As Sherry mentioned, this is the first of a series of six courses that we called the Advanced Pre-Stressed Concrete Design Series. And I will be presenting these on Thursdays. The sequence is every Thursday from now until early December, skipping the Thursday of Thanksgiving. So the objective of this course was to build upon the material that was presented. Some of you may have taken the basic pre-stressed concrete design class. The idea was to prepare material that was more advanced and was contained in Chapter 5 of the PCI Design Handbook. And there are a series of topics that we're going to be covering as part of that. So this gives you a very quick overview of what you have signed up to take. The first thing, this class or this session, I'm going to introduce the example building that we'll be using throughout the course. And we'll focus through the torsion design provisions in Chapter 5 of the PCI Design Handbook. Session 2 refers to ledge design. Session 3 talks about bearing on concrete and dap dens. Session 4 covers columns, axially loaded members subjected to compression and bending. Session 5, corbel design. And finally, Session 6 is a series of topics that are within Chapter 5, again, of the PCI Design Handbook. As I alluded before, we're going to be designing different components of the sample building. The building was selected as a typical parking garage located in a low seismic region. And because we wanted to really be able to just present some topics without having to deal with seismic or lateral loads now, the PCI Academy is currently developing other modules on lateral load design that I encourage you to be on the lookout for future reference. And the way I prepared this course is I'm going to go over, give you a brief background to the design provisions without going too deep into them and not being this kind of a research type of approach. I want it to be as practical as possible. And then I will present then the design equations that appear in the PCI Design Handbook. So after that brief presentation, then I will go into calculations using the design building or the sample building that was selected for this course. As a note, everything that is presented here assumes that people attending have a basic knowledge of design provisions for precast, pre-stressed concrete elements, such as those provided in the first class of this series, the basic pre-stressed concrete design class. And the idea is that we won't really talk much about how to determine the pre-stressing reinforcement or design for shear. We will build upon that knowledge to cover the more advanced topics that are presented in this class. The other point of note is that all of the course is based on ACI 3.18.14. Many of you may know that ACI 3.18.19 just came out, but so the course was developed prior to ACI 3.18.19 coming out. And not only that, but the PCI Design Handbook that we're using, it's the eighth edition, and that is based on ACI 3.18.14. So with that, that is just the basic background of what the class is about and where the general information comes from and the relevant editions of the standards and the handbook. Okay, so this is our first session. I'm going to be presenting the building example and go into details about torsion design. This is what you will expect after finishing this session. You will understand the background of torsion design provisions and be able to apply equations for torsion design contained in the PCI Design Handbook. So first, going into the building, the prototype building that was selected. The committee that was in charge of developing this course, we discussed what would be the best way of illustrating the concepts and design procedures in Chapter 5 of the PCI Design Handbook. And what we decided was that it might be a good idea to use a prototype building, a typical three-story parking garage that could be used for the entire class. So what is typically done is that we're going to be designing different components of these three-story parking garages that I'm about to describe. This is by no means intended to be all-inclusive, so some of you may have slightly different practices and parking garages that you're designing might be slightly different. But the idea is, again, to present concepts in a general way that you can then apply them to your own particular design. The dimensions of the parking garage, by the way, I paused because sometimes if you have questions, you can type them into the chat box, and then Sherry will send them to me and I'll keep checking to my left to see if any questions have come in and maybe I can respond live. So if you have any questions, you should type them in the chat box and send them to Sherry. Again, the parking garage then is 190 by 122 feet in floor plan. It consists of a double T floor system with a three-inch topping. It's supported on spandrels. The floor system is supported on perimeter spandrels and inverted T beams in the crossover base. And I selected just one area of the building that I will show you in the next slide for design. And the idea in this module is that we will design the spandrel beam in that selected area of the building to illustrate the torsion design provisions. The typical floor-to-floor height of this structure of this parking garage is 10 feet, just over 10 feet, or close to 11 feet, I guess. Okay, so this is a very quick view of the parking garage, 122 feet. I mean, 200 feet by 120 feet in the short dimension. And the bay that I'm designing, if you follow the cursor on the screen, that's the bay we're going to be concentrating on in this module. And here's a zoomed up view of that area. There's a series of double Ts that are spanning between a spandrel beam here on the left and an inverted T beam on the right. So this is the edge of the building. The spandrel beam span is 46 feet, 8 inches, and the double T, or the distance between column lines in the double T span direction is 60 feet, 3 inches. So these are some of the general dimensions that we need to remember for the class, or for this particular module. A very quick view on the elevation. I know I don't anticipate that you will be able to see all the details, but the region that is selected is away from the ramps in this parking garage. So you see the cross-section of the double Ts that are in that region, and the 10-foot-8 floor-to-floor height is roughly delineated by this blue box around that slide. It is important to be able to design the spandrel that will be focusing our attention today to understand how the double Ts are supported on it, and what forces are going to be acting on the spandrel. So typically, each stem of a double T applies a vertical force onto the spandrel that is supporting it. There's also a horizontal force that needs to be accounted for in design, and the value of this horizontal force is typically taken as 0.2 times the dead load shear, or if there's a better estimate of that horizontal force, that can be used as well. But what's important in this slide is the support conditions, and this eccentricity that's going to then be used to determine the design actions for the spandrel. So the eccentricity is referred to as E, is the distance between the centerline of the applied force, in other words, the reaction of the double Ts per stem, and the centerline of the spandrel. So this is a sketch of how all these different loads coming from the double T floor system are applied to the spandrel. The spandrel we'll be assigning, by the way, will be used again for the next module when we are looking at ledge design. As you note here, given the way that the spandrel is supported, it's embedded into the column at each of these two column lines. So as a result, the ledge doesn't continue through the column. So the ledge is stopped short of the column face. You'll see a better view in the next slide. And these horizontal forces, or white arrows, indicate the tieback reactions. As these forces are applied eccentric to the spandrel, they create torsion and out-of-plane bending. So the spandrel will want to rotate away from the column. So these are the tieback forces that are at some vertical distance between them. And those are created by having these holes passing and tiebacks passing through the column that supports the spandrel. The blue lines, the blue arrows, represent the reactions from the double T, or the loads coming from the stems of the double Ts. And these are actually what generates the forces on the spandrel that we need to consider for design for torsion. So the first thing we always need to do, and this is going to be fairly simple for those of you that are in precast plants that are used to these types of calculations. We have to look at, of course, dead loads for design. We