Ask photographers and artists about the pictures they've created, and you inherently hear about the emotions and feelings that they were trying to capture as they portrayed the physical object of their work. Pictures in statics provide a different purpose — something a bit more unemotional and unbiased; after all, statics is a science. However, the facts do show that a picture can serve as a very handy and even necessary tool; these pictures are what allow you to create those (objectively) super-awesome equations of equilibrium. Without a properly detailed picture (known as a free-body diagram), the game is over before you even get off the bench.
In this chapter, I describe the four types of forces that must be included on a free-body diagram and discuss the proper technique for displaying them.
The picture that you draw in statics is known as the free-body diagram (or F.B.D. for short) and represents the physical condition of the rigid object you want to analyze, including dimension data and the forces acting on the system. Free-body diagrams can be complex pictures of multiple objects and"systems, or diagrams of a smaller subcomponent of a larger piece within a system. Each representation must still obey all laws of physics. Without a proper F.B.D. sketch, correctly analyzing a problem in any field of engineering and physics is extremely difficult, if not impossible.
I've always found a checklist useful for remembering what to include on any given free-body diagram.
You can classify the majority of forces on an object into four separate categories, each of which becomes an item on the checklist: external forces, internal forces, support reactions, and self weight. Just remember, you need to include the forces themselves as well as all information that locates their point of application (physical location on the object or in space where the vector is acting — see Chapter 4). Check out Chapter 9 for more on how all of these categories affect concentrated loads; Chapter 10 discusses how external and internal forces and self weight relate to distributed loads.
External forces: External forces are the forces exerted on a rigid body (which isn't deformed by the force) by sources outside the body. A ball thrown at a wall exerts an external concentrated force at the point of impact; the weight of snow on your roof exerts a vertical distributed external force on the roof.
Internal forces: Internal forces are the forces exerted within a rigid body. The tension in a rope and the compressive force in the leg of the chair you're sitting on are both examples of internal forces.
Support reactions: Support reactions are the physical restraints, such as door hinges and bridge piers, that prevent a rigid body from moving.
Self weight: Self weight (in both concentrated and distributed form) is the force due to gravitational effects on the mass of the object.
In addition to these four categories of forces, you also need to include the necessary dimensions and angles that help you properly define their lines of action and points of application. I explain more in the coming sections about how you actually draw each on the free-body diagram.
External forces are typically the easiest forces for you to determine because they're often the result of a measureable action — you typically know the sizes and shapes of their distributions. These forces include both concentrated forces (or point loads) and distributed forces (forces over area), as I discuss in Chapters 9 and 10, as well as concentrated moments, which I discuss in Chapter 12.
Consider the example drawing in Figure 13-1, which shows a man pushing horizontally on a crate with a force of 100 Newton. The crate also has a very heavy lid of uniform thickness resting on its top.
If you want to draw the external forces acting on just the crate (without the lid) you have to apply a bit of logic and reasoning to determine the sources of external forces. Figure 13-1 has two external forces — concentrated and distributed — acting on the rigid crate. The following sections show you how to depict those forces in your F.B.D.
Concentrated forces are typically the easiest to portray on an F.B.D. For any concentrated force vector, you know the magnitude, sense, and point of application (or in some cases, the line of action). If you know the point of application, you can use that point directly on the free-body diagram. If you only know the line of action, you need to locate the line on the F.B.D. and then apply the force somewhere along that line.
The first external load acting on the crate in Figure 13-1 is from the force exerted by the man as he pushes horizontally at Point A. The force of the man's hands on the edge of the crate in this situation is depicted as a concentrated load because the force is applied at a single point. Figure 13-2 shows how you can represent this force as a concentrated load.
To properly represent a concentrated force, you must include information about the three basic properties of a vector: magnitude, sense, and point of application, all of which I cover in Chapter 4.
Magnitude: The magnitude is the vector's length. In both parts of Figure 13-2, notice that the magnitude of the force is given as 100 Newton acting horizontally. The 100 Newton in this example is actually the magnitude of the vector you're drawing.
Sense: The sense is the direction in which the vector is acting, which you can help determine by figuring out which direction the object would want to move as a result of the force. Put yourself into Figure 13-2; the crate would want to slide to the right if you pushed hard enough. Therefore, you can reason that the vector's sense is also to the right.
Point of application: The point of application of this force vector is situated at Point A because that's where the man's hands are located. The most conventional means of representing this vector is to apply the tail of the vector at the point of application as shown in Figure 13-2b. However, when forces are pushing on objects, placing the head of the vector at the point of application instead (as shown in Figure 13-2a) can be a more convenient reminder. The principle of transmissibility (covered in Chapter 9) tells you that the two drawings shown in Figure 13-2 behave identically as long as the crate is considered to be perfectly rigid.
