Good evening everyone. Again so as Sherry mentioned this is the last module or the last session of this course and today we are going to finish designing the beam the rectangular beam that we've been discussing over the last five or a few weeks after we started working on the design example. Let me just go through the first few slides that Sherry has already discussed with you and today we're going to finish up as I mentioned part number four the way we divide our design example and these are the expected outcomes as you can see from the first bullet we will focus primarily on shear strength, shear design of a rectangular beam that we've been designing and we will be able to look at we'll cover how to design stirrups for a pre-stress concrete beam. Then the rest of the session we will cover two other additional items that are worth considering for precast pre-stress concrete members. One of them are the additional transverse steel requirements that are that are often used to in the end of the beam to resist bursting stresses induced by the pre-stressing strand when we transfer the pre-stressing force into a beam so there's an equation that we will cover there and then finally the last part will involve some deflection and camber calculations on our pre-stress concrete beam and this will wrap up in a manner you know the basic items that one needs to consider when we design a pre-stress concrete beam as I said you know this is a basic pre-stress concrete course there are other more advanced topics that are going to be covered in the future in a future course that's being developed by PCI. So without that as you can see we are being we have been moving along our design example and the only item that's left here we need to design for shear which is the black bullet at the bottom of this slide. So before we get going I wanted to just pull you on whether you are familiar with a shear or remember shear strength design provisions for reinforced concrete design so I I'd like to ask Sherry to launch the first poll. I am launching the first poll it is the question is are you familiar with shear design provisions and their background for reinforced reinforced concrete beams? The poll is launching now it's on your screen please answer accordingly. We're waiting on a few more people okay I'm going to close the poll now 64% yes 21% yes but have not reviewed them in a while and 14% no. Okay great thank you Sherry. So let's then review very briefly what those provisions are. As you might have noted noted there are several provisions that apply to reinforced concrete and pre-stress concrete members with some modifications for pre-stress concrete members and in fact both pre-stress and reinforced concrete design are termed you know structural concrete design in the ACI code so it's not surprising that there are some overlap in the design provisions. So those of you that do remember shear design you might recall and for those that don't these are the two basic equations that are often used for design in shear for reinforced concrete design members. The basic philosophy is that and let me just jump to the bottom of the slide is that the nominal shear strength V sub N is made up of two components one that is attributed to concrete so the V sub C term is often called the shear that the concrete carries and V sub S is the shear that stirrups carry. These are the two most common equations that are present in the ACI code. There's a more involved procedure to determine V sub C but most designers actually use this equation because of the fact that there's a constant here too that doesn't change along the span the only thing that V sub C is a function of is the width of the beam the the web width of the beam and the effective depth and of course F prime C. Lambda as you might recall is a factor related to lightweight aggregate factor and that always appears when you have a you know the tensile strength of of concrete which is expressed as the square root of F prime C in this equation. The second equation that allows us to determine the shear contribution of the stirrups comes out from the assumption of a truss model. So this equation is based on a truss model that the ACI code assumes it forms in a beam after cracking on a beam after after cracking occurs in the beam after diagonal cracking occurs. So that truss model has a top cord for the compression zone a bottom cord for the tension steel and it has diagonals and those diagonals are compression struts that are formed like they're like web diagonals that come down the truss and vertical elements that represent the stirrups. So the ACI code assumption is that that truss model has those diagonals oriented at 45 degrees from the horizontal axis of the member. So that's an assumption in this a basic assumption in this model. But if we have a beam like we do now we have a 12 by a 14 by 28 part of a beam you can plug in those numbers here you'll see that this is a constant and our only task would be to design stirrups if the ultimate shear exceeds the capacity as shown in this equation. So that's the basis for for the for reinforced concrete design in the ACI code. So this is what I what I mentioned that for reinforced concrete members you have the concrete and steel contributions as it turns out for pre-stressed concrete members you also rely on that basic philosophy. The the main difference is that instead of considering b sub c to be constant throughout the length of the beam if one uses the equation presented in the previous slide for pre-stressed concrete members there are two possible failure mechanisms that have been found through experiments or verified through experiments to occur. One of them is a shear failure in let's say members without transverse reinforcement a shear failure associated with flexure shear and I'll explain in the next few slides what that means. So it's it's it's often also termed a diagonal tension cracking strength so it's it's it's it's a combination of flexure and shear cracks that cause failure. The second potential failure mode in which concrete can fail in shear is through web shear cracking and for that we have this web shear cracking strength equation. So the important part here is that the shear strength will be provided by a combination of concrete and steel steel being represented as steel shear reinforcement or transverse reinforcement. In this graph we will first talk independently we'll first try to explain what the VCI and VCW terms correspond to. So imagine you have a simply supported beam with a concentrated force and you test it to failure. So the first thing that will form our flexural cracks that are nearly vertical but as you can see as we increase this load in the center generating constant shear between the reaction and the load we see that those flexure cracks that were nearly vertical at the level of the reinforcement turn diagonal after they start going up into the beam and that is because of the presence of shear stresses that combine with flexural stresses particularly near the center of the beam. So that's why this is called a diagonal tension crack this is associated with the tensile strength of concrete so that's why the square root of prime C term appears in many of those equations. So the key aspect here is that these diagonal cracks and one may be the one that causes failure let's say let's say here that this crack keeps growing and eventually it generates failure so that is referred to as the the shear critical crack but you'll see that it starts as a flexural crack turns diagonal and eventually forms into a full shear failure. All right so it is important to remember that cracks grow from flexural crack that forms along the beam. Now if that is the case what we need to do then is to first determine when that flexural crack forms and as you might recall we had derived expressions to estimate M sub CR so the moment associated to flexural cracking of the beam. When that occurs at a given section there's a shear that occurs concurrently with this M sub CR so VCR this deep cracking is a shear corresponding to the flexural cracking strength of a beam at a section. With if we can determine MCR we will then know when this flexural crack is likely to occur and then we'll only have to determine how much force it then the beam can take then to fully develop that that diagonal crack that I drew in the previous slide. So what's important here is that this VCR shown in this slide corresponds to the moment at a given section at the cracking moment at a given section. So that's what the V sub CI equation tries to reflect this first term is a term associated with with the flexural cracking of that beam and we then also calculate VCI which is the shear that is that occurs in combination at a given section in