have to consider all the loads acting on a structure. Dead loads, live loads are the two primary ones. And other gravity loads, such as snow, rain, ice. Talking about dead loads, we have to consider self-weight of components. And the system we're using in this example is an un-topped double T that then carries a 3H cast-in-place topping on it. The live loads that are going to be used throughout the design example are 40 PSF live loads for vehicular loading, according to ASE 710. I forgot to mention that that is the standard we're going to be using for loads. And note that for parking garages, ASE 7 does not allow a live load reduction when the garage is supporting vehicular or passenger vehicle, vehicular loads. And the other thing to note is, if you recall the picture with the elevation, I'm not designing a floor system at the rooftop. So snow, rain, and ice is not considered in this design example. Of course, you can incorporate that if you're talking about the rooftop of your parking garage. So this is, again, it's just to exemplify calculations and to show how these are done. So all right. So the material data that is used in this example is F prime C. The 28-day concrete compressive strength is 5,000 PSI. The concrete strength at release is 3,500 PSI. These have to be checked with the producer to make sure that those strengths can be attained. And in terms of reinforcing steel, we have, for mild steel, we're going to be using nominal grade 60 reinforcements with a yield strength of 60,000 PSI. The pre-stressed reinforcement is going to be grade 270, low relaxation, seven-wire, half-inch strength. So all of these are just assumptions in the design example. And the welded wire reinforcement, we're going to be using that. And recall that, in some cases, recall that the maximum permitted yield stress of welded wire reinforcement is 60,000 PSI in calculations, even though the wire may have a higher yield stress. So that's what we use for design calculations. One thing I want to mention is that despite the number of reviews that these slides have received, we, I still found a couple of typos that are very minor that I will be pointing them out to you throughout the presentation. If you have printed your presentation, you can just correct them directly. Your slides show WWF for welded wire fabric. The more appropriate term now is WWR for welded wire reinforcement. So I found a couple of other typos that I'll indicate to you where those are so that you can correct your slides or make note of them as we proceed. And I'll send the revised version to Sherry so that she can distribute if she feels it's necessary. All right. So this is an elevation from the interior of the building. The profiles of the double Ts are dashed. But this view is just to indicate the position of all these concentrated forces that are generated that come from the double Ts stems. And to give you general geometry, span geometry for the spandrel. So you'll note that the centerline to centerline dimensions of the spandrel are 45 feet. So column centerline to column centerline. There are two types of double Ts in the area that we're designing. There's double Ts with 12-inch width. And there's double Ts just to be able to, you know, fit the double Ts within the bay that we're designing. At the end, there's one double T that is 10 feet wide. The length of all double Ts is 59.1 feet, and these are some dimensions we're going to use to calculate the forces that we're going to use for design. You'll see that there are two ultimate or factored forces for those associated with the ultimate loads for the 12-foot-wide double Ts and those for the 10-foot-wide double Ts. And the distances between stems of the double Ts are shown here, 6 feet typical for the double, the 12-foot-wide double Ts. And in the transition between the 12-foot-wide and the 10-foot-wide, you have 5 feet 6 inches or 5 feet between stems of the 10-foot-wide double Ts. The bearing length, I'll show a detail of the bearing length on each side of the span rule so that you have an idea of what I assumed. And again, the span, the clear span of the, or the centerline of bearing to centerline of bearing span of the span rule is 44 foot 6 inches, so 44 and a half feet. The next, this slide shows the two details I mentioned. So 44 and a half feet to centerline of bearings. And on the left-hand side, we have two different bearings with a gap between adjacent spandrels. On the right-hand side, you see that the spandrel extends a little bit farther beyond the centerline of the column. Just this is the corner of a building, so that's why the spandrel extends beyond that. I've neglected this extension. It doesn't affect the details I want to present today significantly. So I've neglected that extension and just focused on the 44 and a half foot length of the spandrel. Notice that a one-inch gap was assumed between the ledge and the face of the column. So that one-inch gap here, where the cursor is moving, is between the face of the column and the edge of the ledge. And if we load a spandrel the way shown here at the eccentricity E, it is likely that the failure of this spandrel will be associated with a diagonal crack. So we have loads and we have tieback forces that are restraining this triangular region at the end of the spandrel. So that generates this crack. Just keep in mind that this is one of the failure cracks that may form by the eccentricity or by torsion of this edge spandrel. Since we need to estimate self-weight of the double T system, floor system, the first thing to do, what I went through is I went and I used tables in Chapter 3 of the PCI Design Handbook to have an estimate of the size of the double T that can satisfy the span and live loading that is going to be applied to the system. So these types of charts that are included in Chapter 3 are only meant to be used for preliminary sizing of your floor system. The procedure to use these charts, the load tables, is described in the first class, the basic prestressed concrete design class. My intent here is just to indicate to you that I went through that process, I looked at the live loads that the floor system needed to support and I estimated what double T system, what size of double T would be able to handle those loads to preliminary design the double T's just for weight estimates because I wanted to have a reasonable weight estimate to use to load the spandrel. So what's important here is we can use the weight without topping is 635 of this 12 by 28 inch double T. The weight per foot is 639 pounds per linear feet. You could include the topping as well. I have separated that in the following slide. And with topping, with a three inch concrete topping, the weight, this would be for normal weight concrete by the way, the weight would be 1089 pounds per linear foot. So if you don't wish to include that initially, if you want to separate the topping from the double T in some calculations that is needed, it is best to do the calculations separately. So the 639 pounds per foot is included here in the self-weight. So before I go into the details of these calculations, please note that there are two widths of double T's as I referred earlier, the 12 foot wide double T and a 10 foot wide double T. So the dead loads are slightly different for each and also the live loads because of the width difference of the double T. So this is the self-weight of the double T times the overall length of 59.1. And these factors, one half and one half, one of them is divided by two stems of the double T and the other is divided by half the span. So it's 59.1 divided by two, that gives me the reaction on one end. And then I divide by two again to get the reaction per stem. So that's the dead load coming from self-weight. Using a similar calculation, the 3-inch concrete, normal weight concrete topping is 37.5 pounds per square foot divided by 1,000 to convert to kips, span, one half of the double T width, so just for one stem, and then the other half is to divide the span over two to get it on one end. So we have 9.4 and 6.7 kips for self-weight and 3-inch topping. The total service dead load is 16.2 kips per stem. And doing a similar analysis for the 10 foot wide double T, we have self-weight, the topping, and now you'll note that the distance here is 10 feet times 0.5. So we have these two loads per stem for dead loads. These are all service loads. And then the live load at 40 pounds per square foot, and using similar calculations as above, you get 7.1 and 5.9 kips per stem. And this is for the 12 foot wide and the 10 foot wide double Ts. So the total factored load for the 12 foot wide ends up being, after we apply load factors from the relevant load combination in ASCE 7, it's 1.2 dead