Statics problems in three dimensions are some of the most intimidating types of problems you encounter. You can easily get lost in all the dimensions, coordinates, and angles in three dimensions. A single force in three dimensions is completely capable of producing moments about three different Cartesian axes (see Chapter 5). And determining whether a rotation is positive or negative by the scalar methods I show in Chapter 17 can be a real challenge (though not impossible with the right-hand rule I describe in Chapter 12). If you choose to work with vectors, a lot of those difficulties are automatically handled for you. And after you've become an old pro at turning forces and distances into vectors, why not let vectors do the work for you?
In this chapter, I show you how to apply all those vector calculations to the free-body diagrams (F.B.D.s) I cover earlier in the book. Then I show you how to write the equations of equilibrium in vector form to solve for unknown support reactions.
The solution method for solving three-dimensional statics problems is very similar to the solution methods for two-dimensional situations that I describe in Chapter 17. Although it requires a bit more work upfront, the actual completion of the equilibrium equations is much easier from a statics point of view; the math, on the other hand, can be a bit lengthy. To solve for support reactions in three-dimensional problems, you follow a similar set of steps to the two-dimensional scalar methods, with a few minor exceptions and additions.
Following is a basic summary of the steps required to find the support reactions for a three-dimensional problem:
Draw a free-body diagram of the object of interest.
Note
You include all concentrated forces, distributed forces, and moments on your F.B.D. Be sure to include the necessary information for describing the lines of action of your forces and the axes of rotations for your concentrated moments. You need to construct unit vectors in this process, so be sure to include any necessary information to construct position vectors in order help you define these important directions. Also, remember that you must provide coordinate data and angles.
Write the equations of three-dimensional equilibrium.
The number of equilibrium equations that are required to solve a three-dimensional problem depends on the method you use to solve the problem:
Scalar methods: If you plan to perform scalar equilibrium calculations on three-dimensional problems, you need to work with a total of six equilibrium equations per F.B.D. — three scalar equations for translational equilibrium and another three scalar equations for rotational equilibrium. (See Chapter 16 for these six equations.)
Working a scalar method in three dimensions is actually no different than it is in two dimensions. The process is just a little trickier in three dimensions because you need to keep track of the directions (whether they're positive or negative) by hand, just like you do with the two-dimensional problems of Chapter 17. However, in three dimensions, a force can create a rotation about any of the three Cartesian axes, and keeping track of those signs and magnitudes can be a bit confusing.
Vector methods: For vector method solutions, the process is actually a lot more straightforward. Instead of needing six scalar equations, you need only two vector equations to establish equilibrium:
Although you're working with fewer equations, you have to do a few more calculations upfront — you need to create vectors out of every force and moment on your free-body diagram. (I describe many of those methods earlier in the book, so if you need a refresher, look to Chapters 5, 7, and 8.)
Note
You use these equations to determine as many of the external reactions as possible. If your F.B.D. has no more than six unknown reactions, these two vector equations can help you calculate the magnitudes of all of them. If your F.B.D. has more than six reactions, you may not be able to actually determine their magnitudes, but you can to write expressions that relate them to each other.
Calculate necessary internal forces (see Part VI for these techniques).
In Chapter 16, I define equilibrium for an object as the state that occurs when all translational and rotational behaviors are balanced at the same time, and in the preceding section, I show you the two basic vector equations needed to define equilibrium. In this section, I explain each of these equations in more detail and show you how the vector equations actually contain the scalar equations in them automatically.
For an object to be in translational equilibrium, the net effect of all forces must be in balance. In statics terms, this setup means that the resultant of all translational behaviors (forces) must be equal to zero.
In Chapter 16, I mention that it takes a minimum of three translational directions to establish equilibrium for a three-dimensional object. So, if you arbitrarily choose the Cartesian x-, y-, and z-axes to represent your three directions, the resultant force F acting on an object can be expressed as
If an object is said to be in equilibrium, the resultant vector F must take the following form:
In order for these two expressions to be equal, the following expressions must be true:
That means that the sum of forces in each of the three Cartesian directions must be equal to zero. Notice that these three equations are the same translational equilibrium equations I mention in Chapter 16.
