TABLE OF CONTENTS

Chapter No.

.. ~...

Al The Work of the Aerospace Structures Engineer.

STATICALLY DETERMINATE STRUCTURES

(Loads. Reactions, Stresses. Shears. Bending Moments, Deflections>

A2A3

j A4A5A6

! ;, A7

rj,ASA9A1QAllA12

Equilibrium of Force Systems. Truss Structures. Externally Braced Wings. Landing Gear.Properties of Sections - Centroids. Moments of Inertia, etc.General Loads on Aircraft.Beams - Shear and Moments. Beam - Column Moments.Torsion - Stresses and Deflections.Deflections of Structures. Castigliano's Theorem. Virtual Work. Matrix Methods.

THEORY AND METHODS FOR SOLVING STATICALLYINDETERMINATE STRUCTURES

Statically Indeterminate Structures. Theorem of Least Work. Virtual Work. Matrix Methods.Bending Moments in Frames and Rings by Elastic Center Method.Column Analogy Method.Continuous Structures' Moment Distribution Method.Slope Deflection Method.

BEAM BENDING AND SHEAR STRESSES.MEMBRANE STRESSES. COLUMN AND PLATE INSTABILITY.

A13 Bending Stresses.A14 Bending Shear Stresses - Solid and Open Sections- Shear Center.A15 Shear Flow in Closed Thin-Walled Sections.A16 Membrane Stresses in Pressure Vessels.A17 Bending of Plates.A18 Theory of the Instability of Columns and Thin Sheets.

INTRODUCTION TO PRACTICAL AIRCRAFT STRESS ANALYSIS

A19 Introduction to Wing Stress Analysis by Modified Beam Theory.A20 Introduction to Fuselage Stress Analysis by Modified Beam Theory.A21 Loads and Stresses on Ribs and Frames.A22 Analysis of Special Wing Problems. Cutouts. Shear Lag. Swept Wing.A23 Analysis by the "Method of Displacements".

THEORY OF ELASTICITY AND THERMOELASTICITY

A24A25A26

The 3-Dimensional Equations of Thermoelasticity.The 2-Dimensional Equations of Elasticity and Thermoelasticity.Selected Problems in Elasticity and Thermoelasticity.

. j 1-3;

;...,.....l_~ ....

TABLE OF CONTENTS Continued

Chapter No.

FLIGHT VEHICLE MATERIALS AND THEIR PROPERTIES

81 Basic Principles and Definitions.82 Mechanical and Physical Properties of Metallic Materials for Flight Vehicle Structures.

STRENGTH OF STRUCTURAL ELEMENTS AND COMPOSITE STRUCTURES

Cl Combined Stresses. Theory of Yield and Ultimate Failure.C2 Strength of Columns with Stable Cross-Sections.C3 Yield and Ultimate Strength in Bending.C4 Strength and Design of Round. Streamline, Oval and Square Tubing in Tension, Compression. Bending,

Torsion and Combined Loadings.CS Buckling Strength of Flat Sheet in Compression, Shear, Bending and Under Combined Stress Systems.C6 Local Buckling Stress for Composite Shapes.C7 Crippling Strength of Composite Shapes and Sheet-Stiffener Panels in Compression. Column Strength.C8 Buckling Strength of Monocoque Cylinders.C9 Buckling Strength of Curved Sheet Panels and Spherical Plates. Ultimate Strength of

Stiffened Curved Sheet Structures.C10 Design of Metal Beams. Web Shear Resistant (Non-Buckling) Type.

Part 1. Flat Sheet Web with Vertical Stiffeners. Part 2. Other Types of Non-Buckling Webs.ell Diagonal Semi-Tension Field Design.

Part 1. Beams with Flat Webs. Part 2. Curved Web Systems.C12 Sandwich Construction and Design.C13 Fatigue.

CONNECTIONS AND DESIGN DETAILS

01 Fittings and Connections. Bolted and Riveted.02 Welded Connections.03 Some Important Details in Structural Design.

Appendix A Elementary Arithmetical Rules of Matrices.

INDEXAccelerated MoUon of

Rigid Airplane.Aircraft Bolts .Aircraft NutsAircrait Wing Sections -

Type,Aircraft Wing Structure -

Truss Type.Air Forces on Wing .Allowable Stresses (and

Interactions) .Analysis of Frame with

Pinned Suppor-ts .Angle MethodApplication of Matrix Methods

to Various StructuresApplied LoadAxis of Symmetry.

Beaded WebsBeam Design - Special Cases.Beam Ftxed End Moments by

Method of Area MomentsBeam Rivet Des IgnBeam Shear and Bending

MomentBeams - Forces at a SectionBeams > Moment Diagrams.Beams with Non-Parallel

FlangesBeams - Shear and Moment

Diagramsaeams > Statically Deter-

minate & Indeterminate .Bending and Compression

of ColumnsBending Moments - Elastic

Center Method.Bending of Rectangular

PlatesBending Strength - Basic

Approach.Bending Strength - Example

ProblemsBendtng Strength of Round

Tube'Bending Strength - Solid

Round Bar.Bending StressesBending Stresses - Curved

BeamsBending Stresses - Elastic

RangeBending Stresses - Non-

hcmcgenecus Sections.Bending Stresses About

pr-mcipal Axes .Bending of Thin PlatesBolt Bending Strength.Bolt & Lug Strength AnalysiS

MethodsEolt Shear-, Tension &

Ber:ding Str-engthsBoundary ConditionsBox Beams AnalysisBrazingBuckling CoefficientBuckling of Flat Panels wiui

ntsetmuer FecesBuckling of Flat Sheets under

Combined Loads.Buckli.ng or Rectangular

Plates

A4. 8D1.2D1. 2

A19. I

A2.14A4.4

Cll.36

A9.l6C7.1

A7.23A4.1A9.4

CIO.16D3.10

A7.32ClO. B

AS.IA5.7AS. 6

en, 9

AS.2

A5.1

AlB. 1

A9. 1

Ala. 13

INDEX Continued

Bl.2

C9.S

C4.5

A2.4

.' :''"it':-"'A

C4.25

C4.26

A8.11

AS. '7C4.5

A.26. 5

Ct. 24

C4.26Cl.I

AS. IS"A9.2

AIO.4

M.tSA9.13

A4.7

Cl1.4CtO.5

Cll.18CIO.10CtO.5

A5.9A19.25

AS. 10A23.11A19.5.'1.19.5

A19.11Al.2

Al5.11A2l.t

A19.14

ClO.17A19.2Al9.1

A19.12 A23.14

.("r-

Trusses With MultipleRedundancy . . .

Trusses With SingleRedundancy . . . .

Tubing Design FactsTwo-Dimensional Problems.Two-Cell Multiple Flange

Beam. One Axis otSymmetry ....

Type of Wing Ribs. .

Wagner Equations ..Web Bending &: Shear StressesWeb Design ...Web Splices ....Web Strength. Stable Webs.Webs with Round Lightening.

Holes ..Wing Analysis ProblemsWtng Arrangements. Wing Effective SecttonWing Internal StressesWing Shear and Bending

AnaLysis ...Wing Shear and Bending

Moments ....Wing - Sbear Lag . Wing Shears and MomentsWing Stiffness Matrix.. Wing Strength ReqUirementsWing Stress Analyl'lis MethodsWing - Ultimate Strength .Work of Structures Group.

tntimate Strength in CombinedBending &: F1exural Shear

tnUmate Strength in CombinedCompression, Bending,Flexural Shear &: Torsion.

Ultimate Strength in CombinedCompression, Bending &ITorsion ...

illtimate Strength in CombinedTension. Torsion andInternal Pressure p in psi.

Uniform Stress Condition. Unit Analysis lor Fuselage

Shears and Moments. . UnsymmetriCal Frame . .Unsymmetrical Frames orR_ .Unsymmetrical Frames using

PrinCi~Axes. . ..'Tnsymmeirical Structures

" ,

\t, . 'Jy - Load Factor" .....

AS.7

A6.10

A6.S

A6.2Cl.5

A7.33A2.9

C4.17

A6.3AS. 4

AS. ISA6.SA5.9

C4.1'7

A8.16

A24.1A16.5

A7.5A8.2

A7.5

AB.39A8.14AB.33

Al9.5A6.l

AS. 10

Bl.5A1B.a

C13.33D3.2

en. 1C11. 2A7.5

AIS. 15

ToUlgent ModulusTangent-Modulus TheoryTaxi Loads ...Tension CHps . . . . .Tension-Field Beam Action.'renetcn- Field Beam FormulasTheorem of Castiglta.no ..Theorem of Complementary

Energy......Theorem of Least Work .Theorems of Virtual Work and

Minimum Potential EnergyThermal Deflections by

MatriX Methods . .Thermal Stresses . . Thermal Stresses . . . Thermoelasttcity - Three-

Dimensional Equations. Thin Walled ShellsThree Cell - Multiple Flange

Beam. - Symmetrical aboutOne Axis ....

Three Flange - Single CellWing

Torsion - Circular Sections.Torsion - Effect of End

Restraint.....Torsion - Non-circular

Sections ...Torsion Open sectionsTorsion of Thin-Wailed

Cylinder having Closed TypeStiHeners .........

Torsion Thin Walled Sections.Torsional Moments - Beams Torsional Modulus of Rupture.Torsional Shear Flow in

Multiple Cell Beams byMethod of SuccessiveCorrections . . .

Torsional Shear Stresses inMulttple4Cell Thin-WallClosed Section - Distribution

Torsional Strength of RoundTubes .....

Torsional Stresses 1nMuitiple4 Cell Thin-WalledTubes .......

Transmission of Power byCylindrical Shaft. ...

TriaXial Stresses . .Truss Deflection by Method

of Elastic Weights Truss StructuresTrusses with Double

Redundancy. . .

A2.7

A2.3

AB.lC2.14

Al2.7

A24.5A7.l

A14.2

A.6.7B1.7

A24.6A22.1

Cll.17

C4.22Cll. 15

All. 29

Cll.32ci, 6A2.2

D3.12

AlS.24

Static Tension Stress-Strain Diagram . . .

Statically DeterminateCoplanar Structures andLoadings .

Statically Determinate andIndeterminate Structures

Statically IndeterminateFrames - Jomt Rotation

Statically IndeterminateProblem ......

Stepped Column - StrengthStiliened Cylindrical

Structures - illUmate3trength ....

Stiffness &I Carry-overFactors lor CUrved Members All. 30

Stiffness Factor. . . All. 4Strain - Displacement

Relations .....Strain Energy . . . . .Strain Energy of Plates Due

to Edge Compression andBending A18.19

Strain Energy In Pure Bendingof Plates. . . . . . . . AlB. 12

Streamline Tubing - Strength. C4. 12Strength Checking and

Design - Problems . . .Stren~_ ".: Round Tubes

_ ..nder Combined Loadings .sn-ess Analysis FormulasStress Analysis of Thin Skin -

Multiple stringer CantileverWing .......... A19.10

Stress Concentration Factors. C13.10Stress Distribution & Angle

of Twist for 2-Cell Thin4Wall Closed Section . .

Stress-Strain Curve ..Stress-Strain Relations .Stresses around Panel Cutout.Stresses in UprightsStringer Systems in Diagonal

Tension ......Structural Design Philosophy.Structural Fittings . .Structural Skin Panel Details.Structures with Curved

Members .....Successive Approximation

Method for Multiple CellBeams .......

