First, one point needs to be clarified. Statics works on any structure in equilibrium. So, if you want to, you could calculate what it would take to support your soup pot if you rested a brick on it (believe me, though, it's not too exciting). This section right here deals with simple trusses, a kind of idealized structure of connected joints. When you use our web programs, you're building simple trusses.

So, to find all the forces inside a simple truss, you have to accept a few conditions (aren't there always some catches?). First, all joints are able to rotate freely, as if the beams are held together at each joint by one, highly greased pin. This means the joint can't do any supporting of its own, you have to position your beams to do this. No squares, in other words; only triangles will work. A square (or any other shape with more than three sides) would just collapse sideways.

Another condition is that a force can only be applied at a joint. You can't sit something on a beam. The combination of this and the previous one means that beams can only get forces directed along their length (compression or tension, both discussed later). It's like if you have a toothpick, one end on each fingertip, where both ends can move freely (no cheating by grabbing an end). Now, try to put a bending force (that's another kind of force on a beam besides compression or tension) on the toothpick. You can push towards a finger (that would be compression), but if you try to push up or down, the toothpick just follows your finger and doesn't bend. You can't do it without holding onto an end or pushing down somewhere in the middle of the toothpick. And these are the two things that you have to accept as not occuring in simple trusses. Finally, the beams don't have any weight of their own. The only load in a simple truss comes from outside.

Now, the method we're going to use (there are a couple of methods, though all very similar) is called the Method of Joints, named because it breaks a large structure into small parts: the joints. For each joint, you draw a Free Body Diagram showing all the forces leading into it. But, looking at each of those free body diagrams, can you even start to solve even one of them? No, because you don't have any information to start from.

Now we need to decide on the force to be applied to our structure. We could even apply more than one force, but the programs on this web site don't support that (sorry, I guess we're not such genius programmers after all). This new force (or forces) is just drawn into the free body diagram of whatever point it connects to, along with its magnitude.

Up until this point we've been looking at this structure as more of a collection of separate points, not as a complete object, which is another completely acceptable view. By doing this, we can find the reaction, or support, forces (remember, Newton's Third Law, about all forces having equal and opposite reactions). These are the external forces that keep the structure from just moving as a whole. So, all the vertical support forces are going to have to equal the load force (and so will the horizontal ones, if there are any horizontal loads). But the support forces aren't necessarily going to be equal to each other. We also have to take moments into consideration. They also have to all add up to zero, and for this to happen, the support forces may have different magnitudes. In general (with only a vertical load force), the support that is closer (in a purely horizontal direction) to the load will have a larger magnitude than the other support. If you've read and understood both Calculating Moments and Calculating Equilibrium, you should now be able to write two equations (one for vertical forces and one for the moments) or maybe three if you also have to deal with horizontal loads.

In our example, two equations would be written. For now, we'll represent the magnitude of the left support force as M1 and that of the right support as M2. The load, we'll decide, is 100 Newtons. So, the equation for the vertical forces is "M1 + M2 - 100 = 0" (or, moving the numbers around a little, "M1 + M2 = 100"). The one for the moments (arbitrarily calculated around the left support) would be "0*M1 + (-10)*M2 + 7*100 = 0" (or "0*M1 + (-10)*M2 = -700"). The -10 is negative because the right support exerts a counter-clockwise moment. Likewise, the 7 is positive because the load exerts a clockwise moment. This convention was covered earlier in Moments, Not What You Thought They Were. These two equations can then be solved with conventional Algebra or with Cramer's Rule (covered in Calculating Equilibrium).

With all the external forces figured out (shown at right), we can turn to the internal ones. In the case of our example, we can start at any joint we want, but in general, you want to start at the joint with the least number of unknown forces (no more than two) and at least one known force. Once you've solved one joint (covered in Calculating Equilibrium), you can fill in some more force magnitudes. I know you've already filled in the ones around the joint you just solved, but there are more. This is a structure, remember, and the parts are all connected together. In the example at right, all the forces known after solving the joint at the lower-left have been filled in. You'll notice that the upper force on the lower-left joint (still following me?) has had it's arrow reversed in direction. This is because a negative magnitude was found for that force when the joint was solved. The arrow could have been left as it was, and the force's magnitude written as a negative number, but this way it's easier to visualize what the forces are actually doing. You'll also notice that we now know the forces at the other end of the two beams connected to our lower-left joint. To find those, all you do is keep the same magnitude as the force at the other end and reverse its direction.

Doesn't make sense?(What's new??) Ok, I'll lead you through a little logic so it does. The force at one end of the beam represents the force exerted on the joint by the beam. If it's pushing on the joint, then, by Newton's Third Law, the joint must be pushing back on the beam. So that means the other end of the beam is going to be pushed up against it's joint, exerting a pushing force on that one, too. Both ends of the beam are pushing on their joints, so the directions of the forces are going to be opposite. If the beam was pulling on the joint, just substitute "pulling" for "pushing" in the previous argument and everything still holds.

Another way of thinking about it is as the beam being squashed at both ends by joints, so it pushes back out. This case is called compression. If the beam is being pulled apart, it's in tension. One important thing to note is that, in a drawing, if the beam is exerting forces out, that's compression (it's reacting to being squashed), and if it's exerting forces in, that's tension (it's reacting to being pulled apart). The left and right sides of our bridge are in compression and the bottom is in tension.

Whew, that was a fair amount to absorb in one sitting, but most of it was just primer to get you used to the concepts. Actually calculating all the internal forces isn't too hard, if you can calculate equilibrium of a single point (covered in Calculating Equilibrium). You just hop from point to point filling in whatever you can. There's no limit to how big a structure you choose to make! Aren't you happy now?

In the next section, we'll discuss the limits of this method. It will also discuss a few shortcuts and other problems you may run into.