Good morning everyone. So as Sherry mentioned, this is the fourth of six sessions of this course on basic pre-stress concrete design. And today we're just going to continue with our design that we started last time. What we're going to do is we're going to continue calculating stresses for the design of the beam that we started last time. Recall that it's a rectangular beam that we determined 10 strands to be necessary to carry the service loads at mid span. So this is what we are going to cover in this module again. We are going to understand ways, we're going to continue computing stresses for different load conditions. We're going to understand ways to control top fiber stresses and design against those for different loading conditions as well. We won't necessarily spend too much time talking about losses as we covered that last time, but we will mention when losses need to be recalculated based on the modifications we made tonight. And we will also talk a little bit about harping. So the idea here is I'm going to spend quite a bit of time talking about the beam with straight strands. And then if we have time towards the end, we'll talk about the effects of harping strands instead of using straight strands for our design. So that comes, those last few slides are going to be somewhat repetitive or seem repetitive, but it's going to be using two different techniques to control top fiber stresses. All right, so this is where we are. The blue text illustrates what we have done before. And right now what we need to do is we need to start checking the service load condition. And for that, we need to first calculate or compare service load stresses against allowable stresses. So the first few allowable stresses that are indicated both in a PCI handbook and the ACI 318 code are classified for our class U member. Remember that we're designing a member that is uncracked or that is to remain uncracked, class U. There are two limits, for compressive stresses and tension stresses. And these are limits for service loads. For compression stresses under the action of all loads, that means all dead and live loads, the limit is 0.6 F prime C. Under compressive stresses under sustained loads only, the limit is 0.45 F prime C. And for tension stresses, it's seven and a half root F prime C. So the calculations based on our material properties are shown here. So we have for all loads, compression stresses all loads is 3000 PSI. For sustained loads fraction, it's 2250 PSI. And for tension stresses is 530 PSI. Again, remember that these are conditions at service. So this limit of 7.5 root F prime C only applies to region of the beam that is termed the pre-compressed tensile zone. And here's the definition of what that means. The pre-compressed tensile zone is essentially the area of the beam that is supposed to be, that would typically be under tension stresses when dead and live load service loads are acting on the pre-stressed concrete beam. So in this particular case, since we're dealing with a simply supported beam, the zone is the bottom of the beam. So all of the bottom of the beam it classifies as a pre-compressed tensile zone. So you'll see that throughout this session, we'll calculate a bunch of stresses using the flexure formula. And the only thing that we'll be modifying are what moments to include in each of the cases. So if we check first service load stresses at mid span, which corresponds to the maximum moment, we can separate because of the allowables, we can separate dead load stresses from live load stresses. So the allowables recall that depend on whether you're considering all loads or only the fraction of sustained loads, at least for the compression stress limit. For the tension stresses, there's a unique limit for both, for all loads, I should say. So notice what these moments are. These come from our first, when we first introduced the problem, the moment induced by self-weight at mid span is 39.4. The moment due to superimposed dead loads is 110.3 kip feet. This is all at mid span. And the distance to the top or bottom fiber, in this particular case, it's indistinct, it's 14 inches. So when we divide by the moment of inertia of the section, we get 1.15 KSI for dead load stresses and 1.72 KSI for live load stresses. This is a corresponding live load moment at mid span. So again, these are top and bottom stresses that are the same because we're only considering the applied loads so far. Then if we consider the effect of the pre-stressing force, we have the two components, P over A and PEY over I that need to be considered. And for this, we're just calculating the value of P that we calculated last time. After determining losses, you might recall, we ended up with 22.5% worth of losses. So these are our 10 strands, their area, the initial pool of 75% of FPU and the fraction of that initial pool that stays in the beam after losses. So we've got 240 kips for P. You might also recall the eccentricity of the 10 strands is 11 inches. That's what this number means. And the 14 inches, again, is to the bottom or top fiber. In this case, it's the top fiber stresses and the bottom fiber stresses. The subscripts here correspond to stresses induced by the pre-stressing force. So these are the top fiber, bottom fiber stresses. The signs reflect whether they're tension stresses or compressive stresses. So this would be stresses only induced by the pre-stressing force. And the previous slide, stress is induced by the applied loads when we add them together using the principle of superposition. The left-hand side illustrates stresses due to all loads, dead and alive. And the right-hand side illustrates stresses only caused by sustained loads. Sustained loads are only dead loads. So you'll see the three components here on the left-hand side and only two components here on the right-hand side for both top and bottom stresses. So we add algebraically with sign. And we have, for bottom stresses, caused by all loads, we have 1.9 compression on the top, 0.47 tension on the bottom. And stresses due to sustained loads, we have 0.18 compression on the top and 1.25 compression on the bottom. So when we only use sustained loads, the beam is an entire compression. When we have all loads, including live load, we have tension on the bottom, compression on the top. So this table summarizes that section at mid-span, right? The top stresses for all loads, sustained loads, bottom stresses for all loads and bottom stresses for sustained loads. The actual stresses we just computed, instead of using signs, we illustrate them with compression and tension. So notice the top is under compression under both all loads and sustained loads at mid-span. For bottom, the bottom of the beam is under tension under all loads. And the bottom of the beam only under sustained loads is under compression. And these are the corresponding limits. You see that the limits are higher in absolute value compared with the applied stresses. So all of our checks, all of our sections satisfy the allowable stresses. The beam is okay at mid-span and this should not be surprising because we designed the number of strands based on the section at mid-span. But now we need to check the entire beam. So what one does is one picks