combination with the moment that is generated at a section. So this ratio here reflects that V sub CI is the let me just go back V sub CI is equivalent to this cracking moment and this VI is the shear associated to the maximum moment that may take place at a given section and M sub CR is the cracking moment at that same section. So this first term reflects the the flexural or the the occurrence of the first flexural crack. The second term is associated with how much extra shear force is required to fully form a diagonal crack that it will eventually generate a shear failure and this is largely empirical so this has been calibrated through experiments so there's no other way to explain it other than it is the force required to go from a flexural crack that's nearly vertical to a diagonal you know to form a full diagonal crack that will generate shear. Okay so how do we calculate the cracking strength? Well the cracking strength can be calculated using the flexural formula where this term here the 6 lambda root F prime C corresponds to the tension strength at the extreme layer of tension fiber of the beam and F sub PE is the effective prestressing stress so in essence this quantity in the parentheses right here is the tension stress that we need to overcome or pardon me the compression stress that we need to overcome to crack a beam and the rest here is the moment of inertia and Y is the distance to the tension fiber so as you can see this is simply the flexural formula where we associate the total stress that we need to crack the beam we include it in the brackets here in the parentheses. All the terms are listed here and for that previous equation again V sub CI is the factored shear force that occurs simultaneously with M max at a given section so notice that BCI here this VI and M max vary section to section along the beam if we have uniform loading perhaps so you have to calculate BCI at different sections along the beam even if MCR were constant right even if we have a constant FPE along the beam this term generally so is varying along the length of the beam. What else is important to point out here D sub P is the effective depth measured from the extreme compression fiber to the center of the prestressing steel and D sub P should not be taken greater than 0.8 H, H being the overall depth of the beam that's one requirement and just to note here that Y sub T is the distance from the neutral axis to the tension fiber and sometimes you know in some other slides you may see it as the distance from the centroid to the top fiber you know T for top in this particular case is to the tension fiber. Okay so the flexure shear strength BCI should not need not be taken less than 1.7 lambda root F prime CBW this is a requirement by the ACI code what happens is that this BCI is going to decrease along the length of the beam but we cannot we should not consider it to be very low this is sort of the bare minimum I mean and try to relate that to reinforced concrete design you know V sub C is a constant 2 root F prime CBW D if we neglect lambda it shouldn't be much less than that for prestressed concrete members. So if you're so this all these equations are used for non-composite sections or members for composite members I would encourage you to just visit ACI 318 this section for more details on how to treat composite members. Okay so now let's talk about the other potential shear failure mode web shear failure mode web shear strength so when we think of this type of failure let me just try to sketch here if I can a beam let's say we have a simply supported beam so web shear failures are associated with the formation of a diagonal crack that all of a sudden appears in the web of a member without growing from a flexure shear crack so as you might imagine these cracks only appear near the end of the beam where there really aren't any flexural cracks if they grew from a flexural crack and they would they would be a flexure shear crack but near the end of the beam there's really very low flexural stresses because we have low moments so there might be a case where these diagonal cracks form and in on the web especially for thin web members like I-beams or astrotype girders things like that and this is called a web shear crack and it is induced by a combination of stresses so if this stress block is look assumed to be located here at the center of the beam at the neutral axis at that center we're going to have the presence of normal stresses from pre-stressing FPC but we are also going to have the presence of these shear stresses if I were to plot let me allow me to do this if I were to plot I'm going to plot the distribution of shear stresses along the depth of the beam so this is a plot of shear stresses their maximum near the center at the center of the beam this is Vmax okay so they are maximum there shear stresses are maximum there and these normal stresses if the beam were not pre-stressed these normal stresses would be zero because we would be at the neutral axis so this is the the element the stress element that's shown here but since we do have pre-stressing right either concentric or eccentric you know we have P there we FPC can be simply calculated as the effective pre-stressing force divided by the cross-sectional area of the beam so that's the normal stress occurring at the at the center of the beam okay so if we could we could try to find what the principal stress direction is for this combination of stresses and and try to find BCW that way however the ACI code and PCI handbook give an alternate way of doing that it tells you they tell you that B sub CW could be estimated as three and a half root F prime C I'm skipping the lambda term since we're not dealing with a lightweight aggregate concrete three and a half root F prime C plus 0.3 FPC to include the presence of pre-stressing force there times the effective concrete area plus V sub P so so this is really what we want to consider in a case of straight a beam with straight strands V sub P is the vertical component of any strands that are oriented diagonally with respect to the horizontal axis of the member so those diagonally oriented strands like the hard strands we've designed in the last couple of sessions have a vertical and a horizontal component that vertical component opposes the shear acting at a section so if we have that hard strand presence we could include that here as an additional strength otherwise we only deal with this first term in this equation so you see the definition of all these parameters down here in this box okay so again ACI has you know gives the designer an alternative approach to the previous equation for VCW and it tells you well if you want to follow a strength of materials approach and find the principal stress direction of this block which would be oriented not as shown here but at some diagonal angle you can do that and if you find the principal stresses you you try to equate the principal tensile stress to 4 root F prime C and call that the VCW term so you are the designers free to do that so this would involve a stress transformation like one does for when we first teach strength of materials or involve more circle analysis and again for composite members all of what's discussed previously is for non composite members for composite members there are equations that are slightly different okay so there are some other caveats to the use of VCW since VCW normally will nominate dominate near the ends of the member we might be within the region of the transfer length region remember that the transfer length is defined as that length required for the strength to develop its effective prestressing stress or force so what this first paragraph says is that if we're at a section that's that's within a distance between h over 2 and the support we were at that section and the transfer length so that part of me that section at h over 2 is closer to the support than the transfer length of the strand we might be having to reduce VCW because the effective prestressing stress is not at its highest or maximum level and that is because we want to include that here so we want to use a smaller value consistent with where we are along the transfer length and the reason that there's this h over 2 quantity here is that the ACI code also allows us to design for the shear at h over 2 from the support from the face of the support instead of designing for the shear at the support which would be maximum according to our analysis we can push our design section to h over 2 that's what the ACI code allows us to do so jumping to the last paragraph which is associated with this first one you know we have looked at equations on how to determine the transfer length but in the case when we're doing shear calculations to determine