c planetrusses

10/8/08 7:29 PM4/2. Plane Trusses (Reading content)

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4/2 Plane TrussesA framework composed of members joined at their ends to form a rigid structure is called a truss. Bridges, roof supports,derricks, and other such structures are common examples of trusses. Structural members commonly used are I-beams,channels, angles, bars, and special shapes which are fastened together at their ends by welding, riveted connections, orlarge bolts or pins. When 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 truss assembly placed on eachside of the structure. A section of a typical bridge structure is shown in Fig. 4/1. The combined weight of the roadwayand vehicles is transferred to the longitudinal stringers, then to the cross beams, and finally, with the weights of thestringers and cross beams accounted for, to the upper joints of the two plane trusses which form the vertical sides of thestructure. A simplified model of the truss structure is indicated at the left side of the illustration; the forces L representthe joint loadings.

Figure 4/1

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



Figure 4/2

Simple Trusses

The basic element of a plane truss is the triangle. Three bars joined by pins at their ends, Fig. 4/3a, constitute a rigidframe. The term rigid is used to mean noncollapsible and also to mean that deformation of the members due to inducedinternal strains is negligible. On the other hand, four or more bars pin-jointed to form a polygon of as many sidesconstitute a nonrigid frame. We can make the nonrigid frame in Fig. 4/3b rigid, or stable, by adding a diagonal barjoining A and D or B and C and thereby forming two triangles. We can extend the structure by adding additional units oftwo end-connected bars, such as DE and CE or AF and DF, Fig. 4/3c, which are pinned to two fixed joints. In this waythe entire structure will remain rigid.



Figure 4/3

Structures built from a basic triangle in the manner described are known as simple trusses. When more members arepresent than are needed to prevent collapse, the truss is statically indeterminate. A statically indeterminate truss cannot beanalyzed by the equations of equilibrium alone. Additional members or supports which are not necessary for maintainingthe equilibrium configuration are called redundant.

To design a truss we must first determine the forces in the various members and then select appropriate sizes andstructural shapes to withstand the forces. Several assumptions are made in the force analysis of simple trusses. First, weassume all members to be two-force members. A two-force member is one in equilibrium under the action of two forcesonly, as defined in general terms with Fig. 3/4 in Art. 3/3. Each member of a truss is normally a straight link joining thetwo points of application of force. The two forces are applied at the ends of the member and are necessarily equal,opposite, and collinear for equilibrium.



The member may be in tension or compression, as shown in Fig. 4/4. When we represent the equilibrium of a portion ofa two-force member, the tension T or compression C acting on the cut section is the same for all sections. We assumehere that the weight of the member is small compared with the force it supports. If it is not, or if we must account forthe small effect of the weight, we can replace the weight W of the member by two forces, each W/2 if the member isuniform, with one force acting at each end of the member. These forces, in effect, are treated as loads externally appliedto the pin connections. Accounting for the weight of a member in this way gives the correct result for the averagetension or compression along the member but will not account for the effect of bending of the member.

Figure 4/4

Truss Connections and Supports

When welded or riveted connections are used to join structural members, we may usually assume that the connection is apin joint if the centerlines of the members are concurrent at the joint as in Fig. 4/5.

Figure 4/5

We also assume in the analysis of simple trusses that all external forces are applied at the pin connections. Thiscondition is satisfied in most trusses. In bridge trusses the deck is usually laid on cross beams which are supported at the



condition is satisfied in most trusses. In bridge trusses the deck is usually laid on cross beams which are supported at thejoints, 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 andcontraction due to temperature changes and for deformation from applied loads. Trusses and frames in which no suchprovision is made are statically indeterminate, as explained in Art. 3/3. Figure 3/1 shows examples of such joints.

Two methods for the force analysis of simple trusses will be given. Each method will be explained for the simple trussshown in Fig. 4/6a. The free-body diagram of the truss as a whole is shown in Fig. 4/6b. The external reactions areusually determined first, by applying the equilibrium equations to the truss as a whole. Then the force analysis of theremainder of the truss is performed.

Figure 4/6

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.

10/8/08 7:31 PM4/3. Method of Joints (Reading content)

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4/3 Method of JointsThis method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for theforces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces,and only two independent equilibrium equations are involved.

We begin the analysis with any joint where at least one known load exists and where not more than two unknown forcesare present. The solution may be started with the pin at the left end. Its free-body diagram is shown in Fig. 4/7. With thejoints indicated by letters, we usually designate the force in each member by the two letters defining the ends of themember. The proper directions of the forces should be evident by inspection for this simple case. The free-bodydiagrams of portions of members AF and AB are also shown to clearly indicate the mechanism of the action andreaction. The member AB actually makes contact on the left side of the pin, although the force AB is drawn from theright side and is shown acting away from the pin. Thus, if we consistently draw the force arrows on the same side of thepin as the member, then tension (such as AB) will always be indicated by an arrow away from the pin, and compression(such as AF) will always be indicated by an arrow toward the pin. The magnitude of AF is obtained from the equationΣFy = 0 and AB is then found from ΣFx = 0.

Figure 4/7

Joint F may be analyzed next, since it now contains only two unknowns, EF and BF. Proceeding to the next joint havingno more than two unknowns, we subsequently analyze joints B, C, E, and D in that order. Figure 4/8 shows the free-body diagram of each joint and its corresponding force polygon, which represents graphically the two equilibriumconditions ΣFx = 0 and ΣFy = 0. The numbers indicate the order in which the joints are analyzed. We note that, whenjoint D is finally reached, the computed reaction R2 must be in equilibrium with the forces in members CD and ED,which were determined previously from the two neighboring joints. This requirement provides a check on the correctnessof our work. Note that isolation of joint C shows that the force in CE is zero when the equation ΣFy = 0 is applied. Theforce in this member would not be zero, of course, if an external vertical load were applied at C.



Figure 4/8

It is often convenient to indicate the tension T and compression C of the various members directly on the original trussdiagram by drawing arrows away from the pins for tension and toward the pins for compression. This designation isillustrated at the bottom of Fig. 4/8.

Sometimes we cannot initially assign the correct direction of one or both of the unknown forces acting on a given pin. Ifso, we may make an arbitrary assignment. A negative computed force value indicates that the initially assumed directionis incorrect.



This New York City bridge structure suggests that members of a simple truss need not be straight.© Stephen Wilkes/The Image Bank/Getty Images

Internal and External Redundancy

If a plane truss has more external supports than are necessary to ensure a stable equilibrium configuration, the truss as awhole is statically indeterminate, and the extra supports constitute external redundancy. If a truss has more internalmembers than are necessary to prevent collapse when the truss is removed from its supports, then the extra membersconstitute internal redundancy and the truss is again statically indeterminate.

For a truss which is statically determinate externally, there is a definite relation between the number of its members andthe number of its joints necessary for internal stability without redundancy. Because we can specify the equilibrium ofeach joint by two scalar force equations, there are in all 2j such equations for a truss with j joints. For the entire trusscomposed of m two-force members and having the maximum of three unknown support reactions, there are in all m + 3unknowns (m tension or compression forces and three reactions). Thus, for any plane truss, the equation m + 3 = 2j willbe satisfied if the truss is statically determinate internally.

A simple plane truss, formed by starting with a triangle and adding two new members to locate each new joint withrespect to the existing structure, satisfies the relation automatically. The condition holds for the initial triangle, where m= j = 3, and m increases by 2 for each added joint while j increases by 1. Some other (nonsimple) statically determinatetrusses, 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 condition for stability but it is not a sufficient condition, since one or more of the mmembers can be arranged in such a way as not to contribute to a stable configuration of the entire truss. If m + 3 > 2j,there are more members than independent equations, and the truss is statically indeterminate internally with redundantmembers present. If m + 3 < 2j, there is a deficiency of internal members, and the truss is unstable and will collapseunder load.