stations along the beam, so sections along the beam to check. And that's what we will be illustrating here. We will do the detailed calculations for sections that are important, and then we'll just show graphs for the rest of the checks along the length of the beam. So first things first, you might remember that we covered this equation or the concept of development length and transfer length earlier, like in the second module or second class of this, the second module of this class. So the equation to calculate development length was given here as having two terms with the first term representing the transfer length, where FSC is the effective stress after losses and FPS are the stresses at ultimate corresponding to the ultimate bending condition of the beam. So this is the equation that one would use for development length. And this is the equation, the first term corresponds to the transfer length. So after illustrating this slide, I wanted to just pull you and see to get a feel for whether you know the distinction between development length and transfer length, physically what it represents. Okay, our first poll for the evening is please indicate if you have a good understanding on the difference between transfer and development length of strand. I am launching the poll now. It is on your screen. Please vote now. Okay. If you haven't voted, please vote. I will be closing the poll now. Sergio, 67% answered yes, 33% answered no. Okay, thank you, Sherry. Okay, so let me, for those of you that answered no, let me just go over it very quickly. The basic difference is that, let me start with transfer length. The transfer length is the length required for the strand to transfer the effective stresses after losses into the effective pre-stressing stresses into the beam. So in other words, when we tension the strand and we release it, the pre-stressing force is transferred into the beam, but it doesn't get transferred at a point. It gets transferred over a length. And that length, so after that force, after we have undergone some, all the losses, we call those stresses the effective pre-stressing stresses, but the length required to transfer those into the beam, which in essence gives me the pre-stressing force that I can use for design, that length is called the transfer length. Now that occurs in a fairly short distance towards the end of the beam, but that only transfers the effective pre-stressing force into the beam. When we then keep loading a beam until it fails in flexure, the stress in the strand needs to keep increasing beyond the effective pre-stressing stress FSE. And that increase, further increase in stress necessitates a longer distance to develop those higher stresses. And that is that second term here that's added to the first term in the development length, the totality of the transfer length, plus that additional length required to develop the additional stresses to FPS is called the development length of the strand. So of course, LD will be longer than LTR. So hopefully that clarified it. So what we need to do is we need to, we'd like to see how long, what the transfer length is for our beam and what the value of the effective pre-stressing stress is. So the effective pre-stressing stress is no other thing than the stress induced by the initial pull, 75% of FPU minus losses. So remember that we pulled the strand to put 75% of 270 as the rate of strand. And this is the value of losses that we calculated, total losses. So we have an effective pre-stressing stress of 157 roughly KSI. The transfer length then, the length required to transfer these stresses into the beam in the form of pre-stressing is FSC over three times DB, DB being the diameter of the strands we're using. So we're using half inch strands and notice that the units that we use here are in KSI. So the transfer length becomes 26 inches or roughly 2.2 feet. Again, note the units of FSC. This is different from previous versions of the PCI design handbook. All right, so at the end, FSC is, at the end of the beam, FSC is zero because simply the stresses start developing at the end and towards the inside of the beam. Their maximum stress, the maximum pre-stressing force or stress is developed after LTRR is reached. So it is important to check the stresses at the point where the pre-stressing force is fully transferred. So at the end of the transfer length. At that point, since we're no longer at the end of the beam, at that point, we have already started adding some dead load and light load to the beam. We're at a short distance, 2.2 feet from the end of the beam, but we're still, we now have some moment on the beam. It's only two feet of span that we're in, but in order to check stresses at that section, we need to include dead loads and light loads acting at that point. So that's what we're going to do. We're going to first recognize that the transfer length is measured from the end of the beam, not from center of supports. So the total length of the beam is 32 feet. We define the center line of our supports, one foot in from either side of the beam, giving us a clear span, a span of 30 feet. So the transfer length occurs 1.18 feet from the center of the support. So that is 2.18 feet minus one foot from the end of the beam to the center of the support. So at that section, we can then calculate the moment induced by self-weight and dead load and superimposed dead load by using a general equation for moment. So this is an equation that we derived or presented in one of our earlier sessions, where simply what we're doing here is we're using a section X here. This represents a coordinate X measured from the center line of the support. So this would be W times X divided by two, where X is any station, and times L, the span of the beam, minus X, again, the station of interest. So LTR is our length, 2.2 feet, and we subtract the one foot to just consider the clear span. So the moment induced by self-weight plus superimposed dead load at that section is 22.6 feet. It's a small moment, but still non-negligible. So if we calculate stresses at that section induced by superimposed dead loads and self-weight, those are the stresses, top and bottom, that would occur on the beam at that section. So compression and tension stress is a 0.17 at that section. Now, we also have live loads at that section. The calculation proceeds similarly, except we use the live loads applied to the beam. We had two kips per foot, and the rest is just the same, the X, the section of interest. And then we get the live load stresses, top and bottom, 0.26 KSI. Compression on the top, tension on the bottom, because these are superimposed loads. And then what we do is we then apply, or we then use the superposition where we use the top and bottom fiber stresses induced by the pre-stressing force, since we have at the section after the transfer length. So at that section, we have the full pre-stressing force. And recall that since we have straight strands throughout the beam, those, these stresses are the same as they, the pre-stressing stresses are the same as they would be at mid span or any other section along the beam, as long as we have the full pre-stressing force after the transfer length. So we then use superposition of