The equilibrium equations I present in Chapter 16 give you the tools you need to begin studying the effects of behaviors on structures. You're well on your way to determining support reactions, calculating internal forces, and solving specific application problems with either scalar or vector techniques. For two-dimensional problems, scalar solution techniques are much more efficient, and that's what I show you in this chapter. But never fear — I show you the vector methods (which you almost always need for three-dimensional problems) in Chapter 18. In this chapter, I show you how to use your free-body diagrams (F.B.D.s — see Part IV) to determine the magnitude and sense of unknown support reactions. I start the chapter by outlining the three basic steps that you follow when working a scalar statics problem and then I show you how to create the translational and rotational equilibrium equations in two-dimensional situations. Finally, I highlight a few considerations for selecting points for moment equilibrium equations.
As you start to analyze a system or object, keep in mind that you're always using the concepts of static equilibrium to determine the unknown forces and behaviors. Although some problems (such as those defined in Part VI) may require specific techniques to unravel them, you quickly discover that the vast majority of statics problems begin with the same basic steps:
Draw a free-body diagram of the object of interest.
Construct your F.B.D. by including all information about the applied external forces, any revealed internal forces, the self weight, and unknown support reactions.
Tip
Even if you don't know the magnitude or sense of a particular action at this point, go ahead and include it on the diagram. Just make sure that the action is located at its proper point of application and oriented with its proper line of action, and give the load a label.
Write the equations of equilibrium for your free-body diagram.
The number of equations you have to write depends on the number of dimensions of your free-body diagram:
For two-dimensional problems: If you've constructed a two-dimensional F.B.D., you write three equations. You need two translational equations (sum of forces) and a rotational equation (sum of moments).
For three-dimensional problems: If you've constructed a three-dimensional F.B.D., you need six scalar equations (three translational equations and three rotational equations).
In this step, your primary goal is to determine the magnitude of as many of the any unknown support reaction forces as possible.
Calculate any necessary internal forces.
After you've calculated as many of the external reactions as possible, your next step is usually to determine internal forces and other more specific information about the object(s) you're working with. Many of the problems you encounter within statics have specific solution techniques once the support reactions have been determined. I highlight these different types of problems in the chapters of Part VI.
In the coming sections, I show you the techniques for computing the support reactions.
When you know the steps (see the preceding section), you're ready to put Mr. Newton's equations to work for you.
Consider the structure of Figure 17-1a, which is loaded with a 0.5-kip-per-inch uniformly distributed load from Point A to Point B, a concentrated load of 10 kip at Point B, and a concentrated moment of 40 kip-inches at Point C. The structure is supported by a pinned support at Point A and a roller support on a 30-degree incline ramp at Point D. In the following sections, I show you how to determine the magnitudes of the external support reactions using this example.
Your first step is to create the system F.B.D. by using the tools in Part III. Make sure to include the following four types of forces on your diagram:
External forces: In the example of Figure 17-1, they're actually provided for you. Just copy the distributed load, the concentrated load, and the moment to their locations on the F.B.D. as shown in Figure 17-1b.
Internal forces: For this F.B.D., you have no internal forces to include because you don't have to actually cut the structure.
Self weight: Neither the mass nor the self weight is given, so you can assume that the self weight is negligible in this problem. This example has no self weight forces.
Support reactions: In this example, the support at Point A is a pinned support, which means that you need to include at least two unknown support reactions. Even though you don't know the sense or the magnitude at this time, you can work around this problem. (I show you how in the "Solving for the unknown reactions" section later in this chapter.)
The roller support at Point D is a bit more complicated. Remember: A roller support must have a single reaction force that is normal (or perpendicular) to the plane of the support. In this example, the support is sloped at 30 degrees from the horizontal, so the line of action of this reaction must be perpendicular to the plane of the support.
In Chapter 16, I introduce you to the equations that you need to develop in order to establish equilibrium conditions. For a two-dimensional problem, you need to write two translational equilibrium equations (sum of forces) and one rotational equilibrium equation (sum of moments):
When using the equations of equilibrium, notice that the equations contain vector expressions. However, you can actually cheat a little if you consistently assume a vector direction. If two vectors, F1 and F2, are both acting in the positive Cartesian x-direction, you can write their vector expressions in terms of their scalar magnitudes and their Cartesian unit vectors:
The equilibrium equation in the x-direction is then written as
Notice how the unit vector doesn't actually affect the outcome in the expression. At the end, you're left with a scalar expression that involves only the magnitudes of the forces acting in that direction.
Tip
I also recommend including a bit of notation before the expression to help you remember to be consistent with the directions that you assume to be positive:
In this case, I like to add an arrow indicating the direction that I assume to be positive. This arrow serves as a constant reminder as I establish the equations of equilibrium.
Tip
It doesn't matter which way you assume is positive when writing the equilibrium equations — just pick a direction as positive and then be consistent with it when you start writing the equations.
Similarly, the moment vector equation can also be simplified to scalar expressions if you choose a consistent direction for the axis of rotation. You can also add a similar reminder before each of your moment equations:
When you're solving for reaction forces, it doesn't really matter which equilibrium equation you write first, so for this example I start with summing forces in the positive x-direction. To write this equilibrium expression, you must include every component of a force (see Chapter 8) that acts in the positive x-direction. For example, consider the reaction at Figure 17-1's Point D, shown in Figure 17-2.
