of six sessions that we're going to be talking about advanced pre-stress concrete and tonight we have a fairly busy session. We're going to be talking, so I'll get moving right away, we're going to be talking tonight about bearing on concrete and adapt and design, those two general topics. And my goal today is I'm going to be presenting you some of the background information for these two topics and towards the end of the session is where I will include the calculations that are pertained to these two topics, going back to our design example. So most of today's session will be on presentation of background information and if we run short in time I think that the calculations are self-explanatory and you can follow along. My goal is to give you enough background so that you can understand those calculations. So without more than this is what you should expect from this session. The first part is to be able to check for bearing strength of unreinforced and reinforced bearing conditions. The second part, which is the longer part, is to understand what the failure mechanisms and adaptants are, what the research, there has been PCI funded research involving adapt and design and I will be presenting you with that research and the most relevant findings and along with that how does that research affect or how that influenced the equations and the methodologies that's currently in the PCI design handbook. As part of that we'll talk about load transfer characteristics and also some detailing requirements of adaptants. So it's a lot to cover as I said. So the first thing that in this session is talking about bearing on concrete. This is the general equation that's given in the PCI design handbook for bearing. Bearing of concrete or bearing on concrete as many of you know is the result of two surfaces becoming in contact and that's very common in precast concrete components because one component say a double T or a floor element may be resting on top of a column through a bearing plate and that generates bearing stresses on the component that then are transferred into the structural element through bearing. So in this general equation the applied load or the contact load is shown here as phi V sub N in general for a shear force but it's really not shear, it's a contact force and the resistance against bearing is given by the equation on the right. I'll explain this term A2 over A1 in the next slide but the bearing strength is given as a kind of a stress so the bearing strength is reached when concrete reaches a stress of 0.85 F prime C bearing on the surface A1. A1 is the contact surface. The C sub R factor is a factor that is as bearing is applied or as a load is applied on the surface of concrete it transfers into a concrete through an assumed model is a truncated frustum so a truncated pyramid or cone and this equation is based on the confinement that is provided by concrete adjacent to the bearing surface. So this magnification factor here A2 over A1 provides a magnification of the bearing surface into the concrete in which the element is bearing and C sub R is a reduction factor that accounts for normal stresses perpendicular to the direction of bearing. This factor was determined through testing back in the 1960s so you'll see the equations that refer to this factor but keep in mind that it's a reduction factor intended to capture if there's some tension stresses perpendicular to the bearing to the direction in which bearing is being transferred. For all bearing calculations PCI and ACI uses a phi factor of 0.65. So this is the equation that's used for plain concrete bearing surfaces and surface the best way to explain where surface to is looking at a figure taken from the PCI design handbook and it's a figure 551 says this is the surface how the frustum surface is applied. The hatch region here is the actual bearing surface of width W and length B and it is assumed that the surrounding the area 2 is drawn as you know projecting that the contact surface downward into the supporting member at a slope of 2 horizontal to 1 vertical and whenever that surface reaches a free surface like on the right left edge here in this figure the level at which it it reaches defines it reaches a free surface defines the area A2. So the frustum in this case is stopped right at the left edge here and the rest is drawn based on the assumption of a spread of 2 to 1. Now one thing that is important to remember is that this 2 to 1 ratio does not represent that the necessarily the the spread of stresses inward into the concrete but rather it's just a factor that is used to define this second area that's used to to take into account the confinement that is concrete around the bearing surface is providing to the bearing. And note here that V sub U is the applied force and a vertical force normal to the surface and N sub U is parallel to the surface. S is the distance from the edge surface that for which the bearing reaches concrete contact to measure to the centerline of the bearing where the load is applied so S is that and W is the dimension parallel to that dimension on the bearing surface. C sub R as I mentioned is a reduction factor that accounts for the presence of that normal force that would create tension stresses perpendicular or to the to the bearing direction so that would have the the effect of reducing the bearing strength. So when we we will see later in the calculations that this factor often becomes less than 1. So in the case where there's no normal force so no horizontal force or force parallel to the plane of bearing this factor would become zero as would be this coefficient or exponent would become zero right so it would be C sub R would become 1. So when there is no horizontal force C sub R is 1 and therefore there's no reduction in in the bearing strength. So again the notation is given here I mentioned it earlier I don't think that we need to repeat it now and N sub U which is the horizontal force is always very often taken as 20% of the factored permanent loads on the bearing. So if in this case this is a the general case of bearing so the for thin stem members which is what we're going to be focusing this this class on most of the session there is a requirement in the PCI 8th edition handbook that thin stem components having a bearing area less than 20 square inches will require a minimum amount of reinforcement to equal to this quantity here but not less than 1 number 3 as a recommendation. We'll see how that applies to thin stem members. Okay so if let's say the reaction force these acting force is greater than the strength provided by the previous equation for plain concrete then reinforcement will be needed to arrest two potential cracks that would form from bearing and I will show you a picture of a figure of where those cracks form. Further if these of you so the acting force exceeds 1.1 f prime CA1 which is the upper limit for equation 561 then confinement reinforcement is additional needed in all directions so as to prevent the concrete from splitting vertically in the direction of the bearing force. I will mention in the next slide the type of reinforcement that's needed but there are two important cracks that may form from bearing one is a nearly vertical crack and reinforcement is needed to resist or to provide to control the whitening of that vertical crack that reinforcement is going to be designed based on shear friction theory and then there's another crack that may form almost parallel or perpendicular to the shear friction crack that will also have to be accounted for and this crack almost propagates to a horizontally towards the support. We'll see that this crack is arrested by bars that I will show you in the next slide denoted as A sub SH. So the two cracks that I'm mentioning imagine that there's a side this is a side elevation