I don't see windows open. No, I don't see windows. Okay. Good. Okay. Let me just change the pointer to what I did last week. So, you can see, can you see a red pointer now? A laser. Yes, I can. Okay. Yes, I can. Okay. Great. So thank you. Good afternoon. Good evening. Everyone. So, today's, as Sherry mentioned, is the second session of the six session course. And as we discussed last time, the, the next topic that we're going to be covering is, is ledge design. What we'll be talking about today is ledge design of the same beam that we designed for torsion last week. And the idea is that after this session, you'll be able to understand the failure mechanisms that are considered in design of ledges. We are also going to start to talk about the design method, the basis for the design method that's included in the PCI design handbook, which is a new design method. And you should be able to apply those design procedures that are included in the eighth edition of the PCI design handbook. And finally, the importance of appropriate detailing of a ledge for adequate performance will also be covered throughout the course. So, first of all, this is a cross section you might remember from last time. This is the cross section of the typical, of the building, the prototype building that we, we selected for, for design. And let me see if I can, I can do something here. I'm going to try something. I'm going to try to drag just to remind you. This is the, can you see the plan view of the building, Sherry? Or I was just trying to refresh people of what we were. This is the view of the building that we had designed last time. And what we're going to be doing is we're also, we'll still, we are still going to be covering this area between column axes A and B. And we're going to be looking at the same spandrel beam we did last time. This spandrel beam that has a span of 46 feet, 8 inches. So, that's the only thing I wanted to show from last week. And then moving on to today's, continuing on to today's presentation. What we want to do is we want to focus on all the possible failure modes that a ledge might experience. And we're going to, we're going to focus on provisions for shear strength. Those are included in Section 561 of the PCI Design Handbook. We're also going to be designing the ledge for what's referred to as transverse or cantilever bending. So, ledges are subjected to shear. They're also subjected to bending moments. That's in 562 of the PCI Design Handbook. We'll be talking about what reinforcement is needed for longitudinal bending of a ledge. And how to connect the ledge to the supporting spandrel beam through hanger steel. And that's in 564. Also, we are going to have a couple of polls today. And the first one will come after this first, this slide that's shown here that talks about shear strength of ledges. The idea behind shear strength, when we refer to shear strength of ledges, one has to satisfy the basic design equation that's listed here at the top of the screen. Where V sub U, the factored shear force in the ledge, and this is local shear forces, have to be less than or equal to the nominal capacity of the design capacity. Nominal times strength reduction factor of the ledge itself. But when we refer to a shear strength, the objective, when we're referring about shear strength, the objective is to avoid what's referred to as local punching shear failure. And what the assumed failure surface for shear punching shear is modified, was modified in the PCI Design Handbook as a result of research that is listed at the bottom of this slide. Research conducted at NC State back in 2016. And where a reduction in punching shear strength was introduced because of the effects of global bending and shear stresses of the spandrel. So I wanted to start a poll just to get an idea of where attendees, whether attendees are familiar with the research that was conducted back in NC State. So Sherry will start a poll in a second. I don't see the poll started, Sherry. Sherry, can you hear me? Okay, I don't see, I don't see the poll starting. We might be having some difficulties with the poll. What I wanted to find out from this poll is whether people were familiar with what punching shear failure represented. This is, it is likely that many of you have experience, have some experience with design against punching shear in flat slabs of reinforced concrete structures. And this is a similar failure mode that is included in the design of ledges as we will be covering in this session. Except that the critical section that results in punching is different, of course. And that's what we'll be discussing in the next few slides. What's important here is that I also wanted to know whether people were familiar with this research, the research that was conducted at NC State. And it was a really comprehensive research program that included a vast variety of tests. And the research was also focused on different parameters for ledge design, including height of ledges ranging from 8 to 18 inches. What the ledge projection was from 6 to 10 inches projecting out from the face of the spandrel. What was also studied are varieties of bearing width of components on the ledge from 4 to 12 inches and also distances between loads and the end of the ledge. The range of values that was tested was between 4 and 36 inches and concrete strengths range from 5000 to 15,000 PSI concrete. So all of the design provisions in the handbook are based on this research. However, the handbook does indicate that these ranges are not intended to exclude design of other ledges that might exceed these limits. But it's just an indication of the variety of dimensions and properties of concrete that were used, dimensions and location of loads and properties of concrete that were used in the research. So as you can see, it was a very comprehensive research program. And as shown here in this slide, you can see that it was a very large testing program. It included 106 tests that were conducted on 30 spandrel beams that had two different lengths, two different spans, 21 short span beams and nine long span beams. And those included in those tests, so the tests were able to be, multiple tests were able to be conducted on the same beam. So that's why you have for 30 beams, you have about 106 tests. So that included more than three tests per beam. And the idea was to capture what the effect of global flexural and shear stresses was on the behavior and strength of ledges. The other objective of the research was to determine what the effect of pre-stressing was. So it was found that pre-stressing did influence ledge strength. Also important, as I mentioned earlier, were ledge dimensions, concrete strength, some details, reinforcing details where the loads were applied and what the spacing between those loads was. So let me just back up for a second, because in terms of background of where this research started, you might recall that in the last session, there was, we talked about the torsion design provisions in the handbook. And one of the methods is the Zia Xu method, but the other one is referred to as the slender spandrel method. That research project was initiated before the ledge project. And when the investigators were starting to come up with testing for these spandrels under torsion, they found that many times ledges were designed to carry the loads that were expected the spandrel could carry. But then they found that ledges were failing at loads that were lower than anticipated. So that prompted this second research project with a ledge research project that included, the observations were that these global flexural and shear stressors or global bending and shear of those beams were the culprits for inducing these premature failures of ledges. So that's a little bit of background of where this program began. And what's very interesting, at least