Structures

STRUCTURES CHAPTER OUTLINE 4 /1 4/2 4/3 4/4 4/5 4 /6

Introduction Plane Trusses Method of Joints Method of Sections Space Trusses Frames and Machines

Chapter Review

4 /1

INTRODUCTION

In Cha pter 3 we studied th e equilihr ium of a single rigid body or a syste m of connected memb ers t reated as a single rigid body. We first d rew a free-body diagram of t he body showing a ll forces external to t he isolated body a nd then we applied the force a nd moment equatio ns of equilibrium. In Chapter 4 we focu s on the determination of the forces internal to a st ructure , that is, forces of act ion and reaction between t he connected memb ers. An engineering struct u re is any connect ed sys te m of members built to su pport or transfer forces and to sa fely withstand the loads applied to it. To determine the forces internal to a n engi nee ring structure , we must d ismember th e st ruc t ure a nd ana lyze se parate free -body diagrams of individual members or combinations of members. This analysis req uir es ca re ful application of Newton's th ird law, whi ch states th at eac h action is accompa nied by a n equa l and oppos ite reaction . In Cha pte r 4 we analyze the internal for ces acting in several types of struct ures, nam ely, trusses, frames, and machines. In this treatment we consider only statically determinate stru ctures, whi ch do not hav e mor e su pporti ng 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 ade quate to determine all u nknown reactions. Th e analysis of trusses, fram es a nd ma chines, a nd beams under concentrated load s cons tit utes a straightforwa rd applicati on of th e mat erial developed in th e previous two chapters. The ba sic procedure developed in Cha pt er 3 for isolating a body by construct ing a cor rect free-body diagr am is essential for the analysis of statica lly deter min ate st ructure s. MalWan and Waseem AI-Iraqi

