Good evening, everyone, and welcome to the fifth of six sessions in this Advanced Pre-Stressed Concrete Design course. Today, we will keep moving along our design example, and we're now, as I mentioned in session five, we'll be talking about corbel design. And this session, unlike others that we have covered, focuses only on corbels. And this is the fifth of six sessions. The last session on special topics, there's a variety of topics that we'll be covering in that session, but that session won't be held next week, but rather two weeks from today because of the Thanksgiving break next week. So with that, just a couple of very brief comments. I wanted to start by showing you what you should expect of this session. The first thing I'm going to be talking about failure modes in corbels, what those mean. And we are going to cover essentially two design methods that are covered in the PCI Design Handbook. So both of those will be covered in this session. The first one, the cantilever beam method, is going to be covered first. And then we will also cover the second of those two methods that are used for corbel design, strut-and-tie method. And in order to do that, I'm going to give you a little bit of a very brief background on what a strut-and-tie model is and how one uses it for design of structural components. So right on to directly into the topic of tonight. As I mentioned earlier, there are two methods that are included in the PCI Design Handbook. The cantilever beam design method, that's covered in Section 571 of the PCI Handbook. And through that method, the designer really is involved in three main activities, as we will talk about in a few slides. The first is to consider flexure and horizontal load that are applied to the corbel. And that will lead to design of a reinforcement called A sub S. Also the designer has to check shear friction, and that will lead to reinforcement denoted as A sub H. And finally, we're also going to be talking about anchorage of reinforcement and the general detailing of the corbel, which is quite important. The second method that we'll be covering is a strut-and-tie method that's covered in Section 752 in the PCI Design Handbook. And that is a method that can be used not only for corbels, but also for other structural components that fall into the categories of the method, mostly for discontinuity regions and structural members. And I'll cover those when the time comes. So I wanted – I had a – just to get an understanding of what – how familiar people are with corbel design, I wanted to start a poll early. And the question is, in practice, how is it that you design corbels in your practice? So I'm going to ask Sherry to launch the poll. The poll – the poll is launched. Please answer on your screen. I will leave the poll open for a few more minutes. Please answer now. If you haven't voted, please vote now. I will be closing the poll very soon. Sergio, 89 percent of the people voted. Six percent said, I use standard details available in my – I can't see the rest of that answer. The first one. Thirty-one percent said, I use software commercial spreadsheet. And 63 percent by hand using one of the methods in the PCI design handbook. Okay. Great. Thank you, Sherry. So that's great to hear, because I think – so what's going to happen is, although it might seem repetitive in this session, that's what we're going to do. We're going to follow calculations used by hand. But I'm going to try to make – my intent was essentially to show you where some of the design equations come from, in case you haven't had a chance to think of over where the design equation might come from, and to give you my perspective of corbel design in a way. So first, the two methods is the cantilever beam design method in Section 571. And for that to be able – for a designer to be able to use this method, the corbel has to meet certain geometric limits and force limits, as shown in the first bullet of this slide. First of all, A over D, A being the distance between the applied vertical force from – that serves as reaction for a supporting member, divided by the effective depth of the corbel, needs to be less than one. And that means that the reaction force needs to be fairly close to the face of the column. And in this next slide, we'll see how the definitions apply. The second is that the horizontal force, if any, applied to the corbel has to be less than the vertical force. And for all calculations that are done on corbels, even if they are flexural design calculations, the strength reduction factor phi will be 0.75. And the last thing that the cantilever beam design method specifies is that the – or specifies that anchorage, effective anchorage of A sub S, the flexural and normal reinforcement needed for the normal force, has to be properly anchored in order to ensure proper performance of the corbel. So the basic notation – and this is – this slide shows typical detailing of a corbel, where we see the main reinforcement for flexure and horizontal force is denoted as A sub S. These horizontal bars, A sub H, are intended for shear friction reinforcement. In all cases, the critical section of a corbel is the intersection of the corbel itself with the face of the column. And you see here that a corbel is normally designed to resist the vertical force from the member it's supporting, but also a horizontal restraint force that might be generated by shrinkage of that large dual member or any other horizontal forces that may take place in the floor system. So this is a horizontal force that needs to also be considered for design. Notice in this slide as well that this horizontal reinforcement has to be distributed over the top two-thirds of the corbel. So in this case, the corbel, the top would be in tension and bottom is in compression. L sub P is the length of the corbel, and L sub DH is the length required to develop this AS reinforcement using a standard hook with an extension of 12 dB into a column or supporting member. What's really important in corbel design is that – to provide adequate details. And you'll see that one of those details specifies that the front face of the corbel cannot be less than D over 2. That's the minimum. And the other one – or the other important detail is that the longitudinal reinforcement A sub S has to be properly anchored to develop the yield force at the critical section. And the critical section can be encountered entering from the column or from the corbel itself. Typically, L sub P is a fairly short distance, and these are straight bars. So the PCI design handbook provides a couple of anchorage details for this longitudinal reinforcement to ensure that it yields at the critical section. One of them is welding the reinforcement that meets at this corner in this supporting plate. Welding both A sub S and this framing bar reinforcement, which is simply there to be able to tie the rest of the reinforcement in the corbel. And that ensures that when this bar is being pulled in tension, there's bearing against this vertical bar, and also the plate is anchoring – or helping anchor this A sub S reinforcement. Another method might be to weld a crossbar. And again, this is to ensure that there's bearing between the crossbar and the concrete surrounding it. And a third method that's not shown here is to perhaps weld the longitudinal reinforcement to a corner angle. Some other details using a corner angle are also used. Notice that A – again, as I mentioned, the vertical force – the distance between the vertical force and the face of the column is shown here. And that unless there are detailed calculations to determine where this reaction is going to take place, the designer can use 75 percent of L sub P, where L sub P is the length of the corbel. So many years – so corbels have been studied for many years. There's a reference by Kriz and Raths from 1965. And in that reference, there were some of the failure modes that had been observed in practice associated with corbel. And there was a statement that I'm just copying here literally from that paper, which states that almost invariably, distress of corbels in the field is a trace to – can be attributed to poor detailing. And one of the comments that was made is that if the tension reinforcement is not effectively anchored close to the outer face of the corbel, the full strength potential of the reinforcement cannot be developed, and therefore, there might lead to undesirable or premature failures of the corbel. So, an example is shown