Symbols for ReactingFitting Units ....

Symmetrica sections -External Shear Loads

Y Stiffened Sheet Panels C7.2C

CHAPTER AlTHE WORK OF THE

AEROSPACE STRUCTURES ENGINEER

AI. 1 Introduction.

The first controllable human flight in aheavier than air machine was made by OrvilleWright on December 17, 1903, at Kitty Hawk,North Carolina. It covered a distance ot 120feet and the duration of flight was twentyseconds. Today, this initial flight appearsvery unimpressive, but it comes into its trueperspective of Lmportance when we realize thatmankind for centuries has dreamed about dOingor tried to do what the ~rlght Brothersa:campllshed in 1903.

The tremendous progress accompliShed in thefirst 50 years of aviation history, with mostof it occurring in the last 25 years, 1s almostunbelievable, but without doubt, the progressin the second 50 year periOd will still be more~~believable and fantastic. As this is writtenin 1964, jet airline transportation at 600 MPHis well established and several types ofmilitary aircraft have speeds in the 1200 to2000 ~ range. ?reliminarJ designs of asupersonic airliner with Mach 3 speed have beenccmpleted ~~d the government is on the verge ofsponsoring the development of such a flightvehicle, thus supersonic air transportationshould become co~on in the early 1970's. Therapid progress i~ ~!ssile design has usheredin the Space Age. Already many space vehicleshave been flown in search of new knowledgewhich is needed before successful explorationof space such as landings on several planetscan take place. Unfortunately. the rapiddevelopment of the missile and rocket powerhas given mankind a flight vehicle when combinedwith the nuclear bomb, the awesome potential toquickly destroy vast regions of the earth.TMhile no person at ~resent ~~ows where or whatspace exploration will lead to, relative tobenefits to ~nkind, we do know that the nextgreat aviation expanSion besides supersonicairline transportation will be the full develop-

~ent and use of vertical take-off and landingaircraft. Thus persons who will be livingthrough the second half century of aviationprogress will no doubt witness even morefantastic progress than oceurred in the first50 years of aviation history.

A!. 2 General Organization of an Aircraft CompanyEngineering Dfvtetcn,

scientific machine and the combined knowledgeand experience of hundreds of engineers andscientists working in close cooperation isnecessary to insure a successrul product. Thusthe engineering division of an aerospace companyconsists of many groups of specialists whosespecialized training covers all fields ofengineering education such as PhySics, Chemicaland Metallurgical, MeChanical, Electrical and,of course, Aeronautical ~~lneering.

It so happens that practically all theaerospace companies publiSh extensive pamphletsor brochures explaining the organization of theengineering division and the duties andresponSibilities of the many sections and groupsand illustrating the tremendous laboratory andtest facilities which the aerospace industrypossesses. It is highly recommended that thestudent read ~~d study these tree publicationsin order to obtain an early general under-standing on how the ~odern flIght vehicle isconceived, deSigned and then prOduced.

In general, the engineering department ofan aerospace company can be broken down into saxlarge rather distinct sections, which in turnare further divided into specialized groups,which in turn are further divided into smallerworking groups of engineers. To illustrate, thesix sections will be listed together with someat the various groups. ThiS is not a completelist. but it should give an idea or the broadengineering set-up that is necessarJ.

I. Preliminary Design Section.

II. TecrJlical _~alysis Section.

(1) Aerodynamics Group(2) Structures Group(3) ~eight and 3alance Control Group(4) Power Plant Analysis Group(5) Materials and Processes Group(5; Centrols AnalYSiS Group

III. Component DeSign Section.

(1) Structural DeSign Group(~lng. Body and Control Surfaces)

(2) Systems Design Group(All mechanical, hydraulic, electricaland ther.nal installations)

IV. Laborato~J Tests Section.The ~odern commercial airliner, militarJairplane, missile and space vehicle is a highly

Al.I

Al.2 THE WORK OF THE AEROSPACE STRUCTURES ENGINEER

(1) Wind Tunnel and Fluid Mechanic5 ~estLabs.

(2) Structural Test Labs.(3) Propulsi~n Test Labs.(4) Electronics ~est Labs.(5) Electro-Mechanical Test Labs.(6) Weapons and Controls Test Labs.(7) Ar~log and Digital Computer Labs.

v. Flight Test Section.

VI. Engineering Field Service Section.

SinCe this textbook deals with the subjectof structures, it seems appropriate to discussin some detail the work of the Structures Group.For the detailed discussion of the other grou~s,the student should refer to the various air-craft company publications.

At. 3 The Work of the Structures Group

The structures group, relative to number ofengineers, is one of the largest of the ~anygroups ot engineers trat make up Section II,the technical analySis section. The structuresgroup is primarily responsible for thestructural integrity (safety) ot the airplane.safety may depend on sufficient strength orsufficient rigidity. This structural integritymust be accompanied with lightest pOSSibleweight, because any excess weight has detri-mental etfect upon the perfo~ce of aircraft.For example, in a large, long range missile,one pound of '~ecessary structural weight mayadd mora than 200 Ibs. to the overall weight ofthe missile.

The structures group is usually divided'into sUb-groups as tollows:-

(1) Applied Loads Calculation Group(2) Stress AnalySiS an~ Strength Group(3) Dynamics AnalYSiS Group(4) Special Projects and Research Group

THE '"ORK OF THE APPLIED LOADS GROUP

Before any part ot the structure can befinally proportioned relative to strength orrigidity, the true external loads on the air-craft must be determined. Since critical loadscame tTom many sources, the Loads Group mustanalyze loads fram aerOdynamiC forces, as wellas those forces from power plants, aircraftinertia; control system actuators; launching,landing and recovery gear; a~ent, etc. Theetrects of the aerOdynamic forces are initiallycalcUlated on the assumption that the airplanestructure 1s a rigid bOdy. Atts: the aircraftstructure is Obtained, its true rigidity canbe used to obtain dynamic effects. Results ofwind t~~el model tests are usually necessaryin the application of aerodynamic principles toload and pressure analYSiS.

The final results of t.he work of thisgroup are formal reports glv~n~ complete a~pliedload design criteria, with ~ny graphs ~nd swu-mary tables. The final results ~y 61v8 com-plete shear, moment and no~l forc~s =e~er=~dto a convenient set of :CY2 axes for major air-c ra.r t units such as the Wing, rus eIage , e t c .

THE WORK OF STRESS ANALYSIS ~\m S~R~~GTH GROUP

Essentially the primary job of :he stressgroup is to help specify or deter.nine the kindof material to use and the :h:c~~ess, size andcross-sectional shape Jt every struct~l ~eQber or unit on the airplane or ~issile, andalso to assist in the deSign of all jOints andconnections for such ~embers. safety with ~ightweight are the paramount str~ctural jesl~ re-quirements. ~he stress group ~ust consta~tlywork closely with the Structural DeSign Sect:Gnin order to evolve the best structural over-allarrangement. Such factors as ~ower ~lants,bUilt in fuel tanks, landing gear retractingwells, and other large cut-outs can d~ctate thetype of wing structure, as for example, a twospar single cell wing, or a multiple spar~ultiple cell wing.

To expedite the initial struct~r~l ~esignstudies, the stress group ~ust s~~ply initialstructural sizes based on approximate loads.The fi~l results of the work by the stressgroup are recorded in elaborate reports whichshow how the stresses were calculated and hewthe reqUired member sizes were obtained to carrythese stresses efficiently. The r:nal size ofa member may be dictated by one or more rae torssuch as elastic action, tne Ias t t c action, ele-vated temperatures, fatigue, etc. To insurethe accuracy of theoretical calculations, thestresS group must have the assistance of thestructures test laboratory in order to obtaininformation on which to base allowable designstresses.

THE WORK OF THE DYNAMICS A~LYSIS GROUP

The Dyna~ics AnalysiS Grou) has rapidlyexpanded in recent years ~elative to number ofengineers required because supersonic airplanes,missiles and vertical riSing ai~craft have pre-sented many new and complex problems in thegeneral field of dynamics. In some airc~ttcompanies the dynamiCS group 1s set up as aseparate group outside the Structures Group.

7he engineers in the dynamiCS group areresponSible for the investigation ot Vibrationand shOCk, aircraft flutter and the establish-~ent of desig~ requirements or c~2nges for itscontrol or correction. Aircraft contain dozensof mechanical installations. Vicration of ~~ypart of these installations or systems ~y beof such character as to cause faulty operationor danger of failure and therefore the dynamic

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A1.3

characteristics must be changed or modified inorder to insure reliable and safe operation.

The major structural units of aircraft suchas the wing and fuselage are not rigid bodies.7hus when a Sharp air gust strikes a fleXiblewing in high speed flight, we have a dynamicload situation and the wing Nill vibrate. Thedynamicist must determine whether this vibration1s serious relative to induced stresses on thewir~ structure. The dynamics group is alsoresponsible for the determination of thestability and performance of miSSile and flightvehicle guidar.ce and control systems. Thedynamics group must work constantly with thevarious test laboratories in order to obtainreliable values of certain factors that arenecessary in many theoretical calculations.

THE ',jaRK OF THE SPEC IAL PROJECTS GROUP

In general, all the various technical

groups have a speCial SUb-group which are work-ing on deSign problems that Nill be encounteredin the near 1r distant future as aviation pro-gresses. For example, in the r.t r-uc tur-ea Group,this sub-group might be studying such problemsas: (1) how to calculate the thermal stressesin the wing structure at super-sonic speedS;(2) how to stress analyze a new type of wingstructure; (3) what type of body str~cture isbest for future space travel and what kind ofmaterials will be needed, etc.

Chart 1 illustrates in general a typical~ke-up of the Structures Section of a largeaerospace company. Chart 2 lists the manyitems which the structures engineer must beconcerned with in insuring the structuralintegrity of the flight vehicle. Both Charts1 and 2 are from Chance-Vought StructuresDeSign Manual and are reproduced with theirpermi saton.

~EIlOE'-'\STlc:'4U..f ,.

~'IlOE'-'\STIC ,..---lcU"", i

,I STWCTVI~S I

""lAKlIlATo'" I' ~~u,11:'I. "'",ru,,, 1'liSI U,,"T

J i, -.c...."I.IC

'NO "Owu' .....1 ""ST

: UN"

, .....c:..'H~COMPUIATICH~'''G'' GIlOU'~~,

Chart 1. Structures Section OrganizationChance- Vought Corp.

