Many common structural systems that are used in buildings are made from numerous members that have been connected together to form a more complex system. When you enter your local hardware store or warehouse shopping center, take a look up at the ceiling. You may see the most popular type of these structural systems: trusses.
Trusses are very lightweight structural systems, capable of spanning very long distances. They're used to span major rivers or to span football fields or basketball courts in arenas. You may even possibly have trusses in the roof of your home. Trusses provide a wide array of shapes and sizes that make them extremely versatile to engineers and architects.
I start this chapter by identifying the major criteria that define trusses so that you can spot a truss when you see one. Then I introduce you to two of the most popular methods of solving for internal forces in the members of the trusses: the method of joints and the method of sections. I conclude the chapter by showing you how to determine zero-force members in trusses without ever writing a single equation.
Trusses are structural systems that are composed of numerous members connected together. You may encounter a wide variety of shapes in a truss, but these shapes always have several common basic properties:
Internal forces: All members in a truss system must be axial-only members — that is, their internal forces are axial (parallel to each member's longitudinal axis) and defined in terms of tension and compression.
End connections: All members in a truss are pinned at the ends at locations known as joints (sometimes referred to as panel points). It's these pinned ends that ensure all internal forces in truss members remain axial.
Load locations: All loads are concentrated loads (see Chapter 9) and can only be applied at the joint locations. Truss systems have no distributed loads or concentrated moments. If a concentrated load isn't applied at the joint, you can't analyze the structure as a truss; instead, you need to analyze it as a frame or machine, which I introduce in Chapter 21.
Trusses are especially useful because of the variety of ways that the members can be put together. Figure 19-1 shows examples of several trusses. Notice that straight members can be used to create a wide variety of roof shapes. Even curved and peaked roofs can be created with trusses. If you can imagine it, you can probably create it with a truss system.
When you're solving a truss problem, the first steps are no different than the first steps of any other statics problem. You always start a statics problem by drawing a free-body diagram (F.B.D. — check out Part IV) of the entire system; you then write the equilibrium equations to determine as many of the support reactions as possible.
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In this chapter, I provide all the support reactions and skip drawing the initial F.B.D. in order to help you focus on the different solution techniques for trusses. Just don't forget to first find the magnitudes of the support reactions when you work your own problems.
When you have the support reactions calculated, you're ready to start solving for internal forces by using one of the two common methods: the method of joints or the method of sections. Each of these methods has benefits and drawbacks. I explain both in the following sections.
The first method for solving a truss system is pretty straightforward. As the name of the method implies, you draw an F.B.D. for each joint in the truss; you then apply the equilibrium equations to those diagrams.
Consider the truss shown in Figure 19-2, which is pin-supported at Joint A and roller-supported at Joint E. The support reactions have been given. With this figure, I show you how to use the method of joints to solve for the internal truss forces in the following sections.
To start the analysis of this truss by the method of joints, first draw F.B.D.s of each joint by drawing an isolation box around each joint. Draw an isolation box to isolate the joint of interest. When you isolate a joint, you also end up cutting each truss member that's connected to that joint. Also include any support reactions or applied loads that show up on the F.B.D. at that location. Figure 19-3 shows the F.B.D. for Joints A, B, and C.
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When drawing the internal forces for trusses, I find it convenient to assume that unknown internal member forces are always in tension. At the joint locations, by assuming a member is in tension, you must draw the internal force as though it's pulling on the joint.
The following list explains the method you use to draw the F.B.D.s for each of the joints at A, B, and C.
F.B.D. for Joint A: At Joint A, the isolation box cuts members AB and AH. To account for these revealed internal forces, you draw FAB with the force arrow point to the right, and you draw FAH with the arrow point up and to the right on the joint of A in the F.B.D. You also have the support reaction Ay = 50 kip acting upward at Joint A. The horizontal reaction Ax = 0 kip equals zero, so I've omitted it from the F.B.D.
F.B.D. for Joint B: At Joint B, the isolation box cuts three members — AB, BH, and BC — so you must also include the internal forces for each of these three members on your F.B.D. For Joint B, the internal force FAB must now be pointing to the left in order to be pulling on Joint B. There is also an applied load of 50 kip acting downward at this joint, so you include that on the F.B.D. of the joint as well.
F.B.D. for Joint C: At Joint C, the isolation box cuts a total of five members — BC, CH, CG, CF, and CD — revealing a whopping total of five internal forces on the same joint. Because you have only three equilibrium equations in two-dimensions, this joint is two degrees indeterminate, which means you need to find several of these forces by other means (from other joint F.B.D.s).