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165

166

Ch ap t er 4

Str uct ures

Stringer Cross beam

\

~~~

L

Figure 4/ 1

Pra tt

Howe

Warren

Common ly Used Brid ge Trusses

Common ly Used Roof Tru sses

Figure 4/2 Marwa n and W aseem AI-Iraqi

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Articl e 4 /2

4 /2

Plane Trusses

B

PLANE TRUSSES

A framework composed of members joined at thei r ends to form a rigid st ru ct ure is called a tru ss. Bridges, roof supports, derricks, an d other such structures are common examples of trusses. Structura l members commonly used are l-bearn s, channels, angles, bars, and special shapes which are faste ned together at t heir ends by welding, r iveted connecti ons, or large bolts or pins. \Vhen the members of the truss lie essentially 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 assembl y placed on eac h side of the st ruct ure. A sect ion of a ty pical brid ge structu re is shown in Fig. 4/1. Th e combined weight of th e roadway and vehicles is tra nsfer red to the longit udina l str ingers, t hen to the cross bea ms, a nd fina lly, wit h the weights of the stringe rs and cross beams accounted for, to the upper joints of the two plane trusses which form the vertical sides of the structure. A simplified mode l of th e truss stru cture is indicated at t he left side of the illust ra tion; the forces L re prese nt the joint loadings. Several examples of commonly used trusses which can be ana lyzed as plane trusses are shown in Fig. 4/2. Simple Trusses

A 'a )

[)

A

' bJ

F

The basic element of a plan e truss is t he t ria ngle. Three bars joined by pins at th eir ends, Fig. 4/3a , cons titute a r igid frame. Th e ter m rigid is used to mean noncollapsible and also to mean that deformation of the members due to induced internal strains is negligible. On the other hand, four or more bars pin-jointed to form a polygon of as many sides constitute a nonrigid frame. \Ve can make the nonrigid frame in Fig. 4/3b rigid, or stable, by adding a diagonal bar joining A a nd D or B and C and t hereby for ming two triangles. We can exte nd the st ructure by add ing additional unit s of two end-connected bars, suc h as DE an d CE or AF and DF , Fig. 4/ 3c, which are pin ned to two fixed join ts. In t his way the entire structure will remain rigid. Structures built from a basic triangle in the manner described are known as simple trusses. \Vhen more members are present than are needed to prevent collapse, t he truss is stat ically ind eterminate. A statically indetermina te truss cannot be analyzed by th e equa tions of equilibrium alone. Additional members or supports which are not necessary for maintaining the equilibrium configuration are called redunda nt. To design a truss we must first determine the forces in the various members and then se lect appropriate sizes and structural shapes to with stan d the forces. Several ass umptions are made in the force analysis of 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 forces only, as defined in genera l te rms with Fig. 3/4 in Ar t. 3/3. Each member of a truss is normally a straight lin k joi ning the two points of application of force. The two forces are applied at the ends of the member and a re necessarily 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 compress ion C acting on the cut section is the same Marwan and Waseem AI-Iraqi

167

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e

A (e )

Figure 4/3

T e

IT

Ie e

Tension

Compression

Two-Force Members Figure 4/4

16 8

Ch ap t e r 4

S truct u res

,

for all sections. We assum e here that t he weight of th e memb er is s mall compared with the force it supports. If it is not, or if we must account for th e s mall effect of th e weight, we ca n replace the weight 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 . Accounting for th e weight of a memb er in thi s way gives t he cor rect result for the average tension or compress ion along the member but will not accoun t for the effect of bendin g of th e member .

,

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

l

Truss Connections and Supports

Figure 4/5

F

E

D

L (a)

\Vhen welded or riveted connections are used to jom st ructural members, we may usually ass ume that the connectio n is a pin joint if the centerlines of the members are concu rrent at the joint as in Fig. 4/5. \Ve also ass ume in the analysis of simple trusses that all external forces are applied at the pin conn ection s. This condition is satis fied in most trusses. In br idge t russes th e deck is usu ally laid on cross beam s which 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 used at one of the supports to provide for expansion and contraction due to temperature changes and for deformation from applied loads. Trusses and frames in which no such provision is made are statically indeterminate, as explain ed in Art. 3/ 3. Fig. 3/ 1 shows exa mples 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 tio ns are usua lly determined first , by applying the equilibrium equation s to the truss as a whole. Then the force analysis of the remainder of t he t ru ss is performed.