dead loads, live loads onto those pre-stressing stresses for the bottom fiber and the top fiber. And we end up at a section 2.2 feet from the end of the beam, or 1.2 feet from the center line of the support. We have a top fiber, a bottom fiber, pardon me, compression stress, 1.97 KSI. And a top fiber of 0.54 KSI in tension. Since we have very little moment induced by superimposed loads at a section at the end of the beam, we end up with a portion of the beam that has tension stresses on the top and compression stresses still on the bottom. So it appears that this stress slightly goes over the limit of this tension stress limit. But remember that this tension stress limit only applies to pre-compressed tension zones. And towards the end of the beam, the end of the beam is not necessarily going to be, pardon me, the top of the beam is not, it doesn't qualify as a pre-compressed tension zone. It's going to be under compression under service loads. So that's what I just said. So I'm just going to continue. So this stress, although it might seem that we've exceeded the stress, the 0.537 and 0.53, first of all, it's very close. And secondly, this is just a condition that will occur. This is a service load condition, but top fiber stresses are generally dealt with at the release condition. They are going to be more critical. So we're going to be assuming at this point that service load stresses at that section are okay because we still need to check release. And then we can come back and check service load stresses if needed at that section at the end of the transfer link. So one could keep continue, so must continue checking stresses along the length of the beam. But rather than showing all the calculations, we are simply using the bending formula and you could program your stress calculations onto a spreadsheet where you have on one column, you have X station, then you plot moments induced by dead loads and live loads at each station. And then you can compute stresses, top and bottom fiber stresses, and given the pre-stressing for stresses. And then you could end up with a plot, in this case of stresses along the bottom of the beam in terms of distance. So remember a 32 foot long beam. So we start with bottom fiber stresses that initially are compressive along the bottom. That's what we just calculate. This peak here corresponds to the end of the transfer length. And as soon as we start dead loads and live loads start acting on the beam in a significant manner, those bottom stresses decrease in their value of compression and then become tensile near the center of the beam. And then they become compressive towards the other end of the beam. This is for all loads. And the dashed line corresponds to only sustained loads portion. So you see that these are the allowable compression stresses for all loads and sustained loads. And these are the allowable tension stresses for the pre-compressed tension zone. So it seems like stresses at the bottom are satisfied. Stresses at the top for all service load levels. In this case, it's only important to look at the top at all loads condition. So again, stresses increase gradually until the end of the transfer length. And then the effect of superimposed loads decreases tension stresses right away and has the top of the beam going to compression towards the middle. And again, the allowable compression stresses are here. Again, for the compression stresses, they are both satisfied and tensile stresses. One could say that we have reached the limit if we were to consider this line, the allowable tension stress under service conditions. And the dashed line for tension stresses does not really, is not representative because we do not check tension stresses for only sustained loads. We only do that for compression stresses. So the beam meets all service requirements, service load requirements. So as mentioned before, we now need to move on and check the release condition and we would need to adjust strands if needed. So there are several things that we need to remember. First of all, when we release the pre-stressing force onto the beam. Actually, I apologize. I ran through in my... Okay, so in my attempt to keep moving along, I forgot to launch the second poll and it would start right here. The second poll is related to these two diagrams that I just discussed with you. Okay, I will launch that poll now. Does it make sense to you that the curves illustrating stresses, top and bottom, along the beam are similar to moment diagram curves? Please answer on your screen now. Okay. We have a few more people that need to answer. I am closing the poll now. Sergio, 89% yes, 6% no, 6% not sure. Okay, thank you, Sherry. The majority of you say yes. And the short explanation that I like to give is, yes, it makes sense that these curves resemble a moment curve for a beam that has uniform loading, simply because the only thing that's changing here is the magnitude of the moment. So the moment follows a parabolic curve and remember that all stresses are calculated based on the bending formula, which in summary, tells us that those stresses are simply following the shape of the moment applied to the beam, because these two other parameters are constant, right? The Y is 14 inches, that's the moment of inertia. So the shape of the stresses, I'm just pointing to these live load stresses, has to follow the shape of the moment diagram for live loads and dead loads, as long as they are uniform loads. Good. So now, as I mentioned earlier, we need to check the release condition. And these are some of the notes that we need to remember. First of all, that when we release the pre-stressing force onto the beam, concrete is not at 28 days, it's probably one day old. So one has to use the initial concrete strength. For our calculations, we've been using 4,000 PSI so far. The other item is that the beam is sitting on the bed, the pre-casting bed. And so there's no, it's not sitting on its supports. So the full length of the beam that actually is bending is the 32 foot length. And one can consider the corners of the beam to rotate once the pre-stressing force is released into the element. And the last item here is that the losses should not account for long-term losses. So the only two losses that are important at the release condition are elastic shortening, because the beam shortens immediately as soon as the pre-stressing force is released into it, and a modified relaxation loss because the pre-stressing strength does relax in stress a little. So there's a typical loss assumed at the time of release of about 10%. Remember that we have calculated 22.5% in our case. So there are specific limits for the release condition. They are both listed in ACI 318 or the PCI design handbook. And you can see them here. For the limit for compression at the ends is 0.7 of F prime CI. Remember F prime CI is the concrete strength at release. For compression at any other locations along the length of the beam, 0.6 F prime CI. And in parentheses here is what's usually done as common practice in the PCI design handbook, seventh edition. For tension stresses, the ends of the beam are limited to six root F prime CI, and at other locations to three root F prime CI, but it's common practice