from a researcher's perspective, and I'm sure from a practitioner's is too, is that these four pictures illustrate some of the failure modes that were observed as part of the test. So going from top left to bottom right, picture A shows the failure pattern that was observed for a ledge that was loaded with a single load near mid span. And you can see here as I hover with the pointer here, pointing to the failure surface that was observed, it was observed that many of these cracks roughly followed a one vertical to two horizontal slope. But you could see that this whole failure surface was mobilized as the loads were applied at this point in the ledge. In the case where loads were applied very close together, you see that instead of forming independent failure surfaces, so here's one load point and here's another load point, those two failure surfaces joined into one. So the failure surface that was observed was this truncated pyramid shape where the two loads were pushing down on the same area of the ledge and causing failure. And these two, the top two, refer to loads applied near mid span or in the middle portion of the spandrel beam. And then when loads were applied close to the edge, the observation was that the failure surface could actually form inward from the application of the force. And then there was no diagonal cracking towards the end of the beam. So in this case, the point load or the simulated stem load was so close to the edge that there was not an opportunity for the failure surface to extend downward into the ledge, but rather just travel along the web ledge interface. And in the case where an edge load was applied with sufficient edge distance, you can see the end of the ledge here. You can see that the failure surface actually did form inward and actually the observation was that it would resemble a load acting near mid span. I just was looking at the chat box, that's why I got distracted and apparently the poll did launch. So I apologize, I wasn't seeing the poll or hearing Sherry on my side, but I'm glad that it was able to launch. So the response was about a third had heard of the program, about 40% hadn't, and about one third had heard but didn't know much about the program. So I'm hoping that with this background that I'm just giving you, it will then make sense to those of you that had not heard or had heard but didn't know much about this research program. So some of the additional results, these two graphs of this research program show results of the ... So these graphs illustrate two important results. The diagonal line here represents the perfect match between what would be measured in the test program, as illustrated in the vertical axis here, and what would be predicted by the applicable design procedures in each of the two editions of the handbook that are shown in this slide. So the left-hand curve or plot shows the seventh edition equation compared with test results, and the right-hand curve or plot shows the eighth edition equations compared with testing on these ledges. So all the tests and finite element simulations are included here, and one can see immediately that there were several cases for the green symbols here, referring to longer beams, longer span rules, that would fail unconservatively if one were to use the seventh edition equations, whereas if we use the updated eighth edition equations, those points now moved over to the conservative side of this plot, indicating that the current method in the eighth edition is much better, not only in terms of being conservative, but also the scatter is less pronounced if you compare both graphs here. So those two are indicators of the importance of this research program as it refers to design of ledges. So that's a brief summary of the research program, and what I'd like to do is to start getting into some of the notation that we'll be using for this course, and I'm going to spend some time in this slide. So the left-hand side of the slide shows an edge, double T, supported on the ledge. So you see some of the overhanging flange here from the end of the span rule. So the edge distance D sub B is the distance between the end of the ledge and the first stem, the centerline of the first stem. S is the spacing between double T stems. B sub T is the web width of one of the stems, and H sub L is the depth of the ledge in the span rule, and H is the span rule depth. Also in this slide is shown some of the notation for vertical and longitudinal reinforcement within the span rule. So this A sub SH and AVW, AVW first is the web shear reinforcement, A sub SH is hanger reinforcement, and the span rule also has longitudinal bars, and they're referred to as AWL. If we zoom in into the ledge region, we have the width of the span rule is B, L sub P is the projected length of the ledge beyond the face of the span rule. We have A, the eccentricity of the applied force, V sub U, this is the force that is coming from the stems, and A is taken from the center of that force to the center of the interior centerline of the interior vertical bars, and N sub U is a horizontal force acting on the ledge. Finally, A sub S is a flexural reinforcement in the ledge, and D sub L and D are effective depths. First, D is measured to A sub S, D sub L is measured to A sub L, and A sub L is longitudinal reinforcement within the ledge. So there's a lot of notation here that I wanted to point out, so this is going to be useful in the equations. And then going back here on the left-hand side, we see that these dark lines, diagonal lines, represent just a graphical representation of where those failure cracks form. They form roughly at a two-to-one slope, horizontal to vertical, and what's going to be assumed as the design perimeter, we're going to draw a sketch, we're going to simplify this failure surface into vertical faces, and those vertical faces are going to be drawn at a distance equal halfway along this diagonal crack that represents failure of the ledge. So you see that if this crack follows a two-to-one ratio, then halfway to it is one HL on either side of a stem, so the failure perimeter or failure surface will be located at a distance of HL from the face of the stem of a double T. So that's what we'll be doing. And finally, the horizontal force, ultimate force for design, is taken roughly typically as 20% of the factor dead load on the ledge. So a lot of notation and things that we need to remember from this slide. First, I'm going to present to you what the common design equations are for shear strength, for punching shear strength of the ledge. So when the edge distance is large enough, when the distance between the load point or the center of an applied load, for example, at the end of the ledge coming from the stem of a double T, is greater than half of the stem width plus the ledge depth plus the projection length of the ledge from the web of the spandrel, when this edge distance is large enough, then the perimeter can form fully as illustrated by the pictures a couple of slides ago. And if that is the case, the design shear, punching shear strength is determined by the smaller of these two equations, equation 576 for loads that are far enough apart that the failure surfaces do not overlap, or equation 577 in the handbook when those failure surfaces overlap. And I'll explain the factors here in the next few slides. You don't need to worry right now for what each of these factors mean. These are the equations that are in the PCI design handbook. So we'll go into detail into those. When the edge distance is less than this quantity here on the right, then the full failure surface cannot form. And the only modification from the previous two equations, 76 and 77, become 578 and 579. The only modification is here on the term within the brackets. And this modification occurs because then the full failure surface cannot form. And