times service dead loads plus 1.6 times service live loads gives me 30.8 kips per stem and 26.6 kips per stem for the 12 foot wide and the 10 foot wide double Ts. So these calculations should be fairly straightforward. All right. So now that we have an estimate, and going back one slide just for a second, these numbers 30.8 and 26.6, we need to remember them, or we need to use them in the configuration of the double T that I showed you, or the spandrel that I showed you before, so that we can estimate all the actions that a spandrel beam is subjected to. So as you note from the loading on a spandrel, the spandrels are a complex structural element because they're subjected to combined moments, shears, and torsions throughout the length. So we have to account for the effect, all these three effects, and that's what we'll be doing here in this class. Mostly what is done really is we design the spandrel for moment, for flexural strength, and we account simultaneously for shear and torsion, because shear and torsion combined affect the strength, or affect shear stresses in the spandrel, so to speak. So the first thing to do is run a structural analysis of the spandrel. Being a simply supported member, one can construct the diagrams that are needed for design fairly simply. The way I did this is I selected different stations along the span of the spandrel, and you'll note here the section that's being referred to at each point, and the distance from that section to the support center line, which is on the left-hand side, beginning on this side of the spandrel. So stem number one through eight are number left to right, and you see here the different distances based on the geometry, starting from the support center line. So this is not the overall length of the spandrel, this is from support center line or bearing center line. And these stations here, since most of the diagrams are fairly symmetric, not entirely symmetric because of the difference in double T width, but they're fairly symmetric, I stopped computing here at the center and just focused on the left-hand side of the spandrel for this table. But as you can see, in this case, the moment diagram is constructed throughout the length of the spandrel, and you'll see the shear and torque diagrams or torsion diagrams in the next slide. So you see there are two stations or two x distances or stations that have the same distance for the stem, and for each stem, I should say, and this is because since each stem corresponds to a location of a concentrated force, we have to look at the left, just to the left of the force, and just to the right of the force, and then compute the shear and the torque, that if there's any changes to the left and the right of these concentrated forces applied throughout the spandrel. So that's why these two rows are repeated. In any case, the ultimate shear diagram looks like this, would look like this. You'll note that it's not perfectly symmetric, right? Because the loads are not symmetric. The torsion diagram, factor torsion diagram, and the moment diagram. Excuse me, I clicked too early. Sherry mentions, I need to make my pointer bigger. I'm afraid I don't know how to do that. So I might be able to... How about that, Sherry? Does that work? A laser pointer. So I'm going to go back. You'll... Okay, great. She says yes. So the row here that's indicated in red is what's going to be our critical section for design. Please bear in mind that this D is not the effective depth of the spandrel. This D is defined in a different way. And I will show you how it is defined later. But for now, the estimates for D and the eccentricity of the shear forces, you know, where the reaction of the double Ts are applied, are estimates for now. So it's taken, D is taken as 5.6 inches for now. Or roughly half a foot. And the eccentricity is three quarters times the spandrel, the ledge length, plus four inches means half of the width of the stem of the spandrel. So that would be 10 inches of eccentricity. This eccentricity is used to compute, to calculate the torsion diagram. So each concentrated shear force or reaction force times the eccentricity gives me a torsion. And that's what's used to construct this T sub u diagram. Okay, so similarly in these slides, you see the shear diagram. The slight slope here, the sudden jumps represent the location of concentrated forces. And the slight slope represents the weight of the spandrel. That's why it's sloped. In the case of torsion, the weight of the spandrel doesn't produce a gradual change in torsion. Just the, or if it does, it's minor. Just the big changes in torsion at the location of each of these concentrated forces is shown here. So we're going to use a section that is right here near the point of maximum shear and definitely within the point of maximum torsion for design. Okay, first thing, since we're going to need to have the flexural design of this spandrel for subsequent calculations, we need to know the longitudinal reinforcement. We have to conduct the design for the spandrel. So the first thing that I'm showing here is considering a spandrel as if it were just a reinforced concrete spandrel without any prestressing. Later on, we will see that prestressing is needed. So I'll recalculate the flexural strength. But for now, let's consider that's just a reinforced concrete section. And we use the material properties that I described before, a width of the spandrel of eight inches and an effective depth. This is now the effective depth of the spandrel of 77.5. The ultimate moment near mid-span is 1,504 kipfews. These calculations, for those of you that know reinforced concrete design, this is one way of designing a reinforced concrete section where you divide the factored moment by the strength reduction factor in BD squared and you get a coefficient of nominal resistance. With that, you can then use that to estimate the reinforcement ratio given the material strengths. For the spandrel, that would be 0.7 percent reinforcement. And that would result, given the cross-sectional dimensions, that would result in four and a half square inches of steel that five number nines can carry. So initially, I assumed that the spandrel was going to be a reinforced concrete spandrel without any prestressing force. And then, so five number nines would be sufficient to carry the moment. So notice that the estimate of D of 77.5 is slightly off. The actual D is 77.7, 76.7, pardon me, given this reinforcing bar arrangement. So it's a little bit lower than I originally assumed but since we have more area of steel that should still work. It should be checked though. So a very important aspect is to determine the critical section for shear and torsion. So ACI 318.14 says that for shear and torsion the critical section should be located at a distance, at a section located distance D from the face of the support if the loads and reactions compress laterally, transversely the beam or the beam that's being designed. Furthermore ACI also says that the critical section is at H over 2 if the section is pre-stressed. Again if loads and reactions compress laterally the beam. However in our case you recall that the loads are coming from the bottom, right? They're applied to a ledge. So there's really no, the loads are not compressing transversely. So in fact they're coming from the bottom and then their reactions are almost at the bottom. So in that case ACI indicates that the critical section has to be taken at the face of the support. So that means right at the face of the column that will be our critical section. And that sections between the face of the support and the critical section in the case if the section were farther from the face of the support the design could be based on values determined at the critical section if the section is farther from the face of the support. On the other hand the 8th edition PCI design handbook, by the way the way that these provisions are indicated you'll see a blue box around PCI design handbook provisions and a black box for ACI provisions that was indicated earlier in the previous slide. The 8th edition handbook says that the critical section for shear and torsion is to be defined as according to ACI 318.14. And again it also indicates that indirectly loaded precast components such as the inverted T's or the alleged beam L beams, they should be taken, the critical section should be taken at the first face of the support conservatively. Section 541 in the handbook also says that the critical section can be located at a distance D where D is a different definition here it's not the effective depth of the span rule but it's this dimension which is measured from the point of application of the load and the tension reinforcement near the