CHAPTER OUTLINE 4/1 Introduction 4/2 Plane Trusses 4/3 Method of Joints 4/4 Method of Sections 4/5 Space Trusses 4/6 Frames and Machines Chapter Review
STRUCTURE
S
4
/1
INTRODUCTIONIn Cha pter3 we studied the equ ilihr ium ofa single rigid body or a syste m of connected memb ers treated as a single rigid body.We first drew a free-bodydiagram of the bodyshowingallforces externalto the isolated body and then we applied th e force and moment equatio ns of equ ilibrium. In Chapter 4 we focu s on the determination of the forces internalto ast ruct u re , thatis, forcesofaction and reactionbetweenthe connected memb ers.An engineering st ruct ure is any connect ed syste m of members built to su pport or transfer forcesand to sa fely withstand the loads applied to it. Todete rmine the forces internal to an engi nee r -ing st ru ct u re , we must dismember the st ruc ture and analyze separate free-body diagrams ofindividualmembersor combinationsof members. Thisanalysis req uir es ca re fu l application of Newton's thirdlaw, which states that each action is accompa n iedbyanequa l andopposite reaction. In Cha pte r 4 weanalyze the internal forcesacting inseveral types ofstruct u res, nam ely,trusses, frames,and machines.In thistreatment we consider on ly statically determinate st ru ct u res, which do not have mor e su pporti ngcons t rain ts than are necessary to maintain anequ ilib-riu mconfigu rat ion.Thus,as wehave already seen, the equat ions of equi-librium are ade quate to determineall unknown reactions .
The analysisof trusses,fram esand machines, and beamsunder con -cent rate d load s cons t itutes ast raightforwa rdapplicati on of themat erial developed in the pre vioustwo chapters.The basic procedu redeveloped in Chapter 3 for isolating a body by constructing a correct free-body diagr am is essent ial fortheanalysisof staticallydeter min ate st ruct u re s.
166 Chap t er 4 Struct ures L Stringer Crossbeam \
~~~
Figure 4/ 1 Pratt Warren HoweCommonly Used BridgeTrusses
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CommonlyUsed Roof Trusses
Figure4/2 WNW.gigapedia.com
Article4/2 Plane Trusses 167 'a) 'bJ B A A [) A framework composed of members joinedat their ends to form a
rigid st ru ct ure is called a truss. Bridges, roof supports , derricks, and other such structuresarecommonexamplesof trusses.Structuralmem -bers commonly used are l-bearn s, channels, angles, bars, and special shapes which are faste ned together at their ends by welding, riveted connecti ons, or large boltsor pins.\Vhen the members of the truss lie essentially ina single plane, thetrussiscalledaplane truss .
For bridgesandsimilarstructures,planetrusses arecommonlyu ti-lized in pai rs with onetrussassembly placed on each side of the st ruc-ture.Asection of a typical bridge st ructu re is shown in Fig. 4/1. The combined weight ofthe roadway and vehicles istransfer red to thel on-gitudinalst r ingers,thentothe crossbea ms, and finally,with the weights of the stringe rs and cross beams accounted for,to the upperjoints of the two plane trusses which formthe verticalsides ofthe structure.A simplified modelof thetruss st ruct ure isindicatedatthe left side ofthe illustra ti on; theforcesL represe nt thejoint loadings.
Several examplesof commonly used trusses which can be analyzed as planetrusses areshown in Fig.4/2.
4
/2
PLANE TRUSSESe
Compression Figure 4/4 Two-ForceMembe rs Ae
(e) Figure4/3T
e
I
T
I
e
F Tension Simple TrussesThebasic elemen tof aplan e trussisthetriangle.Threebarsjoined bypins at their ends , Fig.4/3a,constit utearigid frame. Thetermrigid isusedto mean noncollapsible andalsotomeanthat deformation ofthe members due to induced internal strains is negligible. On the other hand,fouror more bars pin-jointedto formapolygon of as many sides constitute a nonrigid frame.\Ve can make the nonrigid frame in Fig. 4/3b rigid,orstable,by addingadiagonalbar joiningA andDorBand C and thereby for ming two triangles.We can extend the structure by add ing additional units of two end-connected bars,suchas DE and CE or AF and DF, Fig.4/3c, which are pin ned to two fixed join ts. In this way the entire structure will remain rigid.
Structures built from a basic triangle in the manner describedare known as simple trusses. \Vhen more members are present than are needed toprevent collapse, the truss is statically indeterm inate.Astat -ically indetermina te trusscannot be analyzed by the equa tionsofequi -libriumalone.Additional members orsupportswhichare notnecessary formaintain ing the equilibriumconfigurationarecalledredunda nt.
Todesign atrusswe must firstdetermine theforces in thevarious members and then select appropriate sizes and structural shapes to withstand the forces.Severalassumptions are madein theforceanalysis ofsimpletrusses.First, weassume all membersto betwo-forcemembers. A two-force member isoneinequilibriumunder the act ionof two forces only,asdefined in generalterms withFig.3/4inAr t.3/3.Eachmember of a truss isnormallyastraightlink joiningthe twopoints of application of force.Thetwo forcesareapplied at the endsofthe member and are necessarilyequal, opposite,and collinear forequilibrium.
The member may be in tension or compression, as shown in Fig. 4/4.\Vhen werepresenttheequilibriumofaportionof a two-force mem -ber,thetension Torcompression C actingon the cut section isthesame
16 8 Chap t e r4 Struct u res
Truss Connections and Supports
for allsections. Weassum eherethat theweightofthememb er is small compared with the force itsupports. Ifit is not,or if we must account for the sma lleffect of the weight, we ca n replace theweight W of the memb er by two forces, each W/2 if the memb er is uniform,with one force acting at each end of the member. These forces, in effect, are treated as loads exte rnally applied to the pin connections.Accounti ng for the weight of a memb er in thiswaygives thecor rect result for the average tension orcompress ion alongthememberbutwill not account forthe effect ofbendin g ofthemember.