VCW the transfer length can be assumed to be equal to 50 bar diameters for strand or a hundred bar diameters for wire when we're only when we are doing the calculations for V sub C for VCW in fact so for the case our case let's say 50 bar diameters is remember that the strand we're using is half inch so we have 50 times half inch that would be 25 inches would be for the transfer length which is slightly different from what the what the equations we've covered so far okay so what happens when we have de-bonded strands recall that de-bonded strands are strands that are partially so that are blanketed near the end of the beam to to progressively transfer the prestressing force farther into the beam so we have blankets among some strands and then the transfer length starts occurring there some distance into the beam it doesn't start at the end of the beam but rather at the end of where this blanket or sheath is covering the strand so this is what this slide says right that if we are we have de-bonded strands we need to account for for those de-bonded in order to calculate FPC the prestressing stress we have to account for the fact that those that those strands are ineffective within their their de-bonded region and then they start being effective as they start bonding to concrete and again the transfer length for those strands starts from the end of the sheath and ends at 50 bar diameters from from the end of the sheath and not the end of the beam that's all this paragraph says and finally B sub C that contribution of concrete to shear strength is going to be the minimum of these two equations and again both equations are changing with distance right BCI changes because there's the term VI or N max that changes along the span and VCW is also variable because of the region the transfer length region otherwise it would be constant throughout for straight strands and we'll see some plots and I think it'll make it clearer and finally the the nominal strength BC plus BBS is factored by the strength reduction factor and the shear the strength reduction factor for shear is 0.75 okay as we do for reinforced concrete design whenever the ultimate shear the factored shear exceeds half of what the concrete can provide so 50% of Vc then we have to provide at least so this should be at least a minimum area of shear reinforcement should be provided so AV min if we are in a region of the beam where VU is less than 0.5 CVC then in theory we do not need to provide any shear reinforcement that's the gist of it so when we exceed this we have to have at least minimum areas of shear reinforcement but of course if V sub C is much higher then we have to provide more than just a minimum right the only thing is that this particular provision does not apply for these four cases right for very shallow beams beams that are integral with slabs or these other systems like one-way joist systems and the other thing is that that fact or that critical section for shear for pre-stressed concrete members that critical section for shear is at h over 2 from the face of the support so what that means is that the code lets us design for the shear force at h over 2 from the face of the support instead of designing it for the shear acting at the centerline of the support as long as primarily the loads on the beam and the reactions on the beam compress the beam laterally so that lateral or transverse compression kind of restrains that diagonal crack from forming and propagating towards the top of the beam that's the basic reason for this provision so we can start designing for a section at h over 2 and that sometimes helps us because as I said the shear keeps increasing the acting shear or shear demand keeps increasing towards the support but we can design for a smaller value and of course no concentrated loads act between h over 2 and the end and these other two conditions is what I just explained A and B okay so these are the steps that one follows for shear design the first one is to create find the vis-a-vis diagram the demand then we extrapolate or find the vis-a-vis at the critical section compute these two components VCI and VCW and from these two we decide which one governs the minimum one does and then we plot all these components and the difference between BU and what the concrete can carry will be used as the basis for our stirrup design. All right so now we're back to our calculations this is our beam I mentioned 14 by 28 I apologize I couldn't remember exactly it's a 12 by 28 beam 12 by 28 inch beam and we have 10 strands located near the tension face located in a 2 inch on center pattern and this you might recall this equation for shear this this equation for shear is valid as long as we have a simply supported beam with constant loading uniform loading throughout the span okay uniform loading this equation would apply where any section here X measured from the end of the beam the shear at X can be calculated using this equation where L is the span of the beam so this is this could only can only be used for dead load because that's the only load that is applied throughout the span but remember that live load has to be patterned that's what this means must be enveloped so the other way of saying it is is that we need to find the live load pattern that will generate maximum shear at a given section X let's say X1 or any other section XI so we need to find a procedure that will let us determine what's the live load envelope or what the position of live load will give me the highest shear at each and every section along the length of the beam so with this before we we discuss that procedure I'd like to ask Sherry to start the second poll tonight just to see where we stand with this idea of live load pattern okay the second poll do you understand why the live load pattern must be varied to generate maximum shear along the span of a beam I am launching the poll now please answer on your screen we're waiting for several more people to vote we'll give you a few more seconds I am going to be closing the poll now 82% yes 18% no okay majority of you do So that's great. And so I'll go. I'll try to for those that didn't. Let me try to explain it this way. This may not be the best way of remembering why LiveLoad needs to be patterned by the use of InfluenceLine. But the basis of LiveLoad patterning really grows out of the concept of InfluenceLines for shear. In this particular diagram, this is the InfluenceLine. InfluenceLine represents the effect of a unit load as it travels along the span of the beam. So what this InfluenceLine represents is that for this section, we're trying to find what the shear force is as a unit load is placed on the beam and it is moved. So this unit load, a value of one kip, actually moves along the beam. And for each position, we determine the shear at this section. And that's the graph of the shear for that section when the unit load travels from end to end of the beam. But the information that the InfluenceLine gives us is that for any region where that InfluenceLine is positive, like this region, that tells me where I need to place the load, such as LiveLoad, to generate maximum shear. So this area times the unit load will give me the magnitude, but it does tell me at least the place where I would need to place LiveLoad. So to generate the maximum shear force at that section, I would need to place the LiveLoad to the right of that section. What that means is that I would need to investigate, say, another section here. And what would the InfluenceLine look like? Well, it would look like something like this. And it would tell me, well, now you have to place the LiveLoad only over here. And I would need to proceed section by section along the span of the beam, placing LiveLoad to try to determine at each section what the LiveLoad placement has to be to generate maximum shear at that section only from LiveLoad. We know that dead load has to be placed uniformly throughout the length of the beam. There's an easier way, though. We can simply load the beam, first, full dead load and LiveLoad throughout the span. And that would generate a shear diagram that would give me the maximum reaction at the ends. Then, if we analyze the beam only with dead load throughout and LiveLoad over half of the beam, that would give me those two values, which are an approximate envelope of all those other conditions of, you know, analyzing the beam for all the possible LiveLoad placement. So, this is a shortcut approach that gives you a very close way of determining the envelope for shear. Now, this number here is really just the effect of the LiveLoad acting on the beam, because you might recall that when we have uniform loading like dead load, the shear starts, shear force diagram starts at a maximum and then it