-:3

e.13W m =r ETT? 5ZWF

AI. 4 THE WORK OF THE AEROSPACE STRUCTURES ENGINEER

THE LINKS TO STRUCTURAL INTEGRITY ARE NO BETTER THAN THE WEAKEST LINK

STRESSANALYSIS

SKIN PANElSBEAM ANALYSIS

STRAIN COMPATIBILITYSTRAIN CONCENTRATION

JOINT ANALYSISBEARI~G ANALYSIS

BULKHEAD ANALYSI SfITTING ANALYSISi1'ERMAL STRESS

,'oIECHAN1CAL COMPONENTSD:PERIMENTAL STRESS ANALYSIS

MATERIALS OFCONSTRUCTION

FA.STENERSWELDINGBONDING?LATE~ND BAR

FORGINGSCASTINGS

XTRUSIONSSHEET METALSANDWICH

?tASTIC l,OMINAITBEARINGS

MATERIALS ANDQUALITY CONTROL

DUCTILITYSTRESS-STRAIN

HOMOGENEOUS MATERIALRESIDUAL STRESS

HEAT TREAT CONTROLSTRESS CORROSION

STABILITY AT TE~PRATURESPECIFICATION CONFORMANCE

BLUE PRINT CONFORMANCE

ALLOWABLESYIHDINGFRACTUREATlGUE

WEAR, BRINELLINGCREEP

DEFliCTIONSTl'ERMAl O"FECTS

STIFFNESSCOMBINED LOADINGS

3UCKlING

COMPONENTANALYSISUNIT SOLUTIONS

INOffiRMINATE STRUCTURESWING ANALYSISTAIL ,l,NAlYS1S

FUSElJ.GE SHElL A""lYSISTHERMAL ANALYSIS

DfFlICTION ANALYSISSTIFFNESS

STIFFNESSCRITERIA

FLUTTERCONTROL SYSTEM STABILITY

PANEL fLIJTTER-SKIN CONTOURSCONTROL SYSTEM DEfLECTIONS

THERMAL EFFECTSMtCHANICAL VIBRATIONSROLL POWR-O IVERGENCE

AEROtWtAMIC CENTER SHIFTDYNAMIC RESPONSE

LOADS ANDENVIROMENT

FLIGHT LOAD CRITERIAGROUND LOAD eRITER IAFLIGHT LOAD DY~M'CSLAUNCHING DYNAMICS

LANDING DYNAMICSDYNAMIC RESPONSE

aECOVERY DYNAMICS,UGHT LOAD DISTRIBUTIONS

INERTIAL LOAD DISTRIBUTIONSFl..D:IBILITY EFFECTS

GROUND LOAD DISTRIBUTIONSREPEATEO LOAD SPECTRUMS

TEMPERATURE DISTRIBUTIONSLOADS FROM TllERMAL

DEFORMATIONSPRESSURES-IMPACT

Chart 2From Chance- Vought Structures Destgn Manual

CHAPTER A2EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES

EQUILIBRIUM OF S?AC::; ?J..RALLEL FORCE SYSTlli

force system pass through a cammon point. ThereSUltant, if any, must therefore be a forceand not a moment and thus only 3 equations arenecessary to completely define the conditionthat the resultant must be zero. The equat10nsof equilibrium available are therefore:-

A combir~tion of force and moment equationsto make a total of not more than 3 can be used.For the moment equations, axes through the pointof concurrency cannot be used since all forcesof ~he system pass through this point. Themoment axes need not be the same direction asthe directirns used in the force equations butof course. they could be.

In a parallel force system the direction ofall forces is known, but the magnitUde andlocation of each is unknown. Thus to determine~gnitude, one equation Is required and forlocation twa equations are necessary since theforce is not confined to one plane. In generalthe 3 equations commonly used to make the re-sultant zero for this type of ~orce system areone force equation and two moment equatiOns.For example, for a space parallel force systemacting in the y direction, the equations ofequilibrium would be:

} - - - - -(2.2)ooo

m, =m, =m, =

orl:Fx = aZFy = al:Fz =0

A2.2 Equations of Static Equilibrium.

To completely Cefl~e a force, we must knowits ~agnltude, direction and ~olnt of ann1ica-tion. These facts regaTding the ~arce aregenerally refer~ed to as the characteristics ofthe ~orce. So~etimes the more ~eneral te~ ofline Of act~on or lecation is used as a forcecr~racteristic in Place of paint of applicationdesignation.

A force acting in space is completelydefir-ed :: we %now its components in threedirections and its ~oments about 3 axes, as forexample FX J F~, Fz ~nd ~x, tly and Xz ~orequilibrium o~ a force system there can be noresultant force and thus the equations ofequllibri'4n are obtained ~y equating the forceand moment cCill~onents to zero. The equations

~f static equilibrium for the various types offorce systems 'Nill now be suear-tzeo .

A2.1 Introduction. The equations of staticequilibrium must constantly be used by thestress analyst and structural designer 1~ ob-taining unknown forces and reactions or unkno~ninternal stresses. They are necessary whetherthe structure ..or machine be s tmp.Le or complex.The ability to apply these equations 1s nodoubt best developed by solving ~ny problems.This Chapter l~ltlates the application of theseimportant phySical laws to tt.e force and stressar~lysls of structures. It 1s assumed that astudent has completed the usual college coursein engineering mecrznlcs called statics.

SG.UILIBRIUM OF GE:N""'..2.AL CO-PUu'JAR ~ORCE SYSTEMEQ.UILI3RIUi1 SO.UATICNS FOR GSNERALSPACE: urCN-COPLANAR) ?ORCE SYSTE?:

ZFy = 0, ZI1x ::: 0, n1z:= 0 - -(2.3)

Co~curre~~ 7;e~ns that all forces of thesq,UILI3RI'L11 OF' SPACE ::CNCJEP,S::T ropes SYS':'~

rhus :or a general space ~orce system,there are 6 equations of static eqUilibriumavailable. T~ree of these and ~o ~or8 can beforce equations. It is or t en acr-e convenientto ~ake the moment axes, 1, 2 ~nd 3. as any setot X, y and z axes. All 6 equat10ns could beno~ent equaticns about 6 ciffere~t axes. ~~eforce equations are written for 3 ~utually

~er?endiC~lar ~es ~nd need not be t~e x, yand z axes.

l:Fx = 0ZFy = 0ZFz = 0

~ ... = 0m, = am, = 0

} - - - - -(2.1) In this type ot farce system all forces liein one plane ~d it ta~es only 3 equations todeter.Aine the magnitUde, direction and locationor the resultant of such a force system. Eithertorce or moment equations C3n be ~sed. exceptthat a ~aximum of 2 :orce equa~lons can be used.For example, for a force system acting in thexy plane, the follcwiig co~blnaticn of equili-

bri~ equations could be used.

ZFx =0 ZFx = 0 ZFy a zr1z .. =aZFy =0 or L:!":z..= 0 or ZMz... :: 0 or 1:1z II = 0 2.4z:1z = 0 Z!"!Zil= 0 z:M:Z ll= 0 mz" =a

(':'he subsc=-lpts 1, 2 and 3 refer to differentlocations for z axes or moment center-s . )

A2.1

.AC

A2.2 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

reacted 'Jy other ext er-ne; -or-ces , ccmnorLyreferred to as reactions which hold the j:,cwnforces on ~he str~cture in equiliJri~ll. Sl~cethe static equa~ions of equil:bri:2n ~vai:~~:efor the various t~?es ot force syste~ arelinited, the str~c:ural engineer resorts tc theuse of fitting units whic~ establ:sh thsdirection 8f an ~n}~i8~TI fc~ce C~ ~~s ~o:nt Jf

ap~11c8:~on or both, t~~s decreasins th3 ~~berof UIL~owns t8 be determi~ed. ~~e ~l~ureswhich follow illustrate t~e ty;e of ~itt1~gunits employed or o:her gene~~l Ieth~ds ~JrestabliShing the ~orcs c~aracterist1cs ofdirsction and pOint of application.or ZFx

l:Fx : 0

ZFy : a

Since all forces lie in the ss~e plane andalso pass thr8ugh a c~~,on ~oint, the ~a~~itudeand direction of the r~s~ltant of this type offorce system is unknown ~ut the location ~s~own sin~e the ;oint of conc'~iency is on thelin8 of action cf the resultant. ~hus only twoequations 0: equilibri~ are necessary to definethe resultant and ~ke it zero. T~e combin-ations available are,

= a or ZFy = a or ZMz~ =0 } 2.5=0 ZI'1z = O ZI'1z a = O

EQUILIBRIUM: OF CO-PLANAR P.4RALLt:L FCRCE SYS'I'~

(The moment centers 1 and 2 cannot be on thesame y axis)

For any space or coplanar f8rce suc~ as ?and Q acting on the bar, the line o~ action ofsuch forces must act t..u-cucn the cent.cr- 0': theball ~! rotation of the bar is prevented. ~husa ball and socket joint can be used to establ:shor control the di~ection and line action of aforce applied to a struct~re through :~i~ :j~eof fitting. Since the ~oint has ~o rctaticTzlreSistance, no couples in any plane can beapplied to it.

Ball and Socket Fitting

~'lII

- p

-----2.6}4I'1z Ii = al:l1z. : aorl:Fy : 0n1z = 0

Since the direction of all forces in thistype or rorcs system 1s known and since theforces all lie 1n the same plane, it only takes2 equations to define the magnitude and locationof the resultant of such a force system. Hence,there are only 2 equations of equil1bri~ avail-able for this type of force system, namely, aforce and :noment equation or two momentequations. For example, for forces parallel toy axis and located in the xy plane the equili-brium equations available would be: -

(The z axis or moment center locations must beother than through the paint of concurrency)

EQUILIBRITJr1 OF COLINEAR FORCE SYSTEM

A2.3 Structural Fitting Units for Establishing the ForceCharacteristics of Direction and Point of Application.

where moment center 1 is not on the line ataction of the force system

To completely define a force in space re-qUires 5 equations and 3 equations if t~e forceis limited to one ?lane. In ~eneral a structureis loaded by ~own forces ar.d these !orces aretransferred tPIough the struct~re in some~nner of internal stress distribution and then

A collnear torce system 1s one where allforces act along the same line or in otherwords, the direction and location of the forcesis known but their nagnt tudes are unknown, thusonly magnitude needs to be found to define theresultant of a collnear force system. Thusonly one equation ot equilibrium is a'ffiilable,namely

If a bar AB has si~Gle pin f~t~ings ateac~ end, then any :orce P lying in the xyplane and anplied to end B ~ust have a directionand line of action co tnc tc t ng .... ith a line jo tn-.

i~g the pin centers at end ~it~1ngs A ar.d 3,since the :lttings cannot resist a ~oment aboutthe 3 axis.

A ,B-

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.3

Double Pin - Universal Joint Fittings

z Q

L. /-/11 PFh" liillirn

Another general fitting type that is usedto establish the direction of a force or reactionis illustrated in the figure at the bottom of thefirst column. Any reacting force at joint (A)~ust be horizontal since the support at (A) isso designed to provide no vertical resistance.

y

LSince single pin fitting units can resist

applied moments about axes normal to the pinaxiS, a double_pin joint as illustrated aboveis otten used. T~is fitting unit cannot resist~oments about y or z axes and thus appliedforces such as P and Q ~ust have a line ofaction and direction such as to pass throughthe center of the fitting unit as illustratedin the figure. The fitting unit can,. however,resist a moment about the x axis or in otherworas, a universal type of fitting unit canresist a torsional ~oment.

Cables - Tie RodsB

~PSince a cable or tie rod has negligible

bending resistance, the reaction at joint B onthe crane structure' tram the cable must becolinear with the cable axiS, hence the cableestablishes the force characteristics of direc-tion and point of application of the reactionon the truss at point B.

A2.4 Symbols for Reacting Fitting Units as Used inProblem Solution.