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/3
METHOD OF JOINTSThismethod forfindingtheforcesin the membersof atruss cons ists ofsatisfying the condit ions ofequilibr iu m for the forces acting on the connectingpin of each joint. The methodther eforedeal swith the equ i-librium of concurren t forces, and only two independen t equilibrium equationsare involved.
Webegin the ana lysis with anyjoint whe re at least one knownload exists and where not more than two unknown forces are present. The solut ion may be started with thepin at the lell end.Its free-body dia -gram is shown in Fig.4/7.Wit h thejoints indicatedby letter s,weu su-ally designat e theforce in each memb er by thetwo letter s definin g the ends of the memb er .Theproper directions of the forces should be ev i-den t by inspecti on for this simple case.Thefree-bodydiagram s of p or-tion s of memb ers AF and AB are also shown to clearl y indicat e the mechanism of the actionand reaction.The memberAB actually makes contacton theleft side of thepin, although the forceAB isdrawn from the right sideand is shown actingaway from thepin.Thus, ifwe co n-sistently drawthe forcearrowson thesameside of the pinasthe mem-ber ,thenten sion (suchasAB)willalwaysbe indicat edby anar rowaway
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\Vhen welded or riveted connections are used to jom struct ural members, we may usually assume that the connectio n is a pin joint if the cente rlinesofthemembersareconcu rrent atthe jointasin Fig.4/5. \Ve also assume in the analysis ofsimple trusses that allexternal forces are applied at the pin conn ection s.This condition is satis fiedin most trusses. In bridge trusses thedeck is usu ally laid on cross beam s which are supported at thejoints,as shown in Fig.4/1.
Forlarge trusses ,aroller, rocker,orsome kind ofslip joint isused at one of the supports to provide for expansion and contraction due to temperature changes and for deformation from applied load s.Trusses and frames in which no such provision is made arestatically indeter-minate,asexplained in Art. 3/3.Fig.3/1showsexa mplesofsuch joints. Twomet hodsfor theforceanalysisofsimple trusses willbegiven. Each method willbeexpla inedforthe simpletruss shown in Fig.4/60 . The free-body diagram of the truss as a whole is shown in Fig. 4/6b. The exte rn al reactio ns are usually determined first, by applying the equilibrium equations to the truss as a whole.Then the force analysis of the remainder of thetrussisperformed. D Figure4/7 Figure4/6 Tension ~- - -x '-'=---_ ...1 Figure4/5 F E (a) L L R, (b) y I I I AF I
r
AB RIfro m the pin,and compression (such asAF)willalways beindicated by an arrow toward the pin.The magnit ude of AFis obtained from the equa tion':::.Fy = 0 andAB is the n found from ':::.Fx = O.
Joint Fmaybe analyzed next, since it now contains only two un-knowns,EF andBF.Proceedingto the nextjoint having no more than two unknowns, we subsequently analyzejointsB,C,E,andD in that orde r.Fig.4(8 shows the free-bodydiagram of eachjoint anditsco rre-sponding force polygon, which represents graphically the two e quilib-riumconditions'i..Fx
=
0and'iFy=
O.The numbersindicatethe order in which the joints are analyzed.We notethat , whenjoint D is finally reached, the compute d reaction R2 must be in equilibrium with the forces in membersCD andED, whichwere determinedpreviously from the two neighb oring joints.This requi rem en t provides a check on the cor rect ness of our work. Note that isolation of joint Cshows tha t the force in CEis zero when the equa tio n':::.Fy=
0 is applied.The forcein2 EF EF
zl
BF{l
AB JointF 4 t CE=O R1 JointA BC E ) CD JointC 3 BF BC 5 BE BE~
E DE
AB BC L BE EF BF JointE 6~
AB CD CDD~R
,
L R, JointB JointD y I I f->--~---I----'>;D L - - - xArticle 4/3 Method of Joints 169
R1 L
Figure 4/8
R,
170 Chapter 4 Structu re s
MarwanandWaseem AI-Iraqi
this member would not be zero,ofcourse, ifan external vertical load were applied at C.
It is often convenien tto indicate the tens ion T andcompression C ofthe var ious memb ersdirectly on theorigi na ltruss diagram by dra w
-ingarrows away fromthepinsfor ten sion and toward thepinsfor c om-pression.This designation isillustrated at thebottom ofFig.4/8.
Sometimes we cannot initiallyassign the correct direction ofoneor both of the unknown forces acting on agiven pin. Ifso, we may make anarbitraryassignment. A negativecomputedforcevalueindicat esthat the initially assumeddirection is incorrect.
Internal and External Redundancy
If a plan e trus s has more exte rn al supports than are necessa ry to ens ure a stable equilibriumconfigu ration,thetruss as a wholeis sta ti-cally indeterminate, and the extra supports constitute externa l re dun-dancy. If a truss has more internal members than are necessary to prevent collapse whe n the truss isremoved from its supports, then the extra members const itute internal redundancy and the truss is again statically indet erminat e.
