Chapters 13 and 14 equip you with the countless tools and equations to create, calculate, and depict the behaviors of vectors on objects by using free-body diagrams (F.B.Ds). F.B.D.s can be complex, intimidating, and a bit overwhelming because you often have an object that is subjected to distributed loads, concentrated forces, and applied moments applied at multiple locations, all acting simultaneously. (Flip to the chapters of Part III for more on these concepts.) Figuring out where to begin in tackling these diagrams can be quite a task.
One of the first steps you want to perform after making your sketch is to look for ways to simplify your F.B.D. In this chapter, I show you several handy simplifying techniques that you can utilize on a regular basis.
The principle of superposition basically states that multiple actions on an object are equivalent to the sum of the effects of each action applied individually. The principle of superposition allows you to quickly compute behaviors (such as reactions, displacements, and internal forces) from combined multiple load cases by simply adding together the responses of the individual cases.
Note
I assume that all objects in this book are rigid bodies, which are objects that aren't deformed by the forces acting against them. Although the principle of superposition is based on assumptions surrounding non-rigid bodies (bodies transformed by forces), it's still a handy tool in mechanics and statics because it provides good approximations for many rigid bodies.
The fundamental principles behind the principle of superposition have their origins in vector formulations for finding resultants. In Chapter 7, I show you how to take two vector components, F1 and F2, and simply compute the vector sum to find a resultant vector FR. If you have more than one vector component, you can add as many as you like to find the resultant vector. That is:
The idea of determining each part individually and then combining them to find a single combined behavior is the premise behind the principle of superposition.
If you consider all the assumptions from the preceding sections to be true for Figure 15-1, you can apply the principle of superposition by examining the effects of each load individually, creating three separate F.B.D.s that reflect the behavior of each of the individual loads.
The combined loading of this example consists of a simply supported beam subjected to a concentrated load P1 acting at a distance c from the left support. In the same loading, a uniform distributed load with a magnitude of w is applied for the entire length of the beam. Finally, a concentrated moment is applied at a distance of a from the left end of the beam.
For this example, F.B.D. #1 (shown in Figure 15-1a) is a simply supported beam subjected to a uniformly distributed load w over the entire length. F.B.D. #2 (Figure 15-1b) is the same beam and support reactions with a concentrated moment M at a distance a from the left support. F.B.D. #3 (Figure 15-1c) is the same beam and support reactions yet again, with the concentrated load P1 applied at a distance c from the left support. The combined effect of each of these three load diagrams results in the same combined loading.
This principle is extremely useful for complex loadings situations in statics because it allows you to break a problem down into more manageable pieces. In fact, you may be surprised to find that many design handbooks out there have simple load cases already worked out.
Another useful simplification technique is to look for a line of symmetry within a structure or loading diagram. A line of symmetry is an imaginary line that produces a mirror image of one part of a load or structure onto another. For a symmetric condition to occur, you must make sure that each of the following is mirrored by the line of symmetry:
Beam properties: Properties such as mass, geometry, and cross- sectional properties.
Support reactions: Support reactions (restraints that keep a rigid body from moving away when force is applied; check out Chapter 13 for more). A fixed support condition on one side of the mirror line must be reflected on the other. The one exception to this rule is with pinned and roller supports. If all loads are in one direction, a roller support can actually mirror a pinned support as long as the parallel component of the pinned support is zero.
Load magnitude and location: A distributed load must be mirrored by either being centered on the line of symmetry or by having an identical distributed load on the reflected side of the line. Similarly, you can consider a concentrated load or moment symmetric by itself if it occurs on the line of symmetry; otherwise, it must have a matching concentrated load or moment and location that is reflected in the line of symmetry.
If any one of these items isn't properly mirrored about the line of symmetry, you can't consider the loading symmetrical, and the methods in this section won't always work.
In Figure 15-2a, you see a simply supported beam with a partial uniform distributed load of magnitude w, with a distributed length of 2b. Because the load is centered on the beam, it's symmetrically loaded — if you draw a mirror image line (or line of symmetry) exactly in the middle of the beam, the loads, supports, and beam conditions on the left half will be exactly the same as those on the right. In cases involving symmetry, you can usually assume that the reaction's symmetric loading will be one half of the total resultant of the symmetric loading. In this case, the total applied load amounts to a resultant of 2wb, or wb for each reaction RA and RB.