Rollers

-~~-t t

In solVing a str~cture for reactions,member stresses, etc., one must know what foroecharacteristics are unknown and it is cammonpractice to use simple symbols to indicate. whatfitting support or attachment units are to beused or are assumed to be used in the tinaldesign. The following sketch symbols are com-~onlY used for coplanar tares systems.

Lubricated Slot or Double Roller Ty~e of FittingUnit.

In order to pe~it structures to move atsupport paints, a fitting unit involving theidea of rollers is often used. For example,the truss in the figure above is supported bya pin :'itting at (A) which is rur-tner- attachedto a fitting portion that prevents any hori-zontal movement of truss at end (A), however,the other end (8) is supported by a nest ofrollers which provide no hor1zontal resistanceto a horizontal movement of the truss at end (8).The rollers fix the direction of the reactionat (B) as ,erpend~cular to the roller bed.Since t~e fitti~g uni~ is Jol~ed to the trussjOint by a pin, the ,oint of application of thereaction is also known. hence only one :orcecharacteristic, na~ely magnitude, 1s unknownfor a roller-pin type of fitting. ?or thefitting unit at (A), ,oint of application of thereaction to the truss 1s kncvrn because of thepin, but di~ectlon and magnitUde are unknown.

ioi

b..........Kni.fe Edge

bttRy

Rx

A small circle at the end of a member or ona triangle represents a single pin connectionand fixes the point of application of forcesacting between this unit and a connecting memberor structure.

Pin

The above graphical symbols represent areaction in which translation of the attach-ment point (b) is prevented but rotation of theattached structure about (b) can take place.Thus the reaction is lxnknown in direction andmagnitude but the point of applIcation is known,namely through point (b). Instead of using

I direction as an unknown, it is more convenientto replace the resultant reaction by two com-ponents at right angles to each other as indi-cated in the sketches.

~Double RoUer',~

",4",b'RI=Jn-'Lubricated Slot !~

/1~, .

A2.4 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

The above fitting units using rollers fixthe direction of the reaction as normal to theroller bed since the fitting unit cannot resista horizontal force through point (b). Hencethe direction and point of application of t~ereaction are established and only magnitude isunknown.

l~eEdgeRollers

!I

fixedrrrrrt

/

~FA Roller.I

(bJM.,'" RX-III\\\\

Illy

minate, and the degree of redundancy dapends onthe number of ~~nowns beyond that ~umber whichcan be found by the equations o~ static eqUili-brium. A structure can be statically indeter-minate with respect to exter~al reac~ions aloneor to In:ernal stresses alene or to 60th.

The additional eq~ations :hat are neededto solve a statically indeternlnate structureare obtained by conSidering the distortion ofthe structure. This means that the size of allmembers, the ~~terial ~rom which members are

~de must be known since dist~rtions ~ust becalculated. In a statically cete~lnatestructure this information on sizes and ~terial1s not required but only the configuration ofthe structure as a whole. Thus deSign analYSiSfor statically deter,ninate structure is stra~ghtforward whereas a general t~ial and error pro-cedure is required for design analYSiS ofstatically indete~inate structures.

A2. 6 Examples of Statically Determinate and StaticallyIndeterminate Structures.

The graphical symbol above is used torepresent a rigid support which Is attachedrigidly to a connecting structure. The re-action is campletely unknown since all 3 forcecharacteristics are unknown, namely, magnt 'tude ,direction and point of application. It Is con-venient to replace the reaction R by two forcecomponents referred to some paint (b) plus theunknown moment Mwhich the resultant reaction Rcaused about point (b) as indicated in theabove sketch. This discussion applies to acoplanar structure with all forces in the sameplane. For a space structure the reactionwould have 3 further unsnowns , nameLy, Rz, I1xand lly.

A2. 5 Statically DetermLnate and Statically IndeterminateStructures.

A statically determinate structure is onein which all external reactions and internalstresses tor a given load system can be foundby use at the equations of static eqUilibriumand a statically indeter.nlnate structure isone in which all reactions and internal stressescannot be round by using only the equations ofequilibrium..

A statically dete~inate structure is onethat has just enough external reactions, orjust enough internal members to make thestructure stable under a load system and if onereaction or member is removed, the structure 1sreduced to a linkage or a mechanism and istherefore not further capable or reSisting theload system. If the st~ucture has more ex-ternal reactions or internal members than isnecessary for stability ot the structure undera given load system it is statically ir.deter-

The first step in analyzl~g a stTuc~ure 1sto determine whether the str~cture as presentedis statically det.ermtnate . It so, the reactionsand internal stresses can Qe found without know-ing sizes of members or kind of mater1al. Ifnot statically determinate, the elastic :heory~ust be applied to obtain additional equa~ion5.The elastic theory is treated in considerabledetall in Chapters A7 to Al2 inclusive.

To help the student become f~illar withthe problem of determining whether a structureis statically determinate, several ex~pleproblems will be presented.

~ample Problem 1.

w '" 10 lb. lin.t

Fig. A2.1

In the st~~cture shown in Fig. 2.1, the~~Qwn forces or loads are the distri8uted loadsof 10 lb. per inch on member ABD. The reactionsat points A and C are unknown. The reaction atC has only one ~~own Characteristic, namely,magnitude because the point of application of Heis tnrough the cin center a: C and the directicnof Rc must be parallel to line SB because thereis a pi~ at the other end B of ~ember CE. Atpoint A the reaction 1s unknown in 11rectionand ~gnitude but the point of application ~ustbe through the pin center at A. Thus there are2 unknowns at A and one ~~o~vn at C or a total

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.5

I

):0""llC'fMc "'-RnVc

Fig. A2.6Fig. A2.5

.. c 01;A

Example Problem 4.

1 'L-J .

Fig. 2.5 shows a beam AB which carries asuper-structure CED which in turn is SUbjectedto the known loads P and Q. The question iswhether the structure is statically deter.nir~te.The external unknown reactions for the entirestructure are at points A and B. At A due tothe roller type of action, magnitude is the onlyunknown characteristic ot the reaction sincedirection and point ot application are known.At B, ~gnitude and direction are unknown butpaint of application 1s known, hence we have 3unknowns, namely, RA, VB and Hg, and with 3equations ot equilibrium available we can findthese reactions and therefore the structure isstatically dete~inate with respect to externalreactions. we now investigate to see it theinternal stresses can be found by statics arterhaVing found the external reactions. Obviously,the internal stresses will be affected by theinternal reactions at C and 0, so we draw a freebody of the super-st~cture as illustrated in?ig. 2.6 and consider the internal forces thateXisted at e and 0 as exterr~l reactions. Inthe actual structure the members are rigidlyattached together at point C such as a Nelded or

attached to reaction paints ABeD.

At reaction points A, E and D, the react.onis known in direction and pOint of applicationbut the magnitude is ~~own as indicated by thevector at each support. At point C, the re-action is unknown in direction because 2 strutsenter jOint C. Magnitude is also unknown butpaint of application is known since the reactionmust pass through C. Thus we have 5 unknowns,namely, Re, Ra, RO, Vc and He. For a copjanarforce system we have 3 equilibrium equationsavailable and thus the first conclusion mightbe that we have a statically indeterminatestructure to (5-3) = 2 degrees redundant. How-ever, observation of the structure shows twointernal pins at points E and F which meansthat the bending moment at these two points iszero, thus giVing us 2 more equations to usewith the 3 equations ot equilibrium. Thusdrawing tree bodies of the structure to lett atpin E and to right or pin F and equating momentsabout each pin to zero we obtain 2 equationswhich do not includ& unknowns other than the 5

l~OwnS listed above. The structure is there-tore statically determinate.

Fig. A.2.3Fig. A2. 2

Exa'Ilple Problem 3.

p p p P(1)~ t E .~ I-, ~In \~S::

,He fRB .J.VCFig. A2.4

~ig. 2.4 shows a straight ~ember 1-2 carrying aknown load system ? and supported by 5 struts

E:x:ample Problem 2.P

-Pin

~lg. 2.2 shows a structural frame carryinga known load system P. Due to the pins atredction points A and B the paint or applicationis known tor each reaction, however, the magni-tude and direction of each is unknown making atotal of 4 unknowns With only 3 equations ofequilibrium available for a coplanar forcesystem. At first we might conclude that thestructure is statically indeterminate but wemust realize this structure has an internal pinat C which means the bending moment at thispoint 1s zero since the pin has no resistanceto -ceat ton. If the entire structure is inequilibrium, then each part must likewise bein equilibrium and we can cut out any portionas a f~ee bOdy and apply the equilibriumequations. Fig. 2.3 shows a free bOdY of theframe to left of pin at C. Taking momentsabout C and equating to zero gives us a fourthequation to use in deter.nining the 4 unknowns,HA, VA' VB and Hg. The moment equation about Cdoes ~ot include the unknowns Ve and He sincethey have no ~oment about C because of zeroams. As in exa~ple problem 1, the reactionsat A and B have been r-ep Iac ed by H and V com-ponents ins:ead of ~sing an angle (direction)as an unknown characteristic. The struct~e isstatica:ly dete~inate.

of 3. with 3 equations of equilibrium avail-able for a co?lanar force system the structure1s statically determinate. Instead of uSing anangle as an '~nown at A to find the directionof the reaction, it is usually more convenientto re~lace the reaction by components at rightangles to each other as HA and VA in the figureand thus the 3 unknowns for the structure are 3magnf tudes ,

'_ '\{4x5(

A2.6 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

Physically, the structure has t~o ~oremembers than is necessary for the stability ofthe structure under load, as we cou:d leaveout one diagc~l member in each tr~ss panel and

Fig. A2. 12Fig. A2. 11

Suppose we were able to find the stressesin CA, DA, CB, DB in some ~nr.er, and we wouldnow ?Toceed to joint D and treat it as a treebody or cut through the upper panel alongsection 4-4 and use the lower portien as a freebody. The same reasoning as used above wouldshow us we have one mor-e unknown than the number-of equilibrium equations available and thuswe have the truss statically lncete~1~t8 tothe second deg~ee relative to internal memberstresses.

~e now investigate to see ~~ we can ~indthe internal ~ember stresses after :iaving :ow~dthe values of the reactions at A and 8. Supposewe cut out joint B as indicated 8Y section 1-1in Fig. 2.10 and draw a free body as sho'#TI inFig. 2.11. Since the ~embers of the truss ~avepins at each end, the loads in theSe ~enbers~ust be axial, thus direc~ion and 11~e of actionis known and only magnitUde is '~~~~own. Inrig. 2.11 HE and ~~ are known but AS, CE, ~~dDB are ~~own in magnitUde hence we have 3 un-knowns but only 2 equations of equilibrium fora coplanar concurrent force syste~. If we cutthrough the truss in Fig. 2.10 by the section2-2 and draw a free body of the lcwer'portionas shown in Fig. 2.12, we ~ave 4 unknowns,namely, the axial loads in CA, DA, C8, DB butonly 3 equations of equilibrl~ available fora coplanar ferce sys~em.

Fig. 2.10 shows a 2 ~a7 ~~~ss supported atpoints A and B and carrying a known load syst&~P, Q. All members of the truss are conr.ectecat their ends oy a common ~in at each jCi~t.The reactions at A and a are applied trIoughfittings as indicated. The question is wt-etherthe structure is statically dete~inate.Relative to external react~or.s at A and B thestructure is statically det8~inate ~eca~se thetype of support produces only one unkncwn at Aand two unknowns at B, name Iy , VA, 'IS and HS asshown in Fig. 2.10 and we have 3 equations ofstatic equilibrium available.