here, right, when we as a designer might want to detail the reinforcement and say, well, let's just bend the bar – the main longitudinal reinforcing – bend it close to the free surface of the corbel and back into the column or wall that is supporting corbel. But what happens in practice is that because of construction tolerances and radii that are used to bend bars, one might leave the front edge of the corbel unreinforced, and a premature failure surface might generate as this load kind of works – or the cracks that form because of this load work their way around the reinforcement that's bent near the tip of the corbel. Another poor detailing practice would be to end the corbel in a very shallow – at a very shallow distance. And this poor detailing option might trigger a failure mode where a crack would cross the corbel and simply break it in half. So that's why the minimum depth of 50 percent of the effective depth of the corbel is specified so that the load path from this vertical force into the column is ensured. One can create this load path as illustrated by this diagonal line. So this is – I like this picture because it actually shows two very important failure modes in corbels that were identified, as I mentioned, back over 40 years ago now. The first slide I wanted to show you is to give you a summary of all the equations that are used in the PCI design handbook for corbel design. And then we'll go – we're going to go into details of – into their background. So the first is that A sub F is that top horizontal reinforcement for flexure and normal force is calculated as the maximum of A sub F and A sub N, maximum of the addition of A sub F plus A sub N, and two-thirds of A sub VF plus A sub N. The individual components A sub F, AN, and AVF are listed here, or equations for those are listed here. And these are listed as a series of equations in the PCI design handbook. So this – you can think of this term here, and I'll go into details in the explanation of this first component, is to resist bending in the corbel. The second would be to resist the normal force that may be acting, that horizontal force that's acting on the corbel. And the third, AVF, is to resist direct shear from the vertical force V sub U applied vertically to the corbel. So this is to provide shear friction resistance. There's also a minimum reinforcement requirement for A sub S, and it can't be less – so A sub S cannot be less than 0.04 F prime C over FY times V times D, where V and D are dimensions of the corbel, characteristic dimensions of the corbel. So let's take a look at – a critical look at where these equations come from. So imagine you have a corbel with a vertically applied force, a horizontally applied force. These are factored forces. These are VU and N sub U, respectively. And then we take a cut at the critical section. This dashed line is the critical section. And we're looking into deriving an expression that would give us the amount of A sub S that is needed to develop this tension force and that reinforcement. So this is similar to – it is a bending problem, right, where we have to – the corbel acts as a short cantilever, where this V sub U and N sub U create a tension on the top of the corbel, compression in the bottom. And near the bottom of the corbel, we can assume that there's a compression force here generated, horizontal compression force, that sets in equilibrium this tension force at the top. Notice that I've replaced the actual stress distribution in the concrete with an equivalent stress distribution of magnitude equal to 0.85 F prime C. This is not important for this slide. But the depth of that stress distribution is A1 to distinguish it from A, the shear span in the corbel. And again, the key dimensions here are D, the effective depth, and H, the total depth of the corbel. So if we take moments about this point, where my mouse is hovering right now, we can say that, well, the force T, ASFY, times the distance to that point, it would be D minus A1 over 2, has to be equal – that's the internal moment that needs to be developed from the moment that is induced by the acting forces. That has to be equal, then, to VU times A, VU times the distance to the face of the column, plus NU times the distance to this point. And that distance is H minus A1 over 2. So if we do some algebraic manipulation, notice here that the phi factors on the left-hand side of the equation, because we're using a phi factor for design. So this is the nominal resistance, and this is the factored loads on the right-hand side. So if we use some algebraic manipulation, we simply remove or factor out N sub U times H out of the parentheses, and we add equal and opposite terms, NUD negative and NUD positive, and insert that into the parentheses here. This is just going to be better solved for A sub S. So if we do that, then, we can manipulate this equation and solve for A sub S, and the answer would be VU A plus NU H minus D from these two terms here in the numerator, divided by phi times FY times D minus A1 over 2. That's in the denominator for these first two terms, first three terms, pardon me. And then the second term is given here as NU divided by phi times FY, because the D minus A1's over 2 cancel out. So as you can see, this is an equation that's based on equilibrium, and going on to the next slide, I'm copying that same equation here, and we see that in the denominator, this quantity A1 over 2, if we assume this quantity to be negligibly small, we can drop it and assume it's zero. We end up with the equations that two equations, or an equation for A sub S, where it has two terms. The first term is a term required for a moment caused by DU and NU, and the second term is required for direct tension that is caused by N sub U, that horizontal force in the corbel. So in a way, this, the numerator here, can be thought of as an equivalent moment of magnitude equal to VU times A plus NU times H minus D. And if we think of it as that, as an equivalent moment, then we can say, well, if that is the moment that's acting in this corbel because of VU and NU, I can then do the same exercise again and say, well, what is the amount of tension reinforcement that I would need for flexure, right, because it's a moment calculation? Well, taking moments again about point C, we end up with the following, that A sub F would correspond to, or the amount of reinforcement that's required for that flexure, that equivalent moment, is VU times A plus NU H minus D divided by phi FYD, which is the first equation in the PCI design handbook, or one of the equations in the PCI design handbook, equation 586. The second one being more, you know, easily, the horizontal force generates also a tension force in the top reinforcement, where I'm calling this the tension force needed for the normal force, and if we solve from horizontal force equilibrium, we can come up with this equation here, which is the normal, the reinforcement required for the normal force, and again, this is an equation that is in the PCI design handbook, and this other one, 587, is an equation in the PCI design handbook. Notice that there are two simplifications in these assumptions. The first one is that this distance, one half of A1 is small, and the second simplification is that we move here, N sub U is moved down to the level of the tension reinforcement. So the PCI design handbook has other equations for corbel design in this cantilever beam method. These two conditions for A as being the maximum of these two conditions we mentioned earlier, from condition one, we can plug in the terms of the, for the AF and AN, and then just manipulate them algebraically. So you can write this expression by factoring out one over phi FY, and then leaving inside the parenthesis for the first term you leave, for the first term you leave VU divided, and you divide directly by D, so it's VU divided by A over D, and then in the second term here in the numerator, you group it with the second term involving horizontal force, leading us to this second term in this alternate form for condition one equation. So this is given as equation 589 in the PCI design handbook. So this equation might be more convenient because then you group terms related to applied vertical and horizontal force. The same thing could be done for this other condition to determine A as max, where you could group the term associated with V sub U and the second term which is simply associated with N sub U. Keep in mind that this term relates