of a double T this is one stem the reaction force is shown here as V sub U the width the length of the bearing is W so the vertical crack is shown here and this horizontal bar this diagonal bar pardon me this diagonal bar is the bar that is used for to based on shear friction theory that also has to have a component to resist this normal force here that's acting at the bearing. So this let's call it shear friction reinforcement at the end there might there's a possibility that this other crack will form that will turn horizontal as shown here and to resist to restrain growth of this crack A sub SH has to be provided in a reinforced concrete bearing. The plan view of this A sub of this shear friction reinforcement is shown here it has to be anchored well enough on both sides of the crack so that it develops yield. So a view a possible way of anchoring it is by forming a C bar in plan view where B is the width of the stem of this thin stem member. So before I go further I'd like to ask Sherry to open up the first poll of tonight's session related to shear friction theory. Hello I am launching the poll now it is on your screen so please answer the poll as quickly as possible. I will be keeping the poll open for just a few more minutes. Okay Sergio 47% said yes and 53% said no. Okay well great thank you Sherry. So it's about 50% of the people that know about shear friction theory it's a it's a relatively simple concept. So going back one slide again this vertical crack that forms imagine this force pushing the left-hand side up so there would be and there are loads acting on this double T pushing the beam downward so there's a possibility that this vertical crack forms creating a near pure shear condition that's parallel to this vertical crack. So a view of what the concept of shear friction involves is this is equivalent to that vertical force I was showing you it's except it's shown here horizontally and there's a potential this this diagonal or this undulating crack represents the cracking plane. So as you can imagine in this you know this is a concrete piece that with the application of the shear forces that are in equilibrium to the right and to the left that crack may displace laterally and because we are dealing with rough surfaces it's not a perfectly smooth crack there is going to be a relative displacement or vertical displacement between these concrete blocks that that start separating from each other. So the purpose of the function of this shear friction reinforcement is to keep this crack closed as best as possible based on shear friction theory and you might remember from physics that shear the shear force developed in a in a surface is a function of the coefficient of friction times the normal force applied to that surface. If we equate that normal force to the maximum force that can be developed through across the crack as a total area of shear friction reinforcement times the yield stress then we have the shear friction equation that is listed in given in many in many codes including in the ACI code and also the PCI design handbook. So this is a basis for shear friction theory it's fairly simple the shear friction reinforcement is there to provide the normal force that can then allow these two blocks of concrete to develop the shear strength that is required to resist the applied forces horizontal forces. So if we then do that we use that so the reinforcement in bearing regions near such as a support will be needed is if the applied force exceeds the bearing strength for plain concrete and the reinforcement that will be needed will be part of it will be shear friction reinforcement and that is obtained this equation 533 is obtained by solving from the previous slide the value of V sub U equals phi FY mu A V sub F where mu is a coefficient of friction or so PCI adds another equation that uses an effective coefficient of friction instead of just the this mu coefficient of friction for the effective coefficient of friction PCI the PCI handbook provides a table 5 3 1 and I'll show you what that table says in that next slide to for values of mu and mu sub E and since this is a shear problem the phi factor is 0.75 mu sub E is calculated given this equation the effective coefficient of friction where A sub CAR is the plane that is the planing of concrete that is displacing so it's the shearing surface at the crack interface in square inches additionally the additional reinforcement that's needed to complement that reinforcement crossing the vertical crack is if there's any normal force to the crack that is that results in additional reinforcement that needs to be normal to that vertical crack calculated by equating simply the normal force to the maximum force that reinforcement can can resist notice here that for this equation phi would be 0.9 when it generates tension perpendicular to the crack so a crack wanting to open up by the application of N and except for brackets and corbels and we'll cover this in a future module where phi will be still 0.75 so looking at table 5 3 1 from the PCI handbook you see columns here for different conditions so I'm just going to say here concrete to concrete cast monolithically there's a potential of a the potential of a vertical or a crack forming the coefficient of friction is 1.4 all of these lambda factors are for lightweight aggregate concrete so it's 1.0 for normal weight concrete and you I referred you to the section in the handbook for other for lambda values for lightweight aggregate concrete and you'll note that I'm not explaining too much what this mu sub e represents these effective coefficient of friction other than referring you to the original paper in 1977 which talks about this coefficient of friction that was determined from from testing for concrete corbels the other thing to also remember is that there is a maximum force that can be transferred across the crack so one could have the inclination to add a lot of reinforcement to increase the shear strength of the shear friction strength but there's a maximum and that maximum refers on the capacity of concrete to just break through the aggregate say and eliminate those corrugations or that unevenness in the surface so these are the maximum values that we can use to compute v sub n and if these v sub n or the v sub u the v sub n required based on v sub u exceeds these limits then we have to resize the element or increase the surface for shear friction this is not common in the case of double t steps okay so the other component of this bearing problem is designing this vertical reinforcement a sub sh that will then control this horizontal crack that that forms outside of the determination point of the shear friction reinforcement an ace of sh is it's essentially taken imagine this vertical force this reinforcement has to provide a downward vertical force a total downward vertical force that resists this v sub u so a sub sh is is given by by this quantity here so this also has the the amount of reinforcement this reinforcement here has a vertical component as well that needs to be set in equilibrium with this vertical component of a sub sh so the way to calculate a sub sh is adding the two reinforcements or the air the total area of reinforcement crossing this vertical crack times fy to make it a force and then we divide by the coefficient effective coefficient of friction and the yield stress for this vertical reinforcement so that's how a sub sh is is determined and the to determine the effective coefficient of friction it's the same formula as before instead of a CR we use B times H where B is the width of the stem and H is the depth of the stem so it is using the total contact surface the one note the note here indicates that this vertical reinforcement seems like shear reinforcement or reinforcement from diagonal tension this