4 /3 L

R, (b)

Figure 4/6

y I I I I

r

AF

AB

'-'=---_ ...1

~- - - x

Tension

RI

Figure 4/7 MarNan and Waseem AI-Iraqi

METHOD OF JOIN TS

This method for finding the forces in the members of a truss cons ists of satisfying the condit ions of equilibr iu m for the forces act ing on the connecting pin of eac h joint. Th e method ther efore deal s with th e equ ilibrium of concurrent forces, and only two independent equilibrium equations are involved. We begin th e ana lysis with an y joint whe re at least one known load exists and where not more than two unkn own forces are present. The solut 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 letter s, we usually designat e th e force in each memb er by the two letter s definin g the ends of t he memb er . Th e prop er directions of t he forces s hould be eviden t by inspecti on for thi s simple case. Th e free-body diagram s of portions of memb ers AF and AB are also shown to clearl y indicat e the mechanism of the action and reaction . The member AB actually makes contact on th e left side of t he pin, although t he force AB is drawn from th e right side a nd 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 ten sion (such as AB ) will a lways be ind icat ed by an ar row away www.gigapedia.com

Article 4/3

fro m the pin , an d compression (such as AF) will always be indicated by an arrow toward t he pin. Th e magnitude of AF is obtained from the equa tion ':::.Fy = 0 an d AB is t he n 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 ha ving no more than two unknowns, we subsequently analyze joints B, C, E, and D in that orde 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 order in which t he joints are analyzed . We not e that , when joint D is finally reached, th e compute d reaction R 2 must be in equ ilibriu m wit h t he forces in members CD and ED, which were determined previously from the two neighb oring jo ints . This requi remen t provides a check on the cor rect nes s of our work. Note that isolation of join t C shows tha t the force in CE is zero when the equa tio n ':::.Fy = 0 is applied. The force in

2

zl

EF

EF

{l

BF Joint F

AB

4

t CE =O

R1

BC

Joint A

E

)

CD

Joint C

3

BF BE

AB

BC

5

BE

BC

~E DE BE

L

EF

BF

Joint E 6

AB CD L

Joint B

~

CD

D~R,

R, Joint D

y I I

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

R1

L - - -

x

R,

L Figure 4/8

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Method of Joi nts

169

170

Chapter 4

Struct u re s

th is member would not be zero, of course, if an exte rnal vertical load were ap plied at C. It is often convenien t to indicate the tension T and compression C of t he various members directly on t he origi na l truss diagram by dra wing arrows away from the pins for ten sion and toward th e pins for compression. This designation is illustrated at th e bottom of Fig. 4/8. Sometimes we cannot initially ass ign the correct direction of one or both of the unknown forces acting on a given pin. If so, we may make an arbitrary assignment. A negat ive computed force value indicates that the initially assumed direction is incorrect.

Internal and External Redundancy If a plan e t rus s has more exte rn al supports than are necessa ry to ensure 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 to pr event collapse whe n t he truss is removed from its supports, then the extra members constitute internal redundancy and the truss is again statically indet erminat e. For a truss which is stat ically determinate externally, there is a definit e relation between th e number of its members a nd th e number of its joints necessary for inte rnal sta bility without redundan cy. Becau se we ca n specify th e equilibrium of each joint by two scalar force equat ions, th ere are in all 2j such equ ations for a truss withj joints. For th e entire truss composed of m two-force mem bers and having the maximum of three unknown support reactions, there are in all m + 3 unkn owns (m tension or compression forces and three reactions). Thus, for any plan e truss, th e equ ation m + 3 = 2j will be sa tisfied if the truss is sta tically determinate internally. A simple plan e truss, formed by starting with a tri angle and add ing two new members to locate