Example Problem 5

Example Problem 5

~ultiple bolt connection. This ~eans tha: allthree force or ~eaction charact9ristics, na~ely,magnitude, direction and point of a?pllcationare unknown, or in o~her words, 3 ~ownsexist at C. For convenience we will representthese unknowns by three components as sho~m inFig. 2.6, namely, He, Vc and Me. At jo Int D inFig. 2.6, the only unknown regarding the re-action is RD a magnitude, since the pin at eachend of the nember DE establishes the directionand point of applIcation at the reaction RD.Hence we have 4 unknowns and only 3 equationsof equilibrium for the structure in Fig. 2.6,thus the structure is statically indeterninatewith respect to all of the internal stresses.The student should observe that internalstresses between paints AC, 80 and FE arestatically determinate, and thus the staticallyindeterminate portion is the structural tri-angle CEDC.

Figs. 2.7, 2.8 and 2.9 show the samestructure carrying the same known load systemP but with different support conditions atpoints A and B. The question is whether eachstructure is statically indeterminate and ifso, to what degree, that is, what number ofunknowns beyond the equations of statics avail-able. Since we have a coplar~r force system,only 3 equations at statics are available forequilibrium of the structure as a whole.

In the structure in Fig. 2.7, the ~eactionat A and also at B is unknown in ~gnitude anddirection but point of application is ~~own,hence 4 urJcnowns and with only 3 equations ofstatics available, makes the structurestatically indeterminate to the first degree.In Fig. 2.8, the reaction at A is a rigid one,thus all 3 characteristics of magnitude,direction and point of application of the re-action are unknown. At point B, due to pinonly 2 unknowns, na~ely, ~gnitude and di-rection, thus making a total of 5 unknownswith only 3 equations of statics available orthe structure is statically indeterminate tothe second degree. In the structure of Fig.2.9, both supports at A and B are rigid thusall 3 force characteristics are unknown at eachsupport or a total of 6 unknowns which makesthe structure statically indete~inate to thethird degree.

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.7

the strJcture Nould be still stable and all~ember axial stresses could be found by theequations of static equilibrium without regardto their size of cross-section or the kind atmaterial. Adding the second diagonal memberin each panel would necessitate knOWing thesize of all truss members and the kind ofmaterial used before member stresSes could befound, as the additional equations needed mustcome from a consideration involving distortionof the truss. Assume for exa~ple, that onediagonal in the upper panel was lett out. Wewould then be able to find the stresses in the~embers of the_upper ?anel by statics but thelower panel Would still be statically inde-terminate to 1 degree because or the doublediagonal system and thus one additional equationis necessary and would involve a considerationof truss distortion. (The solution ot static-ally indeterminate trusses is covered inChapter A.B.)

A2.7 Example Problem Solutions of Statically DeterminateCoplanar Structures and Coplanar Loadings.

~ine the axial loads in the members and the re-actions on the spar.

Solution: The first thing to decide is whetherthe structure is statically determinate. Fromthe figure it is observed that the wing spar issupported by five struts. Due to the pins ateach end of all struts, we r~ve five unknowns,~~elYJ the magnitude of the load in each strut.DirectIon and location at each strut load isknown because of the pin at each end at thestruts. We have 3 equations of equilibrium forthe wing spar as a single unit support'ed by the5 struts, thus two ~ore equations are necessaryit the 5 unknown strut loads are to be found.It is noticed that the wing spar includes 2 in-ternal single pin connections at points a and 0'.This establishes the fact that the moment of allforces located to one side of the pin must beequal to zero since the single pin fitting can-not resist a moment. Thus we obtain two addi-tional equations because of the ~NO internal pinfittings and thus we have 5 equations to :ind 5unknowns.

~a~ple Problem 8.

Fig. A2.14 shows a much SI~plitied wingstructure, cor~lsting of a wing spar supported'oy lift and cabane struts wmch tie the wingspar to the fuselage structure. The distributedair load on the wing spar is unsymmetrical aboutthe center line of the air!r~e. The wing sparis ~ade in t~ree units, readily disassembled byusing ern fittings at points 0 and 01 Allsu~porting wing struts have Single pin fitting~l~ts a: eac~ end. The problem is to deter-

Although a student has taken a course insta:ics ~efore taking a beginning course inaircraft structures, it is felt that a limited

~eview of )roblems involving the applicationof the equations of static equilibrium is quite

j~stlfied, part~cularly 1: the prOblems areposSibly somewhat more difficult than ~ost or~he problems in the usual begir~ing course instat~cs. Since one ~ust use the equations orstatic equilibrium as ?art ot the necessaryequations in solving statically indeterminate

st~uctures and since statically indeterminatestructures are covered in rather complete detailLn other c napters at :::115 book, only limiteds?ace will be given to ~roblems involVingstatics in this chapter.

Fig. A2.15

1013",(30+ 15)45I 20~ 2

The sense of a force Is representedgraphically by an arrow head on ~he end of avector. The correct sense is obtained from thesolution ot the equations of equilibrium since,a force or ~oment must be given a plus or minusSign in writing the equa~ions. Since the senseof a force or moment is unknown, 1t is assumed,and if the algebraic solution of the equilibriumequations gives a ~lus value to the magnitudethen the true sense is as ass~ed, and oppOSiteto that assumed it the solution gives a minussign. If the unknown forces are axial loads in~embers it is camoan practice to call tensilestress plus and compressive stress minus, thusif we ass~.e the sense of an unknown axial ~oad~s tenSion. the solution of the equilibr!um

Fig. 2.15 shows a tree body of the wingspar to the right of hinge fitting at O.

In order to take moments, the distributedload on the spar bas been replaced by the re-sultant load on each spar portion, namely, thetotal load on the portion acting through thecentroid of the distributed load system. Thestrut .r-eact.t on EA at A tas been shown in phantomas it is ~ore convenient to deal with its com-ponents YA and XA' The reaction at 0 15 un-known tn ~gnltude and direction and for con-venience we will deal with its components xoand YO. The sense assumed is indicated on thefigure.

l5#/in.! ,--,

A PIn

t i f

Fig. A2.l4

401ll/in.

A 8' B 0"Lilt Caban , -+-Struts truts C~ c 36"

Fuselage -' "30":.......i..-E' E

'i 8ym. 'i

" ' . ~Oi/in...I...,....,--r;i t j " Ill'-r i-45"~82" ---iZO'!_60" ---! Hinge

o' , ';0

201ll/in.

A2.8 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

equations will give a plus value for the magni-tude of the unknown if the true stress istension and a ~inus sign will indicate theassumed tension stresses should be reversed orcompression, thus giving a consistency of signs.

To find the unknown YA we take ~cmentsabout point 0 and equate to zero for equilibrium

To find strut load g'C I ~ake m~ents aboutpoint C.

L~ ~ 1325 x 65 + 2000 x 40 + (58eO - 44JO)30 - 1500 x 10 - 993 x 35 - 30 (3'8')30/33.6 =0

whence, BtCI = 6000 lb. with sense as sho~TI.

- 2460 x 41 - 1013 x 102 82YA = 0

Hence YA = 204000/82 = 2480 lb.. The plus signmeans that the sense as assumed in the figure1s correct. By geometry XA = 2480 x 117/66 =4400 lb. and the load in strut ~ equalsv4400 3 + 2480 3 = 5050 lb. tension or asassumed in the :lgure.

To find load in member BIC use equationZFy '::: 0 =. 1325 + 2000 + 1500 + 993 - 6000(30/33.6) - 2720 (30/33.6) - B'e (30/54)= 0

whence, BIC =- 3535 lb. The ~inus signmeans it acts opposite to that sho'Nn in fi~~eor is compression instead of tension.

To find Xo we use the equilibrium equationZFX =0 = Xc - 4400 = 0, whence Xc = 4400 lb.

To find YO we use,ZFy = 0 = 2460 + 1013 - 2480 - YO =0, whenceYO = 993 lb.

To check our results for eqUilibrium wewill take moments at all torces about A to seeit they equal zero.l:11A = 2460 x 41 - 1013 x 20 - 993 x 82 = 0 check

The reactions on the spar can now bedeter.nined and shears, ber.ding moments andaxial loads on the spar could be round. Thenumerical results should ~e checked for eqUili-brium or the spar as a whole by taking momentsot all forces about a dif:erent ~oment centerto see it the result "is zero.

Example Problem 9.

12"

-r-r-'

I

+12"

-LShockI Strut

A

r300

F--

strut i ~ 12 12"i\~ I ~ Brace Strut I24lJ~ \, E' ---+--

i12 ~ I

Fig.A2.17

Hence AIEl. 6750, XOI = 5880, YO' = 1325

Fig. 2.16 shows a tree body ot the centerspar portion with'the reactions at 0 and at asfound preViously. The unknown loads in thestruts have been assumed tens10n as shown bythe arrows.

1500",50x30

On the spar portion OIA' J the reactionsare obViOUSly equal to 40/30 times those :oundfor portion OA since the external loading is 40as compared to 30.

Whence, Be 2720 lb. with sense asassumed.

To find the load in strut Be take momentsabout s'

!MB' = 1325 x 20 - 2000 x 5 - 1500 x 55- 993 x 80 60 (BC) 30/33.6 = a

Fig. 2.17 shows a Simplified airplanelanding gear unit with all members and loadsconfined to one plane. The brace struts ~epinned at each end and the support at C is ofthe roller type, thus no vertical reaction canbe produced by the support fitting at point C.The member at C can rotate on the roller buthorizontal movement is ~revented. A knovm loadot 10,000 lb. 1s applied to axle ~nit at A. Theproblem is to f1nd the load in the brace strutsand the reaction at C.

Solution:Due to the Single pin fitting ~t each end

of the brace struts, the ~eact1ons ~t a and D

Flg.A2.16

"'':In," 3015

, "

C"9----~II IL __ J

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.9

are collnear with the strut axis, thus directionand point of application are known for reactionRB and RD leaving only the magnitude of each asunknown. The roller type fitting at C fixesthe direction and point of application of thereaction Re, leaving magnitude as the onlyunknown, Thus there are 3 unknowns Re, He andRO and with 3 equations of static equilibriumavailable, the structure is statically deter.ni-nate with respect to external reactions. Thesense of each of the 3 unknown reactions hasbeen assumed as indicated by the vector.

To find Rn take moments about point B:-ZME = - 10000 sin 30 x 36 - 10000 cos 300 x 12- RO (12/17) 2~ = 0

whence, RO ~ - 16750 lb. Since the resultcomes out with a minus Sign, the reaction ROhas a sense OPPosite to that shown by thevector in Fig. 2.17. Since the reaction RO iscolinear with the line DE because of the pinendS, the load in the brace strut DE is 16750lb. compression. In the above moment equationabout B, the reaction RO was resolved intovertical and rorizontal components at point D,and thus only the vertical component whichequals (12/17) RD enters into the equationsince the horizontal component has a line ofaction through point B and therefore no moment.He does not enter in equation as it has zerOmoment about B.