You draw distributed loads similar to their concentrated counterparts — they have a sense to define their direction, and a magnitude (defined as its intensity). However, a distributed load has no specific point of application because the load is spread out over a line or region. You must include dimensional information that shows where the distributed load begins as well as where it ends.
The second type of external load on the crate in Figure 13-1 is from the weight of lid as it sits on the crate. When you look at just the lid, this 50-Newton weight is actually the self weight of the lid itself. However, because your F.B.D. is of the forces on just the crate itself, this force becomes an external force on the crate.
Tip
Don't be too alarmed about the difference between self weight and external forces. As long as you include the force at its proper location on the object, the solution process is identical. Just remember that self weight can be either a single value (lumped mass) or a distributed load over a length or area (continuum). For more on self weight, flip to Chapters 9 and 10.
If you consider the weight of the lid as a uniformly distributed load, you can compute that this 50-Newton external force is acting over a length of 1 meter (the dimension of the lid). Thus, the external distributed load from the lid is computed as
To display this distributed force on the F.B.D., you need the same three basic requirements (magnitude, sense, and direction) as for the concentrated loads in the preceding section:
Magnitude: As for the concentrated loads, the magnitude (or intensity) of the load distribution is 50 Newton per meter.
Sense: Because the load is actually coming from a self weight, the sense of the distributed load acts in the direction of gravity. I'm assuming this object is on planet Earth (because scientists haven't found any evidence of crates on Mars, at least not yet), so gravity and therefore the sense are acting downward.
Point of application: The point of application of this load depends on whether the load is distributed or concentrated. If the load were concentrated, it would act at the center of mass of the lid (in this case, the midpoint). But, because you've calculated the weight of the lid as a distributed load and you know that the lid's thickness is uniform (or constant), this load is evenly spread (or uniformly distributed) over the entire length.
In addition to these three requirements, you also need a couple of additional pieces of information: the start and end points of the load distribution. This load's beginning and ending locations occur at the two ends of the lid. As a result, you show the diagram of this load on the crate's F.B.D. as a series of downward arrows acting along the entire area of the crate's lid. Figure 13-3 shows how this distributed load is depicted on the lid of the crate.
Tip
The second part of this figure illustrates how you can also display the resultant for this load distribution. See Chapter 10 for more information about computing the resultant.
Figure 13-4 shows the combined F.B.D., including all the external forces I describe in the preceding sections. I've also added vertical and horizontal reactions at the contact surface to keep the free-body diagram correct. I explain more about these contact surfaces in the "Restricting Movements with Support Reactions" section a little later in this chapter.
The third type of external load you must remember is loads created by concentrated moments (which cause an object to rotate). Concentrated moments are actually fairly easy to depict on a free-body diagram.
Because moments are vectors, too, they must again display magnitude, sense, and point of application. Figure 13-5a illustrates how a concentrated moment can be depicted on a free-body diagram. For both two- and three-dimensional objects, you display the magnitude of a concentrated moment as the numerical value (if it's known), or a label for the vector (if it's unknown).
The point of application and sense are usually depicted slightly differently for moments in two dimensions and moments in three dimensions, although the methods still have similarities.
For a two-dimensional object as shown in Figure 13-5b, the sense and point of application are determined as follows, depending on whether you're using a single- or double-headed arrow:
Single-headed circular arrow:
Sense: The sense of a concentrated moment is determined by the direction of the circular rotational arrowhead. In this example, the moment is acting counterclockwise.
Point of application: You can depict the concentrated moment by drawing the circular rotational arrowhead about a given point. In this case, the point of application is at Point O.
Double-headed arrow:
Sense: The sense of the double-headed arrow is determined by the right-hand rule for moments, which I explain in Chapter 12. In Figure 13-5c, the 1,000 Newton-meters (N-m) cause a counterclockwise rotation when you look at the end of the shaft. This moment can also be transmitted to Point O, as shown in Figure 13-5b, resulting in a counterclockwise applied moment on member AOC.
Note
Remember, you don't usually see the axis of rotation in two dimensions because it's often oriented perpendicular to the plane of the drawing (or out of the page). For two dimensions, you just use the normal circular vector depiction as I show in Chapter 12.
Point of application: Just like for the concentrated loads I describe in "Portraying concentrated forces" earlier in the chapter, you can also apply the double-headed arrow with the tail or head acting at Point B. However, this point must have dimensions to help properly locate this action.
