Good evening, everyone, and it looks like we made it to the last session. And today what we're going to cover, we're going to cover a series of topics that are classified here in the series of items we've covered so far that we refer to as special topics. And these are because they don't fall into one of the categories that we've covered so far. And these are topics that are referred to in Chapter 5 of the PCI Design Handbook. The items that we will cover essentially are summarized here in this slide. There's going to be three general areas or design aspects that we're going to consider tonight. The first one is how to take into account the design of double Ts in the transverse direction. So this is for loads applied on the top flange of the T and not in the longitudinal or in the span direction, but transversely towards the stem. So how to treat those kinds of loads and how to design for those. The second item we'll cover is how to design and detail openings in webs of double Ts. And the last item towards the end of this evening is we will talk about how to deal with or issues related to double T warping that might be needed in parking garages for certain conditions. So without too much more than this, I will just jump right into the first topic, transverse design of double T flanges. So this topic is covered in the PCI Design Handbook in Section 5.13.1. The idea is that any loads, concentrated forces that are applied on the double T flanges have to travel to the double T stems in order for the double T to act as it's intended, as a supporting system, floor system of a parking garage. So there has to be reinforcement that is designed in the transverse direction of the double T to be effective in resisting any loads that are applied outside, away from the stems of the double T. One of the items that we will talk about in more detail is that in the case of concentrated forces, it is not clear what portion of the flange, let's say in the longitudinal direction, resists a concentrated force. And we'll talk about the different possibilities that one could consider. Uniform loads will be treated as a surface load on the flange. So those are easy to consider. But the more challenging one is how to come up with a width of slab, so to speak, that is part of the flange that resists a concentrated force. And particularly important are those that occur farthest away from the stems, or at the flange edge, or essentially at the connection between two stems, right? Between two adjacent double Ts, not two stems, the two double Ts. So if one looks at the literature on slab design, we could find that in the transverse direction a designer could go and try to solve this problem using an elastic plate solution. So essentially try to find what the effect of a concentrated force is on a two-dimensional object like is the flange of a double T. This would work well as long as the double T flange behaves in an elastic range and it doesn't form cracks or reinforcement starts yielding. The other approach could be to use the yield line theory, or in other words, using a plastic solution where the designer is charged to identify a potential way that the slab may crack and therefore look at the mechanism that that forms and use a virtual work approach to find the capacity, let's say, of that slab reinforced to resist a concentrated force. So that approach is more amenable to investigating what the capacity of a slab might be rather than design. And the third approach, which is the simplest of the three, is to identify a width of slab that resists a concentrated force. So how much width of slab engages in supporting a concentrated force in the case we're concerned with is the width of a double T. So the first two items, elastic and plastic solutions, have been found to be conservative in practice. So the PCI design handbook therefore adopts an effective slab width method and that's what we will cover in this course. So again, as I mentioned earlier, this method assumes that there's an effective width that resists a concentrated force and this width of slab is engaged and that's where we have to concentrate the reinforcement to resist the concentrated force. In order to do that, we have to identify that there's a possibility, so that there's an angle of distribution or a spread angle in which this concentrated force engages part of the slab and that spread angle can be estimated from cracking patterns that have been observed in past tests. So two references that are useful for this purpose, and these are both references listed in the PCI design handbook, are references from 91 and 2016 from Aswad and Burnley and Boutrous et al. in 2016. And from those tests, and you can get those by the way through the PCI journal, free of charge if you remember, or actually I think it's free for everyone. You can just log in and find those references. From those tests, it is suggested that these concentrated forces spread at an angle at a ratio of 3 to 1, where 3 is in the longitudinal direction of a double T and 1 is in the transverse direction. This corresponds approximately to a 20 degree spread angle. I'll show you some figures in a second. So this is, for example, a view of, the top view of a test conducted by Aswad and Burnley. And you can see that there were a series of point loads applied at the tips of, at the corners of the double T. This is a plan view again, and you can see the stems dashed here. And these are points of concentration where the loads were concentrated, and you can see that there's cracking along the interface of the top flange and the stem over a certain distance. So this crack indicates how much of the slab is being engaged during the application of this concentrated force in this region prior to failure. So this gives us an indication on how much, what portion of the slab is being engaged by the concentrated force. So these are some conclusions and notes that prior to the 6th edition of the PCI design handbook, there was no guidance on how, what the width of slab should be to resist the concentrated force in the double T. And they found that the distance that is engaged is between 3 and 4 times longer than the clear overhang length. In this particular case, it's about 2 feet 7 inches would be the overhang. So cracking in these tests was used to estimate or approximately determine the width of slab that would be resisting a concentrated force. And again, the reference is here at the bottom of this slide. The second reference I mentioned relates to double Ts that were reinforced with carbon fiber reinforcement. They were also tested, these types of elements were also tested to failure under the application of a concentrated force. This dot represents the application of it. And this spread angle of 20 degrees that I mentioned earlier is with reference to the edge of the slab. And so the effective width could be considered to be the distance between this, where this diagonal line meets the edge of the slab of the double T and the edge of the slab. So this 20 degree angle is how that is referenced. So again, this distance is roughly what could be considered as the effective width of slab that's resisting a concentrated force. This worked well for forces applied, concentrated forces within the span of the double T. However, for concentrated forces in the corner, as shown here on the left hand side of the figure, the failure crack formed at a 45 degree angle roughly from the edge. So therefore, this plane was used to identify the width of the slab that engages or that supports this concentrated force. So the idea behind this approach is that the flange has to be designed for flexure in the transverse direction. And we have to consider the loading that generates the largest moment at the flange web connection. So typically, that would be the governing moment because the flange would behave as a cantilever beam if it's not connected to an adjacent double T. And the loads to consider are, of course, self-weight of the flange, topping, if any, so if there's any casting place topping, any superimposed dead loads, if used in the project, the uniformly distributed force for parking garages, which is specified as 40 PSI for parking garages in ASCE 7. And additionally, these are normal loads that one would consider for a parking garage. But additionally, in the case of transfer spending, one has to consider a 3,000 pound force concentrated at the edge of the double T and assumed to be acting on a four and a half inch by four and a half inch surface on the flange. And just note that this load of 3,000 pounds does not correspond to a wheel load from a truck or any other vehicle. It really corresponds to a jacking force that would be applied for vehicle maintenance, you