reinforcement can be used for to resist diagonal tension as well so it's not additive to the requirement for shear however it will be additive for the requirement for torsion so that is all in bearing it's fairly simple first we check whether the plane bearing can resist the applied force if it doesn't then we provide reinforcement okay so now moving on to DAP design this just a brief introduction you probably all know what a DAP means so when you have a beam that ends in a DAP region you have to examine several potential failure modes and from a theoretical point of view a DAP and or from beam theory a DAP then constitutes what's referred to as a discontinuity region where beam theory so plane sections remaining plane is not applicable so so that is the the key for DAP design that it's you we cannot treat it as a beam in that the end region of the beam cannot be treating treated as plane the plane section remaining plane assumption so if there are one we will see in a few slides certain parameters geometric parameters that are within the PCI design handbook limitations of using the DAP design procedure but but one key aspect is that if you have a DAP that falls outside of the categories that are in the PCI design handbook you can still use strut and tie models to design the DAP region of a beam and I will talk about those more in more detail for corbel design but now I just wanted to show you a typical strut and tie model that was used for many years you know back in 1987 this model was proposed by Schleich and others where you can see the flow of forces you know a shear force and a bending moment moment acting at the right-hand side of this beam and when you get to the DAP region this is a fairly deep DAP you see that these diagonal compressive stresses have to make their way up to the top of the beam before they get to the reaction point so the key is in detailing this region to provide the enough reinforcement and not only enough but with adequate adequate detailing to develop these forces that are illustrated by solid lines these tension forces that are illustrated by solid lines in this strut and tie model so as you can see this vertical tie is going to be important to define the area needed in this vertical reinforcement this horizontal tie is needed for this horizontal reinforcement properly anchored beyond the node and so on so this is one model of a DAP design resulting in this type of reinforcement and there are others so there's this note that that was written in the original paper by Schleich which is going to be important for us when we discuss the research that was recently conducted that it says and I'm just going to quote it is not sufficient to hang the load from the bottom using tie 1 in addition to shear reinforcement for you know tie 2 must be added to permit so this would be tie 2 to permit anchoring this reinforcement that is needed to resist the horizontal force that may be created here at the node so you have to not only hang the reinforcement or provide hanger reinforcement but also this other reinforcement here to provide anchorage of this tie so this was recognized back in 87 already so prior to showing you some of the research results I'm going to just show you this figure that sets some nomenclature here the first thing is that definition of be a D region in the case of DAPs the D region or it's also the discontinuity region is defined as the distance from the end of the DAP to a distance 2h from the end of the DAP inward towards the beam the B region which is we follow normal bending theory for design of it is is shown here in gray that's the beginning of the B region and the reinforcement that's highlighted here is a sub sh that turns and becomes a sub sh prime this is that vertical reinforcement that would be used for a hanger reinforcement a sub h is horizontal reinforcement within the region of this DAP that is referred to as the nib height or the extended region that those are two common notations a sub s is that horizontal bar that's needed to to anchor or to resist n sub u but also to provide anchorage of that strut that forms here and I think that's it so a sub sh AS and a AH are the reinforcement ratio rate areas here there's a question that I just received the question relates to the coefficient of friction mu it says why mu why is it mu 1.4 lambda for concrete to concrete but not 1.0 lambda is there an equation to check the shear between concrete to hardened concrete I'm assuming that the friction coefficient should be less than 1 probably 0.5 or 0.3 to calculate the shear so a good reference for this check so if you look at that table I went through it very quickly I'm gonna go back see just to answer that question quickly the second row 1.4 is concrete to concrete 1.0 is hardened concrete to hardened concrete the the key here is that the coefficient of friction is not necessarily only a coefficient of friction per se coefficient of friction is between two smooth surfaces or relatively smooth surfaces this coefficient so to speak also capture other resisting captures other resisting mechanisms such as the aggregate resistance aggregate interlock and dowel action of reinforcement passing through through the shear friction crack that is why it's 1.4 and keep in mind that this crack has to first form before it becomes a shearing plane so that table actually gives you sorry it's Siri that started talking here I'm speaking too loud I guess that table gives you different values of for the coefficient of friction so going back to where we were which is here these are some definitions for DAP design okay so back to the research there is the current provisions in in PC the PCI design handbook is based on research that was conducted by Klein and others recently finished in 2015 where they looked at because of observed distress in the field they looked at several configurations of that reinforcement and tried to determine what was a good configuration that would resist the loads that were that were being applied so they examined at least conducted research on or the research involved configurations such as the vertical L shape shown here the vertical Z shape so this is the main reinforcement right in the form of Z they called it this reinforcement and inclined L and this reinforcement and inclined C we'll see later that more other patterns were also checked and tested but these are some of the reinforcing patterns that were initially examined by the researchers and so so we'll come back to the research in a second when we look at the PCI design handbook which I the provisions which I said use this research there are two categories of adaptant for DAPN design wide stems or referred to as non-thin stems those those provisions existed in the PCI design handbook in the 7th edition and before and they correspond to double T's with stems exceeding 8 inch width the research that was conducted was to include thin stems in DAPN design and this is brand new for the 8th edition handbook and it includes stems with widths less than 8 inches and it's as I mentioned based on the research by Klein and others and as I've alluded to the method for for thin stem design our DAPN design is based on extensive tests and also finite limit simulations so numerical studies and it was prompted by cracking that was observed in DAPNs in some structures some highlights and I'll go into details of the research in the next few slides or in a few slides ahead the first was that they conducted an industry survey to identify typical reinforcing patterns they tested more than the four DAP reinforcing schemes I showed you earlier they tested six reinforcing DAP reinforcing schemes for that design based on industry and past research they looked at two different strand patterns either placed in a single column in a stem or staggered location of strands