Span Types: 3. Stability & Determinacy Concepts

So now I'll get into the concept of whether a member is stable or unstable, whether it's determinate or indeterminate. And it's a very simple matter of counting the number of reactions on the loading. If you have less than three members, it's called unstable. If you have three members exactly, you have three equations of equilibrium. So if you have three unknown reactions, it's stable, not going anywhere, and it's determinate because we have the same number of unknowns as equations. Good to go. If you have more than three unknown reactions, you only have three equations, so it's called indeterminate. If you have four unknowns, it's indeterminate to the first degree. There's one extra unknown. If you have five unknown reactions, it's indeterminate to the second degree, etc. The number of degrees is basically how many assumptions do you need to make on top of three equations to solve for the number of unknowns. Very good. Let's get into this. What I'm going to do here is put the number of reactions in this cell, and then I'm going to make a conclusion whether it's unstable, stable and determinate, or indeterminate based on the number of unknown reactions. So looking at this piece of wood that we saw the image of earlier, it's sitting on two concrete piers and it's on two rollers as we said, and the roller is one vertical reaction, the second roller is one vertical reaction, there is no horizontal, there is no moment reaction. This one has a total of one, two unknown reactions. So when you have less than three, it's called unstable. So this loading is unstable. Sure enough, if you kick it in the horizontal direction, it moves. The second one is on a pin on the left instead of a roller, and still the roller on the right. So this is one reaction, and this one is a vertical and a horizontal. Now I have one, two, three total unknown reactions. That one is called stable and determinate because I have three equations of equilibrium. Sum of horizontal equals zero. Sum of vertical equals zero. Sum of moments at any point equals zero. So I can solve for up to three unknowns. It's determinate. Very good. Now how does this one deflect? It's going to deflect like so. It rotates at the pin and it rotates at the roller, but it also pulls it in. So now, that's the deflected shape and the top is in compression, the bottom is in tension. It's doing that. That's the type of deflection or the form of the deflection. This is called a positive moment when compression is on the top, tension is on the bottom. Let's go to the next one. The next This one also does that, but it doesn't pull the pin in. This one has 1, 2, 3, 4 reactions. So it is indeterminate to the first degree. One assumption plus three equations would solve this problem. How about we assume that there is vertical loading only, let's say. Let's say there's only vertical loads and a concentrated load, whatever. The loads are all vertical. I'm pretty sure these are zero. That's an assumption. Now we can solve the problem because we got rid of two horizontal unknowns, assuming that they are zero. The load is vertical. So now I went with one degree of indeterminacy. I brought it down to a determinate situation. Very good. In a cantilever, it's supported only on the left in this case. But a fixed end has three reactions, one vertical, one horizontal, and one rotational resistance. So it's going to keep that angle 90 degrees. So when you put a uniform load or concentrated load or whatever it is, this reaction has to equal to these loads and these loads are making clockwise so this guy has to fight counterclockwise. This one is zero. Now how does this one deflect? This one is going to deflect like so. The top is going to go into tension, the bottom is going to go into compression. When you have that, you have compression on the top, tension on the bottom, it's a positive moment. Now we have that. The first one is like a frowny, smiley face. The second one with tension on the top, compression on the bottom, is like a frowny face. And it's a negative moment versus a positive moment. This is preparation for shear and bending moment diagrams whenever we see them. cantilever will have a negative moment diagram versus case number two, a simply supported beam will have a positive moment diagram, no negative whatsoever. The cantilever has no positive whatsoever. It's always tension on the top, compression on the bottom. Looking at an overhang, let's say it's on a roller and a pin. That's a total of three reactions. So this one is stable and determinate. I'm sorry, the cantilever was also stable and determinate because it had three unknown reactions only all on one side. Excellent. So anything that has three reactions is okay to ask you on the architect registration exam with math. If it's more than three reactions, they will not ask the math. They will ask, what is the deformed shape? That's why I'm doing that here. Anything with more than three unknowns needs high math. It needs moment distribution, it needs the portal method, it needs some other system that is on the PE and not on the ARE. Very good. So this one, the overhang is stable and determinant. Let's