know, to lift a heavy vehicle, let's say, in the case of malfunction or, you know, tire replacement in a parking garage. So that's how ASCE 7 came up with this value of 3,000 pounds. And the idea is to consider the possibility of two adjacent flanges being connected through some form of a connector that's cast inside of the top flange. Or in the worst case scenario, consider that the flange is not connected to any other element adjacent to it. Okay, so there's also, in terms of considering what's critical for design, we have to consider that these concentrated force could be applied along the length of the double T at two critical conditions. So one is at the edge of the slab between the ends, and the other one really being applied at the corner. So corner loading ends up being a pretty critical condition that needs to be considered. And it may result in additional transverse reinforcement towards the end of the double T. What also needs to be considered is the fact that there might be free flanges. So in those cases, one has to just count or rely on the load being applied onto a single flange and have that flange be able to resist the values of the load that are being applied. Okay, so let's move on to how we do calculations for this type of problem. So for this case, I chose a double T in the project or the parking garage that we've been using so far for this course. The widest double T was a 12-foot wide double T, and that's what I chose for the example here because the widest double T, of course, has the longest cantilever portion of the flange outside of the stem. In this case, it's three feet to the center of the stem, and that's why this cross-section was chosen. And you might remember that the depth of our double T, at least from preliminary design, we wanted a 28-inch deep double T with a 3-inch casting place topping. The material strengths are 5,000 psi for the double T itself and 4,000 psi for the concrete compressive strength of the casting place topping. And in this course or this part of the course, I'm only going to concentrate on the negative designs or the cantilever design of the double T, but also one should consider any concentrated forces being applied between stems of double Ts to provide bottom reinforcement here. So the cantilever portion will be designed just for illustration purposes, but also the topping and flange between stems has to be considered under positive bending with concentrated forces applied near the center. So with those notes, there are two main conditions that we have to consider when we design or when we consider transverse bending of the top flange. And one of them is during construction, because during construction, there are construction loads being applied to the double T, and the only reinforcement and section that is effectively resisting those forces is the top flange of the precast double T. So for that, what are the loads that are being applied? Well, the flange self-weight, of course, and in this case, it's a 2-inch thick slab. So we multiply 2 inches, convert to feet, and times the density of concrete, normal weight concrete, 100 pounds per cubic foot. That gives us a surface load of 25 pounds per square foot. A 3-inch topping would be also have to be considered 3 inches, and again, with normal density concrete, 150 pounds per cubic foot results in 37.5 pounds per square foot. And I pulled some values from ASCE 37 that describe construction loads. And some people may have different numbers to use in their own practices, but I pulled out a number that for construction where, you know, not too much equipment is going to be used during construction, and there are different categories in this document, ASCE 37 document, I pulled a number of 25 pounds per square foot to consider as loading for personnel loading during construction, just to illustrate calculation of transverse reinforcement. And the factored load that would be relevant for this particular case, the load factors are slightly different from gravity load factors, and these are also delineated in ASCE 37. It's 1.2 times dead loads, plus 1.4 times loads that could be variable in thickness. So for example, this 3-inch topping may not be exactly 3 inches in all places, so there might be some variability. And in fact, ASCE 37 considers weight concrete weight as a variable material load. So the load factor for that is 1.4, and then 1.6 times the personnel load, typical of live load factor. So if we use those values and plug them into our ultimate load combination, we end up with 0.12 kip per foot for the ultimate load considered for construction. So your slide might have a little bit different result because I omitted describing this calculation in detail, and the only combination that was here was 1.2 dead plus 1.6 live. So you might want to edit your slides to cover this. So with these loads in mind, and if we consider a 1-foot width of slab, you know, these are surface loads, right? And this is, again, would be a surface load, except that if we consider 1-foot width of slab perpendicular to the plane of the screen, this load becomes now a line load, but the resulting moment, as shown here in this calculation here, will be a moment per foot of slab. So that's why this load now became a kip per foot instead of a load, surface load of kips per square foot. So we are considering a 1-foot width of slab. So that would be the ultimate moment of 0.43 kip feet per foot for the construction condition. For the in-service condition, we have to investigate two cases, one associated with the 40 pounds per square foot of live load, uniform live load, and the other one is to consider a live load to be a concentrated factor live load. So ASC-7 indicates that you don't need to consider these two concurrently. You can use one or the other. Your design load or your design moment should be computed by assuming dead loads and one of these two live loads, the 40 pounds per square foot or the concentrated force of 3,000 pounds. And of course, when you factor them, the concentrated force becomes 4,800 pounds, whereas the factored uniform load, including dead and live, would be 0.14 kips per foot. So as mentioned in the previous slide, we have to consider whether the concentrated force is critical, is what governs, or whether the surface live load is what governs. So the ultimate moment in any case is, so when we consider a surface dead and live load, it would be 0.14, as stated in the previous slide, that would be the factored load, would result in 0.5 kip feet per foot. Again, we're considering one foot width of slab that is for design purposes. And note here that the effective span, so L here is the effective span measured to the face of the stem. So I'm using 32 and an eighth of an inch, subtracting half of seven and three quarters of an inch from the top width of the stem. So one does not need to go in 36 inches to the center of the first stem, but rather just to the face of the support, as would be the critical section for moment. The other ultimate moment that we need to consider is, well, that associated with uniformly distributed dead load plus the effect of this concentrated force, except that for the concentrated force, we need to, since we're dealing with moments per width of slab, we need to divide the moment induced by the concentrated force, P times L, where L is 32 and an eighth inches, divided by the effective slab resisting that concentrated force. So the effective slab width is calculated as four and a half inches, which is where this load is assumed to be distributed, this concentrated force, and using the three to one spread angle that we mentioned earlier, that's adopted by a PCI design handbook. So that results in 16 feet, roughly, of slab that's resisting this concentrated force. So that's why we divide by 16.4 feet here and end up with 0.66 feet per foot of slab, which is what governs the in-service condition. Okay, so now that we've done that, we use reinforced concrete design because in the transverse direction, the double T is a reinforced concrete member, it's not pre-stressed transversely. So these are equations, or one form of equations, that one might use for a reinforced concrete design. Some people may be familiar with this approach, that's what I normally use. But the bottom line is that you end up by using the, this is, by the way, for the construction condition, we had 0.43 feet per foot. And the element that's resisting those loads is the two inch thick flange. So we need to design the reinforcement within that flange to resist the construction loads. And coming up with the reinforcement ratio for that, it ends up being 0.005, roughly, for a one foot width of slab. The total area per foot of slab