Harbour Bridge in Sydney, Australia.© Photodisc/Media Bakery

Special Conditions

We often encounter several special conditions in the analysis of trusses. When two collinear members are undercompression, as indicated in Fig. 4/9a, it is necessary to add a third member to maintain alignment of the two membersand prevent buckling. We see from a force summation in the y-direction that the force F3 in the third member must bezero and from the x-direction that F1 = F2. This conclusion holds regardless of the angle θ and holds also if the collinearmembers are in tension. If an external force with a component in the y-direction were applied to the joint, then F3 wouldno longer be zero.

Figure 4/9

When two noncollinear members are joined as shown in Fig. 4/9b, then in the absence of an externally applied load atthis joint, the 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 equal andopposite. This conclusion follows from the force summations indicated in the figure.

Truss panels are frequently cross-braced as shown in Fig. 4/10a. Such a panel is statically indeterminate if each bracecan support either tension or compression. However, when the braces are flexible members incapable of supportingcompression, as are cables, then only the tension member acts and we can disregard the other member. It is usuallyevident from the asymmetry of the loading how the panel will deflect. If the deflection is as indicated in Fig. 4/10b, thenmember AB should be retained and CD disregarded. When this choice cannot be made by inspection, we may arbitrarilyselect the member to be retained. If the assumed tension turns out to be positive upon calculation, then the choice wascorrect. If the assumed tension force turns out to be negative, then the opposite member must be retained and thecalculation redone.

Figure 4/10

We can avoid simultaneous solution of the equilibrium equations for two unknown forces at a joint by a careful choiceof reference axes. Thus, for the joint indicated schematically in Fig. 4/11 where L is known and F1 and F2 are unknown,a force summation in the x-direction eliminates reference to F1 and a force summation in the x′-direction eliminatesreference to F2. When the angles involved are not easily found, then a simultaneous solution of the equations using oneset of reference directions for both unknowns may be preferable.

Figure 4/11

Sample Problem 4/1

Compute the force in each member of the loaded cantilever truss by the method of joints.



Helpful Hints

It should be stressed that the tension/compression designation refers to the member, not the joint.Note that we draw the force arrow on the same side of the joint as the member which exerts the force.In this way tension (arrow away from the joint) is distinguished from compression (arrow toward thejoint).

Solution.

If it were not desired to calculate the external reactions at D and E, the analysis for a cantilever truss couldbegin with the joint at the loaded end. However, this truss will be analyzed completely, so the first step will beto compute the external forces at D and E from the free-body diagram of the truss as a whole. The equations ofequilibrium give

Next we draw free-body diagrams showing the forces acting on each of the connecting pins. The correctness ofthe assigned directions of the forces is verified when each joint is considered in sequence. There should be noquestion about the correct direction of the forces on joint A. Equilibrium requires

where T stands for tension and C stands for compression.

Joint B must be analyzed next, since there are more than two unknown forces on joint C. The force BC mustprovide an upward component, in which case BD must balance the force to the left. Again the forces areobtained from



Joint C now contains only two unknowns, and these are found in the same way as before:

Finally, from joint E there results

and the equation ΣFx = 0 checks.

Sample Problem 4/2

The simple truss shown supports the two loads, each of magnitude L. Determine the forces in members DE, DF,DG, and CD.



Helpful Hints

Rather than calculate and use the angle β = 78.8° in the force equations, we could have used the11.25° angle directly.

Solution.

First of all, we note that the curved members of this simple truss are all two-force members, so that the effect ofeach curved member within the truss is the same as that of a straight member.

We can begin with joint E because there are only two unknown member forces acting there. With reference tothe free-body diagram and accompanying geometry for joint E, we note that β = 180° − 11.25° − 90° = 78.8°.

We must now move to joint F, as there are still three unknown members at joint D. From the geometricdiagram,

From the free-body diagram of joint F,

Simultaneous solution of these two equations yields

For member DG, we move to the free-body diagram of joint D and the accompanying geometry.

Then from joint D:



The simultaneous solution is

Note that ε is shown exaggerated in the accompanying figures.

Additional Sample Problem - Method of Joints

Problems

Introductory Problems

4/1. Determine the force in each member of the loaded truss. Explain why knowledge of the lengths of themembers is unnecessary.

Problem 4/1

Answer:

AB = 2000 lb C, AC = 1732 lb T

BC = 3460 lb C4/2. Determine the force in each member of the loaded truss. Identify any zero-force members by inspection.



Problem 4/2

4/3. Determine the force in each member of the loaded truss.

Problem 4/3

Answer:

AB = 3000 N T, AC = 4240 N C, CD = 4240 N T

AD = 3000 N C, BC = 6000 N T4/4. Determine the force in each member of the loaded truss.

Problem 4/4


Problem 4/5



Answer:

AB = 5.63 kN C, AF = 3.38 kN T

BC = 4.13 kN C, BE = 0.901 kN T

BF = 4 kN T, CD = 6.88 kN C

CE = 5.50 kN T, DE = 4.13 kN T

EF = 3.38 kN T4/6. Calculate the force in each member of the loaded truss. All triangles are isosceles.

Problem 4/6

4/7. Determine the force in member AC of the loaded truss. The two quarter-circular members act as two-force members.

Problem 4/7

Answer:

4/8. Determine the force in each member of the loaded truss. Make use of the symmetry of the truss and ofthe loading.

Problem 4/8




Problem 4/9

Answer:

AB = 14.42 kN T, AC = 2.07 kN C, AD = 0

BC = 6.45 kN T, BD = 12.89 kN C4/10. Determine the forces in members BE and CE of the loaded truss.

Problem 4/10

Representing Problems

4/11. Calculate the forces in members CG and CF for the truss shown.

Problem 4/11

Answer:



CG = 2.24 kN T, CF = 1 kN C4/12. Each member of the truss is a uniform 20-ft bar weighing 400 lb. Calculate the average tension or

compression in each member due to the weights of the members.

Problem 4/12

4/13. A drawbridge is being raised by a cable EI. The four joint loadings shown result from the weight of theroadway. Determine the forces in members EF, DE, DF, CD, and FG.

Problem 4/13

Answer:

EF = 3.87 kips C, DE = 4.61 kips T

DF = 4.37 kips C, CD = 2.73 kips T

FG = 7.28 kips C4/14. The truss is composed of equilateral triangles of sides a and is loaded and supported as shown.

Determine the forces in members EF, DE, and DF.



Problem 4/14

4/15. Determine the forces in members BC and BG of the loaded truss.

Problem 4/15

Answer:

BC = 3.46 kN C, BG = 1.528 kN T4/16. Determine the forces in members BI, CI, and HI for the loaded truss. All angles are 30°, 60°, or 90°.

Problem 4/16

4/17. Determine the forces in members AF, BE, BF, and CE of the loaded truss.

Problem 4/17



Answer:

AF = 6.13 kN T, BE = 5.59 kN T

BF = 6.50 kN C, CE = 5 kN C4/18. The signboard truss is designed to support a horizontal wind load of 800 lb. A separate analysis shows

that of this force is transmitted to the center connection at C and the rest is equally divided between D

and B. Calculate the forces in members BE and BC.

Problem 4/18

4/19. A snow load transfers the forces shown to the upper joints of a Pratt roof truss. Neglect any horizontalreactions at the supports and solve for the forces in all members.

Problem 4/19

Answer:

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 T4/20. The loading of Prob. 4/19 is shown applied to a Howe roof truss. Neglect any horizontal reactions at the

supports and solve for the forces in all members. Compare with the results of Prob. 4/19.