DAP height was the DAP height was limited to 50% of the overall height of the stem and they also looked at the horizontal tail of hanger reinforcement what what it did you know how the the horizontal portion how far did it need to extend into the beam so just to illustrate some of the configuration configurations the three-dimensional view I showed you earlier this is the side views and the section view of the vertical reinforcing scheme vertical L vertical Z inclined L on the left bottom right bottom C these are you know these are reinforcing patterns that were tested and the last two was using welded wire reinforcement inside of the DAP just sizing it and see how that performed and they also used this kind of innovative pattern which is a combined C and Z so they called it the vertical CZ reinforcing scheme so if you I'm hoping you have in front of you your those two last slides because the next poll that I'm going to ask a Sherry to open is asking you in your experience what have you most commonly seen of these reinforcing patterns of the six reinforcing are the the reinforcing patterns that will appear in the poll I am launching the poll now please answer on the screen I will be giving you a few more minutes so please quickly read the question and answer it as soon as possible thank you I'm going to give you another few seconds if you have not voted please vote now I am closing the poll Sergio 31% said vertical L 13% said vertical Z 44% said C shape shape 6% said WW our reinforcement and 6% said vertical CZ okay great thank you Sherry um well it's interesting that so the majority said you're familiar with the C shape as you'll see from the research we'll see how that performed so let me just show you some pictures before you know with the different reinforcing patterns so you can see that the vertical L so imagine the reinforcement here in the stem going upward despite having that reinforcement there were still diagonal cracks that form and there was this large diagonal crack here I don't know exactly where that extension ended the horizontal bar ended but interesting to see these cracks forming the inclined L reinforcement if you if you imagine it being here and then bending upward towards the top of the nib I also experienced some of those diagonal cracks what's interesting in the C shape reinforcement if you imagine that reinforcement going down this way coming up and then bending back into the beam you see that the crack that forms in the reentrant corner at the intersection of the DAP and nib diagonal crack can form and actually bypass that C reinforcement and just work its way around it and just form near the top flange and the vertical Z reinforcement so this would be reinforcement it goes like this and down again forms those diagonal cracks where there was a splice here of longitude that Z reinforcement with a longitudinal reinforcement or prestressing strand in this T section so some of the research results these are some of the results and I thought these figures were really telling and interesting of the different reinforcing patterns that were tested you'll notice that the purple line is the welded wire reinforcement a result this is maximum capacity and deformation capacity all of them made it to the factored load but they were not or the design load but but they were not as strong as some of them had a significant over strength and better behavior in particular let's look at the purple is the worst performing so to speak the vertical is next and then the green line represents the C shape which didn't perform as well either the best performance where the CZ shape the vertical Z with a corner angle and the inclined L and another view of this is the the maximum reaction that was achieved by the different reinforcing patterns again this is welded wire reinforcement and all the different reinforcing patterns shown here you know in this bar chart so it's interesting to see how influential the reinforcing pattern is on the capacity of the DAP at the end so the next series of slides one through five I summarized some of the key findings of the research and the recommendations they made that influenced the design of in the PCI design handbook the observation was that diagonal tension failures occur in the D region of numerous specimens those diagonal cracks I was pointing out are evidence of that diagonal tension failure measured loads were lower than calculated using the equations in 318.08 and PCI 7th edition and it seemed like the apparent shear strength BC plus VS considered in the full stem section outside of the DAP was less than 5 root F prime C BD even after increasing the contribution of steel so what the researchers proposed is to use VC equal to 2 or 3 root F prime C and to place at least two strands in the nib in full stem area of the D region and use a maximum VS which is a significant departure from with a maximum that is allowed in the ACI code maximum VS equal to 2 root F prime C because of its diminishing effectiveness so it was in increasing the shear capacity so it was observed that this term did not increase the overall strength significantly so that was the first finding the second finding is they instrumented some of the bars in these in these specimens they found that the force in the hanger reinforcement was typically lower than the total vertical reaction because there are other force-carrying mechanisms they recommend that just to keep designing vertical hanger steel for full VU as is done right now and just to be conservative in that respect the other observation this is another picture from the research this is at the corner of that L bar that that has to be bent and upward into the stem there's a there's a strut that forms diagonally and bears against the radius of that bar so in some tests the sharp radius formed or caused crushing of the concrete so what they the researchers proposed is to increase the radius beyond the standard minimum radius in the ACI code to try to prevent crushing of concrete in this corner as would be you know this is a typical CTT node in a strut and tie model the fourth finding that I'm summarizing is that developing non-pre-stress reinforcement A sub SH prime by splicing it with the strand caused splitting of concrete that resulted in loss of bond and this is because of perhaps the discrepancy in as one releases the pre-stressing force there's a discrepancy in stresses right there's a splitting crack and if we splice A sub SH with the with strand there's the tendency of that incompatibility to cause failure so and by having those splitting cracks form along that A sub SH reinforcement it decreases its bond strength its development length so the idea the recommendation was to prolong extend a A sub SH prime to one and a half the strand transfer length so past where the strand is transferring pre-stressing force into the beam or two times the development length of A sub SH prime and also to provide a minimum cover ratio exceeding 2.5 if possible and finally the last finding and recommendation is that that crack that forms in the re-entrant corner that led to nip failure in some cases was not necessarily arrested by the hanger reinforcement that's vertical because it could that diagonal crack would can't find its way around that vertical reinforcement so particularly in C configurations as we saw in the picture that it could work around it and that the equations to proportion AHS ASH and AS are not good equations to predict that potential nip failure so the researchers recommended to limit the maximum average shear stress within the nip to 5 root F prime C when you have C shape reinforced steel and 6 root F prime C for other hanger reinforcing pattern arrangements. Okay, so now that we know about that research which is the basis for the DAPTEN design provisions in the PCI