Good day. Today we are going to talk about a special class of structures called trusses. Trusses you see often when you look at electricity cable towers, at the roofs of industrial buildings, sudden bridges, etc. So this is an example of a truss, drawing of a truss, showing the thickness of the member. In the drawing below, we show the same truss without showing the thickness. So we represent its member as a straight line segment. The three basic properties that a truss should satisfy are the following. The first is that the members are all straight members. So we don't have any curved members. Second property is that the members are connected with pins at the ends. The connecting points are called joints. The members are shown here. So each member is pinned at its end at the joints. The last property is that any applied loads are applied directly on the joints. There are no loads applied within the span of a member. So in this case, we have 1, 2, 3, 4, 5, 6, 7. We have seven members, and we have 1, 2, 3, 4, 5 joints. So this is the number of joints. Of course, in reality, a member has a weight. So in reality, we have a little force representing the weight of the structure acting at its midpoint. But we assume that this load is very small compared to the applied external loads and usually what we do we distribute this load half to each of its two ends. So finally, we have a free body diagram where the loads we assume are all applied at the joints. Now, the nice thing of trusses, the nice property of trusses is the following. If we consider any member, for example, if we consider member CD and I will show it here as a straight line. This member since it is pinned at the two ends C and D, the only forces acting on it are the forces at its end from the surrounding. When we draw the free body around here, there is no force between C and D. So there is only one force, but here is no couple applied because we assume that this end is pinned, and similarly here, there is some only one force acting at this end. We have said that such a member is called a two force member. That means in reality, we assume that we have four unknown reactions here. But these four reactions, these basically two forces, if we consider the equations of equilibrium of this body, then the three equations of equilibrium, then we can conclude that the only way this body can be in equilibrium is under one of these two scenarios. One scenario is that we have two forces acting at its end along the direction of the member and opposite each other pulling on the member. Let's call this force FCD, or we have again two forces opposite to each other which are in this case compressing the member. So this is tension, the force is pull on the member and this is what we call compression, and so on. So each of these member is either a tensile or compressive member or possibly it could be what we call zero force member if it happens that the force within the member is zero. Now, rather than using the adjective tensile or compressive force, we are going to make a convention, accept the convention that we will call all tensile force as positive and all compressive forces negative. That way, if we give you the force of member FCE to be plus AD, it will mean it is a tensile force of 80. If we state that FBD is minus 150, it means it is a compressive force of 150. Now, when we consider the equilibrium of a body like this which is comprised of seven rigid members, we have said that it is not sufficient to talk about the overall equilibrium. That means writing the three equations of over-roll equilibrium do not satisfy equilibrium of the structure. To satisfy equilibrium, we must make sure that each sub member, each rigid body comprising the structure is under equilibrium. All right. Now, let's try to write all the necessary and sufficient conditions of equilibrium for this structure. So as we said, for each member, we can draw a free body diagram like this. For example, for member CE, we are going to draw and we will assume for now that everything is tensile. So in our drawings, we will assume that all forces are tensile. If the member force turns out to be negative, automatically we will be able to conclude that the force is compressive. So for each of these members, for each of the seven members, we can draw a free body diagram with a tensile force FCE acting on it. That already will have guaranteed equilibrium of each member from one end to another, from one end to another, from this member to this member. FCD from here to here, FDE from here to here, from here to here. So at this point, when we talked about all these free body diagrams using the unknown forces for each member, we already guaranteed equilibrium of the structure of the different members. We've gotten the equilibrium of this, we've got the equilibrium of this, we're going to get equilibrium this member, this member, this member, this member, and this member. So what is left of the structure? What is left as we can see is the joints, the remaining part of the joints. We have not talked about the equilibrium of this joint. So, we have talked about the equilibrium of a member from one end to the other end for each member, but the remaining joints we still need to require their equilibrium. So for example, if we talk about joint B, or let's start with joint A. We will show the free body diagram of this. So here is A. We have at A clearly two reactions, a reaction A_Y and a reaction A_X. So we will have to show this to forces A_Y and A_X. At this point, I will have to show, at this point, the force that this member acts on the joint. Assuming that the member was under tension, it means that this member is pulling on the joint with a force F_AB. Similarly, for member AC, assuming the member itself was under tension, it means the member pulls on the joint with a force F_AC. So this is the three body diagram of joint A. Similarly, we can do the same for each joint. Lets do for example joint B. For joint B, we'll show the three members connected to joint B. We'll, show clearly the forces acting at joint B assuming all members were under tension. We'll show tension forces. Forces pulling on the joint F_AB, F_BC, and F_BD. In this case, there was no external applied load. If we're considering for example the point C, then that group of joint C would look something like this. Four members, there are forces acting from the members on the joint. The force we called F_AC, the force F_BC, the force F_CD and the force F_CE. In this case, since we have an external applied load at C, we will add an external applied load. So we can do that for all joints, and we must satisfy equilibrium of the joints. In general, three equations of equilibrium for every structure. But since all these forces are passing through the joint, the equation sum of the moments is not active, okay? So therefore, we can write only two equations sigma f_x equal to 0 and sigma f_y equal to 0 for each joint. So concluding, we can write two, four, six, eight, ten. We can write two times j, 10 equations. While the number of unknowns are the member forces one, two, three, four, five, six, seven, the member forces, plus the reactions. That means 7 plus 3 we have 10 unknowns. So in this particular case, we can see that we have as many equations as unknowns. We can also show that this structure is stable. We are going to show that a little bit later. Therefore, we have a stable structure with unknowns equal to the number of equations. That means, we can solve these equations, and solve for all the unknowns. Which unknowns are the three reactions A_X, A_Y, and E_Y, as well as the seventh member forces F_AB, F_AC, F_BC et cetera. So let us now go and look, say some words about the stability of a truss structure, and of course, assuming the structure is stable, then if a number of equations equal to the number of unknowns, we claim to have a statically determinant problem. Statically determinant means we can determine not only the unknown reactions but also all unknowns which are also all unknown member forces. Notice that, while we talked about 2 times 5, two, four, six, eight, ten equations, the three overall equilibrium equations, which we could have written at the very beginning, is a matter of fact to solve for the unknown reactions. They are not independent equations, that means once you get into the equilibrium of the joints, and since you already guaranteed equilibrium of the different members, then the overall equilibrium will be automatically satisfied. Therefore, these equations are not independent. If you used already three of the overall equilibrium equations, then three of this joint equations would be dependent. So we would not need to consider them in order to solve for all unknowns. Okay. Let's go now and talk a little bit about stability