Problem 4/20



4/21. Determine the force in each member of the pair of trusses which support the 5000-lb load at theircommon joint C.

Problem 4/21

Answer:

AB = BC = BG = 0, AG = CG = 2890 lb C4/22. Determine the force in member BF of the loaded truss.

Problem 4/22

4/23. The rectangular frame is composed of four perimeter two-force members and two cables AC and BDwhich are incapable of supporting compression. Determine the forces in all members due to the load L inposition (a) and then in position (b).



Problem 4/23

Answer:

a.

b.

4/24. Verify the fact that each of the trusses contains one or more elements of redundancy and propose twoseparate changes, either one of which would remove the redundancy and produce complete staticaldeterminacy. All members can support compression as well as tension.

Problem 4/24

4/25. Determine the forces in members AC and AD of the loaded truss.

Problem 4/25

Answer:

AC = AD = 779 lb C4/26. Analysis of the wind acting on a small Hawaiian church, which withstood the 165-mi/hr winds of

Hurricane Iniki in 1992, showed the forces transmitted to each roof truss panel to be as shown. Treat thestructure as a symmetrical simple truss and neglect any horizontal component of the support reaction atA. Identify the truss member which supports the largest force, tension or compression, and calculate thisforce.



Problem 4/26

4/27. The tower for a transmission line is modeled by the truss shown. The crossed members in the centersections of the truss may be assumed to be capable of supporting tension only. For the loads of 1.8 kNapplied in the vertical plane, compute the forces induced in members AB, DB, and CD.

Problem 4/27

Answer:

AB = 3.89 kN C, DB = 0, CD = 0.932 kN C4/28. Determine the force in member CM of the loaded truss.

Problem 4/28



Answer:

CM = 3.41L T


10/8/08 7:48 PM4/4. Method of Sections (Reading content)


4/4 Method of SectionsWhen analyzing plane trusses by the method of joints, we need only two of the three equilibrium equations because theprocedures involve concurrent forces at each joint. We can take advantage of the third or moment equation ofequilibrium by selecting an entire section of the truss for the free body in equilibrium under the action of a nonconcurrentsystem of forces. This method of sections has the basic advantage that the force in almost any desired member may befound directly from an analysis of a section which has cut that member. Thus, it is not necessary to proceed with thecalculation from joint to joint until the member in question has been reached. In choosing a section of the truss, we notethat, in general, not more than three members whose forces are unknown should be cut, since there are only threeavailable independent equilibrium relations.

Illustration of the Method

The method of sections will now be illustrated for the truss in Fig. 4/6, which was used in the explanation of the methodof joints. The truss is shown again in Fig. 4/12a for ready reference. The external reactions are first computed as with themethod of joints, by considering the truss as a whole.

Figure 4/12



Let us determine the force in the member BE, for example. An imaginary section, indicated by the dashed line, is passedthrough the truss, cutting it into two parts, Fig. 4/12b. This section has cut three members whose forces are initiallyunknown. In order for the portion of the truss on each side of the section to remain in equilibrium, it is necessary toapply to each cut member the force which was exerted on it by the member cut away. For simple trusses composed ofstraight two-force members, these forces, either tensile or compressive, will always be in the directions of the respectivemembers. The left-hand section is in equilibrium under the action of the applied load L, the end reaction R1, and thethree forces exerted on the cut members by the right-hand section which has been removed.

We can usually draw the forces with their proper senses by a visual approximation of the equilibrium requirements. Thus,in balancing the moments about point B for the left-hand section, the force EF is clearly to the left, which makes itcompressive, because it acts toward the cut section of member EF. The load L is greater than the reaction R1, so that theforce BE must be up and to the right to supply the needed upward component for vertical equilibrium. Force BE istherefore tensile, since it acts away from the cut section.

With the approximate magnitudes of R1 and L in mind we see that the balance of moments about point E requires thatBC be to the right. A casual glance at the truss should lead to the same conclusion when it is realized that the lowerhorizontal member will stretch under the tension caused by bending. The equation of moments about joint B eliminatesthree forces from the relation, and EF can be determined directly. The force BE is calculated from the equilibriumequation for the y-direction. Finally, we determine BC by balancing moments about point E. In this way each of thethree unknowns has been determined independently of the other two.

The right-hand section of the truss, Fig. 4/12b, is in equilibrium under the action of R2 and the same three forces in thecut members applied in the directions opposite to those for the left section. The proper sense for the horizontal forcescan easily be seen from the balance of moments about points B and E.

Additional Considerations

It is essential to understand that in the method of sections an entire portion of the truss is considered a single body inequilibrium. Thus, the forces in members internal to the section are not involved in the analysis of the section as a whole.To clarify the free body and the forces acting externally on it, the cutting section is preferably passed through themembers and not the joints. We may use either portion of a truss for the calculations, but the one involving the smallernumber of forces will usually yield the simpler solution.

In some cases the methods of sections and joints can be combined for an efficient solution. For example, suppose wewish to find the force in a central member of a large truss. Furthermore, suppose that it is not possible to pass a sectionthrough this member without passing through at least four unknown members. It may be possible to determine the forcesin nearby members by the method of sections and then progress to the unknown member by the method of joints. Such acombination of the two methods may be more expedient than exclusive use of either method.

The moment equations are used to great advantage in the method of sections. One should choose a moment center, eitheron or off the section, through which as many unknown forces as possible pass.

It is not always possible to assign the proper sense of an unknown force when the free-body diagram of a section isinitially drawn. Once an arbitrary assignment is made, a positive answer will verify the assumed sense and a negativeresult will indicate that the force is in the sense opposite to that assumed. An alternative notation preferred by some is toassign all unknown forces arbitrarily as positive in the tension direction (away from the section) and let the algebraicsign of the answer distinguish between tension and compression. Thus, a plus sign would signify tension and a minussign compression. On the other hand, the advantage of assigning forces in their correct sense on the free-body diagram ofa section wherever possible is that doing so emphasizes the physical action of the forces more directly, and this practiceis the one which is preferred here.



Many simple trusses are periodic in that there are repeated and identical structural sections.© Photodisc/Media Bakery

Sample Problem 4/3

Calculate the forces induced in members KL, CL, and CB by the 20-ton load on the cantilever truss.

Helpful Hints

We note that analysis by the method of joints would necessitate working with eight joints in order tocalculate the three forces in question. Thus, the method of sections offers a considerable advantage inthis case.We could have started with moments about C or P just as well.



We could have started with moments about C or P just as well.

We could also have determined CL by a force summation in either the x- or y-direction.

Solution.

Although the vertical components of the reactions at A and M are statically indeterminate with the two fixedsupports, all members other than AM are statically determinate. We may pass a section directly through membersKL, CL, and CB and analyze the portion of the truss to the left of this section as a statically determinate rigidbody.

The free-body diagram of the portion of the truss to the left of the section is shown. A moment sum about Lquickly verifies the assignment of CB as compression, and a moment sum about C quickly discloses that KL is intension. The direction of CL is not quite so obvious until we observe that KL and CB intersect at a point P to theright of G. A moment sum about P eliminates reference to KL and CB and shows that CL must be compressiveto balance the moment of the 20-ton force about P. With these considerations in mind the solution becomesstraightforward, as we now see how to solve for each of the three unknowns independently of the other two.

Summing moments about L requires finding the moment arm . Thus,

Next we take moments about C, which requires a calculation of cos θ. From the given dimensions we see θ =tan−1(5/12) so that cos θ = 12/13. Therefore,

Finally, we may find CL by a moment sum about P, whose distance from C is given by or . We also need β, which is given by

and cos β = 0.868. We now have

Sample Problem 4/4



Calculate the force in member DJ of the Howe roof truss illustrated. Neglect any horizontal components of forceat the supports.