to, the first term relates to shear friction provisions. So that's it in terms of equations for corbel design based on the cantilever beam method. So I'd like to just go directly to an example. I'm going to do two examples today. The first one, I will design a corbel just based on the cantilever beam method, and the second example, I will design a second corbel, a different corbel, using the two different methods, the cantilever beam method and the strut and tie method. So looking at our building, our prototype building, looking at this area here between axes B and C, column lines B and C, we are going to, these two, these double Ts are supported on corbels along the wall, center wall along axis B here, and also along the spandrel wall on axis column line C. Showing a side view and exterior view of column line C, we can take a section and then we can look at the details of the corbel in that wall panel, the exterior wall panel. And the corbel we'll be designing is shown here. It's 16 inches deep total. The length of the corbel is eight inches deep, wide, pardon me. So this shows a blow-up view, a sketch of our design area. We have between column lines B and C, notice that they're measured to the outside faces of the walls that are bounding these two column lines. So we have 60 foot and a half to those column lines, leading us to the total length of double Ts of 60 feet and half an inch. And the corbel we're going to be designing will be this one here on column line C. And this is a blown up view of that corbel. It has a steel plate and supporting a bearing pad that is five inches by four and three quarters of an inch. The sizing here I estimated based on the width of each stem, double T stem. So that's how I sized that pad. You might have other requirements in your practices. And also A was used or was estimated as three quarters of eight inches without any other more detailed information. One could also do it by knowing what the gap between the double T and the center of a pad. So that's just geometry. One thing that is worth mentioning is that the width of this corbel perpendicular to the screen is 12 inches. So as before, the first thing we need to do is determine the loading data, dead loads, live loads as shown in this slide. We've done this before several times, so I won't spend too much time on this. And the concrete and reinforcing steel that we're using for this corbel design example are shown here. F prime C is 5,000 PSI and grade 60 reinforcement. Notice that before I go on, I'm going to be using for the corbel, I ended up using number four bars. And the reason I did that is I first looked at the wall thickness of eight inches, oops, sorry. Right here, the wall thickness of eight inches. And we quickly see that in order, if we use the PCI design handbook aid 1532, the largest diameter bar that can be hooked to develop yield strength over eight inches is a number four bar. And this is without, so this is to determine, this design aid gives you LVH, the development length for hook bars. And this is without accounting for a horizontal reinforcement that is sometimes placed along the hook, the tail end of the hook that might reduce LVH. So in this case, it's just a plain hook without any benefit of the reduction of reinforcement. So that could also be used to reduce that. So that's why I used number four bars. That's just a brief explanation. So again, calculating the loads per stem of a double T, this is the estimated weight of the 10 foot wide double T. I used the properties of a 12 foot by 28 inch double T from the PCI design handbook. And then I shaved off the flange at the edge to come up with a 10 foot eight inch foot wide double T in order just for load estimates. So the self weight of the double T results in a reaction of 9.10 kips. This is unfactored. The three inch topping results, tributary three inch topping results in 6.1 kips for a total service dead load of 15.2 live loads at 40 pounds per square foot without any reduction because it's a parking garage, resulting 6.4 kips per stem. And the factored vertical load is 28.5 kips per stem. And the horizontal load is estimated as 20% of the factored permanent loads per stem. So that is only the dead loads, 1.2 times 15.2, that gives us 3.7 kips per stem. I carried too many significant figures here. Typically you would express this load as 3.7 kips per stem. So notice this is what we used here, or I used is 20% of the factored permanent load on the per stem. But there are discrepancies in how that horizontal force is defined. And I wanted to just give that, present that to you so that you're aware of these discrepancies. ACI 318 on one hand, this is the 2014 edition, stipulates in 16.534 that the horizontal tensile force in a core ball shall be treated as live load when calculating N sub UC. And that has implications because the load factor for a live load is different from a dead load. So the live load factor, as you all know, it's 1.6. It also stipulates that unless tensile forces are prevented from being applied, N UC, the ultimate or factored load in a core ball shall be at least 0.2 of V sub U. But in this case, V sub U is the total load, not only the permanent load. And again, there's another, the commentary to section 16.534 says, gives a reason for why this force needs to be factored times one or treated as a live load. And it is the important part is that because of the uncertainty in estimating horizontal forces, those forces cannot be determined accurately. So that load factor should be 1.6 to account for that increased uncertainty. That's what ACI does. The PCI design handbook 8th edition stipulates that if bearing pads are used to avoid tensile forces, this N sub U value can, that corresponds to, would correspond to the force that would cause the pad to slip above the bearing plate. And it could be treated as the maximum force that can happen before slips occurs. And unless more or a detailed calculation is done, that horizontal force can be estimated as 0.2 times the factored permanent loads on the core ball. That's the reasoning behind the value in the PCI design handbook. So we're going to use that for the rest of this example, 0.2 times the factored permanent load on the core balls as the value for N sub U. So the first thing to do is to check geometry and reinforcing limits. I'm jumping, I will jump ahead here in this slide, all of the dimensions and loading are given and backing up one slide, we check one by one. So the distance from vertical force to face of column six, the depth of effective depth of the core ball is 15. So that means the requirement of A over D being less than one. Horizontal force is less than the vertical force that checks. And one quick calculation that we must do initially is to determine the minimum area of steel that will be needed for this particular core ball, it's 0.6 square inches. And let's check the maximum shear force that can be applied in the core ball that has to do with shear friction calculations. So the critical area that is used for shear friction calculations in this case is the width of the core ball times the effective depth. That's 12 inches wide times 15. And 1000 lambda times phi is an equation that is given within shear friction calculations for the maximum vertical force or shear force that can be transferred by shear friction through a surface. So if we do that calculation, that results in 135 kips, which is much greater than 28.5 kips that were transferred. So we're okay. So these are some initial quick checks. Notice that this table where this value is, or this equation is pulled out from, is within the section for shear friction in the PCI design handbook. So these are the design data, the vertical force, the normal force, the width, the depth, and the shear span of the core ball. The effective depth is assuming one inch smaller than H, just to account for tolerances. I first estimated a half inch plate here, and half of a number four bar would give me 0.75 instead of one, but I used one just for tolerances. And again, a phi factor of 0.75. Notice that to estimate first, I wanted the first calculation is to estimate A sub S. So I used a second form of the equations that separates the vertical and horizontal forces given all this design data. Everything is just plugging into the equation. So the resulting value is 0.34 square inches for A sub S. That's one of the two requirements. The second requirement is to use this equation for A sub S. So this calculation at the top of the slide is to be able to calculate the effective coefficient of friction, mu sub E, that has to enter into this equation. We're still calculating A sub S here. So we first calculate the effective coefficient of friction using this equation here, where A sub CR is slightly different from before. And this is something that's being discussed currently by the PCI design handbook committee, whether one should use B times H or B times D to make those two calculations I just showed you more consistent. So in this particular case, it doesn't make too much difference, too much of a difference, but I'm going by the book. So 16 inches wide, 12 deep, 12 inches wide is what the corbel is. The phi factor is 0.75. 