therefore, the crack, the failure crack travels along, as I mentioned earlier, that travels along the top surface of the ledge and then goes towards the end of the span row without then coming down again into the ledge. So these two equations, again, are for loads that do not overlap and loads that overlap respectively. So when loads are so close that they can be considered continuous loads or closely spaced loads, equation 580 in the handbook indicates that the design shear strength can be calculated using V times 12 root F prime C times H sub L. So I should point out two things from this slide. The first one is that the research program at NC State did not include very closely spaced loads or somewhat uniform loads on ledges. This equation was derived by Raths in 1984. And the 12 coefficient here is just a conversion factor from inches to feet. So effectively, if one takes out this 12 factor, the shear strength of a ledge would then become V times root F prime C H L. And I should point out that this equation is currently being debated on whether it's too conservative or not, because this is below. This is essentially 1 root F prime C D, which is essentially half of what the 2 root F prime C BWD that's allowed for shear strength of reinforced concrete structure. So the committees, you know, the standard specification committee and the handbook committee are examining the use of this equation to make sure that it's not overly conservative. The research seemed to support that this was a good equation. So for now, it's staying as it is in the PCI design handbook. So another point that's worth making in this slide is that we're only going to concentrate on concentrated forces that are discreet enough that they are not considered to be a uniform load along the spectrum. OK, so let's look at the notation described in two slides ago. So this is the same notation that I was referring to in figures. The only thing here, going back one slide, a couple of items needed to be explained. One of them is the factor beta. Beta is a factor that accounts for global bending of a spandrel, and gamma is a factor that accounts for pre-stressing. So beta is shown here, and I'll come back to this plot later when we solve the design example. And gamma is a pre-stressing factor that is taken or calculated using this expression. And again, we'll use this expression in detail, so no need to worry about that. What is important, though, is that the effect of pre-stressing, as I will show in the next slide, is important for this beta factor. So the presence of pre-stressing is illustrated here on these two graphs. On the left-hand side, a series of finite element analysis that was part of the research conducted at NC State. And on the right-hand side, verification from laboratory testing. What you see here is what is the beta factor. You can think of it as a shear strength factor. So it's a factor that pre-multiplies the square root of f prime c times the critical perimeter that resists shear. So that's beta. So that factor for a reinforced concrete element where gamma is equal to one, so no pre-stressing, would follow this red line here. So it depends on how close the applied moment or shear is to the nominal capacity in moment and shear for a given section. So when, for a reinforced concrete section, that beta factor goes from 2 to 1 depending on whether we are at 20% of the nominal shear or flexural strength and 60% of the nominal flexural and shear strength. When you apply pre-stressing at this level, 1.47, the beta factor increases. So the beam would have a higher shear resistance or the ledge would have a higher shear resistance just because of the presence of pre-stress. So comparing, so this is finite element simulations. These are very detailed finite element simulations with the assumptions listed here. But when we compare those results with test results for a slightly different pre-stressing value, 1.32, or gamma factor 1.32, we also see the same trend. We see that there's an increase in resistance, shear resistance of the ledge because of the presence of a pre-stress factor as indicated by these data points that were determined from testing. So these two figures show that indeed there's a beneficial effect of pre-stressing in ledge resistance, which was one of the objectives of the research program. So as you can see, this horizontal axis depends on ratios of applied moment to flexural strength, nominal flexural strength, and applied shear to nominal shear strength. So these two strengths have to be determined, and also you probably know, well, I'm sure you know that the moment and shear typically vary along the spandrel with applied loading. So what one needs to do in order to be able to determine that R factor that is then used, this horizontal axis is referred to as the R factor. Going back another slide, this is the R factor, which is those two ratios together. In order to determine beta, one needs to determine first the R factor. And the R factor is going to depend on the shear and moment acting at each section, but also the shear resistance and moment resistance at that given section. So one has to check different sections along the span to make sure that the ledge is properly designed. So from the 8th edition handbook, there's a figure indicating that R is determined from the larger of the ratios of VU over VN and MU over MN at a given section. So typically at a section right under the load. And I just wanted to make a note here that not only the factor shear force and moment are changing, but also the resistances are changing along the span because flexural resistance is changing depending on the effective pre-stressing force. And also the shear resistance is changing because we might be at a VCI and VCW values change along the span. So the bottom line is that several sections have to be checked along the span. Okay, so now let's dissect here one of the two equations for shear resistance. Before doing that, I wanted to see if we could poll the audience about the question that I had posed before on whether people are familiar with the punching shear equations for flat slabs. So I'll stop talking for a while and I'll wait for the poll to close. Okay. Okay, so the poll is closed. And I'm waiting for Sherry to type the results for me to get an idea of how many responded yes, familiar with punching shear equations. All right, great. So the majority of you are familiar, 78% are familiar with those equations and 22% are not. So for the benefit of those 22% that don't, let me just go slightly more slowly here, but I'm sure it'll be beneficial for everyone. So punching shear is a failure mode that assumes that a block of concrete is going to be essentially moving vertically, in this case for the ledge case, is going to be moving vertically and creating a failure surface around this perimeter I'm going through with the pointer here. So from the edge of the ledge, a distance LP to the face of the spandrel of the web, and then along the interface between ledge and spandrel, and then outward towards the edge of the ledge. So that creates, this is a free body diagram indicating that along the faces that are exposed here, there are vertical shear stresses that are going to resist this applied force downward. So these vertical shear stresses are acting on the two side faces here of dimensions LP times HL. And they're also acting on the back face that has a dimension B sub T plus 2HL, which is the assumed distance of these failure surfaces forming halfway along those diagonal cracks that form through punching. So that essentially represents the model that's used for punching shear resistance. So what we're doing here is, imagine that this factor here, this box, gamma beta times square root of F prime C represents a stress, a shear stress. So the shear stress resistance is in this box. If we multiply that