Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 2 pot

Calculation of Loads in Diagonal Truss Members:- S because the reaction of this truss at =F would put torsion on the spar and the spar has no appreciable torsional resistance.. The res

A2 18

The wing dDeams due to the distributed air loads acting upon them, are 21so sub1e

to bending loads in addition to the axia

The wing beams thus act as deam-columns

subject of deam-colimn action {ts treated

another chapter of this book

If the wing {5 covered with metal skin instead of fabric, the drag truss can be omitted

since the top and bottom skin act as webs of 2

beam which has the front and rear beams as its

flange members The wing is then considered as

a Dox beam subjected to combined bending and

been made identical to the wing panel of example

problem 1 This outer vanel attached to the cen

ter panel by single pin fittings at points (2)

and (4), Placing pins at these points make the

structure statically determinate, whereas tf the

beams were made continuous through ali 3 panels,

the reactions of the lift and cabane struts on

the wing beams would be statically indeterminate

since we would have a 4-span continuous beam

resting on settling supports due to strut de-

formation The fitting pin at points (2) and

(4) can be made eccentric with the neutral axis

of the beams, hence very little is gained by

making beams continuous for the purpose of de-

creasing the lateral beam bending moments For

assembly, stowage and shipping it is cenventent

to dutld such a wing in 3 portions Ifa

multiple bolt fitting is used as points (2) and

(4) to obtain a continuous beam, not much ts

gained because the design requirements of the

various govermmental agencies specify ‘hat the

wing beams must also be analyzed on the as-

sumption that a multiple bolt fitting provides

only 50 percent of the full continuity 2

EQUILIBRIUM OF FORCE SYSTEMS TRUSS STRUCTURES,

Lengths & Directional Components of Cabane Struts

T

sym Y D s| L

Member Y/L | D/L S/L

Froot Cabane Cp | 30 | 10 | 27 | 41.59 | 721 | 240 | 648 a Diagonal Cabane| Cp | 30 Strut [30 ; 27} 50.17 | 3597 | 597 338 | L Rear Cabane Cp | 23.5; 8 |27140.42 | 731 | 1485| 688

ces exerted oy outer panel on center 2anel at pin point (4) From Table 42.2 of example prob- lem i, this resultant V reaction equais 630 +

62 = 692 1d

The vertical component of the cabane re~

action at joint (8) equals one half the total

beam load due to symmetry of loading or 55 x

drag loac of 336 1b at (3) is due to the rear CRB = - 1510 1b (compression) cabane strut, as is likewise the beam axial load

of - 1510 at (8) The axial beam load of

ID =D, - 2260 x 1485 = 0 - 2281 lb at (7) is due to reaction of front

cabane truss The panel point loads are dus to

forward

De = 336 ld, drag truss reaction

The reaction which holds all these drag

cabane truss at point (7) since the front and

5684 - (1294 ~ 726) = 5684 ciagonal cabane struts intersect to form a rigid

: triangle Thus the drag reaction R equals one

w= 30 26 #/in

_ A 7)

ke 20 R, = 25354 $0 Ry = 25354 20-4 Solving the truss for the loading of Fig

42.44 we obtain the member axial loads of Fiz

Fig A2.43 A2.46

Fig 42.43 shows the V loads on the center front 4457 368 368 1157,

SỞ Š Be Š T

oooh 2) pores System at Joint 7

2535 2535 “15027 -17308 717308 1502

(11)

cp Cr VS Plane C VD Plane Cp L Point 7 oads§ in Cabane Struts Due to Drag Reaction a in Cat Struts Due to Drag Reaction at

Solution for Loads in Drag Truss Members Cr = - 2740 ib (compression)

Cp = 3310 (tension) Fig A2.44 shows all the loads applied to

the center panel drag truss The § and D re- adding these loads to those previously calcu-

actions from the outer 2anel at joints (2) and lated for lift loads:

Applied Air Loads

V component = 7523 (outer oanel) + 65 x 45

= 10448 (check)

~ 1110 (outer ?anel) ~ 65 x

6 = ~ 1500 (error 2 1b.)