each new joint with respect to the exist ing struc ture, satis fies th e relation auto matica lly. Th e condition holds for t he initial tri angl e, where m = j = 3, and m increases by 2 for each added joint while j increases by 1. Some othe r (nonsimple) statically determinate trusses, such as th e K-truss in Fig. 4/2, are arranged differently, but can be see n to satisfy the same relation. This equa tion is a necessary condit ion for stability but it is not a sufficient conditio n, since one or more of the m members can be arran ged in such a way as not to contribute to a sta ble configu ra tion of t he ent ire truss. If m + 3 > 2j, there a re more members t ha n independent equations, a nd the tru ss is statically indeter minate intern ally wit h redundant members present. If m + 3 < 2j , there is a deficiency of internal members, and the truss is unstab le an d will collapse under load. Special Conditions We often encounter seve ral special conditions in the analys is of trusses . When two collinear members are under compression, as indicated in Fig. 4/9a, it is necessary to add a t hird member to main tain Marwa n and Waseem AI- Iraqi

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Ar t icl e 4 /J

/

h,

171

x·

X / /

M ethod of Jo int s

~ x -r - -

1/ L

'F.F;r = 0 require s F1 = }'i rF;r' = 0 require s FJ = F4

~ zr; = 0 requires F 1 = 0 r F;r' = 0 requires F2 = 0

rFy = 0 requires F3 = 0 ! Fx = 0 requires F1 = Iii (c )

(bJ

(c)

Figure 4/9

alignment of th e two members and prevent buckling. We see from a force su mmation in they-direction that the force F3 in the third member mus t be zero and from the x-direct ion that F 1 = F 2' Th is conclusion holds regardless of the a ngle 0 and holds also if th e collinear me mbers are in te nsion. If an externa l force with a component in the y-direction were a pplied to the join t, then F 3 would no longer be zero . Wh en two non ccllinear 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 forces in both members must be zero, as we can see from the two force summations. 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 conclusi on follows 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 indeter minate if eac h brace can support eit he r tension or compression . However, whe n the braces are flexible members incapable of supporting compression, as are cables, then only the tension member acts and we can disregard the other member. It is usually evident from the asymmetry of th e loading how th e pan el will deflect. If the deflection is as indica ted in Fig. 4/lOb, th en member AB should be retained a nd CD disregarded. When this choice cannot be made by inspection, we may arbitrarily se lect the member to be retained. If the assumed te nsion turns out to be positive upon calculation, then the choice was correct. If the assumed tension force tu rns out to be negat ive, t hen the opposit e member mu st be ret ained and th e calculation redon e. We can avoid simultaneous solution of t he equilibrium equat ions for two unkn own 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 a nd F1 and F2 are unknown, a force summation in the x-direction eliminates reference to F1 and a force summation in the x ' -direction eliminates reference to F 2 . When the angles in volved are not easily found , then a simultaneous solution of the equations using one set of reference direct ions for both u nknowns may be pr eferab le. Ma rwan and W aseem AI- Iraqi

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D

B

D

B

wN c

A

c

A

(b)

(a)

Figure 4/10

Figure 4/1 1

172

Cha pt e r 4

St ructures

Sample Problem 4 /1 Compute th e force in each member of the loaded cantilever truss by the method of joints .

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 t he joint at th e loaded end. However, this truss will be analyzed completely, so t he first ste p will be to compute th e exter nal forces at D and E from the free-body diagram of the truss as a whole. Th e equa tions of equilibrium give [~E = [~Fx =

OJ

[~Fy

OJ

=

5T - 20(5) - 30(10)

OJ

0

Ex = 69.3 kN

80 sin 30° + Ey - 20 - 30

0

Ey = 10 kN

80 cos 30

-

30 kN

20kN

T = 80kN

0

Ex

0

A

T y I I

Next we draw free-body diagrams showing th e forces acting on each of t he connect ing pins. Th e correc tn ess of the assigned directions of the forces is verified when each joint is considered in sequence. Th ere should be no question about th e cor rect direction of th e forces on joint A. Equilibrium requires

CD

OJ

0.866AB - 30

0]

AC - 0.5(34.6 )

o o

AB

~

34.6 kN T

Ans.

AC

~

17.32 kN C

Ans.

where T sta nds for tension and C stands for compre ssion. 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 which case BD mu st balance the force to the left . Again th e forces are obtained from

