The versatility of frames and machines makes them extremely useful in day-to-day life for the work and convenience that they provide. In fact, the vast majority of structural systems you encounter are classified as either frames or machines and may require you to use slightly different techniques to analyze them than the methods I outline for working with trusses (Chapter"19) and bending members (Chapter 20). The methods I present in those chapters are a whole lot simpler than the methods I present here, so you should apply the rules in this chapter only as a last resort. Though these techniques aren't difficult, they can require a lot of additional work if you're not careful. However, these techniques do allow you to solve problems that you may not otherwise be able to.
In this chapter, I show you how to identify frame and machine systems and then I tell you how to determine if you even need to use these methods in your solution. Finally, I show you the actual procedure for dealing with frame and machine systems and some of the more common components that are added to them.
You need to be able to identify systems quickly and efficiently. The following sections can help you to recognize that your system is a frame or machine by discussing the basic properties.
The two types of structural systems that I discuss in this chapter are frames and machines and they have several basic differences:
Frames: Frames are structural systems (such as your house or office building) that are constructed of multiple members, at least one of which must be connected by rigid joints (or joints that don't rotate freely).
Machines: Machines are systems that contain moving parts, such as pulleys, blades, and pistons, that are created to perform a certain task. Hand tools such as pliers, vice grips, and scissors are all examples of simple tools that are classified as machines. Machine systems produce the same additional internal forces in the members as frames, but may or may not have the same rigid joint requirement as a frame. Machines are typically used to transmit forces from one location in the structure to another. For example, a piston assembly in your vehicle's engine receives forces from the combustion process and transmits those forces through the transmission, and into the axles or wheels, causing your vehicle to move.
Aside from these differences, frames and machines also have several properties in common:
Member shapes: Members do not have to be straight (or linear). They can also be curved or bent into a wide variety of shapes.
Internal forces: Internal forces on frames and machines have the same type of internal forces as the bending members in Chapter 20 — axial forces, shear forces, and bending moments.
Loads of any type: Loads can be concentrated loads, distributed loads of any shape and order, or concentrated moments, and they can be applied at any location.
Other items included on the structure: Frames and machines can also have other items attached to them. In fact, machines almost always have extra tools attached to them, such as pistons and pulleys.
Statically indeterminate: Frame and machine systems are not necessarily statically determinate (or have as many available equations as there are unknown forces on your free-body diagram — more on those later).
In earlier chapters, I show how to find internal forces and support reactions using the equations of equilibrium with three or fewer unknowns on your free-body diagram (or F.B.D.) for two-dimensional problems and six or fewer unknowns for three-dimensional problems. Unfortunately, many structures often are quite complex, and you can encounter many unknown forces on your free-body diagrams. (Check out Part IV for the complete scoop on working with F.B.D.s.)
Consider the F.B.D. of the two-dimensional machine system shown in Figure 21-1, which shows two members pinned together at Point C and supported by pinned supports at Points A and B. A concentrated load is applied on member BC.
In this F.B.D., there are two unknown reactions at the pin support at Point A, and two more unknown reactions at the pin support at Point B, for a total of four unknowns on the system. Because you have only three equations of equilibrium available for each F.B.D. of the system, this system is said to have one degree of indeterminacy (or the number of additional unknowns present beyond the number of available equilibrium equations). Because Newton's laws of motion (see Chapter 16) provide you with only three equilibrium equations per F.B.D., you must find additional equations elsewhere.
After you've identified the degrees of indeterminacy and confirmed that you dealing with a frame and machine system, the question you face now is exactly how do you solve it?
To solve an indeterminate frame and machine system using only statics requires that your system has internal hinges — this will be the place where you cut the system. In many frames and machines, this point is the point where two or more members are connected, such as Point C in the example of Figure 21-1. Figure 21-2 shows the same system broken apart at the hinges.
Next, draw F.B.D.s for each of the separated members. I arbitrarily assume the pinned forces acting on member AC are such that Cx is acting to the right, and Cy is acting upward on member AC. Member BC has the unknown support reactions and the concentrated load P and forces on the pin at Point C.
