Good day. Today, we are going to talk about two very important concepts in statics and in structural analysis. The concepts of stability and static determinacy. Now, stability is a very important concept, and we will call a structure to be stable. Assuming the structure is made out of rigid members, we will call it stable if it can support an arbitrary loading without changing its geometric configuration. That means without any reflections or deformations of the support, or any other point of the structure. So let us look at some examples of unstable structures. So we can see what we mean. The easiest example would be the example of a beam which is supported by two rollers. Let's assume that this beam is subjected to an inclined loading. It is clear that the only possible reactions are the vertical reactions at the two roller supports. Obviously, the equation Sigma Fx cannot be satisfied. Sigma Fx equal to 0 cannot be satisfied. That is the horizontal component of this applied loading cannot be picked up by any of the reactions and therefore, this body under this loading will accelerate in the x direction. So this is clearly what we call an unstable structure because it cannot sustain this arbitrary loading in its current geometric configuration without any displacements. Of course, a structure which is unstable as this, could possibly be in equilibrium if the applied load is very specific. For example, in this case, if we would have a vertical loading, then clearly, we would be able to satisfy equilibrium because the equation Sigma Fx equal to 0 is a degenerate equation. It's 0 equal to 0. There is no force in the x direction. So it's automatically satisfied. The remaining two equations Sigma Fy and sum of the moments above the point would be sufficient to solve for these two unknown reactions. That is, these two reactions you wouldn't be able to be determined uniquely from the equations of equilibrium, and we could say that this is a case where these reactions are statically determinate. So we have an unstable structure. We have equilibrium. In this case, we have equilibrium, but only for this very specific load. Let's look at a couple more examples. Let say we have a beam which is supported a vertical beam, which is supported by a pin support, by hinge support at the top. It is subjected to an arbitrary loading, let's say at this end. Clearly, this structure cannot be in equilibrium for this arbitrary loading since the possible reactions at this end are the one shown here. Taking moments around this point, we cannot satisfy the equation of equilibrium of moments. That is the sum of the moments about point A for example, is not zero. So this is a case again, which is unstable structure subject to an arbitrary loading, and we cannot satisfy equilibrium. Again, for specific loadings, this structure could be in equilibrium. For example, under such a loading, or under such a loading, because in this case, the equation sum of the moments about point A is automatically satisfied. None of the forces create moments, so it's a degenerate equation. So we have two remaining equations from which we can solve if the horizontal reaction is zero, and the vertical reaction is equal to the applied load. So in these two cases, we have equilibrium. So we have an unstable structure which is under equilibrium. Whenever we have an unstable structure which is in equilibrium, we can classify the equilibrium in one of three categories. This equilibrium would refer to as neutral equilibrium. In the sense that as small perturbation, a small movement from the equilibrium position does not cause the structure to return or to depart from its original position. This structure here, this is a stable equilibrium. In the sense that a small perturbation, a small variation from the original position under this loading, will cause the structure to return to their original position. While this loading, a small perturbation, would cause a structure to depart, and this would be an unstable equilibrium. So these are all unstable structures. All of them. All of them are unstable structures. For general loading, we cannot satisfy equilibrium. For specific loadings, we could satisfy equilibrium, and we could classify into these three categories of equilibrium. Now, in all the examples I showed here, the number of unknowns where two, two, two. So the number of unknowns were smaller than the number of equations because the number of equations in this case of a single rigid body are three. So it is obvious that whenever we have less unknowns than equations, we could not possibly satisfy all equations for an arbitrary loading, only for very specific loadings where some of the equations become degenerate, then we would be able to obtain equilibrium. But in general, whenever this happens, we can say that we have an unstable structure. Another example of whether this is satisfying. We can immediately say we have an unstable structure, would be for example, the case of a structure that looks like this. This is an example of a structure with multiple rigid bodies. We have three rigid bodies, these three members, and the total number of unknowns in this case are two unknowns here, two unknown reactions here, that's four. We have two internal forces. That means when we separate and we look at the individual members, we would have to have two forces at this end, and another two forces at this end. So the number of unknowns in this case is two, four, six, eight, but the number of equations we need to satisfy is three equations for each member, so it is nine. So we fall in this category. Therefore, this structure is unstable. It is clear that under specific loadings, for example, if you would have a vertical loading, then this structure would be in equilibrium. However, as soon as we have a horizontal component, this structure will not be able to sustain itself in the current position, it will have to move. So it has to change its geometric configuration. So this is clearly an unstable structure. So whenever we see the number of unknowns being smaller than number of equations, we can immediately conclude that we deal with unstable selection. The question is, increasing the number of unknowns, increasingly a number of reactions or internal support, would it cause the structure to be unstable? For example, the question is, if this is true, can we conclude that this structure is stable? The answer is, in general, we cannot conclude that. We can show that with a simple example. Let's say if we would have a beam which is supported by three rollers, it is clear that under an arbitrary loading, this equation is not in equilibrium. Sigma Fx cannot be satisfied. We have three vertical reactions. Equation Sigma Fx cannot be satisfied. As a matter of fact, we don't have equilibrium in this case. So we have an unstable structure. Increasing the number of unknowns even more by adding a fourth reaction does not change the situation. We still remain with an unstable structure because the important thing is not the number of unknowns, is the configuration, the setup of the reactions, and the internal connections that will conclude whether the structure is stable or unstable. So the question of stability to answer the question if a structure is stable or unstable. If we have this situation, we can immediately conclude it's unstable. However, if we have a situation where this is equal or the number of unknowns are bigger than the number of equations, then we would have to consider a loading where we could show that equilibrium is not satisfied in order to conclude that the structure is unstable. Let us look at such an example next.