On each of these free-body diagrams, you also want to make sure to include the angles or proportion triangles for all of the forces. (Head to Chapter 5 for more on proportion triangles.) However, if the force is horizontal or vertical, you don't really have to indicate the direction because you already know that the line of action is oriented at 0 degrees for a horizontal force and at 90 degrees for a vertical force.
Now, if you're observant, you'll notice that FAB has shown up on two free-body diagrams. On Joint A, it's pointing to the right, and in Joint B, it's pointing to the left. How can the same internal tension force be pointing in two different directions? Look at an F.B.D. of member AB and its connecting joints, shown in Figure 19-4.
To reveal the tension and compression forces in member AB, you start by drawing an isolation box (which I cover in Chapter 14) that cuts the member at two locations as shown. If you assume that the member is subjected to tension (as I do), in order for this member to be in tension, the force FAB must be drawn as shown — with the left end of the cut member pointing to the left, and the right end of the cut member pointing to the right.
Remember that internal forces must remain in equilibrium on either side of a cut line. That means that the part of the member that's connected at the joint has an internal force that must balance the FAB force on the other side of the cut line from it. For Joint A, the internal force is balancing the force on the left end of the cut member that's pointing to the left. So, on Joint A, it must be pointing to the right, or pulling on the joint as you assumed. Similarly, for Joint B, the internal force at this joint must balance the force FAB that is acting on the right end of the cut member, which is pointing to the right. That means that the internal force of member AB acting on Joint B is pointing to the left, or pulling on the joint.
After you've drawn the F.B.D.s for a particular joint, your next step is to write the equations of equilibrium, which I introduce in Chapter 16.
From this equation, you can solve for the internal force FAH, which turns out to have a negative value. As in all equilibrium problems, this fact means that the assumed direction of the internal force on the F.B.D. was incorrect. Because you assumed it was in tension, the negative sign means that it's actually in compression — that's why the final answer has a (C) after it. This way, on all F.B.D.s that contain the force FAH, you know that force is in compression. If you find that a member is in tension, you write a (T) after it.
Next, you can write the other equilibrium equation for Joint A.
Notice that this equation was written for the exact way that I originally drew the F.B.D. at Joint A. This equation is based on the assumption that the force in member AH, FAH, is still in tension, even though you now know that's not the case. But that's okay — trust me.
All you have to do now is simply plug the signed valued into the equation. If you calculated a positive value, you plug in a positive value, but because FAH was negative, you actually substitute the negative value into the equation. The signed values take care of any errors with your assumed force direction within the equilibrium equation.
Finally, you can also write the moment equation for Joint A:
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As you can see, each force on this F.B.D. is acting concurrently (simultaneously) through Joint A. By summing moments at this point, the perpendicular distance to each force is actually zero. So for the method of joints, the moment equation no longer gives you any useful information about relationships between forces.
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The loss of the moment equation's usefulness isn't a major hindrance. It just means that in order for you to work with a particular joint, you can only completely solve all the forces if there are no more than two unknown forces acting on the F.B.D. of interest.
Your next step is to look for another joint that has only two unknowns. At this time, the F.B.D. of Joint B has two unknown forces (just FBH and FBC, because the preceding section determined FAB), so you can move to that joint and repeat the process.
You keep moving from one joint to the next, learning more about the different member forces, and applying them to each F.B.D. Repeating this for all joints in the truss reveals all the internal forces within each member.
The major drawback with the method of joints is that in order to solve for a member force that's connected to a joint with a large number of other members, such as Joint C in Figure 19-3 earlier in the chapter, you have to first solve for forces at several other places. In fact, to solve for the force FCD, you have two options:
Starting at Joint A: First, solve for forces at Joint A as described in the preceding section. Next, go to Joint B and do the same. Then go to Joint H, followed by Joint G, at which time you could move to Joint C and finally solve for force FCD.
Starting at Joint E: First, solve for forces at Joint E, followed by Joint D.
Depending on which joint you start with, you can greatly increase the number of free-body diagrams you have to work with. In this case, to find FCD by starting at Joint A, you have a minimum of five joints, or ten translational equilibrium equations you have to write. Starting at Joint E has much fewer F.B.D. stops along the way, but you still have to work at two places, or four total translational equilibrium equations.