bottom. So it's a much smaller distance than the effective depth of the span rule. So if we count that could be our critical section from the bottom of the component to the point of application of load. But this value is only used for the purpose of defining the critical section for torsion and shear and not for anything else that shows up in this module. So that D would be 4.7 and you might recall that my original estimate was 5.9. If I were to go back to the slide indicating all the shear and torsion values you'd see that the difference is minimal between a section at 4.7 and 5.9. So I'm going to use the values that were plotted in that slide for design purposes. So our critical section essentially was going to be very close to the face of the support and with the values not changing too much. So that's going to be okay for us. Okay so now into torsion. Now that we've defined the critical section and what our design values are going to be, there are two methods that are in the PCI design handbook. On the first hand the ACI 318 torsion design method is based on thin-walled tube space theory that was introduced back in the 95 version of 318. Prior to that version ACI only contained a torsion provisions that were applicable to reinforce concrete elements and not prestressed. And this former design method was modified by Professor Zia and McGee later and later modified by Zia Xu for use in prestressed concrete members. The two methods that are used in the PCI handbook is one of them is the Zia Xu method. So it's really based on these modification that could then be used for prestressed concrete members. So it's analogous to what's in ACI 318. And the second method is referred to in the PCI design handbook as the slender spandrel beam method. This is brand new for the 8th edition so this is material that that was just incorporated into the 8th edition based on testing. So what we will do is we will present the Zia Xu method and also discuss the slender spandrel beam method which are the basis for spandrel design in PCI. So let me just show you, I'm going to try to show you on the camera, that all of these methods if you're members of PCI you can download the actual references that I'm going to be listing here for the two methods. One of them is shown here. The PCI journal shows the design for torsion and shear in prestressed concrete flexural members. That's from the Zia Xu method and there are other papers that are used, you know, as background for this method. And then the other method is described more recently in the 2011 fall edition of the PCI journal. It's a two-part paper that has actually won several PCI awards and this is the basis or this describes the basis for the slender spandrel beam method. Both of these are referenced in the PCI design handbook. Okay, so the first question is, well, are these methods in violation of ACI 318 which is the standard that is referred to by applicable code, by other codes throughout the country. And when you look at section 9546 in ACI it tells you that ACI allows an alternative design procedure for sections that have a height to web width ratio of greater than 3 which is typical of spandrels. The only thing that you need to be careful about is that the notation differs from the handbook and ACI. ACI calls the web width B sub T whereas the handbook refers to it as B or BW in some sections. So you just, this B or BT or BW refers to the web width. So the idea is that we can use an alternative design provision as long as it is demonstrated by analysis and testing that this works for torsion. And furthermore, section 9547 in ACI 318 allows the use of open web reinforcement. I'll refer to this in later slide. Instead of using closed hoops for torsion design, when the aspect ratio H over B exceeds 4.5 or is equal to 4.5. So both of these methods that we're going to be describing here fall into these two sections of ACI so they're compliant with ACI 318. Okay so the first thing that we're going to cover is we're going to cover the torsion design based on the ZiaZhou method. And the basic design equation for any strength-based design is that the factor torsion in this case cannot exceed the reduced torsional resistance. And the nominal torsional resistance of a section is given by a concrete contribution to torsion and the steel contribution to torsion. Notice that T SEBES here is the steel contribution to torsion. It's transverse reinforcement. And in the case of the ZiaZhou method, they indicate that it has to be provided by closed hoops. So it's also based on the space truss analogy of torsion. In this equation, the strength reduction factor, since these are torsion-generated shear stresses, the strength reduction factor for torsion is 0.75 in accordance with 318. And again, all the calculations that I'm going to be illustrating are based for the torsion and shears applied at a section that is located a distance d, a little from the face of the support. But remember that this d is a small d, not the d for the entire span. Okay so the first thing is to check whether we need to worry about calculating or designing for torsion. So the ZiaZhou method and equations from the PCI design handbook are going to be listed here. This is equation 542, again chapter 5 of the PCI handbook, lists this equation. And the purpose of this equation is to tell you that if the torsion, the factor torsion, is less than what is given in this equation, then you do not have to worry about designing for torsion. Torsion can be neglected. So since we have first assumed that the span rule is non-pre-stressed, this gamma factor, which indicates a factor that is used for to account for pre-stressing and increasing the torsional resistance of concrete, is assumed to be 1. So this gamma factor here you see in the, this was part of the basis of the ZiaZhou method, is that gamma is the square root of 1 plus 10 times the pre-stressing, average pre-stressing force FPC in a cross-section divided by F prime C, the 28-day strength of concrete. In this equation, lambda is a factor for lightweight concrete, phi is a strength reduction factor, and this property x summation of x squared y is a torsional property that is used to, it is based on elastic theory. So that property comes out of analyzing elements, rectangular elements under torsion. There's a torsion property. This is a section property that is important to determine shear stresses induced by torsion. So in our calculation, so in our design example, the TU mean calculation is the first thing we would do. That's the phi factor, 0.5 times 1, lambda is 1, so no lightweight aggregate factor, or a factor of 1 because it's normal weight concrete. 5000 psi, everything in the square root has to be, has to use psi units. And this number 5632 cubic inch, I will demonstrate in the next slide how that is calculated, but for now let's just do this calculation. So TU minimum would be 149.3 kip inch, and we have a T sub u, a factored load at the critical section of 1156 kip inch, which is much greater than 149, so we do have to account for torsion. By the way, all these calculations we're just going to design for one section, but you might recall that the shear diagram and torsion diagram decreases stepwise. So as you get closer to the center of the spandrel, you might fall into a category where you don't necessarily need to account for torsion anymore. So you know when you fall below 149 kip inch, so that may occur near the center of the spandrel. So everything else in this calculation, all the notation is described here, and I want to show you how this x squared y, summation x squared y term is calculated in this slide. So the PCI handbook indicates that if you have a rectangle or a section made up of different rectangles, so the previous slide indicated that x is the short dimension of a rectangle and y is the long dimension of a rectangle. So these are the dimensions of our spandrel, so it's 80 inches deep, the ledge is 8 inches deep, ledge length is 8 inches for an overall width of 16, so the web width is 8 inches. So we subdivide this section into component rectangles and do the calculations shown here for x squared times y. So we sum up, so this is our first rectangle, x is 8, so 8 squared, 72 is 80 minus 8, plus 8 squared, that's x for the second component, and 16 is y for the second component. So we use the minimum dimension, the smallest dimension, in each of the component rectangles for this term. Adding them together gives me 5632. If I were to use the other option, you would get 5632 anyway. So this is including the ledge. Typically, however, you may remember that the ledge does not extend all the way out to the end of the spandrel, so typically in practice one might say, well let me eliminate