CHAPTER OUTLINE

4 /1 Introduction4/2 Plane Trusses4/3 Method of Joints4/4 Method of Sections4/5 SpaceTrusses4/6 Frames and Machines

Chapter Review

STRUCTURES

4 /1 INTRODUCTION

In Cha pter 3 we studied th e equilihr ium of a single rigid body or asyste m of connected memb ers t reated as a single rigid body. We firstdrew a free-body diagram of t he body showing a ll forces external to t heisolated body a nd then we applied the force a nd moment equatio ns ofequilibrium. In Chapter 4 we focus on the determination of the forcesinternal to a st ructure , that is, forces of act ion and reaction between t heconnected memb ers. An engineering structu re is any connected sys te mof members built to support or transfer forces and to sa fely withstandthe loads applied to it. To determine the forces internal to a n enginee r­ing structure , we must dismember th e st ructure a nd analyze separatefree -body diagrams of individual members or combinations of members.This analysis requires ca re ful application of Newton's th ird law, whichstates th at each action is accompanied by a n equa l and opposite reaction .

In Cha pte r 4 we analyze the internal forces acting in several typesof structures, nam ely, trusses, frames, and machines. In this treatmentwe consider only statically determinate stru ctures, which do not havemore supporting cons train ts than are necessary to maintain an equilib­r ium configu ration . Thus, as we hav e a lready seen, the equations of equi­librium are adequate to determine all u nknown reactions.

Th e analysis of trusses, fram es a nd machines, a nd beams under con­centrated load s cons titutes a straightforwa rd application of th e mat erialdeveloped in th e previous two chapters. The ba sic procedure developedin Cha pter 3 for isolating a body by construct ing a cor rect free-bodydiagram is essential for the analysis of statica lly determin ate st ructures.

MalWan and Waseem AI-Iraqi www .gigapedia.com 165

166 Ch ap t er 4 Str uct ures

L

Stringer

Cross beam \

~~~

Figure 4/ 1

Pra tt

Warren

Howe

Common ly Used Brid ge Trusses

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Commonly Used Roof Tru sses

Figure 4/2

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Articl e 4/2 Plane Trusses 167

' a )

' bJ

B

A

A

[)

A framework composed of members joined at their ends to form arigid st ruct ure is called a truss. Bridges, roof supports , derricks, an dother such structures are common examples of trusses. Structura l mem­bers commonly used are l -bearns, channels, angles, bars, and specialshapes which are fastened together at t heir ends by welding, r ivetedconnections, or large bolts or pins. \Vhen the members of the truss lieessentially in a single plane, the truss is called a plane truss .

For bridges and similar structures, plane trusses are commonly uti­lized in pairs with one t ru ss assembly placed on each side of the st ruc­t ure. A sect ion of a typical brid ge structu re is shown in Fig. 4/1. Th ecombined weight of th e roadway and vehicles is transfer red to the lon­git udina l str ingers, t hen to the cross beams, and finally, with the weightsof the stringers and cross beams accounted for, to the upper joints ofthe two plane trusses which form the vertical sides of the structure. Asimplified mode l of th e truss stru cture is indicated at t he left side of theillust ra tion; the forces L represe nt the joint loadings.

Several examples of commonly used trusses which can be ana lyzedas plane trusses are shown in Fig. 4/2.

4 /2 PLANE TRUSSES

e

Compression

Figure4/4

Two-Force Members

A e(e )

Figure 4/3

Te

I T Ie

F

Tension

Simple Trusses

The basic element of a plan e truss is t he t ria ngle. Three bars joinedby pins at th eir ends, Fig. 4/3a , constitute a r igid frame. Th e ter m rigidis used to mean noncollapsible and also to mean that deformation of themembers due to induced internal strains is negligible. On the otherhand, four or more bars pin-jointed to form a polygon of as many sidesconstitute a nonrigid frame. \Ve can make the nonrigid frame in Fig.4/3b rigid, or stable, by adding a diagonal bar joining A and D or B andC and t hereby forming two triangles. We can exte nd the st ructure byadding additional unit s of two end-connected bars, such as DE and CEor AF and DF, Fig. 4/ 3c, which are pinned to two fixed join ts. In t hisway the entire structure will remain rigid.

Structures built from a basic triangle in the manner described areknown as simple trusses. \Vhen more members are present than areneeded to prevent collapse, t he truss is stat ically indeterminate. A stat­ically indetermina te truss cannot be analyzed by th e equa tions of equi­librium alone. Additional members or supports which are not necessaryfor maintaining the equilibrium configuration are called redunda nt.

To design a truss we must first determine the forces in the variousmembers and then se lect appropriate sizes and structural shapes towith stan d the forces. Several assumptions are made in the force analysisof simple trusses. First, we assume all members to be two-force members.A two-force member is one in equilibrium under the action of two forcesonly, as defined in genera l te rms with Fig. 3/4 in Art. 3/3. Each memberof a truss is normally a straight link joi ning the two points of applicationof force. The two forces are applied at the ends of the member and arenecessarily equal, opposite, and collinear for equilibrium.

The member may be in tension or compression, as shown in Fig.4/4. \Vhen we represent the equilibriu m of a portion of a two-force mem­ber, the tension T or compression C acting on the cut section is the same

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16 8 Chap t e r 4 Struct u res

Truss Connections and Supports

for all sections. We assum e here that t he weight of th e memb er is smallcompared with the force it supports. If it is not, or if we must accountfor th e small effect of th e weight, we can replace the weight W of themember by two forces, each W/ 2 if the member is uniform , with oneforce acting at each end of the member. These forces, in effect, aretreated as loads externally applied to the pin connections. Accountingfor th e weight of a member in thi s way gives t he correct result for theaverage tension or compression along the member but will not accountfor the effect of bendin g of th e member .