times an area, which is found by multiplying the perimeter, 2 times LP plus BT plus 2HL, 2 times LP plus BT plus 2HL times the height, that gives me an area, multiplied times the stress gives me a force. So this is essentially the resistance against punching for this particular case of an interior load with sufficient spacing from the next load over. Then these other two factors, of course, are the resistance factor phi and lambda is a factor for lightweight aggregate concrete, which may not be the case in our case. The important factors are these represent the shear strength, you know, in terms of a stress value, and this represents the critical perimeter times the height of the ledge, which forms a surface. So that's the area over which these stresses are acting. And we just talked about gamma and beta. Gamma is a pre-stressing factor and beta is really a factor that captures the global effects of the spandrel as it bends and shears and how that affects the shear resistance of the ledge. So this is one of the several equations I showed you earlier. All of the equations have the same format, and it just really the only thing that changes is the critical perimeter, right? So when we have two loads that are acting fairly close to each other at a very small spacing, we discussed that the two independent critical perimeters cannot form. So the total load of two V sub u coming from two stems that are fairly close together are acting and mobilizing this punching shear failure surface that's shown here in this left-hand corner of this slide, where now the length of the failure surface is BT plus S plus 2HL. And LP is again the projection of the spandrel beyond the web face. So again, the only thing that changes here is this term, which is associated with a critical perimeter, and you can determine it based on what we had been discussing earlier. So we just received a question, and I'm going to read it out loud. The question is, does the ledge design relate to design of precast staircase sections? No. The ledge design that we're presenting here mostly relates to spandrels, the spandrels that are either located at the edge of a building or ledge design for an inverted T, such as in intermediate or base of a structure that carries double T supported on both sides of this inverted T. So that's the ledges that we're now considering. So when you think of the behavior of those elements, those bending elements, you know, as loads are getting applied, the spandrel bends longitudinally, right? And the ledge bends with it. Not only the spandrel or the inverted T, they bend longitudinally, their shear and moments generated globally on that beam, and the ledge bends and could possibly, you know, there could be cracks. There's a reduction in prestressing stresses because of that bending behavior. So that's what we're covering here. Okay. So if you have further questions, just type them in the chat box, and Sherry will relate them to me, and I'll answer them as soon as I get hit with those on my screen. So notice from going from the previous slide, going back one slide, the only difference, there's two differences in these two equations, the failure surface term, and there's also a 0.5 factor that appears in the next equation. And this 0.5 factor appears because, as you see, there's twice as many load being carried when loads are concentrated to each other, that they both generate or are resisted by this same critical perimeter. So when you move this two factor from the left-hand side of this equation to the right-hand side, which is the strength equation, that becomes a 0.5. So that's an important factor to consider and know what's happening with that 0.5 factor. Okay. Okay. So now we're going to start doing some calculations, and I'm going to remind you of some of the relevant dimensions. So this is the spandrel up in the top-left corner of our design building, 45 feet to column centerline. And some dimensions that are going to be important is that the centerline of the column to the first – the centerline of the first load is at 3 feet, and that typical loads are spaced at 6 feet. You might remember that there was a 10-foot-wide double T here on the right-hand side. We won't focus on that for this – we don't need that for this session. We are going to design – or we're going to look at the ledge design for two cases, two locations, an interior location marked by this red arrow here, and an end location marked by this red arrow here. Just to illustrate to you the calculations that are relevant, and we need to – as designers, you need to verify or at least make sure that none of these other load points would be critical in your design. Okay. So that's where we are, 3 feet from the centerline of the column for the edge. And this other section, I'll tell you, since these loads are 6 feet apart, we'll see where that lands in a second. Reminding you what our design looked like after we designed for torsion last time, if we – if we just pick one of the two designs we performed, the Zia Zhu method, for example, this is the reinforcement we had, and it consisted of closed – some of the torsion reinforcement and shear reinforcement consisted of a number for closed stirrups at a spacing that's given at the bottom of this figure, at 12 inches near the center, 6 inches towards the end, and 3.3 quarters of an inch near the end, the 3-foot in inbound from the column face. The pre-stressing was two half-inch top strands and two strands at mid-height and six strands at the bottom, half-inch strands near the bottom. So we covered all this in module number one. Just make – please note that the ledge research was done mostly using the slender spandrel method for the reason I explained before, because that was the research that was ongoing when these ledge failures started occurring. But it's also applicable – the ledge methodology is also applicable to beams that might have been designed using the Zia Zhu method as shown here. So I picked this reinforcement to show you that it can also be used for spandrels designed using the Zia Zhu method. Okay. So some of our geometry and forces. So, again, some of the notation. Here are the six half-inch strands near the bottom of the beam. That was part of our flexural design from a previous module. And these are the loads that now are unfactored, because we need to calculate factored loads here in a second. These are the dead and live loads that are unfactored per stem, so 16.2 kips per stem and 7.1 kips per stem for dead and live loads. And so the factored load for Zia Zhu, including dead and live, is 1.2 times dead plus 1.6 times live. It gives me 30.8 kips per stem. And the horizontal force that's assumed is 20% of the factored dead load. So it's 20% times 1.2 times 16.2. That's 3.9 kips per stem. So we need to consider V sub u and N sub u in the design of this stem. And the typical crack that we – you know, the backside of it would look like this. We had been discussing how the front side of those cracks look like. So this ledge will tend to not only want to shear and, you know, create a punching shear failure, but also might want to separate and, you know, create a flexural failure as shown here. So we have to design for all those actions. The longitudinal and vertical bars consist of number four bars. And here's some of the notation that we described before. So just keep in mind, all the subscripts L refer to the ledge, the width of the ledge from including the stem of the beam and the projection, LP, and B is the width of the web for the spangle. So here are some of the geometry. Eight inch wide stem, B sub L is 16 inches. B sub T, the width of the stem is four and three quarters. L sub P is eight inches here. H sub L is eight inches initially. So the parameter A is calculated as three quarters of L sub P plus one and a half inch cover, plus half the diameter of a number