would then be 0.0049, which is the reinforcement ratio, times 12 inches, that's for one foot of slab, and times the effective depth, 1.25 inches, assuming a three quarter inch cover, top cover for that reinforcement in the flange. So we would end up with 0.073 square inches per foot. And in the longitudinal direction, we have to provide the minimum area of steel required for shrinkage and temperature. And that is given by this equation. I'll describe this equation in more detail and give you the reference section in a second. So we could, the reinforcement demands could be satisfied by using welded wire reinforcement at six inch and 12 inch spacing, where the wires are D4 in both directions. And that minimum area of reinforcement, that's in the previous slide, comes from ACI 318, section 7611, where for non-pre-stressed one-way slabs, the minimum area of reinforcement is this quantity here, 0.0018, so it's a reinforcement ratio. And this 60,000 divided by FY is just simply a factor that for reinforcement grades greater than 60 KSI. And A gross is a gross cross-sectional area. So the maximum of these two quantities, 0.0018 and 0.014 AG. So for the case of, so this is the section that also would apply to the previous slide as shown here, right? And for the case of the topping, we also need to provide at least minimum reinforcement ratio. So that's what I used here to come up with the minimum reinforcement ratio within the topping. And that would be 0.06 square inches per foot of slab. And the reason I did this, I wanted to have an estimate, say, of well, what happens with minimum reinforcement within the topping and within the flange, if the minimum reinforcement is able to resist the applied moment in the topping plus flange, then we don't need to provide any additional reinforcement to resist the in-service ultimate moment generated. So that's how I'm approaching the flexural strength for the in-service condition using first minimum reinforcement and see if we can satisfy the design strength that's needed. So the bottom line is going back again, is we have for minimum reinforcement in the transverse direction within the topping, we would need 0.06 square inches per foot. And that is what's shown here for the topping, 0.06 square inches per foot. Your slide may have a slightly different number. And then within the flange, the welded wire reinforcement that we designed ended up being 0.08 square inches, which is greater than what we needed. So the design would look roughly like this. And I estimated effective depth to the two layers of reinforcement as shown here. So in order to determine the resistance, the capacity of this section, I divided the tension forces into two components, T1 and T2, that would be balanced by the compression force. And I think the notation in your slide was slightly off. T2 is always going to be, or the subscripts 2 are always going to be for the topping reinforcement. Subscripts 1 are going to be always for the reinforcement within the flange. So if we follow that notation, and these are the variables that we're using, so this flange is subjected to negative moment. So all the reinforcement, we need to see whether it's all in tension. And we use that, we do that by estimating the effective stress block depth. It ends up being 0.178 inches from the bottom of the flange from this point. So it results that these two forces are in tension. And the other item that one should check is whether the two layers are at yield. But I will show you in a second that that makes, that it actually, I did check it, I'm not showing it here, but one has to, for these equations to be valid, the reinforcement has to yield. And I did check that and it yields based on the position of the neutral axis. So having done that, we can write an equation, a bending equation, considering the two tension components by taking moments about the centroid of the applied, of the resisting compression force. And that's what As1 results, times Fy results in 5.2 kips. That's the force for T1. The force for T2 is 3.9 kips. And these are the corresponding lever arms for T1 and T2, giving us a nominal moment strength of 1.6 kip feet per foot. And using the strength reduction factor, that's 1.61 kip feet per foot. There's some problem here. I think what I meant is 0.9 times 1.61 is 1.59. That seems a little bit high, but in any case, it's higher. There's some numerical error here that I hadn't identified. In any case, it's greater than the m sub u of 0.66 kip feet per foot. So with this reinforcement, which is the minimum reinforcement required, the top flange and topping have enough capacity. I became distracted because there was a question that popped in. And it says, why is the concentrated force O divided by 2? Thank you, good question. I forgot to mention that. When let's see, go back here. I assumed that this force that was considered, and I didn't mention this when I was presenting this slide. I considered that there would be attachment between these double T and the adjacent double T. So these two flanges would be connected using some connector plate. So any concentrated force at the tip, I assumed that it could be distributed half and half between those two double Ts. Now the worst case scenario would be to consider the full force applied to the double T itself, but that would mean that there wouldn't be any connection between the flanges. That's why I considered it divided by 2. Thanks for the question. So going forward to where we were, now we need to consider a concentrated force at the flange corner. And this is this case shown here. In this case, and this relates to the point that was just being made, it is very common for the first connector, flange to flange connector, to not be located at exactly the corner, but rather some distance inward or inboard from the end of the double T. And typically, I mean, that distance could be as large as 5 feet. So in this case, what we're going to do is we considered, or I considered, that this force was entirely resisted by one flange. The total concentrated force of 3,000 pounds was considered by cantilever action of a single flange of a double T. And for the in-service condition, that led to a 4.4 kip feet per foot moment. And note again here that the effective length of slab resisting that force is much smaller than the 16 feet we're using before. So this is a much more critical case. If we use this moment to design the reinforcement needed, we would end up with 0.28 square inches per foot of slab. And that is a much heavier welded wire reinforcement pattern than we need for the center of the double T with heavier gauge wires. So sometimes what happens is that this is at the edge of a building. So the corners are at the edge of a building. And there might also be some cord reinforcement within the diaphragm. So that might be used as well to resist this negative bending. So this reinforcement here shown as circles, as black circles, could also be coming from a cord reinforcement or any other reinforcement that's needed in a diaphragm. And that could also be used to resist concentrated forces. So I realize I'm not showing you the details of how these numbers came up. But it's using a similar approach as for the concentrated forces within the span. And just modifying the effective length of slab that's resisting the concentrated force. And these are the end results that I came up with. So I would encourage you to just look at those numbers and see if you can come up with them yourselves. All right. So that's all there is for transverse bending of flanges. The next topic I wanted to cover tonight is design of openings through webs of beams. So prior to going into this topic, I wanted to ask Sherry to start the first poll of the night just to get an idea of how many people design openings in webs of double Ts. Okay. Your poll is now launched. Please answer on your screen. Thank you. Please answer as quickly as possible. I will be closing the poll shortly. All right. Nice job, boys. Now you've seen the check. Well, everybody else, tell me. I am closing the poll now. Okay. Sergio, 36% answered I use standard details, 7% answered I use strut and tie, 43% I never design reinforcement, and 14% I use references available in the PCI designer handbook. Okay. Thank you, Sherry. While the poll was going on, there is a question that came in that says, how do you ensure that the topping is acting as a structural topping? Do you have to prepare the flange to bond with the topping? And the answer would be yes. You'd have to ensure that there's composite action in order for there not to be any slippage between the casting place topping and top flange. Otherwise, they would have to be considered acting separately. And