I

l, ,

};~'0--~':"""""/",~.\

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4 /3 METHOD OF JOIN TS

This method for finding the forces in the members of a truss consistsof satisfying the condit ions of equilibr ium for the forces act ing on theconnecting pin of each joint. Th e method therefore deal s with th e equ i­librium of concurrent forces, and only two independent equilibriumequations are involved.

We begin th e ana lysis with an y joint where at least one known loadexists and where not more than two unkn own forces are present. Thesolut ion may be started with th e pin at th e lell end. It s free-body dia­gram is shown in Fig. 4/ 7. Wit h th e joints ind icated by letters, we usu­ally designat e th e force in each member by the two letters definin g theends of t he member . Th e prop er directions of t he forces should be evi­den t by inspection for thi s simple case. Th e free-body diagram s of por­tions of members AF and AB are also shown to clearl y indicat e themechanism of the action and reaction . The member AB actually makescontact on th e left side of t he pin, although t he force AB is drawn fromth e right side and is shown acting away from the pin. Thus, if we con­sistently draw the force arrows on the same side of the pin as the mem­ber , then tension (such as AB ) will a lways be ind icat ed by an arrow away

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\Vhen welded or riveted connections are used to jom st ructuralmembers, we may usually assume that the connectio n is a pin joint ifthe centerlines of the members are concurrent at the joint as in Fig. 4/5.

\Ve also assume in the analysis of simple trusses that all externalforces are applied at the pin connection s. This condition is satis fied inmost trusses. In br idge t russes th e deck is usually laid on cross beam swhich are supported at th e jo int s, as shown in Fig. 4/1.

For large trusses , a roller, rocker, or some kind of slip joint is usedat one of the supports to provide for expansion and contraction due totemperature changes and for deformation from applied loads. Trussesand frames in which no such provision is made are statically indeter­minate, as explained in Art. 3/ 3. Fig. 3/ 1 shows examples of such joints.

Two methods for the force analysis of simple trusses will be given.Each met hod will be expla ined for t he simple tru ss shown in Fig. 4/60 .The free-body diagram of the t russ as a whole is shown in Fig. 4/6 b.The exte rn al reac tions are usua lly determined first , by applying theequilibrium equation s to the truss as a whole. Then the force analysisof the remainder of t he t ru ss is performed.

D

Figure 4/7

Figure 4/6

Tension

~- - - x'-'=---_ ...1

Figure 4/5

F E

(a)

L

L R,

(b)

yIII AFIrAB

RI

from the pin , an d compression (such as AF) will always be indicated byan arrow toward t he pin. Th e magnitude of AF is obtained from theequa tion ':::.Fy = 0 an d AB is t hen found from ':::.Fx = O.

Joint F may be analyzed next, since it now contains only two un­know ns , EF and BF. Proceeding to the next jo int having no more thantwo unknowns, we subsequently analyze joints B, C, E, and D in thatorde r. Fig. 4( 8 shows the free-body diagram of each joint and its corre­sponding force polygon, which represents graphically the two equilib­rium conditions 'i..Fx = 0 and 'iFy = O. The numbers indicate the orderin which t he joints are analyzed . We not e that , when joint D is finallyreached, th e computed reaction R2 must be in equ ilibrium with t heforces in members CD and ED, which were determined previously fromthe two neighboring jo ints . This requi remen t provides a check on thecorrect nes s of our work. Note that isolation of join t C shows tha t theforce in CE is zero when the equa tion ':::.Fy = 0 is applied. The force in

2 EF EF

zl {lBF

AB Joint F

4t CE=O

R 1

Joint A BC E ) CDJoint C

3BF BC 5

BE BE ~EDEAB BC

L BE EFBF Joint E

6

~AB CD

CD

D~R,L R,

Joint B Joint D

yII

f->--~---I----'>; D L - - - x

Article 4/3 Method of Joi nts 169

R1 L

Figure 4/8

R,

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170 Chapter 4 Struct ure s

Marwa n and Waseem AI- Iraqi

th is member would not be zero, of course, if an exte rnal vertical loadwere applied at C.

It is often convenient to indicate the tension T and compression Cof the various members directly on the original truss diagram by dra w­ing arrows away from the pins for tension and toward th e pins for com­pression. This designation is illustrated at the bottom of Fig. 4/8.

Sometimes we cannot initially ass ign the correct direction of one orboth of the unknown forces acting on a given pin. If so, we may makean arbitrary assignment. A negative computed force value indicates thatthe initially assumed direction is incorrect .

Internal and External Redundancy

If a plan e t rus s has more exte rn al supports than are necessary toensure a stable equilibrium configuration, th e truss as a whole is stati­cally indeterminate, and the extra supports constitute external redun­dancy. If a truss has more internal members than are necessary toprevent collapse when the truss is removed from its supports, then theextra members constitute internal redundancy and the truss is againstatically indeterminate.

For a truss which is stat ically determinate externally, there is adefinit e relation between th e number of its members and th e number ofits joints necessary for inte rnal stability without redundan cy. Becau sewe can specify th e equilibrium of each joint by two scalar force equa­t ions, th ere are in all 2j such equations for a truss withj joints. For th eentire truss composed of m two-force members and having the maximumof three unknown support reactions, there are in all m + 3 unknowns(m tension or compression forces and three reactions). Thus, for anyplane truss, the equation m + 3 = 2j will be sa tisfied if the truss issta tically determinate internally.