1,000 is in pounds unit. So that's why we plug in units of pounds. That's why we plug in V sub U in pounds here. And 1.4 is the coefficient of friction for concretes, for monolithic concrete. That's 1.4 taken from ACI 3.18.14. And that results in an effective coefficient of friction of 7.1, roughly. The PCI design handbook lists in Table 5.3.1 that this effective coefficient of friction cannot exceed 3.4. So that's what we use here in the first term in this equation, 3.4. Again, I use the second form of the equations to say that the required A sub S will be 1 over phi FY times 2 thirds V sub U over mu sub E plus NU. And all the other, so the forces here were determined before. So by plugging into this equation, we get 0.21 square inches. And we summarize the three calculations for A sub S from 5.89, from 5.90 shown in this slide. This is for flexure and normal force. This is for shear friction and normal force. And A is minimum. In this case, A is minimum governance. So we have to design our corbel using 0.6 square inches. And that is satisfied by using three number four bars in order to satisfy, as I mentioned, that LDH is enough within an eight inch wide wall. So the next step is to design the horizontal reinforcement for shear friction. And the equation in the PCI design handbook is 5.92. That equation says that A sub H has to be at least equal to 0.5 times AS minus AN. So this is, it recognizes the fact that this AS reinforcement that we have just designed actually also acts in shear friction or to provide some shear friction resistance. So this equation ensures that at least 50% of the total reinforcement required for, that at least 50% of that reinforcement that's provided for shear friction resistance lies down here somewhere farther down in the corbel. So this A sub H is 50% then. And notice that it's not the reinforcement that's provided, in this case, minimum reinforcement, but the maximum of the two reinforcing requirements for A sub S. Going back a previous slide, of these two equations, 5.89 and 5.90, the maximum of these two is the reinforcement that we, or the amount that we plug into this first term. And then the second term is the reinforcement that would be required for that horizontal normal force. That results in 0.13 square inches to be distributed here horizontally over two thirds of D starting from the top here. So to ensure that most of that reinforcement is on the tension side of the corbel. So that results in a spacing of five inches approximately for that reinforcement. So notice here, the box here, I had this note, since we only have 0.13 square inches and this reinforcement as shown in this sketch here is looped around this construction reinforcement here. So we have two legs of a number three. So even one number three bar would satisfy this requirement, but I'm adding a second for constructability and it's really not that expensive and it's better to distribute that horizontal reinforcement more evenly throughout the height of the corbel. So the idea is to, again, going in the spirit of good detailing, provide at least two bars, two layers of bars to facilitate construction. So finally, this is the design as you can, 63% of you responded. You're familiar perhaps with these types of calculations. This is a final design, very straightforward if one uses the cantilever beam method. Our corbel design ended up being three number fours for ASAPS, two number threes distributed this way over the top 10 inches of the corbel starting from the longitudinal reinforcement. These are the geometric requirements and these framing reinforcement, I provided two number four bars just because they're the same size of this longitudinal reinforcement and you'd need at least two to tie to be able to loop around this horizontal reinforcement for shear friction. Notice, by the way, and I failed to mention this, that since this is the vertical, the critical section, this horizontal reinforcement for shear friction also has to be anchored at that critical section. So we have to do it by bending it around this framing bar and also tying it behind the extension of the, for the hook, for the main reinforcement hook. And the detailing, the anchorage detailing shown here for ASAPS is blown up here in this part of the slide. And you see that I selected just the welding the reinforcement to the base, to the plate here, the bearing plate at the top of the corbel. Where do these numbers come from, the minimum size of welds? Well, I pulled those from the PCI design handbook as well. Design 8-6-15-3 and 6-15-5 give minimum weld sizes for this type of arrangement of connection. And also another design aid from the PCI handbook, 15-3-2, tells you that the development length for a number 4 bar is 6 inches as shown here. So we have more than 2 inches behind the tail end of the hook. So that's satisfying. So the second example tonight is a different corbel, and I'm going to design it just for comparison using the two methods that are allowed by the PCI handbook, the cantilever method, cantilever beam method, and the strut and tie method. And for that, I picked a different area in our building. And this is an area where this parking garage has ramps going up to this crossover base and keep going up. So I selected this area because this is a crossover bay where I'm hovering with a pointer here. There's an inverted T that's supporting the floor system at the same level from two sides of this column line B. It is blown up here, shown here. Notice that the double T stems don't exactly match because, especially in this region, the stem of the double T's from the left-hand side lies at a different location along this inverted T because of the double T width. And so yeah, that's it. So these are some of the dimensions that are of interest. I'm going to show you a section here, an elevation section XX in the next slide. And that shows the inverted T from column lines 2 to 1. Going back, 2 is this column line and 1 is the top column line. And looking at that section XX, we're going to be looking at the details of this corbel, which comes out from the column at column line 1. This is a 24-inch perpendicular to the screen column by 33 inches wide. So the corbel is 24 inches in the direction perpendicular to the screen. And these are the loads that are acting on the inverted T beam. Notice that their spacing is different along the span because we have to consider where the stems of the double T's lie. On the backside that is not shown in this slide, also those loads would not exactly coincide. So I took all those features into account when coming up with the total reaction force that we would need here in this corbel for its design. The general dimensions here are much larger now than for that other smaller corbel we just designed. This corbel is 24 inches deep total. Half of the depth is 12 inches. That's the front face of the corbel. It is still 8 inches wide only or long only. And the bearing pad that I estimated is 6 inches by 22 inches in that direction into the screen. And again, I used an A of 3 quarters of 8 just to estimate where the vertical force would be. So the first bullet just tells you how I did the calculations. I'm sure you don't want me to go through all of them. I mean, I just calculated reactions based recognizing that each stem of the double T lies a different spacing from both sides. So I calculated first one side and then the second. It's just a simple reaction calculation. I ended up with 145 kips of dead load and 57 kips of live load for that inverted T beam. And that's only on the right hand side for the corbel that we're designing. So the factored loads on that end of that beam are going to be 1.2 dead plus 1.6 live gives me 265 kips and 0.2 times factored permanent loads gives me 34.8 kips. So these