Diaphragms: 3. Rigid Diaphragms 2

Okay, so let's do some of the math on rigid diaphragms. And we said previously that if the diaphragm, sorry, if the collectors deflect more than twice the diaphragm deflection, then we have a rigid diaphragm. In this case, I'd like to do some of the math. It's not complicated math, I promise you. And it's mostly to figure out if there is torsion or not, rather than how much the torsion is and how it's resisted. So looking at this first example, I have two collectors. The sum of rigidities is 1 plus 1 equal 2. I will get one part of the two, which means I will get 30 kips. You will get 30 kips. And the center of rigidity lines up with the center of mass. Therefore, there is no torsion, but the code requires that we incorporate a 5% accidental torsion, which means move your center of rigidity 5% of this dimension to either side, but you must also move it in the other direction 5%. So just to, it's for safety. Theoretically, there is no torsion, but 5% of each dimension of the diaphragm is considered as an eccentricity for accidental torsion. Looking at the second example, it's a brace frame. I don't care. I just look at the rigidities. I see they're equal. I know the green center is on top of the red center, and there is only accidental torsion. And shear-wise, 60 kips is going equally to either side. Either side gets 30 kips. In the next example, it's 1 plus 2 equals 3. Therefore, I will get 1 third of 60. You will get 2 thirds of 60. And the green center moves to the right. So there will be a torsion in this example. But first, let's do the shear. one-third of 60 and two-thirds of 60. So I will end up with 20 kips and you will end up with 40 kips. I'm resisting 40 kips of shear. You're resisting 20 kips of shear. Together, 40 plus 20 is 60. Therefore, we're fighting the shear, the red shear of 60. We're fighting it with a 20 and 40 equals 60 green. But the centroids are not lined up, and in the previous slide, we figured out how to calculate this dimension. And therefore, there is torsion. But in this case, I just want to understand that the torsion is going to be counterclockwise. Moving down here, I have a rigidity of 1 plus 5. So that's 6. I will get 1 sixth of 60, which is 10. I will get 5 sixths, which is 50. And the green center is way over on the right, and therefore the torsion is going to be in this direction, which is counterclockwise. Next example says 3 plus 1 is 4. 3 plus 1 equals 4. I will take 3 quarters. You will take 1 quarter of 60. And therefore, one quarter of 60 is 15 and three quarters is 45. That's the shear distribution. But then the green center is over on the left. So in that case, I'm going to have clockwise torsion. Clockwise. Again, I don't need to calculate the numbers. We saw it in the previous slide. They're never going to ask you. But just you need to understand what's going on. That's why I had the animations. Excuse me. Looking at the next one, I have 2 plus 1 equal 3. I will get 2 thirds. You will get 1 third. Therefore, I'm going to handle 20 kips. You're going to handle 40 kips. And the green center is over on the left. And therefore, the torsion is going to be clockwise. Is that right? Clockwise, yeah. It's going to do that. Okay, so the torsion is clockwise. Very good. When we have two bays, same logic as before. I have 60 and 60, this time 120 kips. I add the rigidities. That's the main thing. That's why the distribution of load is based on rigidity. So I'm going to get one third, I'm going to get one third, I'm going to get one third, which means 40 and 40 and 40. So the left collector gets 40 the middle collector gets 40, the right collector gets 40, and we have to account for only accidental torsion because it looks like the collectors are symmetrical. And the green center is on top of the red center. Very good. So going down one, I have the sum of rigidities is one plus four plus one equals six. So I'll get one sixth, you get four sixths, you get one sixth, which means I am 20 kips, 1 6th of 120. I am 80 kips, which is 4 6th, and I'm another 20 kips. So left collectors gets 20, the middle collector gets 80, and the right collector gets 20, and there is no, theoretically, there is no torsion, but we have to account for 5% accidental torsion. 5% of this dimension. So you move the green center this way and that way. 5% of 96, almost 5 feet. That's your eccentricity. Okay. Now let's move the wall off to the side. The shear does not change. I'm still doing 80 because this time 80 kips and 20. Oops, undo. Sorry. Okay. So 80, 20, 20. But now all of a sudden you have a lot more of torsion. So the left collector is getting shear worth 80 kips versus the middle one versus the right one. But there is significant torsion because this guy is doing that and this guy is doing that. And there's a lot of spin. So we can calculate the centroids. It's just that's too much math. This one with the collectors being equal, then each one of us gets a third, and therefore a third is 40, 40, and 40. And shear-wise, each collector is getting the same amount, but there is a lot of torsion due to the placement, the non-symmetrical placement of the collectors. So the center of rigidity is somewhere here. The center of mass is somewhere here. And we have a clockwise torsion. Very good. I think that's it. Excellent. Thank you.