D component =

Tne total side load on a vertical plane thru

centerline of airplane should equal the $ com~

ponent of tne a ted loads ‘The applied side

loads = - 394 lb (see problem 1} The air load

on center panel is vertical and thus nas zero $

component

From Table 42.3 for fuselage reactions fave 23 = 16178 From Fig Az.45 the load in

the front beam at £ of airplane equals - 17308

and 568 for rear beam The horizontal component

of the diagonal drag strut at joints 11 equals

216 x 45/57.6 = 169 1b

Then total S components = 16178 - 17308 +

568 + 169 = - 393 1b which checks the side

component of the applied air loads

Example Problem 12 Single Svar Truss Plus

EQUILIBRIUM OF FORCE SYSTEMS

Torsional Truss System

In small wings or control surfaces, fabric

is often used as the surface covering Since

the fabric camnot provide reliable torsional

resistance, internal structure must be of such

design as to provide torsional strength A

single spar plus a special type of truss system

1S often used to give a satisfactory structure

Pig 42.46 tilustrates such a type of structure,

namely, a trussed single spar AEFN plus a tri-

angular truss system between the spar and the

trailing edge 0S Fig A2.46 (a, >, c) shows

the three projections and dimensions The air

load on the surface covering of the structure

is assumed to be 0.5 1b./in.* intensity at spar

line and then varying linearly to zero at the

trailing edge (See Fiz d)

The problem will be to determine the axtal loads in all the members of the structure It

will be assumed that all members are 2 force

members a8 is usually done in finding the

fT panels @ 12" = 84" —-—

Fig, 46b Fig 46c

SOLUTION:

The total air load on the structure equals

the average intensity per square inch times she surface area or (0,5)(.5}(36 x 94) = 756 lb In order to solve a truss system by a method of

joints the distributed load must be replaced by

an equivalent load system acting at the Joints

of the structure Referring to Fig (4d), the total air load on a strip il wide and 36 inches long ts 36(0.5)/2 = 9 1b and its c.g

or resultant location is 12 inches from line AZ

In Fig 46a this resultant load of $ lb./in ts imagined as acting on an imaginary beam located along the line 1-1 This running load applied along this line is now replaced dy an equivalent force system acting at joints OPGRSEDBCA The results of this joint distribution are shown Dy tne joint loads in Fig A2.46 7o 112ustrate

how these foint loads were obtained, the caicu-

lations for loads at foints ESDR will de given

Fig A2.48 shows a portion of the

to De considered For a running load o D re t «a *y ucture

1b./in., along line 1-1, reactions will be for

c D

E——34”————— !?' — Fig A2 48

Simple beams resting at points 2, 3, 4, 5, etc

The distance between 2-3 is & inches The total

load on this distance is 8x9 = 72 1b One

half or 36 1b goes to point (2) and the other

half to point (3) The 36 lb at (2) 1s then

replaced by an equivalent force system at § and

S or (36)/3 = 12 lb to S and (36)(2/3) = 24 to

BE The distance between points (3) and (4) is

8 inches and the load is 8x 9 = 72 lb One

half of this or 36 goes to point (3) and this

added to the previous 36 gives 72 lb at (3)

The load of 72 is then replaced by an equivalent

force system at S and D, or (72)/3 = 24 lb to

S$ and (72)(2/3) = 48 to D The final load at §

is therefore 24 + 12 = 36 lb as shown In Fig

A 2.46 Due to symmetry of the triangle CRD,

one half of the total load on the distance CD

goes to points (4) and (5) or (24 x 9)/2 = 108

1b The distribution to D ts therefore (108)

(2/3) = 72 and (108)/38 = G6 to R Adding 72

to the previous load of 48 at D gives a total

load at D = 120 1b as shown in Fig A2.46

The 108 lb at point (5) also gives (108)/5

36 to R or a total of 72 lb at R The student

should check the distribution to other joints

as shown in F1g A2.46

To check the equivalence of the derived

joint load system with the original air load

system, the magnitude and moments of each

system must be the same Adding up the total

joint loads as shown in Fig A2.46 gives a total

or 756 1b which checks the original air load

The moment of the total air load about an x

axis at left end of structure equals 756 x 42 =

31752 in lb The moment of the joint load

system in Fig 42.46 equals (66 x 12) + (72 x

36) + (72 x S0) + (56 x 84) + 144 (24 + 48) +

(120 x 72) + (24 x 84) =# 51752 ín.lb or a

check The moment of the total atr load about

line AE equals 756 x 12 = 9072 in.l> The

moment of the distributed joint loads equals

(6 + 66 + 72 + 72 + 34)36 = 9072 or a check

Calculation of Reactions

The structure is supported by single pin

fittings at points A, N and 0, with pin axes

parallel to x axis It will be assumed that

the fitting at N takes off the spar load in

2 direction Fig A&.46 shows the reactions

Oy, Og, Ay, Ny» Nz To find O; take moments

about y axis along spar AEFN

IMy = (6 + 66 + 72 + 72 + 36)36 ~ 36 0, =0

whence Og = 252 1b acting down as assumed

To find point (A)