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What happens if the truss you're analyzing is very long and has hundreds of joints? If you're interested in a force in the middle of this truss, you may end up having to write an extremely large number of equations. And if you have irrational numerical values for each force along the way, you may end up incorporating a lot of error into your final answer. I won't even mention the fact that the more steps you take, the more chances you have for writing an incorrect equation or keying a bad numerical value into your calculator. Generally speaking, the fewer calculations you have to do to get a desired answer, the more likely that you won't make a mistake along the way.
To find forces in the middle regions of long trusses, a technique that lets you skip directly to the middle would be pretty handy. That's where the second major method of truss analysis, the method of sections, comes into play (see the following, well, section).
The second major method of truss analysis is the method of sections. As the name implies, in this method you basically slice a truss into sections or, more specifically, into two pieces. But you can't just go hacking up the truss — you need to obey several rules:
Cut the member that you're interested in. Obviously, to calculate the internal force of a truss member, you must first expose the internal force. The only way you can do that is by cutting the member.
Cut the truss completely into two parts. Don't cut partway through a truss and then stop. In fact, to assure yourself, you may consider using an isolation box that's completely closed. Remember: You can include one of the supports inside your box.
Cut a maximum of three members. You're allowed to cut up to three members total. If you cut fewer than three, that's fine. But if you cut four or more, you may have some problems with this method because you may not be able to solve for all of the unknowns on the free-body diagram.
If you cut more than three members, all but two of the members must be on the same line of action. In rare circumstances, you can actually cut more than three members, but you'll need to make sure that all but at least two of them are collinear. In general, it's a very rare occasion when you can actually cut more than three members at a time.
If performed properly, the method of sections can save you a load of time analyzing those members in the middle regions of trusses. In fact, in many cases you can actually calculate the force in a particular member by writing just a single equation.
As with any truss analysis, the first steps require that you draw an F.B.D. and compute as many of the external support reactions as possible. However, with the method of sections, if you have a truss that's cantilevered (or supported in such a way that at least one end of the truss is unsupported), you may not even have to find the support reactions at all — although calculating support reactions is always a good idea, if you can.
Suppose you're tasked with the job of finding the force in member CD of the truss shown in Figure 19-5.
If you use the method of joints (see "The Method of Joints: Zooming In on One Panel Point at a Time" earlier in the chapter) to compute this force, you need no fewer than six free-body diagrams (start at N, A, M, B, L, and finally C) or even seven free-body diagrams (start at G, H, F, I, E, J, D), depending on which end of the truss you started on. That's a lot of diagrams and even more equation writing. (Remember: You have to write two equations per F.B.D. for the method of joints.)
To analyze the truss of Figure 19-5, a likely way to cut this truss would be to cut members LK, CK, and CD (which is your desired member). This strategy obeys all the method-of-sections rules I explain earlier in the chapter.
After you cut the truss, you're left with two pieces. Draw the free-body diagrams, including all reactions, point loads, and the revealed internal forces of members LK, CK, and CD, as shown in Figure 19-6.
The F.B.D. that you choose to work with doesn't really matter, but I usually recommend choosing the smaller of the two remaining pieces because this strategy usually ensures that your equilibrium equations are smaller as well. For this example, I work with the left F.B.D. in Figure 19-6 because it has fewer applied loads. However, if you choose to work with the F.B.D. on the right instead, you still end up calculating the exact same internal forces, so which diagram you select doesn't really matter.
In this step, you write the two translational equilibrium equations.
The translational equilibrium equations give you an internal force for a member in the interior part of the truss, just not the one you're interested in at this time. But never fear — you still have one equilibrium equation left to work with.
The last equation you have to work with is the moment equation. The moment equation is typically the most useful because you have complete control over the point about which the equation is written. In the method of sections, the moment equation truly shines.
To choose the point for the moment equation, you need to look back at the F.B.D. In this system, you have three unknown internal forces acting on the truss after you cut it, one of which is the force you're interested in. So it would be convenient to write an expression that includes FCD but doesn't include FLK or FCK in it. (You know that if you calculate the moment of a force at a point on its line of action, the moment from this force is actually zero.) Thus the only way to make that happen is to write the moment equation somewhere along the lines of action of both member LK and member CK.
Because these two members have different lines of action, only one point in space is common to both members; the intersection of the lines of action of member LK and CK happens to be at Joint K. Both of those two forces intersect at that location. So if you sum moments at that location, both forces will have a perpendicular distance of 0 meters because that point is concurrent with each of the forces. Summing moments at Joint K, you're now left with
Check it out! You've found the force of a member in the middle of the truss by writing only one equation. All this was possible because you chose to calculate your moment at a convenient point where the other forces intersect, and you stuck to the basic rules for creating the F.B.D.