four bar. So these dimensions taken to the center line of the vertical reinforcement. That gives me seven and three quarters of an inch. D sub W right here is eight inches minus cover, minus one number four bar, minus half another number four bar. So that's five and three quarters. D is the distance or the effective depth from the bottom of the beam for the ledge purpose from the bottom of the beam to the center line of this A sub S steel. And D sub L is from the compression face of the ledge to the center line of this A sub L steel, which is also reinforcement for the ledge. So these are some dimensions that we will be using. And then we need to check the shear strength, punching shear strength for the ledge, the interior load point. That stem number four is 20 feet and a half from the bearing center to the end of the ledge. From the bearing center of the load, pardon me, the fourth stem to the end of the ledge. So we have 246 inches, that's 20.5 feet times 12, minus six inches from the bearing center and minus one inch from the face of the column to the start of the ledge. So that gives me that the edge distance is 235 inches from this load to the end of the ledge. So we need to verify whether that edge distance exceeds this quantity right here. And we quickly do that. And that quantity of 0.5 BT plus HL plus LP is 18 and three eighths. And of course, this is much larger than that. So we're dealing with an interior point load for which the critical failure surface is able to form entirely. So the shear strength is the minimum of these two values. The second equation applies if loads were very close to each other. It's not likely that that will be the case in our case. So since we are not using lightweight aggregate concrete, lambda will be one. Gamma and beta need to be calculated. And we'll see how those are calculated. So first, we need to calculate R, the factor R, which is the ratio of factored load divided by nominal strength for shear and bending moment. And these are values taken from part one in this course. So at that position, at the fourth stem location, V sub U is 4.8 kips, M sub U is 1,504 kips. So we have those values corresponding to stem four. We also, even though I'm not showing it here, the shear strength, the nominal shear strength of the spandrel was calculated as the minimum of BCI and BCW the procedure to determine the shear strength of a pre-stress concrete beam is presented in the basic pre-stress design class that's also given by PCI. Doing calculations for BCI and BCW at that section resulted in 72 kips and 160 kips at that section. That gives me 72 kips. Then we need to add to that the contribution of transverse reinforcement. And what I assumed here was that at that section, since it's very close to mid span, it is likely that it would be reinforced, that the reinforcement for shear strength would likely only need minimum reinforcement. And that was also determined in part one of this course. And that was 0.11 square inches per foot. And then using a 60 KSI steel and an effective depth of the spandrel of 68, the contribution of the steel to shear was 37. Adding the concrete and steel contribution to shear would give me roughly 110 kips. And then I calculate the ratio of V sub U divided by V sub N, and that is 0.044. So this ratio is really small because we're near the point of zero shear. So we would be back here and we would have a beta factor if the section were non-pre-stressed of two. The effect of pre-stressing, we're going to capture through the gamma factor, even though I showed you previously the effect of pre-stressing on beta is to jack up that beta factor. It is considered separately in the PCI design handbook. So now the effect of moments. So we have an acting moment, an ultimate moment of 1,504. If you go back to your first module set of slides, the nominal resistance at mid-span for the fully developed strand was 1,670 kip feet. So we calculate those ratios and that gives me a factor of R of 0.9. And the greater one of the larger one of R for shear of 0.044 and moment of 0.9 governs. And that gives me a beta factor of one. So now taking into account the benefit of pre-stressing. So at 252 inches from the end, pre-stressing is very likely to be fully developed. I didn't check it explicitly here, but I'll check it at the end. So we can consider that FPC in this equation is the effective pre-stressing stress. So we assume that it's after losses, it's 170 KSI. We multiply that effective pre-stressing stress by the area of six strands. So, pardon me, eight strands. So six strands near the bottom and two near the mid-height of the beam, considering they are all acting on the tension side of the beam, divided by the gross area of the spandrel gives me the FPC of 0.296. And then plugging that value in here and doing the calculation gives me a pre-stressing stress factor of 1.26. So plugging in all those values into the two equations, 576 and 77, phi factor of 0.75 for shear, punching shear, lambda of one, pre-stressing factor of 1.26, beta of one. And the rest are numbers that we have discussed earlier. So that's the depth of the spandrel. Pardon me, the ledge, et cetera. We get 19.6 kips. That's the resistance. Equation 77 gives me 29.1. So the lower value governs because as mentioned earlier, these loads are widely spread. So this is the value that governs. And when we check, we compare that value with the 30.8 kips per stem, that is the factored load acting at each load point, we see that the ledge would not be able to carry those forces. So it's not good. We need to have a solution. So since we're trying, punching is our problem, H sub L, the depth of the ledge should be increased. And here are two bullet points of a couple of observations. So if you were to say, well, I can increase F prime C to see if I make it, just keep in mind that by increasing, for every 1,000 PSI increase in concrete strength is about a 10% gain in B sub N. So notice that that is within a square root. So if I increase to 6,000, that gives me about 10% additional strength. So that solution is not very efficient. A similar effect happens with pre-stressing. If I include twice as much pre-stressing force that will affect gamma, the resulting gamma would be 1.48. That's about a 17% increase. So also not very efficient. The most efficient solution would be to increase the depth of the ledge so that it doesn't punch. This note, this is just work in progress. Just as an aside, some tests have shown that by concentrating some of the reinforcement within this width, really close to the load point, right? Where the stem is lying or supported on the ledge, concentrating reinforcement within the ledge in that area is giving higher punching capacities and also hanger reinforcement is giving higher punching capacities. And that may be because the punching or the failure surface is intercepted by reinforcement. So that gives you a higher strength. So this is still a work in progress. So more tests are likely needed to corroborate this finding. But for now, we're doing things by the book. So now with our revised dimensions, we only increased H sub L. That has a two-fold effect. It increases the critical perimeter and it also increases the depth of the ledge. So that's why it's the most efficient way of increasing strength. So plugging into the previous two equations, we get now 36 kips for the widely spaced loads and 47 kips for the closely spaced loads. Since we're in a widely spaced load and because this is the lower value that governs and that exceeds the factored shear of 30.8 kips. So now our ledge is okay near mid span. And since it's not common to change a ledge height, right? Along the span of a beam, we use that same depth and now we go check the end. So we first go and check the edge distance. Remember that the first load point is 36 inches inbound. From the end of the