A simple plane truss, formed by starting with a tri angle and add ingtwo new members to locate each new joint with respect to the exist ingstruc ture, satis fies th e relation auto matica lly. The condit ion holds forthe init ial tri angl e, where m = j = 3, and m increases by 2 for eachadded joint while j increases by 1. Some other (nonsimple) staticallydeterminate trusses, such as the K-truss in Fig. 4/2, are arranged dif­ferently, but can be seen to satisfy the same relation.

This equation is a necessary condit ion for stability but it is not asufficient condition, since one or more of the m members can be ar­ranged in such a way as not to contribute to a sta ble configura tion ofthe ent ire truss. If m + 3 > 2j, there are more members tha n inde­pendent equations, and the tru ss is statically indeterminate intern allywith redundant members present. If m + 3 < 2j , there is a deficiencyof internal members, and the truss is unstab le and will collapse underload.

Special Conditions

We often encounter several special conditions in the analys is oftrusses . When two collinear members are under compression, as indi­cated in Fig. 4/9a, it is necessary to add a third member to main tain

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Ar t icl e 4 /J M ethod of Joint s 171

'F.F;r = 0 requires F1 = }'irF;r' =0 requires FJ =F4

x·

~ 1/~ zr; = 0 requires F1 = 0

x -r - - L r F;r' = 0 requires F2 = 0

X/

//

h,

rFy = 0 requires F3 = 0! Fx =0 requires F1 =Iii

(c ) (bJ (c)

Figure 4/9

D B D B

wN

alignment of th e two members and prevent buckling. We see from a forcesu mmation in they-direction that the force F3 in the third member mus tbe zero and from the x-direct ion that F1 = F2' Th is conclusion holdsregardless of the angle 0 and holds also if th e collinear members are intension. If an externa l force with a component in the y-direction wereapplied to the join t, then F3 would no longer be zero .

When two nonccllinear members are joined as shown in Fig. 4/ 9b,th en in t he absence of an exte rnally applied load at t his joint, th e forcesin both members must be zero, as we can see from the two forcesummations.

When two pairs of collinear members are joined as shown in Fig.4/ 9c, the forces in each pair must be equa l and opposite . This conclusionfollows from the force summations indicated in the figure.

Tru ss pan els are frequently cross-braced as shown in Fig. 4/ 10a .Suc h a panel is statically indeterminate if each brace can support eit he rtension or compression . However, when the braces are flexible membersincapable of supporting compression, as are cables, then only the tensionmember acts and we can disregard the other member. It is usually evi­dent from the asymmetry of th e loading how th e panel will deflect.If the deflect ion is as indica ted in Fig. 4/lOb, th en member AB shouldbe retained and CD disregarded. When this choice cannot be madeby inspection, we may arbitrarily select the member to be retained.If the assumed tension turns out to be positive upon calculation, thenthe choice was correct. If the assumed tension force turns out to benegat ive, t hen the opposite member must be ret ained and th e calculationredone.

We can avoid simultaneous solution of the equilibrium equations fortwo unknown forces at a joint by a careful choice of reference axes . Thus,for the jo int indicated schematically in Fig. 4/ 11 where L is known andF1 and F2 are unknown, a force summation in the x-direction eliminatesreference to F1 and a force summation in the x '-direction eliminatesreference to F2 . When the angles involved are not easily found , then asimultaneous solution of the equations using one set of reference direc­t ions for both unknowns may be preferab le.

A

(a)

c A

Figure 4/10

Figure 4/1 1

(b)

c

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172 Chapt e r 4 Structures

Sample Problem 4/1

Compute th e force in each member of the loaded cantilever truss by themethod of joints .

[~E = OJ 5T - 20(5) - 30(10) 0 T = 80kN

[~Fx = OJ 80 cos 300- Ex 0 Ex = 69.3 kN

[~Fy = OJ 80 sin 30° + Ey - 20 - 30 0 Ey = 10 kN

Next we draw free-body diagrams showing th e forces acting on each of theconnect ing pins. Th e correc tn ess of the assigned directions of the forces is verifiedwhen each joint is considered in sequence. Th ere should be no question aboutth e correct direction of th e forces on joint A. Equilibrium requires

A

30 kN 20kN

T5m 300

y 600

II 5mL __

xEx

5 m 5m

30 kN 20kN EyAns.

Ans.

AB ~ 34.6 kN T

AC ~ 17.32 kN C

oo

0.866AB - 30

AC - 0.5(34.6)

OJ

0]

Solution. If it were not desired to calculate the exte rnal reactions at D an d E,the analysis for a cantilever truss could begin with the joint at th e loaded end.However, this truss will be analyzed completely, so the first step will be to com­pute th e external forces at D and E from the free-body diagram of the truss asa whole. Th e equa tions of equilibrium give

CD where T stands for tension and C stands for compression.Joint B must be analyzed next , since there are more th an two unknown

forces on joint C. The force BC must provide an upward componen t, in whichcase BD must balance the force to the left . Again th e forces are obtained from

OJ

0]

0.866BC - 0.866(34.6)

BD - 2(0 .5)(34.6)

oo

BC = 34.6 kN C

BD = 34.6 kN T

Ans.

Ans.

yI I\BIII 600

_ AC - - x

30kN

BD

I\B =~OO34.6 kN 600BC

J oint C now contai ns only two unknowns, and these are found in the sameway as before :

CE = 63.5 kN C

Finally, from joint E th ere results

an d the equation '2:.F:r: = 0 checks.