passes through zero at the center. So, if we have dead load applied over the entire beam, the dead load shear here passes through zero. So, this little jump here is only induced by the fact that the LiveLoad is not acting over the entire length of the beam, but it's only acting over half of it. So, there's a way to calculate that shear. Since it's only induced by LiveLoad, we can say, okay, it's going to be WL, the LiveLoad times L, L being the length of the beam, divided by eight. So, if you were to analyze a beam with only half of it loaded with a uniform load, you would find that the shear at the center of that beam would be WL times, or WL over eight. So, that's an easier, a fairly easy way to find that point. Once we find that point and we know this other point from the first distribution of loads, which is given here, then we can draw a straight line and by symmetry draw the other part of the shear diagram and we have a shear diagram envelope that we can use for design. So, throwing in numbers, these are our actual example numbers. These loads are factored. So, we have the maximum reactions from full dead and live are 72 kips. That's those two points, 72 plus minus. And then, if you have a calculator handy, you might want to check this, 3.2 kips per foot, that's LiveLoad, times L, which is 30 feet, divided by eight, that gives me 12 kips. And those 12 kips are those two points here in the center of the beam. And that gives me the line that becomes my shear force envelope. Okay? So, that's what it is. It's a fairly simple approach and it works really well. So, we can forget about these reactions. We don't need them, right? Because they're always lower than the 72 kips. And then, the only thing here is that instead of having a zero shear force in the center, we'll have a 12 kip shear force that we need to deal with. Okay. So, recall ACI code tells me 942 tells me that I can take the shear force at H over 2 from the face of the support, as long as we have this lateral compression from loads and reactions, and there's no concentrated force between H over 2 and the support. So, we take H over 2 instead of D. D is used for reinforced concrete design as the critical section for shear, because there's the possibility that the effective depth varies if the strand is debonded or harped. So, to avoid complications, ACI code uses H over 2 as a critical section. So, where is that? I'm sorry. There's something just. So, H over 2 here is 28 over 2 is 14 inches. Our critical section there is going to be 14 inches plus 12 inches, which is the distance between the center line of the support and the face of the support. So, to bring us to our critical section, that we would need to add half the bearing length distance plus 14 inches, that's 26 inches. So, what we're doing here is we're subtracting from 72, that slope line that starts at 72, and we subtract, the slope is given by 72 minus 12 over 15. This is the slope, and we're at a section that's 26 inches from the from the center line of the support. So, we get a design shear of 63.3 kips. This is what it looks like, right? This is 12. If I were to continue this line up here, it would end up at 72, but the code, ACI code, allows me to design for this shear force, maximum shear force, which is 63 kips. 63 kips here and 12 kips there. So, we have now the vis-a-vis diagram. All right. So, how do the shear calculations work? We have to do them section by section. So, it would be very repetitive to try to pick the different sections along the length of the beam. So, what we will do is we will just pick one section to illustrate calculations, that section being six feet from the center of the bearing or seven from the end of the beam, and all other sections are checked the same way. And what you will see are plots of shear along the length of the beam that were created in a spreadsheet so that you can see how it all looks like in the end. So, you can program these formulas in a spreadsheet, and you can generate the plots that you'll be seeing in this course. Okay. The other thing is that we also have to find a way to get the maximum moment along the length of the beam, because remember that BCI is a function of the maximum moment occurring at each section along the length of the beam. For a simply supported beam, if you run an influence line analysis, what that would tell you is that placing dead and live load throughout the length of the beam actually generates, gives you the maximum moments at each section along the length of the beam. So, it's fairly simple. Now, associated to each of these moments at each individual section, there's going to be a shear, right? So, you can compute M max, let's say at that section, and I apologize for my writing, you're also going to have at that section, you're also going to have a V sub u, say, you know, V sub u at that section, or M max ultimate at that section. You'll have a shear that's associated with that section, you'll have a different set of values for that section, and so on. Okay. First of all, BCI. How do we calculate BCI? We compute the cracking moment by this equation, and then BCI for each individual combination of VI over M max, all right? So, to calculate VI and M max, we can use these two expressions that correspond to cases for uniform loading throughout the span of the beam. These are equations that only apply when the uniform loading is throughout the length of the beam. And remember that the ultimate loads are 4.8 kip per foot. By the way, I think your slides say unfactored shears and moments, it should say factored shears and moments. So, if we are critical, not critical, our design section that we chose arbitrarily was six feet, we simply plug that into x in these two equations, and we find VU at a section six feet from the centerline of the support to be equal to 43 kips, and MU to be equal to 346 kip feet at a section six feet from the centerline of the support. And notice that the span here is taken as 30 feet because this is an ultimate strength check condition, so the beam is already sitting on its supports. FPE is the effective stress in the extreme, this value right here, in order to calculate end cracking, this is associated with the extreme fiber intention. So, we use the prestressing formula, the effective prestressing force divided by the area, we all have used this formula many times now. So, this calculation is to the extreme fiber intention. So, we have a compressive stress of 2.41 ksi that we need to overcome in order to crack the beam, and the other thing that we need to overcome is the tension strength of the concrete, which is given by 6 root of 5ci, and then the rest is just plugging in, right, plugging in i and plugging in y sub t, which is half of 28, 14 inches. So, solving for that, we have cracking moment of 370 kip feet, and if we plug that in along with the value of vu and mu at the section six feet from the centerline of the support, these two values of 43.2 and 345.6, I think everything else here we've determined, we have vci of 59 kips. Remember that the concrete strength is 5,000 psi. Recall that the ACI code tells us that vci should not be less than 1.7 root f prime c b w d, so we check that limit, that limit is at 36, so we have 59. For the section at six feet from the centerline of the support, or seven from the end of the beam, vci is 59, minimum vci is 36, so this one is governed. So, that's what it is. The reduced or the nominal strength is multiplied by times the strength reduction factor that gives me 44.3 kips, and if I repeat those calculations over and over again, the six foot section is right here, pardon me, that's from seven, the section we're dealing with is here, seven feet from the end of the beam, so we just calculated one point right here. This is the minimum vci of 32 something, right, 36 pardon me, and so again drawing that line here, vertical line, that's the section we're dealing with, and that is the plot of vci, and notice that the minimum value, this curve would continue down, and ACI limits the minimum vci to this level, and this is superimposed to the BU diagram, so it starts telling us something that the BU diagram right here is exceeding what the concrete can provide at that section. We still have to check BCW. Well, for the section at six feet from the end, from the central line of support, I should say, we have to plug into this equation FBC, which is the effective pre-stressing stress at the center, at the neutral axis, and this will be zero in our case. This is zero because we don't have any hard strands, so we simply plug in really P over A, which is 0.717 right here, times 0.3, and this is a simple calculation converted to KSI, 12 by 25, and it gives me a BCW at that section of 139 that when factored is 104, so this is