span rule, we subtract the bearing dimension of the span rule on top of the column and then the one inch clear distance between face of column and end of ledge. I have a bunch of questions that I hadn't realized were here. So let me just pause here. This is a check for the end condition and I'll just pick up from here. So this is the end distance first that we need to compare with this quantity right here. And if the end distance is larger than this quantity, then the full failure surface can form. So this edge distance is 23 inches and this quantity on the right is 22 and three eighths. So again, even for the edge case, the loading point closest to the edge of the ledge, a full failure surface can form. So let me just answer a few of the questions that appeared on my screen. Okay, the first question I have, why are we using values for the span rule as a whole span rule depth capacity to calculate beta factor and R for just the ledge? Okay, that's a good question. That's part of the effect of the global effects, how the global effects, global bending of the beam and global shear of the beam affects the ledge resistance. So imagine that you have, this ledge is attached to the span rule and as the beam bends, so it pulls not only the bottom fiber of the span rule, but also the ledge with it. So it has an effect of counteracting the effect of the pre-stressing force. So it diminishes, there's an effective kind of a tension, it's not really a tension yet, but it's a reduction in the compression that reduces the punching shear strength of the ledge. And that has been verified experimentally. It's often referred to as the global stress effect on the resistance of the ledge. That's why we're using those global parameters for the ledge itself. The question said, wouldn't also increasing HL impact geometry and we may need to do a combination in order to make the depth possible. Yes, I just arbitrarily increased H sub L. So I would suggest that US pre-casters know better than I do. Some of you who are pre-casters, I'm sure. So that has an impact on overall building depth. So one might have to either provide a notch in the double T to have enough head clearances for your parking garage, or modify other aspects of it to avoid increasing the floor to floor height. So yes, the impact of increasing H sub L, I'm just focusing here on the structural aspects, but the operational, architectural, other layout problems might have to be addressed as well. My suggestion is for, get an estimate of what the loads are going to be in your case and do some of these quick calculations prior to determining the depth of the ledge and the other dimensions that are needed. Okay, so thank you for the questions. All right, so picking up again, the end check really is conducted the same way as before. So the only thing that changes here are the V sub U and M sub U over M sub N and V sub N factors. Again, we proceed the same way. You'll notice that V sub U and M sub U values have changed now. The moment is lower, shear is higher. Nominal shear resistance is higher because we have more closely spaced syrups at the end. This is based on the quantities that we determined from the Zia Zhu method at the end. So it was 0.382 squared inches per foot. So the nominal resistance at the end of the spandrel is 452. The ratio now is much larger, 0.3. This is R for shear. Similarly, R for moment is now going to be 0.49. So moment again governs. Let me just back up a little for the moment calculation. So in nominal strength, the nominal flexural strength, since we're at the end of the beam, we need to make sure that we have a full development length of the prestressing strength. So there's a typo in your handout. This plus sign that I'm hovering over was omitted, I omitted it by mistake. The calculation is correct and this second side of the equation is correct. The plus sign here did appear, but I forgot to type in the plus sign here. Guess I just, I don't know. It wasn't there. So the development length of strand is the transfer length plus what's needed to fully develop the strand for flexural capacity. You're probably familiar on how to use this equation. So the development length for our half inch strand in our case is 73, 72.6 inches. There's a typo here again. I apologize. This should be inches, not KSI. This is 72.6 inches. And the distance at which we are, we are at 30 inches. So we're not fully developing the strand. So we can assume that the flexural strength increases with development length of strand. So if we have a 30 inches out of 72.6, we calculate that ratio and multiply that times the 1,670 kip feet that is the flexural strength of fully developed or a beam with fully developed strand. So we have 690 kip feet for the nominal flexural strength. So the ratio for a moment is four nine as I mentioned earlier, governs again, the beta factor in our case, in this case, is now going to be 1.28. So that's the only difference from the previous calculation. And you see the effect of prestressing force again considers the fact that the strand is near the end. So we check whether the strand is fully transferred. And in this case, the strand has fully transferred the prestressing force into the beam. So we can use again, FSE of 170, giving me a gamma factor of 1.26 again. Plugging in values into the same two equations we did before. We have that at the end, the punching resistance is 46 and or 60. So 46 governs, which is greater than 30. So this illustrates further the effect that it's exacerbated near the center of the beam because of the larger moment or larger fraction of applied moment to flexural strength that will, that provides a resistance, ledge resistance that's closer to the punching resistance that is needed. So in this case, the edge again makes it. Okay. So another way, and I was just thinking about this, another way to improve the resistance against punching would be to add more prestressing force. If we don't, a combination of HL and more prestressing force, because that will have the dual effect of increasing the flexural resistance and also increasing the gamma factor that will give me more punching shear. So that might be a nice combination. So that takes care of punching. We know what we need to resist punching. The other failure mode that we need to be concerned about is flexural failure. And that's captured by designing transverse bending of the ledge. So the relevant equation is shown here on this slide. This is equation 581 from the handbook. And these are the variables that are listed in that equation. And I believe we've covered all of them. So we need to account for this vertical force and this horizontal force that essentially generates bending and direct tension on the ledge. So we need reinforcement for that. And the type of reinforcement that we need, let me just focus on this right-hand side, is this red line corresponds to the reinforcement that's going to prevent that failure mode, H sub S. It needs to be anchored around this longitudinal reinforcement and behind the reinforcement in the spandrel itself. So that H sub S is shown in front view along the spandrel. This is the H sub S that we're designing, that we're reinforcing for. And it should be placed within a distance equal to three HL from the face of the stem, but not more than S over two, S being the space in between stems. One, so this is how, so it has to really be concentrated near where the load is acting, right? For it to be effective. Notice that the spacing of that A sub S cannot exceed HL and 18 inches, whichever one is smaller. So S star is to distinguish that spacing from the spacing between stems. Please note that there was an error in the PCI design handbook, H edition. It states six HL on either side of the stem face. It should be three HL on either side of the stem face where that reinforcement should be distributed as