Helpful Hints

There is no harm in assigning one or more of the forces in the wrong direction, as long as thecalculations are consistent with the assumption. A negative answer will show the need for reversingthe direction of the force.

If desired, the direction of CD may be changed on the free-body diagram and the algebraic sign ofCD reversed in the calculations, or else the work may be left as it stands with a note stating theproper direction.

Observe that a section through members CD, DJ, and DE could be taken which would cut only threeunknown members. However, since the forces in these three members are all concurrent at D, amoment equation about D would yield no information about them. The remaining two force equationswould not be sufficient to solve for the three unknowns.

Solution.

It is not possible to pass a section through DJ without cutting four members whose forces are unknown.Although three of these cut by section 2 are concurrent at J and therefore the moment equation about J could beused to obtain DE, the force in DJ cannot be obtained from the remaining two equilibrium principles. It isnecessary to consider first the adjacent section 1 before analyzing section 2.

The free-body diagram for section 1 is drawn and includes the reaction of 18.33 kN at A, which ispreviously calculated from the equilibrium of the truss as a whole. In assigning the proper directions for theforces acting on the three cut members, we see that a balance of moments about A eliminates the effects of CDand JK and clearly requires that CJ be up and to the left. A balance of moments about C eliminates the effect ofthe three forces concurrent at C and indicates that JK must be to the right to supply sufficient counterclockwisemoment. Again it should be fairly obvious that the lower chord is under tension because of the bending tendencyof the truss. Although it should also be apparent that the top chord is under compression, for purposes ofillustration the force in CD will be arbitrarily assigned as tension.



illustration the force in CD will be arbitrarily assigned as tension.

By the analysis of section 1, CJ is obtained from

In this equation the moment of CJ is calculated by considering its horizontal and vertical components acting atpoint J. Equilibrium of moments about J requires

The moment of CD about J is calculated here by considering its two components as acting through D. Theminus sign indicates that CD was assigned in the wrong direction.

Hence,

From the free-body diagram of section 2, which now includes the known value of CJ, a balance of momentsabout G is seen to eliminate DE and JK. Thus,

Again the moment of CJ is determined from its components considered to be acting at J. The answer for DJ ispositive, so that the assumed tensile direction is correct.

An alternative approach to the entire problem is to utilize section 1 to determine CD and then use the method ofjoints applied at D to determine DJ.

Additional Sample Problem - Method of Sections

Problems

Introductory Problems

4/29. Determine the forces in members CG and GH.



Problem 4/29

Answer:

CG = 0, GH = 27 kN T4/30. Determine the force in member AE of the loaded truss.

Problem 4/30

4/31. Determine the force in member BC of the loaded truss.

Problem 4/31

Answer:

BC = 4.81 kips T4/32. Determine the forces in members GH and CG for the truss loaded and supported as shown. Does the

statical indeterminacy of the supports affect your calculation?



Problem 4/32

4/33. Determine the force in member DG of the loaded truss.

Problem 4/33

Answer:

DG = L T4/34. Determine the force in member BE of the loaded truss.

Problem 4/34

Representative Problems



4/35. Determine the forces in members DE and DL.

Problem 4/35

Answer:

DE = 24 kN T, DL = 33.9 kN C4/36. Calculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation

which contains that force as the only unknown.

Problem 4/36

4/37. Calculate the forces in members BC, CD, and CG of the loaded truss composed of equilateral triangles,each of side length 8 m.

Problem 4/37



Answer:

BC = 1.155 kN T, CD = 5.20 kN T

CG = 4.04 kN C4/38. Determine the forces in members BC and FG of the loaded symmetrical truss. Show that this calculation

can be accomplished by using one section and two equations, each of which contains only one of the twounknowns. Are the results affected by the statical indeterminacy of the supports at the base?

Problem 4/38

4/39. The truss shown is composed of 45° right triangles. The crossed members in the center two panels areslender tie rods incapable of supporting compression. Retain the two rods which are under tension andcompute the magnitudes of their tensions. Also find the force in member MN.

Problem 4/39

Answer:

FN = GM = 84.8 kN T, MN = 20 kN T4/40. Determine the force in member BF.



Problem 4/40

4/41. Determine the forces in members CD, CJ, and DJ.

Problem 4/41

Answer:

CD = 0.562L C, CJ = 1.562L T

DJ = 1.250L C4/42. Compute the force in member HN of the loaded truss.



Problem 4/42

4/43. Determine the forces in members DE, DL, LM, and EL of the loaded symmetrical truss.

Problem 4/43

Answer:

DE = 4.80L C, DL = 0.0446L T, LM = 4.54L T

EL = 3.80L T4/44. Determine the forces in members DQ and CQ of the loaded symmetrical truss.

Problem 4/44

4/45. Calculate the forces in members CB, CG, and FG for the loaded truss without first calculating the forcein any other member.



Problem 4/45

Answer:

CB = 56.2 kN C, CG = 13.87 kN T

FG = 19.62 kN T4/46. The hinged frames ACE and DFB are connected by two hinged bars, AB and CD, which cross without

being connected. Compute the force in AB.

Problem 4/46

4/47. Determine the force in member JM of the loaded truss.

Problem 4/47



Answer:

JM = 0.0901L T4/48. Determine the forces in members DE, EI, FI, and HI of the arched roof truss.

Problem 4/48

4/49. Determine the force in member GK of the loaded symmetrical truss.

Problem 4/49

Answer:

GK = 2.13L T4/50. Determine the force in member CL of the loaded truss. The radius of curvature of the upper chord

BCDEFG is 30 m.



Problem 4/50

4/51. Determine the force in member DK of the loaded overhead sign truss.

Problem 4/51

Answer:

DK = 1 kip T4/52. Determine the force in member DG of the compound truss. The joints all lie on radial lines subtending

angles of 15° as indicated, and the curved members act as two-force members. Distance .

Problem 4/52



Answer:

DG = 0.569L C4/53. Determine the force in member CK of the loaded truss.

Problem 4/53

Answer:

CK = 9290 N C4/54. A design model for a transmission-line tower is shown in the figure. Members GH, FG, OP, and NO

are insulated cables; all other members are steel bars. For the loading shown, compute the forces inmembers FI, FJ, EJ, EK, and ER. Use a combination of methods if desired.

Problem 4/54



Answer:

FI = ER = 0, FJ = 7.81 kN T

EJ = 3.61 kN C, EK = 22.4 kN C


Problem 4/2

Determine the force in members: (a) AD, (b) CD, (c) AB, (d) BC, (e) BD of the loaded truss. (Note: the compression force must be negative.)

Determine the force in members: (a) BC, (b) CD, (c) AD, (d) BD, (e) AB of the loaded truss. (Note: the compression force must be negative.)

Sample Problem 4/1

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

Solution. If it were not desired to calculate the external reactions at D and E,the analysis for a cantilever truss could begin with the joint at the loaded end.However, this truss will be analyzed completely, so the first step will be to com-pute the external forces at D and E from the free-body diagram of the truss as awhole. The equations of equilibrium give

Next we draw free-body diagrams showing the forces acting on each of theconnecting pins. The correctness of the assigned directions of the forces is veri-fied when each joint is considered in sequence. There should be no questionabout the correct direction of the forces on joint A. Equilibrium requires

Ans.

Ans.

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

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

Ans.

Ans.

Joint C now contains only two unknowns, and these are found in the sameway as before:

Ans.

Ans.

Finally, from joint E there results

Ans.

and the equation ΣFx � 0 checks.