For a truss which is statically determinate externally, there is a definit erelationbetween thenumber of itsmem ber s and thenumber of itsjoints necessary for inte rnal sta bility without redundan cy.Becau se we ca nspecifythe equilibrium of each joint by two scalar force equa -tions,therearein all 2j such equationsfor atrusswithj joints.For the entiretrusscomposedofm two-force mem bersand having the maximum of three unknownsupport reactions,there are inallm
+
3 unknowns (m tension or compression forces and three reactions).Thus, for any plan e truss, the equation m+
3 = 2j will be satisfied if the truss is sta tically determinateinternally.Asimple plan e truss, formedby starting with atriangle and addin g two new members to locate eachnewjoint with respecttothe exist ing struc ture, satis fies the relation auto matica lly. The condit ion holdsfor the init ial triangl e, where m = j = 3, and m increases by 2 for each added joint while j incr eases by 1. Some othe r (nonsimple) statically determinate trusses, such as the K-truss in Fig.4/2, arearranged dif-ferently,butcan be seen tosatisfythe same relati on.
This equation is a necessary condit ion for stability but it is not a sufficient condition, since one or more of the m members can be ar-ran ged in such a way as not to contribute to a sta bleconfigu ra tion of the ent ire truss. Ifm
+
3>
2j, ther e are more memberstha n i nde-pendent equations, and the truss is statically indeter minate internally with redundant members present.Ifm+
3 < 2j, there is adeficiency of internal members, and the truss is unstableand willcollapse under load.Special Conditions
We often encounter several special conditions in the analys is of trusses . When two collinear members are under compression, as indi-cated in Fig.4/9a, it is necessary to add a third memb er to main tain
Art icl e4/J Methodof Joint s 171 'F.F;r=0 requiresF1=
}
'i
rF;r'=0 requiresFJ=F4 x·~
1/
~
zr
;
=0 requiresF1=0 x-r - -L
rF;r'=0 requiresF2=0 X / / /h,
rFy=0requiresF3=0 !Fx=0 requiresF1=Ii
i
(c) (bJ (c) Figure 4/9 D B D BwN
alignmentof thetwomembers and prevent buckling.We see from aforcesummation in they-directionthatthe forceF3in the third membermust be zero and from the x-direction that F1
=
F2' This conclusion holds regardless ofthe angle0and holds alsoifthe collinear members are in tens ion. Ifan external forcewith a component in the y-direction wereapplied to the join t, thenF3would nolonger be zero.
Wh en two non ccllinear members are joined as shown in Fig.4/9b, then in theabsence of an externallyappliedloadatthisjoint, theforces in both members must be zero, as we can see from the two force summations.
When two pairs of collinear members arejoined as shown in Fig. 4/9c,theforcesineachpairmustbe equa landopposite.This conclusi on followsfromthe force summations indicatedin the figure.
Tru ss pan els are frequently cross-braced as shown in Fig.4/10a. Sucha panelis staticallyindeter minate if each bra ce cansupporteithe r tension orcompression.However, whe nthe braces areflexible members
incapableof supporting compression,as are cables,then onlythe tension member acts and we can disregardthe other member.It isusually evi -dent from the asymmetry of the loading how the pan el will deflect. If the deflect ion isas indica ted in Fig.4/lOb, thenmember AB should be retained and CD disrega rded. When this choice cannot be made by inspection , we may arbitrarily select the member to be retained. Ifthe assumed tension turns out to be positive upon calculation, then
the choice was correct. If the assumed tension force turns out to be negative,thentheopposit emembermust beret ained andthe calculation redone.
Wecan avoidsimultaneoussolutionof theequilibriumequations for twounknown forcesat ajointbyacarefulchoiceof reference axes.Thus, for thejoint indicatedschematicallyin Fig.4/11 whereLisknown and F1andF2are unknown,a force summation in thex-direction eliminates reference to F1 and a force summation in the x'-direction elimina tes reference to F2.When the angles involved are not easily found,then a
simultaneous solutionof theequations using one setof reference direc -tion s for both unknowns may bepreferab le.
A (a)
c
A Figure4/10 Figure4/11 (b)c
172 Cha pt e r 4 Structures
Sample Problem 4
/1
Compute the forcein each member of theloaded cantilever trussby the method of joints .
[~E = OJ 5T - 20(5) - 30(10) 0 T = 80kN
[~Fx = OJ 80 cos300
- Ex 0 Ex = 69.3kN [~Fy = OJ 80sin 30° + Ey - 20 - 30 0 Ey = 10 kN
Next wedrawfree-body diagramsshowing theforces acting oneach ofthe con nect ing pins.Thecorrectn ess of the assigned directionsof the forcesisverified when each joint is considered in sequen ce.There shou ldbeno question about the cor rect directionof theforces onjointA.Equilibrium requires
A 30 kN 20kN T 5m 300 y 600 I I 5m L __ x Ex 5m 5m 30 kN 20kN Ey Ans. Ans. AB ~34.6 kN T AC ~ 17.32 kN C
o
o
0.866AB - 30 AC - 0.5(34.6) OJ 0]Solution. Ifit were notdesired to calculate the external reactions atDand E, the analysis for a cantilever truss couldbegin withthejoint at theloaded end. However, thistrusswill be analyzedcompletely,so the first ste pwill beto com-putethe exter nal forces atDandEfrom thefree-body diagram of the truss as a whole.The equa tionsof equilibriu mgive
CD
whereTsta nds for ten sionand C standsfor compression.Joint B must be analyzed next ,since the reare morethan two unknown forces on joint C.Theforce BC must providean upward componen t, in which caseBD mustbalancetheforce to the left.Againtheforces are obtained from
OJ 0] 0.866BC - 0.866(34.6) BD - 2(0.5)(34.6)
o
o
BC=
34.6kN C BD = 34.6kNT Ans. Ans. y I I\B I I I 600 _ AC- - x 30kN BDI\B
=
~
O
O
34.6kN 600 BC Joint Cnowcontai nsonly twounknowns, and theseare foundin thesameway as before:
CE = 63.5kNC
Finally,fromjointEther eresults
andtheequa tion'2:.F:r: = 0 checks.