To find RE take ZFv = 0ZFv =10000 x cos 30 + (- 16750)(12/17) + HE(24,/26.8) = 0

Whence, HB = 3540 lb. Since Sign comesout plus, the sense is the same as assumed inthe figure. The strut load BF is therefore3540 lb. tension, since reaction HB is collnearwi th line SF.

To find He take ZH : 0ZH = 10000 sin 300 - 3540 (12/26.8) + (- 16750)(12/17) + He =0

whence, Rc = 8407 lb. Result is plus andtherefore assumed sense was correct.

the aluminum alloy tubular truss. Trussed typebeams composed of closed and open type sectionsare also frequently used in Wing beam construc-t1on. The stresses or loads in the members ota truss are commonly referred to as nprimaryWand wsecondarJn stresses. The stresses whichare found under the following assumptions arereferred to as primary stresses.

(1) The members of the truss are straight,weightless and lie in one plane.

(2) The members of a truss meeting at a'point are considered as jOined together by acommon frictionless pin and all member axes in-tersect at the pin center.

(3) All external loads are applied to thetruss only at the jOints and in the plane otthe truss. Thus all loads or stresses producedin members are either axial tension or compres-sion without bending or torsion.

Those trusses produced in the truss mem-bers due to the non-fulfillment or the aboveassumptiOns are referred to as secondarystresses. Most steel tubular trusses are weldedtogether at their ends and in other truss types,the ~embers are riveted or bolted together.This restraint at the joints may cause second-ary stresses in some members greater than theprimary stresses. Likewise it is common inactual practical deSign to apply torces to thetruss members between their ends by supportingmany equipment installations on these trussmembers. However, regardless of the magnitudeat these so-called secondary loadS, it iscammon practice to first find the priwarystresses under the assumption outlined above.

GENERAL CRITERIA FOR DETERl'lINING WHEI'JlERTRUSS STRUCTURES ARE STATICALLY DEI'ERMINATE

WITlf RESPECT TO INl'ERNAL STRFBSES.

The simplest truss that can be constructedis the triangle which has three members m andthree joints j. A more elaborate truss consistsof additional triangular frames, so arrangedthat each triangle adds one joint and two mem-bers. Hence the number of members to insurestability under any loading is:

A truss haVing fewer members tillL~ requiredby Eq. (2.8) is in a state at unstable eqUili-brium and will collapse except under certainconditions of loading. The loads in the membersof a truss With the number of members shown inequation (2.8) can be :ound With the availableequations of statiCS, since the forces in theme~bers acting at a paint intersect at a commonpoint or form a concurrent force system. ForthiS type of force system there are two staticequl11brium equations available.

Thus for j number of joints there are 2j

To check the numerical results take~aments about point A for equilibrium.ZMA = 84,07 x 36 + 354,0 (24,/26.8) 12 - 3540(12/26.8) 36 + 16750 (12/17) 12 - 16750 (12/17)36 = 303000 + 38100 - 57100 + 142000 - 426000:. 0 (Check)A2. B Stresses in Coplanar Truss Structures Under

Coplanar Loading.

In aircraft construction, the t~~s typeof construction 1s quite co~on. The ~cstcommon is the tUbular steel welded tr~sses :h~tmake up the fuselage frame, and less freq~~rr:ly,

m = 2j - 3 -----(2.8)

I ./' ~, .

A2.10 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

500

ZV = -500 - 1250 (40/50) + UaLl (40/50) = 0whence, UaLl = 1875 is. (tens 1on as

asaumed ,.)

Fig. A2. 21

i 2lL"

2,

(1, -L~~ ~_:)oo

\ - ,u,~

L,

500

~ ~HL. , 250u.,

tu. Fig. A2. 20

L,

T\

40

""{'--~---T-'-::,I('-, 30- U U \ U 31 Fig. A2. 18

c 1 3 2f--- 30" -+- 30" -+- 30" -----.jiii

',1000

Fig. A2. 19

40"

-I

-l

Next consider joint Ua as a :ree jody c~tout by section 3-3 in Fig. A2.18 and erawn asFig. A2.2l. The known member s~resses are showr.with their true sense as ~reviouslY found. Thetwo ~own member stresses U2L l and UaU l haveJeen assumed as tension.

For equilibrium of jOint La, ZH and ZV =0ZV =- 500 + LaUa = 0, whence J LaUa = 500 lb.Since the Sign came out plus, the assumed sensein Fig. A2.20 was correct or compreSSion.ZH = 250 - LaLl : 0, wr.ence LaLl =250 IJ.

or tension. In equation (0) the load of 1250in L~L:;l was substituted as a minus value sinceit was foune to act opposite to ~hat sho~m inFig. A2.l9. Possibly a tetter p r-ccedur e wouldbe to change the sense of the ar~ow i~ ~he :reebodY diagram for any solved ~embers Jefore writ-ing further equilibriUW equati~ns. ~e ~ustproceed to joint L:::I instead c r joint U2, asthree ~~~own nembers still exist at jOint U2Whereas only ~NO at jo~nt La. Fig. A2.20 shows

f~ee body of jOi~t La cut ou~ by section 2-2(see ~lg. AS.IS). The sense of the urJL~ownmember s~ress L:;lU:;l has Jeen assw~ec as COID-

pressic~ (pUShing toward jOint) as ~t is ob-Viously act~ng this way to Jalance :he 500 lb.load.

equations available. However three i~dependentequa~ions a~e neceSsa~J to determine the ex~ernal reactions, thus the number of equationsnecessary to solve :or all the loads in themembers is 2j - 3. Hence if the nw~Jer of truss

~embers is that given by equation (2.8) thetruss is statically deter.ninate relative to theprimary loads in the truss ~embers and thetruss is also stable.

whence, L ~U:;l =- 1250 1'0. Since the Signcame out minus the stress is opPOSite to thataSStooed in Fig. A2.l9 or compression.Eli =- 500 - (- 1250)(30/50) - L,L, =0 - -(b)

Whence, LaL~ = 250 lb. Since Sign comesout plus, sense is s~e as assumed in figure

In general there are three rather distinctmethods or procedures in applying the equationsof static equilibrium to finding the primarystresses in truss type structures. They areotten referred to as the method of joints,moments, and shears.

ANALYTICAL IlE:rHODS fOR DE:l'SRtrINI'IGPRIMARY STRESSES IN TRUSS STRUCT1JRES

A2. 9 Method of Joints.

It the truss as a whole is i~ equilibriumthen each member or joint in the tr~ss mustlikewise be in equilibrium. The forces in themembers at a truss jOint intersect in a commonpoint, thus the forces on each joint form aconcurrent-coplanar force system. The ~ethodof joints consists in cutting out or isolatinga joint as a free bOdy and applying the laws otequilibrium for a Concurrent force system.Since only two independent equations are avail-able for this type of system only two unknownscan exist at any jOint. ThUS the procedure isto start at the jOint Where only two unknownsexist and continue ,rogressively throughout thetruss joint by joint. To illustrate the methodconsider the cantilever truss of Fig. A2.18.From obse~lation there are only two membersWith internal stresses unknown at jOint L~.Fig. A2.19 shows a free body of jOint L~. Thestresses in the members L~ La and L~ U~ havebeen assumed as tension, as indicated by thearrows pulling away from the jOint L~.

The static equations or equilibrium forthe forces acting on joint L~ are ~~ and ZV =o.!V = - 1000 _ L,U. (40/50) = a - - - - - - -fe )

It the truss has more members than indi-cated by equation (2.8) the trusS is consideredredund~~t and statically lnGeterminate sincethe member loads cannot be found in all themembers by the laws of statics. Such redundar.tstructures it the members are properly ?lacedare stable and will support loads of anyarrangement.

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.11

ZH = (-1250) (30/50) - 1875 (30/50) - U.U. = 0whence, U..U a = - 1875 lb. or opposite in

sense to tr2t assum~c ana therefore compression.

The algebraic sign of all unknowns came outpositive, thus the assumed direction as shownan Fig. A2.22 was correct.

Note: The student should continue with succeed-ing joints. In this example involving a canti-lever truss it was not necessary to ~lnd thereactions, as it was ~oSSlble to select jointL 3 as a joint involving only VNO unknowns. Intrusses such as illustrated in Fig. A2.22 it 1snecessary to first find reactions R:l. or R a wnicnthen provides a joint at the reaction point in-volving only two ~cwn forces.

Check results by taking &~B = 0~B = 1400 x 150 + 500 x 30 - 500 x 120 - 500 x

30 - 1000 x 90 - 1000 x 60 = 0 (Check)To determine the stress in member Fl , Fa and F"we cut the section 1-1 thru the truss (Fig.A2.22. Fig. A2.23 shows a free bOdy diagram ofthe portion of the truss to the left of thissection.

f, 'f~ _

500 ~;~:1400 1

FIg. A2. 24

o ,

500 1

a

Fig. A2. 23

1400

The truss as a whole was in equl1ibriumtherefore any portion must be in eqUilibrium.In Fig. A2.23 the internal stresses in the mem-bers F l , F~ and F~ which existed in the truss asa whole ~ow are considered external forces inholding the portion of the truss to the left ofsectien 1-1 in e~Jil1brium in combination withthe other loads and reac~ions. Since the ~embers a and b in Fig. A2.23 have not been cut theloads in these ~embers remain as internalstresses and have no i~fluence on the equilib-rium of the portion ot the truss shown. Thusthe ~ortion at the truss to lett ot section 1-1could be considered as a solid block as shown

i~ Fig. A2.24 without affecting the values atFl , F. and F". The ~ethod of ~oments as thename unpIles involves the operation of takingmoments about a point to find the load in aparticular member. Since there are three un-knowns a moment center must be selected suchthat the moment of each of the two unknownstresses will have zero moment about the selectedmoment center, thus leaving only one illL~Own

~orce or stress to enter into the equation tormoments. For exa~ple to cetermine load F 3 i~Fig. A2.24 we take moments about the inter-section of forces F l and Fa or paint O.

10"

Fig.A2.22

ioo""3 2 1100M'F" 500#

50~f

Fig.A2.22

HA----"'(----+-\-"''-+-''--.''-'-'-----''!---''>I,

A2. 10 Method of Moments.

For a coplanar-non-concurrent force systemthere are three equations of statics available.These three equations may 08 taken as momentequations about three different ~oints. Fig.AZ.22 shows a typical truss. Let it be re-qUired to find the loads in the members F l , Fa JF", F. J r , and F o '

The first step in the solution is to find thereactions at pOints A and B. Due to the rollertype of support at B the only unknown element ofthe reaction force at B 1s ~gnitude. At paintA, magnitude and direction of the reaction are

l~own giving a total of three ~~cwns Withthree equations of statics available. For con-venience the wllic~own reaction at A has been re-

~laced by its unknown H ~~d V components.

ZH = 500 - nA = 0, therefore HA = 500 lb.