are our design forces for that corbel now. So proceeding again with a cantilever beam design method, similarly to before, we use A over D and check that it satisfies less than one limit. We know that NU is less than BU because of the way we calculated. The minimum reinforcement, longitudinal reinforcement in this case would be 1.8 square inches because of the dimensions of the corbel, they're much larger than before. And the maximum shear force that can be applied to avoid a shear friction failure, I would say would be 405 kips, which is much greater still than the applied factored shear force of 265 kips here. So all of these initial calculations check again. And then determining our first equation for A sub S. These are the design data. I don't think there's anything very critical here to go over other than the main difference here is that the effective depth now I subtracted one and a half inches recognizing that the bearing plate might be thicker and also the reinforcement might be larger bars so instead of one inch I provided one and a half inches that gave me 22.5 inches of effective depth and recall that for all corbel design calculations involving the cantilever design method the phi factor is 0.75. So plugging in our design data into A sub S gives me a requirement for 2.4 square inches of longitudinal reinforcement near the top of the corbel then checking the requirement involving the shear friction requirement again first calculating an effective coefficient of friction mu sub e going to equation 5.3.5 in the PCI design handbook and remembering that we need to plug in this value in units of pounds because this coefficient is in units of pounds so we end up with an effective coefficient of friction of 2.28 which is less than 3.4 so this is the one we should use now I'm plugging into this equation 5.90 of 2 thirds of V sub u divided by mu sub e we use 265 and 2.28 in the denominator and 34.8 is the normal horizontal force to be resisted. So this requirement now or requirement including the shear friction reinforcement is or two-thirds of the shear friction reinforcement should be located here as part of A sub S we have 2.5 square inches which actually ends up governing in this case above or above the 1.8 minimum reinforcement that's needed so in this case equation 5.90 governs. So we also supplement that shear friction reinforcement we recognize that we've used two-thirds of it for located at the place where A sub S lies so the another 50% is located underneath that so we use 2.5 which is the governing requirement for this corbel 2.5 square inches A sub N is again for the normal force the reinforcement required for normal force which results in 0.86 square inches so instead of using number three bars we use three number four bars each layer so it's going to be two legs of a number four bar each layer gives me 0.4 square inches three layers gives me 1.2 so two layers would be slightly less than what's needed so I went to the third layer here and again the the spacing is not terrible spaced at five inches and distributed over the top two-thirds of D of this corbel which is recall is a D is 22.5 so the top two-thirds is 15 inches so that results in a nice even spacing of five inches starting from A sub S. So similar to before this is our our final design in this case we have a very deep column we have 33 inches we can place five number sevens to satisfy the reinforcement requirement of 2.5 square inches and we have three layers of number four hoops that are tied around or wrapped around the framing bars that are number four bars again the anchorage detail that I that I used is here and we check again make sure that the development of the number sevens is satisfied so we have 31 inches we bring that reinforcement as close to the outside surface of the of the column as possible we don't we don't use 11 inches and call it a day this this is important to develop compression in front of this in front of this bend which is going to be needed it will become better more apparent here when we look at strut and tie models but it is important to bring this these hook as close to the outside face of the column as possible even though the development length for the number hooked number seven bar would only be 11 inches and again the minimum weld distances here are our well lengths are determined using two design aids from the PCI design handbook okay so moving on to the strut and tie design method for corporal design for this second corporal design I wanted to first give you a few references for strut and tie method for the strut and tie method and these are very good references one of them appeared in the ACI journal back in 1985 you can barely see the date here by Peter Marty another one is what I would consider one of the landmark publications and strut and tie modeling and it appeared in the PCI journal back in 1987 is the the reference by Schleich and others these are German researchers were where they devoted this special this is a very long report on how to use strut and tie models for design of many different components you see here that there's a corbel here with some odd shape and describes how to construct a strut and tie model models for different cases including that bends and so on and finally there are two publications from ACI special publications that provide examples one is SP 208 that's the cover that's shown here the second one of those publications is SP 273 and again it has very detailed examples on how to construct and design based on strut and tie models so I wanted to pause here to open to ask the second poll this evening and asking whether you'll use the strut and tie method for design of structural members so I would like to ask Sherry to start the poll the second poll for tonight I will be launching the poll now please answer the poll using your screen thank you okay I will be closing the poll if you haven't voted please vote now okay Sergio 90% voted 28% yes 72% no thank you thank you Sherry all right so it begs that's good because I do have some background on strut and tie models in the next few slides but I should I strongly referred you to in particular this publication from a PCI journal you I believe you can download it for free even so it's a it's a must read it's a long publication but it's a it's a must read so the first is a is a where what's the strut and tie method and where does it come from the background is that if you look at some of the literature in the strut and tie method it many times you will find that it says that the strut and tie method satisfies the lower bound theorem of theorem of plasticity so this is a theorem that is used in structural mechanics and there are three sets of equations in structural mechanics equilibrium equations constitutive relations and deformation compatibility equations so that's the strut and tie method being a lower bound theorem or satisfying the lower bound theorem of plasticity satisfies equilibrium everything must satisfy equilibrium in structures so summation of forces and moments and so on the other a set of equations that the strut and tie method satisfies is the the constitutive relations as defined by a yield criterion what this means in layman's terms is that the forces in the elements of the strut and tie model will not exceed the strength of those elements often referred to as their yield strengths even though it's it's the reinforcing that's yielding also elements that are in compression sometimes people look at them and say well the ultimate strength one could think of them as the yield strength the set of equations that it doesn't satisfy that the strut and tie method does not satisfy is deformation compatibility so you do not it is very difficult to estimate deformations of structures that are this or deflections of structures that are designed using the strut and tie model and not only that but if you try to estimate you'll see that the strut and tie model involves using a truss for design if it if you use the deformations of that truss those are by no means estimates of what the deformation of the structural component will be so what's really important is that even though so by satisfying the lower bound theorem of plasticity it's called the lower bound because any structure you design using the strut and tie method would result in a load carrying capacity that is typically greater than what's estimated using the model the strut and tie model and that's great because that gives us some map margin additional margin