handbook, let's look at what they are. So first of all, the PCI design handbook tells you that you have to provide DAPTEN reinforcement according to one of the reinforcing patterns as long as the depth of the recess exceeds 0.2 H or 8 inches. So the depth is 0.2 H or 8 inches or larger, but not exceeding. This is for all DAPTs. And with fixed and non-fixed DAPTs, the length of the recess, which is the nib length, has to exceed 12 inches. You have to provide reinforcement when the length of the nib exceeds 12 inches. And the nib is not designed for full flexural and shear capacity. And you have to provide DAPTEN reinforcement when there's less than a third of the main flexural reinforcement that extends into the end of the component above the DAPTEN, the nib essentially. So these are requirements for providing DAPTEN reinforcement. And for thin-stemmed members with DAPTs, the stem width cannot exceed 8 inches. The components with hanger reinforcement or inclined reinforcement, or components have to be, have to have vertical hanger reinforcement or inclined reinforcement where the inclination angle cannot exceed 50 degrees. And the DAPTEN height is less than or equal to 0.5 H. This doesn't mean that deeper DAPTs don't work. It just simply means that that's the range of values that was tested. So there might be future tests. I believe there are some tests being conducted with deeper DAPTs, but I'm not 100% sure. So again, this is a familiar figure. The nib region is here, delineated by this vertical line and this dashed line. A sub SH is the horizontal reinforcement that is referred to as the nib flexural reinforcement. A sub H is the nib shear friction reinforcement. So this is a potential shear friction plane here in the nib. A sub SH is the hanger reinforcement that then turns into the horizontal extension or horizontal reinforcement of A prime sub SH. And it's called A prime sub SH. A is the distance between the reaction and the central line of the hanger reinforcement. NU is the horizontal force. And the recess length or the nib length is L sub P. And again, as I mentioned earlier, the nib is also referred to as to the extended region. So those are the definitions. And what I did is I summarized all the failure modes. And there are figures like this that are similar in the PCI Design Handbook. There's a table here that I'm providing based on what I saw in the PCI Design Handbook where the failure mode is indicated here by a number. Its description is in this column. The reinforcement we're designing for the failure mode and the corresponding Design Handbook equation. So say, for example, if we are designing the reinforcement for the reentrant corner at the DAP transition. So that would be, going back one slide, that would be the reinforcement for this reentrant corner. What reinforcement prevents failure of a crack that forms in this reentrant corner? So we look at this table and that would be A sub SH. And that is given by 5, 6, 7. So this is just a summary table. And there's a series of figures that guide you, give some guidance to. So, for example, number one here is flexure and axial tension of the nib. This is the action, flexure and axial tension. It provides a pseudo moment, so an equivalent moment. And the reinforcement to resist that is A sub SH. And it's A sub SH is here. And the equation for it was listed in the previous slide. Okay, so the question is, what if the recess is less than 12 inches for non-thin members? So if I read the PCI handbook literally, it doesn't, going back to the non-thin, here it is. It says that you do not need to provide the DAP reinforcement according to one of the reinforcing patterns, which are illustrated in the following slides. So there are reinforcing patterns in the PCI design handbook, and you don't need to necessarily follow those. You can provide orthogonal reinforcement if LP does not exceed 12 inches for non-thin DAPs. But you have to be careful. You have to design the nib for full flexural and shear capacity in order to use this provision and not provide the reinforcement needed. So you have to design the nib area for full flexural and shear capacity needed at that section. So that results in a lot of congestion. Okay, so going back to, so this is failure modes and the reinforcement. So this vertical crack is shear friction. This diagonal tension in the nib. This is a diagonal crack in the re-entrant corner and so on. And as you can see from the labels here, they refer to values in the table and the corresponding reinforcement that is meant to capture or arrest that failure mode. Notice that where, I want to make another point here in this slide. Well, the dimensions of the DAP are shown here, but where the development length is considered for A sub S and A sub SH prime. This is a different way of anchoring A sub SH prime, by the way. So this diagonal crack is assumed to form at an angle of 45 degrees. So this distance here, this horizontal distance is equal to the vertical distance that this crack travels until it intersects A sub SH. So this reinforcement, A sub S, pardon me, this reinforcement A sub S has to be anchored starting at that point to the right and to the left. And similarly down here, this A sub SH prime has to be anchored to the right by providing LD or, you know, if it's a thin stem member, two times LD starting from here and to the left. That's why this bearing plate or this head reinforcement is needed at the end because you don't have enough distance to anchor that bar. So those are important observations of how the crack location influences anchorage of the reinforcement. For the C bar alternative, you have, you want to try to minimize this cover. This is taken, this is one of the reinforcing patterns in the PCI design handbook. Again, the different failure modes number and the cracks that are associated with a failure mode. So failure mode three is associated with this crack. Failure mode five, shear or diagonal tension in the full section is given as five. Okay. So I think that's all I have to say. You have other figures here for thin stem members. These are some of the failure modes that were observed and the corresponding reinforcement. Another pattern for thin stem members, the inclined L reinforcement with the corresponding failure modes associated with cracks and the notation here. So I think these graphs will be useful for you so that you can, depending on the condition you have in your projects, you can use the table in combination with these figures and the PCI design handbook to identify what type of reinforcement, what reinforcement you're designing. All right. So now into the details. The eighth edition handbook has different sections for DAPT and region design. Section 5531 refers to flexure and axial tension. That's, we know what that means now. 5532 is direct shear, diagonal tension at re-entrant corners captured in 5533, etc. We'll cover each of these sections in detail in the next few slides. So flexure and axial tension in the nib. So as mentioned earlier, this is this action here, this equivalent moment acting in the nib, which is the vertical shear, the shear force times A, A the distance from the centerline of the shear to the centerline of the hanger reinforcement that produces a moment. And the horizontal force N sub U times H minus D, the effective depth of the nib produces another moment. These two moments have to be taken into account within the nib. And this reinforcement, so compression would be in the top of the nib, tension on the bottom. So this reinforcement A sub S is taken as two components, the flexural component from V sub U, V U times A, and the normal component from this horizontal force. So it's calculated as, if you add, so the flexural, there will be equations for A sub F and A sub N, and that has to be equal to this equation here