JointB

BC =34.6 kN

J oint A

Helpful Hint

CD It should be stressed th at t he te n­sion/com pression designa tion refersto th e member , not the joint . Notethat we dr aw th e force arrow on thesame side of th e joint as the memberwhich exer ts th e force. In thi s wayten sion (ar row away from th e joint)is disti ngu ished from compression(ar row toward the joint).

Ans.

Ans.

Ans.

o

o

11.55 kN CDE0.866DE ~ 10

0.866CD - 0.866 (34 .6) - 20

CD ~ 57.7 kN T

CE - 17.32 - 0.5(34 .6) - 0.5(57.7)OJ

[~Fy = OJ

I:EFy = 0]

AC=17.32 kN

DE

600 69.3 kN

CE CE =63.5 kN

20kN 10kN

Joint C Joint E

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Art icle 4/3 Problems 173

PROBL EMS

Introductory Problems

4/4 Calculate the forces in members BE and BD of theloaded trus s .

4/1 Determine the force in each member of the simpleequilateral truss.An s. AB ~ 736 N T, AC ~ 368 N T, BC = 736 N C

A

2 m

B C ~22

8' 8'

A'8' E 8'

2m ' B

10001b

2m Problem 4/4

3 kN C6 kN T6 kN C

An. , AB = 12 kN T, AEBC = 5.20 kN T , BD

BE ~ 5.20 kN C, CD ~ DE

4/5 Determine the force in each member of the loadedtruss.

75 k~

Problem 4/ 1

C

4/2 Determine the force in each member of the loadedtruss. Discuss the elTects of varying the angle of the450 support sur face at C.

A

A 6' BB C

1001b2.5'

E3D'"

3 kN

CD

Problem 4/5

45°- - - - 4/6 Calculate the force in eac h member of th e loaded tru ss.

"'- - - - "'- -2 kNProblem 4/2 E D

3 m

B

Problem 4/6

3 m

4/3 Determine the force in each member of the truss. Notethe presence of any zero-force members.

An". AB = 5 kN T, BC = 5/2 kN CCD ~ 15 kN C, AC ~ 5,/5 kN T, AD ~ 0

Prob lem 4/ 3

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174 Chapter 4 Structures

c

DcB

Ans. AB ~ DE ~ 96.0 kN CAll ~ EF = 75 kN T, BC = CD = 75 kN C

BH ~ CG ~ DF = 60 kN TCH ~ CF ~ 48.0 kN C, GH ~ FG = 112.5 kN T

Determine the force in each member of the loadedtruss. Make use of the symmetry of the truss and ofthe loading.

4/7

4m 'i---- - - ---->:D

5mG

5 mF

5 mE

4 kN 2 kN

Problem 4/ 930 kN 60 kN 30 kN

Problem 4/7 Representative Problems

4/8 Determine the force in each member of the loadedtruss. All triangles are isosceles. 4/10 Solve for the forces in members BE and BD of the

truss which supports the load L. All interior anglesare 60° or 120°.

4 m

~4m

11300

I lO kI'I

B

c

c

A

B

Ans. AC = !:: T2

Problem 4/ 8

4/9 Determine the force in each member of the loadedtruss. All triangles are equilateral.

Ans. AB ~ 9,13 kN C, AE = 5 ,13 kN T

BC = ,?J3 kN C, BD = 3 ,13 kN C, BE = iJ3 kN CCD ~ lfJ3 kN T, DE ~ ¥J3 kN T 4/11

D

F E

GtL

Problem 4/ 10

Determine the force in member AC of the loadedtruss. The two quarter-circular members act as two­force members .

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Article 4/1 Prob lem s 175

L

B

Problem 4/11

4/12 Calculat e the forces in members CG and CF for th etruss shown.

2 kips

2 kN Problem 4/14

Problem 4/12

4/14 A drawbridge is bein g raised by a cable £ 1. Th e fourjoi nt loadings shown result from the weight of t heroadway. Det ermine th e forces in members EFt DE,DF, CD, and FG.

4/16 Determine the force s in members BI , CI, and HI forthe loaded truss. All angles are 30°, 60°, or 90°.

Problem 4/16

Problem 4/15

4/15 Th e equiangular t russ is loaded and su ppo rted asshown. Determine the forces in a ll members in termsof th e horizontal load L .

Ans. AB BC = L T, AF = EF = L CDE = CD = L/2 T, BF = DF = BD = 0

20'B

Problem 4/13

E

20'

4/13 Each member of the truss is a un iform 20·ft barweighing 400 lb. Ca lcu late the average ten sion orcom pr ession in each member du e to the weights of

t he members. Ans. AB = BC 1000/ ,/3 1b T

AE = CD = 2000/ ,13 Ib CBD BE = 800/ ,13 Ib TED = 1400/ ,13 Ib C

D

F

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176 Chap ter 4 Sto uc tur es

Problem 4/17

4/20 Determine the force in each member of the pair oftrusses which support the 5000-lb load at their com­mon joint C.

D 1 1<.."

F 2 m

1 kN1 kl'i

4/17 A snow load transfers the forces shown to the upperjoints of a Pratt roof truss. Neglect any horizontalreactions at the supports and solve for the forces inall members.

Ails. AB ~ DE = BC ~ CD = 3.35 kN CAH = EF = 3 kN T , BH ~ DF ~ 1 kN C

CF ~ CH ~ 1.414 kN T, FG = GH = 2 kN T

1 kN

~c

4/18 The loading of Prob. 4/ 17 is shown applied to a Howeroof truss . Neglect any horizontal reactions at thesupports and solve for the forces in all members.Compare with the results of Prob. 4/17.