to simply use six root F prime CI. I won't mention too much about this right now. We'll cover this later. So this is, as alluded to in the previous slide, one could use the standard design practice that was included in the seventh edition of the PCI design handbook to say simply that for release, one could check compression stresses as 0.7 F prime CI for all locations or six root F prime CI for tension stresses at all locations along the length of the beam. And notice that these limits are a bit less conservative than those in ACI 318. One note here is that even if we only have bottom strands, it is recommended that there are at least two number four bars or two bonded strands to the top of the beam so that we can tie the reinforcing cage, even if the stresses, acting stresses are falling below the tension stresses, I should say, fall below the allowable of six root F prime CI. Let's use, for this problem, let's use the more stringent limits in ACI 318. And for those, these are the calculations, the 0.7 root F prime CI and 0.6 root F prime CI for compression stresses and six root F prime CI and three root F prime CI for tension stresses. And these are the corresponding values, 2,800, 2,400 for compression, 380 and 190 for tension. Now, the ACI code specifies the ends of a simply supported beam, but it doesn't really define what that corresponds to, whether it's exactly just at the end, or is it a region of the beam near the end? So since the code doesn't specify what is the end region of the beam, we will consider just arbitrarily that the limits shown here for the end of the beam will be applicable to the first four feet. And again, this is just arbitrary. We consider the first four feet as the ends of the beam. So it's not clearly a section, it's really a region. Okay, and as mentioned earlier, the loss of prestressing force should not be 22.5%. So we're assuming a 10% loss. So we called our 10 strands with the new loss at release of 10% only. So we have a P force of 279 kips. So the top fiber and bottom fiber stresses induced by this prestressing force in this section is P over A minus or plus PEY over I. Calculations are shown here. A top fiber stress only induced by prestressing force of negative 1.13 KSI. That's a fairly large tensile stress over a thousand PSI. And for the bottom stresses, we have a positive number, meaning compression of 2.8 KSI. Now, remember that this force increases linearly from the end of the beam to the end of the transfer length to its maximum value. So again, we use the equation we used before where we can compute moment at any section, at any X section measured from the end of the beam in this particular case with the magnitude of the applied uniform loading in each condition. So the moment induced by self-weight, we can say, let's compute our X section. Let's say our X section varies along the length of the beam. So we can simply choose whatever X is and compute the moment induced by self-weight along the length of the beam. And once we have the moment, we could compute the stresses, the self-weight stresses. And if we plug in the equation for moment, which ends up being a function of X squared, that's the equation of moment. We have an equation for stresses, self-weight stresses that are also a function of X squared. And when we do that, we can plot the allowable compressive stresses. These are the two limits specified in the ACI code. And we have these stresses, FSW plus those induced by the pre-stressing force. We can plot them again along the length of the beam. And we see that these rise until the end of the transfer length. And then since the only load acting on the beam is self-weight, that's the only thing that counteracts these high compressive stresses. And we see that these are bottom stresses, right? And they're compressive. We see that we're exceeding the allowables mostly throughout the length of the beam. In terms of tension, these are the allowables. And notice the negative sign here on the left. And tension stresses rise along the top of the beam to very high values near just past one KSI as we calculated. And then the self-weight of the beam decreases those top stresses, but it doesn't even bring them close to the allowables here, right? This is the step function that represents the allowables. So we clearly have top tension stress problems throughout the length of the beam, as well as a minor compression stress problems at the bottom of the beam. So we need to do something about that. So here are some comments on why at release the stresses exceed so much the allowables. And just to recall, the only load that's acting on the beam that helps decrease those stresses is the self-weight. So we don't have live load acting on the beam. So applied loads actually help counteract those pre-stressing or stresses induced by the pre-stressing force. And we also have, remember, we also have less loss of pre-stressing force. We only have 10%, whereas we computed 22.5%. So we have a higher P force and we have a lower reduction in moments induced by service loads. And that's why the stresses are way exceeding our limits. So these are just the comments I already made. Tension is exceeded along the top of the beam throughout the length of the beam. And in terms of compression, we also exceed the allowable compression stresses. But note the following, we use the more stringent ACI allowables. If we had used the PCI stress limit of 0.7 F prime CI throughout the length of the beam, we would have been all right. But since we're using the ACI code just to illustrate how we can deal with these problems, we didn't make it, right? So one possibility for compression stresses is to simply increase F prime CI from 4,000 to 4,500. Now it says, from a calculation perspective, that works and that's easier, but it could be problematic from a fabrication perspective because this release strength might not be very easy to attain. The other option would be to change the strand pattern towards the end of the beam to decrease compressive stresses and tensile stresses. And we'll deal with that when we talk about the harped solution. So we could do that by harping, de-bonding, or adding top strands. So let's increase for, to try to take care at least of the compression stresses at this point, let's see if increasing to F prime CI or 4,500 PSI takes care of the problem. Notice the note here. It could be difficult to get a release strength of 4,500. So that needs to be checked with the fabricator. If we do that though, by moving the allowable curves upward, that's essentially what we're doing by increasing 4,500 from 4,000 to 4,500 PSI, we're moving the allowables a little bit higher. And then where we were violating them, we are no longer violating them. So this is now in terms of compression stresses in the bottom of the beam, we're fine. If we compute the tension stresses, they're also a function of F prime CI, but it's a root function, a square root of F prime CI. The allowables are here now, and the actual stresses are still exceeding the allowable. So we still have a problem in terms of tension stresses. So compression now passes, top tension does not. So how can we deal with this from a practical perspective? So these are different ways of