Because the pin at Point C itself must also be in equilibrium, and because I already assume that the Cx component is acting to the right on the F.B.D. of member AB, the Cx pin force on member BC must be in the opposite direction. Likewise, the pin force Cy must be assumed to be acting downward on member BC because I already assumed that its counterpart was acting upward on AC.
Note
Don't be alarmed if you perform your equilibrium calculations and discover that your forces have a negative value. Just as with other equilibrium calculations, if you assume the wrong direction for the forces on your F.B.D., the signs of your calculations will always tell you. As long as you ensure equilibrium of the pin with your assumed directions (by making sure that all forces on the pin are balanced), your calculations will still work.
Tip
For each two-dimensional F.B.D. you're able to draw properly, you're allowed to use up to three equilibrium equations — two translational equations and one rotational equation.
In the earlier example of Figure 21-2, you now have a total of six unknown forces on your two F.B.D.s: the four reaction forces R1, R2, R3, and R4 as well as the two unknown hinge forces Cx and Cy. Because you have two members on two separate F.B.D.s, you now have 2 members × 3 equations per member = 6 total equilibrium equations to work with. This means you now have six unknown forces and six equations of equilibrium which means you have a statically determinate set of free-body diagrams, or a solvable system. Now you just have to figure out where to start.
For most frames and machines, you typically want to start writing your equilibrium equations for the member that has a known applied load on it. For the example of Figure 21-2, I start with member BCbecause it has a known load value acting on it: 10 kip downwards. The two translational equilibrium equations that you write are
The trick for solving most frame and machine problems is in determining the unknown hinge forces because they provide you with the extra information to use on all the other attached members. For the example of Figure 21-2, your goal is to get expressions for the hinge forces Cx and Cy.
You still have a rotational equilibrium equation remaining to write for member BC, so you want to write this equation with Point B as your reference. (After all, you don't want to have to deal with the reactions R3 and R4 yet, so sum moments at their point of application, and they vanish from the moment equations.)
This produces a handy relationship between the hinge forces Cx and Cy. You then switch to each of the remaining F.B.D.s and perform the similar equilibrium calculations. First, you write the translation equilibrium equations for member AC:
The four translational equilibrium equations that you've written for members AC and BC are all related to one of the hinge forces at Point C. Finally, I sum moments at Point A on member AC to produce another equation for the pin forces at Point C.
You now have a second relationship between the same hinge forces Cx and Cy. Substituting the rotational equilibrium equations for member AC into the rotational equilibrium equation for BC:
Now that you know the horizontal force Cx, you can substitute into either equation and solve for the remaining hinge force Cy:
After you know the hinge forces, you can then substitute into the remaining equilibrium equations and solve for the support reactions:
For R1: R1 + 2 k = 0, so R1 = −2 k
For R2:R2 +(–2 k) = 0, so R2 = 2 k
For R3:R3 – 2 k = 0, so R3 = 2 k
For R4:R4 − 10 k − (−2 k) = 0, so R4 = 8 k
Note
To solve statics problems involving frame and machine problems and other indeterminately hinged structures, you must find the hinge forces first. If more than one hinge is on a structure, you must solve for the forces on each hinge.
While the concept of "blow-it-all-apart" is a pretty sound technique for solving frame and machine problems, you quickly discover that deciphering the forces that you're dealing with is sometimes a bit harder. In this section, I give you a bit more insight in handling some of the more common situations that you might encounter.
Some frames and machines have more than two members that are connected at the same hinge location. Consider the example shown in Figure 21-3, which has three axial members connected to the same internal hinge at Point D.
The first figure shows the external pinned support conditions acting at Points A, B, and C, for a total of six unknown forces acting on the combined system. With only three equilibrium equations available to solve for six unknowns, you now know that this structure is statically indeterminate to three degrees. This means that you need an additional three equations to solve for the reaction forces. Yikes!
Your first inclination may be to automatically assume that this system meets all the criteria to be solved as a truss with the method of joints (see Chapter 19 for more information on trusses). After all, each member is pinned at the end, each member is axially loaded, and the point load is concentrated and applied at a hinge point. Clearly, this is a truss, right?
Note
In order for the method of joints to work, you can have a maximum of two unknown forces on any given joint at the time you're solving it. This rule means you can't use the same truss solution techniques to solve this problem.