using the strength reduction factor is 104. So we could repeat those calculations as well, and that would be the plot of BCW for 7. Now you'll note here that BCW increases near the end, so if I were to go back, BCW would seem to be constant. However, it isn't because this is the transfer length, so this region here is L sub TR. So what affects L sub TR? FPE, the effective pre-stressing stress, which is a function, is a number that affects BCW. So BCW will grow until its maximum value after LTR, the transfer length is reached, and then it remains constant throughout the length of the beam. So in this particular section, 7 feet from the end, BCI governs because it's the lowest one of the two. And as you can see, only BCW will govern from this point to the left, right, because BCI keeps growing. And that's what's plotted here. Essentially what we're combining is BCI and BCW in a single plot, and then we take half of that just to check where minimum stirrups would be required. And by plotting these two functions, BU and PVC, we can immediately identify areas that probably will need stirrups, right? This region will need stirrups because BU exceeds PVC. This other small region will need stirrups, and then the rest, in theory, the concrete can take it without stirrups. Concrete can carry the shear without stirrups. Now the other question is where am I okay to not have stirrups at all? So remember that the trigger here is that if BU exceeds 0.5 PVC, then you'll need at least minimum stirrups. So that's this plot here on a darker line. So we can see from this plot that because BU exceeds 0.5 PVC throughout the length of the beam, essentially, we'll need at least minimum stirrups throughout the length of the beam, okay? So that's what plotting all these components means. And that's what this slide says, essentially what I just illustrated in a previous graph. So I'm just going to go ahead and highlight the regions where we need stirrups, and of course we need minimum stirrups everywhere because of BU exceeding 0.5 PVC. So the other requirement before we compute the requirement for minimum stirrups is to look at spacing of stirrups. And there are spacing of stirrups, or maximum spacing of stirrups, is a function of how much will V sub S have to contribute to the overall shear strength. So if V sub S, the contribution of stirrups to shear strength, is less than 4 root F prime CE, less than 4 root F prime CE BWD, the maximum spacing is 3 quarters of H, or 24 inches. That's by code. If V S is higher, greater than 4 root F prime CE BWD, then the maximum spacing is half of these quantities. And that's for high shear demands, right? Where V sub S has to carry a good portion of shear, of the shear force. And if we have a V sub U that exceeds what the concrete can carry, plus 8 times root F prime CE BWD, we have to resize the section. There's no other way around this. For very high shears, or if with the stirrups, we're trying to design stirrups such that they carry more than 8 root F prime CE BWD, that's a bad section. We need to resize it. Most of the time, one of these two limits will govern. So let's calculate them very quickly. Notice that when we calculate the first limit of 4 root F prime CE BWD, that gives me 85 kips roughly. And notice that my maximum design shear in the entire beam is only 63 kips. So it is clear that V sub S will always be less than 63, right? Because we, pardon me, that V sub S will always be less than 4 root F prime CE BWD, because the maximum shear is only 63 kips. So, so we will, the maximum spacing of stirrups is going to be either three quarters of H, which is 21 inches, or 24 inches, which is S max. So the minimum governs. So we have a maximum spacing of stirrups of 21 inches. So now let's look at the minimum stirrup requirement. So you have to comply with these two equations that are the same as for reinforced concrete design. And I think all of the parameters should be known here, where S in our case is going to be the maximum because, you know, it makes sense to try to space stirrups at the maximum value to avoid congestion. And then for pre-stressed concrete members, we have to comply with this third equation. So the one that gives me the highest number of these three equations is the one that I need to work with. Actually, so I think it's the maximum of these two, or the minimum of this and the combination of these two. That's how the ACI code lists these series of requirements. So that's why in this particular case, when we're triggering the pre-stressed concrete beam AV minimum, which is 0.1 square inches, that governs, even though in this particular case for the two requirements, first, we would have to provide 0.22 square inches as a minimum spaced at 21 inches. So we can do with 0.1 square inches at 21 inches on center, but realistically, that's not easy to provide in a rectangular beam. So we will do with a number three stirrup that has 0.22 square inches, and that's what we will provide as minimum transverse reinforcement. And what we do now is calculate V sub S based on the minimum area that we're providing. So that's 0.22 square inches times FY times D divided by the spacing. The minimum contribution of shear strength from stirrups is going to be then 75 percent of 15.7. So that's close to 12 kips that the transverse reinforcement will contribute. Okay, so let's look very quickly to see whether by providing minimum stirrups, which would give me an additional 12 kips, would help me carry the shear throughout the length of the beam. So we're only checking a section at six feet from the central line of the support, but we notice from the V sub U diagram and the V sub C diagram that the maximum difference between V U and C occurs at about nine feet from the support. And let me just go back to that diagram. So this is roughly from the central line of support, we're talking roughly about this distance. So if the minimum stirrups can carry at least this much, we're set, right? Because the rest would be okay. So we have these difference decreases away from that section. So let's calculate at that section, V sub U is 37, CVC is 28.5, CVS required is the difference. So that's 8.8 kips. And we quickly find that since minimum V S is 12, then that helps me. So that is enough to carry the 8.8. So it is clear that with minimum stirrups, number three is at 21 inches, my design would work. Now recall that we need at least minimum stirrups throughout. And the rest we have here is detailing. We want to space stirrups at 21 inches. And for all practical purposes, the 32 foot long beam is not divisible by 21 inches. We can start one foot from the end of the beam. So that leaves us a 30 foot long beam. And we can divide that at 18 inch spacings. And that gives me a nice even spacing of stirrups throughout the length. So in order for me to plot the overall diagram, let me just finish this slide. There is a question I want to answer. In order for me to plot the actual V, CVN and BU diagram, I need the actual contribution from the stirrups now at the new spacing of 18 inches. So they will resist 14 kips throughout the length of the beam because the spacing is not changing. Okay. So the question is, is the design equation for stirrups of a prestressed beam the same as the design equation for punching shear of a footing? Is there a correlation between the two? So no. I'm not sure I'm understanding this. Because we normally, for punching, unless you're talking about headed reinforcement, and if that's the case, I'm not familiar off the top of my head on that equation, but if this is the equation for headed reinforcement to avoid punching failure in a footing, there must be a correlation of it because this comes out from a, like I said, a truss model, assuming diagonals at 45 degrees. And for punching strength, you also assume that a, let's say this footing or flat slab, the cracks that generate punching also form at 45 degrees. So maybe that's what the correlation is. But like I said, I don't remember off the top of my head what that equation might look like. Okay. So then plotting all components, BU, and now phi BN. Notice that by adding a constant contribution of BS of 14 kips roughly, we simply moved the phi BC diagram 14 kips throughout its length. And we see that our design is well above our demand. So this design is satisfactory. And since, you know, we may see, well, we have such, you know, we have this cushion here. Why can't we make it closer to BU? And the reason is that our design is governed by minimum stirrups. Okay. That's why we ended up with such a high difference. If not, we could make it, we could tailor it to be closer to the visa BU diagram. Okay. So there's an other way, another way of calculating BC instead of using BCI and BCW. That other way, that