usually, normal roughening practices would, I mean, there's a requirement of a quarter inch roughening amplitude, but that seems to be excessive. Some research has pointed to a quarter inch being excessive, however, that is the code mandated roughness. So, you know, one has to do what the code suggests. But yes, you do have to ensure that there's composite action between the topping and the top flange. So anyway, I only, so what I wanted to point out, and the reason I asked this poll is design of webs of beams, openings through webs of beams, is I wanted to get an idea of how many of you use the strut and tie models, which we very briefly covered last time in the design of corbels. This is one, that method is one possible approach. It is slightly, you have to actually know or have to come up with a strut and tie model that reflects the load, how the load gets carried around an opening through the web of a double T, and that is not necessarily an easy task, and there's not a unique model for that. The other two, I wanted to get an idea as well as, you know, whether people have standard details, which, by the way, may not necessarily work all the time, and I'm going to point to two topics here that might be of interest to you. And what I followed is I also used the references available in the PCI design handbook and kind of massaged them in a way that could be used for ultimate strength design. So I'll spend some time pointing you to those references and give you my interpretation entirely of how one can use those references for design of openings. And without more than that, I wanted to point out the two references that there are several – I mean, there are several references, and I read a few of them prior to developing this module, and the bottom line is that I found that there's no unique approach available for design of openings if one wants to use a design method such as proposed by Kennedy and Abdallah or Barney and others. So there are different design procedures. The other thing that I found is that mostly when you look at those references, they were developed for loads corresponding to service load level. None of them really studied the ultimate condition in detail. And these are some of the highlights of two of the references. Kennedy and Abdallah developed a design procedure that actually focused on developing – on avoiding cracking in the bottom or the lower cord of an opening, which is perhaps not necessarily very efficient. And Barney and others developed a procedure to determine the distribution of shear in the top and bottom cords. This is actually what I'm going to discuss in more detail later, so I'll defer to that point. But many of the other references that are in the PCI Design Handbook, and I supplemented that with some other references that I found, are here. So you can see that there's – openings through webs have been studied in the past. And some of the findings have been – for example, single T beams have been tested with fairly large openings, and what has been found or identified were cracking patterns and also the idea that the cords around the opening, the top and bottom cords, deform in an S shape or double curvature shape, and we'll see how that is – or why that is important. But what I also found is that the number of – the difference in the different tests that I found, in terms of the shape of the opening, the location of the opening, the size of the opening, was – varied tremendously. So if you have these many different variables in a test program, it's very difficult to establish a uniform or unique design procedure that could apply to any case. So this type of beam might behave more like a truss-type beam, right? So – and this – I don't think these are very common. The most common type of opening, in my mind, for parking garages would be of the kind shown here on the far right, although these openings seem a little bit large, and notice the span length is not that long, so it's 36 feet span. So the PCI design handbook does have some general guidance that I think was pulled from some of these references, and some of the key features of these guidance is that – one thing is that one should locate an opening outside of the strand development region. So you want to avoid – in the strand development region, there's a region of stress transfer and so on that you want to avoid creating stress concentrations by placing an opening in that region. The other part is that openings towards the ends of the girders are under high shear and low moment, so there might be high amounts of reinforcement that could be required in those openings that – or around those openings that would be difficult to place. So I – after speaking with people and making my own judgment, there's – preferably openings should be placed in the middle third of the beam would be my recommendation, although sometimes that's not possible, so I did go and design an opening outside of that region. But then I believe – and I'm not 100 percent sure now that I'm reading this – that placing an opening at a minimum distance from a – from the support at least equal to one-quarter of the span length is a reasonable suggestion, and I don't know whether that comes from PCI or whether it just came from something I was thinking at the time. Try to place the opening near the top of the beam, but not right under the flange. If it's placed right under the flange, it might interfere with any top strands that might be needed to control top tension stresses. So you have to account for where are you – where you're going to place top and bottom strands and try to place the opening to avoid any conflict with that. And the topping, as was mentioned earlier, would not – should not be considered for strength unless there's transverse reinforcement crossing the interface that allows reliable stress transfer. So I mentioned roughening the surface would be enough. In this case, in the case of openings, you might want to have some extending reinforcement into the topping to be more effective in transferring stresses. So the PCI design handbook does have some guidance, and these are just pulled directly from it. You can see that there's guidance on geometric requirements, on the location where it's outside the strand development length, the – where it's located vertically on the web. Also do not locate an opening where you anticipate a concentrated force to be acting. And try to locate it in regions of low shear and below where the compression block for a moment might form. And when you're trying to check service level stresses, you want to avoid cracking because cracking in the section would then interfere with the bottom cord strength. So you want to keep service load stresses for the growth section to be below cracking. And again, I'll point you to references by these three research groups for more detailed recommendations. But the section in the PCI design handbook is 5.13.4. So there – as you can see, I wanted to show you just a couple of references and to see – to demonstrate that there's a variety of suggestions on how the shear is distributed. So imagine an opening in a beam. So you have the web is split, right? The top cord, then there's an opening in the bottom cord. And the total shear at a section has to be resisted by what's left of the section within the opening. So there are recommendations on how much shear is carried by the top and bottom cord. So Barney and others used a proportion which is of shear that is proportional to the ratio of the moment of inertia of each of the cords. The bottom cord in this case would be proportional to the shear in the bottom cord. And likewise for the top cord. In the case of Abdallah and Kennedy, they proposed a distribution that is shown here. And they only referred to the bottom cord. They only said, well, the shear in the bottom cord is determined using this equation where the notation is shown here. Area ABW corresponds to area at the bottom cord, et cetera. But this equation would only be applicable to an uncracked tension cord. So what I decided to do rather than trying to compare all methods and so on, I concentrated on the Barney research and recommendations and illustrated my interpretation on how that could be used for ultimate strength calculations. So that's what we're going to focus on in the rest of this evening. So I wanted to show you in our design example how I treated openings through our double Ts. And for that, again, this is a plan view of the parking garage. I chose a double T that is 10 foot 8 inches wide. And the span is 61 feet 6 inches to the far end of the interior wall or far face of the interior wall and the edge of the building, resulting roughly in these dimensions, right? A total