In this example, you actually have three members (with three internal axial forces). At this point, you're left with only one option: using the frame and machine solution techniques of this chapter.
As you did in the previous example, you blow apart the system at all hinge locations and draw free-body diagrams of each member. The F.B.D. for member ADconsists of the pin support reactions Ay and Ax and the force of the hinge at Point D on the member FDA, for a total of three unknown forces.
Similarly, member DB has two unknown support reactions Bx and By, and the unknown pin force aFDB, for a total of three unknowns for this member. Finally, member DC also has two unknown support reactions, Cx and Cy, and the unknown pinned force FDC, for a total of three more unknowns for this member.
In total, this means that there are three members, with three unknowns per member, for a total of nine unknown forces acting on the exploded frame system. But you have three equations per F.B.D. that you created, so that means that you now also have nine equilibrium equations to work with. Nine unknowns and nine equilibrium equations means that this system is now statically determinate and thus solvable!
A pulley is a mechanical apparatus consisting of a round object attached to a shaft that allows the object to rotate. A belt or a rope is then wrapped around part of the pulley and is used to transmit a force from one location to another. This shaft is then attached to bearings (making it a pin support), which is then attached to an external structure or support.
A pulley serves two basic purposes: It provides a convenient change of angle for a force, and when combined with multiple other pulleys, they can help you lift a much heavier object than you might otherwise be capable of lifting. A pulley is also a very convenient mechanism for transmitting forces between mechanical parts. Just look under the hood of your car — several different parts of your engine are connected by a variety of pulleys and belts.
Pulleys are used to change the direction of force. Figure 21-4 shows a person pulling on a rope wrapped around a pulley, lifting a weight W.
In many statics problems, pulleys are often considered as frictionless, meaning, the motion of the pulley, or the wrapping of the belt around the pulley, does not provide any resistance to the person pulling on the rope. (Don't worry, in Chapter 24, I discuss more about friction problems.)
Like other machine F.B.D.s, working with pulleys requires you to break the system apart into individual pieces. For pulley problems, you first draw an isolation box (see Chapter 14) around each individual pulley and cut any necessary ropes or belts — belts and ropes are always axial tension members, so you always know the direction of those forces on the F.B.D. Be sure to also include the support reactions of the pulley on the F.B.D. You then treat the pulley F.B.D. as any other exploded frame or machine part.
Interconnecting multiple pulleys and wrapping the same rope multiple times around the assembly is known as a block-and-tackle assembly. Block-and-tackle assemblies are used to increase the amount of force lifted on one end of a rope for a given applied load at the other end. Consider the three-pulley system of Figure 21-5, which shows a rope passing around Pulley A, wrapping around Pulley C and over Pulley B, and connecting to the middle of Pulley C again. Suppose you're interested in calculating the force required to lift a 900-pound weight suspended from Pulley C. For this example, I assume that each of the pulleys is frictionless.
As with other machine problems, your first step is to explode the system into as many pieces as possible and then draw your F.B.D.s. In this example, I include the rigid links and supports on my F.B.D. for Pulleys A and B. If I had chosen to, I could have drawn a separate F.B.D. for each of the rigid links as well, but because I'm not interested in finding their internal forces, and because they're axial members connected to a support (I recognize this because the links are pinned at both ends and don't have any loads acting directly on them), their F.B.D.s would provide little useful information with regards to calculating the applied load P. As long as I follow my F.B.D. checklist (see Chapter 13), I can actually cut this system in any way I want. I just have to make sure to get all the forces, support reactions, internal axial forces from the cables, and self weight (if there were any) applied at the proper locations on the F.B.D. For this problem, I have three unique free-body diagrams to consider, one for each pulley that I removed from the system (refer to Figure 21-5).
Note
For multiple pulley problems, you want to start with the pulley that has a known load value attached to it. In this case, I'll start with Pulley C, which is supporting 900 pounds.