DE � 11.55 kN C0.866DE � 10[ΣFy � 0]

CE � 63.5 kN C

CE � 17.32 � 0.5(34.6) � 0.5(57.7) � 0[ΣFx � 0]

CD � 57.7 kN T

0.866CD � 0.866(34.6) � 20 � 0[ΣFy � 0]

BD � 34.6 kN T BD � 2(0.5)(34.6) � 0[ΣFx � 0]

BC � 34.6 kN C 0.866BC � 0.866(34.6) � 0[ΣFy � 0]

AC � 17.32 kN C AC � 0.5(34.6) � 0[ΣFx � 0]

AB � 34.6 kN T 0.866AB � 30 � 0[ΣFy � 0]

Ey � 10 kN 80 sin 30� � Ey � 20 � 30 � 0[ΣFy � 0]

Ex � 69.3 kN 80 cos 30� � Ex � 0[ΣFx � 0]

T � 80 kN 5T � 20(5) � 30(10) � 0[ΣME � 0]

180 Chapter 4 Structures

5 m

30 kN 20 kN

5 m

5 m 5 m 5 m 5 m

5 mC E

B D

A

y

x

T

Ex

Ey

30°60°

30 kN 20 kN

5 m

5 m

5 m 5 m

30 kN

AC

BC

BDAB =

34.6 kN

AB

Joint A Joint B

x

y

60°

60°60°

20 kN 10 kN

DECD

CE

BC =34.6 kN

AC =17.32 kN

CE =63.5 kN

69.3 kN

Joint C Joint E

60°60° 60°

Helpful Hint

� It should be stressed that the ten-sion/compression designation refersto the member, not the joint. Notethat we draw the force arrow on thesame side of the joint as the memberwhich exerts the force. In this waytension (arrow away from the joint)is distinguished from compression(arrow toward the joint).

�

c04.qxd 1/26/06 1:25 PM Page 180

Sample Problem 4/2

The simple truss shown supports the two loads, each of magnitude L. Deter-mine the forces in members DE, DF, DG, and CD.

Solution. First of all, we note that the curved members of this simple trussare all two-force members, so that the effect of each curved member within thetruss is the same as that of a straight member.

We can begin with joint E because there are only two unknown memberforces acting there. With reference to the free-body diagram and accompanyinggeometry for joint E, we note that � � 180� � 11.25� � 90� � 78.8�.

Ans.

We must now move to joint F, as there are still three unknown members at jointD. From the geometric diagram,

From the free-body diagram of joint F,

Simultaneous solution of these two equations yields

Ans.

For member DG, we move to the free-body diagram of joint D and the accompa-nying geometry.

Then from joint D:

The simultaneous solution is

Ans.

Note that � is shown exaggerated in the accompanying figures.

CD � 1.617L T DG � �1.147L or DG � 1.147L C

[ΣFy � 0] �DG sin 2.92� � CD cos 33.8� � 0.601L cos 47.9� � 1.020L sin 78.8� � 0

[ΣFx � 0] �DG cos 2.92� � CD sin 33.8� � 0.601L sin 47.9� � 1.020L cos 78.8� � 0

� � tan�1�2R sin 22.5� � R sin 45�

2R cos 22.5� � R cos 45�� 2.92�

� � tan�1�2R cos 22.5� � 2R cos 45�

2R sin 45� � 2R sin 22.5�� 33.8�

GF � 0.646L T DF � 0.601L T

GF sin 67.5� � DF sin 42.1� � L � 0[ΣFy � 0]

�GF cos 67.5� � DF cos 42.1� � 0.1989L � 0[ΣFx � 0]

� � tan�1� 2R sin 22.5�

2R cos 22.5� � R� � 42.1�

EF � 0.1989L C EF � DE cos 78.8� � 0[ΣFx � 0]

DE � 1.020L T DE sin 78.8� � L � 0[ΣFy � 0]

Art ic le 4/3 Method of Jo ints 181

A

H

F

L

2R

R

L

O E

D

C

B

G

= 22.5°

α ααα

α

11.25°

11.25°β

β

L

Joint E

DE

OD = OE = 2R

EF E

D

O

y

x

11.25°

22.5°67.5°

L

0.1989L

Joint F

DFGF

OF = FE = R

2R

EF

D

O

γγ

22.5°47.9°

Joint D

hor.

DE = 1.020LDF = 0.601L

DG

CD

OG = CG = R

2R

D

C

G

O

β

δδ

22.5°

��

Helpful Hint

� Rather than calculate and use theangle � � 78.8� in the force equa-tions, we could have used the 11.25�

angle directly.

�

c04.qxd 1/26/06 1:25 PM Page 181

Sample Problem 4/3

Calculate the forces induced in members KL, CL, and CB by the 20-ton loadon the cantilever truss.

Solution. Although the vertical components of the reactions at A and M arestatically indeterminate with the two fixed supports, all members other than AMare statically determinate. We may pass a section directly through members KL,CL, and CB and analyze 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 section isshown. A moment sum about L quickly verifies the assignment of CB as com-pression, and a moment sum about C quickly discloses that KL is in tension. Thedirection of CL is not quite so obvious until we observe that KL and CB intersectat a point P to the right of G. A moment sum about P eliminates reference to KLand CB and shows that CL must be compressive to balance the moment of the20-ton force about P. With these considerations in mind the solution becomesstraightforward, as we now see how to solve for each of the three unknowns in-dependently of the other two.

Summing moments about L requires finding the moment arm � 16 �(26 � 16)/2 � 21 ft. Thus,

Ans.

Next we take moments about C, which requires a calculation of cos �. From thegiven dimensions we see � � tan�1(5/12) so that cos � � 12/13. Therefore,

Ans.

Finally, we may find CL by a moment sum about P, whose distance from Cis given by � 24/(26 � 16) or � 38.4 ft. We also need �, which is givenby � � � tan�1(12/21) � 29.7� and cos � � 0.868. We now have

Ans. CL � 5.76 tons C

20(48 � 38.4) � CL(0.868)(38.4) � 0[ΣMp � 0]

tan�1(CB/BL)PCPC/16

KL � 65 tons T20(4)(12) � 1213

KL(16) � 0[ΣMC � 0]

CB � 57.1 tons C20(5)(12) � CB(21) � 0[ΣML � 0]

BL


26′16′

20 tons

F

H I J KL

M

G E D C B A6 panels at 12′

20 tons

y

xP

GC

θ

β

CB

CL

LKL

Helpful Hints

� We note that analysis by the methodof joints would necessitate workingwith eight joints in order to calcu-late the three forces in question.Thus, the method of sections offers aconsiderable advantage in this case.

� We could have started with mo-ments about C or P just as well.

� We could also have determined CLby a force summation in either thex- or y-direction.

�

�

�

c04.qxd 1/26/06 1:25 PM Page 190

Sample Problem 4/4

Calculate the force in member DJ of the Howe roof truss illustrated. Ne-glect any horizontal components of force at the supports.

Solution. It is not possible to pass a section through DJ without cutting fourmembers whose forces are unknown. Although three of these cut by section 2 areconcurrent at J and therefore the moment equation about J could be used to ob-tain DE, the force in DJ cannot be obtained from the remaining two equilibriumprinciples. It is necessary to consider first the adjacent section 1 before analyzingsection 2.

The free-body diagram for section 1 is drawn and includes the reaction of18.33 kN at A, which is previously calculated from the equilibrium of the trussas a whole. In assigning the proper directions for the forces acting on the threecut members, we see that a balance of moments about A eliminates the effects ofCD and JK and clearly requires that CJ be up and to the left. A balance of mo-ments about C eliminates the effect of the three forces concurrent at C and indi-cates that JK must be to the right to supply sufficient counterclockwise moment.Again it should be fairly obvious that the lower chord is under tension because ofthe bending tendency of the truss. Although it should also be apparent that thetop chord is under compression, for purposes of illustration the force in CD willbe arbitrarily assigned as tension.