put a load on top of it. Let's say a uniform load. Well then this one is going to do that. And compression tension. It flips on the overhang. Tension on the top, compression on the bottom. And the compression went from the top to the bottom. And this is called the point of inflection, or a point of contraflexure. And the significant thing about the point of inflection is it has no moment. So the beam is doing this here, and then it's doing that. Positive moment, negative moment, and a negative moment on the other overhang. So an overhang, when it comes to moment diagrams, will have positive between the points of inflection, and then beyond the points of inflection, it will have a negative moment diagram. Very good. Coming down to a restrained beam that is fixed on both ends, a rigid support or a fixed support has three reactions, one, two, three, and another three, and this one has six unknown reactions and highly indeterminate, highly meaning to the third degree. So I need to make three assumptions, plus three equations to solve this one, definitely not on the ARE. But we need to understand how it deforms. That's what's really important. So this one is not simply supported. This angle shall remain 90 degrees. This angle shall remain 90 degrees. So it's not going to rotate. It's not going to do that. It can't. So the support is rigid and takes part of the beam to itself. And what happens here is something like that, with a point of inflection and another point of inflection. And actually, the span is actually from here to here. That belongs to the support. And what I have here is compression, tension, and then it reverses. The tension goes to the top, the compression on the bottom, tension on the top, compression on the bottom, point of inflection, there is no moment. So with a restrained beam, we're going to see negative moments at the support and a positive moment somewhere in the middle. Excellent. Now to get to continuous supports, these are running on top of more than two supports. That's it. I'm not going to discuss this any further. I'm not going to count the unknowns. They're all indeterminate. Okay? Because there's three supports, that's it. There's more than three unknowns. Very good. Still, I would like to do the deflected shape. So for this first one, if you have a uniform load on top of it, then it's going to do something like that. And it's overhanging. I see a point of inflection where the moment is zero. Those two points of inflection, I see a total of three points of inflection. So I have no moment at the pin. It cannot resist. And I have zero moment at the points of inflection. I have compression on the top, tension on the bottom between the supports, but over that roller support, I see that there's tension on the top, compression on the bottom. So the moment diagram for this one will look like a positive moment between the supports, a negative moment over the support, and then on the overhang, I'm going to see a negative moment. Please remember, this is positive, this is negative. Positive, negative. Excellent. Now what happens if we put the uniform load only on this bay and the rest of the beam, this beam, is not loaded? What happens here is, of course, this one has to fight back and carry the load. But if this support were to become detached, maybe it's a column and it's settled or something, then what would the deflected shape look like? If I lost the pin on the right, this is going to go up. So actually this support is responsible for bringing down the beam and keeping it at the support if it's not lost and it didn't settle. Then it brings that point down. This is a downward reaction. These are upward reactions. And the deformed shape is going to look something like this. And it looks like positive moment and then negative moment. Excellent. Now what happens when we take that uniform load on a three bay continuous beam and we put the uniform load only in the middle? We recognize immediately that these are upwards. And if I were to lose the two rollers, if they were to become detached somehow, then this one would go up. This one is going to bow down because of the load. and then it's going to go up again. So this one is downward, this one is downward. And the deformed shape is going to look something like this. With a point of inflection here, and another point of inflection here. Looks like a positive moment. This one is bending backwards, and this one is bending backwards. To understand the math, I'm not going to do math, But to understand the math, this loading is symmetrical. So this guy on the left has to carry that uniform load, and in addition, it has to carry that one. And then this other one on the other end has to carry that much uniform load, and there's a downward reaction, and that's what this guy has to carry. Very good. Looking at condition 10, this one is an important one because anything cast in place, a slab on top of beams or a beam on top of multiple columns, is going to be like condition 10. And it's going to do exaggerated, something like that, definitely indeterminate. But it looks like there's a point of inflection and another one, and another one, and another one, and one