the part of the ledge that extends beyond the spandrel face, so I'll use this large rectangle for x squared y, which is, in this case, would be, this calculation is, that case would be more conservative. The next thing is to make sure that we don't exceed the maximum torsion that is allowed by the Zia-Zhu method. This limit is to ensure that we don't end up over-reinforcing the section and then end up crushing concrete diagonally. So in this expression, as I'll refer to later as well, this includes the effect of simultaneous shear acting on the spandrel. So as it turns out, the simultaneous application of torsion and shear causes the other effect or the other strength to decrease, and what's assumed in theory is that there's a circular interaction between torsion and shear. So as you apply shear, there's, imagine an interaction diagram where torsion is in one direction, in one axis, shear is in the other one. As you start applying, without zero shear, you get the maximum torsional strength. With zero torsion, you get the maximum shear strength, and in between you get a circle, an interaction, a circular interaction, that's how it's referred to. And this square root in the denominator indicates that. So these two factors, so this is the equation for maximum torsion. You see the same type of equation here, summation of x squared y, and K sub T and C sub T are torsion factors and prestress factors. K sub T is the effect of prestressing factor and C sub T is the effect of the cross-sectional factor for torsion, I should say. So the maximum torsion that we could apply in our given spandrel is 1,100 kipinch following these calculations. So this is a simultaneous shear force that occurs at the critical section along with the torsion of 1,169. So if you go through the calculations here, you get 1,100 and we have 1,169. So we must change the section or add prestress, otherwise we can't proceed. This exceeds the maximum torsion that we can apply to this section. So we have to go back to square one, but I wanted to show you how K sub T and C sub T are calculated as well. As I mentioned earlier, K sub T is a factor that's used to use the benefit of prestressing in increasing the resistance of concrete against torsional cracking. Since it's a spandrel that has not yet been being prestressed, we assumed it was reinforced concrete, so we plug in a zero. So K sub T is 12 here and C sub T is a sectional factor for torsion, and notice that it's only section properties that are included here. So we've got 0.11 and going back to the previous slide, the 0.11 is here for C sub T and 12 is K sub T, both here in the denominator and the numerator. So we have to go back. So the possible solutions is we could change the dimensions, but these are normally defined at the beginning of a project, so that's not a feasible or economically feasible solution, I would say. We could increase the concrete strength because the maximum torsion is a function of root F prime C. In our case it might work because, you know, perhaps with 6,000 PSI you've actually did some calculations and it actually worked, but I wanted to illustrate what the effect of prestressing would be since in many cases spandrels are going to be prestressed anyway, so I will use prestressing to illustrate the calculation. So going back to redesigning our spandrel, we now have to revise our flexural design to include prestressing, and these are the material properties that we'll use. The scanless of P factor is for low relaxation strand, and again this builds upon the first module where all these flexural design procedures presented, and I first assumed that the prestressing reinforcement was going to be concentrated or distributed in three layers, so six half-inch strand near the bottom of the beam, two half-inch strand near mid-height, and two top bars distributed along the beam so that there's not a large eccentricity of the prestressing force. And FPS, which is the prestressing stress in the strand at ultimate, is calculated using the approximate equation, equation 5.1 from a PCI design handbook. So everything else here is from flexural design of a prestressed concrete section. Notice that the effective depth to the prestressing force is 68 inches, and I considered since we're at a section at mid-span, I considered only the strand that were on the lower half of the bottom half of the beam. This is right at the half, but I still considered it because the neutral axis depth at failure would probably be up here in this section, so all of these strands would be in tension. So using this equation for FPS is going to be important, and we will define rho sub P in this equation as the area of prestressing strand, which is 8 strand, again only for the bottom half, 6 here plus 2, divided by the width and the depth through the effective depth of the strand gives me that value. FPS is calculated in the next slide. So FPS is now 259 KSI, roughly. That's the equivalent stress block depth of 9.3 inches. So the nominal moment given by the 8 prestressing strand is 1670, when multiplied times 0.9 is 1503, which is roughly what we had at mid-span for a moment, so we say this design for flexure is okay. Now we have to revise the K sub T constant to include the effect of prestressing stress. So this is the effective prestressing stress at the section, and you'll note that at the end of the section, since we're near the end, the prestressing strand is not fully developed, and we might be even within the transfer length. So we first calculate what the transfer length is based on an assumed value of 170 KSI for the effective prestressing stress. That is 28 inches, and the critical section that we're, from the end of the span rule, that we're considering is 5.3 inches plus 12 to the end of the span rule. That's 17 inches, which is less than the 28 inches. So the strand at our critical section are not fully, have not fully transferred the prestressing force, so we need to estimate how much of that force is transferred. So the transfer, the design transfer stress is going to be 104, and then the effective prestressing stress is 104 KSI times the number of strand that we're considering, dividing by the gross area of the section gives me 181 PSI, which I plug in here to get the K sub t factor now of 13.6, slightly higher than 12, and redo the calculation. Remember that K sub t is the only factor that's affected by prestressing. C sub t stays the same other than changing for the effective depth of the prestressing strand now, so we have C sub t of 0.097, K sub t of 13.6. Our maximum torsion is 1190, and the applied torsion is 1170, so we're okay. We can proceed with our design. Okay, so now that we can do that, remember that we are designing shear and torsion simultaneously, so each shear and torsion should be considered jointly. We also need to check whether the maximum shear, the factored shear, does not exceed the maximum shear allowed, also to avoid crushing of diagonal, crushing concrete in the diagonal direction. This equation 544 from the PCI design handbook gives you the maximum shear that can be applied, and again the square root in the denominator indicates that the interaction between torsion and shear is a circular interaction, and this is the upper limit of 10 root F prime C B WD times the the phi factor. So following the calculation, I think there's nothing else to explain here. Now we have the torque at the section, the shear at the section, the new K sub t, and we have 137 kips for maximum shear, which is greater than the shear at that section of 135, so we can proceed with our design. So if we didn't meet this limit, we have to redesign the section. That's the bottom line. Okay, so now we need to consider, as I've mentioned several times before, we need to consider the concrete contribution to torsion and shear simultaneously. These are the interaction equations that are listed in the PCI design handbook. So the first equation, and both of these equations, the primes, TC prime and VC prime, correspond to the concrete contribution to torsion and shear as if they're acting alone. So this is the concrete strength as if there's no shear applied. This is the shear strength of the concrete as if there's no torsion applied. The square root under here again accounts for the interaction between torsion and shear. So this torsional concrete contribution and shear concrete contribution are going to be reduced with the presence of the other action. So this is the square root again accounts for the interaction between torsion and shear. So first we need to estimate these TC prime and VC prime. VC prime is the resistance of the concrete, prestressed concrete beam, for shear when there is no torsion. So these should be some equations that