By the analysis of section 1, CJ is obtained from

In this equation the moment of CJ is calculated by considering its horizontal andvertical components acting at point J. Equilibrium of moments about J requires

The moment of CD about J is calculated here by considering its two componentsas acting through D. The minus sign indicates that CD was assigned in thewrong direction.

Hence,

From the free-body diagram of section 2, which now includes the knownvalue of CJ, a balance of moments about G is seen to eliminate DE and JK.Thus,

Ans.

Again the moment of CJ is determined from its components considered to be act-ing at J. The answer for DJ is positive, so that the assumed tensile direction iscorrect.

An alternative approach to the entire problem is to utilize section 1 to deter-mine CD and then use the method of joints applied at D to determine DJ.

DJ � 16.67 kN T

12DJ � 10(16) � 10(20) � 18.33(24) � 14.14(0.707)(12) � 0[ΣMG � 0]

CD � 18.63 kN C

CD � �18.63 kN

0.894CD(6) � 18.33(12) � 10(4) � 10(8) � 0[ΣMJ � 0]

CJ � 14.14 kN C0.707CJ(12) � 10(4) � 10(8) � 0[ΣMA � 0]

Art ic le 4/4 Method of Sect ions 191

6 panels at 4 m

A G

F

E

D

B

C

L K J I H10 kN

10 kN

10 kN 1 2

6 m

10 kN

18.33 kN

10 kN

Section 1

A

CJ

JKJ

CDC

A GJJK

DJ

DE10 kN

10 kN

Section 2

14.14 kN

18.33 kN

� Observe that a section through mem-bers CD, DJ, and DE could be takenwhich would cut only three unknownmembers. However, since the forcesin these three members are all con-current at D, a moment equationabout D would yield no informationabout them. The remaining two forceequations would not be sufficient tosolve for the three unknowns.

Helpful Hints

� There is no harm in assigning one ormore of the forces in the wrong di-rection, as long as the calculationsare consistent with the assumption.A negative answer will show theneed for reversing the direction ofthe force.

� If desired, the direction of CD maybe changed on the free-body diagramand the algebraic sign of CD re-versed in the calculations, or else thework may be left as it stands with anote stating the proper direction.

�

�

�

c04.qxd 1/26/06 1:25 PM Page 191

Sample Problem 4/5

The space truss consists of the rigid tetrahedron ABCD anchored by a ball-and-socket connection at A and prevented from any rotation about the x-, y-, orz-axes by the respective links 1, 2, and 3. The load L is applied to joint E, whichis rigidly fixed to the tetrahedron by the three additional links. Solve for theforces in the members at joint E and indicate the procedure for the determina-tion of the forces in the remaining members of the truss.

Solution. We note first that the truss is supported with six properly placedconstraints, which are the three at A and the links 1, 2, and 3. Also, with m � 9members and j � 5 joints, the condition m � 6 � 3j for a sufficiency of membersto provide a noncollapsible structure is satisfied.

The external reactions at A, B, and D can be calculated easily as a first step,although their values will be determined from the solution of all forces on eachof the joints in succession.

We start with a joint on which at least one known force and not more thanthree unknown forces act, which in this case is joint E. The free-body diagram ofjoint E is shown with all force vectors arbitrarily assumed in their positive ten-sion directions (away from the joint). The vector expressions for the three un-known forces are

Equilibrium of joint E requires

Rearranging terms gives

Equating the coefficients of the i-, j-, and k-unit vectors to zero gives the threeequations

Solving the equations gives us

Ans.

Thus, we conclude that FEB and FEC are compressive forces and FED is tension.Unless we have computed the external reactions first, we must next analyze

joint C with the known value of FEC and the three unknowns FCB, FCA, and FCD.The procedure is identical with that used for joint E. Joints B, D, and A are thenanalyzed in the same way and in that order, which limits the scalar unknowns tothree for each joint. The external reactions computed from these analyses must,of course, agree with the values which can be determined initially from an analy-sis of the truss as a whole.

FEB � �L/�2 FEC � �5L/6 FED � 5L/6

FEC � FED � 0FEB

�2 �

3FEC

5 � �L

FEB

�2 �

3FED

5 � 0

��L � FEB

�2 �

3FEC

5 �i � ��FEB

�2 �

3FED

5 � j � ��4FEC

5 �

4FED

5 �k � 0

�Li � FEB

�2 (�i � j) �

FEC

5 (�3i � 4k) �

FED

5 (�3j � 4k) � 0

[ΣF � 0] L � FEB � FEC � FED � 0 or

FEB � FEB

�2 (�i � j), FEC �

FEC

5 (�3i � 4k), FED �

FED

5 (�3j � 4k)

Art ic le 4/5 Space Trusses 199

x

y

A

E

C

D

LB

2

1

4 m

3 m

4 m

3

z

3 m

x

yA

E

FEDFEC

FEB

LB

4 m

3 m3 m

4 m

z

Helpful Hints

� Suggestion: Draw a free-body dia-gram of the truss as a whole and ver-ify that the external forces acting onthe truss are Ax � Li, Ay � Lj, Az �

(4L/3)k, By � 0, Dy � �Lj, Dz �

�(4L/3)k.

� With this assumption, a negativenumerical value for a force indicatescompression.

�

�

c04.qxd 1/26/06 1:25 PM Page 199

Sample Problem 4/6

The frame supports the 400-kg load in the manner shown. Neglect theweights of the members compared with the forces induced by the load and com-pute the horizontal and vertical components of all forces acting on each of themembers.

Solution. We observe first that the three supporting members which consti-tute the frame form a rigid assembly that can be analyzed as a single unit. Wealso observe that the arrangement of the external supports makes the frame sta-tically determinate.

From the free-body diagram of the entire frame we determine the externalreactions. Thus,

Next we dismember the frame and draw a separate free-body diagram ofeach member. The diagrams are arranged in their approximate relative positionsto aid in keeping track of the common forces of interaction. The external reac-tions just obtained are entered onto the diagram for AD. Other known forces arethe 3.92-kN forces exerted by the shaft of the pulley on the member BF, as ob-tained from the free-body diagram of the pulley. The cable tension of 3.92 kN isalso shown acting on AD at its attachment point.

Next, the components of all unknown forces are shown on the diagrams.Here we observe that CE is a two-force member. The force components on CEhave equal and opposite reactions, which are shown on BF at E and on AD at C.We may not recognize the actual sense of the components at B at first glance, sothey may be arbitrarily but consistently assigned.

The solution may proceed by use of a moment equation about B or E formember BF, followed by the two force equations. Thus,

Ans.

Ans.

Ans.

Positive numerical values of the unknowns mean that we assumed their direc-tions correctly on the free-body diagrams. The value of Cx � Ex � 13.08 kN ob-tained by inspection of the free-body diagram of CE is now entered onto thediagram for AD, along with the values of Bx and By just determined. The equa-tions of equilibrium may now be applied to member AD as a check, since all theforces acting on it have already been computed. The equations give

�13.08/2 � 2.62 � 3.92 � 0[ΣFy � 0]

4.32 � 13.08 � 9.15 � 3.92 � 4.32 � 0[ΣFx � 0]

4.32(3.5) � 4.32(1.5) � 3.92(2) � 9.15(1.5) � 0[ΣMC � 0]

Bx � 9.15 kNBx � 3.92 � 13.08 � 0[ΣFx � 0]

By � 2.62 kNBy � 3.92 � 13.08/2 � 0[ΣFy � 0]

Ex � 13.08 kN3.92(5) � 12Ex(3) � 0[ΣMB � 0]

Ay � 3.92 kNAy � 3.92 � 0[ΣFy � 0]

Ax � 4.32 kNAx � 4.32 � 0[ΣFx � 0]

D � 4.32 kN5.5(0.4)(9.81) � 5D � 0[ΣMA � 0]


400 kg

3 mA

E F

D

B

C

0.5 m0.5 mR

1.5 m

1.5 m

2 m

1.5 m

Ex

Ex

D =4.32 kN

By

By

Bx

Cx

Cx

12

x

y

Ay

0.4(9.81)= 3.92 kN

Ay = 3.92 kN

3.92 kN

3.92 kN

3.92 kN

3.92 kN

3.92 kN 3.92 kN3.92 kN

Ax =4.32 kN

Ax

D

Ex1–2

Cx1–2

Cx1–2

�

�

Helpful Hints

� We see the frame corresponds to thecategory illustrated in Fig. 4/14a.