alternative approach only applies to members that have, that meet this requirement where the total area of pre-stressing force times the effective pre-stressing stress exceeds 40% of the capacities of the pre-stressing reinforcement and mild reinforcement. If that is the case, B sub C can be calculated as the minimum of these three equations. In my opinion, this doesn't save us much time, right? I mean, one could calculate the only thing that, that actually, it doesn't really save us much time because you also have to compute BU and MU along the length of the beam. So if we were able to use a constant value, say from this equation and that equation with these two governed, then we would be saving some time, but it's just as hard to determine BCI and BCW. So if we do this approach, let me just show you at the section corresponding to six feet from the central line of the support. That's what these three equations plugging in values would look like, right? And B sub C at that particular section is 67 opposed to or compared with 59 kips that we calculated previously. So BCI and BCW approach in this particular case ends up being conservative, right? So if we were to compare the two methods, the green line represents the method where we calculate BCI and BCW individually. This dark line corresponds to this other method that is allowed by the ACI code. So the difference is not significantly different or important in particular for this case, because in any case, our design is going to be governed by minimum stirrups. So because we are exceeding 0.5 fee BC throughout the length of the beam. In any case, our design would still satisfy. With either one of the equations. Okay, so that is all that I have for shear. And now we're going to talk about some final detailing of our beam. And the first thing to take care of is to detail the beam near its ends for what are referred to as bursting stresses. So we are going to first talk about bursting stresses. And I'm going to illustrate what those are, because this is the reinforcement that we need to provide. When we apply the pre-stressing force, we often sketch it here as if it were an external force, external load. But the reality is that if the strand is bonded to the beam, this total force is transferred through the transfer link. But in either case, it's being transferred over a short region near the end of the beam. And it has to then engage some distance away from the end. We have to now create constant stresses, right, P over A. And to those, we would add P E Y over I. And to those, we would add P E Y over I. So in other words, even though the prestressing force is pretty located or nearly located at the bottom of the beam, it has to somehow spread out over a certain distance, which is the transfer length throughout the length of the beam. So there is some transverse tension, right? As the load spreads out, there's transverse tension that might generate horizontal cracks in the concrete as this force is being transferred or laterally spread into the entire cross section. And those are called bursting stresses. And bursting stresses are taken care of by adding bursting reinforcement. So you add more transverse reinforcement near the end of the beam to take care of the region where prestressing strand is transferring stresses into the beam when you release the force into the beam. And there's a very simple equation that the PCI handbook uses to estimate how much additional transverse reinforcement we would need. So you, and I don't know the basis for this equation. I suspect it is a combination of some finite element analysis or elasticity analysis and verified by experiments. So P sub zero here is the prestress force at transfer. So not the effective prestressing force, but rather the prestressing force that we are applying to the beam, say the initial pool of 0.75 FBU. H is the height of the beam, the depth of the beam. F sub S is the stress level that we'd like to keep those additional transverse reinforcement under. So in this particular case, we're going to deal with allowable stresses in those terms. We're going to use half of yield as a parameter there as allowable. And L sub TR is the transfer length. Since it's a phenomenon associated with the distance right over which we transfer the prestressing force. So that's why the transfer length is embedded into that equation. So P zero, remember 10 strands, half inch strand and 75% of FBU, that's 310 kips. As I mentioned for grade 60 steel, the allowable stress in the steel is going to be 30 KSI. H over five is six inches. I failed to mention that in this transverse reinforcement needs to be distributed over a length starting from the end of the beam over a distance H over five. So that's why we need to calculate H over five here. So about six inches and the transfer length, we revert back to our, the equation we had discussed before, which is the effective prestressing stress divided by three in KSI units times the diameter of the strand. That's 26 inches, right? The difference between 50 bar, 15 diameters, and this LTR is minimal in our case, right? It was 25 instead of 26 now. So if we plug in all these numbers into the formula in the previous slide, we have 0.021, 310, 28 inches for our depth, 30 KSI and 26 inches for the transfer length, we have 0.23 square inches. So it exceeds, it goes a little bit over the one double legged number three stirrup. So we have two additional number three stirrups. So, you know, we're not, we're not being, we're not throwing money away. The cost of one stirrup, I think is minimal compared with the possibility of transverse cracking and having to patch up that crack, you know, that would be much more costly. So instead of just one stirrup at two stirrups over the first six inches of the beam, it's not going to be too bad. So remember we have 18 inch spacing. So if we had two additional stirrups over the first six inches, we're going to have three stirrups at a fairly close location, right? And then we're going to spread them out at 18 inches over the length of the beam. We also need to check cover requirements. So we're wrapping everything up here, you know, bursting stresses, we just finished shear. Now we're checking that we satisfy covers. And since a pre-stressed concrete is, is fabricated in the factory. So the requirements for cover are much less stringent than, than for reinforced concrete, primarily because the concrete quality is going to be much better. You know, the product itself is going to be more, have a much more tightly pore structure, space pore structure. So not, not as much, not as permeable. So the cover to the primary reinforcement is reduced compared with reinforced concrete. So you have either the bar diameter as a minimum or five-eighths of an inch, and you do not need to exceed what's required for, for reinforced concrete, which is one and a half inches. This is for an element not exposed to weather. And the coverage of stirrups only needs to exceed three, three-eighths of an inch. So if we check those, we call that our pattern is at every two inches. So we have two inches to the center of the first strand. So from two inches, we subtract half of the strand diameter that leaves us one and three quarter to the center of the first, the outermost strand. And that of course exceeds five-eighths of an inch. So we have more cover than required. And the coverage of stirrups again, we subtract from two inches, half of the strand diameter minus the diameter of the stirrup, which is three-eighths. And that leaves me with one and three-eighths of an inch, which is greater than three-eighths. So we have more, more cover than needed just because we have a rectangular beam. So, so we're, we're satisfying those cover requirements. Okay. So then the last topic today will be how to deal with camber and deflections. And perhaps several of you have done those calculations and I wanted to, to see, just to have an idea of how you feel about those calculations that you've done before, whether you feel they're quite accurate or not. So Sherry, if you could launch the third poll. Okay. The third poll. I believe that I can calculate camber and deflections very accurately if I know the material properties of a beam. And I am launching the poll now. On your screen, please vote yes or no. We have several more people that need to vote. I am going to close the poll now. 35% said yes, 65% say no. All right. So yes, you're absolutely right. The majority of you do not feel very comfortable or do not, do not think that you can calculate deflections and camber very accurately. And that is correct. If you know, even if you, if you think, you know, the properties of a section, it's going to be as the point here is going to try to convince you that camber and