shown in this slide. So just so that when you're reading your PCI design handbook, make that correction. Okay, so what is the design model based on? The equation we just, I just showed you. So this is the reinforcement where, or the model of the forces that are acting on the ledge. And if we take a free body diagram with an assumed flexural crack, and we take moments about a point O here near the base of the ledge, we have the vertical force acting divided by phi to make it nominal. And N sub U divided by phi again to make it nominal. So we take moments about this point O and consider those moments as positive. So we have VU over phi times A. So A is, we're taking, at the critical section, we're taking moments about that point where the pointer is hovering now. Plus N sub U times HL, this dimension for the ledge. Minus TD, T times D. T is the tension force in that reinforcement, which is equal to ASFY. And D is the effective depth to that reinforcement. So with this moment equation, we simply group terms, divide by D, and then solve for AS. And that's the equation that we use for A, to determine A sub S. So you can see that it's a very simple model based on mechanics principles. So doing the calculation for that reinforcement, these are known values we had determined before. And these are the dimensions that we have currently. We plug in those values and dimensions into this equation right here. And we need 0.62 square inches to be distributed over that spacing of three HL plus B sub T B sub T and plus three HL on the other side. That would give me 0.097 square inches per foot. Converting to feet here, two inches per foot here. So this is for the entire width over which I'm distributing that reinforcement, which is three HL on either side of the stem. So we can distribute that, or we can cover this amount using number three bars at 12 inches, that would be 0.11 square inches per foot. So that's number three at 12. And that satisfies the maximum spacing of 18 inches and HL, HL governs in this case. So very simple design. So longitudinal bending of the ledge. So we also have to account for that longitudinal bending of the ledge itself. And that's taken care of by using this reinforcement. Now the equation the PCI handbook provides is shown here, 200 times LP times DL. DL is the depth to the centroid of this reinforcement divided by FY. And AL has to be placed top and bottom of the ledge as shown. So each HL is for one of these two bars right here. So I wanted to show you a little bit of background for that reinforcement. And it was kind of interesting. When Klein proposed a model for that longitudinal reinforcement, and consider a concentrated force acting on a ledge. And one can say, well, continuity of the ledge, we can model that portion of the ledge as a fixed beam with a concentrated force here. The bending moment diagram neglecting any other effects on the ledge would be aligned as shown here. And the values for those bending moments would be equal to V sub U times S divided by H. That's just from your moment tables from a textbook, say for example. If we assume that that moment needs to be carried by the longitudinal reinforcement at yield, and assuming a lever arm between the compression force and tension force of D sub L. So that's a moment equal to a moment. We can solve for A sub L and we would get this equation. And the model proposed by Klein, it said, well, instead of using a phi factor of 0.9 here, we would use 0.85 to kind of offset the effect that the distance between compression and tension is not really DL, but slightly less than that. So that was taken in a phi factor. Now, if we compare that, this equation, compared with the equation that's in the handbook, we notice that this equation resembles much more the equation for minimum A sub S required by the ACI code. The minimum amount of flexural reinforcement in the ACI code is given by 200 times the cross-sectional dimensions of a beam divided by FY. The beam is now the ledge, so BW becomes LP and D becomes DL. We see that this equation is very similar to that. And the only reason I was able to get, or the only explanation I was able to get was from people that know about the derivation or when this equation came into the handbook was that ledges reinforced with their minimum flexural reinforcement. There has been no distress observed in any of those ledges. So this amount of reinforcement would be much larger. So therefore, where the ledges, the longitudinal reinforcement corresponds more or is closer to the minimum reinforcement required for flexure. And even with that reinforcement, the ledges have performed appropriately. That could be the effect of various factors, such as the effect of pre-stressing force that's not accounted in this model, and the effect of these reinforcement contributing to that ledge resistance as well. So some possible explanations. So we need to apply 580, so this equation 582 is for bottom reinforcement within this width, okay? So very quickly, what that means in terms of numbers, 200 times eight times 9.9 is the effective depth. That's 0.26 square inches. By the way, I forgot to mention that the denominator in... Let me just back up for a second. The denominator here in the PCI design handbook had a phi factor here, which is incorrect. There should not be a phi factor because this is a equation that is closer to the minimum flexural reinforcement. So at the end, since we do need to develop this reinforcement throughout the ledge, we might run into trouble. That reinforcement needs to be developed at the location of the first concentrated force. So sometimes you need a U-shaped bar located near the ends of the ledge to be able to develop those bars. So we need number fives, top and bottom, longitudinally within the ledge to carry that reinforcement or to provide that reinforcement. And then the last design check that we need to make, I believe, is hanger reinforcement. So hanger reinforcement is used to transfer the loads that are acting here into the rest of the spandrel. So this vertical reinforcement, which is a continuous tie, so a closed tie, is used as hanger reinforcement. So we only use this part of that hanger, of that reinforcement, as hanger reinforcement. Notice that the requirement for hanger steel is not additive to that for shear and torsion. So that's important. The equation comes from direct shear application, direct vertical force is resisted by A sub SH at yield, and the phi factor for strength. And this modification factor depends on whether we have the ledge on one side of the beam or two sides as an inverted T. A sub SH, again, the handbook is explicit in distributing A sub SH in a distance equal to 6HL on either side of the double T stem. I am not 100% sure that this should be six, it might be three again, but just to set aside. So the notation for hanger steel, keep in mind that V sub U is the applied factor load per stem, as we have done before. It is not shear in the spandrel, even though we're designing this reinforcement, it's not the shear in the spandrel, but this concentrated force coming from the stem reaction. And I believe, again, that all of these terms have been defined earlier. D sub S we hadn't, it's the distance from outside space of the web to A sub SH, hanger steel. M is the modification factor that we'll see in the next slide. And these are properties of the cross section that I'm referring you, of the ledge and the spandrel, that I'm referring you to look at your design handbook because it's well explained there. So we want to save some time here. And X and Y are the shorter or longer sides of each component rectangle of the spandrel, including the ledge. Gamma sub T here, it's unfortunate that we use gamma here. Gamma sub T here has nothing to do with prestress, but rather the type of reinforcement that used as hanger reinforcement. When