hopefully are familiar to you. There's two methods that are allowed to be used to estimate B sub C. The first one has some implication in terms of its limit of applicability, or otherwise if those limits are exceeded you can simply calculate B sub C as the minimum of VCI and VCW, so the diagonal tension strength or the web shear strength of the section. So these, in our case, the limit of applicability of this equation is satisfied because 524, equation 524 here is only applicable if the effective prestressing force exceeds 0.4 times APS FPU plus ASFY. So since we don't have ASFY, we don't have mild steel in our section, we only have prestressing. For APS FPU, 0.4 APS FPU, pardon me, is 132.2. The effective prestressing force is the total area of prestress times the effective prestressing stress and that gives me 208. So since P effective exceeds 0.4 APS FPU, then we can apply this equation with the exception that this term here in the dashed box cannot exceed 1. So I'm going to show you the calculations in the next slide. So we can use that V prime C equation, that's the concrete contribution to shear. These are the values at that section. Notice that MU is much lower, we're not at mid-span, we're near the end. When we are at the end, this ratio is going to be fairly large. That's common for when we're at the end of a spandrel or any beam for that matter. So we're going to use 1 as this ratio and the rest are just factors that we need to plug into the equation. So 0.6 root F prime C, again land ice for lightweight aggregate concrete. So we have 404 kips for V sub C, that would be their concrete resistance to shear without any torsion present, but that cannot exceed the 5 root F prime C VWD, which is 192. So we need to estimate the VC prime as this value, so the smaller of these two governs. Similarly for torsion, the concrete contribution to torsion is given by this equation right here, where gamma is the pre-stressing factor that we calculated before as 1.17 and x squared y we also have calculated it before. So running this calculation we have 454 kip each. So the concrete contribution to torsion and shear combined with the other action acting concurrently is calculated here. You'll see that for the case of torsion, if torsion were acting alone that's that would be the concrete contribution 454, because there's shear acting as well that gets reduced to 438 roughly. For the case of shear there's a much larger reduction. 192 would be the Yeah, right. Yeah, of course. The shear would be ... the concrete contribution to shear would be 100 without presence of torsion would be 192, but when we account for torsion we have 51 kips of resistance. So the difference between what the concrete can take and the applied torque or shear has to be taken by transverse reinforcement, and remember that for the Zia-Zhu method we need to use closed terms to contribute to torsion. So the way to do this is we have a factored torsion, we divide by the strength reduction factor, and that has to be equal to the nominal strength T sub C plus T sub S, and we solve for T sub S, the contribution of transverse reinforcement. That's what we need to determine or that that's what we need to determine to find what the demand or what how much transverse steel we need in the form of closed hoops, and using a space truss model, this equation comes from the Zia-Zhu method where from a space truss model the closed hoop contributes ATFY, that's the force in each leg in the leg of the hoop, times the perimeter, so the product of XY and YY, where these are dimensions taken to the centerline of the perimeter of the hoop. And this is a factor that is normally used for torsion design. It's a torsion factor which is calculated or shown here as 0.66 plus 0.33 y1 over x1. The derivation or the more background about this equation is in that paper that I was showing you earlier. So when we solve for the transverse reinforcement for torsion, this is the equation. So we plug in, we solve for A sub T and plug in for TS using this equation up here. So we get that the required area for torsion is given by that equation. So if we apply this equation, which is listed as equation 547 in the handbook, we have 11, 16, and 9 would be our factor torsion minus whatever the concrete can carry in combination with shear, right? So this is the reduced torsion strength reduction factor. Alpha sub T is given here at the bottom of the slide. It is 6.27 but it can't exceed by going back it can't exceed 1.5. So that's the limit we use. We use 1.5. And the dimensions of the x and y dimension to the centerline of the closed hoop is given here. So from the web we subtract two times cover and half of the transverse bar diameter assuming number four closed hoops. So that's 4.5 for x1 and 76.5 for y1. So these are the ways the way I estimated these two dimensions. So we have a demand of 0.4434 square inches per foot. Notice that the spacing here that I neglected to mention is 12 inches. So this is a per foot calculation. And by the way, in your notes I believe that I omitted the s. So you might want to add the s on the left-hand side here. The calculation is correct because the 12 inches was listed here. So it's only the s that was missing. So going back, I'm sorry, going back this corresponds to one leg of closed stirrup and is needed in addition to the area required for shear. Okay, so the reinforcement required for shear interacting with torsion is determined in a similar manner. So we first solve for the required nominal strength as Vc plus Vs. We then subtract how much the concrete contributes and we compute how much the steel has to contribute. And we use the familiar truss analogy formula for, you know, assuming 45 degree angle truss to calculate the shear contribution from transverse reinforcement. So we follow through the calculations. Notice that this is the V sub c that has been reduced because of the presence of torsion. And we require 0.38 square inches per foot again for shear. Now this corresponds, in our shear calculation, this is the total area for two legs of transverse reinforcement. So we see that this is applied directly and then we multiply it times two of the previously determined a sub t value to get a total square inch area per foot of beam that corresponds to two legs of this hoop that will take the torsion and shear acting at that section. So we can use number four bars at 3.8 inches roughly or 3 and 3 quarters. That gives me 1.28 square inches per foot. Having determined that area of closed stirrups, we have to check whether that exceeds the minimum. The minimum are either this value of 50 VWS over FYT gamma squared or but and does not need to exceed 200 VWS over FYT. So the calculations for minimum reinforcement are 0.11 or 0.32. The smaller one governs and you see from the previous slide that what we need is greater and what we're providing is greater anyway. So these minimum requirements are satisfied and just keep in mind that there's just these are notes that one needs to remember. Torsion reinforcement needs to be continued past to a section past the distance b sub t plus d beyond the point of theoretically that is theoretically required and this is because of the formation of helical cracks. There's also longitudinal reinforcement that is required for torsion. Torsion generates longitudinal stresses as well as transverse stresses and this statement in the PCI handbook indicates that the longitudinal reinforcement is needed to resist the longitudinal component of diagonal tension that forms in the beam because of torsion. This is a relevant equation for that case or this one whichever one gives you the largest value. So following through the calculations if we if we go back one slide again sorry this term right here can be substituted for for this value and if so the value of a sub L calculated from equation 550 should not need not exceed the value obtained from that equation if we substitute this term in that equation for for this value right here 50 BW over FYT. So we first calculate 50 BW over FYT times 1 plus 12 FPC over F prime C that gives me 0.0096 if since 280 over S exceeds that number then this is governing for equation 550 so we use 280 over S in equation 550. So the longitudinal steel requirement is then given by one of the two equations this one below 550 or 549 this gives me a negative value so that doesn't govern the one above governs at 5.86 square inches and this is to be distributed along the height of the beam on both faces and the way I decided to distribute that is use nine number five bars into the span and since we're at the end of the beam we needed more longitudinal reinforcement so I added some U bars at the end and also accounted for three half inch strands in each