Note
In most two-dimensional free-body diagrams, the moment acts about the z-axis (which is the Cartesian axis that seems to be coming out of the page). Double-headed arrows always act along the axis of rotation of the object, so distinguishing the sense of the double-headed arrow in two-dimensional pictures can be hard (because you can't exactly draw the arrow out from the page). For that reason, you should probably use the circular arrowheads I discuss earlier in this section to denote the direction of the applied moment (shown in Figure 13-5b). However, for cases where the moment is acting about a line in the plane of the picture (in the xy Cartesian plane), you can still use double-headed arrows.
Tip
Although you can use double-headed arrows in two dimensions, they're definitely better suited for problems in three dimensions, which I discuss in the following section.
For a three-dimensional free-body diagram, the sense and line of action are determined as follows:
Sense: As with two-dimensional moments (see the preceding section), the sense of a three-dimensional concentrated moment is determined by the right-hand rule for moments (see Chapter 12).
Line of action and point of application: For a three-dimensional rotation, you need to indicate both the line of action and the point of application on the object. The line of action is simply the axis of rotation about which the moment is acting, and it always passes through the point of application (Point O) on the F.B.D. as shown in Figure 13-5c. You need to be sure to include all the necessary dimensions to locate the point of application and the orientation of the axis of rotation.
Although external forces (see the earlier "Displaying External Forces" section) are the easiest to define and draw, a properly drawn F.B.D. must depict all forces, including any revealed internal forces, and their locations acting on the object.
Internal forces only appear on an F.B.D. after you've sliced the object or structure — that is, when you're looking at a part of an object and not the entire object. (In Chapter 14, I explain how to know when to include an internal force.) Typically, you treat internal forces as concentrated loads and concentrated moments and draw them in the exact same manner I describe in the "Displaying External Forces" section of this chapter. Their points of application are assumed to be at the centroid (geometric center) of the cross section. (Chapter 11 gives you more detail on centroids.)
Support reactions are the restraints that keep a rigid body from moving away when a force is applied, and they are typically classified into two categories: two-dimensional (or planar) supports and three-dimensional (or spatial) supports.
Two-dimensional planar supports: Planar support reactions are the restraints for two-dimensional objects. Two-dimensional supports can have as many as two restraining forces and one restraining moment, depending on the type of support reaction.
Three-dimensional spatial supports: Spatial support reactions can be much more complex. Three-dimensional restraints will have as many as six different forces and moments acting on a given support.
Note
When dealing with support reactions, you must take into consideration the restraints to any motions on the object. If a motion is restrained, a support reaction has been created and must be included on your free-body diagram. For all support cases, you sketch the support reactions exactly as they're drawn, as concentrated forces and moments acting at the support location. Refer to "Displaying External Forces" earlier in the chapter for more on drawing concentrated forces and moments.
Note
Restraints for translation (movement along a line in any direction) are always forces, and restraints for rotation are always moments.
In two-dimensional statics, support restraints are categorized into one of three different support conditions: roller supports, pinned supports, and fixed supports, which I dive into in the following sections.
The simplest of the three planar support reactions is the roller support, which is free to move parallel to the support surface and to rotate but is restrained from moving perpendicular to the support surface. Examples of roller supports include a pair of roller skates and the wheels on a car. In most textbooks, roller supports are depicted as either a single wheel, or multiple wheels as shown in Figure 13-6, where the simplest roller support is designated by a simple wheel. The object shown in this figure is free to move in one direction parallel to the support surface (or left and right) but is restrained in moving in the perpendicular. It's also free to rotate about the support location.
The second of the three planar support reactions is known as the pinned or simple support and is a support reaction that restrains translation in two directions but is free to rotate. You commonly encounter two types of pinned supports in statics.
External pinned support: As its name suggests, the external pinned support is a support condition that restrains an object externally. You usually depict this type of support by drawing a triangular support reaction as shown in Figure 13-7.
Internal pinned (hinge) support: The internal pinned support is also known as an internal hinge. At this location, the internal motion of the object is restrained from translating but is free to rotate. I explain a lot more about internal pins when I talk about frames and machines in Chapter 20.
As with the roller supports in the preceding section, the restraint of motion is what creates the support reaction. Because this support is restrained from translating, it must have a restraining force in at least two mutually perpendicular (or orthogonal) directions to help hold it back or prevent it from moving. As long as these two reaction forces remain perpendicular to each other, their overall orientation doesn't matter. However, for convenience, these reactions are typically aligned with the Cartesian x-axis and y-axis of your coordinate system.
