the distance from the 4 reference axis, The general problem is to distribute the

concentrated loads as shown in Fig AS.45 into

an equivalent system acting at the various fuselage station points

Obviously, if a weight item from Table AS.3,

represents a concentrated load such as a pilot, student, radio, etc., the weight can ve dis-

tributed to adjacent station points inversely

as the distance of the weight (c.g.) from these adjacent stations However, for a weight item such as the fuselage structure (Item 2 of Table AS.3) whose c.g location causes it to fall be- tween stations 80 and 120 of Fig A5.43, it

would obviously be wrong to distribute this

weight only to the two adjacent stations since the weight of 350= is for ‘he entire fuselage

This weight item of 3505 shoul’ thus de dis-

tributed to all station points The controlling requirement on this distribution is that the

moment of the distributed system about the ref- erence axes must equal the moment of the orig-

Fig AS.44

inal weight about the same axes

WEIGHT DISTRIBUTION TO FUSELAGE STATIONS STATION

Fig A5.49 Final weight distribution in X direction referred

to X axis plus proper couples

ANALYSIS AND DESIGN OF shows how the dead weight of 350# was distribu-

ted to the various station points considering

the weights to be acting in the Z direction

Table AS.4 shows the results of this sta-

tion point weight distribution for the weight

items of Table A5.3 The values in the hori-

Zontal rows opposite each weight item shows the

distribution to the various fuselage stations

The summation of the weights in each vertical

column at each station point as given in the

third horizontal row from the bottom of the

table gives the final station point weight

These weights are shown in Fig AS.45 for

weights acting in the Z direction The moment

of each total-station load about the Z axis is

given in the second horizontal row from the

bottom of Table 45.4 The summation of the

moments tn this row must equal the total wx

moments of Table A5.3 or 219700"# This check

is shown in the last vertical column of Table

AS.4

The distributed system must also be distri-

buted In the 2 or vertical direction in such a

manner as to have the same resultant c.g lo-

cation as the original weight system which is

illustrated in Fig AS.46, Fig AS.47 illus- trates how the fuselage weight distributed system as shown in Fig A5.44 1s distributed

in the vertical direction at the various

station points so that the moment of this sys-

tem about the X axis is equal to that of the

original fuselage weight of 350# For con-

venience, these distributed fuselage weights can be transferred to the X axis plus a moment

as shown in Fig AS.48,

Table AS.4 shows the vertical distribution

of the various items at the various station

points The bottom horizontal row gives the

moment about the X axes of the loads at each

station point, which equals the individual

loads times their Z distances The summation

of the values in this horizontal row must equal

the total wz moment of Table AS.3 This check

is shown at the bottom of the last vertical

column Fig 45.49 shows the results as given

in Table 45.4 for the weight distribution in

the X direction

TABLE A5,4 PANEL POINT WEIGHT DISTRIBUTION {Monocoque Type of Fuselage}

Skation No * distance x from Ref, 2 Axis

i X ? distance of station from z reference axis which

= = distance above or below thrust Una or x axis

A5.18 Unit Analysis for Fuseiage Shears and Moments,

Since there are many flight and landing

conditions, considerable time can be saved if a

unit analysis is made fcr the fuselage shears,

axial and bending forees The design values in

general then follow as a summation of the values

in the unit analysis times a proper multiplica-

tion factor

The loads on the fuselage in general con-

sists of tail loads, engine loads, wing re-

actions, landing gear reactions tf attached to

fuselage and inertia forces due to the airplane

acceleration which may be due to doth transla-

tional and angular acceleration of the airplane

For simplicity, these loads can be resolved into

components parallei to the Z and X axes

To illustrate the unit analysis procedure,

a unit analysis for our example problem will be

carried out for the following unit conditions:

(1) Unit acceleration or load factor in Z

direction and acting up

(2) Unit acceleration or load factor in X

direction and acting forward

(3) Unit tail load normal to X axis acting

be carried out in detail The others will be discussed in detail in later paragraphs

Solution for Unit Load Factor in Z Direction,

Pig A5.50 shows the dead weight loads

acting in the Z direction as taxen from Table

AS.4 or Pig AS.45 The wing ts attacned to the

fuselage at stations 73 and il

A5.50 The fittings at these points are assumed

as designed to cause all the crag or reaction in

the X direction to be taken entirely at the

front fitting on station 73

To place the fuselage in equilibrium, the

wing reaction will be calculated:

©

SFy = 9, Ry + O=C, hence Ry u la

Meration 0 = 219700 ~ 116 Rg - 73 Rp = O (A)

(Note: 219700 from Table AS.3)

IF, = - 2555 + Rp + Rn m0 Ta Tan nan (8)

Solving equations (A) and (8) for Rp and

Re: R

Rp = 1780 lb., Ra = 775 1b

Table A&.5 gives the calculations for th

fuselage shears and bending moments at the var-

fous station points,

Solution for Unit Load Factor in X Direction

Fig A5.51 shows the panel point dead

weight distribution for loads acting in the X

direction and aft, as taken from Tabie 45.4 or

Fig AS.40, To place the fuselage in equili-

brium the wing reactions at points (A) and (8)

will be calculated,

aFy > 2555 ~ Ry = 0, hence Ry = 2555 1b (forward)

Take moments about point (A)

IM, * 2555 x 17 + 5920 - 43 Rp =

hence Rp = 1147.8 (up) (5920 equals the sum of the couples from Table

AS.4

IF, = 1147.8 - Rp = 0,

hence Rp = 1147.8 1b (down)

Table A5.6 gives the calculations for the shears,|

moments and axial loads for the loading of Fig

FUSELAGE SHEARS AND MOMENTS FOR ONE LOAD FACTOR IN Z DIRECTION (Dead Weight Acting Down)

st Load or ! V = shear} Ax = Dist aM Moment

No Reaction; = Zw ! between = VAx M*

Now w | (ibs.) | stations | (in, Ibs.)| {in, Los.) | ! i

; 1 + 0 0 | |

(1) + refers to aft side of station

- refers to forward side of station

* M = Mat previous Station in col 6 plus AM in col 5,

STA.73 STA.116

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A5 17

TABLE AS.4 TABLE AS.7

FUSELAGE SHEARS, MOMENTS & AXIAL LOADS FOR

ONE LOAD FACTOR IN X DIRECTION FUSELAGE SHEARS & MOMENTS FOR UNIT

i Inerua Loads Acting Alt HORIZONTAL TAIL LOAD IN Z DIRECTION

¡ Sta | load Wx tụ: |[PxYEWX|V z : sỹ M„| AM | 2K” so ; Dist | &Mz =

No [in zor r axial | shear Moment bạn ! và | i 2 3 í 3 5

‡ Load or AX = dist 4 +19 0 a: 0 AM | Moment

35 ot KG: 9 2 9 220 | 9 220 Sta Reaction| = 2 Ww between z : VÀ i lbs

3 5 — 25 m w ibs, lbs stations V4 x |in Ibs

“1 1 x (0565) 2 43435 + refers to aft side of station 9 100 4 ~16150

Col 1 ~ refers to forward side of station Px is plus for tension in fuselage 11 _Í 45.6 -315.6 ! - 400 | _16815g

¡ (Col 9) M = M at previous station In Col 9 plus AML of col & plus 36 :

- 1

Solution for Unit Horizontal Tail Load Acting Down a? 0 -3%5 8 2629 9

-| -375.6 9 9

tail acting in the Z direction, with balancing * 39 —

center of pressure on the horizontal tail is at 1H _ 0 0 )

station 277.5 Fig A5.52 shows the fuselage - 7 ” — 1.6 nh

loading To find wing reactions at (A) and (B): | | Col 6) = OM Moers station in Col, 6 plus

Table AS.7 gives the detailed calculations

for the shears and moments at the various sta-

Rg=375 6

STA T3 fig A5.32 CALCULATION OF APPLIED FUSELAGE SHEARS,

MCMENTS AND AXIAL LOADS FOR A SPECIFIC

FLIGHT CONDITION

Using the results in Tables AS.S, AS.6, and

A5.7, the applied shears and moments for a given

2light condition follow as a matter oŸ propor- tion and addition To i{liustrate, the applied values for one flight condition will be given

It will be assumed that the aerodynamic calculations for this airplane for the (H.A.A.)}

nigh angle of attack condition gave the follow-

ing results, wnich the student will have to

accent without «nowledge of how they were

contained