more, and one more. And the point of inflection is important, the moment is zero, but that's where the compression and the tension are reversing. So I have tension on the bottom here, I have tension on the top. I need to understand what's going on because my rebar goes wherever the tension goes. So there's tension on the top, there's tension on the bottom. Tension on the top, tension on the bottom. So I need to make sure my rebar follows where that T is. And basically all these reactions are upward and the rest of course are compression and the concrete is perfectly happy with the compression. has no problem handling the compression. Excellent! At the point of inflection is where the rebar is reversed from bottom to top or top to bottom. And please remember this is a positive moment because it's like a smiley face. And then at the supports we have a negative moment. And we have zero on the two ends because they're not overhanging. If they were overhanging then it would be a negative moment. Excellent. Now condition 11. This is a tricky one because of the way it's drawn. Let me erase all of this. So the difference between 10 and 11 is loading 10 looks like one beam on top of five supports versus loading 11 looks really like it's three beams. And we saw the picture here which was important and and I discussed it with the picture. And we said that maybe NCARB will put something here, which means that's a pin connection. Or, if they put something like that, that means the flanges are connected, and now it's one beam. So we need to look at those connections and understand what is meant, and it looks like these are three separate beams, B and C. Beam A is simply supported. It has a pin and a roller and it overhangs. Not simply supported, sorry, it's overhanging, but it's determinate. And C is also an overhang with three reactions. I have three equations, no problem. And then B is sitting on the tips. So I need a reaction here to carry whatever load is on this guy and that is going to transfer as a concentrated load on the tip of the overhang A and likewise on the tip of the overhang C. I have three, I have beams A, B, C. Each one is stable and determinate and I solve three of them and the problem solved. This one needs higher math condition 10. Very good. Now understanding the 3 pin versus the 2 pin, it's a similar conversation to this one, because I need to know if this is one piece, as we saw with this frame, sitting on two pins, which is this diagram, or is it like Grimshaw's Waterloo Station, where you have a pin here, a pin at the crown, and a pin at the bottom, this is 3 pin. Excellent. So 2 pin is one piece sitting on top of four reactions, or two pins. This one, is this the two pin? It's four unknown reactions, and it's indeterminate, and I don't know how to solve it. But I understand that this load is going to push down and want to make the legs spread out, so I need a reaction horizontal in this direction. I just don't have enough math to solve it. That's it. But I assure you that whatever uniform load you have is going equally to two vertical reactions. Very good. Let's look at that three pin arch because what this one is, is there is one half and leaning against it is another half. So I look at them separately. It's like I did here, A, B, C separate. Now structures, and they're leaning against each other. So let me analyze one of them. And then whatever I come up with A, I mirror and say B is the same as A. Excellent. So you have no vertical load in the middle. It starts at the pin and goes left and goes right. In the dead middle, there's no vertical. The pin here has two reactions, a vertical and a horizontal. and this thing wants to lean. It's going to topple over. So what we need is somebody to give us a reaction. So how about I support you from falling over and you support me from falling over. And these two will cancel out. In the pin, the pin is capable of providing horizontal support. It cannot prevent rotation. yes, those two pieces will wobble with respect to each other, but that's what happens. Now, if I look at A, I see three reactions, and the sum of verticals equals zero, which means whatever this load is, it's half on the left, half on the right, well, then this guy has to be equal, sum of verticals equals zero. Sum of horizontal equals zero is telling me This reaction is equal to this one. Maybe I need to take that out of the picture. We're looking at one half only. The vertical load is equal to the vertical reaction. The horizontal load is equal to the horizontal reaction. And the sum of moments should equal to zero. So I'm making a moment. You have to make a counter moment. Prevent rotation about this point. That's how I find how much the thrust is. Okay. Excellent. So, this one is causing counterclockwise about this point. This one has no moment. This one has no moment. This one is making clockwise. So, I can find this reaction and then say this is equal. We're not going to do the math. I need you to understand how this works. Very good. So, this one has A and B. Each one has three. Each one is stable and determinate. And these two are equal and opposite. Therefore, these two are equal and opposite. Very good.