� Without this observation, the prob-lem solution would be much longer,because the three equilibrium equa-tions for member BF would containfour unknowns: Bx, By, Ex, and Ey.Note that the direction of the linejoining the two points of force appli-cation, and not the shape of the mem-ber, determines the direction of theforces acting on a two-force member.

c04.qxd 1/26/06 1:25 PM Page 206

Sample Problem 4/7

Neglect the weight of the frame and compute the forces acting on all of itsmembers.

Solution. We note first that the frame is not a rigid unit when removed fromits supports since BDEF is a movable quadrilateral and not a rigid triangle. Con-sequently the external reactions cannot be completely determined until the indi-vidual members are analyzed. However, we can determine the verticalcomponents of the reactions at A and C from the free-body diagram of the frameas a whole. Thus,

Ans.

Ans.

Next we dismember the frame and draw the free-body diagram of each part.Since EF is a two-force member, the direction of the force at E on ED and at F onAB is known. We assume that the 30-lb force is applied to the pin as a part ofmember BC. There should be no difficulty in assigning the correct directions forforces E, F, D, and Bx. The direction of By, however, may not be assigned by inspec-tion and therefore is arbitrarily shown as downward on AB and upward on BC.

Member ED. The two unknowns are easily obtained by

Ans.

Ans.

Member EF. Clearly F is equal and opposite to E with the magnitude of 50 lb.

Member AB. Since F is now known, we solve for Bx, Ax, and By from

Ans.

Ans.

Ans.

The minus sign shows that we assigned By in the wrong direction.

Member BC. The results for Bx, By, and D are now transferred to BC, and theremaining unknown Cx is found from

Ans.

We may apply the remaining two equilibrium equations as a check. Thus,

(30 � 15)(40) � (�20)(30) � 0[ΣMC � 0]

100 � (�20) � 100(4/5) � 0[ΣFy � 0]

Cx � 75 lb30 � 100(3/5) � 15 � Cx � 0[ΣFx � 0]

By � �20 lb50(4/5) � 60 � By � 0[ΣFy � 0]

Ax � 15 lbAx � 15 � 50(3/5) � 0[ΣFx � 0]

Bx � 15 lb50(3/5)(20) � Bx(40) � 0[ΣMA � 0]

D � 100 lbD � 50 � 50 � 0[ΣF � 0]

E � 50 lb50(12) � 12E � 0[ΣMD � 0]

Cy � 100 lbCy � 50(4/5) � 60 � 0[ΣFy � 0]

Ay � 60 lb50(12) � 30(40) � 30Ay � 0[ΣMC � 0]

Art ic le 4/6 Frames and Machines 207

20″12″

30 lb

CA

E

F

B

D50 lb

12″20″

30″

30 lb3

4

50 lb

x

y

AxCx

Ay Cy

3

3

3

4

4

4

x

y

Cx

F

F

Ax

By

Bx Bx

By

Ay = 60 lb Cy = 100 lb

30 lb

D

D

E

E

50 lb

�

�

�

�

Helpful Hints

� We see that this frame corresponds tothe category illustrated in Fig. 4/14b.

� The directions of Ax and Cx are notobvious initially and can be assignedarbitrarily to be corrected later ifnecessary.

� Alternatively the 30-lb force couldbe applied to the pin considered apart of BA, with a resulting changein the reaction Bx.

� Alternatively we could have re-turned to the free-body diagram ofthe frame as a whole and found Cx.

c04.qxd 1/26/06 1:25 PM Page 207

MECH 2110 - Statics & DynamicsChapter S4 Problem 23 SolutionPage 188, Engineering Mechanics - Statics, 4th Edition, Meriam and Kraige

Given: The truss shown below supporting a mass M equal to 500 kg.

Find: The force in each member of the truss.

0. Observations:A. All connections between the members are pins. Each member is connected at two points. Allexternal loads are applied at connection points. The weights of the members are small compared tothe external loads. The system may be classifed as a truss. All members are two force members,transmitting force along the line between the member connection points.

B. In order to determine the directions of all of the members (and hence all of the forces), we need todetermine the coordinates of each of the pins. Setting our origin at point A, X positive to the right, Ypositive upward, all coordinates in meters:A (0,0)B (0,(42-22)1/2) = (0, 121/2)C (2,121/2+2)D (2+121/2,121/2+2)F (2,121/2)E (2+31/2,121/2+1) { midway between D and F, average of those two points)G (XG, 121/2) { where XG is some unspecified negative number. }

Chapter S4 Problem 23 Solution file:///Users/ØDocuments/08_Fall_Statics/NELS_mech2110/p_...

1 of 5 10/8/08 8:52 PM

C. From the above coordinates we can determine the direction of each member (and hence eachforce). This can be done by subtracting the coordinates of two points along the line of action of theforce and then evaluating the unit vector parallel to that direction.X-Direction: BF, BG, and CDY-Direction: AB and CFeEF and eDE = +- { 31/2/2 i + 0.5 j } { 30 degree angle }

eBC = +- { 1/21/2 i + 1/21/2j } { 45 degree angle }

eAF = +- { 0.5 i + 31/2/2 j } { 60 degree angle }

eCE = +- { 31/2/2 i - 0.5 j } { -30 degree angle }

D. We are asked to determine the force in each and every member of the truss. This will require theconsideration of several mechanical systems. One approach would be to successively considerdifferent connecting pins. If we can sequentially identify pins that are connected to no more than twomembers transmitting loads that have not yet been determined, we can readily determine the forces inthose members (method of joints). In so doing, we must exploit the fact that the force transmitted bya two force member is along the line between the two connection points. We must further exploit thefact that the forces acting at the two ends of the member are oppositely directed. We can see thatconsidering the pins in the following sequence will enable us to determine the member forces asfollows:Pin D (Members CD and DE)Pin E (Members CE and EF)Pin C (members CF and BC)Pin F (Members BF and AF)Pin B (Members BG and AF)For convenience we will assume all members to be in tension. In this way any negative results willindicate a member transmitting a compressive force.

1. Mechanical System - Successive consideration of 5 mechanical systems, beginning with pin D andfollowing sequentially with pins E, C, F and B (each considered individually). Note that themechanical system including pin D also includes the supported mass and the cable that connects pin Dto the mass. The other mechanical system include only the pins.

2. Free Body DiagramThe figure provides the free body diagrams ofthe required five mechanical systems. Thecoordinate axes used are shown. Note that asevery member of a truss is a two force member,each of the forces exerted by a truss member ona pin, is parallel to the member. Further notethat the forces exerted by either end of a twoforce member must be equal in magnitude andopposite in direction. Both of these facts arereflected in the free body diagrams. Additionally, all forces are shown pulling the


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pins closer together. This reflects an assumptionthat each and every member of the truss is intension (being stretched) and is exerting aresisting force pulling the attached pins towardone another. This is convenient as a positiveresult for any member force indicates tension,

while a negative result indicates compression in that member. This approach is reflected in the signsof the various terms appearing in the equilibrium equations for each of the pins. As noted above, bysequentially considering pins D, E, C, F, and B, all member forces can be evaluated.