JointB BC = 34.6kN JointA Helpful Hint
CD
It should be stressed tha t the ten-sion/compression designa tion refers to the member, not the joint. Note that wedrawtheforce arrowon the same sideofthejointasthe member
which exerts the force. In this way
tension (ar row awayfrom thejoint) is distinguished from compression (ar row toward the joint).
Ans. Ans. Ans.
o
o
11.55 kN C DE 0.866DE ~ 10 0.866CD - 0.866(34 .6) - 20 CD ~ 57.7 kNT CE - 17.32 - 0.5(34.6) - 0.5(57.7 ) OJ [~Fy = OJ I:EFy = 0] AC= 17.32kN DE 600 69.3kN CE CE= 63.5kN 20kN 10kN Joint C Joint E Marwan andWaseemAI-Iraqi www.gigapedia.comArticle 4/3 Problems 173
PROBL
EMS
Introductory Problems
4/4 Calculate the forces in members BEand BDof the
loadedtruss .
4/1 Dete rmine the force in each member of the simple equ ilateral truss. Ans.AB ~ 736N T,AC ~ 368 NT,BC = 736NC A 2m B C ~2 2 8' 8' A' 8' E 8' 2m 'B 10001b 2m Problem 4/4 3kN C 6 kN T 6kN C An. ,AB = 12kN T,AE BC = 5.20kNT,BD BE ~ 5.20kN C,CD ~ DE
4/5 Determine the force in each memberof the loaded truss.
75k~
Problem 4/1 C
4/2 Dete rmine the force in each member of the loaded truss. Discuss theelTects ofvarying the angleof the 450 suppor tsur faceat C. A A 6' B B C 1001b 2.5' E 3D'" 3kN C D Problem 4/5 45°
- - - - 4/6 Calcu late theforceineach memb eroftheloadedtruss.
"'
- - -
-
"'
-
-2
kN Problem 4/2 E D 3m B Problem 4/6 3m 4/3 Dete rminetheforcein eachmember ofthetruss.Notethe presence of any zero-force member s.
An".AB = 5 kNT,BC = 5/2kN C
CD ~ 15kN C,AC ~ 5,/5kNT,AD ~ 0
Problem4/3
174 Chapter4 Structures
c
Dc
B Ans.AB ~ DE ~96.0 kN C All ~ EF = 75 kNT,BC = CD = 75kN C BH ~ CG ~ DF = 60kNT CH ~ CF ~ 48.0kN C,GH ~FG = 112.5kNTDet ermin e the force in each member of the loaded truss.Makeuse of the symmetryofthe truss andof the loading. 4/7 4m 'i---- - - ---->:D 5m G 5m F 5m E 4kN 2kN Problem4/9 30kN 60kN 30kN
Problem 4/7 Representative Problems
4/8 Det ermine the force in each member of the loaded
truss.Alltrianglesareisosceles. 4/10 Solve forthe forces in membersBE andBD of the truss which supports theloadL.Allinterior angles are60°or120°. 4m
~
4m 1 1300 I lOkI' I Bc
c
A B Ans.AC =!::
T 2 Proble m 4/84/9 Determine the force in each member of the loaded truss.Alltriangles areequilateral.
Ans.AB ~ 9,13kNC,AE = 5,13kN T BC =
,?J3
kNC,BD = 3,13kN C,BE =iJ3
kN C CD ~lfJ3
kN T,DE ~¥J3
kN T 4/11 D F E Gt
L Problem4/10Determine the force in memberAC of the loaded truss.Thetwoquarter-circularmembersactastw o-forcemembers .
Article 4/1 Probl em s 175
L
B
Problem 4/11
4/12 Calcul at ethe forcesin members CGand CFfor the
truss show n. 2kip s
2 kN Problem 4/14
Problem 4/12
4/14 A drawbridgeisbein g raised byaca ble £1.Thefour join t loadings shown result from the weight ofthe roadw ay. Det erminethe forcesin membersEFtDE, DF,CD,andFG.
4/16 Deter minethe forcesin membersBI,CI,andHIfor theloaded truss.All angles are30°,60°, or 90°.
Problem 4/16
Problem 4/15
4/15 The equiangular truss is loaded an d su pport ed as show n.Det ermin e theforcesinall membersin terms ofthe horizontalloadL.
Ans.AB BC = L T,AF
=
EF=
L C DE = CD = L/2 T,BF = DF = BD = 0 20' B Problem 4/13 E 20'4/13 Each member of the truss is a uniform 20·ft bar weighing 400 lb. Calcu la te the average ten sion or
compression in each membe r due to the weights of
themembers. Ans.AB = BC 1000/,/3 1b T AE
=
CD = 2000/,13Ib C BD BE = 800/,13IbT ED = 1400/,13IbC D F176 Chap ter 4 Stouctur es
Problem 4/17
4/20 Determine the force in each member of the pair of trusseswhichsupportthe 5000-lb load attheir com-monjointC.
D 11<.."
F 2m 1kN 1kl'i
4/17 Asnow loadtransfersthe forcesshown to the upper
joints of a Pratt roof truss.Neglect any horizontal
reactions at the supportsand solve for the forces in
all members.
Ails.AB ~ DE = BC ~ CD = 3.35kN C
AH = EF = 3 kNT,BH ~ DF ~ 1kN C
CF ~ CH ~ 1.414 kN T,FG = GH = 2 kN T
1kN
~
c
4/18 TheloadingofProb.4/17is shown appliedto aHowe
roof truss . Neglect any horizontal reactions at the supports and solve for the forces in all members. Comparewith theresul ts of Prob.4/17.