deflection calculations are part and art and part of science, even though we'll use, we'll use equations that come out from structural analysis. And even though we assume we're dealing with a beam in its elastic range, those calculations can be quite tricky or, you know, comparing with what actually happens. So here are the first two items to consider. Initially, when we release the pre-stressing force into the beam, there will be, if the pre-stressing strands are eccentric, like we have designed our beam, the initial result will be that there will be an upward deflection called camber due to the pre-stressing forces. Now that is going to be counteracted by self-weight of the beam because that is acting on the beam as we release the pre-stressing force into it. So the first two items that we need to consider in camber and deflection calculations are these two effects. And it is important to note that these items, deflections or camber are affected by concrete and steel properties. And several of these properties, particularly the concrete properties, are very hard to predict accurately or to estimate. Steel properties are not as much. I mean, steel is fabricated in a controlled environment. So these are the two sources of camber and deflection that occur at pre-stress force transfer. The first one, as you know, occurs from the eccentricity of the pre-stressing force, PI. And notice that PI here is the initial pre-stressing force or the force at release. And that generates this upward deflection, which we call camber. And notice though that we are subjecting this beam to constant negative moments at the end. So the moment diagram is going to be constant. The downward deflection varies along the length of the beam and is induced by this self-weight component. And you may remember from when you took perhaps structural analysis that there were formulas to estimate deflections based on what the loading conditions were on simply supported beams. And two of those formulas are for uniformly distributed loads. This is a formula to estimate deflections based on the magnitude of the uniform load on the beam, the span length, the modulus of the elasticity of the material, and the moment of inertia. And these 5 over 384 is a factor that depends on the loading and on the support conditions. Now, the camber caused by the pre-stressing force, a formula for it is the moment times the span length squared divided by 8 modulus of elasticity times I. So notice that the moment that we're inducing is P times E. So we replace that here. It's the effective pre-stressing force and the rest remains the same. So the hardest part here, in my opinion, to estimate accurately is the modulus of elasticity, right? That has some inherent variability. Now, this equation here only refers or is only applicable to the case of straight strands where we have constant eccentricity. So I encourage you to visit appendix chapter 15 in the PCI design handbook. This is where all the design aids are located. So there are other equations for other loading conditions that might become useful for you when you have complex strand patterns like carved or draped strands. So you can find deflection equations for those cases as well. So again, ECI is perhaps the hardest parameter to determine accurately. We think we know that concrete strength at release is 4,500 PSI. And let's say we do, let's say we test cylinders and it's 4,500 PSI. The fact of the matter is that this equation is largely empirical. So ECI may be higher or lower than what's given by this equation. So that's one key aspect. And ECI, depending on the modulus of elasticity, camber and deflection will change. The other item here is that the effective prestressing force that we need to consider only needs to account for losses that occur at release. And remember that we estimated those as being about 10%. So we do that here, but also remember that in this loss calculation, there's a modulus of elasticity involved, again, with uncertainty. And the design, the loss equations also have certain uncertainty. So this is only an approximate value as well, 279. So the other thing that I should call your attention to is the span of 32 feet. That's end to end of the beam because the beam is on the bed. And unit consistency is very important. If we're in kips here and inches, we need to be in kip and inches throughout. So if we have 32 feet, we need to multiply times 12 to convert two inches. That may seem silly, but I often make a mistake. And when our mistakes in these equations, if I'm not careful with units and I end up with deflection calculations that make no sense to me, and then I find my error and correct it, of course. So the pre-stressing force would generate roughly a 0.7 inch upward deflection or camber. And that is counteracted by self-weight. This is the self-weight of the beam per foot. And we use the appropriate formula here. So that's a downward deflection. We're carrying three significant figures here to the right of the decimal point or three digits, I should say. But the reality is that that's over too much accuracy or implied accuracy. We have 0.579. I mean, the best we can do is, you know, we can't carry three decimal places. The best we can say, well, the camber after the pre-stressing force is released is about 5 eighths of an inch. Okay. So then the PCI design handbook has a series of multipliers for these short-term deflections and cambers. When you move the beam to storage in the yard, it'll grow. The camber will grow. And this has been determined through years of monitoring how beams camber grows in the yard. And these multipliers are applied separately to the pre-stressing force component and to the self-weight component. So we end up with a long-term growth, so to speak, in the storage of one inch. So initially the beam will have about 5 eighths of an inch as in the bed. As we move it out into the yard, it will end up with approximately one inch of upward camber before shipping to the site. And this again depends on how long the beam is sitting in the yard. Okay. So once we erect the beam, there are other multipliers. So as time goes by, creep takes over, right? Creep of concrete. And creep is a phenomenon that occurs due to permanent loads. So one way that the PCI handbook deals with it is just by modifying those multipliers to the pre-stressing camber result, to the self-weight result, and to the superimposed dead load result, we use a multiplier as well for the beam in its place and where only the permanent loads are active. So we need to compute then the superimposed dead load deflection, and we use the same formula as before, the five over 384 WL fourth over EI. But now the span length is 30 feet. That's the value of the superimposed dead load. That's from the first session. And notice the weird units, kip per inch. This is to avoid problems down here with our deflection. Notice also that we're dealing with F prime C after 28 days and not F prime CI. So that affects EC or that results in EC and not ECI. So we have a downward deflection under permanent loads of 0.2 inches and applying the multipliers shown in the previous slide, we end up with an upward camber for the beam in place of 0.78 inches or three quarters of an inch would be a more accurate way or a more, not accurate, a better way to indicate how much camber we're thinking the beam will have on site. And what's important then is to then compare or add live loads to this beam and compare those to allowable limits. Now let me make a parenthesis here. And, you know, the class of member that we are dealing with, class U, uncracked, we base our deflection calculations according to properties that are listed in this table of ACI, 24.521. And what that table tells you is that for class U pre-test members, you should use gross section properties. So gross moment of inertia is used in those calculations for uncracked members. And that there are live load deflection limits also listed in ACI. And in table 24.22, it tells you that for superimposed loads, for floor loads, the live load deflection limit is L over 360. So we need to compute that. So we have left a three quarter inch camber. What we really need to compute, I mean, we can make adjustments if you want. But the reality is that what we need to do is we need to compute the live load deflection to the allowable limits. So the live loads for our beams are two kips per foot. So that's the conversion to kip per inch. The live load deflection then, when we consider F prime C, is 0.4 inches downward. The L over 360, we could either compare the length over 360 here and compare that with the allowable, that would give me the allowable deflection that I could compare with 0.4 inches. Another way of doing it is to find what factor 0.4 inches