Takir~ momer..ts about ~oint A,L~A = 500 x 30 + 100e x 60 + 1000 x 90 T 500 x

30 + 500 x 120 - 150 VB = aHer-ce Vg = 1600 lb.Take ZV = 0

z.v := '/. --,

foreTake

1000-1000-500-500VA. = 1400 lb.ZH = 0

+ 1600 = a there-

Thus Z~O =1400 x 30 - 18.97 F, =0II ... = 42000 = ")21::: ,,, (il.ence "3 18.97 ~ '-' ~..;. compr-eas t cn or

acting as assumed)To ~ind the a~ of the force F3 tram :hemomen~ center a involves a s~~ll a~o~~t of cal-Clllat!on. thus in general i~ 13 Simpler to re-solve the ~~nown force into H and V componentsat a ~oi~t In ~ts 11ne Ji action such ~hat oneof these ~omDonents passes thru the ~omentcenter and the arm of the other cJmponent can~sually be dete~1~ed Jy inspecticn. Thus in

A2.12 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

Fig. A2.25 the ferce F3 is resolved into itscomponent F 3V and F3H at point 0'. 7hen taking

/IF,500 ...."::,,,-----7"- F ,

1000lii

!

Fig. A2. 271400lii

the other two ~Jcnowns F~ and ~s lles at i~flnity. Thus for conditions where two of the 3 c~tmembers are parallel Ne have a ~ethod of solving

~or the web ~ember of the t~uss ccmr.,only re-ferred to as the ~ethod cf shea~s, or the sum-mation of all the forces nO~ial to the tNOparallel ~{nown chord members ~us~ equal zero.Since the parallel chord ~embers have no com-ponent in a direction no~al to their line ofaction, they do not enter the above equatton ofequilibrium.

50'"'

Fig. A2. 2S

Fig. A2. 26

moments about point 0 as before:-500 3

11400Fig. A2. 28

whence, FeH ~ 2100 lb. and therefore

F. =2100 (21.6/30) =2215 lb. as pre-Viously obtained.

The load Flo. can be found by taking momentsabout point Ill, the intersection at torces F.and F. (See Fig. A2.23).

ZMm = 1400 x 60 - 500 x 30 - 500 x 30-30Flo.=Q

whence, Flo. = 2800 lb. (Tension as assumed)To tinct rcrce F. by using a moment equation,

we take moments about point (r) the inter-section ot rorces F J, and F 3 (See Fig. A2.26).To eliminate solving for the perpendiculardistance trom paint (r) to line at action ofF., we resolve Fa into its n and V componentsat point 0 on its line at action as shown inFig. A2.26.

ZMr = - 1400 x 30 + 500 x 60 + 60 F.V = 0

whence, FaV = 12000/60 = 200 lb.

Therefore F. = 200 x ~ =282 lb. com-pression

A2. 11 Method of Shears

In Fig. A2.22 to tind the stress in memberF. we cut the section 2-2 giving the free bOdyfor the left portion as shown in Fig. A2.Z7.

The mebhod ot moments is not sufficient tosolve tor member F. because the intersection ot

Referring to Fig. A2.27

ZV =1400 - 500 - 1000 - '.I1/v'2): awhence F. = - 141 lb. (tension or OPPOSite

to that assumed in the ~igure.

To find the stress in illember F? , we cutsection 3-3 in Fig. A2.22 and draw ~ free bOdydiagram of the left portion in Fig. AZ.28.Since Flo. and Fs are horizontal, the member F?must carry ~he shear on the truss on this sectior.3-3, hence the name ~ethod of shears.

ZV =1400 - 500 - 1000 F, =0

l~ence F? = 100 lb. (compression as assumed)Note: The student should solve this example il-lustrating the methods of moments and shearsusing as a tree body the portion of the truss tothe right of the cut sections instead of thelett portion as used in these illustrative ex-amples. In order to solve for the stresses inthe members of a truss most advantageously, oneusually makes use of more tha.~ one cf the abovethree methods, as each has its advantages forcertain cases or members. It 1s important torealize that each is a method of sections and ina great many cases, such as trusses with paral-lel chords, the stresses can ?Tactically befound mentally without writing down equations ofequilibrium. The following statements in gen-eral are true for parallel chord trusses:

(1) The vertical camponent of the stress ~nthe panel diagonal members equals the verticalshear (algebraic sum of external forces to ones1de of the panel) on the panel, since the chord

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.13

members are horizontal and thus have zero verti-cal component.

(2) The truss verticals in general resistthe vertical component of the diagonals plusany exter~l loads applied to the end joints otthe vertical.

(3) The load in the chord members is dueto the horizontal components ot the diagonalmembers ana in general equals the summation ofthese horizontal components.

To illustrate the simplicity at determiningstresses in tITe members of a parallel chordtruss, consider the cantilever truss of Fig.A2.29 with supporting reactions at points A andJ.

since no external vertical load exists at jointE. Similarly, by the same reasoning for LH ~ 0,load in DE = O. The load in the diagonal FOequals the value on the diagonal of the panelindex triangle or 167 lb. It is tenSion byobservation since the shear in the panel to theright is up and the vertical component of thediagonal FD must pull down for eqUilibrium.

ConSidering Joint F. ZH = - FG - FOR = 0,which means that the horizontal component of theload in the diagonal OF equals the load in FG,or is equal to the value of the horizontal sidein the index triangle or - 133 lb. It is nega-tive because the horizontal component of DFpUlls on Joint F and therefore Fa must pUShagainst the joint for equilibrium.

Considering Joint D:-

550

150

36"-+- 40"

ZV = DFv + DO = O. But DFv = 100 (vertical sideof index triangle) ", DG = - 100LH =DE + DFR - DC =0, but DE =and DFa =133 (from. index triangle)

DC = 133

Considering Joint G:-Fig. A2.29

~lrst, compute the length triangles ineach panel of the truss as shown by the dashedtriangles in each panel. The other trianglesin each ~anel are referred to as load ar indextriangles and their sides are directly pro-portional to the length triangles.

The shear load in each panel is first writ-ten on the vertical side at each index triangle.~hus, in panel EFGD, considering forces to theright of a vertical section cut thru the panel,the shear is 100 lb., which is recorded on thevertical side of the index triangle.

For the second panel from the tree end, theshea~ is 100 + 150 ~ 250 and for the third panel100 + 150 + 150 = 400 Ib , , and in like manner550 ~or fourth panel.

The loads in the diagonals as well as theirhorizontal components are directly proportionalto the lengths of the diagonal and horizontalside at the length triangles. 'Thus the load indiagonal member DF = 100 (50/30) = 167 and ~ormember CO = 250 (46.8/30) = 390. The hori-zontal compor.ent of the load in OF = 100 (40/30)=133 and :or CG = 250 (36/30) =300. Thesevalues are shown on the index triangles toreach truss panel as shown in Fig. A2.29. Westart our analysiS for the loads in the ~embers8f the tr~ss by considering joint E tirst.

USing LV = 0 gives EF = 0 by obser~ffition.

ZH=-GH - GF - GCR = O. But OF =- 133, and GCR= 300 from index triangle in the second panel.Hence OH =- 433 lb. Proceeding in this manner,we obtain the stress in all the members as shownin Fig. AZ.29. All the eqUilibrium equationscan be solved mentally and with the calculationsbeing done on the slide rule, all member loadscan be written directly on the truss diagram.

Observation at the results at Fig. AZ.29show tr~t the loads in the truss verticals equalthe values of the vertical sides or the indexload triangle, and the loads in the truss di-agonals equal the values of the index trianglediagonal side and in general the loads in thetop and bottom horizontal trJSs members equalthe summation of the values ot the horizontalsides of the index triangles.

The reactions at A and J are found whenthe above general procedure reaches joints Aand J. As a c~eck on the work the reactionsshould be determined treating the truss as awhole.

Fig. A2.3C shows the solution for thestresses in the members of a Simply supportedPratt Tr~ss, symmetrically loaded. Since allpanels have the same width and height, only onelength triangle is drawn as shown. Due tosymmetry, the index triangles are drawn forpanels to only one side at the truss centerline. First, the vertical shear in each panelis written on the vertical side of each indextriangle. Due to the symmetry of the truss and

~ ,/.,/;c

A2.14 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

Fig. A2. 32 Champion Traveler

Fig.A2.31 Piper Tri-Pacer

The reac:icn R~ equals t~e val~e on thevertical side of our i~cex triang:e in the 91.jpanel, or 375. This should oe c~ecked usi~gthe truss as a whole and ta~lng ~c~ents ~joutR,

A2.12 Aircraft Wing Structure. Truss Type with Fabricor Plastic Cover

TI:e metal covered carrt t Lever' wing 'N1t;'1 i tsbetter overall aerodyna~lc e~:iciency and s~f

ficien~ torsion~l ri 6! : l t y has ~ractlcal1y re-placed t~e externally braced wing excep~ for lewspeed cornmerc~al or ?rivate )i:ot aircraft asillustrated by the aircra:t 1~ tigs. A2.31 a~d32. The wing covering is usually fabric andtherefore a drag tr~ss inside the wi~g isnecessarJ to resist loads in the drag t7USsdirection. Figs. A2.33 and 34 shows the gen-eral structural layout a: such wings. The twospars or beams are ~etal or wood. Instead ofUSing double wires in each d~g truss bay, aSingle diagonal strut capable ot t~ 0;.= 1'.9 .s '1';~~

312 ~L~ 499.5 p..3100 100 1006 Panels @ 25" = 150"

100312

Length~5" Triangle30"33" 50 50-499.5 U

center panel = (100 + 50) 1/2 = 75. The verti-cal shear in panel U:I,UgL1L a equals 75 pl~s theexternal loads at Ua and La or a total of 225and Similarly for the end panel shear = 225 +50 + 100 = 375. With these values kTIQwn, theother two sides of the index triangles are di-rectly proportio~41 to the sides of the lengthtriangles tor each panel, and the results are as

sho~m in Fig. AZ.30.

The loads in the diagonals are equal to theva.Iues on the hypotenuse 0 f the index triangles.The sense, whether tension or campression, isdeter.nlned by inspection by cutting ~entalsections thru the truss and noting the directionot the external shear load which ~st be bal-anced by the vertical camponent of the diagon-als.

The general procedure fram this paint is tofind the loads in the diagor21s, then in theverticalS, a~d finally in the horizontal chordmember's

Thus for jOint U3 , ZV = - 50 - U3L 3 = aor U3L3 = - 50.

For jOint U~, ZV = - 50 - UaL 3 v - U~L~ = 0,but U~13V =75, the vertical cornpo~ent of Ua L3rr-cm index t r tangLe. . ", U~L~ =- 50 - 75 =- 125. For JOint L~, ZV = - 100 + L:I,U1 = 0,hence L1U1 = 100.

Since the horizontal chort me~bers receivetheir loads at :~e joints due :0 horizontalcomponents of ~he diagonal members of the tr~ss'lwe can start a: La and add u9 these horizontalcomponents to obtain the chord stresses. Th~S, ILoLl = 312 (from index triangle). L1L 2 = 312 Ifrom ZH = a for joint L1 At joint U"I.' the i

The loads in the verticals are deter.ninedby the method of joints and the sequence ofjOints is so selected that the stress in thevertical member is the only unknown in theequation ZV = 0 for the joint in question.

loading, we know that or-e t~l~ of the exter~alloads at jOints U3 and L3 is supported a~ re-action R:I, and 1/2 at r-eact.t cn Ra , or shear- in

i!R,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2.l5

~a~~le Problem 10. ~~e~nally Braced Mono-plane ';ing Structure

Drag Truss

Fig. A2. 36

RearBeam

Drag

FrontBeam

,

Center ofPressure

/iL ift~--,...-,-Airstream-

litt and drag forces on the drag truss direction,the fonvard projection ~ue to the lift 1s great-er than the rearNard projection due to the airdrag, which dl~ference ~n our exa~ple problemhas been assumed as 6 Ib/in. In a low angle ofattack the load in the drag truss directionwould act rearward.