In order for an object to be in a state of equilibrium, all the rotational behaviors (or the moments) must also be balanced — that is, the resultant of all rotational effects must also be zero. To establish rotational equilibrium, you need to ensure that the rotational effects about three different non-coplanar axes are also balanced. Following the logic for translational equilibrium in the preceding section, you can write a similar expression for the resultant moment vector M acting on an object as the following:
For an object to be in rotational equilibrium, the resultant moment M must be given by
This resultant implies that the scalar components of the moment about any three axes must be equal to zero to ensure rotational equilibrium:
Notice that these three equations are the same rotational equations that must be satisfied using the scalar methods.
As with two-dimensional equilibrium problems, you always want to try to compute all of the magnitudes of the unknown support reactions — or as many of the magnitudes as possible — after drawing the F.B.D. of the system. Three-dimensional problems are no different.
In Figure 18-1, a uniform 2-meter-x-3-meter plank with a mass of 10 kilograms is supported by a three-dimensional roller at Point A and a three-dimensional pinned support (or ball and socket) at Point D. Both Points B and C are tied with cables to a point on the wall at Point E. A concentrated point load of 100 Newton is applied on side AB at an angle of 30 degrees below the horizontal, and parallel to the yz plane. A moment of 300 Newton-meters is applied along edge BC. In the following sections, I walk you through the basic process for computing the magnitudes of the reactions for a three-dimensional statics problem. But first, as always, you need to create a proper free-body diagram first.
With any statics problem, regardless of whether it's two-dimensional or three-dimensional, your first step in the solution process is always to sketch a free-body diagram of the forces and moments that are acting on the object. Then you need to find as many of the unknown support reactions as possible.
In the example of Figure 18-1, if you draw an isolation box (which lets you zoom in on a specific feature of a larger object) around the plank by cutting the two cables, you can expose the support reactions that you're seeking. (See Chapter 14 for more on using isolation boxes.)
External forces: Typically, you know the external forces acting on a system. The external forces for this problem are the 100-Newton concentrated point load alongside AB and the 300-Newton-meter concentrated moment along edge BC.
Internal forces: Internal forces appear anytime a physical structure is cut. In this example, you're cutting both of the cables BE and CE in order to isolate the F.B.D. of the plank. You know that cables are axial members only, so you need to include a force for cable BE, TBE, along the line from Point B to Point E. Similarly, you include another force for cable CE, TCE, acting along a line from Point C to Point E. Even though you don't know the magnitude of these forces, you must include them on the free-body diagram.
Support reactions: The external support reactions in this example are a three-dimensional roller at Point A, which has a single point load Ay acting upward in the positive y-direction, and a three-dimensional pinned support at Point D that has three component forces (Dx, Dy, and Dz) acting parallel to each of the Cartesian axes.
I should point out that in this example, you can also combine these scalar reactions into a single resultant (see Chapter 7) that represents all of the concentrated forces for the support reaction. For RD, the reaction resultant vector at Point D:
where Dx, Dy, and Dz represent the scalar magnitudes of the components in the x-, y-, and z-directions respectively. Similarly, for the reaction at Point A:
However, because the support at Point A is a three-dimensional roller, you automatically know that Ax = 0, and Az = 0, so the reaction at Point A can be simplified to
Self weight: In this problem, you also know that the plank has a mass of"10 kilograms, which means that you need to include the self weight on"your free-body diagram. Compute the weight as W = mg = (10 kg) · (9.81 m/s2) = 98.1 N of the plank and apply it at the center of mass of the plank acting downward (in the direction of gravity), or in the negative y-direction. (For the lowdown on self weight, head to Chapters 9 and 10.)
Note
When using vector methods for solving, be sure to convert every force and moment on the F.B.D. to an appropriate vector form — Cartesian forms are usually the most common — by using the techniques I describe in Part I of this book. The following list helps you break down these forces for the example in Figure 18-1, much like the list in the preceding section does for the isolation box.
External forces: To convert the moment along edge BC, the double-headed arrow notation (refer to Chapter 12) tells you that the direction of the moment is about the negative z-axis. In vector form,
You must also convert the 100-Newton concentrated load along edge AB to a vector. You can use direction cosines (which I cover in Chapter"5) to determine the unit vector and then simply multiply by the 100-Newton magnitude:
Internal forces: The internal forces of the cables are unknown, but you still need to establish expressions for them in terms of their unknown magnitudes. For cable BE, you need to define a position vector from Point B (2,0,3) to Point E (0,4,3) in order to create a unit vector:
Similarly, you can create another tension vector for cable TCE using a position vector from Point C (2,0,0) to Point E (0,4,3).