4/21 The rectangular frame is composed of four perimetertwo-force members and two cables AC and BD whichare incapable of supporting compression. Determinethe forces in all members due to the load L in position(0) and then in position (b) .

AilS. (0) AB = AD ~ BD = 0, BC ~ I. C5L 4L

AC = '3 T, CD ~ '3 C

Ibl AB ~ AD ~ BC ~ BD ~ 0

AC ~ 51. T CD ~ 41. C3 ' 3

L

Problem 4/20

1 kND

1 kN

2m

Problem 4/ 18

B

1 kN

2 m H 2m G 2m F 2m

1 kN

3d

4d

Problem 4/21

nl'----------~l C

H b)

tL

(a)

10'

20001b

20001b

C

D

10'

10'

pe-----"'----~B +10'

f--- - - 26' - - - -..j

4/19 Calculate the forces in members CF, CG, and EF ofthe loaded. truss.

Ans. CF ~ 15381b C, CG = 41 70 lb T, EF = 0

Problem 4/ 19

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Article 4/3 Problems 177

Ans. FD ~ 24,500 Ib T

f"-------:24 '----~r

A FD

(0 ) (b)

F

C D

B

(e) (d l

Problem 4/24

16'

4/25 Analysis of the wind act ing on a small Hawaiianchurc h, which withstood th e 165-mi/hr winds ofHurricane Iniki in 1992, showed th e forces t ra nsmit ­ted to each roof truss panel to be as shown. Treat thestructure as a symmet rical simple truss and neglecta ny horizontal component of th e support reaction atA . Ident ify the truss mem ber which supports th elargest force, te nsion or compression, and calcu lat ethis force.

2450 lb

4/24 Ver ify the fact that each of the trusses contains oneor more eleme nts of redundancy and propose twoseparate cha nges, either one of which would removethe redundancy and produce complete stat ical deter­minacy . All members can support compression aswell as tension .E

/

U2LL

Problem 4/22

LU 2

I16 m

E'"

Problem 4/ 23

4/23 T he movable ga ntry is used to erect and pre pare a500·Mg rocket for firing. The primary structure ofth e gantry is approx imated by t he symmetrical plan et russ shown, which is sta tically indeterm inate. As thegantry is positi oning a 60-Mg section of th e rocketsuspended from A , strain-gage measurements indiocate a compressive force of 50 kN in member AB anda tensile force of 120 kN in member CD due to the60-Mg load . Calculate th e corresponding forces inmembers BF and EF .

Ails. BF ~ 188.4 kN C, EF ~ 120 kN T

H C D

4/22 Determine the forces in members AB , CG, and DE ofth e loaded truss.

Problem 4/25

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178 Chapter 4 Struc tur e s

1-- - - - - 6 panels at 5 m - --- - -

1.8 kN

5 panels at 3 m ----iJ D E IA

Problem 4/27

Problem 4{28

10 kN

I 20kN

E 20kN

20kN

IO kN

HI'3m I

I '

T-IJ 1--'~-7:~3 m 15°

-j--3 m

i5 m

J_ - .J5._ ..A..

.... 4/28 Find t he forces in members EF, KL, an d GL for th eFink truss shown.

Ans. EF = 75.1 kN C, KL 40 kN TGL 20 kN T

4/26 The 240· ft st ructure is used to provide various sup­port services to launch veh icles pri or to liftoff. In atest , a re-te» weight is suspended from joints F andG, wit h its weight equally divided. between the twojoints. Determine t he forces in members GJ and GI.What would be you r path of jo in t analysis for mem­bers in the vert ical tower, such as AB or KL?

Problem 4{26

.. 4/27 The tower for a transmission line is modeled by th etruss shown. The crossed members in the cen ter sec­t ions of t he t russ may be ass umed to be capable ofsupporting tension on ly. For the loads of 1.8 kN ap­plied in the vertical plan e, compute th e forces in­du ced in members AB, DB, and CD.Ans. AB ~ 3.89 kN C, DB ~ 0, CD ~ 0.932 kN C

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Articl e 4 / 4 Me th o d of Secti on s 179

4 /4 METHOD OF SECTIONS

When ana lyzing plane trusses by the method of joint s, we need onlytwo of the three equilibrium equations because the procedures involveconcurrent forces at each joint. We can take advantage of the third ormoment equation of equilibrium by selecting an entire section of thetruss for the free body in equilibrium under the action of a nonconcur­rent sys tem of forces. This method of sections has the basic advantagethat the force in a lmost any desired member may be found directly froman ana lysis of a section which has cut th at member . Thus, it is notnecessary to proceed with th e calculation from joint to joint until t hemember in question has been reached. In choosing a sect ion of the truss,we note that, in general, not more than three members whose forces areunknown should be cut , since t here are only three available independ entequilibrium relations.

Illustration of the Method

F \ E

AC r\

R, L R,

(a)

F E E/ EF

II REI

1//

C

R, L y 8 ,

IIL __ x

(b )

Figure 4/12

Th e met hod of sect ions will now be illustrated for th e truss in Fig.4/ 6, which was used in th e explanation of t he met hod of joints. Thetruss is shown again in Fig. 4/ 12a for ready reference. The externalreactions are first computed as with the method of joints, by cons ideringthe truss as a whole.