Conversely, Figure 15-2b shows the same load applied nonsymmetrically. In this example, no line of symmetry meets all the requirements for mirroring of supports, loads, and geometry.
In some cases, lines of symmetry can also occur at different angles from the rest of the structure, as shown in the loading of the L-shaped object of Figure 15-2c. In this example, the loads and geometry aren't mirrored by a vertical or horizontal line of symmetry but rather on a line oriented at 45 degrees passing through the corner of the L shape.
Tip
Symmetry allows you to use the principle of superposition to quickly determine information about an F.B.D. without having to write any of the more complex equilibrium equations I show you in Part V. For example, consider the beam of Figure 15-2a again. With a bit of simple logic and Chapter 10's resultant techniques for distributed loads, you can quickly determine the unknown support reactions RA and RB. In this example, you have a partially uniformly distributed load acting in the middle of a beam. If a beam is symmetric, and the loads acting on it are also symmetrically positioned, you know right away that each support of the beam carries exactly half of the symmetric load. In this example, the total load (or resultant) of the distribution is equal to R = 2wb. If the force is shared equally by the supports because of symmetry, you know that the support reactions are
Another method of simplifying a structure is by determining equivalent behaviors through the relocation of forces and moments. (In fact, this principle is what drives a lot of the discussion on equilibrium beginning in Chapter 16.) An equivalent system is a system of forces and/or moments that you can replace with a different set of forces and/or moments and still achieve the same basic translational and rotational behaviors. (For more on these behavioral concepts, take a look at Chapter 12.)
To start the discussion of relocating a force, I now introduce you to the space potato.
The space potato is a unique, albeit somewhat nonsensical, method of referring to an arbitrary three-dimensional object in space. Suppose you have your very own space potato with a force FA acting at a Point A somewhere on the object, as shown in Figure 15-3a.
Now suppose you want to compute the effects of this force at a different location, such as Point B, on the space potato. To determine these effects, you need to compute the equivalent system at that point.
To determine the required translational effect, add a new negative force −FA at Point A, and an additional force FA at the force's new location (B). These two forces are illustrated by the dashed arrows in Figure 15-3b. The resultant of these newly added forces is FA + (–FA) = 0, or zero net translational effect.
The original force FA plus the new negative force −FA also cancel each other at the original Point A. That is, the forces acting at Point A result in a net translational effect of zero at Point A. With that in mind, you can see that the only force that now remains on the space potato is the newly relocated force FA, which is now acting at Point B. Congratulations! You've relocated your first force. Unfortunately, you've also managed to introduce a new behavior to the potato in the process.
Now that you've relocated the force, examine the two additional forces that were added in Figure 15-3b. For now, check out each of the two additional forces and their points of application as shown in Figure 15-3c. In this sketch, you see that these two new forces are a negative force −FA acting at Point A, and a positive force FA acting at Point B. The two forces are parallel, acting in opposite directions and separated by a distance. This setup is the very definition of a moment couple that I discuss in Chapter 12. Remember, moments and couples cause rotation in an object, and the added rotational effect of this couple is what you also need to include when you move a force.
To compensate for this couple, you need to compute an equivalent moment at Point B, which requires a position vector from your point of interest to a point on the line of action (usually taken as the point of application) of the original force. Compute the equivalent moment at Point B from the basic moment vector formula:
The space potato analogy shows you that to relocate a force, you simply need to take the original force and apply it at the new location, plus compute a newly applied moment and apply that at the new location. The final result of the relocated force and newly created moment are shown in Figure 15-3d.
Note
Equivalent systems must have both the same translational behavior and the same rotational behavior at a given point. Forces provide the translational behavior, and moments provide the rotational behavior.
Tip
To relocate a distributed force, you simply need to convert the distributed force to a single concentrated resultant, determine the location of that resultant on your space potato, and then follow these same rules for relocating a force.
In the preceding section, I illustrate how you can convert a couple into a single concentrated moment at a new location. But how do you handle an applied concentrated moment? As I discuss in Chapter 12, a concentrated moment is an action that causes a rotational behavior but doesn't affect the translation of an object. Moment vectors are a type of vector known as a free vector, which applies the same rotational behavior regardless of where on the object it's acting (see Chapter 4). As a result, you can freely move any moment (both concentrated and couples) around the object as long as the magnitude and sense of the moment vector remain unchanged (see Figure 15-4). The point of application of a moment or couple doesn't matter when creating an equivalent system. If you want to move a moment, just move it!
