Hello, in this video, we will talk about statically determinate, labile and statically indeterminate systems. I begin this lecture by a quite catastrophic picture, which illustrates the danger that the structures which we build can represent for their users. We definitely want to avoid what happened the first of August 2007, in Minneapolis, Minnesota, in the United States. During peak hours, this highway viaduct collapsed without any warning indicators, after having been in service for about twenty years, and while it was being undergoing maintenance work. This is a little reminder to say that our structures can be dangerous and that it is important to be careful to their behavior. So what we want to see in this video, it is about the questioning I had let you at the end of the previous video: when can we be sure that our structures can be calculated by the methods which we study in this course, and is a collapsing of the structures we are looking at possible? For this, we will visit the concepts of support, of node, of bar and of hinge, and I will introduce to you a general method of calculation to determine if a structure can be calculated, respectively, if it is labile. A bar. By the way, I will use in this lecture the red and blue colors without making a reference to the convention of tension and compression, simply because I need these colors. A bar is a continuous element between two nodes, so here, we have a bar, we have another bar, we have another bar, and we have another bar. Or else here a bar, another bar, another bar, another bar, so on the left, we have four bars, on the right, we have five bars. A node is the end of a bar, whatever if there is a support or not, so we have a node here, a second node here, a third one here, a fourth one, and actually, it is exactly the same for the structure on the right. A support is a point of support, respectively, we want to look at the number of forces at the support which are necessary to entirely determine these supports. Here, we are going to have a force with a vertical component, which means only one contribution, while here we can have a vertical force but also a horizontal component, which thus means two contributions. Likewise here, we will have two contributions, and here we will have one support contribution. A hinge is a rotation point between two bars, it indicates the end of a bar Let's look at the different types of supports and at the unknowns which are related to them in details. Leftmost, we have a mobile support, for which we can only have a vertical reaction, so we have the number of reactions, which is counted as one. Then, we have here a fixed support, for which, as we have already seen, we have two components, a horizontal component, and a vertical component, which means that a fixed support counts as two. Here, we have a clamping, that is to say a type of support which is like a fixed support and which, in addition, prevents any rotation. The rotation. We do not have two but three parameters, which can be visualized as both the vertical and the horizontal components of the force at the support, and the eccentricity since this support reaction does not pass anymore by the support which is here, but at a certain distance from it. And on the far right, in an illustrative way, we have a hinge, which is thus a end of bar. We have on the left the bar one, and here, the bar two, and here, we have the hinge. If, on the contrary, it is not hinged but welded, we have only one bar. You have already encountered e a variety of symbols in this cours to represent supports, so I drew some here. Here, we have fixed supports, so with two reactions, here, we have a support which is horizontally mobile, that is to say that it can move leftwards or rightwards, but it cannot move vertically, which thus gives us only one reaction. We can also have supports which are vertically mobile, with also one reaction. Actually, we could even have a support which is mobile on a slope, but we will not deal with this topic in this course. And finally, we have clampings, with three contributions. Let's note that these three drawings here are very similar. The difference is that here we have a hinge, which makes the rotation of the bar possible, while here we do not have a hinge and therefore the rotation is not possible. But be careful, both these symbols are very similar. So, how to know if a structure is labile, isostatic or hyperstatic - I am going to come back to these terms thereafter - and we are going to do it by writing an equation for which we will use the number of reactions coming from the supports, we will add to it the number of bars, and we will compare that sum to twice the number of nodes. Let's start here, on the left. We have here two unknowns at the support, one unknown at the support, that is to say three, plus one, two, three, four bars and then, we have one, two, three, four nodes. We multiply four by two, and if look at this equation, three plus four is equal to seven, which is smaller than two times four which is equal to eight, thus the structure is labile. That means that it is maybe unstable, but it is not necessarily unstable, that will depend on the type of structure. But you have to be careful. If we look at the example in the center, we still have two support reactions on the left and one one the right, for a total of three, plus one, two, three, four, five bars. We compare this to one, two, three, four, two times four nodes, three plus five is equal to eight, which is also equal to two times four, this structure is statically determinate, so it is computable according to our system. We will see a lot of statically determinate structures throughout the remained of this course. And on the right, we still have two reaction forces plus one, which means three, and afterwards, we have one, two, three, four, five, six bars, and we have one, two, three, four nodes, so two times four is equal to eight, and three plus six is equal to nine, thus this structure is called statically indeterminate. We are going to say "not directly computable", we will see later that we can calculate some types of statically indeterminate structures Let's note that what we can see now here, is that there is just one additional bar, if there was one bar less, it would be a statically determinate structure, so in this case, we have a structure which is one time indeterminate. So it would be enough to take off one bar, (it is necessary to choose to right one because we could do something stupid) but if we take off a bar, we could make the structure statically determinate. For example, if we took off bar number three here, we would be back to the case of this structure, and the structure would be statically determinate. Imagining a statically indeterminate structure, - you will by the way see some several times in the exercises - we could then take them off one, two, three, four bars. On the opposite, this structure is one time labile, since it would be enough to add one bar for it to be stable. A statically indeterminate structure, we cannot directly calculate it, however it has the advantage that if for example, further to an accident, let's say for example the impact of a truck on a bar, by losing one bar, the structure would become statically determinate at this moment but still can stand, so it is an advantage. Let's look at these three examples of structures, on the left, we have a cable, we have in all the cases, for all these structures, each time, two support reactions, on the left and on the right, and then here, we have one, two, three, four bars. And one, two, three, four, five nodes. Let's come back to our equation, we have four plus four, and then we