that corresponds to the equivalent moment acting in the nib. And if you can imagine passing these two terms phi FY to the left-hand side, that's the total force that is being developed in this tension reinforcement. In this series of slides, what I've done is I've highlighted the call by callers using the equation and the parameters that are needed that are involved in the equation and the corresponding reinforcement that's being designed for each case. So this is for flexural and actual tension in the nib. So for direct shear, you see that it's a shear crack, vertical shear crack, that forms within the nib. The reinforcement that's effective in restraining that crack, or in keeping it from widening, is the horizontal reinforcement and this A sub S. So A sub S is calculated as, using this equation, using the, again, a similar to direct shear friction theory, except it has a factor of two-thirds. And we add to that normal tension, normal area of reinforcement that would be needed for A sub S to resist this normal force. And that's simply a normal force and that's the resistance, AN times phi FY is the resistance provided by that reinforcement. And A sub H is designed as this reinforcement, A sub S, minus what's required to resist the normal force. So this is effectively taking the remaining portion of what's needed for shear friction, essentially, without accounting for the normal force. In the previous slide and in this slide, the strength reduction factor is 0.75. And as we've discussed previously, the friction coefficient and the effective friction coefficient are taken from table 5.3.1. I have a question. Can you go over A again? I think it's referring to this distance perhaps. It's cluttered here in this slide and in the previous slide as well. So A is the distance from the center line of where the force is applied to the center line of the hanger reinforcement, to the resultant where the hanger reinforcement resultant would be. So if you have a single bar, that would be to the center of the bar. But you have a multiple series of bars, it's to the center of those bars. Okay, so this equation we've seen before as well. Shear in the reentrant corner, this is this diagonal crack here. And the reinforcement that arrests that crack is this vertical reinforcement. It could be in the form as shown here in this slide or as a bent L reinforcement. And it is simply designed as a direct shear. So vertical force is in equilibrium with this downward vertical force in the reinforcement that is going to be obtained by multiplying A sub sh times phi fy. So very simple equation here again. As mentioned earlier, the research found that crushing of the concrete existed sometimes or occurred sometimes in this corner of the L or C bars. So it was proposed to increase the radius using this equation. I honestly don't know, I don't have the background for this equation. I'm sure it's in the report of how it's, what was the derivation for it. I'm not sure if it was based on, I don't know. I don't think it was based on strong time modeling techniques, but this is the equation that was proposed. Okay. Now for the extended end, there's also another failure mode that is observed in some cases. It is this diagonal tension failure. And to arrest or control this crack, this vertical reinforcement within the nib or extended end and that horizontal reinforcement, so A sub sh and A sub v within the nib is effective. And so the resistance, the shear resistance for that type of failure mode is given by two forces developed in A sub v and A sub h, and plus a shear resistance that is inherent in the concrete, right? So we have a concrete contribution and the shear, the contribution from vertical and horizontal reinforcement that cross that diagonal crack. And there's an extra requirement that it says that half of the reinforcement must come from this vertical reinforcement. And the second provision ensures that by saying that the minimum A sub v is what is required. This is the nominal that's required, subtracting, so dividing by two and subtracting the contribution of concrete. So that's how the diagonal tension within the nib is taken. As it refers to the extended end or the nib as well, so the shear strength has to be limited given the observed failures of the nib. So it was proposed that the shear strength of the nib was limited by five root f prime c b of the nib times d, or six root f prime c b of the nib times d for configurations that don't include the c-shape configuration. So for c-shapes is five root f prime c, and for other configurations is six root f prime c. Those are equations 5.71 and 5.72 in the PCI design handbook. And you know the phi factor and lambda factor are, as before, the strength reduction factor and the lightweight aggregate factor. There are other requirements. You have to look at what the critical section, as I mentioned earlier, for anchorage of this type of, each type of reinforcement is. It depends on the critical crack. So there's a discussion in section 5535 in the PCI design handbook. And there's also additional reinforcing detailing requirements in that region for normal and thin stem members. So I would encourage you to look those over. We don't have time to look at everything. The other requirements refer to the limitations of that we've, some of them we've already mentioned before. The depth of the extended end cannot exceed 50 percent of the overall depth. That is because of the testing limits that were performed. I mentioned earlier that the vertical reinforcement, the hanger reinforcement, it is not additive to shear. So you can design for shear and see what shear reinforcement you have at that section and use that for hanger reinforcement. But you have to detail it accordingly so that it conforms or behaves as expected. And then the, you have to also check the, from the end of the beam to H, V sub C is taken as the minimum of V sub CI and V sub CW as determined at a section H over 2 from the end of the full depth web. So this is another requirement for critical section to determine V sub C. And this next slide, which is the last one on theory, refers to additional requirements when you have, you have to check a section that is a distance 2H from the end of the depth. You have to look at the acting shear at that section and you want, you have to make sure that V sub C, or you use V sub C as 2 root F prime C BWDP. This is based on the research results. It was, it was observed that a diagonal crack could form here and then you would need to provide reinforcement into arrest this diagonal crack, which is theoretically outside of the DAP, but still affects the, you know, it's kind of the, in the DAP region as well. The other requirement is that the transverse steel contribution at that section cannot exceed 2 root F prime C BWB. I discussed this earlier. And also look at this figure that tells you how long this horizontal bar needs to be. It's twice its development length or one and a half times the transfer length for pre-stress reinforcement. So that needs to be extended a little bit more than what one might do. And that is because of the splitting cracks that form here when you have pre-stressing, when you're transferring pre-stressing. Okay, so I, we've covered in detail all the equation, equations and background to the equation. So now we're going to do, go back to the design example. And I feel like the numbers, I'm going to go fairly quickly in the numbers because it's just plugging numbers into the equations that we've discussed earlier. So we don't have much else to discuss. So going back to the design example, we had designed this area or we had looked at this area of the typical of the prototype building. Now we're going to move to this area and in particular we're going to be looking at this double T that has a 12 foot