OJ

0.866BC - 0.866 (34.6)

0]

BD - 2(0 .5)(34.6)

o o

= 34.6 kN

C

Ans.

BD = 34.6 kN T

Ans.

BC

L __

OJ

0.866CD - 0.866 (34 .6) - 20 CD

OJ

~

57.7 kN T

Ans.

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

o Ans.

Finally, from joint E th er e results 0.866DE

~

10

DE

x

5m

5m

30 kN

20kN

Ey

y I\B

I I

I I 600 _ AC - - x

BD

I\B =~ OO

34.6 kN 600

BC

30kN

J oint A

Joi ntB

Helpful Hint

o

CE = 63.5 kN C

I:EFy = 0]

5m

Ex

J oint C now contai ns only two unknowns, and these are found in the same way as before : [~Fy =

30 0 60 0

5m

11.55 kN C

an d t he equation '2:.F:r: = 0 checks.

A ns.

CD It

should be stressed th a t t he te nsion/com pression designa tion refers to th e member , not the joint . Note that we dr aw th e force arrow on the same side of th e joint as the member which exer ts th e force. In thi s way ten sion (ar row away from th e joint) is disti ngu ished from compress ion (ar row toward the joint). BC =

34.6 kN

DE 60 AC= 17.32 kN

CE

Joint C www .gigapedia.com

69.3 kN

CE =

63.5 kN 20kN

Marwan and Waseem AI-Iraqi

0

10kN

Joint E

Art icle 4/3

PROBL EMS

173

Problems

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

Introductory Problems 4/1 Dete rmine the force in each member of the simple equilateral truss. An s. AB ~ 736 N T , AC ~ 368 N T , BC = 736 N C

B

2

A

8' 2m

2m

~2

C

A'

8'

8'

8'

E

'B 10001b

2m

Problem 4/4 75

k~

4/5 Determine the force in each member of the loaded

C

truss.

An. , AB = 12 kN T , AE BC = 5.20 kN T , BD BE ~ 5.20 kN C, CD ~ DE

Problem 4/ 1 4/2 Dete rmine the force in each member of the loaded truss. Discuss the e lTects of varying the angle of the 450 suppor t sur face at C. A

3 kN C 6 kN T 6 kN C

A

B

6'

B

C 3D'"

E

2.5'

1001b

3 kN

D

C

Problem 4/5 45°

- - - -

4 /6 Calculate th e force in eac h memb er of th e loaded tru ss. E "'- - - "'-D -2 kN

Problem 4/2

4/3 Dete rmine the force in each member of the truss. Note the presence of any zero-force members.

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

3m

3m B Problem 4/6

Prob lem 4/ 3

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174

Ch apter 4

Structures

4/7 Determine the force in each member of the loaded truss. Make use of the symmetry of the truss and of the loading. Ans. AB ~ DE ~ 96.0 kN C All ~ EF = 75 kN T, BC = CD = 75 kN C BH ~ CG ~ DF = 60 kN T CH ~ CF ~ 48.0 kN C, GH ~ FG = 112.5 kN T

c

B

c

D

'i---- - - ---->:D

4m

5m

5m

E

5m

G

F

60 kN

30 kN

4 kN

2 kN Problem 4/ 9

30 kN

Problem 4/7

Representative Problems

4/8 Determine the force in each member of the loaded truss. All triangles are isosceles. 1 300 1

I

lO kI'

I

4m

4/10 Solve for the forces in members BE and BD of the truss which supports the load L. All interior angles are 60° or 120°.

A

B

~

c c

4m

B

D

F

t

G

Problem 4/ 8 4/9 Determine the force in each member of the loaded truss. All triangles are equilateral. Ans. AB ~ 9,13 kN C, AE = 5 ,13 kN T kN C, BD = 3 ,13 kN C, BE = kN C BC = kN T, DE ~ kN T CD ~

,?J3

lfJ3

E

L

Problem 4/ 10

iJ3

¥J3

4/11 Determine the force in member AC of the loaded truss. The two quarter-circular members act as twoforce members . Ans. A C = T 2

!::

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Article 4 /1

Prob l em s

175

L

B

Problem 4/11

4/12 Calcul at e the force s in members CG a nd CF for th e truss shown.

2 kip s

Problem 4/14

2 kN

4 /15 Th e equiangular t russ is loaded a nd su ppo rted as shown. Determine the for ces in a ll members in terms of th e horizontal load L . Ans. AB BC = L T , AF = EF = L C DE = CD = L/2 T, BF = DF = BD = 0

F

Problem 4/12

4/13 Each member of the truss is a un iform 20·ft bar weighing 40 0 lb. Ca lcu late the average ten sion or com pr ession in each membe r du e t o the weights of

t he members.

Ans. AB AE BD ED

=

=

2000/ ,13 Ib C BE = 800/ ,13 Ib T 1400/ ,13 Ib C =

B

Problem 4/15

4 /16 Determ ine t he force s in members BI , CI, a nd HI for t he loaded truss. All a ng les are 30°, 60°, or 90°.

D

E

20'

1000/ ,/ 3 1b T

BC

= CD

20'

Problem 4/13

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

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Problem 4/16

176

Ch ap ter 4

Sto uc tur es

4/17 A snow load transfers the forces shown to the upper joints of a Pratt roof truss. Neglect any horizontal reactions at the supports and solve for the forces in all members. Ails. AB ~ DE = BC ~ CD = 3.35 kN C AH = EF = 3 kN T , BH ~ DF ~ 1 kN C CF ~ CH ~ 1.414 kN T , FG = GH = 2 kN T

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

1 kN

~c

1 kl'i

1 kN 1 1