possibly of solving the problem of top tension stresses. We could add top steel, non-pre-stressed to take care of those tension stresses. It has to satisfy certain requirements. We could change the eccentricity of the strand throughout the length of the beam by adding top strands. So we could decrease the eccentricity. The top tension stresses are being created by that eccentricity as the beam cambers upward. We could change the eccentricity by harping the strand along the length of the beam. If we change the eccentricity by adding top strand, then we're essentially working against ourselves because at mid-span, we do need the strand to be at its maximum eccentricity. By harping the strand, we're changing the depth of the strand where we want to reduce those tension stresses. So we could do that. We could also decrease the pre-stressing force and gradually transfer it, not only directly at the end of the beam using the transfer length, but by debonding selected strands. So transferring the force, the pre-stressing force in steps into the beam where the weight of the beam now helps me. In this particular case, by debonding wouldn't work because the tension limit was exceeded throughout the length of the beam. So look at this. If we debond, we would only debond from the end, but we're also exceeding the top tension stresses in the center of the beam. So we would essentially have to debond everything and that doesn't work of course. Or we could try to really change the release strength again but this is not a very efficient solution. So we're going to adopt two possible ways, adding mild top steel and changing the eccentricity but we first will deal with top steel only. Solution number one is to add top steel which is non-pre-stressed and we will calculate at the transfer length of the station corresponding to transfer length of the beam because that is where the highest stress occurs and the least amount of self-weight moment is acting. So that is if we satisfy that section then we'll satisfy any other section towards the center of the beam because self-weight of the beam is helping me out. So the stress is induced by self-weight of the beam at that station and notice the X equals the transfer length of 2.18 feet from the end of the beam. The top and bottom fiber stresses, pardon me, these are the, yeah, these are induced by self-weight. Top and bottom are 0.076 KSI and we superimpose those stresses to the top and bottom stresses induced by the pre-stressing force at 2.2 feet from the end of the beam. So we had a tension stress recall of 1.1 KSI negative. We add a compressive stress induced by self-weight. So that helps me decrease the top fiber stress to 1 KSI instead of 1.1 and the bottom stresses to 2.7 KSI. So this is essentially the elastic stress distribution at the section corresponding to the end of the transfer length negative 1.05 KSI on the top and positive part, yeah, negative being tension and positive meaning compression at the bottom of 2.71 KSI. So if we use a linear distribution of stresses, we can find the point where the stresses go from tension to compression, the point of zero stress, that's 7.82 inches. And the philosophy behind designing this non-pre-stressed top steel is that we need to provide rebar to develop this entire force in tension. So that's what this says, ACI 318-14 in this section specifies that the top steel to control those top fiber stresses need to resist the entire tensile force. So we need to integrate or find the force resultant of that triangle. And the other limit is that the steel is designed at 0.6 FY or 30 KSI, not at full FY. Now the PCI 8th edition handbook allows you to use FY up to 60 KSI where the top of the beam will be in compression under service loads and where the top of the beam will not be exposed to weather. But if we go back and remember our service load check, we found that the top of our beam was always in tension towards the end. So the PCI handbook exception does not apply in our case. So we'll have to go by the stresses that are allowed in ACI. So how do we get the total tension force that we must resist? So we calculate the area of this triangle and then multiply it times the width of the beam. So the area of that triangle is one half times the top stress, times the depth of the triangle and times the width of the beam. That gives me a total tension force of 49 kips that my mild reinforcement must carry at a stress of 60% of FY. That's the FY for mild grade 60 reinforcement, 36 KSI or 30 KSI, whichever one is the minimum. So we have to use 30, divide the force by the allowable stress. I need 1.6 square inches of steel. And you can imagine, you know, in order to cover this much steel, these are possible bar combinations that we need. So it's a lot of steel, top steel to resist this top fiber stress of 1.05 KSI. Furthermore, even though there is still there to carry the force, there are going to be possible problems with congestion, even though this is an allowable solution because the amount of steel that's needed is significant. We also, since we are trying to carry or take care of top fiber stresses that are near the end of the beam, the bars need to be fully developed. So we need to either hook them at the end or provide some other means of anchoring them and having them developed at the section I need them. And still the beam is going to exhibit cracking. There are going to be cracks. The reinforcement there is going to be there to arrest those cracks, but the crack widths may not be very appealing. So regardless, everything else checks. The beam with top steel now meets the requirements at release. Since we changed from F prime CI from 4,000 PSI to 4,500, this should say PSI, not KSI, we would need to repeat some of the last calculations. Now, the only thing that's affected by F prime CI is elastic shortening, because you might recall that the modules of elasticity of concrete uses F prime CI in elastic shortening calculations. Relaxation is also affected because elastic shortening is affected, but shrinkage and creep losses are not affected. So once we change the release strength, academically, one should go back and verify losses. You'll see that the differences are minimal. We have a total loss of 44.8 compared with 45.6 KSI. As the release strength increases, the losses are less because there's less elastic shortening. So our total loss now is 22.1% instead of 22.5, and we have a higher effective pre-stressing force. Instead of 240, we now have 245. The change, frankly, is insignificant. So we can just move on. And after adjusting, using the new concrete strength and loss calculations, these are the plots I've been showing you all night. So these are the stresses at the bottom generated along the bottom of the beam for service conditions. We need to adjust them. So again, we check bottom fiber stresses. They are compressive at the end. They're in compression until we reach 10 feet into the beam, and then they become tensile for all loads and then sustained loads. And again, the allowable limits, both for tension stresses and compression stresses, this is all fine. And the top fiber stresses, again, for allowable compression stresses, we satisfy them and recall that the only limit that we