Fig A2.49 shows a diagram of this spar with its joint external loading The axial loads produced by this loading are written on the truss members (The student should check these member loads.)

the trailing edge member ts negligible, the

A2, 22

load of 36 1b, at Joint S in order to be trans-

ferred to point O through the diagonal truss

system must follow the path SDRCGBPAO in like

manner the load of 72 at R to reach O must take

the path RCOBPAO, etc

Calculation of Loads in Diagonal Truss Members:-

S because the reaction of this truss at =F

would put torsion on the spar and the spar has

no appreciable torsional resistance

Considering Joint S

as a free body and writing

iFy = - 159 x 943 + 943 DR = 0, hence DR =

158 1b, aFz = - 159 x ,118 + 159 X 118 - Ty =O

the diagonal Shear load on

whence T, = 0, which means truss produces no Z reaction or

spar truss at D

aFy = - 314 x 159 - 314 x 159 -T, = 0

whence Ty = - 100 Ib

If joint G 1s investigated in the

the results will show that Tz = 0 same manner, and Ty = 100

The results at joint D shows diagonal truss system produces no that the rear Shear load

EQUILIBRIUM OF FORCE SYSTEMS,

bottom chord, Consider Joint R

Joint 2 Load to be transferred to truss qBL = 72 +

72 + 36 = 180 lb

Hence load in 9B = (180 x 0,5)(1/.118) =

- 762

whence QL = 762, 8P = 762, LP = ~ 762 Joint P

Load = 180 + 6&6 = 246 Load in PA = (246 x 0.5)(1/.118) = - 1040 Whence PN = 1040

Consider Joint (A) ZPy = - 1040 x ,943 + 960 AO = 0, AO =

1022 15

In like manner, considering Joint N, gives NO

= ~ 2022 105,

as a free body

Couple Force Reactions on Spar

As pointed out previously, tne diagonal torsion truss produces a couple reaction on the spar in the y direction The magnitude of the force of this couple equals the y component or the load in the diagonal truss members meeting

at a spar joint Let Ty equal this reaction load on the spar

At Joint C:-

Ty = - (487 + 457),314 = ~ 287 15, Likewise at Joint J, Ty = 287

At Joint 8:-

Ty = - (762 + 762).314 = - 479 Likewise at Joint L, Ty = 479

At Joint A:-

ty F- (1040 x 314) = - 326

These reactions of the torsion truss upon

the spar truss are shown in Fig A2.50 The

loads in the spar truss members due to this

loading are written adjacent to each truss

member Adding these member loads to the loads

in Fig A2.49, we obtain the final spar truss

member loads as shown in Fig AgZ.5Sl

If we add the reactions in Figs A2.50 and

A2.51, we obtain 3528 and 504 which check the

reactions obtained in Fig A2.46

A2.13 > Landing Gear Structure

The airplane ig both a landborme and air-

borne vehicle, and thus a means of operating

the airplane on the ground must be provided

which means wheels and brakes Furthermore,

provision must be made to control the impact

forces involved in landing or in taxiing over

Tough ground This requirement requires a

special energy absorption unit in the landing

gear beyond that energy absorption provided by

the tires The landing gear thus includes a

so-called shock strut commonly referred to as

an oleo strut, which is a member composed of

two telescoping cylinders When the strut is

compressed, oil inside the air tight cylinders

is forced through an orifice from one cylinder

to the other and the energy due to the landing

impact ts absorbed by the work done in forcing

this ofl through the orifice The orifice can

be so designed as to provide practically 4

uniform resistance over the displacement or

ravel of the alec strut

An airplane can land safely with the air-

plane in various attitudes at the instant of

ground contact Fig A2.52 {llustrates the

three altitudes of the airplane that are

Specified by the govermment aviation agencies

for design of landing gear In addition to

these symmetrical unbraked loadings, special

loadings, such as a braked condition, landing

On one wheel condition, side load on wheel, etc

are required In other words, a landing gear

can be subjected to 4 considerable number of

different loadings under the various landing

conditions that are encountered in the normal

The successful design of landing gear for present day aircraft is no doubt one of the most difficult problems which is encountered in the structural layout and strength design of air~