3. EquationsNote the use of the various unit vectors in expressing the force components. The force componentsare obtained by multiplying the magnitude of the force by the corresponding component of theassociated unit vector. The choice of the positive or negative sign is made based upon the free bodydiagrams of the pins.Pin D:S FX = -CD - 31/2/2 DE = 0S FY = - 1/2 DE - M g = 0

Pin ES FX = 31/2/2 DE - 31/2/2 EF - 31/2/2 CE = 0S FY = 1/2 DE - 1/2 EF + 1/2 CE = 0

Pin CS FX = CD + 31/2/2 CE - 1/21/2 BC = 0

S FY = -1/2 CE - CF - 1/21/2 BC = 0

Pin FS FX = -BF + 31/2/2 EF - 1/2 AF = 0

S FY = CF + 1/2 EF - 31/2/2 AF = 0

Pin BS FX = BF - BG +1/21/2 BC = 0

S FY = -AB + 1/21/2 BC = 0

4. SolveFrom the Y equation for pin D:DE = - 2 M g = - 2 * 500 kg 9.81 m/s2 = -9810 NDE = 9.81 kN compression


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From the X equation for pin D:CD = - 31/2/2 DE = 8500 NCD = 8.50 kN tension.

Multiplying the Y equation for pin E by the square root of 3, and then subtracting the X equation forpin E from that result, we observe that all of the terms except the one involving CE vanish:31/2 CE = 0CE = 0

Using this result in the y equation for pin E:EF = DE = -9810 NEF = 9.81 kN compression

From the X equation for pin C:BC = 21/2 CD + (3/2)1/2 CE = 12010 NBC = 12.01 kN tension

From the Y equation for pin C:CF = -1/2 CE - 1/21/2 BC = -8500 NCF = 8.50 kN compression

From the Y equation for pin F:AF = 1/31/2 EF + 2/31/2 CF = -15470 NAF = 15.47 kN compression

From the X equation for pin F:BF = 31/2/2 EF - 1/2 AF = -759 NBF = 0.759 kN compression

From the X equation for pin BBG = BF + 1/21/2 BC = 7740 NBG = 7.74 kN tension

From the Y equation for pin B:AB = 1/21/2 BC = 8500 NAB = 8.50 kN tension

ResultsDE = 9810 N compressionCD = 8.50 kN tensionCE = 0EF = 9.81 kN compressionBC = 12.01 kN tensionCF = 8.50 kN compressionAF = 15.47 kN compression


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BF = 0.759 kN compressionBG = 7.74 kN tensionAB = 8.50 kN tension


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MECH 2110 - Statics & DynamicsChapter S4 Problem 35 SolutionPage 196, Engineering Mechanics - Statics, 4th Edition, Meriam and Kraige

Given: Given the truss shown below supporting the load P of magnitude 8 kN.

Find: The force in members DE and DL

0. Observations:A. All connections between the members are pins. Each member is connected at two points. Allexternal loads are applied at connection points. The weights of the members are small compared tothe external loads. The system may be classifed as a truss. All members are two force members,transmitting force along the line between the member connection points.

B. As we are only interested in the forces in two of the many members in the truss, the method ofsections appears attractive.

C. The only force we know is the applied load P. The two forces we want to know act in membersnear the upper left corner of the truss. If we consider the mechanical system to be that part of thetruss to the right of a section that passes through members DE, DL, CL, and LM; said mechanicalsystem will be acted upon by the known force and both of the two forces of interest. The drawback isthat four unknown forces will act on this mechanical system and we will only be able to write threeindependent equations (2 force and 1 moment). We will not be able to solve for all 4 unknownforces. However, three of the four unknown forces (DL, CL, and LM) all act through the point L. Bysumming moments about L, we should be able to generate a single equation involving only one


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unknown, DE. As this is one of the unknowns of interest, this is very promising.

D. If we can determine the force in member DE, we can then next consider a mechanical systemconsisting of pin D. The forces from members DE, DL, and CD act on this pin. Only two of thesemembers exert force components in the horizontal direction, DE and DL. The sum of the forces in theX will involve only these two member forces. As we will have already determined the force inmember DE, we should be able to determine the second force of interest, the force in member DL.

E. As each of the member forces is parallel to the associated member, the geometry of the truss is ofsome interest. The left part of the truss consists of three square sections stacked one on top ofanother. Each of these square sections has a vertical span, h, of 2 m. Each square has a horizontalspan, w, also equal to 2 m. The truss includes members for each edge of each of these squares and forone diagonal of each of the squares. These diagonal members run from lower right to upper left. Asdiagonals of squares, they make an angle of 45 degrees with the horizontal. The overhanging (rightportion) of the truss is of a slightly more complex geometry. This portion also includes threesections. Each section is a quadilateral, with truss members on each side and along one diagonal. Again the diagonal member runs from lower right to upper left. Two sides of each quadrilateral arevertical. These vertical sides are seperated by identical distances. The horizontal span of each sectionis the same as that for the squares, w. The upper and lower sides of these sections each form straightlines. These two lines converge in a symmetrical way from an initial separation of h (2 m) where theyjoin the left part of the truss, to a separation of h/2 (1 m) at the end of the third overhanging sectionwhere the load is supported. These characteristics could be used to determine the coordinates of eachconnection point. From those coordinates the direction of each member could be established. However, based upon our choice of mechanical systems, the only attribute of interest will be the totalhorizontal span of these three sections. This horizontal span (3 h) is equal to the perpendiculardistance from the proposed moment point, L, to the applied vertical load, P.

1. Mechanical System - Two mechanical systems will be considered (as discussed above). We willfirst consider that portion of the truss to the right of a section passing through members DE, DL, CL,and LM. We will then consider a mechanical system consisting of pin D.

2. Free Body Diagram

The figure includes thefree body diagrams of bothmechanical systems. Thecoordinate axes used areshown. The right portionof the figure includes all ofthe truss to the right of asection through membersDE, DL, CL, and LM. The forces acting in eachof these members havebeen assumed to be tensile(stretching the members)


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and are shown on the free body diagram. The applied load P is also shown. The left portion of thefigure includes a free body diagram of the pin D. Again all of the members have been presumed to bein tension. The three member forces acting on the pin D, DE, DL, and CD, are all shown pulling thepin D toward the member. The forces DE and DL appear in both free body diagrams, reflecting thenature of two force members and the law of action and reaction. Positive member force results willindicate that the member is indeed experiencing tensile forces. A negative result will indicate that themember is actually under compression.

3. EquationsSection System:S ML = -3 w P + h DE = 0 { The applied load is vertical. The horizontal (perpendicular) distancefrom the moment point L to the load is three times the horizontal span, w, of the truss members. Therotational tendency about point L produced by the applied load is clockwise or negative, The force inmember DE is horizontal. The vertical (perpendicular) distance from the moment point L to the forcein member DE is equal to the vertical span, h, of one of the truss members.Pin D System:S FX = DE + 1/21/2 DL = 0 { Exploiting the direction of DL as a diagonal of a square, its

component in the X direction must be equal to the cosine of 45 degrees (1/21/2) multiplied by itsoverall magnitude. }

4. SolveSolving the moment equation for DE:DE = 3 (w/h) P = 3 (2 m/2 m) 8 kN = 24 kNDE = 24 kN tension

Solving the force equation:DL = -21/2 DE = -33.94 kNDL = 33.9 kN compression

ResultsDE = 24 kN tensionDL = 33.9 kN compression


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