already have taken care of, these top fiber stresses through the presence of mild top reinforcement that's not shown here in this cross-sectional view, but these top fiber stresses are going to be not under service condition, but they're more critical for the release condition. So we'll take care of them using top reinforcement. So that seems to be the end, but as I mentioned, what I'd like to do is I'd like to repeat some of these calculations for the case of harped strand configuration. But before we do that, I would like to ask Sherry to launch the third poll of the night. Okay, we will be launching the third poll. Before each session, I have gone back to review slides or concepts that were not clear to me during the presentation. Please answer on the screen now. I will be closing the poll in a few seconds. If you have not voted, please vote now. I am closing the poll. Sergio, 47% said yes, 21% no, 5% have not had to review. I remember all that was covered and 26% not for all of the sessions. Okay, great. So it's a pretty fairly uniform distribution. I mean, I'm glad that the majority of you are going back and looking at the slides so that you have questions from previous sessions. But the other thing that's important is just to keep the, it's important to remember what we are doing, right? What we're designing and to keep in mind the problem we're trying to solve so that when we do some of these calculations, one could get lost with the numbers if we don't remember what it is that we're designing. So hopefully, you've got the time to go back and review or at least just refresh your memory of what we're doing. All right, so now we're going to try, as I mentioned earlier, we're going to try to work with a harp strand solution. And remember that the critical aspect was those stresses at release. So that's what we're going to do. We're not going to go back and revise everything based on service load conditions, but we're going to check the release condition and then go back and plot service load stresses once we have determined the strand pattern. Okay, so this plot is the same as we had before. It's just repeated here for your convenience. These are the checks for bottom stresses at release. You recall that before we increased F prime CI to 4,500 PSI, we had problems along the length of the beam, along the bottom, compressive stress problems. And primarily we had tension stress problems. So we had to do something to deal with these problems. So one possibility, this is repetitive. If we had decided to use the PCI stress limits instead of the ACI stress limits. So this slide, I'm going to skip. And we also discussed that by increasing the release strength to 4,500 PSI, this would take care of perhaps, take care of the compression stresses exceeding the limits. So we're going to instead, now we're going to concentrate on changing the strand pattern. So this is our initial straight strand pattern with 10 strands. Their centroid is at three inches from the bottom with an eccentricity of 11 inches from the center of the beam. And let's say, and this could be done by trial and error. So for this case, let's say that we choose to leave four strands at the bottom, along the bottom of the beam with two layers of three strands in the center of the beam. So let's use this new pattern. What we're doing here is we're going to decrease the eccentricity at the center of the beam from 11 inches to 10.2 inches. So we designed the 10 strands for an eccentricity of 11. If we decrease that, then we're decreasing the effectiveness of the section, right, because we have a smaller eccentricity. But I don't know if you remember, we needed an effective prestressing force that was lower than what we ended up with because we rounded up to 10 strands. So we need to check the service condition at the center of the beam to make sure that with this reduced eccentricity, we're still okay. So let's, we need to check the service condition and we'll do that after we check the release condition. So we have our 10 strands now at, and we're using a 10% loss because we're checking release. So we have 279 kips of prestressing force. Notice now, however, that we are, if we maintain this pattern, straight strand pattern, pardon me, the eccentricity stays at 10.2 inches throughout the length of the beam. So what we're doing is first exploring, just changing the strand pattern and see whether that helps me with the compression stresses and attention stresses. So the prestressing force bottom stresses with this new strand pattern that it's constant throughout the length of the beam creates a bottom stress of 2.6 KSI compression and 0.99 KSI tension along the top of the beam. So notice that the top fiber stress is still very high throughout the length of the beam induced by prestressing. So if we plot again, the bottom fiber stresses, we could say that we're close here, but we're still violating the allowable compression stresses here in these regions. By simply changing to this pattern, we almost satisfy the compression stresses, but we clearly don't satisfy tensile stresses. So we could have to, we have to do something else. So we're still, in terms of tension stresses, we're still way above the limits. So, and the compression stresses are near the end of the beams are still over the limit. So we need to do something else than just changing the strand pattern. So let's try harping now. And by harping, we're going to decrease the eccentricity only at the end of the beams, at the beam, at the ends of the beam. So we're going to go from a constant strand pattern that looks like that in the center. So these three lines, the two red and one orange correspond to the different layers of a strand. And at the end of the beam, the strand pattern is going to look like this. The top three strands are going to be moved towards the top of the beam. And we're going to stop them two inches from the top face of the beam. So the eccentricity at the center is going to be 10.2. The eccentricity of the strands is now going to be only 4.2 inches at the end of the beam, given this strand pattern. And we're going to choose the harping point to be at 0.4 of L, where L is the span length of the beam. All right. So the first thing is that we need to look at the geometry of the harped strand. First is 0.4 L, that's 0.4 times 30 feet. And since we're drawing the total length of the beam, we are going to add the one foot to take us from the center line of the support to the end of the beam. So the horizontal distance of 0.4 L is 0.4 times 30 plus one foot, that's 13 feet horizontally. That's how much these harped strands travel horizontally. And vertically, they travel 26 inches minus six inches at this section. So where does the 26 inch come from? From this pattern, we have 28 inches minus two inches. That's where they are now at the top. And their original placement was six inches from the bottom. So the vertical amount that these go up is 26 minus six. Divide that by the horizontal distance they travel. We get an angle theta that is defined by the inverse tangent of these rise over run. So we have 7.3 degrees is what these harped strands are defining. So we'll need this dimension to determine horizontal and vertical force