length of 60 feet for the double T, a clear span of 59 feet or a center line of bearing to center line of bearing span of 59 feet, I should say. Those are the general dimensions. And of course, we need to start with loading on the double T. So self-weight, 3-inch topping, you know, dead loads are 1.01 kip per foot, live loads, the total service load is 1.4 and the factored uniform load is 1.9. So all these are common numbers we've been using throughout this course. So the first thing to do would be to identify where and how large the opening might be. So I chose the length of the opening and the height of an opening that I will give you in the next slide. But I wanted to make sure that the openings were located at least a distance greater than the development length of the strand that was being used for this double T. And this also illustrates a notation, right? So L0 is the length of the opening, H0 is the height of the opening, and X is the distance from the centerline of the support to the center of the opening. That's what we're using. So in order to illustrate the calculations, I located one opening with a face. The opening dimensions, by the way, were 12 inches high, 36 inches wide, so a fairly long three-to-one aspect ratio of an opening. Seven feet to the face of the opening from the centerline of the support. So that gives us, with these dimensions, eight and a half feet to the center of the first opening and 21 and a half feet to the center of the second opening. This second opening actually is in the middle third of the beam. The first opening is not, but it's a way outside of the strand, just outside of the strand development length. So this would be the closest section where I would feel comfortable placing an opening closest to the support. So using moment and shear equations for uniformly distributed load, for each of the two sections, the section at 8.5 feet and 21.5 feet, we can determine the factored moment and shear at that section. And just note that the section or the moment and shear are being calculated at the center of the opening for each of the two opening locations illustrated here. So that's what's shown here in this slide, where the section X is 8.5 feet and L is the span of the beam from center of bearing to center of bearing. So the rest is just calculations for W sub U, the factored load on the double T beam. So 39.9 BU and 407 kip feet for the section closest to the support. And as expected, a lower shear of 15 kips for the section at 21.5 feet and a much higher moment at that section. It's a uniformly distributed load, so that's what we would expect. So Barney and the method by Barney suggests that there has to be vertical reinforcement located on both sides of the opening, and that vertical reinforcement must resist the entire shear computed at the center of the opening. That's 39.9 kips for the opening located 8.5 feet from the left bearing. So notice that the 39.9 kips divided by, this is for the double T, divided by two stems that the double T has, divided by the phi factor, that would be the nominal demand. And this reinforcement, the steel reinforcement is not, that's designed for both places or both sides of the opening, resists the entire shear. So that's why we divide directly by the 60 KSI, assuming a grade 60 reinforcement. So we end up with fairly heavy reinforcement on both ends or edges of the opening. So that's two number three u stirrups on each side of the opening. So that's what this line would represent. Notice that I, you know, I'm not drawing the curved radii around the opening, because of what I wanted to do here is illustrate the model that Barney uses to estimate demands around the opening. So what they do is they simplify the opening and say, well, what is the, we want to resist this shear located at the center of the opening. And it's assumed that the shear is going to be constant throughout the opening. That's an assumption, right? We know that that's not true for the case of uniformly distributed load, but that's an assumption in the model. So the shear on the left and right ends of the opening is the same. The only thing that varies is the moment. So from zero moment at the center, it's assumed that the moment, the cords are under double curvature. So there's a point of inflection here. There's a zero moment here. So there's a moment equal to M minus the change in moment. So the moment that acts at that section minus the change in moment here due to the opening and plus the change in moment due to the opening. So that the total shear resisted at the top and bottom cords, this is the top cord, this is the bottom cord, times the distance L over two is equal to that change in moment from this section to the edge of the opening. The other thing that's worth mentioning is that these are dimensions that are important. D sub s is the distance between the top face of the beam and the centroid of where the bottom cord. And delta d, the change in depth, is from the centroid of the bottom cord to where the effective pre-stressing force is located. That's why it's illustrated here as P effective. So if we split, if we take a free body diagram right here at the center of the opening, these are the internal forces. There's a shear force at the top cord, shear force in the bottom cord. The total shear, top plus bottom, has to be equal to the shear at that section. There's no moment, there are no moments here because it's a point of inflection. And so this moment that's on the left hand face of the opening is resisted through this force couple, C and T. So by setting a moment equilibrium about point A and point B here, as you can see from these two equations, you can come up with equations to define the compression force in the top cord, the tension force in the bottom cord, which are a function of the applied moment, or the moment at the center, I would say, and the the shear at the center of the opening. And also, but importantly, the effective pre-stressing force here. Notice that it's the effective pre-stressing force that enters into these equations. And I note that because we'll have to modify that for the ultimate condition. And they suggested that if the tension force here is greater than six times root F prime C times the area of the tension cord, or bottom cord, that means that this bottom cord would be fully cracked. So this method applies not only to service load, but also cases where that section would be fully cracked. Okay, so looking at our problem, we have again, the effective force. And I wanted to show you how the distribution of top and bottom cords, of shear and moment diagram. These are the shear and bending moment diagrams that are within the cords. And the relationship between the moment in each cord and the shear in each cord is illustrated by this equation. So for example, if we find the shear in the top cord using one of the Barney et al. equations, we can then find the moment in the top cord at the face of the cord, not at the center. We know that at the center, it's assumed to be zero, right? So we want to find this moment at the end, which produces double curvature bending. And from those equations that I had before shown you, we can say, well, that Barney et al. suggested that the top and bottom shear was a function of the applied shear at the center of the opening, and times this ratio of moments of inertia of the top and bottom cord. So we need the properties of the top and bottom cord to compute that. So let's say, looking at one stem of our double T, this would be the cross-sectional dimensions at the place where the opening is located for one stem only. The center of gravity of a prestressing strand is five inches from the bottom, so that's one number to remember. And the opening starts 12 inches from the bottom, is 12 inches deep, and it's located two inches below the bottom of the top flange. So when we take a section through the center, of this opening, we see clearly that this is the shape of the top cord. It's a T-shape. The bottom is a tapered or a trapezoid. But to simplify our calculations, so I calculated the average width of this trapezoid to be 5.4 inches, to calculate area and moments of inertia. So what we need to do is, we have to calculate the moment of inertia of this shape about its own centroid, to come up with I top, I bottom about the centroid of this shape as well, which I assume to be at mid-height, which is not entirely true for a trapezoid, but it's close enough, so it's six inches. And that gives us a moment of inertia for the bottom cord of 783. And then once we have those two properties, we can plug those in for the I top and I bottom, divide them by the summation of both moments of inertia, and that's the fraction of