Let us determine the force in the member BE, for example. An irnag­inary section, ind icat ed by th e dashed line, is passed through th e truss,cutting it in to t wo parts, Fig. 4/ 12b. Th is sect ion has cut three memberswhose forces are initially unknown. In order for the portion of the trusson each side of the section to remain in equilibrium, it is necessary toapply to each cut member t he force which was exerted on it by themember cut away. For simple trusses composed of two-force members,these forces, either tensile or compressive, will always be in the direc­tions of th e respective members. Th e left-hand section is in equilibriumunder the action of t he ap plied load L, t he end reaction R 1> and th et hr ee forces exerted on the cut members by t he right-hand sect ion whichhas been removed.

We can usually draw the forces with their proper senses by a visualapproximation of the equilibrium requirements. Thus, in balancing themomen ts about point B for th e left-hand section, t he force EF is clearlyto the left, which makes it compressive, because it acts toward the cutsection of member EF. The load L is greater than the reaction R11 sothat the force BE must be up and to t he right to supply the neededupward component for vertical equilibrium. Force BE is therefore ten­sile, since it acts away from the cut section.

With the approximate magni tudes of R , and L in mind we see th atth e balan ce of moments about point E requires that BC be to the right.A casua l glance at the truss should lead to t he same conclusion whenit is realized that t he lower horizontal member will st retch under thetension caused by bending. Th e equat ion of momen ts about joint B elim­inates three forces from the relation, and EF can be determined directly.The force BE is calculated from the equilibrium equa t ion for they-direction. Fin ally, we determine BC by balancing moments about point

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18 0 Chapter 4 Structure s

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E. In this way each of th e three unknowns has been determined inde­penden tly of the ot he r two.

The right-han d section of t he tru ss, Fig. 4/126, is in equilibriumund er the action of R2 and the same three forces in the cut membersapplied in the direction s opposite to t hose for the left sect ion. The propersense for the horizontal forces can easily be seen from the balance ofmoments about points B and E.

Additional Considerations

It is essen tial to understand t hat in t he method of sections an entireportion of t he tru ss is considered a single body in equilibr ium. Thus, theforces in members internal to the section are not invo lved in the analysisof the sect ion as a whole. To clarify the free body and the forces actingexte rnally on it , th e cutt ing sect ion is preferably passed through themembers and not the join ts. \Ve may use either portion of a truss forthe calcula tions, but the one involving the smaller number of forces willusually yield th e simpler solution.

In some cases the methods of sections and joints can be combinedfor an efficient solution. For example, suppose we wish to find the forcein a central member of a large truss. Furthermore, suppose that it isnot possible to pass a section through thi s member without passingthrough at least four unknown mem bers. It may be possible to deter­mine t he forces in nearby members by the method of sections and th enprogress to th e unknown memb er by the method of joints. Such a com­bination of the two methods may be more expedient than exclus ive useof either meth od.

The moment equations are used to great advantage in the meth odof sections . One should choose a moment center, either on or ofT thesection, through which as many unknown forces as possible pass.

It is not always poss ible to assign the proper sense of an unknownforce when the free-body diagram of a section is ini tially drawn. Oncean arbitrary assignment is made, a positive answer will verify the as­sumed sense and a negative result will indicate that the force is in thesense opposite to that assumed. An alternative notation preferred bysome is to ass ign all unknown forces arbitrarily as positive in the tensiondirection (away from the section) and let the algebraic sign of th e an swerdistinguish between tension and compression. Thus, a plus sign wouldsignify te nsion and a minu s sign compression. On th e othe r hand , theadvantage of assigning forces in their correct sense on the free-bodydiagram of a section wherever possible is that doing so emphasizes thephysica l act ion of the forces more directly, and th is pra ctice is t he onewhich is preferred here.

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Arl icle 4/ 4 Mel h o d of Sectio n s 181

Sample Problem 4/2

Calculate the forces induced in members KL, CL, and CR by the 20-too loadon the cantilever tru ss.

C CB

8KL L

~--~""7~""'ZfCL

Ill 'I

p

yII

G fl-....:------'--"~_ - - .r

20 tons

20 tons

1116'

i-Gf---.'F:---:E~-D!':--C~,--+-+---,.·t:J

- - - 6 pa nels at 12'- - -

H

Helpful Hinls

CD We note that ana lysis by the methodof joints wou ld necessitate workingwith eight joints in order to calculat et he t hree forces in question. Th us,the method of sections offers a con­siderable advantage in t his case.

Ans.

Ans.KL = 65.0 Ions T

CB ~ 57.1 Ions C20(5)(12) - CB(21) ~ 0

20(4)(12) - ¥SKL(16) = 0[LAIc = OJ

Next we tak e moments about C, which requires a calculation of cos 9. From thegiven dimensions we see 9 = tan - 1{5/12) so that cos 8 = 12/13. Th er efore,

Solution. Although the vertical components of th e re actions at A and M arestat ically ind etermi nate wit h the tw o fixed supports, all memb ers oth er than AMare statically determinate. We may pass a section dir ectly th rough member s KL,CL , and CR and ana lyze the portion of the truss to the left of this section as astatically determinate rigid body.

The free-body diagram of the portion of the truss to the left of the sectionis shown. A momen t sum about L qu ickly ver ifies the assignment of CR as com­pression, and a moment sum about C quickly discloses that KL is in tension. Thedirection of CL is no t quite so obviou s un til we observe that KL and CB inter sectat a point P to the right of G. A moment sum about P elimina tes re fere nce toKL an d CB and shows that CL mu st be compressive to balance the moment ofthe 20-ton force about P. With these considera tions in mind th e solution becomesstraight forward, as we now see how to solve for each of the three unknownsinde pendently of the other two.