4/21 Therectangular frame is composedoffourperimeter
two-forcemembers andtwo cables AC andBDwhich are incapable of supporting compression.Determine
the forces in all membe rsdue totheloadLin position
(0)and thenin position(b) . AilS.(0)AB = AD ~ BD = 0,BC ~ I. C 5L 4L AC=
'3
T, CD ~'3
C IblAB ~ AD ~ BC ~ BD ~ 0 AC~ 51.T CD ~ 41.C 3 ' 3 L Problem 4/20 1kN D 1 kN 2m Problem 4/ 18 B 1 kN 2m H 2m G 2m F 2m 1kN 3d 4d Problem4/21 nl'---~lC Hb)t
L (a) 10' 20001b 20001b C D 10' 10'pe-
----"'----~B
+
10' f--- - - 26'- - - -..j4/19 Calculatethe forces in membersCF,CG,andEF of
the loaded.truss.
Ans.CF ~ 15381b C, CG = 4170 lbT,EF = 0
Problem4/ 19
Article 4/3 Problems 177 Ans. FD ~ 24,500IbT f"---:24'----~r A F D (0) (b) F C D B (e) (dl Problem 4/24 16' 4/25 Analysis of the wind act ing on a small Hawaiia n
churc h, which withstood the 165-mi/h r winds of
HurricaneInikiin 1992, showedthe forcestra nsmit
-tedto each roof truss panel tobe as shown. Treatthe
struct ure as a symmetricalsimple truss andneglect
any horizontal componentof the support reactionat
A. Ident ify the truss member which supports the
la rgest force, tension or compression,and calcu lat e
this force.
2450 lb
4/24 Ver ifythe fact that each of the trussescontains one
or more eleme nts of redunda ncy and propose two
separate changes,eitherone of which would remove
the redundancyand produce completestat ical det
er-minacy. All members can support compression as
well as ten sion. E / U2 L L Problem4/22 L U2 I 16 m E
'"
Problem 4/234/23 The movable gantry isused to erect and preparea 500·Mg rocket for firing.The primary structure of
the gantryis approx imated by thesymmetrical plan e
trussshown,which is sta ticallyindeterm inate.Asthe
gantry is positioning a 60-Mg section of the rocket
sus pended fromA, st rain-gage measurements indio
cateacompressive for ce of50 kNin memb erABand
a tensile forceof 120 kN in memb er CD duetothe
60-Mg load . Calcu late the corres ponding forces in
memb ersBFandEF.
Ails.BF ~ 188.4kN C,EF ~ 120kN T
H C D
4/22 Determinethe forcesin membersAB,CG,andDEof
theloaded truss.
Problem4/25
178 Chapter4 Structures 1-- - - 6panelsat 5 m- - -- - -1.8kN 5pa nels at 3 m
----i
J D EI
A Problem4/27 Problem 4{28 10kNI
20kN E 20kN 20kN IOkN HI '
3m I I 'T-
IJ
1
--'
~-7:
~
3m 15°-
j
-3mi
5mJ
_ -
.J5._ ..A..
.... 4/28 Find theforcesin membersEF,KL,and GL for the Fink truss shown.
Ans.EF = 75.1kN C,KL 40 kN T
GL 20 kN T
4/26 The240·ft st ruct ure is used to provide various sup-port services to launch veh icles prior to liftoff.In a test ,a re-te»weight is sus pende d from jointsFand G,with its weight equally divided. between the two joints.Det erm inethe forcesin membersGJ andGI.
Wha t woul d beyour pa th ofjoin t analysis for mem -bersin the vertica l tower, such asABor KL?
Problem 4{26
.. 4/27 Thetower for a transmi ssion line ismodeled by the truss show n. The crosse d membersin thecenter sec-tions ofthe truss may be assumed to be capable of supporting tension only. For the load s of 1.8kN ap-plied in the vertical plan e , compute the forces in-duced in membersAB,DB,and CD.
Ans.AB ~ 3.89 kN C,DB ~ 0,CD ~ 0.932kN C
Article4/4 Meth o dof Secti ons 179
4
/4
METHOD OF SECTIONSWhen analyzing planetr ussesbythe methodofjoints,weneedonly
two of the three equilibriumequations because the proceduresinvolve
concur re nt forces at each joint. We can tak e adva nta ge of thethird or
moment equation ofequilibrium by selecting an entire section of the truss forthe free body inequilibrium under the actio n ofanoncon cur-rentsystem of forces.Thismethod of sections hasthe basic advantage
thatthe forceinalmostanydesired member maybefounddirectlyfrom
an analysis of a section which has cut that member. Thus, it is not
necessary to proceed wit h the calculation from joint to joint until the
member in questionhasbeen reached.Inchoosingasectionof thetruss, wenotethat,ingeneral,not morethanthreememberswhoseforces are
unknownshould becut, sincethere are onlythree availableindependent
equilibrium relations.
Illustration of the Method
F \ E A C
r
\ R, L R, (a) F E E / EF I I RE I 1// C R, L y 8, I I L __x (b) Figure4/12 Themethod of sections willnow beillust rated for thetr ussin Fig.4/6, which was used in the expla nation of the method ofjoints.The
truss is shown again in Fig. 4/12a for ready reference. The external
reactionsare first computedaswiththemethodof joints,bycons ide ring
the trussas a whole.