components of these harped strands, the three harped strands. So the easiest way to deal with the harped strand pattern is to separate those strands that are harped from those that are straight. So we have, in all our calculations, we can separate first the straight strands. We have seven strands that are straight, and we only have three strands that are harped. That's their cross-sectional area. They are all subjected to the same initial pull. They all are subjected to the same loss of 10% at release. And notice here that the horizontal component of the harped strand at the end of the beam is defined by using the cosine of that angle we just calculated, the 7.3 degrees. So the horizontal component of the harped strand is 83 kips. The straight strands give me 195 kips of pre-stressing force. The eccentricity of the straight strand is 11.14 inches. Let's go back a couple of slides. That's for this pattern right here. No, pardon me, for the pattern of these three strands and four strands at the bottom. That gives me 11.14. The eccentricity of the harped strand, let's look at them at the center between 13 feet and 19 feet. So that means that we're between this section and that section. The eccentricity only of the harped strand, the three harped strand is eight inches because they are 14 inches minus six inches. So the eccentricity of these three strands is eight inches measured positive below the neutral axis. And then a little bit more explanation is needed for the eccentricity of the harped strand at the end of the beam. So for sections X less than 13 feet. So within the sloping region of the harping, the harped strand. So the eccentricity varies point by point. And the way it varies is we start with an eccentricity of negative 12 inches. That is 14 inches minus two and negative because it's above the neutral axis. And then for any X, the vertical distance that the strands travel are any X distance times the tangent of 7.3 degrees. So we can use X as any variable. So one, two, three, four, five feet and determine point by point how the harped strand eccentricity changes from zero to 13 feet. And it is constant from 13 to 19 feet. And we also have a constant eccentricity of a straight strand throughout the length of the beam at 11.14 inches. So we then again use the pre-stressing formula, the general pre-stressing formula where we separate the harped and the straight strand components. We have P over A. So these can be grouped together because P over A doesn't change. And then we have to separate the PEY over I or PEC over I terms from straight strand and harped strand. We have to separate them for the condition that the harped strand eccentricity is changing with distance. And then we add to these three components representing the effect of the pre-stressing force. We add to that the superimposed moments from dead load, superimposed dead loads and superimposed light loads. So this presents how one can calculate stresses with when we have a harped strand configuration. It is not tremendously difficult. It's just a little bit of bookkeeping and being careful about signs and eccentricities and so on. So let's look, say for example, at the section corresponding to the transfer length, 2.18 feet from the end of the beam. As I mentioned, the eccentricity of the harped strand changes point by point. So we started negative 12 inches. We have traveled 2.2 feet in or 2.18 feet in times the tangent of 7.3 degrees. At the end of the transfer length, we have an eccentricity of only the harped strands of negative 8.65 inches. We then plug in all the corresponding numbers into this equation. We have P harped, P straight, A of the section. P straight, E straight, H over two of the section divided by I of the section, P harped. E harped is number of negative 8.65, 14 inches. Again, to the top or bottom of the section, I of the section. And we have the moment induced by self-weight at the section corresponding to 2.2 feet in from the end of the beam. And that we end up with negative 0.014 KSI. The negative still corresponds to attention stress, but notice now that it is a much lower stress than we had before. We had close to one KSI. So would this harping pattern be enough to keep the bottom compressive stresses below the ACI stress limits? So this plot represents that. These are the allowables. And the reason we have now a step, kind of a tent-like function plotting compression stresses is that we first start transferring the stress, the bottom, the pre-stressing for stress. But at the same time, the eccentricity of the harp strand is varied. So we start now, the weight of the beam starts helping me. And we have these changes, the shifts, because these points here correspond to the point where we started harping strands towards the end of the beam. This is the other point on the right of the beam where we started harping strands. So the fact that we're harping the strands helps me decrease the compression stresses along the bottom of the beam. Otherwise they would rise again and exceed the limits as we had before. Okay? In terms of top fiber stresses at release, these are the allowables, the 0.6 root F prime C near the end, the 0.3 root F prime C near the middle. And these are the top fiber stresses now. They still exceed the allowables. And the problem is that we were exceeding them by a lot. Harping did help me. We decreased the top fiber stresses. We now essentially don't have the problem at the time of release at the ends of the beam, but still we have not taken care of the center of the beam. And you might recall that we also had problems of exceeding the top fiber stresses near the center of the beam. So this should not surprise us. We haven't done much at the center of the beam. Harping has taken care of top fiber stresses at the end of the beam, but not near the center. So we have simply shifted the problem towards the center, which is not a bad situation to have because as I mentioned earlier, if we decide to add top reinforcing bars, we can now start developing them at this section where we are going to need them. We have a little bit more space to develop them to the left and to the right of that section. So we have dealt with bottom compressions along the length of the beam, but we can again focus on the tension at the top of the beam at release. We could fix this again by adding mild steel to the top of the cross section. And remember now that we're dealing with the top of the beam near the center of the beam and the center portion of the beam. We need to design that top reinforcement to carry the entire tension force. And again, that top reinforcement should not be at a stress that's higher than 0.6 FY or 30 KSI. We've calculated these limits before, but now the conditions are different. Notice now that the top tension stresses are most severe at a section between 13 feet and 19 feet. That's between this point and that point where essentially we have the harp strand near the bottom. This is where the highest top fiber stresses are occurring. So let's calculate at that section what those stresses are so that we can design the top reinforcement to carry those stresses. So if we do that, we end up with this stress distribution. I don't think we need to calculate it