In both of these cable forces, even though you don't know the magnitude of the force at this time, you can still treat it as a variable and continue with your calculations. In this case, TBE and TCE represent the unknown magnitudes of their respective force vectors.
Support reactions: For the roller support at Point A and the three-dimensional pinned support at Point D
Just as with the cable forces, you still have to include the magnitudes of the support reactions in the vectors even though they're unknown. Just leave them as variables for now — I show you how to deal with them in the following section.
Self weight: The self weight is acting in the direction of gravity (assumed downward), so the self-weight vector is expressed as
In Chapter 16, I introduce you to the equations that you need to develop in order to establish equilibrium conditions in three dimensions. In three-dimensional problems, you must calculate the vector resultants for forces (translational effects) and the vector resultants for moments (rotational effects), which requires a total of six equations (three each for forces and moments).
You establish translational equilibrium of the plank in Figure 18-1 by computing the resultant force vector of all the force vectors on the system and setting that sum equal to zero. For this example:
Substituting the appropriate vector equations that you computed in the previous section, and gathering all the terms with common i, j, and k directions, you can generate the equations of translational equilibrium as shown in Table 18-1, which uses units of Newton.
By summing each of the columns of this table, you can create the three translational equilibrium components:
i direction: −0.447(TBE) – 0.371(TCE) + Dx = 0
j direction: −50 + 0.894(TBE) + 0.742(TCE) – 98.1 + Ay + Dy = 0
k direction: 86.6 + 0.557(TCE) + Dz = 0
At this point, you have six unknown magnitudes in this problem — TBE, TCE, Ay, Dx, Dy, and Dz — but only three equations with which to solve them. You need to get your remaining equations from summing moments about some reference point.
Summing moments is probably the most labor-intensive step of solving problems with vector methods because, chances are, you'll be computing a lot of cross products (those pesky calculations I show you in Chapter 12). In fact, you need to compute one cross product for every force vector on the object. (Imagine if you had 100 forces on an object — that's a lot of matrices!) So, this stage is where you have to be extra careful in your selection of a reference point for your equivalent moment calculations.
Tip
As a general rule, select your equivalent moment reference points such that they produce the simplest mathematical expressions possible. (After all, why make your work any harder than it already is?) Places where multiple unknown forces intersect at a common point or points where dimensions are easy to compute (such as corners of objects) are often prime candidates for selection of your equivalent moment reference point. At the end of the day, the point you choose really doesn't matter from a mathematical standpoint, because your solutions produce the same numerical value — your equations just end up a lot more complex if you don't choose a convenient point.
For Figure 18-1, at Point D, you have three of your six unknowns acting simultaneously (or concurrently). Based on the criteria, this spot would be a good candidate for your reference because you can eliminate all three reactions at Point D from the moment equation.
Now you're ready to start computing equivalent moments. Consider each force on a case-by-case basis. You need to compute position vectors for each force on the F.B.D., so you create those by drawing your vector from the reference point (Point D) to the point of action of each force. To help you see these vectors, I've sketched each position vector for you in Figure 18-2.
The next step is to compute the equivalent moment vector equation based on the forces and moment vector of Figure 18-2. You need to perform a cross product calculation for each, and then add each of those calculations together. For this problem, the moment about D (MD) is written as
You compute the position vectors for each expression just as I explain in Chapter 4.
Note
The first letter of the subscript represents the tail point of the position vector, and the second letter of the subscript represents the head point. The moment vector M doesn't require a position vector because it's already a moment and it can be moved freely from one point to another on the object. Also, the term containing the position vector rDD vanishes because the length of that position vector is 0. (The head and tail of your position vector are at the same point.) I selected this point for that reason — it greatly simplifies the math.
Table 18-2 shows the results of the cross product computations for each of the quantities required for MD, the equivalent moment at D. The units for the table are Newton-meters.
As you do with the forces, you sum each of the components in the i, j, and k Cartesian unit vector directions. From these results, you can then derive the final three equations for rotational equilibrium:
At this point, all you're left with is a mathematics problem that involves solving for the six unknowns. For this example, I would probably start with the last two expressions of the rotational equilibrium equations and solve those simultaneously for TBE and TCE. After you've computed those values, you can easily substitute into the remaining equations and solve for the other unknowns. The final results for this problem are
Note
As with two-dimensional problems, the positive signs on the magnitudes of this example indicate that the directions of those respective forces on the F.B.D. were in the correct directions. Negative magnitudes tell you that the forces were drawn backward.