of safety all right so what does it represent a strut and tie model is really represent it represents it's a truss that represents the resultants in stresses that occur in complex regions of structural members and I'll show you some examples in a couple of slides the strut and tie a strut and tie model is constructed using as I mentioned truss members some of them are called struts those are the compression members others are called ties those are the tension members and they are connected together through nodal zones through the connections between compression and tension members so those are referred to as nodal zones in the strut and tie model and the first step in constructing a strut and tie model is to break up it for a structural member is to break that structural member into B regions and D regions B regions in the strut and tie literature are defined as regions where typical bending theory plane sections remain in plane is applicable and D regions are regions were that people call disturbed regions or discontinuity regions where the assumption of plane strains or plane section does not hold true and there's a there's there's an irregularity in the strain profile because of the presence of changes in geometry or changes in the loading you know concentrated forces that occur in structural members so some typical examples of B and D so beam regions and discontinuity regions so notice here that a deep beam for example that's loaded in the center by a concentrated force constitutes a D region throughout the length of this member a column and a corbel itself part of the column constitutes a B region and part of it constitutes a D region as highlighted in blue here so the D region ends here where the geometric discontinuity has traveled let's say a distance H beyond where it occurred H being this dimension of the column the largest dimension cross-sectional dimension of the structural member in a longitudinal in a beam in a simple beam with uniform loading these concentrated reactions constitute a D region that that's why beams with that dense are also a D region the small portion within 2D regions here that is the second D region in this beam actually is the region adjacent to a geometric change this is an opening or a hole so all of these blue areas constitute D regions and again the critical dimension here is H the largest dimension of a beam so these D regions are defined from the discontinuity a distance H from it and this is a joint in a beam column frame again the joint itself is a D region okay so those those show some examples and this is a proposed method that I'm this is not the only method by the way that one can use to construct a Stratton-tie model the first step that I've used in the past that is quite useful is to construct a truss model or a Stratton-tie model using knowledge of what the load path is so for example in a joint here we know that the load path this tension force that's generated from this closing bending moment on the beam and column has to travel around the corner of this joint so that's what I would say is the load path for this tension force it has to exit the joint also as tension on the far end of the call that's a typical example of load path the other one would be you know if we can do a structural stress analysis and elastic stress analysis using a finite element model perhaps we could then understand the stress field in a component and then try to orient the members of the Stratton-tie model to follow that stress field this first step is important because it's not the only way to construct a Stratton-tie model but the closest we are to the elastic stress field in a member the less problems we will encounter when we load that member from service loads to ultimate if our Stratton-tie model is not very close to that what's needed from elastic stresses there is the possibility of course of redistribution but then we might encounter problems with cracking and other issues that we could prevent by selecting a Stratton-tie model using this first step the second step is now that we have defined the D region and we have constructed let's say the model or have an idea of what the model might look like we use the forces at the boundary of the model what does that mean these are forces for example again using the example from the beam column joint this is a shear force on the bending moment in the B region so this would be our boundary forces and we need to carry them from one end one member here to the other that's what this point means and then we check using those forces we analyze the structure and I'll have details of this and we determine forces of the elements of the structure of the model we check the strength of struts those compression members we check the strength of the nose and we also check this the the strength of the ties or we size the reinforcement to resist the forces in ties those are points three and four so I wanted to show you an example of what stress fields might look like for a corbel so I asked one of my students to do a stress analysis of a hypothetical corbel here under some applied load the magnitude of the load is not really important but you start seeing here this is the tension stress field these lines represent magnitudes of tension stresses so you see how tension stresses travel here horizontally first and then downward into the column the larger the arrow the larger the stress so this gives me an idea that we have to provide some tension forces here this is the compression stress field which is mostly oriented diagonally inward but also in the direction from the load and down into the front face of the column so these arrows in different colors illustrate those and if I superimpose a potential strut and tie model for these two stress fields it might look like this where I have this load acting here I follow the tension stresses here not exactly but very closely horizontal tension stresses then I turn them vertical into the column and following the compression stresses I use these struts these are compression members and I also use another strut here to follow those compression members and I want to make sure that each of these nodes this looks like a truss right so we need to have enough members connecting at each node for it to be potentially in equilibrium so I need to have a vertical component here I also need to have a horizontal component and I can start checking whether my truss model has the potential for being in equilibrium so that's an important check you'll note here that solid lines in red here represent ties or those tension members and dashed lines in blue represent compression members okay so I mentioned already that this is one way using an elastic stress analysis of the corbel to you construct a strut and tie model and I mentioned also that this is not the only way I can try to use a load path method which would lead to something slightly different but in theory would work as well okay so let's go back to just using the design example for corbel b1 but now we'll design it using the strut and tie model so the first thing is I would like to ask you to compare this model right here which is a simpler version of the model I illustrated here so the model that we're going to use corresponds to a single panel model similar to this where these two nodes are extended down all the way at the interface between the corbel and the column and that's an acceptable model in fact one of the PCI design examples I believe uses this model as well so excuse me the first thing I did here is that I converted the vertical force acting in the corbel and the horizontal force I computed the resultant force and I did that by by simply using a vector vector sum right the vector sum of the square root of the sum of the squares of the components gives me r sub u and I calculated the angle of inclination be it have it be 7.5 degrees so the first thing in any strut and tie model is to establish its geometry so the first note that I established was point C and notice that this force goes in diagonally so I cannot locate this node this is a fictitious point in the model it's 1.1 inches down and I assumed it was that location because I assumed a half-inch plate here and number nine bars eventually a number nine would be used to carry this tension force so we need to know roughly what we are going to be using for design of this corbel so having said that if you look at this force this node does not like perfectly vertical vertically six inches from the face