face of the beam. So the final reinforcement all these calculations by the way were done only for the critical section here I did a plug-in calculations for other points along the beam and the resulting reinforcing pattern is as follows the end of the beam over 38 inches roughly spacing of transverse reinforcement is our hoops is at 33.75 inches and and that's the number for closed hoops and then we have these longitudinal reinforcement for torsion and near the end we need to double that up once we get to low enough torsion and shear we can we can do without it for the rest of the beam and this is plotted to centerline so that's the distribution of or pad reinforcing pattern according to the Zia Zhu method and a cross-sectional view of this is shown here and again as I mentioned earlier this is just the section at maximum torsion or a critical section and this hasn't accounted for hanger reinforcement. Okay so the next step is the slender spandrel method and we have only about 20 minutes left I hope I can cover it all. So the idea is that there was a proposal to try to simplify these torsion provisions that I just went through to to be able to design spandrels using a more simplified approach and also to simplify the detailing. The detailing is quite complex and to try to avoid having these closed hoops. So a research study was commissioned by PCI with the objectives to design slender spandrel and having the option of having open web reinforcement instead of these closed hoops and the limits of the research was that the spandrels were simply supported and loaded along the bottom and that the web was laterally restrained through these tieback points at the end of the spandrel and some of the research included only fairly slender spandrels h over b of four and a half or greater. I know that there are some research studies that are being conducted to try to extend this limit a little bit to values a little bit lower so shallower spandrels as well. And this is the reference I was just showing you earlier. This is the complete reference which you can also download from PCI and they have published papers on this approach as well. And the process consists in the following. So the idea is to divide the spandrel into three regions. An end region that extends a distance h from the end face of the spandrel, so depth of the spandrel is h, and where cracks are assumed to form at 45 degrees. Then a transition region where cracks are assumed to form at 30 degrees and that extends from h to 2h. And then a flexural dominated region near the mid span of the spandrel. So in the slender spandrel method that the notation differs slightly from the Zia-Zhu method, d sub w is the distance between the exterior face of the spandrel and the intersection between longitudinal steel and inside web steel. ASL is the longitudinal steel. AS0 or O is the vertical steel located parallel to the outer face of the spandrel. ASI is the vertical steel located inside or along the inside face of the spandrel and it can consist of an L bar as shown here or a C bar near the ends of the beam. And this, you know, eliminates the need of a closed hoop. Flexural reinforcement is here. This is the ledge. So that's what the notation is. The basic concept is that looking at a diagonal crack at failure of the section, this is the torsion that is applied at the section and if we take a side view of it, the diagonal crack forms at an angle equal to theta. And if those of you that are familiar with vector notation of moments, we can represent those moments as a double arrow. And since moments are a vector, we can decompose them into a tangential and to a normal direction on that failure section. So the tangential component is the ultimate torque times cosine theta. The normal component is the ultimate torque times sine theta. So these two moments, torsional moments, produce the first one that TU along the tangential direction produces bending of the spandrel. So in fact the spandrel has a couple that is generated compression here and tension on the inside face of a spandrel. And this tension on the inside face is going to be used to determine the need for internal reinforcement along that face. Where T sub u is determined as the bending resistance of this spandrel is calculated as the total area of steel located in here times Fy to give me a force and times the distance between compression and tension, which is the d sub w dimension I was indicating that was the notation for the spandrel method. So one can then solve for the need of A as double prime, where this T sub u cosine theta corresponds the demand. So this is the component of the torsional moment acting parallel to the T direction. So that's how we would determine A as double prime. And so that is referred to as the plate bending component of torsion and this is why this method is referred to as the plate bending method. So for that A as double prime is an area of steel acting at an angle perpendicular to the formation of this diagonal crack. So this area has two components, right? A vertical component also and a longitudinal component. The vertical component is A as double prime cosine theta, which then plugging in the equation from the previous slide gives me T sub u cosine squared theta divided by phi sub f Fy dw. Phi sub f is by the way 0.9 because this is a flexural problem, not a shear problem. And this just indicates, you know, this distance L sub CH, you know, the horizontal projection of this diagonal crack is determined from geometry. You can follow the geometry. That would be H cosine over sine. And plugging that into trying to distribute this vertical reinforcement over that distance, we divide by L. That gives me this equation right here. And in the next slide you'll get finally the equations in the PCI design handbook for the end region or the transition region depending on the angle of inclination of those cracks. You'll see equations of this form, 552 and 553, for the end region and the transition region. And just keep in mind that this is the vertical. These are vertical bars that are needed to resist the flexure of the spandrel. Similarly, the longitudinal bars that would be needed as longitudinal components, going back a couple of slides quickly, that would be the area in this direction, so longitudinally along the spandrel. You would arrive at an equation like this, where when you plug in 45 degrees or 30 degrees, depending on whether you're in the end of the spandrel or the transition, you would get factors that, sorry, that would be either 2 or 2.3 for A sub SL. So the only difference is that there's no H appearing in this equation right here, because A sub SL is distributed over the height. Now the other component of the torsion applied to the spandrel truly generates torsional stresses, and the method assumes that this torsion applied in the normal or perpendicular to the diagonally cracked beam is resisted internally through shear stresses that are assumed to vary linearly with depth. So these shear stresses are integrated to come up with this resultant horizontal force that produce a couple. You'll notice that there's a force couple here. So the derivation of H of NT is this maximum shear stress, X times root of F prime C, that was determined experimentally, that's this term right here, times the length of that triangle, that is half of H over sine theta, half of H over sine theta, times one-half, that gives me the area of the triangle, and times DW, that gives me the total stress over that region. That's the total force of the force couple, multiplied at times, so the torque is this equation, H sub NT times the lever arm, which is between those two forces, two-thirds of H over sine theta. So this X factor is determined experimentally, and going back to the right-hand side here, the X factor was found experimentally to be 2.4, and again the theta angle here depends on the angle of inclination of that crack. So you'll find an equation that looks like this in the PCI design handbook that's based on this derivation. Okay, so let's go into the calculations quickly before we finish here. So we have a T sub u in the direction, in the normal direction, T sub u T, which is truly a T sub u T here, is the torsional component of that torque, is 97.4 kip feet at our critical section. So P sub s here is the phi factor for shear. The acting shear force cannot exceed phi sub s times T sub NT. Phi sub s times the equation we just derived is shown here. So for the end region, we apply a 45 degree angle. So phi sub s TNT, plugging in numbers, gives me 2493 kip inch. So the section here is adequate. What's presumed, what is assumed here, is that the torsion cannot exceed, so the section resists the torsional component without the