(1) A constant spanwise lift load of 45Ib/in from hin~e to strut point and then taper-ing to 22.5 Ib/tn at the wing tip.

ASSL~D AIR LCADING:-

(2) A [erHard uniform distributed dragload of 6 lb/in.

The above airloads represent a high angleof attack condition. In this condition a for-ward load can be placed on t~e drag truss asillustrated in Fig. A2.36. Projecting the air

in all ~enbers of the lift anc drag tr~sses willbe dete~,ine~. A Si~plified air loac:ng Nill beass~~ed, as the ?ur~ose of this problem is to;iv8 the st~dent practice in solving sta:icallydeterminate space t~uss str~ctur8S.

Anti-Drag WireAileron RibAileron HingeDrag Strut orCompression Tube

Forming or Plain RibTrailing Edge

Drag wtre

Butt RibBeam or Spar

Fig. A2. 34

Wing Tip (End) Bow11e /;~ (0. Plywood Tip Fairing

AileronAileron SparCompression Rib

, ,

,~

, t: ~\,

X ~'III / \.,r:=r

--l........;'1. "----.ll

"! -/:1:~' \

:1 V 11'1 ,\,i. : / I: ,

'1\ " I ,': I ~

" / i :-+rII ~: A I Ir--I,

"

"I

"/':~

LeadingEdge-

Wing HingeFitting

Fig. A2. 33

Plywood LeadingEdge Fairing

D"gwire~Fitting \ ::

~ig. A2.35 shows the structural d~mensionalj~agram of an externally braced nonoplane wing.The wing is fabric covered between wing beams,ace thus a drag truss composed of struts andtie rods is necessary to prov~de strength and

r~gid1ty in the drag direction. ~he axial loadsDRAG TRUSS

SOLUTION:

The ~~ing loads on the front and rearbea~s will Je calculated as the first step inthe solution. ?or our flight condition, thecenter of pressure or t~e air~orces ~ill beassumed as shown in Fig. A2.37.

Wlng Chor-d e 72"

N '

, It ""Airplane I?. -i-(5lfi.6,..--

Fuselage Fig. A2.35

=:i"':-'";;1 3

--,---

3 Dimedral

(3)II

[SRI1

-!(6)--l

1------- 72" !IlO~'+- 36" ---1 ICre . p~F.Bl R.B.

1.1 24.21" I22.52" Fig. A2. 37

T~e running load on the front bea~ will Je 45 x24.2/36 =30.25 Ib/:n., a~d the re~alnder or45 - 30.25 = 14.74 Ib/in gives the load on the::-ear beam.

I 1..- ' ,

A2 16 EQUILIBRIUM OF FORCE SYSTEMS TRUSS STRUCTURES

To solve ~or loads in a t~uss system by amethod of jOints, all loads ~ust be transferredto the truss joints. The wing bear,s are sup-

~orted at one end by the f~sela~e and o~tooardby the two 11ft st~Jts. Thus we calc~late thereactions on each beam at the strJt a~d ~lngepaints due to the ~~~l~g lift load on eachbeam.

Front Seam

4 R. B. 3'---_71'I" /1' /!\ /1" /1A x I x X 11/ \.,/ '\.k '\.1/ "'-I

:;;/.::B.,rF Fl"Qot Lift

5 Truss

Drag Truss

3

/;

R. B.

R RearLiftTruss

~ember lengthS L and :he ~ar.poner.tfellow by Simple ca:culat~on.

ra.t ; os

Table .\2.1

RJ. =3770 lb.Take ZV =0 where V direction 1s taken normal to

beamZV =- R~ - 3770 + 30.26 x 114.5 +

(30.26 + 15. 13 ) 70.5 = 02

I D drag direction,S ~ side direction,

L ~ Vv2+ D2 + 32

11 114.34 128.791.45011.08541.8878 i

t, I V;1. I D;1. I ';1.

114.501.05231 0 1.9986 I

114.501.05231 0 1.9986

o 114.34128.00i.44861 0 1.89301

o 114.34

o 114.34

D ,lIember gym.: VFront Be" FB I 5.99Rear Beall. RB , ~.99

Front Strut'F 15 7 9 9

Bear Stru.t I '. j 57.49V ... vertical direction,

pofnt (2)x 30.26 x 114.5/2 - 15.13 x15.13 x 35.25 x 138 = O.

hence

Taking moments about114.5R .. - 114.570.5 x 149.75 -

The rear beam has the same span dimensions butthe loading is 14.74 Ib/in. Hence beam re-actions R~ and R3 will be 14.74/30.25 = .4875tilnes those for front beam.

hence R~ =1295 lb.(The student should always check results bytaking moments about pOint (1) to see it ZM ..=0)

Rear Beam

(4) 1 I !R.

(W'" 14. 7#/in. ~Ilftltll;,,/2

114. 5" ija 70. 5,,--1R,

We stare the solution of joints jy sta~i~gw~th joint (1). Free bOdy sketches cf joint (I)are sketchel below. All members are consideredtwo-for-c e r.ember-s or havtng pf ns at each end,thus magnt .ude is the cn.Iy unknown character-istic of ~~ch member load. The jrag truss ~embers coming in to jofnt (l) are replaced :,y asingle reaction called Jl.' Ar t er- Dl. is r ounc ,its influence in causlr.g loads in drag truss~embers can then be found when the drag :russ asa whole is treated. In :he joint solution, thedrag truss has been assumed parallel to dragdirection which is nct quire true frc~ Fig.A2.35, but the error on member loads is negli-gible.

JOINT 1 (Equations of ~quilibri~~)

Table A2.1 gives the V, 0 and S ~rojecticnsor the lift truss ~embers as dete~ined frominformation given in Fig. A2.35. ~he true

The next step in the solution is ~hesolving tor the axial loads in all the members.We will use the ~ethod of joints and considerthe structure made up of trIee truss systemsas illustrated at the top of the next col~.namely, a front li!t truss, a rear lift trussand a drag truss. The beams are co~on to ~othlift and drag trusses.

hence Rs = .4875 x 3770 =R~ = .4875 x 1295 =

1838 lb.631 lb. ~770(1) D,B ~v

SF 4+D

zv =3770 x .9986 - .0523 FE - .4501 S" =0- ( izs = - 3770 x .0523 - .;966 F&-

.36'7'8 SF = 0- ( iZD =-

.0854 SF + 0, = 0 - - - - - - - - - - ! ,ISolVing equa t ions 1, 2 and 3, we obtain

F3 =- 8513 lb. (compression)SF = 9333 l~ . (tensior.)0, = 798 lb. (aft i

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A2 17

Fig. A2.40

Fuselage Reactions

Combined Loads on Drag Truss

254

(3770 + 1295 + 1838 + 631).9986 = 7523 lb. (error 3 lb.-185 x 6 =1110 lb. (er-r-or e 0)-(3770 + 1295 + 1838 + 631).0523 =394 lb. (error 6 lb.

Applied Air Loads:V component =

D component =S component =

As a check on the work as well as to obtainreference loads on ~lselage from Wing structure,the fuselage reactions will be checked againstthe exterr~llY applied air loads. Table AZ.2gives the calculations in table form.

Adding the two load systems of Figs. A2.38and A2.39, the total drag truss loading is ob-tained as shown in Fig. A2.40. The resultingmember axial stresses are then solved for by themethod of index stresses (Art. AZ.9). Thevalues are indicated on the truss diagram. Itis customary to ~ake one of the fittings attach-ing wing to fuselage incapable of tranSferringdrag reaction to fUselage, so that the entiredrag reaction from wing panel on fuselage 1sdefinitely ccnrtnec to one paint. In this ex-ample paint (2) has been assumed as point wheredrag is resisted. Those drag wires which wouldbe in compreSSion are assumed out of action.

~~3G---f1-36~--i39.5 37.5 58.5

118.5 231 225 281.51191 114\ I j 4189 j

~!~::o il~ ~I~ ij~8' ~-13 893 \~~~II 162il "'~J 13 1 ""I~~~

i(2) 13,893 -11,933 -10,313 8513~ -4131908 '" Drag Reaction 798

points (2) and (4). In the design of the beamand fittings at this point, the effect of ~heactual conditions of eccentricity should ofcourse be considered.

Table A,2.2

.Point lIeliber Load V D sIII

-13893 - 726 0 _138702 Dn,

- 1908 0-1908 0

Reactio..Jt2(React1on)

- 1295 129'" 0-

67RB 1191 62 0 1190

R.. (R.eaeUon) - 631 630 0 - 33s F, 9333 4205 "8 8290

, 6 .. 4579 2055 0 4090j Totals 7520

-1110-

.00

Thus the outboard panel paint concentrationof 254 lb. is Jetermined by taking moments about(3) of the drag load outboard of (3) as follows:

It was assumed that the air load componentsin the drag direction were 6 lb./in. of wingacting rorwar-d,

~/in.

P =70.5 x 6 x 35.25/58.5 =254 IJ.

To Si~pllfY the drag truss SOlution, the dra~strut and drag wires in the inboard drag trusspanel have been modif~ed to ~ntersect at hinge

The distributed load of 6 lb./in. is re-placed by concentrated loads at the panel pointsas shown in Fig. A2.39. Each panel point takesone halt the distributed lead to the adjacentpanel point, except for the two outboard panelpOints which are affected by the overhang tipportion.

Drag Truss Panel Point Loads Due to Air DragLoad.

Fig. A2.38 shows the reactions of the liftstruts on the drag truss at Joints (1) and (3)as found above.

Solving equations 4, 5 and 6, we obtainRB = - 4189 lb. (compression)SR = 4579 lb. itension)0 3 = 0

Joint (3) (Equations of equtLrbr-i ua)

t1838 (drag truss~183 8 reaction onRL.-J(3l (3) ---D, ptn (3\)~ RBR V-Splane SR V-Dplane:V =1638 x .9986 - .0523 RB - .4486 SR =0 -(4):5 =-1838 x .0523 - .9986 RB -.8930 SR =0 -(5)ZD = D, + 0 = 0 - - - - - - - - - - - - - - -(6)

,J>

A2.18 EQUILIBRIUM OF FORCE SYSTEMS. TRUSS STRUCTURES.

Lengths k Directional Components of Cabane Struts

Fig. A2. 42

Solution of force system at Joint 8

(630-;.62)",692#

'8

1650D1~C D

90'

w'" 14. 7%/in.

8'

1650

C~/L:..CR S

692#

(4) ~~~~~~~~~~~r~~(4)~20 20-1Ra",1650if R a",1650.f

Member Sym. V D I s L VIL DIL 'ILFront Cabane CF 30 10 127 1 41 . 59 1 .721 .240 .648,Diagonal Caba,ne CD 30 30 i27150.17 .597 .597 .538

Strl.\t

Rear Cabane CR 23.5 6 27 I 40.42 .731 .1485 .668Strut

L ;./ vO: + D'+ "

The vertical component of the cabane re-action at joint (8) equals one hal: the tetal

be~ load due to symmetry of 1cad1~g or 55 x14.74 + 692 = 1650 lb.

Sol~tion