of the column but rather it travels a little bit to the left so that's why we have that dimension as 6.14 inches that's the location of node C not the application of the load here at the top of the plate so this dashed line represents the compression force traveling downward into the corbel so that's an important point the second point that's shown in this slide in order for us to establish the remaining dimensions is that I located this point D at the intersection between the corbel itself and the column and point A lies horizontally to the right of point D the thing we don't know is where this point D lies with respect to this corner we need to provide enough compression zone resistance here so that the force developed here the compression force in the column that's developed here does not cross the concrete that is going to be represented by this vertical strut so how do we do that well we take moments we use moment equilibrium equation so we take moments about this node A and that eliminates all forces that go through A right so V is the vertical component of R sub U that doesn't go through A so we use these acting here at node C so we use 6.14 plus 30.5 inches is its lever arm N is also assumed to be acting at node C so this is node C here on the right hand side N is 22.9 inches from this level to where this 1.1 inches from the top of the corbel is so it's 22.9 and then we subtract the compression force the reduced compression compression force in the column phi CN times its distance to node A so it's 30.5 inches minus the back cover which is CC and the calculation for CC is 2.5 inches shown here so that is minus CC 30.5 inches oh pardon me that's not the back this is 30.5 inches minus the width of the compression block here divided by 2 so that's WC over 2 so that's the variable we're trying to define so we rewrite that equation and that gives me an equation for C sub N but it has also the unknown of WC then we check the strength of this vertical strut here at node D so we want to make sure and this is one of the equations that I'll discuss later in one of the subsequent slides where the strength of the maximum force I can apply to this strut cannot exceed the strength reduction factor phi for the strut in time model times 0.85 F prime C times an efficiency factor for that strut beta sub S and I'll tell you now that it's 0.80 and we'll see where that comes from later so it's 0.80 we need to know where that comes from times F prime C which is 5 KSI and AS is the I should have said a struts to avoid confusion with AS for reinforcement but a strut here is the width of the strut and times its length perpendicular to the screen so that is the width and times 24 inches so we have that equation for C sub N we have equations 1 and 2 that we can solve simultaneously and we can obtain 6.27 inches so the width here is 6.27 inches therefore WC over 2 is half of 6.27 inches and that way we can locate node D after locating node D we can then have the total geometry this is like a truss right so it's a truss with a vertical force and a horizontal force applied here at node C and this vertical force in the column and this also vertical tension force in that is would be in the column so we solve the truss using geometry and each of these members I'm going to label as forces of the truss FAB corresponds the force of the truss here between nodes A and B and respectively we have all the other forces that are calculated in that truss notice that the compression force here C sub N is equal to 383.7 from this equation which comes from the previous slide 61.2 times WC we can simply plug in WC and get 383.7 cubes so the rest we would just proceed by the method of joints and we can come up with all the forces these two are compression forces these two are tension forces these three are tension forces actually all right so we'll use each of these forces I'm going to show you some examples on how to check whether our model is correct so we have the simplified truss model shown here with the corresponding forces and the geometry that's needed 39.9 degrees of inclination of this strut 22.1 from the vertical of this other strut CD but the reality is that each of these members has a width as well so illustrated here within the confines of the column and the corbel we see that for example this tie CB has a width right the upper limit cannot exceed the top surface of the corbel and the so to place it symmetrically between the top surface so one would multiply the 1.1 inches times 2 to get the thickness of this tie the geometry of these struts is going to be defined first by by the depth let's say of ties and but also by geometry of the members that are framing into each individual node so you'll see what I mean in a second when we start working out some of the geometry okay so the first things first design of ties and this is perhaps the simplest calculation that you will see ACI by the way ACI 318 has chapter 23 defines the strength of strut and tie when using strut and tie models it tells you that the nominal strength of a tie is given by Phi times F and T Phi is the strength reduction factor for the strut and tie model which is always 0.75 and times the strength the forces in the tie which comes from mild reinforcement and the possibility of pre stressing for reinforcement if there is any so this is a nominal strength the nominal strength has to exceed the acting force so each of these tie forces 142 and 34.48 and this other vertical force that I haven't listed here has to be resisted by reinforcement the reason I don't list this tie this is the continuation you might remember of that bar that bends into the column so this force is higher so the tie in the force so the strength of tie BC this is tie BC is given by AS times FY that has to be equal at least to 142.4 kips so we can solve for AS and we get 3.2 square inches we can use 4 number 8 bars for this tie this horizontal tie results in 2 number 4 bars this is 4 legs this by the way is already at the column level so this would be ties within the column the next thing is so that just summarizes tie so like I said this is really very straightforward the next thing is perhaps the more complicated and it's complicated based on geometry the strength of nodal zones the equation looks simple but the nominal strength of a nodal zone is the effective concrete strength times the area of the nodal zone perpendicular to the compressions the compression force that's entering that nodal zone where the effective strength in compression is given by an equation like this 0.85 times an efficiency factor beta sub n for nodes times f prime C the table listing values for beta sub n depends on the type of nodal zone look for example a nodal zone that's anchoring only one tie would give you a beta sub n of 0.8 for this equation so that would be a nodal zone such as the one shown here whereas a nodal zone such as the one in the back it anchors two ties so the efficiency factor would be 0.60 so how does one define a sub ns the area of the nodal zone so we let we have to look into details of the node itself so this illustrates a strut entering into a node where also a tie is leaving so definitions within the ACI code are such that if a tie is given by multiple layers of reinforcement or is developed using multiple layers of reinforcement the the tension force is assumed to occur at the centroid of all these bars and we have to develop that reinforcement as it exits the what's referred to as the extended nodal zone the extended nodal zone is given by the this would be the nodal zone for this particular condition this shaded triangle the extended nodal zone is the area within the strut that that that extends beyond the nodal zone itself so in this case it's this lightly shaded area here shown here so you extend lines horizontally here and then follow the border of the of the strut to define the extended nodal zone the width of the strut is given by the boundaries of the zone so this W sub T is defined by the width of the tie this LB is defined by the bearing area of this compression force so this length and that length define how wide the strut is using geometry and I'll let you figure this one out and we'll have some other examples that I'll try to explain in more detail for the width of the strut okay so let's look in detail at this slide and then we can go faster in the next one so let's say we're checking node B here so the node B is anchoring two ties and one a strut so this is a view of a blown-up