In drawing the isolation box around Pulley C, I had to cut three cables, which means I have to include one internal axial tension force for each of the cables that were cut—Pulley A and its link to its support, Pulley B and its link to its support, and Pulley C and its support weight of 900 pounds. Applying the equations of equilibrium in the y-direction:
Tip
This equation has three unknown tension forces currently acting on it. But, because this F.B.D. contains a single rope wrapping multiple times around multiple frictionless pulleys, I know that the tension in each of the rope segments must be equal. That is, TAC = TBC1 = TBC2 = T, where T is the tension in each rope segment.
Rewriting the previous equation, I can now solve for the tension T in the rope:
Next, to determine the amount of force applied I need to look at the F.B.D. for Pulley A. This F.B.D. reveals that only a single rope is acting around Pulley"A. So now I know that the tension in the rope on one side of the frictionless Pulley A must be equal to the tension on the other side. This means that:
The force required to hold the 900-pound force on this tackle assembly in equilibrium is 300 pounds.
Note
Because of the mechanical advantage of the way this system is set up, I only need one-third (or 33 percent) of the load in the rope to balance the applied weight. In fact, the more times you wrap the same rope around the same system of pulleys, the less weight will be required in the rope to balance a suspended load. Talk about advantage!
More-complex machines may include unique attachments that provide you with a specific functionality. Some examples of these assemblies that you use in statics are pistons and slotted connections.
The common piston is a tool that you may find in your car engine or an industrial stamping machine (see Figure 21-6).
In your car, the piston experiences a pressure on one edge (or face) due to the combustion of gasoline in the engine. This pressure resultant is a single force, PEQUIV, that is applied to the piston at the midpoint (assuming uniformly distributed pressure). The natural response of the piston is to move in the direction of the applied force, which then exerts a force into the bar AB, which can be oriented at any given direction (in this case, at an angle θ).
If the forces are all concentric (or concurrent), ask yourself this: With just these two forces on the F.B.D., is the piston in equilibrium? Without either of the normal forces NBOT or NTOP, the piston wouldn't be in balance in the perpendicular (or vertical) direction for this example. That means that you'd need at least one contact force of the piston against the sidewall in at least one direction. But which one needs to be included?
In free-sliding assemblies, such as pistons and slider mechanisms, there are actually multiple cases that you need or want to investigate. For the simplest cases, you need to consider the events of the first two of the following cases. For more complex cases, something like the third one would be a possibility as well.
Case 1: If the force FAB were to pull on the piston in the direction shown (up and to the right), the response of the piston would be to move upward and to push against the top edge of the chamber. But as it makes contact with the top edge, it loses contact with the bottom edge (making NBOT = 0).
Case 2: If the force FAB reverses direction such that it is now pushing on the piston in the opposite direction, the piston will want to move down in response. As it moves down, it makes contact with the bottom wall and loses contact with the top edge (making NTOP = 0).
Case 3: If the forces become nonconcurrent, which can occur if the location of PEQUIV varies (such as would happen with an improper combustion firing) and develops an eccentricity, or if the applied pressure isn't uniformly distributed, the resultant location would no longer be at the midpoint. The force acting at an eccentricity from the point of application of force FAB would actually cause a rotation of the piston within the chamber and produce separate normal forces on both the top and bottom walls at uniquely different locations. These forces would then partially act as a couple to resist the rotation due to the eccentric applied load.
The major issue with these types of problems is that you often don't know which case to start with. This situation means that you have to choose a case by making a guess and then verify that the numerical values and their signs from your calculations all logically make sense. (For more on this idea, turn to Chapter 24.)
Slotted holes are another feature commonly encountered in engineering designs. A slotted hole is an elongated hole (often several inches in length) that allows one piece to move relative to another in just one direction while maintaining strict contact in the other.
Note
The purpose of the slot is to remove the restraint in a given direction. Without restraint, support reactions can't develop.
Consider the example shown in Figure 21-7. This apparatus contains a slotted hole that connects two members of a frame assembly. The analysis of this is the same as if the members were connected with an internal hinge, with one exception. Notice that on the F.B.D. of member ACD, the horizontal pinned force Cx is no longer present because the restraint in that direction has been removed by the slot. Likewise, the opposite force on member BCis also missing. Force Cy can be assumed to be acting in any given direction on ACD just as before. Remember: The opposite force direction must be applied on member BC.