Summing moments abou t L require s finding the mom ent arm BL = 16 +(26 - 16)/ 2 21 ft. Thus,

CD

Final ly, we may find CL by a moment sum about P, whose distance from Cis given by PC/16 ~ 24/ (26 - 16) or PC ~ 38.4 ft . We also need fl, which isgiven by fl ~ tan- 1(CB/ BL) ~ lan - 1(12/ 21) ~ 29.7' and cos fl ~ 0.868 . Wenow have

@ [LAIp = OJ 20(48 - 38.4 ) - CL<O.868)(38.4)

CL = 5.76 Ions C

oAns.

@ We could have started with mo mentsabout C or P ju st as well.

@ We could also ha ve determined CLby a force summation in either the x­or y-directio n.

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182 Chapter 4 Structure s

Sample Problem 4/3

Section 1

2I

1

' VI

10 kN

lO kN

c

18.33 kN

JK

~K I I J J H

10 kN6 panels at 4 m----I

l OkN

B

A _

lO kN

CV If desired, the direction of CD maybe cha nged on the free-body diagramand the algebraic sign of CD reversedin the calculat ions , or else th e workmay be left as it stands with a notestating th e proper direction.

Helpful Hints

CD There is no harm in ass igning one ormore of the forces in th e wrong di­rection as long as the calculati onsare consiste nt with the assumption.A negati ve answer will show theneed for reversing th e directio n ofthe force.

o

CJ = 14.14 kN C0.707CJ(12) - 10(4 ) - 10(8) = 0

0.894CD (6) + 18.33(12) - 10(4) - 10(8)

CD = - 18.63 kN

The moment of CD about J is calculated here by considering its two componen ts@ as act ing through D. The minus sign indicates that CD was assigned in th e wrong

direction .

Calculate the force in member DJ of the Howe roof truss illustrated. Neglectany horizontal components of force at the supports.

In thi s equation t he moment of CJ is calculate d by considering its horizontaland vertical components acting at point J . Equilibrium of moments about Jrequires

Solution. It is not possible to pass a sect ion through DJ without cutting fourmembers whose forces are un known. Although three of these cut by section 2are concurrent at J and therefore the moment equati on about J could be usedto obtain DE, the force in DJ cannot be obtained from t he remaining two equi­librium principles. It is necessary to consider first the adjacent sect ion 1 beforeanalyzing section 2.

The free-body diagr am for section 1 is drawn an d includes the reaction of18.33 kN at A , which is previously calcu lated from th e equilibrium of the trussas a whole. In ass igning th e proper directions for the forces acting on the th reecut members, we see that a balance of moments about A eliminates the effectsof CD and JK and clearly requires th at CJ be up and to the left . A balance ofmoments about C eliminat es th e effect of the three forces concurrent at C andindicates that JK must be to the right to supply sufficient counterclockwise mo­ment. Again it should be fairly obvious that th e lower chord is under tensionbecause of the bending tendency of the truss. Although it should also be apparentth at the top chord is und er compress ion, for purposes of illust rat ion th e force in

CD CD wi ll be arbitrarily ass igned as tension.By t he ana lysis of sect ion 1, CJ is obtained from

From th e free-body diagram of section 2, which now includ es th e knownvalue of CJ , a balance of moments about G is seen to eliminate DE and JK. Thu s,

Hence, CD = 18.63 kN C

12DJ + 10(16 ) + 10(20) - 18.33(24 ) - 14.14<0.707)(12) = 0

lOkN

Sect ion 2<,

<,

14.14 kN ....... ....... .......~-~~~~'~J - - - - - - - ~G

DJ = 16.67 kN T Ans. 18.33 kN

Again th e moment of CJ is determi ned from its components cons idered to beact ing at J . The answer for DJ is posit ive. so that the assumed te nsile directionis correct.

An alternative approach to the entire problem is to utilize section 1 to de­termine CD and th en use the method of joints applied at D to determine DJ .

@ Observe th at a sect ion th rough mem­hers CD, D.J, and DE could be takenwhich would cut only three unknownmembers. However, since th e forcesin t hese three member s are all con ­current at D, a mome nt equationabout D would yield no informationabout them. Th e remaining two forceequations would not be sufficient tosolve for t he three unknowns.

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Articl e 4 /4 Probl ems 183

PROBLEMS

Introductory Problems4 m

J 4 m H 4m G

4 m B 4 m C 4 m D

50kNE

14.14 kips T

10'F10'G10'H

4/29 Determine the force in member CG.Ans. CG

Problem 4/31

10'

AB c Dl

5 kips 5 kips 5 kips

Problem 4/29

4/30 Determine the forces in members BC, CF, and EF ofth e loaded t russ.

4/32 Determine t he force in mem ber DO of the loadedtruss.

I 5 pan els at 4' lr L L

L VE C3'

4'

A F

12'

A

BB

Problem 4/32

4/33 Determine the forces in members BC, BE, and BF.Th e tr ian gles are equilateral.

An s. BC ~ BE

c

E

B

F

A

Gi}--------<L----~----_';D

4/31 Determine th e forces in members OH and CO fort he truss loaded and supported as shown. Does thesta tical indet erminacy of the supports affect yourcalculation?

An s. CG ~ 70.7 kN T , GH ~ 100 kN T , No

Problem 4/10

L

Problem 4/33

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184 Chapt e r 4 Structu res

Representative Problems

4/34 Determ ine th e forces in members DE and DL.6'

Problem 4{34

H

-~"----x--~F

Problem 4{36

4/37 The truss is composed of equilate ral t riangles of sidea and is supported and loaded as shown. Determinet he forces in members BC and ca.