width. A question that just came in says, can you point out which is the transverse reinforcement you are referring to? I'm not sure since I get these questions delayed because of the system I guess. This, I'm assuming it refers to this comment, transverse reinforcement. This limit of transverse reinforcement applies to the transverse reinforcement or vertical bars or shear reinforcement. It's also called transverse reinforcement at that section, needed at that section. I don't know if if that's what the question. So it's shear reinforcement that would be needed at that section outside of the DAP. So we'll be looking at this section going back to the design example. This is a double T and I'm going to show you an elevation. This double T has actually a DAP and that's why I changed the location. If we look at an exterior view, there's an entrance to the parking garage and you can see the profile of the double T's in that first floor. And if you see a detail of the bearing, you see that there's a DAP, an eight and a half inch DAP there that is needed for clearance because you have to have enough clearance here for cars to enter and to have the necessary clearance for vehicles to enter. So that's why this and to avoid changing. I'm sure there were architectural reasons for this to avoid changing the depth of the first floor with respect to the others. A DAP was built into these double T's and not others. Okay so loading data as before. I will skip through these slides. So we look at material data. Concrete is at 5,000 at 28 days at release 3,500. Mild reinforcement at 60. Grade 60 steel. Grade 270 low relaxation strand. And this is a a sketch of that double T that I was referring to with just bigger numbers. So we have the span length of 59 feet and eight and a half inches and this is the DAP geometry. The only thing I want to point out here is the bearing length or width I should say is five inches and there's a gap of one inch to when the DAP begins. It's kind of a diagonal DAP here that the DAP depth is eight and a half inches. Okay so the loading per double T stem follows from the same approach we did in the first module I believe. We calculate dead loads including self-weight 3-inch topping. We calculate live loads on a 40 pounds per square foot loading for parking garages and that gives me the total load per stem of 31 and a half. Factored load per stem of 31 and a half kips per stem. We also look at the horizontal force that we need to account for in design and that's 20 percent of the factored permanent load and that's four kips per stem. And the properties of the double T it's a 12 by 28 12 foot wide double T f prime c 5000 psi and I did a preliminary design with 16 half inch seven wire strands eight per stem just for detailing purposes and to estimate some of the pre-stressing forces if needed. And the way I did it is using the double the load tables in chapter three. And this is the the load table. Recall that the span is roughly 60 feet and for the superimposed live load that we need or that this beam can carry is 44. This is the the service live load that that beam can can can carry 44. 44 pounds per square foot. Remember that the service load in the parking garages is 40 pounds per square foot. So that's just for preliminary purposes 16 strands eight per stem. Okay so now into the calculation. So we have v sub u. This is the first the nip flexure and axial tension and again I've highlighted the reinforcement that we're designing. It's the best in this case for flexural and axial tension the relevant equation the parameters that we're using 31 and a half kips for for shear per stem normal force of four and the rest are dimensions that are shown here in this slide. So a is calculated as two and a half inches of the bearing length and this figure is not exactly the same as the figure I showed you earlier of the actual building but we have two and a half inches for bearing plus one inch and plus one inch cover here. I assume the one inch cover and two inches to the center of a sub sh gives me six and a half inches for a. One other dimension so d here is the effective depth of the nib. So we have the overall depth was 28 inches. So subtracting eight and a half inches we have 19.5 inches for the nib depth minus three-eighths of an inch for this bearing plate and minus assuming a number six bar for a sub s divided by two that gives me 18.75. So those are the numbers I used in this calculation you may assume other numbers of course. So with this result we can use one number six bar for a sub s which gives me more area than I need from that calculation. The next action to resist is direct shear. Shear friction this is this vertical crack so we design a sub s and a sub h. A sub s is taken as using this equation again a sub v f is the full 35 kips per stem and a sub n is given as n u over phi f y and u being the normal the horizontal force. I'd like to point out the calculation for the effective coefficient of friction here. It is down here this is an equation that appears in the PCI design handbook that we've discussed earlier. Strength reduction factor one thousand this is for psi units by the way the one thousand and that you can look at the original research to realize that this is a thousand ps it has units this coefficient has units. Lambda for lightweight aggregate concrete the width of the nib so I estimated the width of the nib as the average of the width at the bottom this is a tapered member right so it has a width at the bottom a width at the top I use the average for the width of the nib h of 19.5 is the overall nib width and mu is 1.4 for concrete cast monolithically divided by v sub u and then turning 1000 kips 1000 pounds per kip so that gives you a coefficient of effective coefficient of friction of 4.36 which exceeds 3.4 from table 531 so we use 3.4 that's what I used right here I think it was worthwhile explaining how it came up with that number. So h sub sh is the remaining portion for that shear friction reinforcement so subtracting what we obtained from the previous slide which is this reinforcement of 0.23 so we we use 0.23 here what would be needed for normal reinforcement for the normal force and that gives me the remaining of 0.08 square inches so we can use a c bar as shown here a number three c bar that gives you much more than needed for shear friction. The check why do we use a c bar here why do we loop it because we have to develop that bar to the left and right of this vertical crack so this is a development length equation in the ACI code where I do use the factor for top bar effect of 1.3 because notice that this bar this horizontal bar here has more than 12 inches of concrete cast below it so this is a standard equation for reinforced concrete design a development length of mild reinforcement so we need 17 inches to the right and then we develop that bar with a hook on the other side. How do we take the shear in the re-entrant corner so that is taken by h sub sh and the relevant equation is here as shown the shear is a vertical force here and the h sub sh is calculated directly from that vertical force what we would need is 0.7 square inches given this equation and we can use two number four bars bent around the top so two bars so each number four gives me 0.2 square inches two bars two legs give me 0.8 square inches which exceeds 0.7 square inches the minimum radius from the equation that was derived experimentally gives me a bar radius of four inches roughly so that's exceeds the minimum radius for a number four bar but we need that to avoid the crushing failure that was observed in the tests. And then we look at the extension of ash prime we calculate the development length of a number four recall that this is a number four bar so we look at how far it needs to extend it has to extend one and a half times