The third type of support is known as the fixed support reaction in all three possible directions, as shown in Figure 13-8. Because this support restrains both of the translations in addition to the rotation, your drawing must have three separate support reactions.
As I mention with the pinned supports in the preceding section, the two translation restraints produce two restraining forces. However, for fixed supports, the rotational behavior of the support is also restrained, and thus you must also include a concentrated moment restraint.
Inclined supports can be either roller, pinned, or fixed supports; the supports are just no longer horizontal or vertical but rather inclined along a slope. Keep these rules in mind when you're dealing with inclined supports in two dimensions:
If a support is free to move, no restraining force is developed.
If the object is free to rotate, no restraining moment is developed.
If the object is restrained from rotation, the restraining moment that's created is always about an axis perpendicular to the plane of the object.
Roller supports don't always have to be aligned horizontally or vertically. In fact, rollers can be oriented in any direction, so you want to pay careful attention to which way the supporting surface is oriented with respect to the free-body diagram. Figure 13-9 shows a roller support oriented on a surface inclined at an angle θ measured from the horizontal.
In this example, the roller support is free to move down the incline, so no force is acting parallel to the incline. However, the restraining force R of the supporting surface prevents the motion perpendicular to the surface. With these two directions accounted for, the behavior in the xy plane is now properly defined.
Because the supporting surface prevents the object from moving in that direction, a hidden support force (or a support reaction) must be created to prevent the object from moving. In this case, the support reaction is oriented perpendicular (or normal) to the support surface.
Three-dimensional supports can be a bit more complicated than their two-dimensional counterparts (see the preceding section). In two-dimensional analysis, three support reactions develop at most: two translational forces (x- and y-directions) and one rotational moment (z-direction).
To fully define translation in three dimensions, you must define three mutually perpendicular forces, typically with respect to the Cartesian x-, y-, and z-axes. Similarly, you must also account for three mutually perpendicular moments to define rotational restraint in three dimensions.
To model three-dimensional support reactions, you have to once again apply a bit of logic about how the object is capable of moving. If any restraint is present (either translation or rotation), you must include a support reaction to represent that restraint. Although countless three-dimensional supports are possible, I describe a couple of the more common ones in the following sections.
A three-dimensional pinned support is commonly referred to as a ball-and-socket connection and is shown in Figure 13-10. (See "Freeing up rotation with pinned supports" earlier in the chapter for pinned-support basics.) Examples of common ball-and-socket supports are the hip and shoulder joints on your body. If you try moving your arm around, you notice that with some effort, you should be able to rotate your arm in any of three directions, but your shoulder stays in the same place. That is, it's free to rotate in any direction but is fully restrained from translation (or popping out of its socket). Because restraint causes support reactions, you know that your shoulder has three translational support reactions but zero rotational support reactions.
Another common support reaction that occurs in mechanics is the slider or collar assembly. This type of support consists of a sleeve that wraps around a rod or shaft. In this connection, the sleeve is free to translate parallel to the axis of the rod, and is capable of swiveling (or rotating) about the axis of the shaft. All other motion (both rotational and translational) is restrained. Figure 13-11 shows a common collar assembly and its corresponding free-body diagram.
As drawn in this example, the shaft is oriented along the y-axis. Because this assembly is free to move in the direction of the shaft, it's considered unrestrained in that direction and should have no force for the support reaction in that direction. Similarly, the sleeve is capable of rotating about the axis of the shaft (the y-axis) and consequently has no moment for the support reaction about the y-axis. The remaining motions are all fully restrained and consequently have both moments and forces as support reactions in the x- and z-directions.
The fourth force category deals with the representation of the self weight of the object. In Chapter 9, I explain how to calculate self weight for a lumped mass (concentrated system), which is a concentrated force acting at the object's center of mass. However, if you're dealing with a distributed mass system (as I describe in Chapter 10), the self weight is actually a distributed force. In statics, you commonly neglect the self weight of an object unless the problem explicitly states that the object has mass. This omission is usually acceptable because the nature of many structural systems is such that the force from self weight is usually only a small percentage of the total force acting on a system. For example, a small construction crane that weighs only a few thousand pounds is often capable of lifting loads of many hundreds of thousands of pounds. That being said, you absolutely want to consider the exact self weight of the object in addition to the external loads if you're performing a final design of the crane.
Note
For simplicity in this book, I often neglect self weight. If I don't mention the mass of the object, I also exclude the self weight from the F.B.D. and subsequent calculations.