view of that node the nodal zone itself is defined by the width of this tie here the back face of the nodal zone is the width of that tie the top face of the nodal zone is the width of this tie and then the front face is the width of this strut how do we define the width of that strut well we look back here we see that this tie the width of this tie was defined as five inches which is twice the cover you might recall that CC was two and a half inches so we say well let's try to center this reinforcement about that location of this tie force so that's five inches I'll come back to WBC here in a second so it's let's see here well sorry this WBC is specified or from geometry of where this tie lies so I use 3.1 inches accounting for the thickness of the plate and the location to the top of the corbel so it's twice the distance from where this tie would lie this is this comes from geometry of the strut and tie model so it's 3.1 inches from the top of the tie to the bottom of or that's the width let's say of tie BC and using this equation which comes from the previous slide here so we can use trigonometry this angle alpha is the angle of inclination of the strut which you might recall we had determined here from the geometry of the of the of the strut and tie model so five is the width of the top face of the node the width of the vertical face of the node we can use geometry to define this angle alpha that angle alpha and then we can these projection this solid line is five times cosine alpha this other solid line is five times pardon me 3.1 times cosine alpha again or sine alpha sorry this is five sine alpha 3.1 cosine alpha and these two lines added define the total width of the strut so that's how the width of the strut is the tech determined as 5.59 so we need to check the nodal zone strength using the equation provided by the ACI code with the effective compressive strength defined as 0.85 beta and f prime C 0.85 times 0.60 because this node anchors two ties and times 5 KSI is the strength of reinforcement of concrete that's 2.55 KSI effective compressive strength the strength of the node has to be greater the factor strength of the node has to be greater to to the factored loads so we check the loads entering from each direction right the tie would bear against the back side of back face of the node this other tie would bear against this top face and the strut would bear against this surface so that's what we check we check let's check BD which is the strut so 0.75 is a fee factor 2.55 is the effective concrete strength 5.59 is the width that we just determined here and 24 inches is the the width of the the corbel perpendicular to the screen that gives me 256 kips which is greater than the force acting in that strut and that's okay that that that's fine we also check forces like I said for the two ties on the back end of the of the of the node and that would be for force BC and force AB just plugging in these values I use WBC here just to indicate that WBC I misspoke enough I misspoke before pardon me because WBC is determined from this calculation where we have to have a node that has a 3.1 inch depth at least to be able to develop this force needed for for the main reinforcement so as I mentioned most of it is geometry and it could get complicated the other item the other node I checked and I'm just gonna hit the highlights here because we are running out of time I checked this other note that it's a bit complicated this note is easier than this bottom one and what I did and the reason it's more complicated is because it has four forces two compressions coming from struts one compression coming vertical and one horizontal tie so I converted these two compressions I calculated the force resultant and I turned it into a single compression to check the node with the resultant force here of 385 kips at an angle of 5.2 degrees and then the geometry would work the same way as before and the checks would be similar in that we would have to check this force acting against the the corresponding face in the node and this horizontal tie force bearing against is one and a half inches on the back side of the node and that is as you see that that that's this I this would the force FAD would be we would be okay or this side face would be okay the more important check is the combined effect of BD and CD which is shown here as expressed by this the resultant component and then the final step in a strut and tie check is to check the strength of struts and this follows a similar approach to to nodal check where the equation is very similar we have an effective compressive strength we have the effective strut area and we use an efficiency factor based on one of these conditions beta goes from beta sub s for the strut goes from 0.75 where the strut is reinforced transversely to 0.60 when it's not or when there's enough where a strut passes through a tension zone it's 0.4 so it starts the concrete has less efficiency to transfer compression in those cases so checking the strut CD as it enters into this node we would have to calculate its width using a similar equation as before so the width of that strut is 7.31 the force it's carrying is 286 kips and this is a strut that has space to expand so it could go you know expand wider than what these lines define and assuming that it doesn't have any transverse reinforcement even though it might let's be conservative we use a efficiency factor of 0.6 so we end up with an effective concrete strength of 2.55 we plug that into the general expression for strength of the strut and that's 336 kips exceeding the acting force of 286 which is okay so one question that's often asked is well if we design a core ball using the strut and tie model instead of using the the cantilever method is do we need horizontal reinforcement and the answer is is yes absolutely for two reasons one reason is that this strut will generate transverse tensile stresses that will then cause cracks to form almost vertically in this region if we do not provide reinforcement within this region and we only use A sub S then we run the risk of having a wide a crack widening through the middle of the core ball which again calls for improper detailing of the core ball and I also wanted to point you to a provision in HCI 318-14 that specifically says that brackets and core balls that are designed using the strut and tie method still have to satisfy the distribution the reinforcement in 16.5.2 and 16.5.6 excuse me which talks about A sub H and that A sub H has to be distributed in two-thirds over D so in other words we do have to provide that steel and we have to use the same equation as the cantilever method so we use that here for the design of our core ball using the strut and tie model in this case we use A sub S from the force that was calculated in tie BC up here that was 3.2 square inches roughly so we end up with 1.2 square inches and it ends up being the same amount of horizontal reinforcement as we had before so our final design is shown here where instead of using four number I believe we had four number sevens we have now four number eights we have five number sevens pardon me we have now four number eights we have slightly more reinforcement in this particular case when we use the strut and tie method for design but other than that the design ends up being almost literally the same as before and with that I think I'm done I wanted to remind you that there's no session next week but two weeks from today we're going to be talking about sessions and special topics in session six so there's a couple of minutes left for any questions that people might have Sherry are you there yes if you have any if you have any questions please put them in the chat box now we have no questions so Sergio okay so I would like to thank everyone for attending tonight and I look forward to seeing you in two weeks for our next our next session if you have any questions or if you think of a question after you leave this session please email it to me and I will send it to Sergio for answering again thank you so much for attending and please take your quizzes and we will get your information into our CEP as soon as possible thank you good night

advanced pre-stressed concrete design course

corbel design

cantilever beam method

strut-and-tie method

flexure analysis

horizontal load analysis

shear friction analysis

reinforcement anchorage

reinforcement requirements

strut-and-tie method application

corbel performance

failure modes