the transfer length of the prestressing strand from the face of the depth or the development length from the diagonal crack the development length of a number four so we have to check both conditions is the development length of the number four bars here it's 17 inches in this case we don't have more than 12 inches of concrete below these bars so we don't use the top bar factor and the transfer length for the strand i assumed an effective prestressing stress of 150 000 psi which might be low but we have 20 25 inches you know this is a number that that one could estimate more accurately but the idea is to illustrate the calculations so the extension is one and a half times lt or two times ld and you'll notice that 37.5 is the one that governs so i'm extending 38 inches from the face of the depth to the end of that extension that's part of the detailing that's needed the diagonal tension limit is applies to the nib and this is not for c configuration this is a normal configuration and l configuration so the nib shear cannot exceed 6 root f prime c b n d so we check that limit and we have a 40 kip capacity greater than 35 point uh 31.5 which is applied so uh the shear limit is is satisfied and then we have to go to that other section that i was referring to a section at 2h away from the deck this is the additional um checks that are needed for thin stem components we have to check the end region as defined for thin stem components this is part of the d region before it becomes the b region so at that section at the end of the d region a b sub c is 2 root f prime c web width and times the depth uh uh to the prestressing reinforcement so v sub c is calculated as 19.8 kips b sub s max is required not to exceed 2 root f prime c b w d so that's equal to 19.8 if we add those two we have um uh 20 uh pardon me um um so yeah so the addition of these two would have to exceed the acting shear so one way of finding out what the required shear would be is to say well how much do i need nominally so v u over phi and i subtract the concrete contribution so that's what what i do 31.5 divided by 0.75 minus what the concrete contributes and the required shear reinforcement at that section would be 22 kips but the maximum that i can provide is 19.8 so that's not good so we don't meet we don't satisfy that requirement uh so we would have to in order to not resize the section i just proposed increasing f prime c to 6000 psi if we do that then v sub c grows vs max also grows and we do the same calculation i have 20.3 kips which is less than vs max of 21.7 so it's good i can now design reinforcement and i don't need to resize the section and the reinforcement i design is based on uh reinforced concrete design theory or you know the traditional um 45 degree angle truss model where v sub s is calculated as the area required for shear times f y d divided by s or written in a different way solving for a v over s that's the equation so we have the demand vs that we need is 20.3 kips divided by f y d f y and d to the pre-stressing strand gives me 0.18 square inches per foot i also check the minimum reinforcement requirement at that section and it is much less than that you know using the two equations shown here and the needed reinforcement i can provide with two layers of welded wire reinforcement with this pattern and with these wire calibers you can see the pci design handbook 1543 to verify that this wire reinforcement meets that area that's needed and just as a final set of calculations i have two more slides or three more slides one is just to check the bearing strength and this is just for illustration purposes we're going to check the bearing without reinforcement as recall is the distance from the edge of the member to the center of the where the bearing force is applied so we have two and a half inches plus two that's four and a half inches the length of the bearing is five inches as mentioned earlier so the product of s times w is 22.5 square inches which exceeds nine square inches so in calculation of this reduction factor s times w cannot exceed nine square inches so we plug that in here divided by 200 and we take the ratio of nu over vu 4 divided by 331.5 that gives me a reduction factor of 0.674 i plug that into here the bearing equation for non-reinforced bearings i calculate the area two in the frustum as follows so the frustum from the bearing area it expands until it hits a free surface and the way the the stem you might recall it's four and three quarters inches wide and we have a bearing width of of three three point three seven three three inches and three-eighths of an inch i don't i may have not mentioned that but the bearing plate was three and three three and three eighths and so the distance that was left from the the bearing surface to the edge of the stem was 1.375 inches so that's how area two is defined in this case in order for us to calculate the square root of a2 over a1 it's a magnification factor in essence that that that because of the confinement that the concrete provides 1.34 and this is a reduction factor because of the presence of this transverse force that decreases bearing strength so plugging in numbers again the phi factor for bearing is 0.65 so we have 50 kips which far exceeds the demand which is 31.5 kips for shear and however even though we do need we just with direct bearing without any reinforcement needed we're okay since this is a thin stem member it is recommended to provide this amount of reinforcement to resist the the horizontal force of four kips so we use that and so we calculate this number it's 0.1 square inches we can use one number four bar to capture that so the final reinforcing pattern is this after for this case we have not one number six for as two number fours here that are looped around let's see how how they loop in the cross section so we have four legs of number fours the same two number fours extend into the stem and then we i didn't show you what happened here at the end of the the end section of the beam an alternative pattern which is i'm just going to show you very briefly would be a c pattern for the cz pattern for this case the total hanger reinforcement this 0.7 reinforcement that we needed and we provided with number four bars we can also provide with one number five and one number six bar and it gives me 0.75 square inches exceeding the 0.7 square inches that was for i think there's a typo in your notes i think it's 0.7 square inches and not not 0.65 square inches and that's the only change here for the cz reinforcement instead of a single number six i'm using two number four bars here for the reinforcing detailing in this case so similar two different patterns that that would work and you'll also notice that the pre-stressing strand pattern is staggered it's just for illustration purposes remember that i had preliminarily designed eight strands per stem so these are only six strands so with that i think this is it we come to the end and i'd like to open it up for further questions if there's any if you have a question please type it in the chat box and send it to the organizer thank you well sergio so far i have no questions okay there were a couple along the way so perhaps people have already had their questions answered maybe well if you have any questions after we log off you can always send it to me via my email at snodden at pci.org i will then forward it on to sergio who will then give me the answer and i will forward it to everybody so thank you very much for attending this evening thank you sergio and we are going to leave

advanced pre-stressed concrete

bearing strength

DAP regions

design provisions

PCI Design Handbook

shear friction theory

horizontal reinforcement

DAP design

diagonal tension failures

DAPTEN reinforcement

double T beam

reinforcing detailing requirements