On the F.B.D., I assume that the force is acting up and to the left. Based on the assumed direction of RD, the corresponding x-component RDx must be acting to the left, and the y-component RDy is acting upward. You must get these directions correct, with respect to your assumed direction for RD. A mistake in the directions of force components will often prevent you from getting a final correct answer.
The scalar equilibrium equation for translation in the x-direction is
In this equation, I've indicated that the positive direction is in the direction of the positive x-axis (or to the right). RAx is entered into the equation as a positive value because it is acting in the same direction as the assumed positive direction. Likewise, the x-component of RD is negative because it acts in the opposite direction. This is one of the expressions required for translational equilibrium. At this point, you can't actually solve this equation because there are two unknown values in this equation. However, you have created a relationship between RAx and RD that satisfies Newton's laws of equilibrium.
Next, you can write the translational equilibrium in the y-direction:
In this equation, the vertical reaction RAy is entered as a negative value because a positive direction was assumed upward, and this force was assumed to be acting downward. RDy is positive because the vertical component of the assumed RD is acting upwards. Finally, all the external forces are included. The 10-kip concentrated force is negative because it is acting downward, and the resultant of the distributed load (0.5 k/in) · (40 in) is also acting downward.
Note
Notice, however, that I haven't included the concentrated moment in either of these translation equations. Concentrated moments show up only in the rotational equilibrium equation because they're rotational behaviors — they don't show up in the translation equations. However, as you see in the following section, both concentrated and distributed forces show up in the moment equilibrium equations.
The third equation that you need to write is the rotational equilibrium equation. This equation behaves a bit differently because, unlike the sum of forces equations that you just wrote, the moment equation requires you to pick a specific point about which you calculate the equivalent resultant moments for the system.
For the sake of this example, I arbitrarily choose Point B from Figure 17-1 as the location and see what happens. (If you want to see an even easier way to solve this problem, skip ahead to the "Choosing a Better Place to Sum Moments" section later in the chapter.) You calculate the equivalent moment of each of the actions on the structure about Point B as shown in Figure 17-3.
Assuming a counterclockwise moment to be positive, the rotational equilibrium equation about Point B is
Each term in the moment equation is the equivalent moment of an action about Point B (which is the point you're summing moments about). Each force (or components) is included in the expression and includes the perpendicular distance to Point B. If it causes a rotation in the opposite direction, you include a negative sign before that term. A concentrated moment (40 kip per inch, in this example) is included in the equilibrium equation as a single value. It is added (or subtracted depending on its direction of rotation) to the equation directly.
Note
If you have a concentrated moment, this value does not require an associated distance. A moment is a free vector, and its equivalent moment at any location is the same as the original moment. The sign before this term in the equation is based on the assumed positive direction of rotation.
Simplifying, you can derive the final rotational equilibrium equation with respect to Point B:
By examining each of the equilibrium equations, you can see that each of the three unknown reactions appears in at least one of the equilibrium equations (and in the case of RD, multiple appearances):
With a few basic algebra skills, simultaneous solution of these equations produces the values of the magnitudes for RD, RAx, and RAy. Of course, the math isn't exactly friendly, but the statics is complete, and you're now left with solvable equations that produce the following reactions:
Note
After the algebra is complete, if you find that a particular numerical result has a positive value, you know that the direction you assumed on the F.B.D. was correct for that load. Notice that RAy had a value of −17.0 kip, which tells you that instead of acting downward as you originally assumed, it's actually acting in the opposite direction (or upward).
Tip
As long as you're consistent in assigning directions for all your scalar values in the equations, the signs will tell you whether your assumptions are correct.
Warning
When you work with scalar equilibrium problems, if you make even a single mistake with one sign or assumed direction, you're probably done for. It's not because scalar methods are overly that difficult — it's just human nature to make simple sign error mistakes from time to time.
When you're writing the rotational scalar equilibrium equation, you can choose any point about which you sum moments, even a point not physically on the structure. Referring to the Figure 17-1 example in the preceding sections, if you were to sum moments about Point A rather than Point B, you would eliminate the reactions at Point A from the equilibrium equation altogether, because at this summation location, the perpendicular distances from the point of interest to the reaction forces at Point A are zero.
Now this equation only has one unknown, RD, and you can solve for it directly, making the math significantly easier than summing moments at Point B.
After you have the value and sense for RD, you can then compute the reactions RAx and RAy from the translational equilibrium equations.
Tip
By carefully selecting the points about which you sum moments, you can greatly simplify your equation writing. When you're looking for possible locations for rotational equilibrium points, you may want to consider the following:
Pinned support locations and internal hinges: Pins and hinges always have two unknown forces associated with them.
Locations where lines of action of multiple unknown forces intersect: This point is known as an instantaneous center or a point of concurrency.
Locations that eliminate troublesome forces with inconveniently oriented lines of action: As the last example shows, summing moments at Point D eliminates some nasty trig calculations from your equations. If you're not sure how to handle an unknown force, sum moments at that point so that you can eliminate it from the rotational equilibrium equation altogether. But remember, just because you eliminate it from the sum of moments equation doesn't mean you can eliminate it from the sum of forces equation.
For example, if you had chosen to sum moments at Point D:
Notice how choosing Point D eliminates all the trigonometry terms that showed up in the other equations. However, the problem with summing moments at Point D, however, is that terms for RAx and RAy both appear again, which requires solving simultaneous equations again. Regardless, the final solution is the same regardless of which moment summation point you choose.