craft In general, the gear for aerodynamic efficiency must be retracted into the interior

of the wing, nacelle or fuselage, tms a re- liable, safe retracting and lowering mechanism system is necessary The wheels must be braked and the nose wheel made steerable The landing gear is subjected to relatively large loads, whose magnitudes are several times the gross weight of the airplane and these large loads must be carried into the supporting wing or fuselage structure Since the weight of land- ing gear may amount to around 6 percent of the weight of the airplane it is evident that nigh strength/wetght ratio is a paramount design requirement of landing gear, as inerficitent structural arrangement and conservative stress analysis can add many unnecessary pounds of welght to the airplane and thus decrease the pay or useful load

A2.14 Example Problems of Calculating Reactions and Loads on Members of Landing Gear Units

In its simplest form, a landing gear could consist of a single oleo strut acting as a cantilever beam with its fixed end being the upper end which would be rigidly fastened to the supporting structure The lower cylinder

of the oleo strut carries an axle at its lower

North American Aviation Co,

Douglas DC-7 Air Transport

Fig, A2.54 Nose Wheel Gear Installations

À2 26

end for attaching the wheel and tire This

cantilever beam is subjected to bending in two

directions, torsion and also axial loads Since

the gear is usually made retractable, it is

difficult to design a single fitting unit at

the upper end of the oleo strut that will

resist this combination of forces and still

permit movement for a simple retracting mechan-

ism Furthermore, it would be difficult to

provide carry-through supporting wing or fuse-

lage structure for such large concentrated

load systems

Thus to decrease the magnitude of the bending moments and also the bending flexibility

of the cantilever strut and also to simplify

the retracting problem and the carry-through

structural problem, it 1s customary to add one

or two braces to the oleo strut In general,

effort {1s made to make the landing gear

structure statically determinate by using

specially designed fittings at member ends or

at support points in order to establish the

force characteristics of direction and point of

planes Fig A2.56 1s a space dimensional

diagram In landing gear analysis it is common

to use V, D and S as reference axds instead of

the symbols Z, X and Y This gear unit is

assumed as representing one side of the main

gear on a tricycle type of landing gear system

The loading assumed corresponds to a condition

of nose wheel up or tail down (See lower

sketch of Fig A2.52) The design lead on the

wheel is vertical and its magnitude for this

problem {s 15000 lb

The gear unit is attached to the supporting structure at points F, H and G Retraction of

the gear is obtained by rotating gear rearward

and upward about axis through F and H The

fittings at P and H are designed to resist no

bending moment hence reactions at F and H are

unknown in magnitude and direction Instead of

using the reaction and an angle as unknowns,

the resultant reaction is replaced by its V and

D components as shown in Fig A2.56 The re~

action at G is unknown in magnitude only since

the pin fitting at each end of member GC fixes

the direction and line of action of the reaction

at G For convenience in calculations, the

reaction G is replaced by its components Gy and

Gp For a side load on the landing gear, the

reaction in the S direction is taken off at

point F by a special designed unit

EQUILIBRIUM OF FORCE SYSTEMS

TRUSS STRUCTURES

SOLUTION The supporting reactions upon the zear at points fF, H, and G will be calculated es a beginning step, There are six unknowns, namely

FS, Fy, Fp, Hy, Hp and G (See Fig A2.56) With

6 equations of static equilibrium available for

@ Space force system, the reactions can be found

by statics Referring to Fig A2.56:-

A2.55)

With Gy known, the reaction G equals (6500) (31.8/24) = 8610 lb and similarly she compon- ent Gp = (6500)(21/24) = 5690 1b

Fig, A2.57 summarizes the reactions as found

the structure as a whole by taking moments about

D and V axes through point A

BMy(p) = - 10063 x 14 + 6500 x 6 + 11109 x 8

= - 140882 + 52000 + 88882 = O(check) IMacy) = 5690 x 8 - 433 x 8 - 3004 x 14

45520 - 3464 ~ 42056 = O (check)

The next step in the solution will be the calculation of the forces on the oleo strut unit Fig 42.58 shows a free body of the oleo~

strut-axle unit The brace members BI and CG are two force members due to the pin at each end, and thus magnitude is the only unknown re- action characteristic at points Band C, The fitting at point E between the oleo strut and the top cross member FH is designed in such a manner as to resist torsional moments about the oleo strut axis and to provide D, V and S$ force reactions but no moment reactions about D and § axes The unknowns are therefore BI, CG, Eg,