close ties are used in the ledge, we use gamma of one, zero of otherwise. What are those close ties? This reinforcement would be considered a close tie within the ledge, and it isn't in our case. Okay, so the M factors, again, I said, I mentioned to you that that was included for a modification factor, sort of for hanger reinforcement, depending on whether the ledge sticks out from one side or both sides. So these two equations have to be checked. In our case, this is the only applicable equation. This term, since we don't have a closed tie in the ledge, this term will go to zero, as you will see in the calculations. And I should point you to the original resource from Klein in 1986 that you can get from the PCI journal if you're curious to see where this equation comes from. There's no time to explain it here in detail, but it comes from a model like this, isolating a piece of ledge and spandrel subjected to a concentrated force, VU, and what the resisting mechanism is. So the ledge will bend and twist, and the spandrel is resisting it, and the hanger forces are pulling upward, providing part of the resisting mechanism. So the spandrel and ledge are under shear forces and bending moments and also torsional moments. Again, the derivation is found in this reference, and we only have time to apply the equations. These are our data, and nothing has changed here. Again, 30.8 kips, and these are all the dimensions that we have been using so far. D sub S is the only dimension that we need to include here that I don't think had been calculated. I believe it was calculated before. I didn't spend enough time on it, but it's the distance from the outside face of the spandrel to the centroid of the hanger steel, which is on the inside face of this spandrel. That's A sub SH, and that's D sub S, and A we had calculated before. We know what that is from the point load application to the centroid of that steel, that hanger steel. So the M factor, as I mentioned earlier, this second term, because gamma sub T is zero, will disappear, so we simply plug in geometric properties into this equation, and we get an M of 2.16, which exceeds the lower limit of 0.6, so we use 2.16 for M, and we simply determine, plug that into the hanger steel reinforcement equation. V sub U is from the factored load here, and we get 1.48 square inches that divided into the 72 inches of 6HL, or the spacing between stems. In this case, I'm using the spacing between stems, because if we were to use a 6HL on either side of the spandrel, that would exceed the 72 inches. That's why I think it should be 3HL on either side of the spandrel, also in the handbook. So the resulting value for hanger steel is 0.24 square inches per foot, which becomes number four bars at 10 inches, so that's 0.24 square inches per foot, distributed over six feet, centered about the double T stem. So this value will govern within the center portion of our design. We used to have some closed hoops, closed stirrups at 12 inches within this region, but because of hanger steel requirements, we now need to place that reinforcement at 10 inches, and again, it's not additive to shear and torsion. The rest of the spandrel and ledge are fine because the spacing of that transverse reinforcement is less than 10 inches, and we have number four closed stirrups throughout the beam, so we only need to modify the spacing here to account for hanger steel. So now we have looked at the design of the ledge, including global effects, bending and shear effects, and we modified summarizing, we modified the ledge depth to 12 inches, and we added, we determined the transverse bending reinforcement requirement for the ledge and hanger reinforcement requirement for the ledge, and we also added the longitudinal reinforcement, so number five bars top and bottom as our final design. So with that, I think we conclude this module with about roughly 10 minutes to go. So I would like to open it up again for questions if there are any. So Sherry is probably ready for any questions, and she'll relay those to me as soon as they appear. If not, she'll take controls over from me and conclude the session. Okay. Okay. So there are no questions appearing on Sherry's screen, so I'll just turn it over to her for conclusion of this session, and again, if you have questions, feel free to type them throughout the week. Question, can you go over slide 40 again? Sure. Let's go back and see what slide 40 is. Okay, so this is a model that I had proposed. I had, this is, so the model that was proposed by Klein showed the ledge bending in this fashion. This is for longitudinal reinforcement in the ledge. So the ledge bending in this fashion. So a simplified model for that, this is like a continuous beam under a series of concentrated forces where you go from positive bending here near mid-span to negative bending here. So it goes from tension on the bottom to tension on the top. So because of continuity, you can assume, okay, let's consider this part of the ledge as if it were a fixed-fixed beam under the concentrated action of a force V sub u. If you analyze this fixed-fixed beam, you would get these moments, a positive moment at mid-span of Vus divided by eight, where S is the span length, and negative moments of Vus divided by eight. So the spandrel reinforcement, this longitudinal, sorry, the ledge reinforcement, this longitudinal reinforcement is top and bottom reinforcement that needs to resist this moment. So if we equate that force of top or bottom longitudinal reinforcement times the depth between tension and compression forces, that should be equal to a moment. So we can solve for A sub L, and we would get this value for A sub L. And so with a small adjustment for a phi to account for the slight difference between L sub N and the actual moment arm between compression and tension. So that's all that that model assumes. But if we compare that model with the requirement for longitudinal steel, we see that if you were to do the calculations, this would give you much higher steel amounts longitudinally than this calculation. And the reason for that is perhaps that even with minimum reinforcement, flexural reinforcement, the spandrel, the ledge has performed, or ledges have performed adequately, so we don't need to reinforce with this, the reinforcement required for a model like this. After all, this is just an assumed model. So that's the reason behind it. Hopefully that further explained what I assumed in this model. I think there's another question, let's see. Shouldn't hanger rebar requirements be additive to applied shear in spandrel as a whole? No, the PCI handbook explicitly states that hanger steel is not additive to shear or torsion. So the idea is that if you have, so how can I explain this? If you have a shear force, so it is, so that shear or that transverse reinforcement is being effective to not only carry the shear that's acting in the beam and torsion, but also being able to move the load, the load that occurs near the bottom of the beam towards the top. And that's the purpose of hanger reinforcement. So the effects are not additive. Otherwise we would need to have hanger steel conditions or hanger steel would normally control over most of the length of the beam. It normally just governs over the center of the beam. So the PCI design handbook is actually quite explicit about that hanger steel not being additive or not added to shear steel. Okay. I think that's it. There are no more questions. So Sherry will take it over from here and conclude the session, the session part. I hope. Thank you.

concrete ledge design

reinforcement requirements

pre-stressed beam

A parameter calculation

D parameter calculation

DW parameter calculation

DL parameter calculation

shear strength requirements

edge distance calculation

R factor calculation

gamma factor calculation

beta factor calculation