assistance of transverse reinforcement. The transverse and longitudinal reinforcement are sized based on plate bending. So for the interior longitudinal, for the interior reinforcement, plate bending reinforcement, I should say, this is the equation in ACI. The interior reinforcement is that required for shear, plus whatever is required for shear, plus whatever is required for torsion, pardon me, plus whatever is required for shear alone. So this first term corresponds to the flexural component of torsion, so plate bending component. We derived this expression before. And to that we need the interior steel. We need to add how much is needed for shear. We should not forget how much is needed for shear. So the amount needed for shear is simply based on our normal shear design provisions. And in this slide, I just wanted to highlight repeating the previous equation up here. So the amount required for torsion is shown here. We derived this expression before. The tool here that appears here comes from, assuming a 45 degree angle near the end region, that's the demand for the applied torsion. This is the demand for shear. And your slide, I believe, had a tool here. The tool that divides the amount needed for shear should be applied in this bottom equation, not here, where we add ASV to AV over 2. So adding that required for plate bending to the amount required for shear gives me the total amount that's needed for the interior reinforcement. And we can use number 4 at 6 inches for the end region. And notice that it can be an L. It does not need to be a closed hoop. The transition region reinforcement is determined in a very similar manner. The only change here is that you have 2.3 and that the shear force is less. So I use the shear force at the start of the transition region. So we can use number 4 at 10 for the beginning of the transition region. For the exterior vertical reinforcement, if we add torsion and shear for interior, we need to subtract because the bending moment would produce, or the torsion would produce tension on, compression on the outside face. So we, in theory, could subtract from torsion, but the method instead neglects this second term and only accounts for what's needed for shear for the outside reinforcement. So this is exclusively what's needed for shear and that's assumed to be conservative. So the exterior vertical reinforcement, again doing the calculations, this is the critical, the shear at our critical section, what's resisted by the concrete. And then without combining it with any torsion acting, this is what we require for exterior reinforcement. We need to check that we have at least the minimum. So yes, we do. And then we use welded wire reinforcement. This is a much lower amount. We can accommodate it by using welded wire reinforcement with these spacings and diameter. Longitudinally, these two are the equations that were derived earlier. And the longitudinal reinforcement, again, is a function of theta, the cracking direction near the end region of 45 degrees and 30 degrees. These are the two relevant expressions. And you'll see that there's the lever arm between tension and compression for plate bending is dw, and torque induces this bending moment that longitudinal reinforcement will carry. So applying the calculations for, given the torque, the maximum torsion that induces this bending component, 97.4 kip feet, the two is for the, comes out of an angle of 45 degrees. And notice here, again, that the phi factor is 0.9, not 0.75, because this is a bending calculation. So for longitudinal reinforcement, we have 1.66 square inches. This is in the end region. We have 1.07 square inches. So we can distribute that longitudinally over the height of the beam, and we can add a little bit, so we need to distribute this throughout the length of the beam, and we can add a little bit more reinforcement to take care of the end region. So this is the reinforcement that would be needed for the slender spandrel method. You see that it is much less congested than before. We have vertical bars that are not closed at 6 inches, and we have a much wider spread of reinforcement. And we do need a little bit more longitudinal reinforcement near the ends, in the end region, than we do towards the center of the beam. So you, this is just in summary what the reinforcement pattern would look like if we use the slender spandrel method. There are a couple of final checks. We have only about five minutes left. There's a couple of final checks that have to be done. One has to check for end region cracking under service loads, eccentric service loads, and that's done using this equation, where we take the cracking strength of concrete as 4 root F prime C, E is the eccentricity of the applied load, and BW is the web width. And again, doing this calculation, the cracking shear would be 51.7 kips, or that's the cracking resistance, the shear that would, eccentric shear that would cause cracking. The service loads per double T stem are dead and alive for the 12 foot double T 7 kips, and 14 and 6, roughly, are the dead and alive for the 10 foot double T. So the service, the total service load shear would be a 108 kip at the left end reaction, which exceeds the cracking strength. So the only calculation here is that we, the only message that this equation gives us is that we should expect cracking near the end based on this calculation. And if you want to have a measure for crack control, you might add a little bit more reinforcement, but there's enough reinforcement near the end that it will arrest these diagonal cracks that form under service conditions. And we also need to check for out-of-plane bending near the beam end, and this out-of-plane bending is induced by the tieback forces, where the reaction is not collinear with the applied loads. That means that the reaction is moved in from the ledge and the applied loads are eccentric with the reaction. So that induces this torque that creates this out-of-plane bending near the beam end, and we have to provide reinforcement, vertical and longitudinal reinforcement, to take that out-of-plane bending. We use this torque at our critical section, and the detail in this calculation is that we have J sub u is the effective depth factor that's applied to this D sub w to estimate the distance between tension and compression as lever arm for this additional, this reinforcement that's needed to resist that out-of-plane bending near the ends. So with this demand, we can check whether the bars that we have near at the end of our beam is sufficient. So we check and we see that the number four bars, the arrangement of number four bars that we have work, and that vertically the number four bars at six inches also work at the end to resist this out-of-plane bending near the beam end. So these are two additional checks that need to be done prior to finishing. We'll go back a couple of slides to see the reinforcement. These are the number four bars that I was referring to that are sufficient to resist that out-of-plane bending that is generated by this tieback forces at the end that will cause a diagonal crack to form here near the end of the beam. Okay, and with that, we barely made it to the end. I know that I was rushing to the end. I apologize. This is the first time this is taught, so I hope that there's a couple of minutes for questions, and if there are, if there's not enough time, you can write them to Sherry and she'll share them with me, and I'll write a response back, and you can get your answers by email. Yes, thank you very much, Sergio. I am going to now take back the controls from you, so thank you very much. Again, thank you all for being here this evening. We really appreciate it, and we are looking forward to seeing you in a week. Now, your Certificate of Continuing Education will appear on your account at www.rcp.net within 10 days. Remember, education credit is given for each individual session. You must take and pass the exam to receive that corresponding credit. If you have any questions, please email me at snodden.pci.org with the subject line Online Academy. You will receive an email link to the session exam. If you are unable to take the exam within the time frame of availability, please contact me immediately. If you have any trouble following that link or do not receive the email within 24 hours, again, email me, and again, I want to thank you all for attending and look forward to seeing you next Thursday. So, good night and have a wonderful evening. Thank you, Sergio. Thank you.

Advanced Pre-Stressed Concrete Design

spandrel beam

design calculations

dead loads

live loads

self-weight

structural analysis

shear

torsion

ACI 318

Zia-Zhu method

Slender Spandrel method

reinforcement patterns