Ey, Ep and Ty or a total of 6 and therefore statically determinate The torsional moment

Tr is represented in Fig A2.58 by a vector with a double arrow The vector direction represents the moment axis and the sense of rotation of the moment is given by the rignt hand rule, namely, with the thumb of the right hand pointing tn the same direction as the arrows, the curled fingers give the sense of rotation,

To find the resisting torstonal moment Tp take moments about V axis through £

Tig checks the value previously obtained when the reaction at G was found to be 8610

The D and V components of CG thus equal,

CGp = 8610 (21/31.8) = 5690 1b

CGy = 8610 (24/31.8) = 6500 lb

To find load in brace strut BI, take moments

about D axis through point Z

A2.28

To find Ep take 2D = 0 2D = S690 ~ 3119 ~ Ep = 0, hence Ep = 2571

Hp The loads or reactions as found from the

analysis of the olso strut unit are also re-

corded on the figure The equations of

equilibrium for this free body are:-

a3 = 0 = - 3920 + 3920 + Fg = 0, or Fg = 0

IMp(p) = 22 Hy - 3920 x 2 - 7840 x 20 -

13332 x6 =0 Whence, Hy = 11110 lb This check value obtained previously, and therefore is a check

whence, Fp = 3004 lb

Thus working through the free bodies of the oleo strut and the top member FH, we come

out with same reactions at F and H as obtained

when finding these reactions by equilibrium

equation for the entire landing gear

The strength design of the oleo strut unit and the top member FH could now be carried out

because with all loads and reactions on each

member known, axial, bending and torstonal

stressea could now be found

The loads on the brace struts CG and BI are axial, namely, 8610 lb tension ana 8775

1b compression respectively, and thus need no

further calculation to obtain design stresses

TORQUE LINK

The oleo strut consists of two telescoping tubes and some means must be provided to trans-

mit torstonal moment between the two tubes and

still permit the lower cylinder to move upward

into the upper cylinder The most common way

of providing this torque transfer is to use a

double-cantilever-nut cracker type of structure

Fig A2.60 illustrates how such a torque length

could be applied to the oleo strut in our

The reaction R, between the two units of

the torque link at point (2), see Fig A2.40, thus equals 24952/9 2 2773 lb

The reactions R, at the base of the link at point (3) = 2773 x 8.5/2.75 = 8560 1b With these reactions known, the strength design of the link units and the connections could be made

Sxample Problem 14

The landing gear as illustrated in Fig

A2.61 1s representative of a main landing gear which could be attached to the under side of a wing and retract forward and upward about line

AB into a space provided by the lower portion

of the power plant nacelle structure The oleo strut Of has a sliding attachment at HE, which prevents any vertical load to be taken by member AB at £ However, the fitting at E does transfer shear and torque reactions between the oleo strut and member AB The brace struts

GD, FD and CD are pinned at each end and wiil

be assumed as 2 force members

An airplane level landing condition with unsymmetrical wheel loadiiug has been assumed as shown in Pig A2.61,

SOLUTION The gear is attached to supporting struc- ture at points A, Band C The reactions at

first, treating Fig A2.62 these points will be calculated

the entire gear as a free body

ANALYSIS AND DESIGN OF

40000

Fig A2 62

shows 4 space diagram with loads and reactions

The reactions at A, B and C have deen replaced

by their V and D components

To find reaction Cy take moments about an

S axis through points AB

The reaction at C must have gv,

a line of action along the line é

CD since member CD is pinned at 28 D

each end, thus the drag compon- 24

ent and the load in the strut

CD follow as a matter of geometry Hence,

Cp = 66666 (24/28) = 57142 lb

Cp = 66666 (36.93/28) = 87900 lb tension

To find By take moments about a drag axis

through point {A}

AZ 29 FLIGHT VEHICLE STRUCTURES

#2(p) = - €0000 x 9 ~ 40000 x 29 - 66666

x 19 + 38 By = 0

whence, By = 78070 1b

To 3v

#ind Ay, take 5V = O

19 = 0 (check) REACTIONS ON OLEO STRUT OB Fig A2.63 shows a free body of the oleo-

strut OE The loads applied to the wheels at

Mo(y) = (18000 - 10000) 10 = 50000 in.1b and

Mo(p) = (60000 - 40000) 10 = 200000 in.lb

Th 2) ng are indicated in Fig A2.63 by the vectors with double arrows The sense of the mement 1s determined by the right hand thumb