Letusdeterminetheforcein thememberBE,forexample.Anirnag -inary section,indicat edby thedashed line,ispassedthroughthetruss,
cutting itintotwoparts,Fig.4/12b.This sectionhascutthreememb ers
whoseforcesare initiallyunknown.Inorder fortheportion of the truss on eachsideof thesection to remain in equilibrium,it isnecessary to
apply to each cut member the force which was exerted on it by the
membercut away.Forsimple trusses composedoftwo-force members, these forces, either tensile or compressive ,will alwaysbe in the dire
c-tions of therespectivemembers.Theleft-handsection is inequilibrium
under the action of the applied load L, the end reaction R1> and the
threeforces exertedon thecutmembersbythe right-handsection which hasbeen removed.
We can usuallydrawthe forceswiththeir proper senses bya visual approximationof the equilibriumrequirements.Thus ,in balancing the moments about point Bfor theleft-hand section,theforceEFisclearl y
to the left,which makes it compressive,because it acts toward the cut section of memberEF.The load L is greater than the reaction R11so
that the force BE must be up and to the right to supply the needed
upwardcomponent for verticalequilibrium. Force BEis therefore ten -sile,since it acts away from the cutsection.
With the approximate magni tudesofR, and Lin mind we seethat thebalan ceof momen tsabout point ErequiresthatBCbe totheright. A casual glance at the truss should lead to the same conclusion whe n
it is reali zed that the lower horizontal memb er will stretch under the
tensioncaused bybending.The equationof momen tsaboutjoint Beli m-inates three forcesfromtherelation,andEFcanbedetermined directly.
The force BE is calculated from the equilibrium equation for the
y-direction.Finally,wedeter mineBCbybala ncingmomentsabout point
18 0 Chapter4 Structures
Marwan and WaseemAI-Iraqi
E.In this way eachof the three unknownshas been determined i nde-penden tlyofthe othe rtwo.
The right-han d section of the truss, Fig. 4/126, is in equilibrium
under the action of R2 and the same three forces in the cut members appliedin thedirection s oppositetothosefortheleftsection.Theproper
sense forthe horizontal forces can easily be seen from the balance of moments about pointsB andE.
Additional Considerations
It is essentialtounderstandthat inthemethodof sectionsan entire
portionof thetrussiscons idereda single bodyinequilibr ium.Thus,the forces inmembersinternaltothesectionarenot involved intheanalysis
ofthe sect ionas a whole.To clarifythe free body and the forcesacting
exte rnally on it, the cutt ing sect ion is prefera bly passed through the
members and not thejoints.\Ve may use either portion of a truss for thecalcula ti on s,but the one involvin gthesmaller number offorceswill
usually yield thesimplersolution.
In some cases the methodsofsections andjoints can be combined foranefficientsolution.Forexample,suppose we wishtofind theforce in a centr al member of a large truss.Furthermore, suppose that it is
not possible to pass a sectio n through this mem ber without passin g
throug h at least four unknown members. It may be possible to d
eter-mine the forces in nearby member sbythemethod ofsectionsand then
progressto theunknown memb er bythe method of joints.Sucha com
-bination of the twomethodsmay be more expedient than exclusive use ofeither meth od.
The moment equationsare used to great advantage in the meth od of sections. One should choose a moment center, either on or ofT the
section,throughwhich as many unknown forcesas possible pass. It is not always possible to assign the propersenseof an unknown force when the free-body diagram ofa section is initi ally drawn.Once an arbitrary assignment is made, a positive answer will verify the as-sumedsense anda negative result willindicate that the force is in the sense opposite to that assumed. An alternative notation preferred by
someisto assign allunknownforces arbitrarilyaspositiveinthetension directi on (away from the sectio n)and letthe algebraicsign of theanswer
disti nguish betw een tension andcompression. Thus,a plus sign would
signify tension and a minu s sign compression. On theot he r hand ,the
advantage of assign ing forces in their correct sense on the free-body diagramof asection wherever possible is that doing so emphas izes the
physicalaction of the forces moredirectly,and this practice istheone
which is preferredhere.
Arlicle 4/4 Mel h o d of Sectio n s 181
Sample Problem 4
/2
Calculate the forcesinducedin member sKL, CL,and CRbythe 20-too load onthe cantilevertruss.
C CB 8 KL L ~--~""7~""'ZfCL
I
ll
I'
p y I I G fl-
....:---
'--"
~
_
- - .r 20tons 20tons1 1
16'i-
Gf---.'F:---:E~-D!':--C~,--+-+---,.·t:J - - - 6pa nelsat12'- - -H HelpfulHinlsCD
Wenot etha t ana lysisbythe methodofjoints would necessitate working wit heightjoints in orderto calcu lat e the three forces in question. Thus, the met hod ofsections offersa con
-sider able advantageinthis case. Ans. Ans. KL = 65.0IonsT CB ~ 57.1 Ions C 20(5)(12) - CB(21) ~ 0 20(4)(12) - ¥SKL(16) = 0 [LAIc =
O
J
Next wetakemomentsaboutC,wh ich requires acalculation ofcos9.Fromthe
given dimensionswesee 9= tan-1{5/12)so thatcos 8
=
12/13.Ther efore,Solution. Although the vertical compone nts of the reactions at A and M are
statically indetermi nate with thetwofixedsupports, all memb ersotherthanAM
are staticallydeterminate.We maypassasection directl ythroughmembersKL,
CL,andCR andana lyze theportionof thetrusstothe left ofthis section asa
statically determinaterigidbody.
Thefree-body diagr am oftheportionofthe trussto theleft ofthesection is shown. Amoment su maboutL quickly ver ifies theassignment ofCR asco m-pr ession,andamomentsumaboutCquicklydisclosesthatKLisin ten sion.The directionofCLisnotquiteso obviousuntilwe observethatKL andCB intersect
at a pointPto the rightofG.Amomen t sumabout P elimina tes refere nce to KLandCB andshows thatCL must be compressiveto balancethemoment of
the 20-tonforceaboutP.With theseconsi de ra ti onsin mind the solution becomes
straightforward, as we now see how to solve for each of the three unknowns
inde pendently of theother two.
