Now, let's look at an example of a structure that is supported by three rollers. It can be shown that this structure is stable. Now, how to show it's stable it's another story. Actually, as long as we don't run into any situation of any loading that would cause a situation where we cannot have equilibrium, then we could say it's stable. But we're going to show later an alternative approach where we can directly show that the structure stable. For now, let's assume this structure is stable, which it is. We show clearly all possible reactions. In order to solve for these reactions, we write the equations of equilibrium, Sigma Fx. Cx is the only force in this direction, so this is zero, this the equation Sigma Fx. Sigma Fy, we get Ay plus Dy minus 100 is equal to 0, and the sum of the moments, we take sum of the moments, let say about point C. I choose point C or I could choose point B because two of these forces, their moments are eliminated when we take moments about C. So what we get is minus Ay times 4 plus 100 times 2 is equal to 0. From this equation, we show for Ay, Ay star to be 50 and Dy clearly also 50 from solving this equation. So we see that, this is a stable structure that under a given loading, we can solve uniquely for the three reactions. Therefore, we say that this is a statically determinate structure. That means, we can uniquely determine not only the reactions but as we're going to see later, if we want to calculate any internal forces, for example, internal forces at this point, we could cut and consider the substructure entitled determined what this internal force as well. So for now, let's just say that we can uniquely calculate the reactions from the equations of equilibrium, it is a stable structure. So this is a statically determinate problem. Now, let's look now at the same problem, a problem with three rollers. The only difference between these two structures is that the support at D while before it was a roller that was rolling on a flat horizontal surface. Here it is rolling along a surface, which has an inclination of 45 degrees. In this case, the three reactions are clearly Ay, a clearly Cx here. At this point, at point D, the reaction is in the direction perpendicular to the surface along which the roller is rolling. Let's call this force D. This force is an inclination of 45 degrees. It clearly has two components. Component D square root of 2 over 2, and D square root of 2 over 2. Let's write the equations of equilibrium Cx to the right, minus D square root of 2 over 2 is equal to 0. Sum of the forces in the vertical direction, Ay plus this component which is D square root of 2 over 2, minus 100 is equal to 0. Finally, let's take the sum of the moments. Let's take sum of the moments about point B. In this case, Ay and Cx cancel out, they both pass through B. The horizontal components creates a moment which is negative D square root of 2 over 2 times the distance of 4, the vertical components creates a positive moment which is plus the square root of 2 over 2 times 4, and the applied load 100 create the moment minus 100 times the distance of 2. We want this to be all zero. Now, from this equation, we see that these two terms cancel out. So we end up with an equation minus 200 is equal to 0, which of course it is not. It means that the sum of the moments about point B is not equal to 0. We cannot satisfy the sum of the moment equation, we cannot satisfy one equilibrium equations. Therefore, this structure is unstable. So a case where the number of unknowns is equal to the number of equations, but the structure is unstable due to the specific orientations of this supports. Now adding a fourth support, adding for example another roller at B, we can easily show that this would not change dissipation because simply we would have another reaction here, let's say Bx. In this equation, sum of the moments force would not contribute in anyway. So this equation would remain unchanged, and still it would not be satisfied. So having a number of unknowns which is equal to the number of equations or larger than number of equations in this case, still we ended up with an unstable structure. Here on the other hand, if we added another fourth roller at this point we would have another force Bx. We can easily show that what would change here is that this equation would become Cx plus Bx is equal to 0. The other two equations remain as a matter of fact unchanged. So we can see that in this case, we have four unknowns 1, 2, 3, 4, we have three equations of equilibrium. We cannot write any fourth equation remember. By taking moments about any other point, that would lead to new information. So we have four unknowns with three equations, therefore we cannot solve for all unknowns. We still consult for Ay and By, but we cannot solve for Cx and Bx. These are still unknown. So this is a case of a stable structure which is, in this case, statically indeterminate because we cannot determine all possible reactions. When we say statically determinant, it means we cannot determine everything, but it doesn't mean that we cannot recover some of their laws. For example, in this case, we're able to determine the two unknowns Ay and Dy.