the total shear that would be resisted by the top and bottom cord respectively. And so we can compute that for the eight and a half foot location of the opening and 25 and a half foot location of the opening. So you see that because the cross-sectional dimensions don't change, it's only the shear force that changes. The distribution of top and bottom shears are as shown here for the two sections. Using those shears, V top and V bottom, we can then determine the moments in the top and bottom cords. We also have to estimate the top compression and tension properties. A question came in. For the top portion, are you using the topping properties? I have to go back here. It does not seem like I did. I used the only the properties of the bare double T because I used a two inch thick flange. So this is the bare double T without using the topping properties. So to determine the top and, I mean, the axial tension and compression in the cords, we use the equations that were derived a few slides back. And notice what I'm doing here. The method that was proposed was proposed for service level loads, but we normally design, you know, for pre-stress concrete, we have to design reinforcement for not only for service but also ultimate. So instead of using an effective pre-stressing force, I use the ultimate or the anticipated pre-stressing force at ultimate conditions. So I estimated that to be a pre-stressing stress of 266 KSI for the ultimate bending condition. To use that instead of P effective, the ultimate pre-stressing force. But the rest of the equation stays the same because I'm using the factored moment here and I'm going to compute the ultimate tension and compression forces in the cords. So there's a follow-up question after my answer of using only the double T section. It says, but we are using the live load and dead load loads on the double T. But I guess I assumed and wanted to be conservative in the case of an opening that those loads would be superimposed on the pre-cast pre-stress section. That section, even if the topping is not made composite, has to, the section has to resist the total loading, right? So it's the dead load includes the topping, but the topping is not assumed to resist any of the loads at the opening because there's no reinforcement that that goes from the cross-section into the topping to guarantee shear transfer. Because this is a region of concentrated shear that we would need more than roughening. My interpretation is one would need more than roughening. One would need effective, you know, friction, shear friction reinforcement there to engage the topping as part of the, to engage it structurally. So that's why I neglected it and wanted to just determine the reinforcement for the bare section. But yes, we're carrying the total loads, that's correct. And that is reflected in this m sub u and the v sub u that we were using. So notice one thing. The first thing, this tension force, the signs in the equation assume that tension was actually a tension force, so it was pointing outward from the section. However, after you doing the calculation here, we end up with a negative number, meaning that there's at the section located at eight and a half feet from the end, the bottom cord is still in compression because of this large pre-stressing force near the end, right, and there's very little moment to offset that. The second equation for compression in the top cord was derived again, assuming that indeed that was a compressive force. The positive sign here indicates that our assumption was correct. So both cords are in compression in this particular case. The negative sign gives that away for the tension cord, theoretically tension cord, and the positive sign gives that for the compression cord. So we can now determine, and this is still for the entire, these calculations are still for the entire double T. The fraction of total shear that the double T, of the double T that's, the double T shear that's resisted by the top and bottom cords is shown here. We divide by two to convert that from total double T to shear per stem. So we have a 2.4 and a 17.6 kips per stem. We can then calculate the moment by multiplying the top shear and bottom cord shear times L0 over 2, which by the way is taken from the center of the opening to the centroid of those stirrups located next to the opening. So that gives me a top moment of 48 kip inch per stem and 352 kip inch per stem for the bottom moment. And once we do that, notice what I did here. I designed the flexural reinforcement for the bottom cord as if it were a plain or a reinforced concrete section. And I'll mention that I neglected the effect of any pre-compression in the, produced by the by the prestressing strand. A strut and tie model would actually reflect the compression effective there. I neglected it in this case. So using again reinforced concrete theory and using the properties of the double T, using the average width of the bottom stem, and the ultimate moment of 352 kip inch for the bottom portion of the bottom stem, the resulting calculation resulted in two number six top bars for the bottom cord that would be needed to resist that moment. The shear strength was checked as follows. Again, using the assumption that the cords are non-pre-stressed, we then fall into the category of the normal shear equations without pre-stressing or the non-pre-stress shear equation from the ACI code. But that I did include the effect of the compressive force generated in the bottom cord that is coming from the pre-stressing force and the moment that acts in the section. So I use this equation, it's like it's the equation pertaining to or applying to a member that's subjected to shear and axial force. And this equation would have to be applied for both the top and bottom cord. And the reason for for using equations for non-pre-stressed reinforcement, non-pre-stressed members, is because of this continuity that the opening generates. It wasn't clear to me that the cords would behave as a pre-stressed, fully pre-stressed member, right? And if they were, you know, how much of that pre-stressing stress would be located in the bottom cord and how much of it in the top cord? So it's a difficult question to ask or to answer, I should say. So in any case, perhaps we're being a bit conservative here, but applying the equations and again noting that the normal force in the bottom cord corresponds to that normal force that I indicated to you was Tu, but in fact it was a compressive force. That's why it has the positive sign here in this shear equation. The contribution of concrete to shear is 21.2 kips. Using the load factor, the resistance factor is 16 kips for the bottom cord. And I tried the minimum area of transverse reinforcement for the bottom cord. What would that be and how much additional that would provide? Notice that we're very close for the concrete to carry all the shear and by providing minimum reinforcement in the bottom cord, that gives us an additional 2.1 kips per stem. So with that, we exceed the demand of 17.6, which would be sufficient. So we could use welded wire reinforcement in a three inch by three inch layout and these are the wire diameters or our gauges, diameters. So what happens with the opening in the center or near or closer to the mid span in the middle center of the beam? I won't go into as much detail as I have now, as I have so far for the end opening. Suffice it to say that you'll see that the tension and compression forces are the tension force, which is again compression is smaller. The compression force is larger in the center. The shear demand is lower, so therefore the the moment demand in the top and bottom cord is lower. And then instead of needing two number six bars, we need two number four bars. And by the way, this should be, so this should be for the bottom cord. The top cord reinforcement would be calculated in a similar manner. And again, using the shear strength equation, we could come up with the amount of steel, the reinforcing steel and the welded wire reinforcement that would give us what we need. And you know, we still need in this case phi v sub c is sufficient to carry the shear generated in the bottom, but we still need to provide minimum transverse reinforcement. And that's why we end up with this welded wire reinforcement needed below the opening. So what I proposed was this type of arrangement for top and bottom reinforcement around the opening. These are the vertical stirrups needed on both ends. And this only illustrates the reinforcing pattern for the opening at the end, near the end. I should say, the welded wire reinforcement at a three by three inch layout is shown here. And notice that the two number six bars have to be