again. This comes out from taking into account the beam self-weight at a station equal to 13 feet. As we have done before, we did it for 2.2 feet. We now do it at 13 feet or 19 feet. It should give us the same stress. And we also consider the top and bottom fiber stresses induced by the pre-stressing force. In that particular case, the pre-stressing force is at an eccentricity corresponding to this strand pattern because we're near the center of the beam. So this is the resulting stress distribution. We have a top fiber stress of negative 0.65, much lower than before, right? Of the negative one KSI that we had near the ends. We've now reduced it at the center to 0.65 negative. So tension compressive of 2.3. We now have a new crossing between the tension stress and the compressive stress. And we can then calculate the total tension force that we must use to design the top reinforcement by using, well, first of all, the 6.15 is, the calculation for it is shown here. It's just adding that stress, the tension stress at the top, plus the compression stress at the bottom and dividing by the height of the beam. That's the slope of this line. And then we use that slope to divide the 0.65 and that gives me 6.15 inches from the top. The total top tension force is again calculated using this triangle now, 0.65 times 6.15 times one half for the triangle times 12 inches. That's the width of the beam. We now have to design the top reinforcement for 24 kips at a stress of 30 KSI. We now only, so to speak only, need to provide 0.8 square inches of mild reinforcement. And these are the possible combinations that one can use. All right, so eight number threes, four number fours, or three number fives. Three number fives seems to be, between four number fours and three number fives seems to be the best solutions in my mind. So again, the same problems we had before, congestion. We have to develop the bars at the section that's critical. So at a section of 13 feet and 19 feet on both sides of that section. And since this is crack control reinforcement, the beam will still crack at release, but at least the top reinforcement will be there to keep those cracks tight. And eventually if this beam carries a topping or has this hollow core that's sitting on top of it, those cracks are not going to be visible in the finished structure. Okay, if we were allowed to use the PCI higher stress of 60 KSI, we would still end up with the necessity of having 0.4 square inches. And that's the combinations that we would have to take those, or to develop the beam the total force that's needed, the 24 kips that we need. So this is if the PCI design handbook limit of 60 KSI were used. Okay, so since we, and again, I don't want to go back and show you the detailed calculations for all the sources of losses, but we changed the strand pattern, right? We changed it from two layers of five strands to now one four strand layer and two, three, three strand layers. We need to recalculate losses again because the eccentricities have changed. And furthermore, there's, so the effective prestressing stress at the end of the beam is different. So we have to go back and calculate elastic shrink, elastic shortening losses, creep and relaxation that are affected by the change in strand pattern and shrinkage is not affected. So if you go back and look at the detailed calculations for shrinkage, elastic shortening, creep and relaxation, you'll see why these three sources of losses change because of the strand pattern, but they don't change. It doesn't change for shrinkage. So notice again that the change in losses is not significant. We originally had for the straight strand case, we had 16.6 KSI for elastic shortening with a harp strand configuration, we now have 15 and also similarly for creep and relaxation. So harping the strand helps us decrease the total loss. We had 45.6 KSI and we end up with 41.9. And with that, we have now, instead of a 22.5% loss, we have a 20.7% loss and the effective prestressing stress is now 160. So it's higher than before. The effective force is 240, effective prestressing force after losses is 246 and we needed at least 240. We needed less than 240. So we have more than that. So we now have an effective prestressing force that's higher. So we need to go back and readjust the service stress plots to make sure that we now have not exceeded any limits with these new losses and with the strand pattern that's now harped. And this is the summary. We won't go through any calculations. The calculations, the philosophy behind them is the same. And these are the plots now that should be familiar to you. These are stresses. These are now service conditions that we need to check after we have satisfied release. Bottom stresses at service conditions. Again, for sustained loads only and for all loads. And these are the three limits, tension and two compressions. All of our stresses throughout the length of the beam are okay. We're getting close here near the center, but we're still okay. And top fiber stresses at service conditions, they're all fine. These are the allowable compressions for total loads and for sustained loads only. So the top stresses for all loads are here, well below the allowable. And the same thing happens with sustained loads. And now that's it. These are the differences between a straight strand pattern and a half strand pattern. And I guess I went fast tonight. So we might be able to finish early tonight. If there are no questions. If you have questions, again, please type them into the chat box and I will pass them on to Sergio. As usual, if you don't think of a question tonight, you can always send them to me via email and I will pass them on to Sergio and get them to you, the answers to you. But again, please type your questions into the chat box. Sergio, I don't have any questions. Good. Probably, it was not too complicated. Oh, sorry,ads. Okay, they're all there. Stop. Yes, we don't. So far, no questions. So we can end tonight if you would like. And again, if they think of questions during the week, I can send them to you and then send the answers to the person who's asking. Okay. Okay, well, I would like to thank you all for attending tonight's session. Again, continuing education certificates will appear on your account at www.rcep.net within 10 to 12 days. Remember, continuing education credit is given for each session. You must take and pass the exam with a 70% to receive the corresponding credit. If you have any questions, please email me at snodnetpci.org with the subject line online academy. You will receive an email link to the course exam. The exam will be available for the next 72 hours. So please take it within that timeframe. If you are unable to take the exam within that timeframe, please contact me immediately at my email again, snodnetpci.org. Or if you have any problems with the link or do not receive the email within 24 hours, again, email me again. It's been a pleasure and a good night and see you next Thursday for session five. See you in a week.

harping

strand pattern

pre-stressed concrete beam

stress distribution

compression stresses

tension stresses

top steel

reinforcement design

losses calculation

service load conditions

structural integrity