have two times five, so our structure is twice labile. Is this structure unstable ? In any case, if I place loads on my structure, it is not instable. It is called a cable, it is a structure which is stable, but with large displacements. Let's now look at the structure of the middle, actually, the numbers are exactly the same, so I write them quickly. We have again four plus four, which is smaller than two times five. The structure is twice labile, and if we build it on this way, this is an arch which unstable, because we enable the rotation between the different elements, because of the hinges. So we can see two configurations, for loads which would act in this way on the structure. We now move to the right structure. First of all, for the left support, I had noted two reaction forces but it was a mistake, since it is a clamping, we cannot have rotation, so we will have three unknowns at the level of the left support, two at the level of the right support, because we have this hinge here. We are going to have one, two bars, this bar here is curved, but there is no hinge, there is not any other bars coming, and this is not an end of structure, so there is no node here. Likewise, we will have here one, two and three nodes. This means that we are going to have five unknowns for the supports, plus two bars, and on the other side, we are going to have two times three nodes, five plus two is equal to seven and is greater than six, this structure is statically indeterminate, and since seven is just one more than six, it is one time statically indeterminate. So, this is a stable structure, but not directly computable. So we will see a bit later in this course, examples of statically indeterminate structures that we can still calculate, or approximate, using the applet or a graphical construction, but in this case, it is not directly computable. Alltogether, you have this line with the degrees, so here, we have a degree for twice labile, for the two first structures, and then one time statically indeterminate, for the last one. It also means that if we wanted to stabilize this structure keeping exactly the same number of nodes, we would have to add two bars, we can think about adding two bars to stabilize these structures. On the opposite, here, the structure being one time statically indeterminate, we could take off one bar. To take off one bar, it would be necessary to choose the right one, for the structure to remain a structure. We are quickly going to visit some types of trusses, here the most simple, we did not even do the model, with a fixed support on the left, a right support on the right, that gives us two support reactions on the left, one on the right, for a total of three. We have here one, two, three bars, and then one, two, three nodes ; two times three is equal to six, this is a statically determinate truss. We can now move to a truss which is a bit more complicated, it is the one we have made at the beginning, with four nodes, a fixed support on the left, a mobile support on the right. We then have two plus one, which is equal to three support reactions, plus one, two, three, four, five bars, and then one, two, three, four, two times four nodes, this truss is thus one more time statically determinate. We move to another truss that we will shortly see in details, with five nodes with still a fixed support on the left, a mobile support on the right, so here, two and one, which gives us three support reactions, plus one, two, three, four, five, six, seven bars, and then, we have one, two, three, four, five nodes, two times five is equal to ten, which is also equal to three plus seven. The structure is thus also statically determinate. We can take another geometry for the first example, a triangle which is a rectangle triangle with a fixed support on the left, a mobile support on the right, like before, so two support conditions here, one here, for a total of three, plus one, two, three bars, and then we have one, two, three nodes, so two times three, three plus three equal to two times three : still statically determinate. Then, we move to the next variant of this truss, where we are simply going to add a second element above, a fixed support, a mobile support, so here, two and one, for a total of three. Here, we have one, two, three, four and five bars, three plus five, to be compared with one, two, three, four, two times four nodes, it is still three plus five equal to two times four, so we still have a statically determinate structure. We now can move to a structure which is the one made with a mirror, so we are going to place the diagonals in this way, and we thus get a bigger structure with a fixed support on the left, a mobile support on the right, so two and one, for a total of three, plus one, two, three, four, five, six, seven, eight, nine bars, and then, we have one, two, three, four, five, six, two times six nodes, three plus nine equal to two times six : twelve. We still have a statically determinate structure, we can also depart from the basic structure of this triangle, with a fixed support and a mobile support, and refining in this way, we still have a structure only constituted by triangles, we have two supports plus one, three, plus one, two, three, four, five, six, seven, eight, nine, bars, and we are going to compare this with one, two, three, four, five, six, two times six nodes : still a statically determinate structure. A bigger example that we will soon be visiting is a structure with four panels, and diagonals inclined in the other direction, so a fixed support on the left, a mobile support on the right, which makes us two and one : three, plus one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen. And, we are going to compare this with one, two, three, four, five, sixe, seven, eight, nine, ten, two times ten nodes, three plus seventeen equal to twenty, which is equal to two times ten, therefore this structure is still statically determinate. What we can notice in this equation, I am going to write it down again : the number of reactions for a statically determinate structure plus the number of bars must be equal to twice the number of nodes. So if we add two bars on the left, and one node on the right, we have increased the size of a structure keeping it statically determinate. We can do that for example on this structure here, I am now going to add two bars and one node. So now, I am going to have ten, eleven bars, with seven nodes. If I redo the calculation here, I have three supports plus eleven bars, which is equal to fourteen, also equal to two times seven nodes, then it also works. In this video, we have seen what a bar is, a node, a support, a hinge. We have talked about the different types of supports and their contribution to the functioning of the structure, we have defined the labile structures, which are structures which do not have enough bars to be perfectly stable, but which may be stable, particularly if they are cable-shaped structures. We have seen that we can calculate statically determinate structures within the framework of this course, so twice the number of nodes is equal to the number of supports plus the number of bars. Statically indeterminate structures have too many bars or support conditions compared to twice the number of nodes. These structures cannot be directly calculated within the framework of this course, but they have the advantage that we can take them off one or several bars, while they remain stable. We have finally seen a general method, which enables, on the basis of a structure, counting the number of supports, of nodes and of bars, to determine if this structure is labile, statically determinate or statically indeterminate.