developed at the face of the opening. So we have to provide their development length. Typically, it's recommended to provide one and a half times the development length of those bars. So I didn't calculate that, but one, you could do that fairly easily to determine how long these bars need to be. You could even extend them into the support for practical purposes. And then the top cord does not need as much reinforcement for bending, so I just provided two number three bars to tie stirrups and all the details, the rest of the details needed. And these are, these show two cross sections, right through the center of the opening and right at the end of the opening, what the reinforcement might look like. So I don't, I'm not showing you the reinforcing pattern at the center, but you could do the same and, you know, given the forces and demands in the center opening. The bottom line is that locating the, I had a little bit of trouble placing or fitting all this reinforcement towards the end of the beam, then that's why it's better to have openings located farther into the span in regions of lower shear. Okay, the last and third, the third and last topic tonight is warping of double Ts, and that is in the PCI Design Handbook in section 15.3.7. And just a little bit of background and why warping is needed or twisting of double Ts. It is sometimes done to provide drainage in parking garages, and the idea or the procedure is to set one of the four support points lower than the other three, and just by self-weight the double T will warp and accommodate the slope. And it will, in fact, the combination, the warping term is, there's a discussion of that in a couple of slides, but the double T will actually twist and bend transversely to accommodate that differential position of the supports. I wanted to, before going into this topic in detail, I wanted to just ask Sherry to launch the second poll just to get a sense for how many times, or how often people consider warping in their designs. Okay, your poll is now launched. Please vote on your screen. What is it? Please vote as quickly as possible. I will be closing the poll soon. I don't think I can hear you. I am going to be closing the poll now. Okay, Sergio, seven percent answered in 100% of parking garage projects. 14% answered in more than 50%, but not 100% of parking garage projects. 21% said between 10 and 49% of parking garage projects, and 57% responded never. Warping decisions are left up to the director. Okay, great. Thank you, Sherry. I think that's interesting. I will show you, so we'll go through the some considerations for warping, and I think you'll find them instructive because we'll see that, you know, cracking could occur because of these warping decisions. So it might be worthwhile in some particular projects for the designers or the fabricator to actually be involved, or at least know whether there might be a need for reinforcement if it is expected that double Ts will be warped extensively. So the idea is that there are some references, of course, for this problem. And there's a good reference, Mack and others, that conducted a survey to see what the common practices were. And the idea was that, you know, typically to provide drainage there is a need to provide, you know, some slopes in the cross slope direction or transverse direction and longitudinal direction to provide effective drainage in parking garages. And given these slopes, in order to allow for these slopes, one of the offsets, let's say, one of the stems needs to be offset between three quarters of an inch and one inch to provide, to result in these types of slopes. But other sources recommend more than two or two percent in any direction, as reported by Mack and others. But what's really important in these values, and this is, as you can see, there's a high variability of these. Canberra also affects, you know, what the actual slope is. So if you have a member that ends up with a positive canberra, that means that the slope might be greater in some areas, typically downstream from where the canberra location, and lower in the higher points. There's also an issue to consider in terms of warping and slopes that ADA accessibility requirements have to be considered. So the amount or magnitude that one tries to warp or offset one support over the other has to also consider the effects of maximum slopes that are permitted for the, to satisfy a medical American, the American Disabilities Act. So the PCI Design Handbook has a couple of statements on warping, but not, which are considered practical, but it hasn't, doesn't have too much detail on how to account for it. It has statements like, you know, what the amount of warping or how much warping has been actually applied in practice successfully. So there's values like a quarter of an inch per foot of width of pre-topped four-inch flange, or three-eighths of an inch per foot. Note that, for example, this upper level here, three-eighths of an inch per foot for a six-foot spaced stem, double T corresponds to two point, two and a quarter inch of vertical stem offset, which exceeds the numbers that were being found by, you know, by Mack and others in their 2003 publication. And here are two publications that are relevant that I used to do some of the calculations that I'm going to be presenting to you in a few slides. The important part is that warping may cause cracking near the flange stem intersection. That has been observed in practice, and according to a survey done by Mack, and warping can result from these four different phenomena, uneven lifting, racking, twisting of a double T during handling, or uneven supports to accommodate transfer slope for drainage. And I wanted to just give you some findings of a couple of references in the PCI Design Handbook, and you can read them on your own. What's important here is that the Banks reference here worked on a rigorous finite element analysis to try to incorporate or to understand the phenomenon of warping. And this other reference really concentrated on a more practical aspect of the warping problem. So I would refer you to these references if you want to read more details into what they found. One thing that I wanted to clarify is if you pick up a book, you know, I work in a university, so I read books and I try to, you know, clear up terms with students. When you talk about torsion, there are definitions that are used. Twisting is the action of a section rotating, a rotation of the cross section. Warping is the longitudinal deformation because of an applied torsion. So the way that we define it in the PCI Design Handbook, we call warping the result of an applied torsion. So the idea is the real fact is that when we apply torsion to an open section like a double T, it's not only going to warp longitudinally, it's also going to twist. So just to be consistent, and there's also cross-section distortion. What that means is the change in general shape or relative shape of the section after the application of torsion. However, we will use the term warping in this module to be consistent with the PCI Design Handbook at practice to signify the combined effects of twisting and warping and cross-section distortion. So what I'm saying is that when you apply a torque or a torsional moment to a cross-section, there will be different components of deformation. One which is a rigid body rotation. The total twist will be given by a combination of a rigid body rotation, as shown here, and a cross-sectional distortion that, as you can see, will generate moments within the flange. It is this deformation component that causes cracking in the flange. So the far end, one end of the section is warped or twists and distorts, and the far end stays undeformed. So that's because one of the supports is the only one that that shifts in location. So what I'm going to attempt is, knowing that, I'm going to use one of our double Ts, so where the the notation is given here. B is the width, you know, the top of the top thickness of one of the stems is T top, T average, T bottom, as shown here. T is the thickness of the flange and H is the distance from the bottom of the double T to the bottom of the flange. And when we apply a torsional moment, the section will twist and also distort. There will be a net vertical deflection on one side of the double T, and we are trying to estimate what the effect of this torsional moment will be, or how much, how much, I should say, how much a vertical displacement, how much moment that vertical displacement induces in the section, so that we can find a way to estimate how much moment is being induced in the top flange. So given this notation here, these are our general dimensions of one of our double Ts. The way I