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Choosing a convenient point about which to write your equations isn't just useful in statics but is also a very handy concept in many other areas of study, such as structural analysis, dynamics, and mechanics of materials, among others. So keep your eyes open for those special locations.
Instantaneous center is a term used to describe those locations on an object where special things happen. In dynamics, the term refers to the point in space (or on an object) about which all other points are moving, resulting in a velocity of zero. Although this terminology isn't 100 percent correct for statics, the term can still be used to describe the point through which multiple forces acting result in zero moment.
For the method of sections, the instantaneous center is the point you look for in order to ensure the internal force you're looking for appears, while making the other unknown forces disappear from your moment equation. In many cases, especially with trusses of horizontal chords, the instantaneous center is often at one of the joints of the truss. However, in practice, this point doesn't have to be at a joint. In fact, it doesn't have to be within the truss at all. Consider the simply supported truss, known as a Gambrel truss, shown in Figure 19-7. (For more on Gambrel trusses, see the nearby sidebar.)
Suppose you wanted to determine the force in truss member DG of the gambrel truss shown in Figure 19-7. Because this member is in the middle of the truss, your first instinct should be to look at the method of sections as a possible solution and to slice the truss on the cut line through member DG. One possibility is to cut through members FG, DG, and CD.
Taking the F.B.D. on the right half, shown in Figure 19-8, you can see that the lines of action for members GF and CD actually intersect at a point outside the boundary of the truss. I define that "extra" distance beyond Joint E as x.
To find the distance x, you need to construct a set of similar triangles that incorporate the slopes of the two members you want to eliminate. In this case, you want to know the force in member DG, FDG, which means that you want to eliminate forces FFG and FDC from your moment equation. These are the two forces that you need to work with to find the instantaneous center at Point O. To accomplish this, you set up a relationship using similar triangles.
For member force FFG, you know that its line of action has a 6:12 proportion (height-to-length ratio) to describe the slope. This line of action is also the hypotenuse of nODF.Similarly, member force FDC is horizontal and forms the lower edge of the same triangle. nODF has a height of 12 feet (side DF) and a length of (x + 12) feet for side DO, or a ratio of 12:(x + 12). Knowing that these ratios must be the same, you can then set up a relationship and solve for x:
Thus, the instantaneous center at Point O is located at a distance of 12 feet to the right of Point E. When you know this location, you can write the sum of moments equation at Point O and solve directly for the force in member FDG:
For some very specific cases, you can determine the forces within a truss without ever performing a single calculation. You just need to be able to identify a couple of conditions, and if all those conditions are met, you can draw conclusions about whether a force is considered a zero-force member, without a doubt. To identify a zero-force member, you have to look at each joint on a case-by-case basis and count the number of members, external reactions, or applied point loads that are acting at the current joint being investigated. Figure 19-9 and the following list illustrate the three specific cases that allow you conclude whether a member is zero force.
One member: If a joint has only one member and no point loads acting at the joint, the member must be zero force.
Two members: If a joint has two members acting at it, and no point load is applied at the joint, both are zero force as long as the members aren't collinear. If the angle between the members is either 0 degrees or 180 degrees, you don't know for certain that both are zero, so you have to leave them in your computations.
Three members: If a joint has three members acting at it, two of which are collinear (and one of which isn't), and no point load is applied, the noncollinear member is zero force. You don't know the internal forces for the two collinear members for sure, so you must retain them in your calculations.
If you have a joint that has two members and a point load, a special case may exist. If the point load is collinear with a member, the noncollinear member is zero force. This case is useful for members at support locations. Often, the members are perpendicular or parallel to support reactions.
When you've concluded that a member is a zero-force member, remove it from your F.B.D. of the entire system, and then go back and look at the system again. Sometimes, removing one member from the F.B.D. helps you conclude that another member is also zero by these same basic definitions.
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Also, you only remove a member from the analysis if you're absolutely sure it's zero force. If the joint doesn't meet the requirements in this section, or you're still unsure, don't remove it. Your calculations in the end will verify the member is zero force anyway.
In fact, you usually want to identify whether a member is zero force before you even start employing the method of joints or the method of sections. You typically start looking for zero-force members after you have calculated the support reactions. Being able to identify zero-force members can be especially useful because you can remove members from the analysis altogether, which can greatly simplify your free-body diagrams.