Ans. BC ~ CG ~ 1-{3 T

8kN

2m 2m

4/35 Calculate the forces in mem bers BC, BE , and EF.Solve for each force from an equilibrium equationwhich contains that force as the only unknown.

Ans. BC ~ 21 kN T , BE ~ 8.41 kN TEF = 29.5 kN C

14 kN

G L

a

a FB

a E

C »:30'

D

Problem 4{37

4/38 The tru ss shown is composed of 45° right triangles .The crossed members in the cente r two pan els areslende r tie rods incapable of supporting compression.Retain the two rods which are und er tension andcompute the magnitudes of t he ir tensions. Also findth e force in member MN.

Problem 4/35

4/36 Determine th e forces in members BC and FG of theloaded sym metrical truss. Show t ha t thi s calculationcan be accomplished by using one section and twoequa tions, each of which contains only one of the twounknowns. Are th e resul ts affected by the statical in­dete rminacy of the supports at t he base? J

80 kN

A

H

K

G

L

F E

N

100 kN

D

o

Problem 4{38

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Articl e 4 / 4 Pro ble ms 18 5

LI2L

Problem 4/41

r- 6 panels at 3'

MI L K J

1-- - ---8 panels at 3 m--- - - -4

4/42 Determine the forces in members CD, 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 th e loadedtru ss.

Ans. GM = 0

LL L

L L

L L F L L2 E G

2D H

E

4 m

3 m

G

D

8 kN

H

10 kN

3 m C 3 m

Problem4/40

I

B

4 kN

3 mA

~--+'---I--:-:----->:r:::--....::,qFJ

9 kips1

3 kips i FI - EIr-- 5'-j-'-- 5'--+-- 5'---4

A

B

Problem 4(39

6 kN

4/39 Determine th e force in member BF.Ans. BF ~ 2.66 kip s C

4/40 The members CJ and CF of the loaded truss crossbu t are not con nected to members BI and DG. Com­pu te the forces in members BC, CJ , CI , and HI.

C

D

4/41 Th e t russ su pports a ramp (shown with a dashedline) which extends from a fixed approach level nearjo in t F to a fixed exit level nea r J . Th e loads shownrepresen t th e weight of th e ramp. Determine theforces in members BH and CD.

Ails. BH ~ 0.683L T, CD ~ 1.932L C

Problem 4(43

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186 Chapte r 4 Struc tures

3

26.4 kN T75.9 kN T

F

Dimen sion s in met ers

Problem 4/47

K

C

4/47 Determine the forces in members DE, EI, FI, and HIof the arc hed roof tru ss.

Ans. DE = 297 kN C, EIFl = 205 kN T, HI

LL

LL

JL

L

L

Problem 4/44

1------ - 8 panels at 3 m------

4/45 Determine th e forces in members DJ and EJ of theloaded t russ.

4/44 Compute th e force in member HN of the loadedtruss. Compare your answer with the stated resu lt ofProb. 4/ 43.

Ails. DJ = 0.45L T, EJ 0.360L T

1--- - - - 6 panels at 8 m----~

Problem 4/45

4/48 Find the force in member JQ for t he Bal tim ore trusswhere all angl es are 30°, 60°, 90°, or 120°.

LL

E

L

D

L

c

L

4/46 Determine th e force in membe r HP of the loadedtruss. Members FP and GQ cross without touchingand are incapable of support ing compression.

All s. DK = 1 kip TL2 L L L L L L L

LL 2

100 "''I 100 kN

Problem 4/48

.... 4/49 Determine the force in member DK of the loadedoverhead sign truss.

u

I6 pan els at 8' lcB C D E F

A -,

0 p Q R S T j'II

5'N

tL tM J I9 panels at 20'

Problem 4/461 kip 2 kip s 4 kips

Problem 4/49

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Article 4/4 Problems 187

... 4/51 Determine the force in member DC of the compoundtruss. The joints all lie on radial lines subtending an­gles of 15° as indicated, and the curved membersact as two-force members. Distance DC = OA =

~ 4/50 In the traveling bridge crane shown all crossed m em ­

be rs are slender tie rods incapable of supporting com­pression . Determine the forces in members DF andEF and find the horizontal reaction on the truss atA. Show that if CF = 0, DE = 0 also.

Ans. DF = 768 kN C, EF = 364 kN CAI = 101.1 kN

25 kN 6m

1-6m

1-6m

4m l4m 14 m , 4 m 14 m 14 m I

I K~· I c-+--+-+-----.-J I

6m

H N j60 0 / .... 60' I

C 0 6 mI

5.4 m

25 kN

A

... 4/51 A design model for a transmission-line tower isshown in the figure. Members GR , FG, OP, and NOare insulated cables; all other members are stee l bars.For the loading shown, compute the forces in mem­bers FI, FJ, EJ . EK, and ER. Use a combination ofmethods if desired.

Ans. FJ = ER = 0, FJ = 7.81 kN TEJ = 3.6 1 kN C, EK = 22.4 kN C

- x

22m

Problem 4/50

A . I!-

II)'

.;-- - - - 5 panels at 8 m - - - - -jI

--.-----<;<;------;i/c----;>;<c---,"'"----;i/I f6 m

1-200 kN

DB = R.Ans. DC = 0.569L C Problem 4/52

Problem 4/51

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