Su mmingmomentsabou t L requiresfindingthemomen t armBL = 16 +
(26 - 16)/2 21ft. Thus,
CD
Final ly,wemayfindCL byamoment sumaboutP,whosedistance fromC
isgiven byPC/16 ~ 24/(26 - 16)orPC ~ 38.4 ft.Wealso need
fl
,
which isgiven by
fl
~ tan-1(CB/BL) ~ lan-1(12/2 1) ~ 29.7' andcosfl
~ 0.868.We nowhave @ [LAIp = OJ 20(48 - 38.4 ) - CL<O. 868)(38.4) CL = 5.76 IonsCo
Ans.@
Wecould have sta rtedwithmoments about CorPjust aswell.@ We could also have det ermined CL byaforce summati on ineit herthex
-ory-di rec tion.
182 Chapter4 Structure s
Sample Problem 4
/3
Section1 2 I 1 'IV 10kN lOkNc
18.33kN JK~
KI I J J H 10 kN 6panelsat4m- -- - I
lO kN B A _ lOkNCV
If desired, the direction of CD maybe cha ngedon thefree-bodydiagram and thealgebrai csignofCDreversed
inthe calculat ions ,or else thewor k
maybe left as it stands witha not e
stating theproperdirection. HelpfulHints
CD
The re isnoharm inassigni ng one ormor e of the forces in the wrongdi
-rection as long as the calculati ons are consiste nt with the assumpt ion.
A negati ve answer will show the
need for reversing the direction of the force.
o
CJ = 14.14 kN C 0.707CJ(12) - 10(4) - 10(8) = 0 0.894CD(6) + 18.33(12) - 10(4 ) - 10(8) CD = - 18.63kNThe momentofCDabou tJ iscalculated herebyconsidering itstwocomponen ts
@
as acting throughD.Theminu s sign indicatesthatCDwasassignedinthewrongdirection.
Calculatetheforce in memberDJof the Howerooftrussillustrated.Neglect
anyhor izon talcomponents offorce at the supports.
In this equa tion the momen t of CJ is calcu late d by considering its horizontal
and vertical components acting at point J. Equilibrium of moments about J
requ ires
Solution. Itisnot possible topassasection through DJ without cutting fou r
member s whose for ces areunknown. Although three ofthese cut bysection2
are concu rrentat J andthereforethe moment equati onaboutJ couldbeused
to obtainDE ,theforceinDJcannotbeobtained fromtheremainingtwoeq
ui-librium pri nciples.Itisnecessarytoconsider first theadjace ntsection1befor e analyzingsection 2.
The free- body diagr am for section 1 is drawn andincludesthe reaction of
18.33 kN atA,which ispreviouslycalculated fromtheequilibri um ofthetruss as awhole.Inassign ingtheproperdirectionsfor the forces acting onthe three
cut mem bers,wesee that a balance of moments about A eliminatesthe effects
of CDand JKand clearly requires that CJ beup and to theleft.A balan ce of
momentsabout Celiminat esthe effect of thethree forcesconcurrent atC and
indicates thatJKmust betotherighttosupplysufficientcounterclockwisem
o-ment. Again it shou ld befairlyobvious that the lower chord is under tension
becauseof the bendingtendency of the truss.Althoughitshouldalsobe apparent
that the top chordisunder compress ion,forpurposesof illustrat ionthe forcein
CD
CDwillbe arbitrarilyassignedas tension.By theana lysisof sect ion 1,CJis obtained from
From the free-bodydiagr am ofsection 2, which nowincludes the known
valueofCJ ,abalanceofmomentsaboutGis seen to elimina teDEand JK.Thus,
Hence, CD = 18.63kNC 12DJ+ 10(16)+ 10(20) - 18.33(24) - 14.14 <0.707)(12)= 0 lOkN Sect ion 2 <, <, 14.14kN ... ... ~-~~~~'~J--- - -~G DJ = 16.67 kN T Ans. 18.33kN
Again the moment of CJ is determi ned fro m its compone nts cons idered tobe
act ingatJ.Theanswer forDJ ispositive.sothattheassumedtensile direction
is correct.
An alter na tive approachto the entire problemis toutilize section 1tod
e-termineCDandthenuse themethod ofjoints applied atDtodetermineDJ.
@
Observethatasect ion thro ughmem-her sCD, D.J,andDEcouldbetaken whichwouldcutonlythreeunknown
memb ers. However,since theforces
in these three members are allcon -current at D, a moment equation aboutDwou ldyield no information
aboutthem .Theremaini ngtwoforce
equations would not be sufficient to
solvefor thethreeunknowns.
Article 4/4 Problems 183
PROBLEMS
Introductory Problems 4m J 4m H 4m G 4m B 4m C 4m D 50kN E 14.14 kipsT 10' F 10' G 10' H4/29 Determin etheforceinmemberCG.
Ans.CG Problem4/31 10' A B
c
D
l
5 kips 5kips 5 kips Problem 4/294/30 Deter mine the forcesin membersBC,CF,andEFof theloadedtruss.
4/32 Determine the force in member DO of the loaded
truss.
I
5 panelsat4'l
r
L LL
V
E C 3' 4' A F 12' A B B Problem 4/324/33 Deter mine the forcesin memb er sBC,BE, andBF. Thetr ianglesareequilateral.