corresponds to. So if we do that, it would correspond to a denominator of 872. That of course is much less than L over 360. So we're okay. We're satisfying the live load deflection limit. So all these are only to classroom members. So when we apply all service loads on the beam, dead loads are permanent, live loads are instantaneous and they happen and then they cease from happening. So we have an upward camber of three-quarters of an inch. We subtract the live load deflection. We would still end up with an upward camber of about three-eighths of an inch, even with full live load acting on the beam. So don't believe this number. It will not be very accurate. It will vary for several of the reasons that I'm showing here that I won't read one by one. But most of it is creep is highly variable because modulus of elasticity e, notice could vary 10 to 20 percent and creep can vary, arc estimate of creep can vary up to 50 percent because it is a function of humidity, of how early the permanent loads are applied to the beam, etc. So it's, I'll leave that for you to look into and see if this slide convinces you that all those calculations we seem to do very accurately are not that accurate. So this would be our final design. You know, wrapping up, we have 10 bottom strands, half-inch strands, a rectangular section, the top steel for number sixes. This is mild reinforcement. This is to control top tension cracking. Goes back a few slides, a few sessions, pardon me. And that's all. These are our material properties, f prime c of 5000, f prime ci of 4500. Again, one has to check with a producer if we can be this close. And remember that we modified these to comply with compressive stress limits. That's the section. So I have a few more slides just as complement. This won't take too long. It just, since we've been dealing with class u members, what happens if we have a member that is, how do we deal with deflections when we have a member that's within, in transition or a class t member? Remember that that member is a class t when the tension stress in the bottom fiber exceeds the cracking stress, but it's below 12 root f prime c. This is the cracking stress, 7.5 root f prime c. So what we could do is we can say, well, let's divide the behavior into two regions. The first region being uncracked. So the beam will behave as uncracked until I apply enough load to create a stress equal to 7.5 root f prime c. And once the beam cracks, the increment in load is going to be applied to a beam that is now cracked. So with reduced moment of inertia. So that introduces the concept of a bilinear load deflection curve as shown in this slide. So this first region corresponds to the uncracked beam where full dead loads are applied. All right. And now the beam becomes cracked under full service loads under the application of dead plus live, but part of it remains uncracked. And when I reach 7.5 root f prime c, then the beam cracks and I follow this second slope. So that's the bilinear load deflection approach to deflection calculation. So notice that the dead load deflection is this amount and the live load deflection part of it would happen in the uncracked beam deflection. And then most of it would occur in the cracked beam region. So the way to do it is to estimate how much load I need to bring the beam to cracking. Then estimate how much of that total load corresponds to live load. So this would be P live load with the uncracked beam, and then the difference would be P live load on the cracked beam. So we would have to use the cracked moment of inertia. So we would use I gross for this region and I cracked for that region. So the bilinear methodology is an approximation to a real load deflection curve. This is not anything having to do with our design, but it's just conceptual and I wanted to show it to you in case you're dealing with a beam that actually does crack. And that's what I just described. So break up the behavior into an uncracked behavior and cracked behavior. In the uncracked behavior, you use gross moment of inertia. And then for the remaining portion of the live load, after I've reached 7.5 root f prime c, we use the cracked moment of inertia. And then we add the deflections to compute the total deflection of the beam. All right. So that's in mathematically what that means. The total live load deflection will be part of it that is uncracked under I gross. And then the other part is the cracked under I cracked. And these factors are the same. They just come out from the equation for uniform loading on a beam. So the total live load deflection is going to be given by this equation. And that's it. So this is all I had for today. We're ending up about six minutes before 8.30. So we have a few minutes for questions and I'd be happy to entertain those. If you have any questions, please type them in the chat box and I will share them with Sergio and the audience. We have a question. Okay. Hold on. And I will share it with everyone. Besides the PCI design handbook, what other books or resources on pre-stressed concrete would I recommend? Well, I do recommend having the PCI design handbook because it has a lot of resources. It summarizes very nicely pre-stressed concrete design, but it has different chapters dealing with a variety of aspects related to precast as well. It has also design aids that are helpful and it points you to sections in the ACI code. There are other textbooks that I do recommend because depending on how deeply you want to delve into the problem. If you want to take it from an academic point of view, I recommend two textbooks. One from Lynn and Burns. It's one of the landmark textbooks. It's called, let me just bring them. Hold on for 10 seconds. So one textbook is from Lynn and Burns. It's like I said, it hasn't been updated in several years, but it deals with the basis of pre-stressed concrete design very nicely. It's called Design of Pre-stressed Concrete Structures, Lynn and Burns. So again, all these, it's one of the books that is often referenced. And also the book by Collins and Mitchell from Toronto, Pre-stressed Concrete Structures. Those are two textbooks that are quite useful to learn very deeply into pre-stressed concrete design. So these are in addition to all the other resources that PCI has available. There are many other, you know, the PCI design handbook also points you to papers that appeared in a PCI journal. And those are free. So those are available to members and even non-members. So I also recommend that you read those papers from which the basis of pre-stressed concrete design was formed. So long answer to a short question. So there's, in addition to PCI, are bursting stresses discussed in ACI? Not specifically, not specifically. The place where these bursting stresses, and they're not necessarily called that, but they are often quite important, is for post-tension construction where you have a large number of strands in a single location being applying a high, a large pre-stressing force that generates quite a bit of bursting stresses. So in those cases you have the hardware already includes some of the reinforcement that's needed. So these trumpets, I don't know if you've seen them, you probably have, you have a bunch of strands coming into this device and you have a trumpet that looks, you know, a device that looks like a trumpet, and around it there's a coil or a spiral that restrains that trumpet from, but it tries to to confine the concrete around the trumpet so it doesn't burst. So that's, so the PTI, the Post-Tensioning Institute, might have some additional provision for, provisions for these bursting stresses. There was an old book I think I ran across from the 1940s or 50s from a French engineer, and I forget his last name right now, it's one of the first pre-stressed concrete books, and that's when it was, that that subject was first approached. Any other questions? We have about two minutes left. If you think of something, please, as always, send the question to me. I'll pass them on to Professor Brynia, or it's been a long day, and I'll get the answers out to everyone. Well, otherwise, I would like to say thank you all. Thank you, Sergio, for another wonderful course. We will please be offering some further courses in the fall, so I would invite you all to come back, check out the list of courses available on our website, and again, thank you. If you missed anything, please email me, and we hope you enjoyed the course. Good night, and have a wonderful evening. Thank you, Sergio. Thank you, everyone. Thanks for your patience.

rectangular beams

shear strength

shear design

pre-stressed concrete beams

reinforced concrete members

equations

failure modes

concrete contribution

steel contribution

live load pattern

Influence Lines

BCI

BCW