calculated the average thickness, average width of one stem, was I subtracted from the total area, I subtracted the flange area, B times T, and then I divided by two stems, and by the height of the stems, to come up with an equivalent width that will give me the same total cross-sectional area of the double T. So what contributes to this torsional moment? Well, it depends on how stiff the section is, right? And torsional stiffness is given by this parameter J times G. G is the shear stiffness, the shear modulus, I should say, of the material, of concrete. And J is a torsional constant that is calculated using this equation here. It's the product of, so the summation sign here indicates that we're going to discretize the section into component rectangles. Each rectangle has a length B and a thickness T, and the J contributor of each of these rectangles is one-third times B times T cubed. Now this one-third here corresponds to a constant that's shown here as Ci, and this value of one-third corresponds to a fairly long rectangle. So we need to check whether the component rectangles are long, or otherwise use different values for the C sub i's to estimate distortional stiffness. So our section is going to be made up of different component rectangles, so that would be the torsional stiffness. The torsional stiffness times the twist angle per length of double T will then result in the torsional moment applied per length of of double T. So that C value in the summation equation for J is determined, as I mentioned, from the component rectangle. So I'm discretizing the double T section into three different component rectangles. Number one is the flange, two and three are the two stems, and for each of the stems, instead of using just H for the length of them, we're going to use H plus two times T, the thickness of the of the flange. I'll come to that point in a second. So this would be the B value divided by the thickness value for each component rectangle for the stem. So that gives me 4.78 for the aspect ratio of these component rectangles. Given this number, we interpolate between these two rows and come up with a C value for the two stems of approximately, you know, almost 0.29, I would say, so 0.289. For the actual flange, we have a width B of 128 inches, a thickness of two inches, so that approximates this infinity value here. It's a much larger value, so we can use about one-third for the C value there. So to come up with a total J, we then use the B over T for the flange, the H over two T over T average for the stems, and we compute the individual J components to the total J for the entire section. And numerically speaking, this is for the flange. This is the C factor, the width and the thickness for the flange, the C factor, the depth plus two times the slab thickness for each stem, and the average thickness of each of the stems gives me a J for each stem of 2,137 inches to the fourth power. The total J is then 4,615. The torsional stiffness is calculated as G times J. G is the shear modulus, which can be computed using this relationship from mechanics, which divides Young's modulus by two times one plus Poisson's ratio. Poisson's ratio is normally taken as 0.2 for concrete, so G is 1,679, giving me a torsional stiffness of 7.75 million kip inch squared. I should go back to mention here that this, I forgot to mention that this, instead of using H, we used H plus two T for the stems, and this, adding this term from one of the research groups that I cited earlier, adding this term resulted in a much better match with those finite element results that they were conducting, so that's why they proposed using to calculate the J contribution from the stems to use H plus two T in this term. Okay, so what is the value of the torsional moment? Well, we assume, let's say, one and a half percent cross slope for drainage, so that means that we are going to have an offset of about 1.1 inches. The angle of twist per unit length over those six feet is 1.5 inches divided by six feet, and divided by the 59 feet of double T, that means the angle, the average angle from the undistorted end to the distorted end, this is 2.16 times 10 to the negative 5 radians per inch, so that's the twist angle per unit length of double T. The torsional moment, then, per unit length would be multiplied times that phi over L times the torsional stiffness would give me 167 kip inch per foot of double T. So, if we now use a very simple model, and here's the the simplest model I could think of, say, well, let's think that that torsional moment is applied to a flange. It generates double curvature bending in the flange, and let's assume that that flange is supported on knife-edge supports. It would bend the flange like that, and I assumed that the portion that was being carried by the flange was proportional to its J stiffness, so I used the J of the flange divided by the total J of the cross section. That generated a moment in the flange of 12.4 kip inches in the flange that would have to probably generate, would be the source of cracking, if any. And then I looked at test results. I said, well, what is the section that is effective in resisting that moment? And there have been tests that have observed cracking in the top flange, and this line here corresponds to mapped cracks at the end of the double T. By the way, from these warping of the double T, cracking is typically observed at the ends, and only at the ends of the double T. And the crack length that I considered was approximately five feet, and notice the source I'm citing here for these tests. So using that five foot length to resist that moment that is generated in the flange, I came up with a moment per foot of flange that would need to be, that would have to be resisted by the flange itself, and to determine whether the flange would crack or not. So the moment per foot of flange is then the moment within the flange divided by the length of that crack that was observed through testing. That's a moment per unit length of a flange, and the flange, if we consider it, if we want to compute the stress as if it were an elastic section, we would divide by the modulus, the section modulus of that section, the width of it being five feet, the length of the crack, the thickness being the thickness of the flange, and that would be the stress of 0.31 KSI. And then if we divide this number by the square root of f prime c, that would give us a coefficient that would tell me, give me an indication of whether it's likely that the top flange will crack or not. And the idea is, well, if I have a coefficient here, and I compare that with what we normally take as the tensile strength of concrete, you know, something between four to six root f prime c, you know, my resulting value of 4.4 indicates to me that there's a possibility that for a vertical displacement of one inch, the top flange might not crack. But still, we might need minimum reinforcement within the flange. We still need minimum reinforcement for temperature and shrinkage. So even if it were to crack, there's reinforcement there to control the cracking. So that's how I propose using some of the references here to try to come up with an idea of whether one could expect cracking in the top flange given a displacement of one of the stems to accommodate drainage requirements. So with that, I conclude barely in time for today. And I wanted to turn it back to Sherry for any questions people might have. And I wanted to thank you before leaving for attending all of our sessions. Again, thank you so much for attending tonight's session. We do have about four minutes, so if you have a question, very quickly please type it into the chat room. But I wanted to thank you all again for attending all the sessions. And your certificate of continuing education will appear on your rcp.net account within 10 days. Remember your continuing education credit is given for each individual session. You must take and pass the exam for each to receive the corresponding credit. If you have any questions, please email me at snon.pci.org with the subject line online academy. You will receive an email link to the session exam. So if you cannot take it within the time limit, please contact us immediately. If you're having any problems following the link or don't receive your email, please contact me. Again, thank you so very much and thank you Sergio. Thank you Sherry for coordinating. Good night everyone and I hope you enjoy the rest of your evening. Thank you. Good night.

design and analysis

double-T beams

transverse design

openings in webs

warping

PCI Design Handbook

elastic plate solutions

effective slab width

detailing openings

geometric requirements

shear distribution

calculations