Ans.BC ~ BE
c
E B F A G i}---<L----~----_';D4/31 Deter mine the forces in members OH and CO for
the truss loaded and supportedas shown. Doesthe statical indet erminacy of the supports affect your
calculation?
Ans.CG ~ 70.7kN T,GH ~ 100 kN T,No
Problem 4/10
L
Problem 4/33
184 Chapte r4 Structur es Representative Problems 4/34 Determ ine theforces in membersDEandDL. 6' Problem 4{34 H -~"----x--~ F Problem 4{36
4/37 Thetrussiscomposed of equilate raltria nglesofside a and is supported and loaded as shown. Determine the forces in mem bersBC and
ca
.
Ans.BC ~ CG ~ 1-{3T
8kN
2m 2m
4/35 Calcu la te the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that forceas the only unknown.
Ans. BC ~ 21 kN T,BE ~ 8.41 kN T EF = 29.5kNC 14kN G L a a F B a E C
»:
30' D Problem 4{374/38 The truss shown is composed of 45°righttriangles . The crossed members in the cente r twopanels are slende r tierodsincapable ofsuppor tingcompression. Retain the two rods which are under ten sion and computethe magnitudes ofthe ir ten sions.Alsofind theforcein memberMN.
Problem4/35
4/36 Determinetheforcesin membersBC andFG ofthe loaded sym met rical truss.Show tha t thiscalculati on can be accomplished by using one section and two equa tions,each of which contains onlyone ofthetwo unknowns.Are theresul tsaffected by thestatical in
-determinacyof the supports atthe base? J 80kN A H K G L F E N 100 kN D
o
Problem 4{38Articl e 4/4 Pro ble ms 18 5 LI2 L Problem 4/41
r -
6 panelsat 3'M
I
L K J 1-- - - - -8panelsat 3m- -- - - -44/42 Determinethe forces in membersCD,CJ,and DJ.
5'
B C D E F G
A
L L L L L L
Problem 4(42
4(43 Compute the force in memb er GM of the loaded truss. Ans.GM = 0 L L L L L L L F L L 2 E G 2 D H E 4m 3m G D 8kN H 10kN 3m C 3m Problem 4/40 I B 4kN 3m A ~--+'---I--:-:--->:r:::--....::,qF J 9ki ps 1
r--
3kipsiF
I
-
EI 5'- j - ' - -5'- - + - -5'---4 A B Problem 4(39 6kN4/39 Det erminetheforcein memberBF.
Ans.BF ~ 2.66kips C
4/40 The members CJ and CF of the loaded truss cross
but are not connecte dto memb ersBIandDG.C
om-putetheforcesin membersBC,CJ,CI,andHI.
C
D
4/41 The truss su pports a ramp (shown with a dashed
line)which ext ends from afixed approach levelnear
jointF toa fixedexit level nea rJ.Theloads shown represen t the weight of the ramp. Det ermine the forces in memb ersBHandCD.
Ails.BH ~ 0.683L T,CD ~ 1.932LC
Problem 4(43
186 Chapter 4 Structur es 3 26.4 kN T 75.9 kN T F Dimension sin met e rs Problem 4/47 K C
4/47 Determinethe forcesin membersDE, EI,FI,andHI
of thearched roof truss. Ans.DE = 297kN C,EI Fl = 205 kN T,HI L L L L
J
L L L Problem 4/44 1--- - -- - 8panelsat3m-4/45 Determine the forces in members DJand EJ of the loaded truss.
4/44 Compute the force in member HN of the loaded truss.Compareyour answerwiththesta tedresultof Prob.4/43.
Ails.DJ = 0.45LT, EJ 0.360LT
1--- - - - 6 panelsat8m----~ Problem 4/45
4/48 Find theforceinmemberJQfor the Bal tim oretruss wher e all anglesare30°,60°,90°,or 120°. L L E L D L
c
L4/46 Determine the force in member HP of the loaded truss. Me mbers FP and GQ cross without touching and areinca pableofsupport ing compression.
Alls.DK = 1kipT L 2 L L L L L L L L L 2 100
"''I
100kN Problem 4/48 .... 4/49 Deter mine the force in member DK of the loaded overhead sign truss.u
I
B C 6panDelsat8'E Fl
c
A- ,
0 p Q R S Tj'
II 5' Nt
Lt
M J I 9 panelsat 20'Problem 4/46 1 kip 2kips 4kips
Problem 4/49
Article 4/4 Problems 187
... 4/51 Det erminetheforcein memberDCof the compound
truss.Thejoin tsall lieonradi allinessubte ndinga
n-gles of 15° as indicated, and the curved members act as two-force members. DistanceDC = OA = ~4/50 In the travelingbridgecraneshownallcrosse dmem
-be rs areslendertie rodsincapableofsupporting com
-pression. Determine the forcesin membersDF and
EFandfindthe horizontal reaction onthetruss at
A.Showthat ifCF = 0,DE = 0 also.
Ans.DF = 768 kN C,EF = 364 kNC AI = 101.1kN 25 kN 6m
1-6m1
-6m 4ml4m 14 m,4 m14m14 mII
K
~
·
I
c-+--+-+---.-J I 6m H Nj
600/ ....60' I C 0 6m I 5.4m 25 kN A... 4/51 A design model for a transmission-line tower is shown in the figure.MembersGR,FG,OP,andNO areinsu lated cables;allother membersare steel bars.
For theloading shown,compute theforces in m
em-bersFI,FJ,EJ.EK,andER. Use a combinati onof methodsifdesired. Ans.FJ = ER = 0,FJ = 7.81kNT EJ = 3.61 kN C,EK = 22.4kNC - x 22m Problem 4/50 A .
